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2025-03-21T14:48:29.728197
| 2020-01-29T01:43:07 |
351390
|
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"David E Speyer",
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|
Stack Exchange
|
Euclid Book 1 Proposition 4
In Euclid's The Elements, Book 1, Proposition 4, he makes the assumption that one can create an angle between two lines and then construct the same angle from two different lines. I do not see anywhere in the list of definitions, common notions, or postulates that allows for this assumption. When teaching my students this, I do teach them congruent angle construction (with straight edge and compass), but do not understand why Euclid is allowed to make this logical leap. Thanks for any help you can give.
This question doesn't belong on Mathoverflow, but might get a good reception on matheducators.stackexchange.com
Posted on MESE here.
|
2025-03-21T14:48:29.728279
| 2020-01-29T02:38:21 |
351392
|
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"Uri Bader",
"https://mathoverflow.net/users/89334"
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|
Stack Exchange
|
Orbit representatives for the action of the maximal compact subgroup
Let $F$ be a non-Archimedean local field and $O$ be the ring of integers in $F$. Take $G=GL(2,F)$ and $K=GL(2,O)$. Consider the action of $K$ on $G$ by conjugation. Is it possible to explicitly write down the representatives for the K orbits? Is this known?
Use $KA^+K$ decomposition. The orbit space for the left $K$ action is $A^+K$ and the map $kak' \mapsto kak'k^{-1}$ intertwines the left $K$ action with the conjugation $K$-action. This map is the identity on $A^+K$, so this set is also a set of representatives for the conjugation action.
|
2025-03-21T14:48:29.728358
| 2020-01-29T05:22:16 |
351396
|
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"url": "https://mathoverflow.net/questions/351396"
}
|
Stack Exchange
|
Completely positive maps and abelian C* algebra
This is a problem I encountered in Jesse Peterson's Notes on Von Neumann Algebras. I want to show the following: given C* algebra A, suppose for any C* algebra B, every positive map from B to A is completely positive, then A is an abelian C* algebra.
I have no idea how to show this. Any help is greatly appreciated.
Thank you very much.
See
Tomiyama, Jun, On the difference of n-positivity and complete positivity in C*-algebras, J. Funct. Anal. 49, 1-9 (1982). ZBL0497.46039.
Thank you very much.
|
2025-03-21T14:48:29.728430
| 2020-01-29T05:51:36 |
351398
|
{
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"Martin Brandenburg",
"Sebastien Palcoux",
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"https://mathoverflow.net/users/44191",
"user44191"
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}
|
Stack Exchange
|
Existence of an addition bifunctor for the category of groups
Let $\mathsf{Grp}$ be the category of groups. A bifunctor $A: \mathsf{Grp} \times \mathsf{Grp}\to \mathsf{Grp}$ is an addition bifunctor if:
$A(C_n,C_m) \simeq C_{n+m}$,
$A(C_0,G) \simeq A(G,C_0) \simeq G$,
for every group $G$ and every $n,m$, with $C_n$ the cyclic group of $n$ elements if $n>0$, and $C_0 \simeq \mathbb{Z}$.
Question: Is there an addition bifunctor for the category of groups?
(or for the subcategory of countable groups)
Stronger question: Is there an addition bifunctor providing a monoidal structure?
This post is inspired by that one (without knowing whether there is an explicit link).
Multiplicative analogous: Existence of a multiplication bifunctor for the category of groups.
How are you defining your category of cyclic groups (what are its morphisms), and how are you defining your addition and multiplication functors (what does each do to those morphisms)?
@user44191 Your comment is relevant!
@user44191 I should replace "category" by "set" and "functor" by "map" because my motivation here is not in category theory and I don't see a natural way to define these addition and multiplication functors; do you?
I'm...strongly doubtful that there's a natural "addition" functor, and weakly doubtful that there's a natural "multiplication" functor. But if you do go with "set" and "map", then from what I can see, there's no interesting "structure" to $\tilde{A}$ - the extension can be defined completely arbitrarily.
@SebastienPalcoux When your motivation is not category theory, I suggest to rephrase the question accordingly and also remove the ct-tag.
@MartinBrandenburg I am in a dilemma: On one hand I don't see an easy/natural way to define such an addition (or multiplication) functor, but it can still exist. On the other hand, removing the category structure would allow arbitrary answers, which is not interesting. Something more subtle is required, and the category theory could help.
It was not really true to say << my motivation here is not in category theory >> because my motivation is in fact an eventual extension to subfactor theory, which is stronlgy related to tensor/fusion category theory.
@MartinBrandenburg: We could ask whether the category $\mathcal{Grp}$ (or the subcategory of countable groups) admits a monoidal structure with unit $I \simeq C_0$ and $C_n \otimes C_m \simeq C_{n+m}$ (or with unit $I \simeq C_1$ and $C_n \otimes C_m \simeq C_{nm}$).
Ok, I understand. A monoidal structure is even stronger than your initial requirement, but also more interesting. Now you got me ... ^^
The answer is no. Notice that $A(-,C_1)$ is a functor $F : \mathsf{Grp} \to \mathsf{Grp}$ with $F(C_n) \cong C_{n+1}$. But there is no such $F$. There is a split monomorphism $C_1 \to C_2$, hence $F$ would induce a split monomorphism $C_2 \to C_3$, contradiction.
Multiplicative analogous: Existence of a multiplication bifunctor for the category of groups
|
2025-03-21T14:48:29.728630
| 2020-01-29T06:06:33 |
351400
|
{
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"Bruce Blackadar",
"François G. Dorais",
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"Gerry Myerson",
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|
Stack Exchange
|
Approximations to $\pi$
Is there a way to efficiently solve the following problem besides brute-force calculation?
Fix $n\in\mathbb{N}$ (say $n=100$). Find the integers $p,q,r,s$ with $0\leq p,q,r,s\leq n$ such that
$$\pm\frac{p}{q}\pm\sqrt{\frac{r}{s}}$$
most closely approximates $\pi$.
Some cases can be handled by finding the continued fraction expansion of $(\pi-\frac{p}{q})^2$ for various $p,q$. Playing with this method I found the approximation $4-\sqrt{\frac{14}{19}}$, which is really quite good, but may not be best for $n=20$. Note that the solution will be unique (for a given $n$).
3 + sqrt(1/50) isn't too shabby. Gerhard "That's Without Pencil And Paper" Paseman, 2020.01.28.
4-sqrt(14/19) is more than an order of magnitude better (but not without pencil and paper!)
Such numbers have eventually periodic continued fractions. Maybe something can be done comparing periodic continued fractions to the continued fraction of $\pi$?
I guess the key to that approach would be to give an upper bound for $\max(p,q,r,s)$ in terms of the continued fraction, enabling a (hopefully) efficient search over periodic continued fraction approximations.
Below $100$, only one stands out: $71/28+\sqrt{29/79}$. Next best is $-47/46+\sqrt{52/3}$, far behind.
Nice! How did you find this?
@BruceBlackadar: gawk -v n=100 'function abs(x) {if(x>0) return x; else return -x;} function gcd(a,b, x,y,t) {x=a; y=b; if (x<y) {t=x; x=y; y=t;} while (y>0) {t=x%y; x=y; y=t;} return (x);} BEGIN {tol=1/n^2; pi=3.141592653589793238462643383276; for(a=-n; a<=n; a++) for(b=1; b<=n; b++) if(gcd(abs(a),b)==1) {r=pi-a/b; rr=r*r; for(d=1; d<=n; d++) {c=int(d*rr+0.5); if(c<=n && gcd(c,d)==1) {pp=a/b+sqrt(c/d); if(abs(pi-pp)<tol) print tol/abs(pi-pp), a "/" b "+sqrt(" c "/" d ")"; pp=a/b-sqrt(c/d); if(abs(pi-pp)<tol) print tol/abs(pi-pp), a "/" b "-sqrt(" c "/" d ")";}}}}' | sort -n -k1 | tail
@BruceBlackadar: beware gawk has only about 15 digits of precision, so n=1000 is almost as far as you can go with this hack.
@BruceBlackadar: I just noticed that the comment syntax deleted the STAR symbol ("*"), so that above rr=rr should be rr=r STAR r and drr should be d STAR rr. Sorry for any inconvenience.
OK, so essentially brute force. I was hoping for a procedure similar to the continued fraction method of finding best rational approximations.
@YaakovBaruch: Use backticks to surround code: https://stackoverflow.com/editing-help#comment-formatting
Some versions of gawk have -M for multiprecision. Gerhard "So Check Your Gawk Version" Paseman, 2020.01.30.
@GerhardPaseman: yes indeed. And the roots and floats can easily be avoided allow to work with integers only.
This is not exactly an answer to the stated question, but it's too long for a comment. Rather than the form given in the question, one could represent a number in the form $\frac{a + b \sqrt{d}}{c}$, where $d$ is a squarefree positive integer,
and this form lends itself to finding good approximations to $\pi$ using lattice reduction.
Fix a bound $n$. For each squarefree $d$ with $1 \leq d \leq n$,
choose a constant $X \approx n^{2} \sqrt{d}$ and create the lattice in $\mathbb{R}^{4}$ spanned by
$$ v_{1} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ X \pi \end{bmatrix}, v_{2} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ -X \end{bmatrix}, v_{3} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ -X \sqrt{d} \end{bmatrix}. $$
A short vector in this lattice with respect to the $\ell_{2}$ norm is a linear combination $a v_{1} + bv_{2} + cv_{3}$ and because the fourth coordinate of these vectors are so large, this forces $\frac{a+b \sqrt{d}}{c}$ to be a close approximation to $\pi$.
Finding the shortest vector in a lattice is a hard problem, even in small dimensional lattices. However, one can get within a constant multiple of the true minimum using the LLL-algorithm. With this, the above algorithm would run in time $O(n \log^{3} n)$ and find "some good solutions", but isn't guaranteed to find the optimal representation (even in this modified form).
I ran this with $n = 10^{6}$ and $X = n^{2} \sqrt{d} \log(n)$ and obtained (after about a minute and a half)
$$
\pi \approx \frac{-327031 + 7075 \sqrt{224270}}{962406}.
$$
The approximation differs from the truth by about $8 \cdot 10^{-22}$.
An approximation using $22$ digits, and good to $10^{-22}$. About what one might expect, no?
To answer Gerry Myerson's question - yes, this is about as good (or even a little worse) an approximation as one would expect. Perhaps I wasn't clear above - for a fixed $d$ the lattice reduction is polynomial time and takes about $O(\log^{3} n)$, and so looping over all $d$, the runtime is something like $O(n \log^{A} n)$ for some $A$ depending on how quick the arithmetic is.
A brute force search to find $\pi \approx (a+\sqrt{b})/c$ can be done with a loop of the form for(a=-n; a<=n; a++) for(c=max(1,int((a-sqrt(n))/pi)); c<=min(n,(a+sqrt(n))/pi); c++) {b=...} and therefore $O(n^{3/2}\log^A)$ operations for some small $A$ that I'm not sure about. Not too much worse than the clever but non-optimal scheme suggested here. However notice that I have $\sqrt{b}$ here rather, than the $b\sqrt{d}$ of this answer.
In under 3 minutes, for $n=10^6$, the best, unexciting approximation $\pi\approx (108062-\sqrt{99097})/34297$ is found, good to about $4 \cdot 10^{-16}$.
|
2025-03-21T14:48:29.729086
| 2020-01-29T06:43:08 |
351401
|
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|
Stack Exchange
|
An exercise from Loday and Vallette about Koszul morphism
I tried to solve the following exercise from Loday and Vallette's Algebraic Operad. The first three parts are straightforward, however I have no idea how to solve the last part. I can't find any statements in that chapter which can be applied directly. Any help or hints would be really appreciated.
Recall: $\alpha$ being Koszul means that $C\otimes_\alpha A$ is acyclic, menaning that the augmentation map $C\otimes_\alpha A\overset{\epsilon\otimes\epsilon}{\longrightarrow}\mathbb{K}$ is a quasi-isomorphism.
$\underline{\Rightarrow}$
Observe that $\epsilon\otimes\epsilon$ is the image of $\xi$ through the functor $\mathbb{K}\otimes_A-$, which is exact on free $A$-modules.
Therefore, if $\xi$ is a quasi-isomorphism, then so is $\epsilon\otimes\epsilon$.
$\underline{\Leftarrow}$
Using the weight grading on (the left most copy of) $A$, you can put a filtration on $A\otimes_\alpha C\otimes_\alpha A$, so that $\xi$ is filtration preserving. Then consider the associated spectral sequence. On the level of the $E_0$ page (i.e. associated graded) you get the following morphism:
$$
A\otimes_\epsilon C\otimes_\alpha A\overset{id\otimes\epsilon\otimes \epsilon}{\longrightarrow}A
$$
Therefore, if $\epsilon\otimes\epsilon$ is a quasi-isomorphism, then we get a quasi-iomorphism at the level of $E_0$. Hence the spectral sequence (of the cone of $\xi$, lets say) degenerates and is $0$ at $E_1$. Thus the cone of $\xi$ is acyclic, which means that $\xi$ is a quasi-isomorphism.
|
2025-03-21T14:48:29.729211
| 2020-01-29T07:14:19 |
351402
|
{
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|
Stack Exchange
|
What happens to eigenvalues when edges are removed?
I am stuck at the following :
Let $G$ be a graph and $A$ is its adjacency matrix.
Let the eigenvalues of $A$ be $\lambda_1\le \lambda_2\leq \cdots \leq \lambda_n$.
If we remove some edges from the graph $G$ and form the graph $H$ keeping the number of vertices same, is there any result how the smallest eigenvalue of $H$ is related to the smallest eigenvalue of $G$?
I know Cauchy Interlacing Theorem which gives the relation between eigenvalues of a graph and its induced subgraph when some vertices are removed.
I want to know what happens when edges are removed keeping the number of vertices same. Can someone help please?
The question stands as:
Let $G$ be a graph and $A_G$ is its adjacency matrix.
Let the eigenvalues of $A_G$ be $\lambda_1\le \lambda_2\leq \cdots \leq \lambda_n$.
Let $H$ be a subgraph of $G$ which has $n$ vertices as $G$ but some edges have been removed from $G$ to form $H$.$A_H$ is its adjacency matrix.
Let the eigenvalues of $A_H$ be $\mu_1\le \mu_2\leq \cdots \leq \mu_n$.
Is $\mu_1\ge \lambda_1$ or $\mu_1\le \lambda_1$?
If someone can give any reference like book or paper , I will be grateful.
The smallest eigenvalue can go up or down when an edge is removed.
For "down": $G=K_n$ for $n\ge 3$.
For "up": Take $K_n$ for $n\ge 1$ and append a new vertex attached to a single vertex of the original $n$ vertices. Now removing the new edge makes the smallest eigenvalue go up.
Both of these follow from the fact that the smallest eigenvalue of a connected graph with $n\ge 2$ vertices is $\le -1$ with equality iff the graph is complete.
It has something to do with whether the two corresponding eigenvector entries have the same or opposite sign, but I don't know if that relationship can be made precise.
Here's a statement from the book "Spectra of Graphs" by Brouwer and Haemers concerning the largest eigenvalue of the adjacency matrix. It implies that $\lambda_n \geq \mu_n$.
Proposition 3.1.1 The value of the largest eigenvalue of a graph does not increase when vertices or edges are removed.
But my question is about the smallest eigenvalue
This might be helpful, from Eigenvalues and structures of graphs
The result gives $0\le \theta_0\leq \lambda_1$ but I want to know how is $\lambda_0$ related to $\theta_0$
Is my question clear?
This thm is for Laplacian eigenvalues, not adjacency-matrix eigenvalues.
|
2025-03-21T14:48:29.729392
| 2020-01-29T08:59:44 |
351410
|
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"A Stasinski",
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|
Stack Exchange
|
On components of centralisers in unipotent groups
Let $k$ be an algebraically closed field with $\mathrm{char}(k)=p>0$. Let $U$ be a connected unipotent algebraic group over $k$.
Question: When $p$ is big enough, is it true that $Z_U(u)$ is connected for any $u\in U$, or at least $u\in Z_U(u)^o$ for any $u\in U$?
Remark: This is true when $U$ is the unipotent radical of a Borel subgroup of a reductive group with $p$ being good (so that a Springer homeomorphism exists).
Did you mean to ask whether, for a unipotent group scheme $U$ over $\mathbb{Z}$, the fibre $U_{\mathbb{F}p}$ has connected centralisers for all sufficiently large $p$? This seems to me to be the more interesting question because one usually expects the fibres $U{\mathbb{F}_p}$, for large $p$, to behave like groups over characteristic $0$ fields. There is no reason to expect this if $U$ is allowed to vary with $p$.
@AStasinski Yes, $p$ is expected to be irrelevant to the defining equations of $U$. The question stated in the current form is a bit misleading.
In that case kneidell's answer is not an answer to the intended question.
If your element and your group are globally defined (say, over $\mathbb Z$) then both $Z_U(u)$ and $Z_U(u)^\circ$ would be $\mathbb Z$ defined, hence so would the quotient group. Taking $p$'s larger than the exponent of this finite group, shouldn't this finite algebraic group have no points over $k$?
The following is a counterexample which can be defined for arbitrarily large $p$'s.
Consider $U=\left\{ \begin{pmatrix}1&a&b\\&1&a^p\\&&1\end{pmatrix}:a,b\in k\right\}\subseteq\mathrm{GL}_3(k)$ and take $u=u_\lambda=\begin{pmatrix}1&\lambda\\&1&\lambda\\&&1\end{pmatrix}$ with $0\ne \lambda\in\mathbb{F}_p$ (i.e. $\lambda=\lambda^p$). Then one easily computes that $$\begin{pmatrix}1&a&b\\&1&a^p\\&&1\end{pmatrix}\in Z_U(u)\quad\iff\quad a=a^p, $$
and thus $Z_U(u)\simeq \mu_p\ltimes \mathbb G_a$ and $Z_U(u)^\circ=\begin{pmatrix}1&&*\\&1&\\&&1\end{pmatrix}\simeq \mathbb G_a$ (here $\mu_p$ is the group of $p$-th roots of $1$). In particular $u\notin Z_U(u)^\circ$.
Note that this gives a family of groups $U$ depending on $p$. See also my comment on the question.
|
2025-03-21T14:48:29.729568
| 2020-01-29T09:37:08 |
351412
|
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"Benjamin Steinberg",
"Dominic van der Zypen",
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|
Stack Exchange
|
Does the lattice of partitions map onto the lattice of subsets?
Let $X\neq \emptyset$ be a set and let $X^X$ denote the collection of all functions $f:X\to X$. We put a binary relation (reflexive and transitive), the composition preorder on $X^X$ by setting for $f,g\in X^X$:
$$ f\leq_{\text{comp}} g \text{ if and only if } \exists h\in X^X (f = h \circ g).$$
It is easy to see that this relation is reflexive and transitive, but not anti-symmetric, so we say for $f,g\in X^X$ that $f\simeq_{\text{comp}} g$ if and only if $f\leq _{\text{comp}} g$ and $g \leq_{\text{comp}} f$.
So $X^X/\simeq_{\text{comp}}$ is a poset with the composition preorder applied to equivalence classes. The class consisting of constant functions is the smallest element, and the class containing the identity function is the largest element of $X^X/\simeq_{\text{comp}}$.
Given a non-empty set $X$, is there a surjective order-preserving map $f:X^X/\simeq_{\text{comp}} \to {\cal P}(X)$, where ${\cal P}(X)$ is ordered by $\subseteq$?
The order seems determined by refinement of the pre-image partitions.
Thanks @JoelDavidHamkins for your comment! So if I remember correctly, there is a surjective homomorphism from the collection of partitions on a set $X$ (ordered by refinement) onto ${\cal P}(X)$?
If $X$ is a singleton, then $X^X$ is a singleton and there cannot exist a surjective map to the power set $P(X)$ (which has cardinality $2$).
This is Green's L-preorder on the full transformation monoid. It is easy to see that two functions are equivalent iff they give the same partition of X into fibers over the image and hence this is the set partition lattice of X but with the order reversed (since the identity has singleton fibers and constant maps universal fibers).
Sorry I see Joel already said this.
Comments:
I. This question reduces to the question of whether the lattice $\textrm{Eq}(X)$ of equivalence relations on $X$ has an order-preserving map onto the lattice ${\mathcal P}(X)$ of subsets of $X$.
II. The answer to this question is Yes if $X$ is infinite.
III. The argument in the infinite case can be applied in the finite case to show that $\textrm{Eq}(n+1)$ has an order-preserving map onto ${\mathcal P}(n)$ when $n\in\omega$, but it does not resolve the question of whether $\textrm{Eq}(n)$ has an order-preserving map onto ${\mathcal P}(n)$ when $n\in\omega$.
Reasons.
I. $f\leq_\textrm{comp} g$ iff $\ker(f)\subseteq \ker(g)$, as observed in the comments. Thus, $(X^X/\simeq)\;\;\cong\;\; \textrm{Eq}(X)$.
II. Let $X=\kappa$ be an infinite cardinal. View it as a graph with edge set consisting of all $(\alpha,\alpha+1)$ for $\alpha< \kappa$. Call $(\alpha,\alpha+1)$ a successor pair and call $\alpha$ the initial element of the pair. Define a map
$$
\Phi: \textrm{Eq}(\kappa)\to {\mathcal P}(\kappa): \theta\mapsto \{\alpha\;|\;(\alpha,\alpha+1)\in\theta\}.
$$
That is, $\Phi$ maps an equivalence relation to the set of initial elements of successor pairs that are in relation.
$\Phi$ is order-preserving, since if $\theta\leq \theta'$, then $\theta'$ has at least as many successor pairs as $\theta$, hence $\Phi(\theta')$ has at least as many initial elements of successor pairs as $\Phi(\theta)$.
It is easy to construct a right inverse to $\Phi$, namely
$$\Psi: {\mathcal P}(\kappa)\to \textrm{Eq}(\kappa): U\mapsto \textrm{equiv. reln. generated by} \{(\alpha,\alpha+1)\;|\;\alpha\in U\}.
$$
To see that $\Psi$ is a section of $\Phi$ you must convince yourself that an equivalence relation generated by successor pairs will contain no successor pairs other than the generators.
III. This argument can be applied in the finite case to show
that
$$\Phi: \textrm{Eq}(n+1)\to {\mathcal P}(n): \theta\mapsto \{i\;|\;(i,i+1)\in\theta\}$$
is an order-preserving surjection.
To make the argument work, we need that every element $i\in n$ is the initial element of a successor pair, so our domain lattice needs to be $\textrm{Eq}(n+1)$ rather than $\textrm{Eq}(n)$.
My guess is that there is no order-preserving surjection $\textrm{Eq}(n)\to {\mathcal P}(n)$ when $n\in\omega$. Surely there can be no $\wedge$-preserving surjection, nor can there be any $\vee$-preserving surjection. The reason there is no $\wedge$-preserving surjection is: for any $\wedge$-preserving surjection $\Phi: \textrm{Eq}(n)\to {\mathcal P}(n)$, there is a $\vee$-preserving section $\Psi: {\mathcal P}(n)\to \textrm{Eq}(n)$ defined by $\Psi(U)$ = least element of $\Phi^{-1}(U)$. But then $\Psi$ is an order-preserving injection, and no such map exists from ${\mathcal P}(n)$ to $\textrm{Eq}(n)$, as was pointed out on this page.
|
2025-03-21T14:48:29.729867
| 2020-01-29T09:58:07 |
351413
|
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|
Stack Exchange
|
Is strong convergence of measures equivalent to convergence in measure of the Radon Nikodym derivatives?
Let $X$ be a measure space, and suppose $\mu_i$ are probability measures on $X$ that are absolutely continuous with respect to another probability measure $\mu$. Is strong convergence of $\mu_i$ to $\mu$ equivalent to convergence in measure (wrt $\mu$) of the Radon nikodym derivatives $\frac{d\mu_i}{d\mu}$ to $1$?
From wiki: "For example, as a consequence of the Riemann–Lebesgue lemma, the sequence μ_n of measures on the interval [−1, 1] given by μ_n(dx) = (1+ sin(nx))dx converges strongly to Lebesgue measure."
@ Fedor Petrov - it depends on what the OP meant by strong convergence of measures. I was surprised to learn that wiki (apparently this is the article you quote) introduces a rather artificial notion of "strong convergence" of measures (I have never come across it in real life) and distinguishes it from the convergence in total variation ($\equiv$ convergence in the strong topology on the space of measures). This is not the first time I come across highly dubious claims in wiki.
I did indeed mean convergence in the sense stated by wiki..
If it’s meant in the other sense is it true?
It seems like it is indeed true, if I didn’t make any mistakes.
Wikipedia's "strong convergence" is called "setwise convergence" in Bogachev's measure theory. It'd be helpful to have the definition in your question.
Note, however, that under your assumptions the Radon-Nikodym derivatives in fact converge to 1 in $L^1(\mu)$. If I'm not mistaken, that is equivalent to convergence in total variation.
Let $A_n:=\{x\colon f_n(x)\le1\}$ and $B_n:=\{x\colon f_n(x)>1\}$, where $f_n:=\frac{d\mu_n}{d\mu}$. Then the total variation of $\mu_n-\mu$ is
$$\|\mu_n-\mu\|=\int_{B_n}(f_n-1)d\mu+\int_{A_n}(1-f_n)d\mu=2\int_{A_n}(1-f_n)d\mu\to0$$
by dominated convergence if $f_n\to1$ in measure wrt $\mu$; the latter displayed equality is the key, even though simple, observation here.
Vice versa,
$$\|\mu_n-\mu\|=\int_X|f_n-1|d\mu.$$
So, as noted by Nate Eldredge, $\|\mu_n-\mu\|\to0$ means that $f_n\to1$ in $L^1(\mu)$, which implies, by Markov's inequality
$$\mu\{x\colon|f_n(x)-1|>\epsilon\}\le\frac1\epsilon\,\int_X|f_n-1|d\mu$$
for all $\epsilon>0$, that $f_n\to1$ in measure wrt $\mu$.
Thus, $\mu_n\to\mu$ in total variation iff $\frac{d\mu_n}{d\mu}\to1$ in measure wrt $\mu$.
|
2025-03-21T14:48:29.730065
| 2020-01-29T10:22:04 |
351415
|
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"Yauhen Yakimenka",
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|
Stack Exchange
|
Interpolation nodes for linear spline (piecewise-linear) interpolation of $x \ln x$
I need to approximate $x \ln x$ on $[0,1]$ as a piecewise-linear function. If $P(x)$ is a piecewise-linear approximation, I want to minimize
$$
\max_{0 \le x \le 1} |P(x) - x \ln x| \rightarrow \min_P.
$$
I tried interpolation over evenly spread points $x_i = \frac{i}{n}$, for $i=0,1,\dotsc,n$. I also tried Chebyshev nodes together with additional nodes $x=0$ and $x=1$. The latter works much better. For my numerical problem, it was enough but I am interested in optimal solution (mathematical mind, you know :)
Is there a way to optimally spread points $x_0, x_1, \dotsc, x_n$ over $[0,1]$ for piecewise-linear interpolation/approximation?
Update
To illustrate the difference between uniform and Chebyshev nodes, here is the error of approximation with $n=20$ points.
My guess is that for optimal nodes, the "bumps" should be of the same height.
Update 2
I have also the following idea of a numerical method.
First, let us see what is the maximum error of approximation between points $x_n$ and $x_{n+1}$.
We can find equation of an approximating (red) line $y = kx + b$, where
$$
k = \frac{x_{n+1} \ln x_{n+1} - x_n \ln x_n}{x_{n+1} - x_n}, \quad b = -k x_n+x_n \ln x_n \,.
$$
Next, for $x_n \le x \le x_{n+1}$, we can define the error of approximation at point $x$:
$$
\delta(x) = k x + b - x \ln x \,.
$$
To find the maximum error, we just solve $\delta'(x) = 0$:
$$
0 = (kx + b - x \ln x)' = k - \ln x - 1.
$$
Therefore, the maximum error on $[x_n,x_{n+1}]$ is achieved at $x_c=e^{k-1}$ and it is equal to
$$
d(x_n, x_{n+1}) = k x_c + b - x_c \ln x_c \,,
$$
where $k$, $b$, and $x_c$ depend on $x_n$ and $x_{n+1}$ as described above. Note also that if we fix $x_n$, the $d(x_n, x_{n+1})$ is increasing function in $x_{n+1}$.
Now, the method itself. Assume we want to ensure that approximation error is not larger than $\epsilon$. Then the following iterative procedure can be applied to construct the list of points.
Set $x_0 = 0$.
For $n=0,1,2,\dotsc$ find (e.g. by binary search) $x_{n+1}$ that ensures $d(x_n, x_{n+1}) = \epsilon$.
Stop when some $x_{n+1} \ge 1$. Change $x_{n+1}$ to $x_{n+1} = 1$ (not necessary but to make things nicer).
But is there an analytical solution?
For example, if we set three points $x_0=0, x_1, x_2=1$, then it seems the optimal position for $x_1$ is $0.2869781560930248$...
There is an analytical solution to the problem in the following sense:
Given a number $N$, the optimal interpolation points
$x_0=0, x_1, ..., x_{N-1}=1$
are the roots of an $(N-2) \times (N-2)$ system of non-linear equations described below.
The main insight to construct this system is the following proposition (using the same notation as in the question):
The interpolation points that minimize the maximal error are achieved when each interval $[x_i, x_{i+1}]$ attains exactly the same maximal error $d(x_i, x_{i+1}) = k_i x_{c_i} + b_i - x_{c_i} \ln x_{c_i}$.
This can be proved using the fact that, since $x \ln x$ is convex, the maximal error on any interval can be reduced by reducing the length of the interval.
Thus, if we assume that the maximal interpolation error is attained at some interval $[x_i, x_{i+1}]$,
then if the error on the neighboring interval $[x_{i-1}, x_i]$ was smaller, we could have moved $x_i$ to the right and reduce the maximal interpolation error. Therefore the error on $[x_{i-1}, x_i]$ cannot be smaller and the same argument can be made further to prove that all intervals attain the maximal error.
So, the minimal maximal error is attained when all the intervals have the same maximal error.
Using this proposition, and the $d(.)$ function developed in the question, we can build a set of non-linear equations of the form:
$$
d(0, x_1) = d(x_1, x_2) \\
...\\
d(0, x_1) = d(x_{N-2}, 1)
$$
and use any non-linear solver to solve it (in the results below I used python's scipy.optimize.fsolve(), with an initial guess of equally spaced $x$-values).
For $N=3$, we get a single equation:
$$
d(0, x_1) = d(x_1, 1)
$$
and the root is (as @yauhen-yakimenka notes in his comment) $x_1 = 0.28697816$ with an optimal error of $0.10557336369191296$.
In the error plot below, we can see this optimal value attained in two places.
For $N=20$, we get the roots at:
$$
0, 0.00369174, 0.01286418, 0.02747082, 0.04751028, 0.0729823, 0.10388683, 0.14022384, 0.18199332, 0.22919527, 0.28182968, 0.33989654, 0.40339587, 0.47232766, 0.5466919, 0.62648861y, 0.71171777, 0.80237939, 0.89847347, 1
$$
With the optimal error of $0.001358114333332766$, see error plot below.
|
2025-03-21T14:48:29.730490
| 2020-01-29T10:52:12 |
351418
|
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|
Stack Exchange
|
Concentration of sum of concentrated random variables
I have a sum of positive random variables, they are not identically distributed, but even that case would be interesting. They are not necessarily independent, but I already have a concentration bound for the individual random variables (that I got using one of the standard methods, such as Chernoff bound, Method of bounded differences, Kim-Vu inequality or similar). I care about the relative size of the deviations (that is, as a fraction of the expectation).
Is there anything that I can say about the concentration of the sum in general? Is there a method how to solve this?
Namely, I would expect that for the number of random variables going to infinity (and maybe even if not) if I have upper bound on the probability that $(1-\epsilon)E[X_i] \leq X_i \leq (1+\epsilon)E[X_i]$ then the worst of these bounds also holds for the sum. Is this so? (In the sense that the sum would be with given probability between $1-\epsilon$ and $1+\epsilon$ multiple of its expectation).
I could estimate the probability that each random variable is deviated by a $\epsilon$-fraction and then use union bound. However, this seems rather wasteful and I cannot think of an example where this would be tight (or even nearly tight)
The worst case scenario seems to be when all variables are just identical (so you have $nX$ instead of $X$ with obvious rescaled concentration bounds). Will that be good enough for you or you want more? (In the latter case some information about the nature of dependence would be necessary). It would also be nice to tell us a bit more about the setup: are your variables identically distributed, for instance?
Thank you for the response. I have updated the question. I also updated it to make clear that I care about the relative deviations. This means that in the case of identical random variables, I can just get bound on one of them and the same bound applies to their sum.
You need some quantitative control on how the dependence between your variables. As @fedja's example shows, without such assumptions you get trivialities.
As I explained in my comment, the example by @fedja does not work when I look at relative deviations (if X is between $(1-\epsilon )E[X]$ and $(1+\epsilon )E[X]$, so is $nX$).
@MattF. there is at least one obvious answer lurking nearby and I made it obvious by an edit. Please reconsider the vote to close :-)
I have done so, as well as removing a part that was wrong.
Let us continue this discussion in chat.
Are $X_i$ posiitve?
Yes, I have updated the question
There is a bad news and a good news. The bad one is that if you have no information other than that the probability of the $\varepsilon$-deviation is at most $p$ for each variable, then you can hardly do anything better than the union bound. Indeed, consider the random variables $X_i$ that are exactly $E[X_i]$ with probability $1-p$ and something else (say, with continuous distribution) with probability $p$. Then you have your $\varepsilon$-concentration for every $\varepsilon>0$, but, of course, the only estimate you may have for the probability that the sum hits the sum of expectations exactly is the union bound. The good news is that if you obtained your concentration bounds by the same method, i.e., by estimating
$$
E\Phi\left(\frac{X_i-EX_i}{EX_i}\right)\le C
$$
with the same convex function $\Phi$ (Chernov, for instance, corresponds to $\Phi(x)=e^{Ax}+e^{-Ax}$), then, by the Jensen inequality, you have
$$
E\Phi\left(\frac{X-EX}{EX}\right)\le C
$$
for the sum $X=\sum_i X_i$.
If you have obtained your concentration bounds by different methods for different variables or if you used different convex functions with different right hand sides, then the story gets more complicated, but then you'll need to tell us more to have a meaningful discussion.
This is an amazing answer! I have had this question a couple of times before, now I know the answer.
I did some more digging and found an answer. In many cases, only the bounds on deviations suffice to get the bounds I wanted on the sum.
The answer is the theory of sub-gaussian random variables and, more specifically, the theorem on sums of (dependent) sub-gaussians.
To learn more, I can recommend these lecture notes as well as a (sub)chapter on this topic in the book "High-Dimensional Probability"
Could you share which theorem you mean? I have read the Chapters in Vershynin and Rigollet, but they only talk about sums of independent subgaussians...
|
2025-03-21T14:48:29.730936
| 2020-01-29T11:56:54 |
351422
|
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|
Stack Exchange
|
Matrix Completion SDP Relaxation and Duality
I am studying the matrix Completion problem, as well as its SDP relaxation. However, I am having trouble deriving the final SDP form of the matrix completion problem. I will give some background, which mainly comes from (ZHA) below.
Nuclear Norm Minimization
The matrix completion rank minimization form can be relaxed as a nuclear norm minimization problem, as the nuclear norm is the convex envelope of the rank of a matrix (ZHA).
$$
\label{eq:nuc_min_raw}
\underset{X \in \mathbf{R}^{m \times n}}{\text{minimize}} \quad ||X||_* \\
\text{subject to} \quad X_{i,j} = M_{i,j} \; \forall i, j \in \Omega
$$
We can rewrite the nuclear norm using its dual norm, the operator norm
$$
\label{eq:op_norm}
\underset{X \in \mathbf{R}^{m \times n}}{\text{min}} \quad \underset{Y \in \mathbf{R}^{m \times n}}{\text{max}} \langle X , Y \rangle \\
\text{subject to} \quad ||Y||_{\text{op}} \leq 1 \\
\text{and} \quad X_{i,j} = M_{i,j} \; \forall i, j \in \Omega
$$
The operator norm constraint can be written as the Linear matrix inequality $\begin{pmatrix} I_m & Y \\ Y' & I_n \end{pmatrix} \succeq 0$. To understand this, consider the following equivalencies, and then using a schur complement like argument.
$$||Y||_{\text{op}} \leq 1 \iff \\
\frac{||Yx||}{||x||} \leq 1 \quad \forall x \iff \\
\langle Yx, Yx \rangle \; \leq \;
\langle x, x \rangle \; \forall x \iff \\
\langle(YY^* - I)x, x \rangle \; \leq 0 \iff \\
(I - YY^*) \succeq 0
$$
Now we need to consider the dual of the resulting SDP:
$$
\label{eq:matcomp_sdp}
\underset{X, Y \in \mathbf{R}^{m \times n}}{\text{max}} \quad \text{tr}( X Y ) \\
\text{subject to} \quad \begin{pmatrix} I_m & Y \\ Y' & I_n \end{pmatrix} \succeq 0 \\
\text{and} \quad X_{i,j} = M_{i,j} \; \forall i, j \in \Omega
$$
Zhao claims the dual of this SDP is
$$
\label{eq:matcomp_dual}
\underset{W_1, W_2 \in \mathbf{R}^{m \times n}}{\text{min}} \quad \frac{1}{2} (\text{tr}( W_1 ) + \text{tr}( W_2 )) \\
\text{subject to} \quad \begin{pmatrix} W_1 & X \\ X' & W_2 \end{pmatrix} \succeq 0 \\
\text{and} \quad X_{i,j} = M_{i,j} \; \forall i, j \in \Omega
$$
How is this dual problem derived? What am I missing? It would be greatly appreciate if you could either reference a derivation, or explain the dual variables that are introduced, etc.
[ZHA]: Zhao, Yun-Bin, [An approximation theory of matrix rank minimization and its application to quadratic equations]
(http://dx.doi.org/10.1016/j.laa.2012.02.021), Linear Algebra Appl. 437, No. 1, 77-93 (2012). ZBL1242.65086.
What is $\mathbf{tr}(X,Y)$?
sorry that should just be $\textbf{tr}(XY)$
The derivation of this dual is provided in example 8.8 "Sum of singular values revisited" of section 8.6 "Semidefinite duality and LMIs" of Mosek Modeling Cookbook 3.2.1.
Thanks for the reference, however I am still having trouble deriving the relationship. Would you mind explaining how they "go back from the dual" and what variables are introduced?
Read from the beginning of section 8.6 so that you understand how to take the dual of a semidefinite program (LMI problem). It is then a straightforward matter.
|
2025-03-21T14:48:29.731155
| 2020-01-29T12:01:47 |
351423
|
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|
Stack Exchange
|
An analogue of the exponential function by replacing infinite series with improper integral
For every positive real number $x$ we define $$E(x)= \int_0^{\infty} x^t/t!\,\mathrm dt$$
where $t!=\Gamma(t+1)$. This is motivated by classical exponential function.
Is this function well defined (the problem of convergence)? Is there a real analytic extention of $E$ to all real numbers? What about a holomorphic extention to complex numbers? How can we compare $E(x+y)$ with $E(x)$ and $E(y)$? What kind of differential equation can be satisfied by $E$? Is $E$ one to one on real numbers?What can be said about its possible inverse?
for small $x$ the integral vanishes as $1/\log x$.
In what sense is this a "generalization of the exponential function"? I understand that one is of the form $\sum_nf(n,x)$ while the other is $\int f(t,x),dt$, but that doesn't really make it a generalization, does it? Is $(n^2+n)/2$ a generalization of $x^2/2$?
@GerryMyerson very interesting point! May be the word "generalization" should be replaced with another word. But I wonder to relate your example to situation of this question since we have 2 variable $t, x$ but you have 1 variable. Any way may you kindly suggest an alternative terminology(for "generalization")? Thank you in advance!
@GerryMyerson Yes, in a way it is. Pre-Fermat's (as far as I know) approaches to integration of polynomials rely on summing up n^k over n and thus closed summation formulas (n is later taken to infinity). So, the term generalisation is very appropriate, at least under a historical perspective.
This function was discussed on math.stackexchange some years back ... link
(Some obvious properties of $E$; too long for a comment, though).
The holomorphic extension of $E$ to $\mathbb{C} \setminus (-\infty, 0]$ (in fact, to the entire Riemann surface of the complex logarithm) is given by $$E(x) = \int_0^\infty \frac{\exp(t \log x)}{\Gamma(t+1)}\, dt,$$ where $\log$ denotes the principal branch of the complex logarithm. This follows from a standard application of Morera's theorem, involving Fubini's theorem, the estimate $$|\exp(t \log x)| = \exp(t \log |x|) = |x|^t \le a^t + b^t$$ when $a \le |x| \le b$, and integrability of $(a^t + b^t) / \Gamma(t + 1)$ over $(0, \infty)$.
In particular, for $x > 0$ we have $\log(-x + 0 i) = \log x + i \pi$, and hence
$$ E(-x+0i) = \int_0^\infty e^{i \pi t} \frac{x^t}{\Gamma(t+1)} \, dt $$
is not real-valued in any neighbourhood of $0$. Thus, there is no real-analytic extension of $E$ to $(-\epsilon, \infty)$ (as already follows from Carlo Beenakker's comment).
By dominated convergence theorem, the integral can be differentiated under the integral sign, so
$$E^{(n)}(x) = \int_0^\infty \frac{x^{t - n}}{\Gamma(t+1 - n)}\, dt = \int_{-n}^\infty \frac{x^t}{\Gamma(t+1)}\, dt .$$
This does not seem to lead to any interesting differential equation.
Since $E'(x) > 0$, clearly $E$ is increasing on $(0, \infty)$, with $E(0) = 0$ and $E(\infty) = \infty$.
One can easily find the Laplace transform of $E(x)$: when $\operatorname{Re} \xi > 1$, we have
$$ \int_0^\infty e^{-\xi x} E(x) dx = \int_0^\infty \frac{1}{\xi^{t + 1}} \, dt = \frac{1}{\xi \log \xi} . $$
(EDIT: This was meant to be an extended comment only, but since it has received a number of upvotes, let me add further remarks, inspired by Nemo's answer.)
The Laplace transform $\mathcal{L} E$ of $E$ has a simple pole at $\xi = 1$ with residue $1$, and a branch cut along $(-\infty, 0]$. Since it decays (barely) sufficiently fast at infinity, one can (carefully) write the usual inversion formula and then deform the contour of integration to the Hankel contour to find that
$$ E(x) = e^x - \frac{1}{\pi} \int_0^\infty e^{-t x} \operatorname{Im} (\mathcal{L} E(-t + 0i)) dt .$$
This leads to the formula given in Nemo's answer: since $$\mathcal{L} E(-t + 0i) = -\frac{1}{t \log(-t + 0 i)} = -\frac{1}{t (\log t + i \pi)} \, ,$$ we obtain
$$ E(x) = e^x - \int_0^\infty \frac{e^{-t x}}{t (\pi^2 + \log^2 t)} \, dt .$$
As a consequence, $e^x - E(x)$ is completely monotone, and
$$ E(x) = e^x - \frac{1 + o(1)}{\log x} $$
as $x \to \infty$. Further terms can be obtained in a similar way.
The function $E(x)$ itself is the Mellin transform of $1 / \Gamma(t + 1)$. Thus, $1 / \Gamma(t + 1)$ can be written as the inverse Mellin transform:
$$ \frac{1}{\Gamma(t + 1)} = \frac{1}{2 \pi i} \int_{c + i \mathbb{R}} t^{-1 - x} E(x) dx , $$
or, equivalently,
$$ \frac{1}{\Gamma(t)} = \frac{1}{2 \pi i} \int_{c + i \mathbb{R}} t^{-x} E(x) dx . $$
The definition of $E(x)$ looks a little bit like Mellin–Barnes integral, but the contour is wrong.
Finally, the (fractional) integral of $E(x)$ of order $\alpha$ is given by
$$ I_\alpha E(x) = \frac{1}{\Gamma(\alpha)} \int_0^x E(t) (x - t)^{\alpha - 1} dt = \int_0^\infty \frac{x^{t + \alpha}}{\Gamma(t + 1 + \alpha)} \, dt , $$
and so
$$ I_\alpha E(x) = \int_\alpha^\infty \frac{x^t}{\Gamma(t + 1)} \, dt . $$
This agrees with the expression for the derivatives of $E$ (which correspond to negative integer $\alpha$).
"...not real-valued in no neighbourhood of zero." I'm having trouble parsing this.
@Gerry Myerson: "not real-valued in any (real) neighborhood of zero".
@GerryMyerson: We don't need no education... Thanks!
Thank you very much for you very interesting answer and the concepts you introduced in your answer.
This is particular case of a classic integral studied by Ramanujan. See Chapter 11 in Hardy's book, "Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work", where it is shown that
$$
\int_{-\xi}^\infty\frac{x^t}{\Gamma(1+t)}\,dt+\int_0^\infty t^{\xi-1}e^{-xt}\left(\cos\pi \xi-\frac{\sin\pi \xi}{\pi}\ln t\right)\frac{dt}{\pi^2+\ln^2t}=e^x, \quad (x\ge 0, \xi\ge 0).
$$
From this it follows that your integral can be represented as an integral of elementary functions as follows
$$
\int_0^\infty\frac{x^t}{\Gamma(1+t)}\,dt=e^x-\int_0^\infty \frac{e^{-xt}\,dt}{t(\pi^2+\ln^2t)},\quad (x\ge 0).
$$
Also, see Fransen-Robinson constant
$$
C=\int_0^\infty\frac{dt}{\Gamma(t)}= 2.8077702420285...
$$
Thank you very much for your very interesting answer and the references you provided in the answer.
|
2025-03-21T14:48:29.731828
| 2020-01-29T12:09:43 |
351424
|
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|
Stack Exchange
|
Conjectures and open problems in representation theory
Are there very famous open problems or conjectures in representation theory, or in enumerative geometry, like the volume conjecture in topology?
Do you want representation theory problems or enumerative geometry problems?
Dear Ben, one of them or both will be ok. Since I am interesting in both fields and they have some connections to each other. Thank you so much for your advice!
I guess that for representation theory you want to be more specific since different people mean different things when they say "Representation Theory". For some open problems in the Representation theory of finite dimensional algebras and quivers, see e.g. http://www.math.uni-bonn.de/people/schroer/fd-problems.html.
The work of J.M. Landsberg on matrix multiplication draws from representation theory and complex algebraic geometry, but maybe not much enumerative geometry, and is driven by conjectures from computer science about the speed with which computers can multiply matrices.
I'm kind of amazed this question is still open, given how strict math overflow usually is. Regardless of what happens, just a bit of advice Khanh: this question is way too broad. Even if your question was faithful to the title, it would be far too broad. The only real answer is yes, there are many conjectures and open problems in representation theory. The more thought you put into your question, the better answers you will get.
The Clemens conjecture in enumerative geometry: a general quintic threefold has only finitely many rational curves in each positive degree.
There are many open, and seemingly deep, conjectures in modular representation theory (or block theory) in connection with enumerating representation-theoretic invariants:
a start of a list might be : Brauer's $k(B)$-problem, the Alperin-McKay Conjecture, the Alperin Weight Conjecture, Dade's conjectures, Isaacs-Navarro conjecture. Gabriel Navarro has several recent survey papers discussing these and other conjectures.
In a different part of (modular) representation theory, with perhaps a more geometric flavor, there are problems such as the Lusztig Conjecture (now known to be false in its original formulation), and work of Geordie Williamson.
As noted by Julian Kuelshammer, Representation Theory is a vast subject, and it might be helpful to point out which specific areas you are most interested in (I only mention two facets of the subject which are most familiar to me).
For some specific conjectures from the Lusztig-Williamson school, see: https://arxiv.org/abs/1703.05898
|
2025-03-21T14:48:29.732028
| 2020-01-29T12:19:34 |
351425
|
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"url": "https://mathoverflow.net/questions/351425"
}
|
Stack Exchange
|
Vector bundles on complete smooth variety $X$ in char $p$ and Frobenius
In "Vektorbündel auf Kurven und Darstellungen der algebraischen Fundamentalgruppe." Math. Z. 156 (1977), no. 1, 73–83. by Herbert and Ulrich, the authors consider a complete smooth variety $X$ defined over an algebraically closed field $k$ of positive characteristic.
They show that: for a vector bundle $E$ of rank $n$ over $X$, the following three conditions are equivalent: (1) $E$ is induced by a continuous representation $π_1(X) \rightarrow GL(n,k)$ (where $GL(n,k)$ is endowed with the discrete topology); (2) $E$ becomes trivial over some étale finite covering of $X$; (3) the pull-back of $E$ by a suitable power of the Frobenius morphism is isomorphic to $E$. Here $\pi_1(X)$ is the algebraic fundamental group in the paper.
I don't know much German, can anyone explain the idea between equivalence of (2) and (3)? And how do we classify Frobenius-unstable vector bundles?
There is a mistake in the paper in that (2) does not always imply (3) but does if E is stable. See the paper of Biswas and Ducrohet "An analogue of a Theorem of Lange and Stuhler"
|
2025-03-21T14:48:29.732133
| 2020-01-29T14:18:50 |
351434
|
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|
Stack Exchange
|
Fourier Mukai kernel which gives an equivalence only in one direction
If $X$ and $Y$ are two schemes and $F \in Perf(X \times Y)$, then we can define a functor from $Perf(X)$ to $Perf(Y)$ as the Fourier Mukai transform $\Phi^{X \rightarrow Y} = q_{\ast}(F \otimes p^{\ast}-)$, where $p$ and $q$ are the projections from $X \times Y$ to $X$ and $Y$ respectively. If $X$ and $Y$ are smooth we can change $Perf(-)$ to $D^b(-)$, while if choose $F$ only to be a complex with quasi coherent cohomologies we change $Perf(-)$ to $D_{qc}(-)$, the unbounded derived category of complexes with quasi coherent cohomologies. Obviously, we could exchange the role of $p$ and $q$ and get a functor from $Y$ to $X$. My question is whether the fact that $\Phi^{X \rightarrow Y}$ is an equivalence implies that $\Phi^{Y \rightarrow X}$ is an equivalence, and if it doesn’t I would like a counter example.
The reason why I believe it not might be the case that the implication holds is that there is no apparent relation between the two functors I described above, even though they share the same kernel. However, I was not able to come up with a counterexample.
|
2025-03-21T14:48:29.732240
| 2020-01-29T15:14:54 |
351439
|
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|
Stack Exchange
|
Lyndon basis of free Lie algebras
Let $A = \{a,b,c,d\}$ be a set of totally ordered alphabets, a Lyndon word over $A$ is a word $w$ in $A^*$ such that if $w=uv$ is a factorization of $w$ into non-empty subwords, then $u<v$ in lexicographic order. This is equivalent to being the unique minimum word (in lexicographic order) among all its rotations.
Let $\mathcal{L}(A)$ be the free Lie algebra on the set $A$. It is well known that Lyndon words in $A$ indexes a basis for $\mathcal(L)(A)$.
Assume $a$ is the least alphabet and let $w=a_1a_2\cdots a_k \in A^*$ (free monoid over $A$) define $L(w) = [a_1,[a_2,[\cdots[a_{k-1},a_k]\cdots]$ to be the right associative Lie monomial associated to $w$ in $\mathcal{L}(A)$.
By rotation, I always mean another word $w^{'}$ which is obtained from $w$ by rotating the alphabets cyclically and further we also want this word to start with $a$.
Let $w = abcad$ and let $w^{'}=adabc$ be a rotation of $w$.
My question is $L(w)$ and $L(w^{'})$ are linearly dependent as Lyndon words indexes a basis and $w$ and $w^{'}$ are rotationally equivalent.
More generally, I want to understand the linear dependence relation between the Lie monomials $L(w),L(w_1),\dots,L(w_k)$ where $w,w_1,w_2,\dots,w_k$ are all possible rotations of $w$ in the above sense.
How to understand the linear dependence relation between general right-associative Lie monomials and what makes the Lyndon words the indexing set for the basis of free Lie algebras? How the rotation of words affects the Lyndon words.
Kindly share your thoughts.
Thank you.
I am not sure I understand what you mean. But $\left[a,\left[b,\left[c,d\right]\right]\right]$, $\left[b,\left[c,\left[d,a\right]\right]\right]$, $\left[c,\left[d,\left[a,b\right]\right]\right]$ and $\left[d,\left[a,\left[b,c\right]\right]\right]$ are (in general) linearly independent.
It's been too long since I've thought about this stuff, but I want to recommend the book "Free Lie Algebras" by Christophe Reutenauer. It was an absolute pleasure to read, and I'm certain that the book has a good discussion of why there is a basis in correspondence with Lyndon words.
@darijgrinberg Can you please explain to me, how those monomials are linearly independent? I have edited the question, kindly reconsider this. Thank you.
Not much to explain -- just compute them in the free Lie algebra and check that there is no nontrivial relation. I did it with SageMath and https://github.com/darijgr/sage-subspace .
@darijgrinberg Thanks for the code. It is very nice. Everything has been worked out for free Lie algebras. Can you please tell me how to check the linear dependence of $\left[a,\left[b, \left[c,d\right]\right]\right]$,$\left[b,\left[c,\left[d,a\right]\right]\right]$,$\left[c,\left[d,\left[a,b\right]\right]\right]$ and $\left[d,\left[a,\left[b,c\right]\right]\right]$ if we impose some commutation relations among the generators $a,b,c$ and $d$ using this sage codes?. Thank you.
The sage-subspace code works in any algebra, as long as you know a basis of said algebra. I have used the free algebra, since it is the universal enveloping algebra of the free Lie algebra. If I imposed some commutativity relations, I'd need more subtle algebras, which Sage may or may not have. Feel free to experiment with any sufficiently up-to-date version of Sage (the development ones from 2020 should work).
|
2025-03-21T14:48:29.732456
| 2020-01-29T15:33:07 |
351441
|
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|
Stack Exchange
|
Relation between the number of spanning trees and the chromatic number of a graph
The number of spanning trees $\tau(G)$ of a simple graph $G$ is seen to satisfy the deletion-contraction recurrence:
$$\tau(G)=\tau(G-e)+\tau(G.e),$$
where $e$ is an edge of the graph $G$ and $G-e$ denotes the graph $G$ with edge $e$ deleted and $G.e$ denotes the graph $G$ with the edge $e$ contracted.
Now, the chromatic polynomial $P(G)$ also satisfies a similar deletion-contraction relation:
$$P(G)=P(G-e)-P(G.e).$$
This gives rise to the question that is there some sort of a relation between the chromatic properties and the number of spanning trees in a graph $G$. Specifically,
Is there a relation between the number of spanning trees and the chromatic number of a graph?
Are they at least asymptotically related?
Thanks beforehand.
Well there's the https://en.wikipedia.org/wiki/Tutte_polynomial
@SamHopkins thanks, but can we have a relation between the chromatic number and the number of spanning trees as such?
No, for example the bipartite graph K_{n,n} has many spanning trees, whereas a 2n vertex graph obtained from a clique on k-1 vertices plus 2n-k isolated vertices, then add a vertex adjacent to all others, has only k^{k-2} spanning trees. Sam Hopkins’ answer is really correct, for note what deletion-contraction is not really finding optimal colourings, it’s counting not necessarily optimal colourings.
|
2025-03-21T14:48:29.732580
| 2020-01-29T15:55:00 |
351445
|
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|
Stack Exchange
|
Looking for a "cute" justification for a Catalan-type generating function
The Catalan numbers $C_n=\frac1{n+1}\binom{2n}n$ have the generating function
$$c(x)=\frac{1-\sqrt{1-4x}}{2x}.$$
Let $a\in\mathbb{R}^+$. It seems that the following holds true
$$\frac{c(x)^a}{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{a+2n}nx^n.$$
QUESTION. Why?
You should consult known textbooks before asking a question https://dlmf.nist.gov/15.4.E18
Please provide some "cute" or "clever" proof.
It's known that $$c(x)^a = \sum_n \frac{a}{a+2n}\binom{a+2n}{n}x^n.$$
it is straightforward to get a differential equation of order 2 for the RHS (based on ${a+2n\choose n}(a+n)=(a+2n-1)(a+2n){a+2(n-1)\choose n-1}$) and check that LHS satisfies it and appropriate initial conditions. Not very clever, but quite a universal method.
Another routine proof: observe that ${1 \over \sqrt{1-4x}}=(x,C(x))^\prime $, and use Bürmann-Lagrange.
Here's a way to do it:
Recall that $C_n$ counts the number of lattice paths from $(0,0)$ to $(2n,0)$ taking only steps of the form $(1,\pm 1)$ that never goes below the $x$-axis; call this a Dyck path. Further, $$\frac{1}{\sqrt{1 - 4x}} = \sum \binom{2n}{n}x^k$$ which counts the total number of paths from $(0,0)$ to $(2n,0)$; call this a bridge. Also, $\binom{a+2n}{n}$ is the number of lattice paths (with the same step set) from $(0,0)$ to $(2n+a,a)$, since we have $a + 2n$ steps total with $n$ down steps (and thus $a + n$ up steps); call this an upward path.
Every upward path can be decomposed into:
A bridge (up to the last time it hits $0$).
A single up step
A dyck path (up until the last time it hits $1$).
another single step
a dyck path
and so on.
This provides a bijection from a single bridge with an $a$-vector of Dyck paths. Since the generating function for a single bridge with $a$-vector of Dyck paths is exactly the left-hand-side of your equality, it must equal the right-hand side.
As it stands such an argument would prove it only for integers $a>0$ — but that's enough because for each $n$ the equality of $x^n$ coefficients is a polynomial identity in $a$.
I'd say there's no question this meets the cuteness criterion.
Combining comments of @esg and myself, we have
$$\frac{c(x)^a}{\sqrt{1-4x}} = c(x)^a(xc(x))' = \frac{1}{(a+1)x^a}((xc(x))^{a+1})'$$
and thus
$$[x^n]\ \frac{c(x)^a}{\sqrt{1-4x}} = \frac{1}{a+1}[x^{n+a}]\ ((xc(x))^{a+1})'=\frac{n+a+1}{a+1} [x^n]\ c(x)^{a+1} $$
$$= \frac{n+a+1}{a+1}\frac{a+1}{a+1+2n}\binom{a+1+2n}{n}=\binom{a+2n}{n}.$$
Let $C_a(x)=\frac{c(x)^a}{\sqrt{1-4x}}$ and $B_a(x) =\sum_{n=0}^{\infty}\binom{a+2n}nx^n.$
The identity
$c(x)=1+xc(x)^2$ implies $C_{a+1}(x)= C_{a}(x)+x C_{a+2}(x).$
The recursion for the binomial coefficients implies
$B_{a+1}(x)= B_{a}(x)+x B_{a+2}(x)$.
If we show that $B_a(x)=C_a(x)$ holds for $a=1$ then it holds for all positive integers.
This follows from $B_1(x)=\frac{1}{2} \sum_{n=0}^{\infty}\binom{2+2n}{n+1}x^n=
\frac{1}{2x}(B_0(x)-1)=C_1(x).$
|
2025-03-21T14:48:29.732786
| 2020-01-29T16:13:09 |
351446
|
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|
Stack Exchange
|
Does $\sum_{i\le k}\mathrm{frac}(n\alpha_i)<1$ hold infinitely often?
For each $t \in \mathbf{R}$, let $\mathrm{frac}(t)$ be its fractional part.
Question. Fix reals $\alpha_1,\ldots,\alpha_k \in (0,1)$ such that $\sum_{i\le k}\alpha_i<1$. Do there exist infinitely many positive integers $n$ such that
$$
\sum_{i\le k}\mathrm{frac}(n\alpha_i)<1\,\,\,\,?
$$
The answer is affirmative if $\{\alpha_1,\ldots,\alpha_k\}\setminus \mathbf{Q}$ is a set which is linearly independent over $\mathbf{Q}$ (applying the multidimensional equidistribution theorem). But I don't know in general.
Yes, because you can make $n\alpha_i$ very close to integers simultaneously, so at the very next step you'll get the fractional parts almost the same as the original numbers.
Yes by Poincaré recurrence.
@fedja Probably I am missing your point: also $0.99$ is close to an integer..
Yes, but $frac(0.99+0.23)$ is close to $0.23$ (next step means $n+1$ instead of $n$)
|
2025-03-21T14:48:29.732998
| 2020-01-29T16:22:59 |
351447
|
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|
Stack Exchange
|
On a type of equations that involve certain multiplicative functions and polynomials, in relation to their number of solutions
Past weekend I was interested in the sequence A058891 from the On-Line Encyclopedia of Integer Sequences, from this, inspired by the equation due to Benoit Cloitre (2002) that shows the comments, I stated the following claim where for an integer $n\geq 1$ we denote the Euler's totient function as $\varphi(n)$, its number of divisors $\sum_{1\leq d\mid n}1$ as $\tau(n)$ and the product of distinct primes dividing to $n$ denoted by $$\operatorname{rad}(n)=\prod_{\text{primes }p\mid n}p$$
with the definition $\operatorname{rad}(1)=1$ (all these arithmetic functions are multiplicative).
Claim 1. Let $x=O_n=2^{2^{n-1}-1}$ (be a term of the sequence A058891), then for each integer $n>1$ we've that $x=O_n$ satisfies $$\operatorname{rad}(x)^{\tau(x)}=4^{\varphi(\tau(x))}.\tag{1}$$
Motivated by simple experiments with Pari/GP scripts, I wrote the following claims that seem similar, and that involve other sequences of integers: the known as Mersenne primes and Fermat primes.
Claim 2. Let $x=M_n=2^n-1$ be a Mersenne prime, then $x$ satisfies $$\operatorname{rad}(x+1)^{\tau(x+1)}=2\varphi(x)+4.\tag{2}$$
Claim 3. Let $x=F_n=2^{2^n}+1$ be a Fermat prime, then the identity $$\operatorname{rad}(x-1)^{\tau(x-1)}=2\varphi(x),\tag{3}$$
holds.
Question. I would like to know what is more plausible, if the existence of different $P(x)\neq Q(x)$ and positive polynomials $P(x)>0$ and $Q(x)>0$ for all integer $x\geq 2$, with integer coefficients $P,Q\in\mathbb{Z}[X]$, and say the same degree $1\leq\deg P=\deg Q$ such that there exist positive constants $A\geq 1$ and $B\geq 0$ for which you can to prove, for one of such suitable choices, that the equation
$$\operatorname{rad}(P(x))^{\tau(P(x))}=A\varphi(Q(x))+B,\tag{E}$$ have infinitely many solutions, or well, for the same set of conditions for our polynomials and constants, the equation $\text{(E)}$ should to have finitely many solutions. If you want you can study this, answering, from a theoretical point of view, or providing examples of equations $\text{(E)}$ that have infinitelty many or finitely many solutions.
Many thanks.
Thus with the question what is more plausible (my English is bad) I am asking about an answer that shows ditect examples, or well a discussion using your knowledges in number theory about if the equation of the form $\text{(E)}$ should to have infinitely many solutions for a a suitable proposal of constants $A$ and $B$ and polynomials $P(x)$ and $Q(x)$, or well if this problem should this problem should have only a finite number of solutions for any choice of constants and polynomials that satisfy previous conditions.
Since the extension of this post is a bit large, I hope that there aren't typos. In this post I prefer the notation $\tau(n)=\sum_{1\leq d\mid n}1$ for the multiplicative function that counts the number of positive divisors of an integer $n\geq 1$ (instead of the notation $\sigma_0(n)$ or $d(n)$). Many theorems about the size and asymptotic of this function and for the Euler's totient function are known.
All, if someone wants to provide me feedback about the question/post I am waiting, many thanks.
|
2025-03-21T14:48:29.733210
| 2020-01-29T16:40:34 |
351449
|
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"url": "https://mathoverflow.net/questions/351449"
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|
Stack Exchange
|
Case of equality in entrywise spectral radius bound
Let $A,B$ denote square matrices such that $\lvert A_{ij}\rvert\le B_{ij}$ for all $i,j$, and denote the spectral radius by $\rho$. From the Gelfand spectral radius formula it is easy to see that
$$\rho(A)\le\rho(\lvert A\rvert)\le\rho(B),$$
where $\lvert A\rvert$ is meant entrywise.
Question: What is known about the case of equality? Is it true that equality can only occur if $A_{ij}=e^{i\phi}B_{ij}$ for some $\phi\in[0,2\pi]$?
If $B$ is reducible, then all but one block can be replaced by zero blocks, with the largest eigenvalue unchanged. So you require $B$ to be irreducible, and likely primitive. In the latter case, if $0 \leq A \leq B$ entrywise, and $A \neq B$, then $\rho(A)$ is strictly less than that of $B$. Thus if $B$ is primitive and $|A_{ij}| \leq B_{ij}$, then $\rho(|A|) = \rho (B)$ implies $B = |A|$. This answers only part of the question; likely the rest ($\rho(A) = \rho(|A|)$ implies what you want) is similarly not difficult.
|
2025-03-21T14:48:29.733302
| 2020-01-29T16:45:51 |
351451
|
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|
Stack Exchange
|
Computing local residues of traces of bilinear forms on algebraic number fields
This is a question about computing the local residues of a general symmetric bilinear form over $\mathbb{Q}$. I've been using Lam's Introduction to Quadratic Forms as a reference, but I'm stuck on the following.
Let $\mathbb{Q}(\alpha)$ be a degree $d$ extension of $\mathbb{Q}$. Given a non-zero element $q=\sum_{i=0}^{d-1}b_i\alpha^i\in\mathbb{Q}(\alpha)$, let $\langle q\rangle\in\mathrm{W}(\mathbb{Q}(\alpha))$ denote the Witt class of the symmetric, non-degenerate bilinear form $\mathbb{Q}(\alpha)\times\mathbb{Q}(\alpha)\to\mathbb{Q}(\alpha)$ given by $(x,y)\mapsto qxy$. Define $\operatorname{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\langle q\rangle\in\mathrm{W}(\mathbb{Q})$ by post-composing $\langle q\rangle$ with the field trace $\mathbb{Q}(\alpha)\to\mathbb{Q}$. For each prime $p$ (including $p=\infty$), we have a local residue map $\partial_p:\mathrm{W}(\mathbb{Q})\to\mathrm{W}(\mathbb{F}_p)$ (or $\mathrm{W}(\mathbb{R})$ for $p=\infty$), and the weak Hasse-Minkowski principle says that the Witt class of $\operatorname{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\langle q\rangle$ is determined the values $\partial_p\operatorname{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\langle q\rangle$ for all $p$.
Question: Can I compute $\partial_p\operatorname{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\langle q\rangle$ in terms of $q$ and the minimal polynomial of $\alpha$ over $\mathbb{Q}$? Or is a general formula too much to ask for?
For example, when $q=1$, this form is a trace form, and I believe a theorem of Taussky-Todd implies that $\partial_\infty\operatorname{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\langle 1\rangle\geq 0$ (Corollary I.5.2, A survey of trace forms of algebraic number fields by Conner and Perlis).
Motivation: I'm writing up an (algebro-geometric) result, and I have a few equations involving terms of the form $\partial_p\operatorname{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\langle q\rangle$. I can leave them this way, but I'm obviously not a number theorist and would thus like to record these values more explicitly if possible.
|
2025-03-21T14:48:29.733443
| 2020-01-29T16:51:48 |
351452
|
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"Adam",
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}
|
Stack Exchange
|
Algorithm for computing external angles for the Mandelbrot set
Let $M$ be the Mandelbrot set: there exists a unique series
$$
\psi(z) := z + \sum_{m=0}^{+\infty} b_m z^{-m} = z - \frac{1}{2} + \frac{1}{8} z^{-1} - \frac{1}{4} z^{-2} + \cdots
$$
which defines a conformal bijection between the complement $\{z\in\mathbb{C} : |z|>1\}$ of the closed unit disk in $\mathbb{C}$ and the complement of the Mandelbrot set. This function $\psi$ is sometimes known as the “Jungreis function”, see the answers to this question for more. The argument $\arg(\psi^{-1}(w))$ for a point $w\not\in M$ is called the external angle of $w$.
There exists an easy way to “almost” compute $\arg(\psi^{-1}(w))$, or $\psi^{-1}(w)$ itself for that matter: indeed, if we let $p_0(w) := w$ and $p_{i+1}(w) := p_i(w)^2 + w$, then
$$
p_n(\psi(z)) = z^{2^n} + o(1)
$$
as $z\to\infty$, so $\psi^{-1}(w)$ can be “almost” computed as the limit of the $(2^n)$-th root of $p_n(w)$ as $n\to+\infty$. The reason for the “almost” is that, while this indeed allows for computation of the modulus $|\psi^{-1}(w)|$, it leads to an indetermination between $2^n$ values on the argument. The formula
$$
\psi^{-1}(w) = w \mskip3mu \prod_{n=1}^{+\infty} \left(1 + \frac{w}{p_{n-1}(w)^2}\right)^{1/2^n}
$$
is no better (it also requires computing $(2^n)$-th roots and one cannot simply take the principal determination; my understanding is that one needs to find a determination of $\Big(1 + \frac{w}{p_{n-1}(w)^2}\Big)^{1/2^n}$ that is continuous outside of $M$, which seems computationally intractable).
So, is there a way to lift this square root indeterminacy and compute external arguments for arbitrary $w\not\in M$? Is there an algorithm that does this in a reasonably efficient way (which excludes, e.g., trying to trace external rays outwards towards infinity)?
I was unable to find anything relevant in the literature. There is a 1986 paper by Douady titled “Algorithms for computing angles in the Mandelbrot set” which seems promising, but it seems to concerns the computation for points of $M$, not points outside $M$. This web page about the Mandel program actually discusses the issue (in the section called “Computation of the external argument”), but the description is vague (e.g., where it speaks of a “modified” function $\arg(z/(z-c))$), and the conclusion that “the discontinuities are moved closer to the Mandelbrot set” is not too promising.
https://gitlab.com/adammajewski/parameter_external_angle
I have implemented some algorithms based on Kawahira's paper, which as presented goes $\theta \to c \not\in M$, but can be adapted to go $c \to \theta$. $\theta$ is conveniently expressed in turns as a binary expansion. When tracing inwards, one peels off the most-significant bit (aka angle doubling) each time the ray crosses a dwell band (integer part of normalized iteration count increases by 1). The trick when tracing outwards is to prepend bits when crossing dwell bands, depending if the outer cell was entered from its left or right inner cell. A picture may make it clearer:
The exterior grid is generated from the fractional part of the smoothed iteration count, and the argument of the final (first to escape) $z$ iterate. One can see that approaching the $\frac{1}{2}$ bond point, the cells alternate left/right corresponding to the binary expansion $.(01)$ or $.(10)$. The argument of the first iterate to escape (typically floating point), within the cell of the starting point, can be used to get a few more least significant bits for the accumulated angle, but the prefix is found by accumulating bits one by one when tracing the ray outwards.
Source code:
m_d_exray_in.c $\theta \to c$, machine double precision
m_r_exray_in.c $\theta \to c$ in arbitrary (dynamically changed as necessary) precision
m_d_exray_out.c $c \to \theta$, machine double precision
m_r_exray_out.c $c \to \theta$, arbitrary (but fixed, dynamic is possible but still TODO) precision
m-exray-in.c command line driver program showing usage of the library functions
m-exray-out.c command line driver program showing usage of the library functions
git clone https://code.mathr.co.uk/mandelbrot-numerics.git
However, tracing external rays to/from dwell $n$ takes $O(n^2)$ time (even ignoring that higher $n$ needs higher precision which costs more), which may make it too slow in practice. I am also looking for faster methods since some years, but I haven't found any yet. See my related question: fast algorithms for external angle computations
Googling for the paper of Douady, I found this paper by Tomoki Kawahira: he explicitly computes the quantity $\psi^{-1}(w)$ for each $w\in\Bbb C\setminus M$ by using Newton's method (he considers $\Phi=\psi^{-1}$), and this allows also an error estimate and a quite fast convergence rate: could it be of some help?
This paper, although interesting, appears to try to compute $\psi$ (which, as you point out, it calls $\Phi^{-1}$) rather than $\psi^{-1}$ as I want. I can't see a way to adapt it to computing $\psi^{-1}$ (I tried, inter alia, applying Newton's method to invert $\psi$ which can be approximated by a power series, but there are all sorts of horrible instabilities which seem to make this approach doomed).
http://math-functions-1.watson.jp/sub4_math_020.html#section030
|
2025-03-21T14:48:29.733783
| 2020-01-29T17:01:31 |
351454
|
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"Bashar Saleh",
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|
Stack Exchange
|
Conjugation map on $\mathbb CP^n\smallsetminus \mathring{D}^{2n}$ that preserves boundary, $n\geq 4$ even
The conjugation map
$$\mathrm{conj}\colon\mathbb{C}P^{n-1}\to \mathbb{C}P^{n-1}$$
extends to an endomorphism of $\mathbb CP^n\smallsetminus \mathring{D}^{2n}$ (unique up to homotopy), since $\mathbb CP^{n-1}$ is a deformation retract of $\mathbb CP^n\smallsetminus \mathring{D}^{2n}$.
I'm wondering whether an extension of the conjugate map is homotopic to an endomorphism that preserves the boundary pointwise, when $n\geq 4$ is even.
When $n$ is odd, it is impossible to make a boundary-preserving extension since it is impossible at the level of the rationalizations of these spaces (which can be proven algebraically, using algebraic models). However, when $n\geq 4$ is even, boundary-preserving extensions are possible rationally, and I am wondering whether this can be realized at the level of the non-rationalized spaces.
I'd be thankful for any suggestion of literature that may be relevant for this type of questions.
Sure, the map induced on the boundary sphere is the restriction of complex conjugation in $T_p \Bbb{CP}^n = \Bbb C^n$ to the unit sphere. This is orientation preserving precisely when $n$ is even, and orientation preserving diffeomorphisms of the sphere are isotopic. So start with your given map and add a small collar $[0,1] \times S^{2n-1}$ where you perform this isotopy.
@Mike Miller, thank you :-)
|
2025-03-21T14:48:29.733896
| 2020-01-29T17:34:09 |
351456
|
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"url": "https://mathoverflow.net/questions/351456"
}
|
Stack Exchange
|
Homology of a semisimplicial scheme
This is a question about the homology of a complex made of algebraic varieties. Consider the following subgroups of $\mathrm{SL}_3$ (defined over $\mathbb{Z}$).
$$
P_{1,2} = \left\{\left(\begin{smallmatrix}*&*&*\\&*&*\\&&*\end{smallmatrix}\right)\right\} \quad
P_{1} = \left\{\left(\begin{smallmatrix}*&*&*\\*&*&*\\&&*\end{smallmatrix}\right)\right\} \quad
P_{2} = \left\{\left(\begin{smallmatrix}*&*&*\\&*&*\\&*&*\end{smallmatrix}\right)\right\} \quad
P_{\emptyset} = \left\{\left(\begin{smallmatrix}*&*&*\\*&*&*\\*&*&*\end{smallmatrix}\right)\right\}
$$
$$
U = U_{1,2} = \left\{\left(\begin{smallmatrix}1&&\\*&1&\\*&*&1\end{smallmatrix}\right)\right\} \quad
U_{1} = \left\{\left(\begin{smallmatrix}1&&\\&1&\\*&*&1\end{smallmatrix}\right)\right\} \quad
U_{2} = \left\{\left(\begin{smallmatrix}1&&\\*&1&\\*&&1\end{smallmatrix}\right)\right\} \quad
U_{\emptyset} = \left\{1\right\}
$$
For each $I \subseteq \{1,2\}$ let $V_I = UP_I/P_I$. This can be identified with $U_I$ and is therefore a variety (a $\mathbb{Z}$-scheme, I suppose). For $I \supseteq J$ the map $gP_I \mapsto gP_J$ defines a morphism $V_I \to V_J$. The $V_I$ together with these maps form a semisimplicial object in the category of schemes over $\mathbb{Z}$.
So there should be an associated homology consisting of functors $H_i \colon \mathcal{Fields} \to \mathbb{Z}\mathcal{-mod}$, $F \mapsto H_i(V_*(F);\mathbb{Z})$.
My (vague) question is this: is there a way to work with this homology (e.g. compute it) that makes use of the variety structure rather than evaluate it field by field? (There is an obvious generalization to $\mathrm{SL}_n$ for other $n$ and a slightly less obvious one to other Chevalley groups; these are implicitly included in the question.)
I would be grateful for relevant references and clarifications about imprecisions in the exposition.
Context: For any field $F$ the associated complex is the augmentation of the complex opposite a chamber in a spherical building and Abramenko shows that its homology vanishes in all degrees except for the top degree provided $F$ is (explicitly) large compared to $n$. My motivation for the question is twofold:
Could it be possible to recover Abramenko's result (without the explicit bound) for abstract model-theoretic reasons? (Like: it is true for infinite fields so it has to be true for large enough finite fields.)
Abramenko's bound on the size of $F$ is exponential in $n$, yet no counterexample is known for $F \ne \mathbb{F}_2,\mathbb{F}_3$. So conceivably there is a bad prime phenomenon that is uniform to all complexes. Can one guess where in the above description such a bad prime phenomenon arises?
I guess $P_I$ should be $B_I$?
|
2025-03-21T14:48:29.734068
| 2020-01-29T17:40:39 |
351457
|
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|
Stack Exchange
|
The asymptotic of $|\{1\leq n\leq x|\gcd(n,S(n))=1\}|$, with $S(n)$ the sum of remainders, and get idea for other miscellany problem
Let $n\geq 1$ be an integer. In this post we denote the sum of remainders function as $$S(n)=\sum_{k=1}^n n \bmod k,$$ for example $S(1)=S(2)=0+0$ and $S(5)=0+1+2+1+0=4$. In the literature there are problems that were studied related to the condition $\gcd(n,f(n))=1$, for a given arithmetic function $f(n)$.
Question.
A) Is it possible to provide roughly a cheap bound for the cardinality $$\#\{1\leq n\leq x|\gcd(n,S(n))=1\}$$
as $x$ grows to $\infty$?
B) The sequence of primes $p$ that satisfy the condition $$\gcd(p,S(p))>1$$ starts as $2,11,17,2161,\ldots$. Can you provide us any idea about if this sequence has finitely many terms?
Just to emphasize, since I'm asking two questions, only is required that you provide a cheap bound for A) and a suitable reasoning/heuristic for B), to get idea for these problems.
Computational evidence and documentation for Question B. We've the following script in Pari/GP showing the first few terms
for(n=1, 10000, if(gcd(n,sum(k=1,n,n%k))>1&&isprime(n)==1,print(n)))
that you can evaluate on the website Sage Cell Server choosing as Language GP. Here the string sum(k=1,n,n%k) is our sum of remainders $S(n)$ with n%k coding $n \bmod k$ for each integer $1\leq k\leq n$.
Thanks you very much @Lspice
|
2025-03-21T14:48:29.734183
| 2020-01-29T18:32:39 |
351459
|
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"Fernando Muro",
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|
Stack Exchange
|
Tensor over projective $R$-modules remain projective
It is known that if $R$ is a commutative module and P_1, P_2 are $R$ projective then $P_1\otimes P_2$ is again projective. Now, if $P$ is projective and $X$ is any, and $P\otimes X$ is projective, does it follow that $X$ is projective? Any countexamples?
Take $X$ non projective and $P=0$.
(Edited after Johannes Hahn's comment.)
We need to assume that $P$ has positive rank at every maximal ideal to rule out examples like Fernando Muro's.
1) Suppose $R$ is local. Then $P$ and $P\otimes X$ are free. Write $P\approx R^n$ with $n>0$. Then $X^n\approx P\otimes X$ is free. Because $n>0$, $X$ is a direct summand of the free module $X^n$ and hence free.
2) More generally, $X$ is projective if and only if it becomes free after localizing at any maximal ideal, so $X$ is projective by 1).
Wait, why is $F\otimes X = (P\oplus P')\otimes X$ projective? Only $P\otimes X$ is assumed to be projective, $P'\otimes X$ isn'.
@JohannesHahn: I think this is fixed now.
|
2025-03-21T14:48:29.734293
| 2020-01-29T19:05:53 |
351462
|
{
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"authors": [
"Dominic van der Zypen",
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|
Stack Exchange
|
The "contrary" of an isomorphism
Roughly, my question is: is there a standard name for functions which one might characterise as the "contrary" of an isomorphism?
Here is a more precise version of my question. Working model-theoretically, consider the following definition.
Definition. Let $\mathscr{L}$ be a relational signature, and let $\mathcal{A}, \mathcal{B}$ be $\mathscr{L}$-structures. An $\mathscr{L}$-anti-embedding $\pi : A \longrightarrow B$ is any injection such that:
$(a_1, \ldots, a_n) \in R^\mathcal{A}$ iff $(\pi(a_1), \ldots, \pi(a_n)) \notin R^\mathcal{B}$, for each $n$-place predicate $R \in \mathscr{L}$ and all $a_1, \ldots, a_n \in A$
A bijective $\mathscr{L}$-anti-embedding is an $\mathscr{L}$-anti-isomorphism. An $\mathscr{L}$-anti-isomorphism from a structure to itself is an $\mathscr{L}$-anti-automorphism.
My question is: are there standard names for the kinds of functions I just defined? Indeed, are they discussed anywhere?
I have only encountered one such function "in the wild". I was reading Thomas Forster on Church-Oswald set theory; he called his $\{\in\}$-anti-automorphism an "antimorphism". (NB that, in the Church-Oswald setting, self-membered sets are perfectly ok.)
Update: Googling the phrase "anti-isomorphism" teaches me that, in group theory, an anti-isomorphism is standardly defined as a bijection $\pi : \mathcal{G} \longrightarrow \mathcal{H}$ such that $\pi(x \cdot^\mathcal{G} y) = \pi(y) \cdot^\mathcal{H} \pi(x)$. That's obviously a totally different idea. So I should use a different name! If there is no standard name, I welcome suggestions!
Hi Tim - welcome to MathOverflow! I would say in a "set setting", the contrary of an isomorphism is a constant map. Maybe there is some categorical or model theoretical setting in which we can say that a morphism is "constant".
@DominicvanderZypen How are constant functions contrary to isomorphisms in the sense specified by the OP?
@DominicvanderZypen - maybe my mention of set theory was helpful, as Church-Oswald set theory is quite unusual. In the context of set theories, an anti-isomorphism is a bijection $\pi$ such that $a \in^\mathcal{A} b$ iff $\pi(a) \notin^\mathcal{B} \pi(b)$. Trivially, there are no anti-isomorphisms between models of ZFC, since they'd need to send the empty set of one model to a universal set in the other. Here's an easy way to obtain anti-isormorphic models though: take an ${\in}$-structure $\mathcal{A} $, and define $\in^\mathcal{B}$ as $(A \times A) \setminus\mathord{\in^\mathcal{A}}$.
@MarkSapir: thanks. I mentioned this in an update, maybe I should make it more explicit (to avoid confusion). And certainly I should avoid using the name "anti-isomorphism", given it has a standard meaning different from what I'm after. That said: I think isomorphism and anti-isomorphism [as you define them] of groups are equivalent only when the target group is abelian? (Also, presumably there is a more general category-theoretic idea, of a functor from $\mathcal{A}$ to $\mathcal{B}^\textbf{op}$?)
@MarkSapir: OK your point was: groups are isomorphic iff they are anti-isomorphic (with the witnessing functions differing as described). Gotcha. Thanks! Well this only further confirms that I must avoid the name "anti-isomorphism" for what I am defining, since being isomorphic is not the same as being anti-isomorphic [in the sense I defined it]!
Tim has drawn my attention to this (Thank you, Tim!) Perhaps i should provide what Old Lags do in this kind of setting, namely provide some ancient history.
The context is Quine's NF. Let us say the dual of a formula of the language of set theory is the result of replacing all occurrences of $\in$ by `$\not\in$'. It's routine to show that the dual of any axiom of NF is a theorem of NF. Thus if we take $<$M,$\in>$ a model of NF and consider M equipped with the complement (in MxM) of the membership relation we get another model of NF! Are these two structures elementarily equivalent? Not reliably! Might they be isomorphic? If they are, then the isomorphism is an antimorphism. Of course - since this is NF - there is the possibility that (not only might there be an antimorphism but) the antimorphism is a set of the model! It's unknown if this can happen. The best i can do is show that every model of NF is elementarily equivalent (with respect to stratifiable formul{\ae}) to one with two (set) permutations $\sigma$ and $\tau$ satisfying
both
$$(\forall x y)(x \in y \leftrightarrow \sigma(x) \not\in \tau(y))$$
and
$$(\forall x y)(x \in y \leftrightarrow \tau(x) \not\in \sigma(y))$$
Arranging for $\sigma = \tau$ is beyond me at the moment. The word
`antimorphism' is a coinage of your humble correspondent, tho' i can't give you chapter and verse.
Sorry if some readers find this perhaps off-piste, but this is the context of Tim's question, and might be helpful.
Thomas Forster www.dpmms.cam.ac.uk/~tf
In the case of an anti-isomorphism we could identify $B$ with $A$ in which case $f$ is the identity function but each predicate for $B$ is the negation of that predicate in $A.$ This is essentially what you did in a comment with $(A \times A) \setminus\mathord{\in^\mathcal{A}}.$
For the unary case of each predicate indicating belonging or not belonging to a certain subset, one gets the set complements. If the model was a simple graph with loops allowed then we have the complementary graph. There the set is the vertices. If one wished to avoid loops then the set could be the pairs of vertices though that reduces to the previous example. Similarly a linear order and the reverse order. Again the pairs $(a,a)$ need care so either take the set to be ordered pairs of distinct elements or else semi-strong linear orders meaning no assumption of reflexive or non-reflexive.
What does Church-Oswald set theory does with the possibility that $a \in a?$ I suppose that again it could be permitted on an element by element basis.
Graph theory is one place I'd expect to encounter these maps, for exactly the reasons you say: the construction which deletes all the edges on a graph, then adds new edges (allowing for loops) where there are none before, gives you an anti-isomorphic [in my terminology] graph. This works for directed graphs too. Then set theory can be considered as a theory of directed graphs. Most set theories shun the possibility of self-membered sets, but Church-Oswald deliberately allows for them. (So $a \in a$ is totally unproblematic, and one can have an anti-automorphism [in my terminology].)
|
2025-03-21T14:48:29.734791
| 2020-01-29T19:51:55 |
351469
|
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|
Stack Exchange
|
Sign of expectation value
Consider a multivariate Gaussian-type measure $$d\lambda(x):=\nu_{\mu,\Sigma} e^{-\langle (x-\mu), \Sigma^{-1}(x-\mu) \rangle - \vert x \vert^2} $$
with vector $\mu \in \mathbb R^n$ and $\Sigma$ positive definite and $\nu_{\mu,\Sigma}$ a normalizing constant to turn $d\lambda$ into a probability measure.
Let $m$ be the vector-valued expectation value $m:=\int_{\mathbb R^n} x d\lambda(x).$
We then consider the expectation value for $X$ distributed according to the measure $\lambda:$
$$\mathbb E \left( \langle X-m, \Sigma^{-1} y \rangle^2 \langle X-m, \Sigma^{-1} \mu \rangle \right).$$
Question: Can we say anything about the sign of this expectation value for general vectors $y \in \mathbb R^n$?-From how I obtained this expression I conjecture that this expression is never strictly positive, but I cannot see it right away.
Isn't $\lambda$ simply a Gaussian measure with covariance matrix $\tfrac{1}{2} (\Sigma^{-1} + \operatorname{Id})^{-1}$?
@MateuszKwaśnicki well, there is also the $\mu$...
But $\mu$ only shows up in the expression for the mean $m$, does it not? I mean, the exponent is a quadratic function of $x$ with prinicipal term $\Sigma^{-1} + \operatorname{Id}$.
I definitely agree with the last sentence you wrote.
The expectation is indeed never strictly positive: it is equal to zero.
The density of $\lambda$ is proportional to $\exp(-\tfrac{1}{2} \langle (x - m), A^{-1} (x - m)\rangle)$, where $A = \tfrac{1}{2} (\Sigma^{-1} + \operatorname{Id})^{-1}$ is a positive definite matrix. Thus, $Y = X - m$ is a centred Gaussian vector (with covariance matrix $A$). It follows that
$$\mathbb{E} ((\langle (X - m), \Sigma^{-1} y \rangle)^2 \langle (X - m), \Sigma^{-1} \mu \rangle) = \mathbb{E} ((\langle Y, a\rangle)^2 \langle Y, b\rangle)$$ for appropriate vectors $a = \Sigma^{-1} y$, $b = \Sigma^{-1} \mu$. Now it is straightforward to note that the last expectation is zero: $Y$ is equal in distribution to $-Y$, and the expression under the expectation is an antisymmetric function of $Y$.
|
2025-03-21T14:48:29.735079
| 2020-01-29T20:47:03 |
351473
|
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|
Stack Exchange
|
One parameter change of a section of $T^*M \otimes End(TM)$ on an affinely flat manifold
Let $\nabla$ be a flat symmetric connection in the tangent bundle of a smooth manifold $M.$ Let $A$ be a global section of $T^*M \otimes End(TM).$ Let $\phi:B(0,1) \to M$ be a local affine chart on $M$ defined on the ball of radius $1$ centered at the origin in $\mathbb{R}^n.$ Let $(\partial_i)_{i=1,n}$ be the local frame induced in $TM.$
Consider the local matrix of $1$ forms defined by
$$ A(\partial_i(x))=A_i^j (x) \otimes \partial_j(x),$$ where $x \in B(0,1).$
Is it true that for a fixed $t$ ($0 < t \leq 1$), there exist a unique global section of $T^*M \otimes End(TM),$ denoted $A^t$ that satisfies $$ A^t(\partial_i(x))=tA_i^j (tx) \otimes \partial_j(x),$$ for all $x \in B(0,1)?$
What is your question? To prove the last statement?
|
2025-03-21T14:48:29.735188
| 2020-01-29T22:20:17 |
351476
|
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|
Stack Exchange
|
Composition with an exact functor
Suppose $F:C\longrightarrow D$ is exact and that $G:D\longrightarrow E$ is another functor such that $GF$ is exact. Is $G$ necessarily exact? Counter examples?
I think the question as currently stated is no for obvious reason that $GF$ do not care about the behavior of $G$ outside of the image of $F$. The example Fernando Muro gave below also shows this, as $G$ is exact on the image of $C$. It would be more interesting to seek for a counterexample where $F$ is essentially surjective.
Thanks for the observation, you are right to point that out.
Any suggestion for the counter example in this case?
The simplest counterexample is $F=0$.
No. Let $C$ be the category of $\mathbb{Q}$-vector spaces and $D=E$ the category of abelian groups. Take $F$ to be the forgetful functor, which is fully faithful and exact, and $G=-\otimes\mathbb{Z}/2$. The latter is not exact but the composite $GF=0$ is the trivial functor, which is exact.
|
2025-03-21T14:48:29.735298
| 2020-01-29T22:39:19 |
351478
|
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|
Stack Exchange
|
Homologically trivial fibre
Let us consider a homotopy fibre sequence of connected spaces $A\rightarrow B\rightarrow C$
and let $K$ be a fixed field. Assume that the homology $H_{\ast}(A, K)$ is trivial and that $C$ is a nilpotent space (but not simply connected).
Does $H_{\ast}(B, K)\rightarrow H_{\ast}(C, K)$ have to be an isomorphism ?
Yes, by the Serre spectral sequence.
Using infinity-categorical language this become very clean (and require no spectral sequences, which in this degenerate cases I consider as a good thing). The fibration is classified by a functor $C\to \mathcal{S}$ classifying the fibers. The assumption tells us that the projection to the constant functor on a point is an $H_*(-;K)$-equivalence and so it is an equivalence on the homology of the colimit, since homology (i.e. tensoring with $K$) is colimit preserving.
@FernandoMuro It will be nice if you transform your comment to an answer with more details .
@S.carmeli I would love to understand your comment, but I'm not used to the infinity categorical language...
@TTip thats ok, ill try to explain. In infinity category theory, a topological space $X$ is a legitimate category, in which points are objects and path are morphisms, and homotopies are "morphisms between morphisms". In this language, there is an equivalence between the homotopy category of fibrations over $X$ and the category of functors from the category $X$ to the infinity category of (nice enough) topological spaces. This associattion takes a fibration to the "functor" of the fibers of the fibration. The total space of the fibration then realizes as the colimit of this functor.
In your situation, the Serre spectral sequence looks as follows
\[H_p(C,H_q(A,K))\Rightarrow H_{p+q}(B,K).\]
The left hand side is the $E^2$ term. There, $H_q(A,K)$ carries an action of $\pi_1(C)$ induced by the fibration, and $H_p(C,H_q(A,K))$ is the homology with local coefficients. Now, if $A$ has the homology of a point then then $H_q(A,K)=0$ for $q>0$ and $H_0(A,K)=K$ with the trivial action of $\pi_1(C)$ (any self-map of a connected space induces the identity on $H_0$). Therefore $E^2_{p,q}=0$ for $q>0$ and $E^2_{n,0}=H_n(C,K)$ is the ordinary homology. Moreover, the spectral sequence ends in the second step and the edge morphism $H_n(B,K)\twoheadrightarrow E^\infty_{n,0}\subset E^2_{n,0}=H_n(C,K)$, which is induced by the map $B\rightarrow C$, is an isomorphism.
Here is a compromise between Fernando's answer and S. Carmeli's comment. Also, we may as well use $E_*$ any homology theory and there's no need to assume $C$ is nilpotent, just that every fiber is $E_*$-acyclic (so, e.g., if $C$ is connected and you have your assumption then we're okay.)
The statement is okay with filtered colimits and equivalences, so we may as well assume $C$ is a finite dimensional CW-complex and induct on the dimension. For the inductive step, observe that $\mathrm{sk}_{n-1}C \to \mathrm{sk}_nC$ is obtained by cobase change from $\coprod \partial D^n \to \coprod D^n$, and it follows that $B\vert_{\mathrm{sk}_{n-1}C} \to B\vert_{\mathrm{sk}_{n}C}$ is obtained by cobase change along $\coprod(B\vert_{\partial D^n} \to B\vert_{D^n})$. These two pushout diagrams are homotopy pushouts, and the projection map from one to the other is an $E_*$-equivalence on the cospans by induction and the assumption on the fiber since $B\vert_{D^n} \to D^n$ is equivalent to $A \to \bullet$ after trivializing the fibration.
(This is a compromise of S. Carmeli's answer because you can compute the `homotopy colimit over $C$' by decomposing $C$ along a cell diagram and computing the colimit one piece at a time; this is a compromise of Fernando's answer since one often constructs the Serre spectral sequence using the above pushout diagrams anyway.)
Yet another rephrasing. Homology/chains can be computed in terms of the functoriality of local systems of complexes. In this language, any homotopy type $C$ has a constant local system $k_C$, and the chains $C_{*}(C,k)$ are computed as $p_!(k_C)$, where $p_!$ is the left adjoint to the pullback functor of local systems for $p: C \rightarrow *$. The important, relevant fact is that local systems satisfy base-change for homotopy fibre products, such as your fibre sequence. For connected base $C$, your hypothesis that the homology of the fibre is trivial is equivalent to the hypothesis that the canonical map $f_{!}(k_B) \rightarrow k_C$ of local systems is an isomorphism, since the fiber of $f_{!}(k_B)$ is just $k$, by base-change and the hypothesis on the homology of the fibre $A$. Now apply $p_{!}$ to $f_{!}(k_B) \rightarrow k_C$ to obtain the isomorphism $C_{*}(B,k) \rightarrow C_{*}(C,k)$.
(Remarks. Really, such an argument has to take place $\infty$-category land, or at least in derived categories. Also, the same argument works for any other homology theory, by changing the system of coefficients.)
|
2025-03-21T14:48:29.735639
| 2020-01-29T23:21:33 |
351480
|
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|
Stack Exchange
|
Example of connected factor of symbolic system that is not a rotation
I am looking for an example of a factor $f\colon (X,T) \to (Y,T)$ between topological dynamical systems, where $(X,T)$ is a minimal subshift and $Y$ a connected topological space such that $(Y,T)$ is not a group rotation. Note that any eigenvalue $\lambda = \exp(2\pi i\alpha)$ of $(X,T)$ gives us a factor $(S^1,+\alpha)$ wich is connected and a group rotation; can there be another class?
Let $\alpha$ be irrational, let $Y$ be the two-dimensional torus equipped with the map $S(u,v)=(u+\alpha,v+u)$. Then the action of $S$ on $Y$ is minimal. Now partition the torus into two pieces, say $A_0=S^1\times[0,\frac 12)$ and $A_1=S^1\times[\frac 12,1)$ and let $j(y)=0$ if $y\in A_0$ and $j(y)=1$ if $y\in A_1$. Let $X$ be the set of bi-infinite sequences with symbols 0 and 1 such that each sub-word occurs as a sequence $j(y),\ldots,j(S^{n-1}y)$ for some $y\in \mathbb T^2$ and some $n$.
Define for $x\in X$,
$$
\pi(x)=\bigcap_{n\in\mathbb Z}\overline{S^{-n}(A_{x_n})}.
$$
This gives a single point of the torus (since the partition is generating).
This map intertwines the shift and $S$, and is surjective. The minimality of $S$ acting on $Y$ implies that of the shift on $X$.
|
2025-03-21T14:48:29.735803
| 2020-01-30T00:23:55 |
351481
|
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|
Stack Exchange
|
Bounding number of k-cycles in a graph
Fix any $k \geq 3$, and suppose I have a simple undirected graph $G=(V,E)$. I want a bound on the number of $k$ cycles in $G$ as a function of $|E|$. In particular, I would like to prove the following "conjectured" statement:
W.T.S: Given any $G=(V,E)$, the number of $k$-cycles in $G$ is at most $O(|E|^{k/2})$.
Clearly $\binom{|V|}{k}$ is tight if we only wanted a bound in terms of $|V|$, which comes from the complete graph on $|V|$ vertices. However, it is not as clear how to prove a tight upper bound in terms of $|E|$. The intuition for the conjecture is that, up to a constant, it seems that it should be optimal to arrange your edges in a complete graph (if $k$ is even, you do win a constant by taking a complete bipartite graph instead of a complete graph).
There seem to be a few proofs of this fact for triangles (k=3), for instance https://math.stackexchange.com/questions/823481/number-of-triangles-in-a-graph-based-on-number-of-edges. However, they don't generalize in any clear way to larger $k$. Moreover, I don't know of any k-cycle enumeration algorithms for k>3 that are parameterized by $|E|$ (the runtime of which would provide an upper bound).
For my application I actually only need bounds for odd $k$, but it seems that any such proof should generalize to any $k$. Also, to be clear, I'm only interested in asymtotic bounds here in terms of $|E|$ (getting tight constants seem much more difficult). Any suggestions or references that I missed would be much appreciated.
For $k=4$ the appropriate complete graph has nearly twice as many $k$-cycles as the complete bipartite graph, not that that changes things. For example $K_{50}$ and $K_{35,35}$ and each have $|E|=1225$ but the first has $690900$ $4$-cycles and the latter has $354025.$ The same should be true for larger ever $k$ as well.
Ooops, right you are. I miscounted the number of 4 cycles in the complete graph, I forgot that four vertices give rise to multiple 4-cycles. Thanks for catching this!
Yes, this is addressed in the paper:
Rivin, Igor, Counting cycles and finite dimensional (L^{p}) norms, Adv. Appl. Math. 29, No. 4, 647-662 (2002). ZBL1013.05042.
Incredible -- I had a feeling that bounding Tr(A^k) would be a good approach, but I didn't see how to do it. The fact that Tr(A^2) = |E| is a great catch. Thanks for the beautiful proof!
@RajeshJayaram Thanks for the kind words!
|
2025-03-21T14:48:29.736028
| 2020-01-30T02:24:23 |
351487
|
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|
Stack Exchange
|
$\pi_0(G)$ as a subgroup of a Lie group $G$
While doing some exercises in Lie groups, I see that the Lorentz group $O(1,3)$ has four connected components and $\pi_0(O(1,3))$ is the Klein four-group $\mathbb{Z}/2 \times \mathbb{Z}/2$. Not only that, but I can find explicit elements representing each connected component $\{1,a,b,ab\}$ which form a subgroup isomorphic to $\pi_0$.
My question then: is it true that for any Lie group $G$, the group $\pi_0(G)$ embeds as a subgroup of $G$? I believe it would follow that any such embedding would take distinct elements of $\pi_0$ to distinct components of $G$.
Not sure about the general case , but ${\pm I_2}\subset\mathrm{GL}_2(\mathbb{R})$ lie in the same component. There are two distinct questions: the existence of sections of $G\to\pi_0(G)$ and the existence of subgoups of $G$ isomorphic to $\pi_0(G)$.
In the case where $G$ is the normaliser of a maximal torus in a reductive Lie group, your question is asking about a lift of the normaliser of the Weyl group, which does not exist in general. (The easiest example is when the ambient group is $\operatorname{SL}_2$, in which case the normaliser is a non-split extension of $\operatorname{GL}_1$ by $\mathbb Z/2\mathbb Z$.) @MikeMiller's example is the compact form of this example.
See also https://mathoverflow.net/questions/150949/in-any-lie-group-with-finitely-many-connected-components-does-there-exist-a-fin (already served up by MO as a related question), which offers a weaker version of this property.
The claim "I believe..." is not true. Consider the quaternion group $Q$ (of order 8) and $S$ the 1-dimensional circle group; let $z_Q$ and $z_S$ be their unique elements of order $2$ and $z=(z_Q,z_S)\in Q\times S$. Define $G=(Q\times S)/\langle z\rangle$. Then the quotient $C_2\times C_2$ indeed embeds into $G$, but cannot be mapped as a section (i.e., for each embedding $C_2\times C_2\to G$ there exists two distinct elements mapping into the same component).
No. The first example that comes to mind is $\text{Pin}(2) \subset S^3$, given as $S^1 \cup jS^1$. This group has two components, but every element of $jS^1$ squares to $-1 \in S^1$. Thus every element of the non-identity component has order 4.
So there is no section of the map $\text{Pin}(2) \to \Bbb Z/2$ which sends the non-identity component to 1.
|
2025-03-21T14:48:29.736263
| 2020-01-30T03:48:45 |
351495
|
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|
Stack Exchange
|
When do $\phi^2$ and $\phi’^2$ have the same expectation under a Gaussian random variable?
I am looking for a function $\phi(x)$ such that
$\mathbb{E}_{x\sim\mathcal{N}(0,1)}[\phi(x)^2] = \mathbb{E}_{x\sim\mathcal{N}(0,1)}[\phi'(x)^2]$.
Obvious solutions are $\phi(x) = x$ and $\phi(x) = \exp(x)$. But do you know any other non-trivial solution?
Do you get something useful if you expand $\phi$ in terms of Hermite polynomials?
Interesting point! Can you explain why Hermite polynomials are relevant here?
They are an orthogonal basis for $L^2$ of standard Gaussian measure, or orthonormal after rescaling. So if ${h_n}$ are the orthonormal Hermite polynomials, we can write $\phi(x) = \sum_n a_n h_n(x)$, and $\mathbb{E}{x\sim\mathcal{N}(0,1)}[\phi(x)^2] = \sum_n a_n^2$. Moreover we have $h_n' = \sqrt{n} h{n-1}$, if I compute correctly, and so at least formally, $\mathbb{E}_{x\sim\mathcal{N}(0,1)}[\phi'(x)^2] = \sum_n n a_n^2$. Thus in order for your equation to hold, we need the coefficients to satisfy $\sum_n a_n^2 = \sum_n n a_n^2$.
In fact, every reasonable function can be made into an example by adding an appropriate constant.
I'll write $Z$ for a standard Gaussian random variable. Recall the Gaussian Poincaré inequality:
Theorem. For every $f \in C^1(\mathbb{R})$ we have $\operatorname{Var}[f(Z)] \le E[f'(Z)^2]$.
Equivalently, this is the fact that the Ornstein-Uhlenbeck "number" operator has spectral gap equal to 1. Perhaps the simplest way to prove the Poincaré inequality is via Hermite polynomials; see Bogachev, Gaussian Measures, Theorem 1.6.4. The statement generalizes directly to absolutely continuous functions (in the appropriate Sobolev space over Gaussian measure)
Corollary. Let $f \in C^1(\mathbb{R})$ with $E[f(Z)^2], E[f'(Z)^2] < \infty$. There exist either one or two real numbers $c$ such that $\phi(x) := f(x) + c$ satisfies $E[\phi(Z)^2] = E[\phi'(Z)^2]$.
Proof. Set
$$\begin{align*}\psi(c) &:= E[\phi(Z)^2] - E[\phi'(Z)^2] \\ &= \operatorname{Var}[\phi(Z)] + (E[f(Z)] + c)^2 - E[\phi'(Z)^2] \\
&= \operatorname{Var}[f(Z)] + (E[f(Z)] + c)^2 - E[f'(Z)^2]\end{align*}.$$
Now $\psi(c)$ is a quadratic in $c$ with $\psi(c) \to +\infty$ as $c \to \pm \infty$, and by the Poincaré inequality we have $\psi(-E[f(Z)]) \le 0$. So $\psi$ has either one or two real roots.
Indeed, the only way for the constant $c$ to be unique is if $f$ is a linear function, because that is the only case in which the Poincaré inequality saturates. Again, this can be seen via Hermite polynomials.
What do you mean by Poincaré inequality saturates? $\phi(x)=\exp(x)$ is also such a function where $c$ is unique, right?
@user3750444: I mean that we have $\operatorname{Var}[f(Z)] = E[f'(Z)^2]$ if and only if $f$ is linear. $\phi(x) = \exp(x)$ does not have a unique $c$; following the calculation above shows that $\phi(x) = \exp(x) - 2e^{1/2}$ also works.
@NateEldredge Thank you very much! I will cite your answer in my paper.
A family of such functions is given by
$$\phi(x)=\phi_c(x):=e^{t_cx+cx^2}$$
for all real $x$, where $c\in[-\frac{1+\sqrt2}2,\frac{-1+\sqrt2}2]$ and $t_c=\pm\sqrt{1 - 8 c + 12 c^2 + 16 c^3}$; then both expected values equal
${e^{{2 t_c^2}/(1-4 c)}}/{\sqrt{1-4 c}}={e^{{2 - 8 c - 8 c^2}}}/{\sqrt{1-4 c}}$.
In particular, choosing here $c=0$, we get two members of this family, given by $\phi_0(x)=e^{\pm x}$ for all real $x$.
Another two members of this family are given by $\phi_c(x)=e^{cx^2}$ for $c\in\{-\frac{1+\sqrt2}2,\frac{-1+\sqrt2}2\}$ and all real $x$.
Another family of examples, now parametrized by an arbitrary function in a certain general class of functions:
Let $g\colon\mathbb R\to\mathbb R$ be any bounded continuously differentiable function with a bounded derivative such that $g(0)\ne0$ and $g'(u)^2\ge1$ if $|u|\le1$. Let us then show that a function $\phi$ of the form $f_c$ for some real $c>0$ will do, where
$$f_c(x):=g(cx)$$
for all real $x$; that is, for some real $c>0$ we will have
$$Eg(cZ)^2=c^2Eg'(cZ)^2, \tag{1}$$
where $Z\sim N(0,1)$.
Let
$$D(c):=Eg(cZ)^2-c^2Eg'(cZ)^2.$$
Then, by dominated convergence, $D(c)$ is continuous in $c\ge0$. Also,
$$D(0)=g(0)^2>0.$$
On the other hand, for $c\to\infty$
$$Eg'(cZ)^2\ge Eg'(cZ)^2 1_{|Z|\le1/c}\ge E1_{|Z|\le1/c}=P(|Z|\le1/c)\gtrsim\frac2{c\sqrt{2\pi}},$$
whereas $Eg(cZ)^2$ stays bounded (since $g$ is bounded). So, $D(c)\to-\infty$ as $c\to\infty$.
Therefore and because $D(c)$ is continuous in $c\ge0$, we see that $D(c)=0$ for some real $c>0$, which indeed yields (1).
Clearly, the condition $g'(u)^2\ge1$ if $|u|\le1$ here can be relaxed just to $g'(0)\ne0$.
E.g., we may take $g(x)=1+\sin x$ for all real $x$, and then
$D(c)=\frac{3}{2}-\frac{1}{2} e^{-2 c^2}-e^{-c^2} c^2 \cosh \left(c^2\right)$, and the equation $D(c)=0$ has two roots, $c\approx\pm1.73$.
|
2025-03-21T14:48:29.736578
| 2020-01-30T04:46:16 |
351496
|
{
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"Puzzler",
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"fedja",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/151588",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351496"
}
|
Stack Exchange
|
General Hoeffding inequality with two uncorrelated random vectors
Let $X_1,\ldots, X_n$ be independent sub-Gaussian variables with sub-Gaussian norm $K$. Let $a_1,\ldots,a_n$ be a random vector such that $\mathbb{E}[a_j X_i]=0$ for all $i,j\in \{1,\ldots ,n\}$ and such that $\left\Vert a \right\Vert_2^2 =1$ almost surely. I'm interested in whether there is a Hoeffding-type bound
\begin{align}
\mathbb{P}[\sum_{i=1}^n a_iX_i \geq t] \leq \exp \left(-\frac{ct^2}{K^2} \right)
\end{align}
as in Theorem 2.6.3 in this book, just with random $a$. It would be easy if $a$ was independent of $X$. But my question is whether we can have a similar bound when $a$ and $X$ are only uncorrelated.
Condition on the value of $a$. Then the inequality holds from the fact that $\sum a_i X_i$ is subgaussian, then remove the conditioning
How do we know that $\sum_i a_i X_i$ is sub-Gaussian conditional on a? It's not obvious to me because $a_i$ and $X_i$ are not fully independent. But I agree that if you can show the sub-Gaussianity with the right sub-Gaussian norm (independent on n), then the inequality follows.
Consider the case when $a$ is a normalized multiple of $X$ with probability $1/2$ and minus that with probability $1/2$ where the sign flip is independent of $X$. That kills all correlations but is technically the worst case scenario because the sum is as large as it can be half of the time. What are you getting then?
Thanks, @fedja! Can you please elaborate/ give a reference for why this is the worst case? It would be nice because it would give me a lot of structure for the $a_i$'s I can use to estimate the sub-Gaussian norm.
|
2025-03-21T14:48:29.736731
| 2020-01-30T06:16:49 |
351498
|
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"Guest123412341234",
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|
Stack Exchange
|
Index of Dirac operator and Chern character of symmetric product twisting bundle
I am having trouble understanding a couple of lines of computation from Theorem 13.30 in Besse's Einstein Manifolds text
We are twisting the spinor bundle (on Einstein 4-manifold) $\Sigma$ with an auxiliary bundle $S^3\Sigma^-$. We form the Dirac operator $\mathscr{D}^D$ formed by twisting the Levi-Civita connection on $\Sigma$ with a copy $D$ acting on $S^3\Sigma^-$ (and composing with the Clifford action acting trivially on $S^3\Sigma$). Besse evaluates the index of this operator using the APS theorem as an integral over Chern and Pontrjagin classes.
I understand that the $(1-\frac{1}{24}p_1)$ terms comes from the $\widehat{A}$-genus of the manifold, and further the signature theorem, $\tau=\frac13 \int_M p_1(M) $. However, I do not understand the evaluation of the Chern character of the twisting bundle as $(4-10c_2)$ and the subsequent evaluation in terms of Euler characteristic and signature.
Any insight would be greatly appreciated.
Edit: I am still not sure about the final calculation relating the index to the Euler character and signature, however, working backwards it seems we require $c_2(\Sigma^-)=\frac12e(M)-\frac14p_1(M)$ (or perhaps something a bit different if there are lower degree terms which could multiply with with the $\frac{1}{24}p_1$ term from the $\widehat{A}-$genus), where $e(M)$ is the Euler class of $M$.
Edit#2: Solved by Michael Albanese.
Your first question can be answered by using the splitting principle.
If $V \to X$ is a complex vector bundle of rank two, then $c_1(S^3V) = 6c_1(V)$ and $c_2(S^3V) = 11c_1(V)^2 + 10c_2(V)$.
Proof: By the splitting principle, there is a map $p : Y \to X$ such that $p^*$ is injective on integral cohomology and $p^*V \cong L_1\oplus L_2$, so $p^*(S^3V) \cong S^3(p^*V) \cong S^3(L_1\oplus L_2)$. In general, we have $S^n(E_1\oplus E_2) \cong \bigoplus_{i+j=n} S^i(E_1)\otimes S^j(E_2)$, so
\begin{align*}
&\, S^3(L_1\oplus L_2)\\
\cong&\, S^3(L_1)\otimes S^0(L_2)\oplus S^2(L_1)\otimes S^1(L_2)\oplus S^1(L_1)\otimes S^2(L_2)\oplus S^0(L_1)\otimes S^3(L_2)\\
\cong&\, L_1^3\oplus L_1^2\otimes L_2\oplus L_1\otimes L_2^2\oplus L_2^3.
\end{align*}
It follows that $c_1(S^3(L_1\oplus L_2)) = 6c_1(L_1) + 6c_1(L_2) = 6c_1(L_1\oplus L_2)$. So $$p^*c_1(S^3V) = c_1(p^*S^3V) = c_1(S^3(L_1\oplus L_2)) = 6c_1(L_1\oplus L_2) = 6c_1(p^*V) = p^*(6c_1(V)).$$ By the injectivity of $p^*$, we have $c_1(S^3V) = 6c_1(V)$.
Similarly, one can compute that
\begin{align*}
c_2(S^3(L_1\oplus L_2)) &= 11c_1(L_1)^2 + 11c_1(L_2)^2 + 32c_1(L_1)c_1(L_2)\\
&= 11(c_1(L_1)+c_1(L_2))^2 + 10c_1(L_1)c_1(L_2)\\
&= 11c_1(L_1\oplus L_2)^2 + 10c_2(L_1\oplus L_2)
\end{align*}
and hence $c_2(S^3V) = 11c_1(V)^2 + 10c_2(V)$. $\quad\square$
In this case, $\Sigma^-$ is an $SU(2)$ bundle and hence $c_1(\Sigma^-) = 0$. So we see that $c_1(S^3\Sigma^-) = 0$ and $c_2(S^3\Sigma^-) = 10c_2(\Sigma^-)$. Therefore
$$\operatorname{ch}(S^3\Sigma^-) = \operatorname{rank}(S^3\Sigma^-) + c_1(S^3\Sigma^-) + \frac{1}{2}(c_1(S^3\Sigma^-)^2 - 2c_2(S^3\Sigma^-)) = 4 - 10c_2(\Sigma^-).$$
To find $c_2(\Sigma^-)$, or $c_2(\Sigma^+)$, we can proceed as follows.
As $\Sigma^{\pm}$ is an $SU(2)$-bundle which is a lift of the $SO(3)$-bundle $\Lambda^{\pm}$, there is a relationship between $c_2(\Sigma^{\pm})$ and $p_1(\Lambda^{\pm})$, namely $p_1(\Lambda^{\pm}) = -4c_2(\Sigma^{\pm})$; see Appendix E of Instantons and Four-Manifolds by Freed and Uhlenbeck, also this question. So now we just need to know $p_1(\Lambda^{\pm})$, but this is given by $\pm 2e(M) + p_1(M)$; see Chapter $6$, Proposition $5.4$ of Metric Structures in Differential Geometry by Walschap. Therefore
$$c_2(\Sigma^{\pm}) = \mp\frac{1}{2}e(M) - \frac{1}{4}p_1(M).$$
Note, as $M$ is assumed to be spin, its signature is a multiple of $16$ by Rohklin's Theorem, so $\frac{1}{4}p_1(M)$ is an integral class. As the signature of $M$ is even, so is the Euler characteristic, and hence $\frac{1}{2}e(M)$ is also an integral class.
Finally, as $c_2(\Sigma^-) = \frac{1}{2}e(M) - \frac{1}{4}p_1(M)$, we see that
\begin{align*}
\int_M(10c_2(\Sigma^-) - 4)\left(1 - \frac{1}{24}p_1(M)\right) &= \int_M 10c_2(\Sigma^-) + \frac{1}{6}p_1(M)\\
&= \int_M 5e(M) - \frac{5}{2}p_1(M) + \frac{1}{6}p_1(M)\\
&= \int_M 5e(M) - \frac{7}{3}p_1(M)\\
&= 5\chi(M) - 7\tau(M)
\end{align*}
as claimed.
Perfect, thank you. Although, I am unsure why symmetric product of line bundles reduces to the standard tensor product.
If $V$ is a one-dimensional vector space, then every tensor on $V$ is symmetric, so $\bigotimes^nV = S^nV$.
|
2025-03-21T14:48:29.737039
| 2020-01-30T08:13:05 |
351502
|
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"Martin Brandenburg",
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"url": "https://mathoverflow.net/questions/351502"
}
|
Stack Exchange
|
Existence of a multiplication bifunctor for the category of groups
For $\mathsf{Grp}$ the category of groups, a bifunctor $M: \mathsf{Grp} \times \mathsf{Grp}\to \mathsf{Grp}$ is a multiplication bifunctor if:
$M(C_n,C_m) \simeq C_{nm}$,
$M(C_1,G) \simeq M(G,C_1) \simeq G$,
for every group $G$ and every $n,m>0$, with $C_n$ the cyclic group of $n$ elements.
Question: Is there a multiplication bifunctor for the category of groups?
(or for the subcategory of countable groups, or of finite groups)
Stronger question: Is there a multiplication bifunctor providing a monoidal structure?
This post is a multiplicative analogous of that additive one.
It's almost the same argument as in my answer in the additive case, as can be seen from Jeremy's answer.
@MartinBrandenburg: Yes, I considered this example, but I had the false belief that a subgroup isomorphic to a quotient is a retract...
We finally get the following funny result: let $F:\mathsf{Grp} \to \mathsf{Grp}$ be a functor, then $F^n(C_1)$ is a retract of $F^{n+1}(C_1)$. For example, if $F(C_1) \simeq C_2$, then $C_2$ is a retract of $F(C_2)$, so in particular, $F(C_2) \not \simeq C_3, C_4$.
No. $C_1$ is a retract of $C_2$, so $M(C_2,C_1)\simeq C_2$ would have to be a retract of $M(C_2,C_2)\simeq C_4$, which it isn't.
|
2025-03-21T14:48:29.737285
| 2020-01-30T09:49:46 |
351507
|
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|
Stack Exchange
|
Infinite directed paths in tournaments on $\omega$
Does every tournament on $\omega$ contain an infinite directed path that doesn't visit any vertex twice?
If $x\to y\to z\ldots$ and $x\leftarrow y\leftarrow z\leftarrow\dots$ are both called infinite paths, then yes. For two vertices $x<y$ color an edge $xy$ of the complete graph on $\omega$ red or blue in dependence of the direction of $xy$ in the tournament. By infinite Ramsey theorem, there exists an infinite monochromatic subgraph which contains an infinite path.
If $x\leftarrow y\leftarrow z\leftarrow\dots$ is not called an infinite path, then no: consider the tournament in which all edges are directed towards lesser number.
|
2025-03-21T14:48:29.737395
| 2020-01-30T10:03:06 |
351509
|
{
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],
"authors": [
"Abdelmalek Abdesselam",
"Kung Yao",
"Yoav Kallus",
"fedja",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/150549",
"https://mathoverflow.net/users/20186",
"https://mathoverflow.net/users/7410"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351509"
}
|
Stack Exchange
|
Matrix positive semi-definite
We construct a non-random matrix using random variables as follows:
We fix the vector $v=(1,1).$
Let $X$ be a $\mathbb R^2$-valued random variable such that $X$ is distributed according to
$$d\mu(x) \propto e^{-\Vert x-v \Vert^2-\vert x_1 \vert^4-\vert x_2 \vert^4} \ dx.$$
We then define the following matrix for $Y=X-\mathbb E(X)$
$$ \langle x,Ay\rangle = -\mathbb E\left(\langle Y,x \rangle \langle Y,v \rangle \langle Y, y\rangle\right).$$
Using wolframalpha I find that the normalization constant of this measure is $0.62,$ the expectation value of $X$ is $\mathbb E(X)=0.27 v$ and the eigenvalues of the matrix $A$ are non-negative.
Observations:
It seems like any non-zero $v$ does the job, i.e. $A$ will be positive semi-definite but I am only interested in this particular choice.
What seems to be important for this to be true is that we have the power $4$ appearing in the probability measure, as the expectation would vanish in the purely Gaussian case by symmetry.
My question therefore is: How can I show the eigenvalues of $A$ are non-negative?
Please let me know if you got any questions, I am happy to investigate further ideas. At the moment, I am a bit puzzled by this problem.
ADDENDUM to the comments:
To respond to a conjecture made in a comment:
I used the following Mathematica code
Plot3D[If[Sign[x1*1 + x2*1] >= 0, (Exp[-((-x1 + 0.27) - 1)^2 + ((-x2 + 0.27) - 1)^2 - (-x1 +
0.27)^4 - (-x2 + 0.27)^4] -
Exp[-((x1 + 0.27) - 1)^2 + ((x2 + 0.27) - 1)^2 - (x1 +
0.27)^4 - (x2 + 0.27)^4])/0.62, 0], {x1, -2, 2}, {x2, -2,2}, PlotRange -> All]
This produces the following plot for $\mu(-y+\mathbb E(X)) -\mu(y+\mathbb E(X))$ which shows that it is almost everywhere (up to some very small region) true that $\mu(-y+\mathbb E(X)) >\mu(y+\mathbb E(X))$ if $\langle y,v \rangle >0.$
$A = -\int_{y\in R^2} \langle y,v\rangle (y\otimes y) d\mu(y+\bar{x}) $ $= \int_{y\in R^2, \langle y,v\rangle > 0} \langle y,v\rangle (y\otimes y) (d\mu(-y+\bar{x})-d\mu(y+\bar{x}))$. I suspect that $d\mu(-y+\bar{x})-d\mu(y+\bar{x}) > 0$ when $\langle y,v\rangle > 0$. Is that true?
@YoavKallus what you write could be true and is interesting, however it is difficult to say since $\overline{x}$ is not explicit. I will try some numerics.
@YoavKallus so it is mostly true but not everywhere
@KungYao: You shouldn't call $A$ a random matrix because it is not random.
@AbdelmalekAbdesselam you are right of course.
Something is strange (about the question) because, due to symmetry, the expectation of $X$ is just $(c,c)$ where $c$ is the expectation of the random variable with the density $p(x)$ proportional to $e^{-x^2-x^4+2x}$. Thus, the off-diagonal matrix elements are merely
$$
-\iint (x_1-c)(x_2-c)[(x_1-c)+(x_2-c)]p(x_1)p(x_2)dx_1dx_2=0
$$
because each of two terms splits into a product in which one of the factors is $\int (x-c)p(x)dx=0$. So why to ask about eigenvalues if the matrix is diagonal?
Similarly, the diagonal entries are
$$
-\iint (x_1-c)^2[(x_1-c)+(x_2-c)]p(x_1)p(x_2)dx_1dx_2
\\
=-\int(x-c)^3p(x)dx,
$$
which is shown to be non-negative in this post
Am I missing anything?
wow, yes you seem to be right. Let me digest this post you are mentioning first, though. But just for my own understanding, would this other post also explain why it is true if you do not go into $(1,1)$ but any other non-zero direction, if it is even true?- Going through it once, it seems so and we would just have some $c_1,c_2$, would you agree?
@KungYao Yes, you'll still have the product structure though it will be $(c_1,c_2)$ and the diagonal elements will be different and have $v_1$ and $v_2$ in them, so the signs of the third central moments will flip together with the signs of $v_1,v_2$, i.e., with the sign of the $y$-parameter in that other post.
I see, so what made my question trivial, as you say, is that the individual components factorize. So let us consider the measure with density $p(x) = e^{-\langle x-v, \Sigma^{-1} x-v \rangle - \vert x_1 \vert^4-\vert x_2 \vert^4}$ where $\Sigma$ is positive-definite. Then your approach suggests to study for every $z$: $\int \langle x-c, z \rangle^2 \langle x-c, v \rangle p(x) \ dx.$ Proceeding as in your other answer leads the condition that for all $a>0$ we have $\int_{\vert \langle x,z \rangle \vert \ge a} \langle x,v \rangle p(x+c) \ dx \ge 0.$
So we can differentiate this thing with respect to $a$ and co-area formula yields a surface integral proportional to $-\int_{\vert \langle x,z \rangle \vert = \pm a} \langle x,v \rangle p(x+c) \ dx. $ This is, as far as I can tell very similar to what you were doing in the other answer. However, in the multi-dimensional case I don't think I can close the argument. Do you see how to do it?-Of course since you showed more than necessary in the other answer, this might be too much to ask for. Maybe the result is also wrong in the multi-dimensional case. Do you have any intution about this?
given your answer to the other post, I guess that is the problem you have been looking for in my post ;) I would be very curious to hear about the answer
@KungYao OK, I'll take a look, but tomorrow :-)
Thank you fedja, I appreciate the effort
so based on some numerical tests, it seems like the statement that the matrix is still positive-definite seems to be true. I am not sure though as to what extend the proof in your other post extends to the multi-variate case....
To be very precise, I am struggeling getting anywhere understanding the integral $$\int_{\langle x,z \rangle =a} \langle x,v \rangle (p(c-x) - p(c+x)) \ dx. $$ What you did in the other answer was that you argued that $c-x$ was closer to the origin that $c+x.$ This seems to be unclear in this setting, as there is no "sign" anymore. The problem seems to be that $z$ can point in any direction and does not see the direction of $v.$ The only way this could work somehow was if the sign of $\langle x,v\rangle $ would somehow correct this "lack of a sign". Do you see any "way out"?
@KungYao Do you see any "way out"? Not yet :-(. However, I think that the scalar product of $Y$ with $v$ is a bit unnatural now because in the density we have $\Sigma^{-1}$ on the same scalar product, so I would search for counter-examples a bit more specifically looking at the cases when $\Sigma^{-1}v$ is very different from $v$ (not sure yet in what respect exactly).
good point, I checked a bit more carefully now how you arrived at the expression in the other post. It seems that in this case, the natural generalization would be to look at $$\mathbb E(\langle Y,z \rangle^2 \langle Y,\Sigma^{-1} v \rangle)$$ if I am not mistaken (instead of what I wrote). In this case we would find $$\int_{\langle x,z \rangle=a} \langle x, \Sigma^{-1} v \rangle (p(c-x)-p(c+x)) \ dx.$$ I agree it looks more "symmetric" (the weight of course is irrelevant for $\langle x,z \rangle$ as this one has to work for all $z$. Do you find this one easier to treat?
My main problem is the asymmetry between $\langle x,z \rangle=a$ and $\langle x, \Sigma^{-1} v \rangle$ in the final expression. But I will check some numerics for the new form you suggested as well and keep you posted. Thanks for thinking about this problem.
okay, i asked it as a separate question https://mathoverflow.net/q/351829/150549
I really checked this carefully numerically now, it seems to be true with the normalization you proposed, I will add the mathematica script I was using to the other question
this is to say, if one studies $\mathbb E(\langle Y,z \rangle^2, \langle Y, \Sigma^{-1}v \rangle)$ then this should be positive (according to numerics), whereas indeed $\mathbb E(\langle Y,z \rangle^2, \langle Y, v \rangle)$ fails to be. An example where the second one fails is $v=(2,-1)$ and $\Sigma^{-1}=\left(\begin{matrix} 4 & 1.5 \ 1.5 & 1 \end{matrix} \right).$ If you want me to check something numerically for you, please let me know. I am really eager to understand this better.
Dear fedja, I was just wondering whether you had developed any feeling about the question in the other thread? I would also be curious to hear some vague feelings. Unfortunately, so far nobody has commented on it...
|
2025-03-21T14:48:29.738262
| 2020-01-30T10:12:11 |
351511
|
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"Trevor Wilson",
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|
Stack Exchange
|
Suslin representation of sets and limits to Shoenfield's Absoluteness
For any natural number $k \in \omega$ and any set $X$ the set $$T \subseteq \bigcup_{m \in \omega} (\omega^m)^k \times X^m $$ is a tree on $\omega^k \times X$ iff
$$(t_o, \ldots, t_k) \in T \: \Rightarrow \forall m \leq |t_0| \: (t_0 \upharpoonright m, \ldots, t_k \upharpoonright m) \in T,$$
i.e. the elements of $T$ are $k+1$-tuples of finite sequences of the same length.
It is known that any tree $T$ on $\omega^k \times X$ is well-founded if there is an order-preserving function $\rho: T \rightarrow Ord$.
Let $x = (x_0, x_1, \ldots x_{l-1}) \in (\omega^\omega)^l$. The set $T(x)$ is a tree on $\omega^{k-l} \times X$ defined as follows:
$$(t_0, \ldots, t_{k-l}) \in T(x) \: \Leftrightarrow \: \exists m \: (x_0 \upharpoonright m, \ldots, x_{l-1} \upharpoonright m, t_0, \ldots, t_{k-l}) \in T.$$
It is also well-known that if $a \in \omega^\omega$ and $A \subseteq (\omega^\omega)^k$, then $A$ is $\Pi^1_1(a)$ iff there is a tree $T$ on $\omega^k \times \omega$ such that for any real $x \in (\omega^\omega)^k$:
$$x \in A \: \Leftrightarrow \: T(x) \text{ is well-founded,}$$
where the relation $\{(i, x): s_i \in T(x)\}$ is computable from $a$, for some computable enumeration $(s_i)_{i \in \omega}$ of the set of finite sequences such that $\forall i \: |s_i| \leq i$.
Now the crucial definition: for a set $A \subseteq (\omega^\omega)^k$ and any set $X$ we say that $A$ is $X$-Suslin if there i a tree $T$ on $\omega^k \times X$ such that$A = p[T]$, i.e. for any $x \in (\omega^\omega)^k$ it holds that
$$ x \in A \: \Leftrightarrow \: T(x) \text{ is ill-founded}.$$
To illustrate the definition, note that every analytic set is $\omega$-Suslin, and observe that (under CH?) any set $A \subseteq (\omega^\omega)^k$ is $2^{\aleph_0}$-Suslin, but an important thing is that the key lemma in the (standard) proof of Shoenfield's Absolutness is the fact that
any $\mathbf{\Sigma^1_2}$ set is $\omega_1$-Suslin (a remark from ref. Kanamori - The Higher Infinite p.174: the reverse implication is independent of ZFC). The way you prove the key lemma is you take an appropriate well-founded tree of triples (by the definition of $\mathbf{\Sigma^1_2}$), using the above-mentioned fact that there is an order-preserving map $\rho$ from $T$ to $\omega_1$ and you define a new tree $T(x)$ on $\omega^2 \times \omega_1$, ewentually transforming it into a tree with a branch that gives us the representation - if necessary, I can put in more details into the question.
My question is as follows (I guess the answer is in the negative, but cannot find an argument):
$$\text{is it true that every $\mathbf{\Sigma^1_{n+1}}$ set is $\omega_n$-Suslin?}$$
If not, what step from the proof of the case of $n=1$ does not generalize? Is it simply (that's my intuition) the fact that when we go higher in the projective hierarchy, we simply lose the ability to use the definition of our set to form an appropriate tree in the first place (in the sense in which having a $\Sigma^1_2(a)$ predicate gives us a formula 'there exists $r$ such that for all s some $\Pi^0_1$ formula holds' and this easily translates into a tree that we can handle quite feasibly)? If so, is there a proper formal argument for that?
I am obviously also interested in any counterexample, but the most important thing for me is to understand which steps of the argument brake if they do?
The reason for the question is the Shoenfield's Absolutness. We can give nice counterexamples of $\mathbf{\Sigma^1_3}$ formulas (like 'there is a non-constructible real') that are not absolute for (transitive) inner models, so we know $\mathbf{\Sigma^1_2}$ (and $\mathbf{\Pi^1_2}$) is the limit, but when asked 'which step of the proof cannot be generalized?', my reply can be only: well, I know how to handle $\Pi^1_1(a)$ properties (and quantify them existentially to handle $\Sigma^1_2(a)$-ones), but I don't see a way what I could do for some $\Sigma^1_{n}(a)$ for $n>2$. Is it just this or there is some deeper, say, 'combinatorial' reason within some of the further steps of the proof that prevents us from the generalization?
I don't know the answer in ZFC, but in ZF + DC the answer is consistently no. If $V = L$ and there is an inaccessible cardinal, then in the associated Solovay model the $\Pi^1_2$ relation $\{ \langle x,y \rangle : y \notin L[x]\}$ has no uniformization, so it is not Suslin, meaning that it is not $X$-Suslin for any wellordered set $X$.
(Moreover, in the Solovay model of a slightly stronger large cardinal, every Suslin set of reals is ${\bf\Sigma}^1_2$ by Theorem 1.8 and Remark 1.9 of Wilson, Generic Vopěnka cardinals and models of ZF with few $\aleph_1$-Suslin sets: https://link.springer.com/article/10.1007/s00153-019-00662-1.)
Consequently, the answer to a version of the question for definable Suslin representations in ZFC is also no: if $V = L$ and there is an inaccessible cardinal, then in the generic extension by its Levy collapse there is a $\Pi^1_2$ set with no Suslin representation definable from a countable sequence of ordinals.
To go from $\omega$-Suslin representations of $\Sigma^1_1$ sets to $\omega_1$-Suslin representations of $\Pi^1_1$ sets, we use the fact that if $T$ is a countable tree, then rank functions witnessing well-foundedness of its sections $T(x)$ can be coded as $\omega$-sequences of countable ordinals and therefore as branches of order type $\omega$ through a tree of cardinality $\omega_1$. The most straightforward attempt to imitate this argument at the next level would seem to involve branches of order type $\omega_1$ through a tree of cardinality $\omega_2$. But the kind of tree that we are talking about can only have branches of order type $\le \omega$ by definition (and this restriction is necessary for such trees to be relevant to uniformization, etc. of sets of reals.)
It should be mentioned here that if projective determinacy holds, then all projective sets have definable Suslin representations by the second periodicity theorem of Moschovakis.
Thank you, Trevor! I guess there will be no better answer here in the closest future, but I'll let the question fly for the next day or so before I accept, ok?
@mtg Sure, no problem.
|
2025-03-21T14:48:29.738669
| 2020-01-30T10:34:25 |
351513
|
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|
Stack Exchange
|
On the relation between categorification and chromatic redshift
In the introduction to the paper Higher traces, noncommutative motives, and the categorified Chern character, Hoyois, Scherotzke and Sibilla write the following.
An important insight emerging from topology is that climbing up the chromatic ladder is related to studying invariants of spaces that are higher-categorical in nature.
The only example that I am aware of is the theory of categorified vector bundles, and its relation with chromatic-level-$2$ cohomology theories, including elliptic cohomology, and the algebraic K-theory spectrum of complex K-theory. See for instance this nLab page and the references therein.
Question. Are there other examples of this phenomenon, and is there a conceptual reason why categorification should be related to chromatic redshifting?
The paper https://arxiv.org/abs/1312.5699 by Torlief Veen involves arbitrarily large chromatic redshifts via higher topological Hochschild homology. I know of no other examples like that.
|
2025-03-21T14:48:29.738779
| 2020-01-30T10:38:55 |
351514
|
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"Alex B.",
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|
Stack Exchange
|
On quadratic forms in four variables
Let $F$ be a non-singular integral quadratic form in four variables. Then a result of Heath-Brown from the 90's states for $m \to \infty$, $$|\{ x \in \mathbb{Z}^4 \,:\, F(x) = m \}| = C_F\sigma(F,m)m + O_{F,\varepsilon}(m^{\frac{3}{4} + \varepsilon}),$$ where $C_F$ is a positive constant depending only on $F$ and $\sigma(F,m)$ is the singular series.
I am interested to know, whether up till today there are improvements on the error rate of this result. More precisely, is there any hope to improve the error rate to $O_{F,\varepsilon}(m^{\frac{1}{2} + \varepsilon})$. If this is unknown, what is the conjectured error term?
For the application I have in mind, it would be already interesting to understand the case $$F(x) = x_1^2 + x_2^2 + x_3^2 + x_4^2.$$ Are there any results for this case or any other specific quadratic form?
Here is the original paper. https://core.ac.uk/display/77442361 Jayce Getz refined the error: https://doi.org/10.1112/jlms.12130
@MyNinthAccount: As far as I understand, Getz is only proving results for the case where the sum is over F(x) = 0.
You should ask separate questions as separate questions, don't make your question a moving target. If the original question has been answered to your satisfaction, then accept the answer; and if you want to ask a different question, do that separately.
Ok – here is the new question: https://mathoverflow.net/questions/351915/on-question-on-quadratic-forms-in-four-variables
It follows from Deligne's bound for the Hecke eigenvalues of weight $2$ holomorphic cusp forms (which is really Eichler's theorem in this special case) that the error term is $O_{F,\varepsilon}(m^{\frac{1}{2} + \varepsilon})$. Indeed, as proved by Siegel, the main term is the Eisenstein contribution of the underlying theta series (which is a holomorphic modular form of weight $2$), so the error term is the cuspidal contribution.
In particular, for $F(x) = x_1^2 + x_2^2 + x_3^2 + x_4^2$ the error term is zero, since the main term in this case reproduces Jacobi's well-known formula $8\sum_{\substack{d\mid m\\4\nmid d}}d$ for the number of representations.
|
2025-03-21T14:48:29.738960
| 2020-01-30T10:53:01 |
351519
|
{
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"Lviv Scottish Book",
"Manfred Weis",
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|
Stack Exchange
|
The complexity of cutting hackers in a computer network
Let $l_1,l_2,\dots,l_m$ be parallel lines in the plane, say $l_k=\mathbb R\times\{k\}$. On the $k$th line fix a set $V_k$ consisting of $n_k$ points.
Let $(V,E)$ be a directed graph whose set of vertices is $V=\bigcup_{k=1}^mV_k$ and set of edges $E\subset\bigcup_{k=1}^{m-1}V_k\times V_{k+1}$.
For an edge $(u,v)\in E$ let $I(u,v)=\{tu+(1-t)v:0<t<1\}$ be the convex segment without the endpoints $u,v$.
We shall assume that for any distinct edges $(a,b),(c,d)\in E$ the open intervals $I(a,b)$ and $I(c,d)$ are disjoint.
The graph $(V,E)$ represents a computer network. Now assume that some set $I\subset V$ of vertices is infected by a virus.
Question. What is the smallest number of cuts, necessary for isolating the ``infected'' set $I$ from its complement $V\setminus I$?
By a cut we understand deleting all edges that start at some vertex or end at some vertex. More precesely, we say that a subset $E''\subset E$ is obtained by a cut from a subset $E'\subset E$ if $E''=E'\setminus C$ where $C$ is equal to $\{(x,y)\in E':x=v\}$ or $\{(x,y)\in E':y=v\}$ for some vertex $v\in V$.
Let $c(V,E;H)$ be the smallest length $l$ of a sequence $E_1,\dots,E_l$ such that $E_1=E$, each $E_{i}$ is obtained by a cut from $E_{i-1}$, and $E_l$ contains no directed edge $(x,y)$ with $x\in H$ and $y\in V\setminus H$.
Problem 1. Find (a reasonable upper bound for) $\max_{E,H}c(V,E;H)$ as a function of $n_1,\dots, n_m$ (where $n_k=|V_k|$).
Problem 2. Elaborate an efficient algorithm for cutting a set $H\subset V$ from its complement $V\setminus H$ in a given digraph $(V,E)$.
This problem (in a bit modified form) was posed on 27.11.2019 by Vladislav Shapiro (from Bedford, Massachusetts) on page 34 of Volume 3 of the Lviv Scottish Book.
Prize. 2 tickets to Boston Bruins.
@MattF Thank you for the comment. I corrected the definition of $n$ using (i) and (ii), because bulleting yields too wide horizontal spaces between the items.
I still have some difficulty in understanding, what the setting if the question really is: -first of all: is it a geometric problem, where the location coordinates of the As and Hs and of the other knots play a role or is it a graphtheoretic one where the As have to be separated from the Hs by deleting a minimal set of edges that yields two connected components, such that there is no path from an A to a H? -Second: how does the direction of edges come into play? This are not all questions that remain open. I'd appreciate a precise definition of the problem.
Just checked the Lviv book and the open questions reduce to "what is the definition of a level" in that problem? Otherwise the formulation of the problem is much clearer: given a directed graph with one source and (I guess) $m$ parallel lines(not drawn) on which $n_1, \cdots,\ n_m$ of the nodes must be placed, what is the maximal number of paths in a straight-line drawing, that can be intersected by a single line. The formulation via admins and hackers obscures more than it elucidates.
@ManfredWeis Maybe you are right. Before writing down this problem I discussed with Vlad Shapiro what exactly he had in mind and he said me that the problem was motivated by the network security and the problem of cutting the hackers, as I wrote. Since the graph represents a genuine computer network, the problem has a geometric flavour. The question is how to reduce this real-life problem to a graph-theoretic one.
@LvivScottishBook I'm seeing forward to a revised statement of the problem,or maybe two: one for a purely graph-theoretic setting and one for the geometric setting with fixed node positions and a crossing-free sraight-line embedding of the graph in the Euclidean plane.
@ManfredWeis Maybe I (as an administrator of the Lviv Scottish Book) am insufficiently competent, but Vlad Shapiro (who posed the original problem) promised to join to the discussion. Indeed, it can happen that the geometric model is not the optimal one. In the real life situation, the administrator just cuts one cable (which can represent many edges of the graph).
As I see it, the problem doesn't require directed arcs, because in real live the connections will be bidirecttional and it will also not be required that the hackers be in the same connected component after cutting connections, whereas it may be necessary that the communication between admins be kept up as much as possible. Assuming such a solution is possible, it may be the objective to find the minimum edge cut that renders the hackers as disconnected as possible and the admins as highly connected as possible.
|
2025-03-21T14:48:29.739288
| 2020-01-30T10:55:41 |
351520
|
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|
Stack Exchange
|
Real Rank of $M_n(A)$
The real rank for C*-algebras was defined by Brown-Pedersen in [1] as a noncommutative analog of covering dimension. Given a unital C*-algebra $A$, its real rank $\mathrm{rr}(A)$ is the smallest natural number $n$ (possibly $n=0$) such that for every $n+1$-tuple $(x_0,\ldots,x_n)$ of selfadjoint elements in $A$ and every $\varepsilon>0$ there exists an $n+1$-tuple $(y_0,\ldots,y_n)$ of selfadjoint elements in $A$ such that $\|x_k-y_k\|<\varepsilon$ for $k=0,\ldots,n$ and such that $\sum_k y_k^*y_k$ is invertible. If no such $n$ exists, one sets $\mathrm{rr}(A)=\infty$. If $A$ is nonunital, one sets $\mathrm{rr}(A)=\mathrm{rr}(\widetilde{A})$ for the minimal unitization $\widetilde{A}$.
Question: Do we have $\mathrm{rr}(M_n(A))\leq\mathrm{rr}(A)$?
It seems that this ought to be true, but I couldn't find such a statement anywhere. For the stable rank (which is defined analogous to the real rank, but using not necessarily selfadjoint tuples) Rieffel proved the formula $\mathrm{sr}(M_n(A))=\lceil \frac{\mathrm{sr}(A)-1}{n}\rceil+1$. Rieffel's method does not seem to generalize to the real rank. In fact, I can show that for the real rank there cannot be a formula computing the real rank of $M_n(A)$ in terms of the real rank of $A$. But one would at least hope for the possiblity to bound $\mathrm{rr}(M_n(A))$ in terms of $\mathrm{rr}(A)$.
[1] Brown, Pedersen. C*-algebras of real rank zero, J. Funct. Anal. 99 (1991)
[2] Rieffel. Dimension and stable rank in the K-theory of C*-algebras, Proc. Lond. Math. Soc. 46 (1983)
You are almost surely aware of this, but the answer is known to be affirmative in case $\mathrm{rr}(A) = 0$ by Theorem 2.10 in Brown & Pedersen.
Yes, one can handle specific cases of low rank. If rr(A)=0, then rr(M_n(A))=0. If rr(A)=1 and sr(A)=1, then sr(M_n(A))=1 and thus rr(M_n(A))\leq 1 (and actually =1), But even the general case of rr(A)=1 is not clear to me.
|
2025-03-21T14:48:29.739436
| 2020-01-30T11:24:50 |
351522
|
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"Carlo Beenakker",
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|
Stack Exchange
|
How are number theory and C*-algebras connected?
I came across this research profile where under Research Overview, it states that
These days C*-algebra theory is a very active area of mathematical research in its own right, and enjoys deep connections with other areas of mathematics such as symbolic dynamics, ergodic theory, group theory and even number theory.
My question is how C*-algebras and number theory are connected and if this is an active area of research?
$C^∗$-algebras
associated with the $ax+b$-semigroup over $\mathbb{N}$ (2006)
Ring $C^*$-algebras (2009)
$K$-theory for ring $C^*$-algebras (2012)
$C^*$-algebras from actions of congruence monoids on rings of algebraic integers (2019)
On $K$-theoretic invariants of semigroup $C^*$-algebras from actions of congruence monoids (2019)
These papers associate C*-algebras to rings and study the inner structure of these ring C*-algebras. The
generators of the corresponding $K$-theory turn out to be determined
by prime numbers. Several connections to algebraic number theory are worked out in section 6 of the 2009 paper.
While I appreciate your list was meant to be indicative not exhaustive, worth pointing out that there has been more recent work strengthening some of the earlier results: see https://arxiv.org/abs/1906.00445 which you could add to the list
thanks for the pointer; two 2019 references added, this might also go some way to answering the second question of the OP: "is this is an active area of research?"
I think the most prominent example of this is Connes' work on the Riemann hypothesis from a C*-algebra perspective.
|
2025-03-21T14:48:29.739605
| 2020-01-30T11:25:53 |
351523
|
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"Aryeh Kontorovich",
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|
Stack Exchange
|
Gaussian concentration inequality
Recently I found a concentration inequality for infinite dimensional Gaussian r.v.s in this paper. Specifically, Lemma 4 on page 307 states (without a proof) that
There exists a universal constant $M$ such that for each Banach space valued Gaussian random variable $X$ (having zero mean):
$$
\mathsf{P}(\|X\|\ge u)\le \exp\left(-\frac{u^2}{M\mathsf{E}\|X\|^2}\right).
$$
The authors refer to an older paper which is not available online. So I'm wondering how one proves this result.
As a first step, applying the generic Chernoff bound, one gets
$$
\mathsf{P}(\|X\|\ge u)\le e^{-su}\mathsf{E}e^{s\|X\|}
$$
for any $s>0$. Then the desired inequality holds if $\mathsf{E}e^{s\|X\|}$ is bounded by $e^{Cs^2\mathsf{E}\|X\|^2}$.
See Lemma 3.1 in the Ledoux-Talagrand book, available online.
Indeed, but it’s a mean-zero random variable, so one should be able to control the median (using concentration)?..
Yes, I see now. I thought it would be absorbed into the universal constant M.
I tried to derive this bound for Gaussian vectors in $\mathbb{R}^d$. For $X\sim N(0,\Sigma)$, the best I could obtain so far is
$$
\mathsf{P}(|X|_2\ge u)\le 2\exp(-u^2/(2\operatorname{tr}(\Sigma))),
$$
and I don't see a way to get rid of the factor of $2$ before the exponent.
There's an unspecified universal constant $M$ in the OP -- doesn't that make any factor before the exponent meaningless?
@AryehKontorovich For small values of $x$, $2e^{−x}>e^{−cx}$ for any $c>0$.
This inequality is false. E.g., consider the random vector $X_n:=(Z_1,\dots,Z_n)/\sqrt n$ in $\mathbb R^n$ with the Euclidean norm $\|\cdot\|$, where $Z_1,Z_2,\dots$ are independent standard normal random variables. Then $E\|X_n\|^2=1$ and, by the law of large numbers,
$$\|X_n\|^2=\frac1n\,\sum_1^n Z_i^2\to1$$
in probability (as $n\to\infty$), so that $P(\|X_n\|\ge u)\to1$ for any $u\in(0,1)$. So, for any real constant $M>0$, any $u\in(0,1)$, and all large enough $n$
$$P(\|X_n\|\ge u)\not\le \exp\Big(-\frac{u^2}{M}\Big)=\exp\Big(-\frac{u^2}{M\,E\|X_n\|^2}\Big).$$
On the other hand, according to formula (3.5) in the Ledoux--Talagrand book, one has e.g. the inequality
$$P(\|X\|\ge u)\le 4\exp\Big(-\frac{u^2}{8E\|X\|^2}\Big)$$
for $u\ge0$.
The constants here can be a bit improved by using Corollary 3, which states that
$$P(\|X\|-E\|X\|\ge x)\le \exp\Big(-\frac{x^2}{2E\|X\|^2}\Big)\tag{1}$$
for $x\ge0$, which implies e.g. that
$$P(\|X\|\ge u)\le\sqrt e\,\exp\Big(-\frac{u^2}{8E\|X\|^2}\Big)\tag{2}$$
for $u\ge0$.
Details on how to get (2) from (1): If $u^2<4E\|X\|^2$, then the upper bound on $P(\|X\|\ge u)$ in (2) is $>1$ and thus trivial. So, without loss of generality, $u^2\ge4E\|X\|^2\ge4(E\|X\|)^2$, so that $E\|X\|\le u/2$. Letting now $x:=u-E\|X\|\ge u/2$, we see that the left-hand of (1) becomes that of (2), and the right-hand of (1) is less than that of (2).
Interestingly, the same inequality is stated in another paper by Basu.
@Iosif Can you kindly explain how you went from Corollary 3 of your paper to this form in terms of u ?
@d.k.o. : By "the same inequality", do you mean the false one? Maybe, this is how false results propagate. Hopefully, this omission of a constant factor does not affect most of the applications.
@gradstudent : I have added details on how to get (2) from (1).
|
2025-03-21T14:48:29.739970
| 2020-01-30T11:41:02 |
351524
|
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|
Stack Exchange
|
Index of a particular subfactor
If a compact group $G$ acts on vN algebra factor $M\subset B(L^{2}(M))$, what would be the index of subfactor $[M^{G}:M]$? KIndly explain the answer.
Dear Desperate mathematician, it would be useful if you could explain what are the things that you've already studied and understood, that go in the direction of the question asked. For example, do you understand the case when $G$ is $\mathbb Z/2 \mathbb Z$? If yes, do you understand the case when $G$ is finite... etc.
No, I do not understand in any case. The definition of index to me is the dimension of $L^{2}(M)$ over $M^{G}$ module.
Are you asking if it could be infinite? This is indeed the case: consider the very first example (which led to Vaughan Jones' results), the product type action of SU(2) on the $2^{\infty}$ UHF C*-algebra, extended to the action on the hyperfinite factor. If the group is connected and the action is nontrivial, presumably the index is always infinite.
Case $G=\mathbb Z/2$, generated by $g\in G$:
In that case, the action of $G$ on $M$ induces a grading $M=M_0 \oplus M_1$ where $M_0=M^G=\{m\in M: gm=m\}$ and $M_1=\{m\in M: gm=-m\}$.
Similarly, the action of $G$ on $L^2M$ induces a grading $L^2M=(L^2M)_0\oplus (L^2M)_1$. The subspace $(L^2M)_0$ is isomorphic to $L^2(M^G)$.
And both $(L^2M)_0$ and $(L^2M)_1$ are isomorphic as $M^G$-modules.
So
$$
\dim_{M^G}(L^2M) = \dim_{M^G}(L^2M)_0 + \dim_{M^G}(L^2M)_1 = 2\cdot \dim_{M^G}(L^2(M^G))=2.
$$
|
2025-03-21T14:48:29.740122
| 2020-01-30T12:15:44 |
351527
|
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|
Stack Exchange
|
A ``1-soft'' improvement of the Parovichenko theorem
This is a ``1-soft'' modification of this problem. We start with the necessary definitions.
Definition 1. A compactification $c\mathbb N$ of the discrete space $\mathbb N$ is called 1-soft if for any subset $A\subset\mathbb N$ with $\overline{A}\cap\overline{\mathbb N\setminus A}\ne\emptyset$ there exists a homeomorphism $h:c\mathbb N\to c\mathbb N$ such that $h(x)=x$ for all $x\in c\mathbb N\setminus\mathbb N$ and the set $\{x\in A:h(x)\notin A\}$ is infinite.
Definition 2. A compactification $c\mathbb N$ of the discrete space $\mathbb N$ is called 2-soft if for any disjoint sets $A,B\subset\mathbb N$ with $\bar A\cap\bar B\ne\emptyset$ there exists a homeomorphism $h:c\mathbb N\to c\mathbb N$ such that $h(x)=x$ for all $x\in c\mathbb N\setminus\mathbb N$ and the set $\{x\in A:h(x)\in B\}$ is infinite.
Before the formulation of a question, let us recall some known results.
Theorem (Parovichenko). Each compact Hausdorff space $K$ of weight $\le\omega_1$ is homeomorphic to the remainder $c\mathbb N\setminus \mathbb N$ of some compactification $c\mathbb N$ of $\mathbb N$.
Theorem (Hart). Under CH, each compact Hausdorff space $K$ of weight $\le\omega_1$ is homeomorphic to the remainder of a 2-soft compactification of $\mathbb N$.
Example (Dow). Under (NT) the compact space $K=\omega_1+1+\omega_1^*$ is not homeomorphic to the remainder of a 2-soft compactification of $\mathbb N$.
Question 1. Is the compact space $K=\omega_1+1+\omega_1^*$ homeomorphic to the remainder of a 1-soft compactification of $\mathbb N$?
Question 2. Is each compact Hausdorff space of weight $\le\omega_1$ homeomorphic to the remainder of a 1-soft compactification of $\mathbb N$?
Added in Edit. The answer to Question 1 is affirmative. So, only Question 2 remains open.
I guess the $B$ in the definition of $1$-soft should be $A$?
Yes, exactly! Thank you (I am glad that you noticed this question).
@KPHart I have just asked another set-theoretic question (https://mathoverflow.net/q/351555/61536) related to Question 1.
Could you include the construction for 1?
@KPHart I added a link to the paper in arXiv.
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2025-03-21T14:48:29.740267
| 2020-01-30T12:25:30 |
351528
|
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|
Stack Exchange
|
Ideal of rational normal curve of degree $d$
Let $A$ consist of the columns of the $2\times (d+1)$ matrix
$$A=\begin{pmatrix}
d & d-1 & \cdots & 1&0\\
0 & 1 & \cdots & d-1 &d
\end{pmatrix}$$
Then consider the map
\begin{array}{lcl}\theta_A:\mathbb{(C^*)}^2\rightarrow\mathbb{P}^d\\
(s,t)\rightarrow[s^d:s^{d-1}t:\dots:st^{d-1}:t^d]\end{array}
We know that $C_d$ is Zariski closure of image of the $\theta_A$ and $C_d$ called the rational normal curve of degree $d$.
Now I want to show that $I(C_d)=\langle x_ix_{j+1}-x_{i+1}x_j: 0\le i <j\le d-1\rangle$.
Consider $$[s^d:s^{d-1}t:\cdots:st^{d-1}:t^d]=[1:\frac{t}{s}:\frac{t^2}{s^2}:\cdots:\frac{t^{d-1}}{s^{d-1}}:\frac{t^d}{s^d}]=[1:u:u^2:\cdots:u^{d-1}:u^d]$$
Then we get $B=\begin{bmatrix}0&1&\cdots&d-1&d\end{bmatrix}$. And also it is clear that $B$ is in the row space of $A$.
This gives the map \begin{array}{lcl}\theta_B:\mathbb{(C^*)}^2\rightarrow\mathbb{P}^d\\
(s,t)\rightarrow[1:t:\dots:t^{d-1}:t^d]\end{array}
And also I know $C_d$ is Zariski closure of image of the $\theta_B$. But I don't know how can I show that $I(C_d)=\langle x_ix_{j+1}-x_{i+1}x_j: 0\le i <j\le d-1\rangle$. I need a hint for this.
I try to use Proposition 2.1.4 (Cox, Little, Schenk-Toric varieties) but I cant see yet.
The columns of $A$ give the Laurent monomials (you see image of the map $\theta_A$) defining the rational normal curve $C_d$.
Okay, but it is not clear to me. How can I show the ideal is equal to ideal of the rational normal curve. I need to hint.
It can probably be done by clever tricks/brute force but to really understand why this is true one needs some representation theory of $SL_2$. If I have time I will explain.
To "close" voters: it is true that the question as is appears as an exercise in many algebraic geometry books, but suitably enlarged (see the end of my answer) it becomes a completely open problem.
I added the "plethysm" tag because this could have easily made it in the book by Fulton and Harris on representation theory as another "A little geometric plethysm" section. See in particular Section 11.3 of the 1991 edition.
The ideal of the rational normal curve
Let $V$ be a two-dimensional complex vector space and let $V^{\vee}$ be the dual space of linear scalar-valued functions on $V$. Although representations of $SL(V)$ are self-dual, I will keep the distinction between vectors and covectors in order to avoid confusion.
Let $W={\rm Sym}^d(V^{\vee})$ be the space of homogeneous polynomial functions of degree $d$ on $V$. It's good to think of the projective space $\mathbb{P}^d$ here as $\mathbb{P}(W)$.
The rational normal curve $C_d$ is the variety of projective classes of nonzero $F$'s in $W$ of the form $F=L^d$ for some linear form $L$ in $V^{\vee}$.
The ideal $I$ is homogeneous and therefore admits a graded decomposition $I=\bigoplus_{N\ge 0}I_N$. Now let me take a homogeneous polynomial $P$ of degree $N$ in the ideal. Namely,
$$
P\in I_N={\rm Sym}^{N}(W^{\vee})\simeq {\rm Sym}^{N}({Sym}^{d}(V))\subset V^{\otimes Nd}\ .
$$
For $F\in W$, one can see the evaluation $P(F)$ as a duality pairing
$$
P(F)=\langle P,F\otimes\cdots\otimes F\rangle
$$
between $V^{\otimes Nd}$ on the left and $(V^{\vee})^{\otimes Nd}$ on the right, with $F$ appearing $N$ times.
Consider the $SL(V)$ module $W^{\otimes N}$ and the identity map $I_{W^{\otimes N}}$ from this module to itself. Decompose this module into irreducibles. This gives a decomposition of the identity of the form
$$
I_{W^{\otimes N}}=\sum_{\alpha\in A}\iota_{\alpha}\circ\pi_{\alpha}
$$
where for $\alpha\in A$, $\pi_{\alpha}:I_{W^{\otimes N}}\rightarrow {\rm Sym}^{k_{\alpha}}(V^{\vee})$ is an equivariant projection on an irreducible indicated by the nonnegative integer $k_{\alpha}$, and $\iota_{\alpha}:{\rm Sym}^{k_{\alpha}}(V^{\vee})\rightarrow I_{W^{\otimes N}}$ is a "reinjection". The complete symmetrization corresponding to $k_{\alpha}=Nd$ occurs only once, say for $\alpha=\alpha_0$.
We now have
$$
P(F)=
\langle P, \iota_{\alpha_0}\circ\pi_{\alpha_0}(F\otimes\cdots\otimes F)\rangle
+\sum_{\alpha\in A, \alpha\neq \alpha_0}
\langle P, \iota_{\alpha}\circ\pi_{\alpha}(F\otimes\cdots\otimes F)\rangle\ .
$$
But the contribution of $\alpha_0$ vanishes because $P$ is in the ideal.
In other words $\langle P, L\otimes\cdots\otimes L\rangle$, with $Nd$ tensor factors equal to a linear form $L$, is always zero. This is because powers linearly span the symmetric power. See, e.g.,
A basis of the symmetric power consisting of powers
Other terms can be analyzed as follows, but one needs to know the explicit structure of the Clebsch-Gordan decomposition. Namely, not just the list of irreducible submodules, but the decomposition of the identity with explicit intertwiners $\pi$ and $\iota$. Moreover, this Clebsch-Gordan decomposition is here iterated $N-1$ times.
This involves symmetrizations as well as contractions with a nonzero $SL(V)$ invariant alternating form $\epsilon\in \wedge^2 V$ and also with its dual. Let $e_1,e_2$ be a basis of $V$ and $x_1,x_2$ be the dual basis of $V^{\vee}$.
One can take $\epsilon=e_1\otimes e_2-e_2\otimes e_1$.
An element $F\in W$ is thus a binary form
$$
F(x_1,x_2)=\sum_{i=0}^{d}\binom{d}{i} f_i x_1^{d-i}x_2^{i}
$$
with complex coefficients $f_0,\ldots,f_d$.
The key observation is: the terms with $\alpha\neq \alpha_0$ always involve at least one copy of $\epsilon$.
So such a term $\langle P, \iota_{\alpha}\circ\pi_{\alpha}(F\otimes\cdots\otimes F)\rangle$ is a linear combination of terms of the form $\langle \Gamma, F\otimes\cdots\otimes F\rangle$
where $\Gamma$ looks like
$$
e_{i_1}\otimes \cdots \otimes e_{i_{a-1}}\otimes
e_1\otimes
e_{i_{a+1}}\otimes \cdots \otimes e_{i_{b-1}}\otimes
e_2\otimes
e_{i_{b+1}}\otimes \cdots \otimes e_{i_{Nd}}
$$
$$
- e_{i_1}\otimes \cdots \otimes e_{i_{a-1}}\otimes
e_2\otimes
e_{i_{a+1}}\otimes \cdots \otimes e_{i_{b-1}}\otimes
e_1\otimes
e_{i_{b+1}}\otimes \cdots \otimes e_{i_{Nd}}
$$
for $1\le a<b\le Nd$. Namely, $\epsilon$ has been inserted in the pair of spots given by positions $a$ and $b$ in $V^{\otimes Nd}$, while the other spots are given by basis vectors.
Now with a little computation, one sees that $\langle \Gamma, F\otimes\cdots\otimes F\rangle$, if nonzero, is a monomial of degree $N-2$ in the $f$'s times a $2\times 2$ minor of the matrix
$$
\begin{pmatrix}
f_0 & f_1 & \cdots & f_{d-1} \\
f_1 & f_2 & \cdots & f_{d}
\end{pmatrix}
\ .
$$
QED.
To get an idea of how the decomposition of the identity looks like when $N=2$, see this picture:
which is taken from my article "On the volume conjecture for classical spin networks" in JKTR 2012. Depending on what one decides is $V$ or $V^{\vee}$, the lines with arrows at the bottom are $\epsilon$'s, i.e., $e_1\otimes e_2-e_2\otimes e_1$ while the ones on top are their duals $x_1\otimes x_2-x_2\otimes x_1$. The rectangular boxes correspond to symmetrizers. This "microscopic" graphical notation can be abbreviated to a "macroscopic" one as in the picture (from the same article) given by:
This is needed for iteration $N-1$ times.
More generally for an integer partitions $\lambda$ of $d$, one can study the defining ideals of coincident root loci, i.e., varieties $X_{\lambda}$
of forms $F$ which factor as $L_1^{\lambda_1}L_2^{\lambda_1}\cdots$ for suitable linear forms. The rational normal curve corresponds to the one part case $\lambda=(d)$. For partitions with two parts, the ideal was determined by Chipalkatti and myself in the two articles:
"Brill-Gordan Loci, transvectants and an analogue of the Foulkes conjecture" in Adv. Math. 2007.
"The bipartite Brill-Gordan locus and angular momentum" in Transform. Groups 2006.
For a more recent update on the ideals of such loci $X_{\lambda}$, see the article by Lee and Sturmfels, "Duality of multiple root loci", in J. Algebra 2016.
Variant:
Since the explicit Clebsch-Gordan decomposition (of course known to Alfred Clebsch and Paul Gordan in the 1870's) is not well known today, one can finish the proof above in a different way. Let $m=Nd$ and consider the action of the group algebra $\mathbb{C}[\mathfrak{S}_m]$ with identity $e$, on $(V^{\vee})^{\otimes m}$. The complete symmetrization corresponds to the element $s=\frac{1}{m!}\sum_{\sigma\in\mathfrak{S}_m}\sigma$. One needs to show that $s-e$ is a linear combination of terms of the form $\rho\ (\tau-e)$
where $\rho$ is a permutation and $\tau$ is a transposition. This reduces to showing the same kind of decomposition for $\sigma-e$, for arbitrary permutations $\sigma$. Write $\sigma$ as a product of transpositions and expand as in the example
$$
\tau_2\tau_1-e=\tau_2((\tau_1-e)+e)-e=\tau_2(\tau_1-e)+(\tau_2-e)
$$
etc.
Because of the Grassmann-Plücker relation, $\tau-e$ produces, in the big tensor product ${\rm Hom}((V^{\vee})^{\otimes m},(V^{\vee})^{\otimes m})\simeq (V^{\vee})^{\otimes m}\otimes V^{\otimes m}$, the insertion of
$$
(e_1\otimes e_2-e_2\otimes e_1)\otimes (x_1\otimes x_2-x_2\otimes x_1)
$$
as wanted.
Remark: The variant works for more general Veronese embedings with ${\rm dim}(V)>2$. Then, no need then to invoke Grassmann-Plücker, but the distinction of $V$ and $V^{\vee}$ is essential. Interestingly, the explicit Clebsch-Gordan decomposition is not well understood in that case, except for ${\rm dim}(V)=3$
as worked out in this article. Instead of considering the varieties $X_{\lambda}$, another direction for generalization is to look at defining ideals of secants of the Veronese. This also full of open problems which although of classical nature, have recently received much attention as central questions in Geometric Complexity Theory. See, e.g., the books "Tensors: Geometry and Applications" and "Geometry and Complexity Theory" by J.M. Landsberg. Another good reference is, the lecture notes "Introduction to geometric complexity theory" by Bläser and Ikenmeyer.
|
2025-03-21T14:48:29.740856
| 2020-01-30T12:27:52 |
351529
|
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|
Stack Exchange
|
finding a good term order for grobner basis
What are the tricks to pick a "good" monomial order to find a Grobner basis for a given ideal?
By good I mean one in which the final Grobner basis has a simple expression in terms of the coefficients of the original polynomials defining the ideal, and/or in which the computation of the normal form of a polynomial with respect to the final basis is fast.
Also, any reference which addresses the above is welcome!
|
2025-03-21T14:48:29.740925
| 2020-01-30T12:35:28 |
351531
|
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"Omar Antolín-Camarena",
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"sort": "votes",
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}
|
Stack Exchange
|
Horizontal categorification: Two questions
According to the nlab, horizontal categorification is a process in which a concept is realized to be equivalent to a certain type of category with a single object, and then this concept is generalized to the same type of categories with an arbitrary number of objects. The prototypical example is the concept of a group, which horizontally categorifies to the concept of a groupoid. It is well-known that in certain parts of algebraic topology groupoids are much more convenient than groups. Similarly, monoids categorify to categories, rings to linear categories, etc.
I have two questions about this. Let C be a concept which horizontally categorifies to a concept D.
1) In many examples, C and D have been known and developed independently from each other. Or at least, D was not introduced as the horizontal categorification of C, but rather this connection was realized afterwards. For example, I am pretty sure that k-linear categories were not introduced as a horizontal categorificiation of k-algebras; instead they were introduced because of the abundance of examples of (large) k-linear categories which appear in everyday mathematics. Although representation theory seems to be in a current progress of a generalization from k-algebras to small k-linear categories, the concept of a k-linear category was already there before that. Are there examples where D was developed with the purpose to categorify C, say in order to solve some problems which deal with C but which are not solvable with C alone? Perhaps C*-categories (categorifying C*-algebras) could provide such an example, but I am not familiar with the history of this concept. And maybe there are other examples as well?
2) In all the examples I know of, it is trivial that C has a horizontal categorification and that it is D. Are there any more interesting examples where, when you look at C, it is not even clear how to interpret C as a type of category with one object? I would like to see examples where the connection between C and D is deep and surprising. These examples should illustrate why horizontal categorification is an important and useful concept in practice.
I could also ask a more provocative question: if any mathematician working outside of category theory reads the nlab article in its current form, why should he/she even care, since, after all, both concepts C and D have been there already without the categorification process?
Good question, I hadn't realized I wanted to know the answer to this!
Often when a horizontal categorification is introduced, the author has intended examples, which the author gives as the motivation. It thus makes it a little tricky to determine when a structure has been motivated explicitly by horizontal categorification. However, here are a few examples where (1) the concept had not previously appeared in the literature, and (2) the authors make it clear horizontal categorification was intended (though this precise language may not appear).
In Street's Elementary cosmoi I, the concept of extension system is introduced, which is explicitly intended to be to closed categories what bicategories are to monoidal categories. Intuitively, the analogue of closed structure is proved by right-extensions.
Two more examples are the multi- and poly-bicategories of Cockett–Koslowski–Seely's Morphisms and modules for poly-bicategories, which are horizontal categorifications of multicategories and polycategories respectively.
My favourite example of a surprising horizontal categorification is that of symmetric monoidal categories. A symmetric bicategory (not to be confused with a symmetric monoidal bicategory) is a bicategory with an involution. Explicitly, a bicategory $\mathcal K$ is symmetric when equipped with a biequivalence $\mathcal K \simeq \mathcal K^{\mathrm{op}}$ that is a bijection on objects. A symmetric monoidal category is precisely a one-object symmetric bicategory for which the involution is the identity. This is interesting in two ways: the generalisation from symmetry to involution is a highly non-obvious step; and one is forced to make a generalisation to the original structure in order to horizontally categorify. There is an analogous concept of closed symmetric bicategory, which is a symmetric bicategory with left- and right-extensions. (Closed) symmetric bicategories are defined and studied in May–Sigurdsson's Parametrized Homotopy Theory.
I think these examples give some indication that studying horizontal categorifications in their own right can lead to interesting structures that may then be motivated by concrete examples.
|
2025-03-21T14:48:29.741232
| 2020-01-30T12:46:44 |
351533
|
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"Sam Hopkins",
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|
Stack Exchange
|
Relations between double coinvariants and affine Springer fibers
Diagonal coinvariants have an interpretation from https://arxiv.org/abs/math/0201148 in terms of the Hilbert scheme.
There are two recent papers https://arxiv.org/pdf/1801.09033.pdf and https://arxiv.org/pdf/1808.02278.pdf which both detail (different and seemingly unconnected) relations between diagonal coinvariants and affine Springer fibers (another paper that does this is https://arxiv.org/pdf/1203.5878.pdf). In the first paper diagonal coinvariants act on suitable (co)homology of affine Springer fibers and in the second paper diagonal coinvariants are (sometimes conjecturally) identified with certain (co)invariants in suitable (co)homology of certain affine Springer fibers. Why should there be such relations between diagonal coinvariants and affine Springer fibers?
Is there any relation of the Hilbert scheme interpretation of diagonal coinvariants to the affine Springer fiber interpretations of diagonal coinvariants?
Also, can one explain intuitively the connections of the above relations between diagonal coinvariants and affine Springer fibers with the results on the action of DAHA/Cherednik algebra on the suitable (co)homology of the affine Springer fibers in https://arxiv.org/abs/1407.5685 and https://arxiv.org/abs/0705.2691 ?
This seems like a very good/interesting question to me; but to play devil's advocate: you cite many papers here- maybe some of them contain an explanation of their motivation/intuition?
@SamHopkins I had the questions I am asking here after looking through all these papers
|
2025-03-21T14:48:29.741366
| 2020-01-30T13:32:10 |
351534
|
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Stack Exchange
|
Motivating the coefficient field of $\ell$-adic cohomology
It was already known to Weil that a sufficiently reasonable cohomology theory for algebraic varieties over $\mathbb{F}_p$ would allow for a possible solution to the Weil conjectures.
It was also understood that such a cohomology theory could not take values in vector spaces over either the rational numbers $\mathbb{Q}$ or the $p$-adic numbers $\mathbb{Q}_p$. The classical argument that I know is that the cohomology of a supersingular elliptic curve should admit an action by a quaternion algebra structure of the endomorphism group of the elliptic curve, and representation theory rules out such an action. (Side question: Due to whom is this reasoning, and was it known to Weil's generation?)
This reasoning does not, however, rule out the possibility of cohomology with coefficients in $\mathbb{Q}_\ell$ with $\ell \neq p$, and indeed $\ell$-adic cohomology would end up being found.
Question. Were there any other historical reasons to believe that looking at cohomology valued in $\mathbb{Q}_\ell$-vector spaces would lead to something fruitful?
N.B. This question was posted on MathSE, but I got advised to post it here instead.
The argument with supersingular elliptic curves and quaternion algebras is due to Serre, I believe.
(See 1.7, p. 315 in Grothendieck's "Crystals and the de Rham cohomology of schemes", "As Serre has pointed out...")
It is not true that etalé cohomology with coefficients in $\mathbb{Q}_\ell$ is a "good" cohomology. $\ell$-adic cohomology is something more involved that just "cohomology with coefficients": its is the projective limit of some with finite coefficients (and tensoring with $\mathbb{Q}$).
@Xarles That is of course what I meant. It takes values in $\mathbb{Q}_\ell$-modules, anyway. I'll edit the question accordingly.
One historical reason for considering $\ell$-adic cohomology, not completely disconnected from the example you introduce, is that for a curve over a field, we get a natural Galois representation by taking the $\ell$-adic Tate module of the Jacobian (i.e., the projective limit of $\ell$-power torsion). Furthermore, if such a curve is defined over a subfield of the complex numbers, then the rank of the Tate module is equal to the rank of the classical degree 1 cohomology of the complexified curve. We now know that the Tate module is naturally dual to étale cohomology in degree 1.
One might then reasonably hope for a similar relationship between the higher degree classical cohomology of higher dimensional varieties and certain $\ell$-adic Galois representations naturally attached to the variety.
I think an important motivation for the $\ell$-adic theory comes from the Riemann existence theorem/the Grauert–Remmert theorem. This says that a finite (topological) covering space $Y \to X$ of a normal complex variety can again be equipped with an algebraic structure, which is more or less what you need to prove to obtain
$$\pi_1^{\operatorname{\acute et}}(X) = \widehat{\pi_1^{\operatorname{top}}(X(\mathbf C))}.$$
So finite covering spaces can be detected using the étale topology (except it's not really a topology, but that's not stopping Grothendieck!).
In particular, you expect to get a good theory of étale cohomology with finite coefficients. But to run the arguments that Weil was dreaming of, you need characteristic $0$ coefficients, so what do you do? Just take a limit!
I think that's really the explanation of why the adic formalism enters the picture. But it doesn't quite explain why it's different at the prime $p$. There are a few ways to look at this:
Serre's argument shows that a $\mathbf Q_p$-valued Weil cohomology theory cannot exist (over any ground field $k$ containing $\mathbf F_{p^2}$; so in particular for $k = \bar{\mathbf F}_p$).
The basic results on $\ell$-adic étale cohomology take the Galois cohomology of function fields of curves over algebraically closed fields as a starting point, and these behave differently at the prime $p$.
Already for elliptic curves, the $p$-adic Tate module behaves a little different from the $\ell$-adic one.
In the end it doesn't really matter, because all they needed was one Weil cohomology theory.
Serre only shows that there is no $\mathbf{Q}$-linear Weil cohomology, at least if we restrict to algebraic varieties over a field which contains $\mathbf{F}_{p^2}$. If we work over $\mathbf{F}_p$, crystalline cohomology actually is a $\mathbf{Q}_p$-linear Weil cohomology, and I am not aware of an argument which rules out the existence of $\mathbf{Q}$-linear one. Your assertion that "it does not matter" can also be discussed: independence of $\ell$ problems have a long history which seem to point in a rather different direction.
It might also be worth mentionning that the standard conjectures together with tannakian arguments imply the existence of a $\bar{\mathbf{Q}}$-linear Weil cohomology.
@Denis-CharlesCisinski thanks for your comments. Serre's argument depends on representations of the quaternion algebra over $\mathbf Q$ that splits at $p$ and $\infty$, so it equally rules out $\mathbf Q_p$- and $\mathbf R$-valued Weil cohomology theories (as soon as $k \supseteq \mathbf F_{p^2}$).
Also, the original question is about "historically, why $\mathbf Q_\ell$", and my point is that if you're trying to prove the Weil conjectures you only need one cohomology theory. (Of course I am aware of the interesting "independence of $\ell$" questions ― I even have a paper on it!)
Crystalline/rigid cohomology for $k$-varieties take values in $W(k)$. So $\mathbf{Q}_p$ is a possible field of coefficient for $k=\mathbf{F}_p$.
Therefore, if, as you wrote, we only need one Weil cohomology, we may take a $\mathbf{Q}_p$-linear one defined over $\mathbf{F}_p$... I still think your post is not completely accurate.
@Denis-CharlesCisinski agreed (but note that this imprecision was already present in the original question).
@Denis-CharlesCisinski Is there a place the Tannakian argument based on the standard conjectures is written?
@lemiller If $k$ is an algebraically closed field, then any $k$-linear tannakian category is neutral. That is due to Deligne for $k$ of char. $0$ and to Coulembrier in general as Theorem 6.4.1 in this paper: https://arxiv.org/pdf/1812.02452.pdf Since the standard conjectures imply that pure motives form a tannakian category (see e.g. the proceedings of the Seattle conference of Yves André's book on motives, or Milne's papers on motives), they imply the existence of a $\bar{\mathbf{Q}}$-linear Weil cohomology.
|
2025-03-21T14:48:29.741942
| 2020-01-30T13:54:35 |
351535
|
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|
Stack Exchange
|
Characterization of bounded variation
For a function $f:[0,1]\to\mathbb{R}$, define
$$ V(f)=\sup_{0=x_0<x_1<\ldots<x_n=1}\sum_{i=1}^{n}|f(x_n)-f(x_{n-1})|.
$$
For $f$ with integrable derivative, the definition coincides with
$V(f)=\int_0^1|f'(x)|dx$.
Now every continuous $f:[0,1]\to\mathbb{R}$ is uniformly approximable by a $C^\infty$ function (Weierstrass); denote by $C_{\epsilon}(f)$ the collection of all $g\in C^\infty[0,1]$ such that $\sup_{x\in[0,1]}|f(x)-g(x)|\le\epsilon$.
I conjecture that for every continuous $f:[0,1]\to\mathbb{R}$, we have
$$
V(f) = \limsup_{\epsilon\to0}\inf_{g\in C_\epsilon(f)}V(g).
$$
Is this true? Known?
This is quite simple: if $\|f - g\|_\infty \leqslant \epsilon$, then clearly
$$ \sum_{i = 1}^n |f(x_i) - f(x_{i-1})| \leqslant \sum_{i = 1}^n |g(x_i) - g(x_{i-1})| + 2 n \epsilon \leqslant V(g) + 2 n \epsilon.$$
Therefore,
$$ \sum_{i = 1}^n |f(x_i) - f(x_{i-1})| \leqslant \inf_{g \in C_\epsilon(f)} V(g) + 2 n \epsilon,$$
and consequently
$$ \sum_{i = 1}^n |f(x_i) - f(x_{i-1})| \leqslant \liminf_{\epsilon \to 0^+} \inf_{g \in C_\epsilon(f)} V(g) .$$
Take the supremum over all possible partitions to get
$$ V(f) \leqslant \liminf_{\epsilon \to 0^+} \inf_{g \in C_\epsilon(f)} V(g) .$$
EDIT: I forgot about the converse inequality:
On the other hand, $f$ can be approximated by functions from $C_\epsilon(f)$ in the total variation norm (see below), so that
$$ V(f) \geqslant \inf_{g \in C_\epsilon(f)} V(g) , $$
and hence
$$ V(f) \geqslant \limsup_{\epsilon \to 0^+} \inf_{g \in C_\epsilon(f)} V(g) . $$
Regarding approximation: if $f$ is non-decreasing, then there is a smooth function $g \in C_\epsilon(f)$ such that $V(g) \leqslant V(f)$ (just mollify $f$ appropriately). In the general case, if $V(f) < \infty$, then $f = f_+ - f_-$ with $f_+$ and $f_-$ non-decreasing and $V(f) = V(f_+) + V(f_-)$. Find $g_+$ and $g_-$ as above; then $g = g_+ - g_-$ is in $C_{2\epsilon}(f)$, and $$V(g) \leqslant V(g_+) + V(g_-) \leqslant V(f_+) + V(f_-) = V(f).$$
|
2025-03-21T14:48:29.742083
| 2020-01-30T14:33:04 |
351538
|
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|
Stack Exchange
|
Applications of linear algebra in the design of aircraft
David Lay mentioned one application of linear algebra in the design of aircraft in the introductory part of chapter 2 of his book:
[...] A computer creates a model of the surface by first superimposing a three-dimensional grid of “boxes” on the original wire-frame model. Boxes in this grid lie either completely inside or completely outside the plane, or they intersect the surface of the plane. The computer selects the boxes that intersect the surface and subdivides them, retaining only the smaller boxes that still intersect the surface. The subdividing process is repeated until the grid is extremely fine. A typical grid can include more than 400,000 boxes.
The process for finding the airflow around the plane involves repeatedly solving a system of linear equations $Ax=b$ that may involve up to 2 million equations and variables. The vector $b$ changes each time, based on data from the grid and solutions of previous equations.
Where can I find details about the boldfaced part? Can someone briefly give the general idea?
1). https://en.wikipedia.org/wiki/Computational_fluid_dynamics
2). https://ocw.mit.edu/courses/mechanical-engineering/2-29-numerical-fluid-mechanics-spring-2015/lecture-notes-and-references/MIT2_29S15_Lecture8.pdf
a Google search for "finite element aircraft design" should give you a good set of pointers; for a useful MO question you would want to focus on a particular advanced math issue, I think.
|
2025-03-21T14:48:29.742231
| 2020-01-30T14:49:45 |
351539
|
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|
Stack Exchange
|
Proper analytic embedding of $\overline{\Bbb C}$ minus a Cantor set into $\Bbb C^2$
I am a PhD student in several complex variables.
I am reading this paper by Orevkov proving that there exists a proper analytic embedding of $\overline{\Bbb C}$ minus a Cantor set into $\Bbb C^2$.
I am really lost and I am writing here to ask some hint/tips on how to proceed reading it.
It is not clear why $K$ is a Cantor set, since in order to find out (analytically) its elements one should find roots of $n$-degree polynomials (for all $n\ge1$, and no recursive formula seems to appear).
Someone suggested me to argue geometrically, but it is very hard and, anyhow, it doesn't seem to help (for what I tried so far).
Also I cannot see how to prove points 2), 3) and 4) and how does are the exploited in the rest of this short but very dense paper.
Thanks.
Of course I don't ask a clear proof of this, but just some guidelines I can follow to going out of this labyrinth.
I have no access to the paper, but the formulation is ambiguous: does it mean $\mathbf{C}$ minus some Cantor subset, or minus every Cantor subset? while topologically these are all "the same", up to biholomorphy it's far from unique.
You should be able to find anwers to your question in a well witten book by Franc Forstnerič: Stein Manifolds and Holomorphic Mappings -The Homotopy Principle in Complex Analysis, https://www.springer.com/gp/book/9783319610573
In particular look at sections 9.10, 9.11 and Theorem 9.11.5. The book contains all the background material you need and also contains a rewritten proof of Orevkov's result.
Thank You very much for your answer. Nonetheless let me point out a couple of things. 1)Forstneric's book is very very unfriendly, at least for me. Rather than a reading book, it seems he's just collected all the result/papers in several complex variables, to resume the state of art of the subject. 2) I got a solution by myself, without needing all the complex tools in sections 9.10 and 9.11; the difficult part was to interpretate the composition of rational shears. Doing that, making also a sort of graph, one can understand the effect of the composition, leading to the final result.
|
2025-03-21T14:48:29.742407
| 2020-01-30T15:36:03 |
351544
|
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|
Stack Exchange
|
Relative projectivness and tensor products
Let $R$ be a commutative ring, $H<G$ a finite group pair. Let $P$ be a $RG$ module that is $RH$ projective. It is known that for any $RG$ module $X$ the tensor $P\otimes X$ is $RH$ projective. This in some sense has to do with Frobenius reciprocity.
Is there a similar result for a more general case other than finite dimensional Hopf algebra extensions?
|
2025-03-21T14:48:29.742473
| 2020-01-30T15:36:37 |
351545
|
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|
Stack Exchange
|
Continuity of solution of a parabolic PDE w.r.t. system parameters
If we have a system of PDE of the form:
$$\begin{cases} \dfrac{\partial y}{\partial t}(t,x)=D\Delta y+F(t,x,f(x),y) ,\ (t,x)\in (0,T)\times\Omega\\ \dfrac{\partial y}{\partial \nu}(t,x)=0,\ (t,x)\in (0,T)\times\partial\Omega \\ y(0,x)=y_{0}(x),\ x\in\Omega\end{cases}$$
with a unique solution $y=y^f$(mild solution). All derivatives are considered in the distributional sense (weak derivatives).Here $\Omega\subset\mathbb{R}^N$ is an open and bounded set with $C^1$ boundary, $y=(y_1,y_2,...,y_m)$, $F=(F_1,F_2,\dots,F_m)$, $D=diag(d_1,d_2,\dots, d_m),\ d_i\in\mathbb{R},\ \forall\ i=1,2,...,m$ and $f:\Omega\to\mathbb{R}$ a function in $L^1(\Omega)$. We consider the following approximation of the system:
$$\begin{cases} \dfrac{\partial y}{\partial t}(t,x)=D\Delta y+F(t,x,f_n(x),y) ,\ (t,x)\in (0,T)\times\Omega\\ \dfrac{\partial y}{\partial \nu}(t,x)=0,\ (t,x)\in (0,T)\times\partial\Omega \\ y(0,x)=y_{0}(x),\ x\in\Omega\end{cases}$$,
with the unique solution $y^{f_n}$.
How can we prove that if $f_n\to f$ in $L^1(\Omega)$, then $y_i^{f_n}\to y_i^f$ in $L^p((0,T)\times\Omega),\ \forall\ i=1,2,\dots,n$, for some $p\in [1,\infty)$? Are there some results in that sense? Maybe with a more general setting...?
Are there some estimates?
$$||y^{f_n}-y^f||\leq C||f_n-f||_{L^1(\Omega)}$$.
What are conditions on $F$?
We can assume any type of conditions (Lipschitz-type probably). I only want a method or some references about that type of parabolic problems.
Do you have the same question for an equation rather for a system?
It will be welcomed.
This is only a sketch of an argument that can be used. Assume that $F$ is Lipscthitz and let $y_1,y_2$ be the solutions corresponding to $f_1, f_2$. If $v=y_2-y_1$, then $$|v_t-\Delta v|=|F(f_2,y_2)-F(f_2,y_1)+F(f_2, y_1)-F(f_1,y_1)| \le L(|v|+|f_2-f_1|)$$ with zero bc and initial value. If $T(t)$ is the semigroup generated by $\Delta$ with the appropriate bc, then $$|v(t)| \le L\int_0^t T(t-s)|v(s)|ds +L\int_0^t T(t-s)(|f_2-f_1|)ds.$$ Next use the fact that $T(t)$ is contractive in $L^p$ and maps $L^1$ to $L^p$ with norm less that $Ct^{-N/p'}$. If $t \le \delta$, taking $L^p$ norms (in space) we get $$\|v(t)\|_p \le L\delta \sup_{0 \le t \le \delta}\|v(t)\|_p+CL\|f_2-f_1\|_1\int_0^\delta t^{-N/p'}dt.$$ Now take $p<N/(N-1)$ so that $t^{-N/p'}$ is integrable near $0$ and take the supremum of the left hand side over $[0, \delta]$. If $L\delta \le 1/2$ this gives $$ \sup_{0 \le t \le \delta} \|v(t)\|\le C_1 \|f_2-f_1\|_1.$$ Now the argument can be iterated to larger intervals.
Isn't that a common question on PDE's? Continuity of the solution for an approximated problem? I didn't find anything on the literature. Do you know some references? Thank you very much for the answer!
Yes it is, and the use of semigroups, though not strictly necessary makes it clean. Note that in argument I wrote you can use directly Gronwall lemma instead of taking a small delta. If you work in the same Banach space then continuous depends on data is exatly what a semigroup gives. In your situation some more effort is needed to go from 1 to p. Concerning references, I think that a book dealing with applications of semigroups to nonlinear equations gives all the ideas (e.g. Haroux-Cazenave).
|
2025-03-21T14:48:29.742699
| 2020-01-30T15:38:35 |
351546
|
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|
Stack Exchange
|
Are all maps $\mathbb{R}^2 \to \mathbb{R}^2$ with fixed singular values affine?
Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a smooth map whose differential has fixed distinct singular values $0<\sigma_1<\sigma_2$ and an everywhere positive determinant (which is the product $\sigma_1\sigma_2$).
Must $f$ be affine?
My assumption is equivalent to $df_x \in \text{SO}(2) \cdot \text{diag}(\sigma_1,\sigma_2) \cdot \text{SO}(2)$ for every $x \in \mathbb{R}^2$.
If we were only allowing a copy of $\text{SO}(2)$ from one of the sides of $ \text{diag}(\sigma_1,\sigma_2)$, then the answer would be positive. (This reduces to the case of isometries).
Similarly, if we had $\sigma_1=\sigma_2$, the answer would also be positive.
Answer modified on 1 February 2020:
It's not true 'locally' in the sense that non-affine $f$'s satisfying this system of PDE can be constructed on some open sets in $\mathbb{R}^2$. This first order, determined PDE system is hyperbolic, so there are many local solutions. However, it turns out (see below) that all $C^3$ solutions with domain equal to $\mathbb{R}^2$ are affine. (The proof I give below does not work for solutions of lower regularity.)
Let $D\subset\mathbb{R}^2$ be a $1$-connected open domain on which there exists a $C^3$ mapping $f:D\to\mathbb{R}^2$ whose differential $\mathrm{d}f$ has constant, distinct singular values $0<\sigma_1<\sigma_2$. Because $D$ is simply connected, one can choose an orthonormal frame field $E_1,E_2$ on $D$ such that, at each point $p\in D$, the image vectors $F_i(p) = \mathrm{d}f\bigl(E_i(p)\bigr)$ are orthogonal and satisfy $|F_i(p)|=\sigma_i$.
Let $\omega = (\omega_1,\omega_2)$ be the dual coframing on $D$, which is of regularity type $C^2$. The $1$-forms $\eta_i = \sigma_i\,\omega_i$ for $i=1,2$ have the property that $(\eta_1)^2+(\eta_2)^2$, being the $f$-pullback of the flat metric on $\mathbb{R}^2$, must also be a flat metric.
Let $\omega_{12}$ be the connection $1$-form associated to the coframing $\omega$, i.e., it satisfies the structure equations
$$
\mathrm{d}\omega_1 = -\omega_{12}\wedge\omega_2
\qquad\text{and}\qquad
\mathrm{d}\omega_2 = \omega_{12}\wedge\omega_1\,.\tag1
$$
Write $\omega_{12} = -\kappa_1\,\omega_1 + \kappa_2\,\omega_2$. The function $\kappa_i$ is the curvature of the $E_i$-integral curve. Since $\omega_{12}$ is $C^1$, so are the functions $\kappa_i$. A straightforward computation shows that the $1$-form $\eta_{12}$ that satisfies the corresponding structure equations
$$
\mathrm{d}\eta_1 = -\eta_{12}\wedge\eta_2
\qquad\text{and}\qquad
\mathrm{d}\eta_2 = \eta_{12}\wedge\eta_1\,.\tag2
$$
is given by
$$
\eta_{12} = -(\sigma_1/\sigma_2)\,\kappa_1\omega_1
+ (\sigma_2/\sigma_1)\,\kappa_2\omega_2\,.
$$
Since $\sigma_1\not=\sigma_2$, the conditions $\mathrm{d}\omega_{12} = \mathrm{d}\eta_{12}=0$ (which hold because the domain metric and the $f$-pullback of the range metric are both flat) are equivalent to
$$
0 = \mathrm{d}(\kappa_i\,\omega_i) = \bigl(\mathrm{d}\kappa_i - {\kappa_i}^2\,\omega_{3-i}\bigr)\wedge\omega_i\,\qquad i = 1,2.\tag3
$$
Proposition: If $D = \mathbb{R}^2$, then $\kappa_1 \equiv \kappa_2 \equiv 0$, and $f$ is an affine map.
Proof: Suppose that, say, $\kappa_1$ be nonzero at some point $p\in\mathbb{R}^2$ and consider the value of $\kappa_1$ along the $E_2$ integral curve through $p$, which, since $E_2$ has unit length, is necessarily complete. Let $p(s)$ be the flow of $E_2$ by time $s$ starting at $p = p(0)$. Then (3) implies that the function $\lambda(s) = \kappa_1\bigl(p(s)\bigr)$ satisfies $\lambda'(s) = \lambda(s)^2$. Consequently,
$$
\kappa_1\bigl(p(s)\bigr) = \frac{\kappa_1\bigl(p(0)\bigr)}{1-\kappa_1\bigl(p(0)\bigr)s}.
$$
Hence $\kappa_1$ cannot be continuous along this integral curve, which is a contradiction. Thus, $\kappa_1$ and, similarly, $\kappa_2$ must vanish identically when $D = \mathbb{R}^2$. In particular, $\mathrm{d}\omega_i = 0$, from which one easily concludes that $f$ is affine. QED
More interesting, locally, is what happens near a point where $\kappa_1\kappa_2\not=0$. (There is a similar analysis when one of $\kappa_i$ vanishes identically that can safely be left to the reader, but see the note at the end.) One might as well assume that $\kappa_1\kappa_2$ is nowhere vanishing on $D$. Then one can write
$$
\kappa_1\,\omega_1 = \mathrm{d}u
\qquad\text{and}\qquad
\kappa_2\,\omega_2 = \mathrm{d}v
$$
for some $C^2$ functions $u$ and $v$ on $D$, uniquely defined up to additive constants.
Writing $\omega_1 = p\,\mathrm{d}u$ and $\omega_2 = q\,\mathrm{d}v$ for some non-vanishing functions $p$ and $q$, one finds that the structure equations (1), with $\omega_{12} = -\mathrm{d}u + \mathrm{d}v$, yield the equations
$$
p_v = - q \qquad\text{and}\qquad
q_u = -p.
$$
In particular, note that $p_v$ is $C^1$ and $p_{uv}-p = 0$.
Conversely, if $p$ be any nonvanishing $C^2$ function on a domain $D'$ in the $uv$-plane that satisfies the hyperbolic equation $p_{uv}-p=0$ and is such that $p_v$ is also nonvanishing on $D'$, then the $1$-forms
$$
\omega_1 = p\,\mathrm{d}u,\quad \omega_2 = -p_v\,\mathrm{d}v,\qquad
\omega_{12} = -\mathrm{d}u+\mathrm{d}v\tag4
$$
satisfy the structure equations of a flat metric, and so do
$$
\eta_1 = \sigma_1\,p\,\mathrm{d}u,\quad \eta_2 = -\sigma_2\,p_v\,\mathrm{d}v,\qquad
\eta_{12} = -(\sigma_1/\sigma_2)\,\mathrm{d}u+(\sigma_2/\sigma_1)\,\mathrm{d}v.\tag5
$$
Indeed, one now sees that the $1$-forms
$$
\begin{aligned}
\alpha_1 &= \cos(u{-}v)\,p\,\mathrm{d}u +\sin(u{-}v)\,p_v\,\mathrm{d}v\\
\alpha_2 &= \sin(u{-}v)\,p\,\mathrm{d}u -\cos(u{-}v)\,p_v\,\mathrm{d}v
\end{aligned}
$$
are closed, and therefore can be written in the form $\alpha_i = \mathrm{d}x_i$ for some $C^3$ functions $x_i$ on $D'$.
$$
(\mathrm{d}x_1)^2 + (\mathrm{d}x_2)^2 = (\alpha_1)^2 + (\alpha_2)^2 = (\omega_1)^2 + (\omega_2)^2
$$
and, hence, they define a $C^3$ submersion $x = (x_1,x_2):D'\to\mathbb{R}^2$ that pulls back the standard flat metric on $\mathbb{R}^2$ to the metric $(\omega_1)^2 + (\omega_2)^2$ on $D'$.
Similarly, setting $\rho = \sigma_2/\sigma_1$ and
$$
\begin{aligned}
\beta_1 &= \cos(u/\rho{-}\rho v)\,\sigma_1\,p\,\mathrm{d}u
+\sin(u/\rho{-}\rho v)\,\sigma_2\,p_v\,\mathrm{d}v\\
\beta_2 &= \sin(u/\rho{-}\rho v)\,\sigma_1\,p\,\mathrm{d}u
-\cos(u/\rho{-}\rho v)\,\sigma_2\,p_v\,\mathrm{d}v,
\end{aligned}
$$
one finds that $\mathrm{d}\beta_i = 0$ and hence there exist $C^3$ functions $y_i$ on $D'$ such that $\beta_i = \mathrm{d}y_i$. Set $y = (y_1,y_2)$.
Restricting to a subdomain $D''\subset D'$ on which $x$ is $1$-to-$1$ onto its image $D = x(D'')$ yields a domain on which $x^{-1}:D\to D''$ is a $C^3$ diffeomorphism. Now set $f = y\circ x^{-1}:D\to\mathbb{R}^2$, and one has a $C^3$ solution of the original PDE system.
This completely determines the structure of the 'generic' local $C^3$ solutions.
The case when one of the $\kappa_i$, say, $\kappa_1$, vanishes identically (so that the corresponding integral curves are straight lines) and the other is nonvanishing can easily be reduced to the normal form
$$
\omega_1 = \mathrm{d}u,\qquad \omega_2 = \bigl(p(v)-u\bigr)\,\mathrm{d}v,\qquad
\omega_{12} = \mathrm{d}v\tag6
$$
where, now, $p$ is a $C^2$ function of $v$, and the rest of the analysis goes through essentially unchanged.
@AsafShachar: I'll try to find some time in the next day or so to explain more fully. Interestingly, the PDE you need to solve to construct solutions can be locally transformed into the equation $p_{uv}+p=0$ for a function $p$ on the $uv$-plane. Since this linear equation is hyperbolic, the original system must be of hyperbolic character.
Thank you very much. This is a superb analysis. I have followed carefully most of the technical details; However, I am afraid that I am missing something fundamental here: How do you know that $d\eta_{12}=0$? If I understood correctly your argument, this is so because $\eta_{12}$ is the connection one-form of some flat symmetric connection (which should be the Levi-Civita connection of the pullback metric?), w.r.t a suitable (orthonormal?) frame. How can we see that this is the case? (what is the frame? Is it the one dual to $\eta_1,\eta_2$ which is orthogonal but not orthonormal?)....
Do Cartan's first structure equations uniquely determine the connection form?... (I am trying to understand how the verification in equation $(2)$ in your answer implies that $\eta_{12}$ is the connection one-form). I am apologise if my question seems a bit unclear; I am not entirely comfortable with the application of Cartan's second equation here to $\eta_{12}$, and I feel that I don't have a clear understanding of what is the chain of reasoning here. Thank you again for this detailed and thorough analysis.
@AsafShachar: You ask whether the structure equations uniquely determine the connection form. Yes, it does. The $f$-pullback metric, ${\eta_1}^2+{\eta_2}^2 = \sigma_1^2,{\omega_1}^2+\sigma_2^2,{\omega_2}^2$ is flat, so $\mathrm{d}(\eta_{12})=0$. Since the formula I claim for $\eta_{12}$ satisfies the Cartan structure equations $\mathrm{d}\eta_1 = -\eta_{12}\wedge\eta_2$ and $\mathrm{d}\eta_2 = \eta_{12}\wedge\eta_1$, it follows that $\eta_{12}$ must, indeed, be the Levi-Civita connection form for the pullback metric relative to the (orthonormal) $\eta$-coframing.
OK, thank you. I think I got it. Just to be sure-this "unique determination" of the connection form from the first structure equation is special to dimension $2$, right? The equations determine the "wedge of $\eta_{12}$ against the basis elements $\eta_1,\eta_2$", hence determine it completely. Now, you can just use the unique $\eta_{12}$ satisfying the equations (which, in this case, is given by the formula you have prescribed) to define a connection in terms of the $\eta$-orthonormal frame dual to $\eta_1,\eta_2$. The resulting connection would be metric w.r.t $\eta$, and torsion-free...
exactly by the same structure equation. Thus, it would be the Levi-Civita connection of the metric ${\eta_1}^2+{\eta_2}^2$. At this stage, you finally use the crucial fact that ${\eta_1}^2+{\eta_2}^2$ is a pullback metric of a flat metric, hence it is also flat. Does my view fit your reasoning? (in particular, it seems to me that while the unique determination of the connection form by the structure equation is unique to dimension $2$, we do not use this fact directly...
In general, we could use any anti-symmetric matrix of one-forms $\eta_{ij}$ which satisfies the first structure equations without torsion, to define the Levi-Civita connection of the corresponding metric. So, it suffices to know that your formula for $\eta_{12}$ satisfies the equation, we do not actually use the uniqueness in any essential way. (If I am not mistaken). Of course, this uniqueness is aesthetically appealing anyway, and shows that your formula was "forced" upon you, and not the result of some funny choice I guess...
I would like to propose a simple local example:
Consider the map in polar coordinates, $\mathbb C\to \mathbb C$ that takes a complex number $z=e^{2\pi i \theta}r$ to $e^{(\sigma_1/\sigma_2)\cdot 2\pi i \theta}r\sigma_2$.
(apologies for the previous wrong example, I confused in it singular values with eigenvalues...)
The $df$ of your map does not have the form of a product of rotation matrix, diagonal 1,2, and rotation matrix, as required in the problem.
I see... I misread, and confused singular values with eigenvalues....
To harmonize the the notation in the original question, maybe in the exponent you want $\sigma_1 / \sigma_2$ instead of just $\sigma_1$?
Thank you! You are right :)
For appropriate $\sigma_1,\sigma_2$ I guess this gives a smooth example with domain $\mathbb R^2$ minus a point.
@WillSawin Don't you need to remove a complete ray for this solution to be well defined (the "$0=2\pi$" problem).
Asaf, the map is well defined on the complement to $0$ if the ratio $\sigma_1/\sigma_2$ is integer. So one needs to rename $\sigma_1$ as $\sigma_2 $ and $\sigma_2$ as $\sigma_1$ in order to have such maps defined on $\mathbb C\setminus 0$ for which $\sigma_1<\sigma_2$
It's interesting to note that Dimitri's solution corresponds to the special case $p(v) = 0$ in the special case I list at the end of my analysis that covers when one of the two eigencurve foliations is a foliation by straight lines.
|
2025-03-21T14:48:29.743678
| 2020-01-30T15:52:19 |
351549
|
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|
Stack Exchange
|
Rough paths theory for Non-Markovian processes
I would like to know whether there is a suitable extension of the theory of rough paths that could be useful to solve Non-Markovian problems.
I would appreciate any example or also any other theory (not necessarily rough paths) that could give a formal framework to deal with non-Markovianity.
My main motivation is to solve 1D or 2D stochastic differential equations driven by non-Markovian processes, arising in physics (e.g., non-equilibrium thermodynamics).
Thanks!
It really depends on what sort of non-Markovian equations you have in mind, but it does certainly allow you to give solution theories for SDEs driven by fractional Brownian motion with Hurst parameter $H>1/4$.
Thanks for your response. I was thinking in a process driven by a non-Gaussian noise following the Tsallis q-Gaussian distribution. Approaches are generally made to transform the noise into a Markovian one, and I wanted to know if it is possible to use the theory of rough paths to be able to preserve the information regarding the non-Markovian part of the process.
|
2025-03-21T14:48:29.743790
| 2020-01-30T16:04:20 |
351552
|
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}
|
Stack Exchange
|
Convergence of Quasi-Newton method with fixed derivative
Consider the Newton iteration
$x^{(k+1)} = x^{(k)} - DF( x^{(k)} )^{-1} \cdot F( x^{(k)} )$
to find a zero of a function $F : \mathbb R^k \rightarrow \mathbb R^k$. If we freeze the first derivative, we get the quasi-Newton method
$x^{(k+1)} = x^{(k)} - DF( x^{(1)} )^{-1} \cdot F( x^{(k)} )$
This way, we do not need to invert a matrix at every step, at the cost of possibly worse convergence.
Question: What is known about the convergence of the second iteration? This trick seems to be quite common in practical computations, but I have had a hard time finding any discussion of its convergence. Even a non-optimal estimate would interest me.
You should be able to get a feel for it by "looking at the standard picture" and tweaking it. For example, try to find the root of y=exp(x) starting at x=0. (Good luck!) You take much smaller steps toward a "zero" with the quasi version. You can imagine a plethora of failures of the quasi method this way. For many practical cases, I imagine the behaviour will be similar to the above failure. Gerhard "Maybe Evaluate The Derivative Periodically?" Paseman, 2020.01.30.
That's how the inverse function theorem is proved in most textbooks. It is just the Picard iteration for a contractive mapping given by the RHS which converges only as a geometric progression, so if you need really high precision and really fast, the trick is a bad idea unless you take a few standard Newton iterations first to get the contracting coefficient very close to $0$. Not sure if there is anything else to say about it.
I have restated the question to illustrate the motivation in a better way.
|
2025-03-21T14:48:29.744041
| 2020-01-30T17:31:41 |
351555
|
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|
Stack Exchange
|
Compactifications with remainder $[0,\omega_1]$ and convergent sequences
Is the following statement consistent?
$(\star)$ Let $K$ be a separable compact Hausdorff space containing the space $[0,\omega_1]$ so that the complement $K\setminus[0,\omega_1]$ is discrete. Then there exists an open neighborhood $U$ of $[0,\omega_1)$ in $K$ such that $U$ contains no sequences convergent to $\omega_1\in[0,\omega_1]$.
Remark 1. The statement $(\star)$ does not hold under $\omega_1<\mathfrak p$. That is why I am asking only about the consistency of $(\star)$.
Remark 2. If $(\star)$ is consistent, then Question 1 in this MO-post has negative answer.
Here's an example, suggested by Alan Dow. Take a Hausdorff Gap: a pair of sequences $\langle a_\alpha:\alpha<\omega_1\rangle$ and $\langle b_\alpha:\alpha<\omega_1\rangle$ of infinite subsets of $\mathbb{N}$ such that $a_\alpha\subset^*a_\beta$ and $b_\alpha\subset^*b_\beta$ whenever $\alpha<\beta$ and $a_\alpha\cap b_\alpha=^*\emptyset$ for all $\alpha$, and with the property that whenever $A$ is such that $a_\alpha\subseteq^*A$ for all $\alpha$ there is an $\alpha$ such that $A\cap b_\alpha$ is infinite.
Topologise $K=\mathbb{N}\cup[0,\omega_1]$ by the subbase consisting of $\{n\}$ and $K\setminus\{n\}$ for all $n\in\mathbb{N}$, as well of the sets $a_\alpha\cup[0,\alpha]$ and $(\mathbb{N}\setminus a_\alpha)\cup(\alpha,\omega_1]$ for all $\alpha$.
This yields a compactification of $\mathbb{N}$ with $[0,\omega_1]$ as its remainder in which every set $b_\alpha$ converges to $\omega_1$.
If $U$ is open in $K$ and $U\cap[0,\omega_1]=[0,\omega_1)$ it then follows that $a_\alpha\subseteq^*U\cap\mathbb{N}$ for all $\alpha$. But then for some $\alpha$ the set $U\cap b_\alpha$ is infinite and it converges to $\omega_1$.
|
2025-03-21T14:48:29.744174
| 2020-01-30T17:50:32 |
351557
|
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|
Stack Exchange
|
factorization morphism between projectives spaces
Please help me with this doubt:
Let $f:\mathbb{P}^1 \rightarrow \mathbb{P}^2$ be a non-constant morphism. Is there any factorization of $f$ as $$\mathbb{P}^{1} \overset{h}{\rightarrow}\mathbb{P}^{2}\overset{g}{\rightarrow}\mathbb{P}^{2}$$ where $h$ is an embedding and $g$ is the finite type such that $c_{1}(f^{*}T_{\mathbb{P}^2})=c_{1}(h^{*}T_{\mathbb{P}^2})$ or maybe $f^{*}T_{\mathbb{P}^2}=h^{*}T_{\mathbb{P}^2}$..($c_1$ is the first Chern class).
Or is there a general result for morphisms $f:\mathbb{P}^1 \rightarrow X$ where $X$ is a projective variety. I am looking for conditions, for example if $X$ is smooth along the image ... I don't know. Please help me, thanks.
I guess that P1 and P2 are the COMPLEX projective spaces? Please specify which kind of "morphisms" and "embeddings". Isn't that a question in algebraic geometry rather than differential topology?
In general, no such factorization exists. Consider a morphism given by three general homogeneous polynomials of degree three.Then, the image is of degree 3 and $f$ is not an embedding. For any factorization as above, one sees that $\deg g$ divides 3, but $\deg g$ is a square and thus $g$ must be an isomorphism and thus so $f$ is an embedding, a contradiction.
|
2025-03-21T14:48:29.744274
| 2020-01-30T17:51:38 |
351558
|
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|
Stack Exchange
|
Real rootedness of a polynomial
Let's consider $m$ and $n$ arbitrary positive integers, with $m\leq n$, and the polynomial given by:
$$ P_{m,n}(t) := \sum_{j=0}^m \binom{m}{j}\binom{n}{j} t^j$$
I've found with Sage that for every $1\leq m \leq n \leq 80$ this polynomial has the property that all of its roots are real (negative, of course).
It seems these roots are not nice at all. For example for $m=3$ and $n=10$, one has $$P(t) = 120t^3 + 135 t^2 + 30t+1$$ and the roots are:
$$ t_1 = -0.8387989...$$
$$ t_2 = -0.2457792...$$
$$ t_3 = -0.0404217...$$
Is it true that all roots of $P_{m,n}(t)$ are real?
Your observation follows trivially from the properties of zeroes of Jacobi polynomials.
@Nemo, That doesn't seem completely obvious enough to say it is trivial. Can you maybe expand on your reasoning?
Though the question already has answers, it might still be helpful to know that there is a handy criterion for the real-rootedness of a polynomial by checking the non-negativity of quadratic form that can be built from its coefficients.
If you have two polynomials $f(x)=a_0+a_1x+\cdots a_mx^m$ and $g(x)=b_0+b_1x+\cdots+b_nx^n$, such that the roots of $f$ are all real, and the roots of $g$ are all real and of the same sign, then the Hadamard product
$$f\circ g(x)=a_0b_0+a_1b_1x+a_2b_2x^2+\cdots$$
has all roots real. This was proven originally in
E. Malo, Note sur les équations algébriques dont toutes les racines
sont réelles, Journal de Mathématiques Spéciales, (4), vol. 4 (1895)
One can make stronger statements, such as the result by Schur that says that $\sum k!a_kb_k x^k$ will only have real roots, under the same conditions. Schur's theorem combined with the fact that $\{1/k!\}_{k\geq 0}$ is a Polya frequency sequence, implies Malo's theorem.
I'm not sure what the best reference to learn the theory of real rooted polynomials, and the associated operations that preserve real rootedness is. One textbook I know that discusses some of these classical results is Marden's "Geometry of Polynomials".
According to the representation for Jacobi polynomials https://en.wikipedia.org/wiki/Jacobi_polynomials#Alternate_expression_for_real_argument
$$
P^{(0,n-m)}_m(x)=\sum_{j=0}^m \binom{m}{j}\binom{n}{j}\left(\frac{x-1}{2}\right)^{j}\left(\frac{x+1}{2}\right)^{m-j}
$$
OPs polynomial equals
$$
P_{m,n}(t)=(1-t)^mP^{(0,n-m)}_m\left(\frac{1+t}{1-t}\right).
$$
Since zeroes of Jacobi polynomials are real valued, all roots of the polynomial $P_{m,n}(t)$ are also real valued (see D.Dominici, S.J.Johnston, K.Jordaan, Real zeros of
hypergeometric polynomials).
|
2025-03-21T14:48:29.744462
| 2020-01-30T17:57:00 |
351559
|
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|
Stack Exchange
|
Very basic question about category theory
I have a symmetric monoidal category, and go to another one by replacing objects with isomorphism classes of objects. What's that called in category theory language?
To give an example, consider the symmetric monoidal category of finite-dimensional vector spaces (with tensor product). The isomorphism class of a vector space is given by it's dimension. So the objects of the new category would be given by non-negative integers. The morphisms between two numbers $a$ and $b$ would be given by linear maps between two canonical representatives $\mathbb{C}^a$ and $\mathbb{C}^b$.
Or, consider the category of finite sets (with functions and cartesian product). Again, the objects of the new category would be positive integers corresponding to the cardinality of the sets. The morphisms would be functions between canonical representatives $\{0,\ldots,a-1\}$ and $\{0,\ldots b-1\}$.
Does that operation make sense in category theory? And what's it called? I heard the term "strictification" and that sounds a bit like what the above operation is doing, but I didn't find any understandable explanation or examples for it.
This is called the skeleton. It is not the same as the strictification.
https://en.wikipedia.org/wiki/Skeleton_(category_theory)
(Note that this notion of "skeleton" is completely independent of the notion of $n$-skeleton for simplicial sets.)
|
2025-03-21T14:48:29.744573
| 2020-01-30T19:32:32 |
351565
|
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|
Stack Exchange
|
Functions that are almost (left-) continuous almost everywhere
Denote the Lebesgue measure on $[0, T]$ as $\lambda(\cdot)$. Call a measurable function $f : [0, T] \to \mathbb{R}$ almost left-continuous almost everywhere if there exists an $A \subseteq [0, T]$ with $\lambda(A) = 0$ such that for all $t \in A^c$, for all $\epsilon > 0$ there exists a $\delta > 0$, and $N \subseteq [0, T]$ with $\lambda(N) = 0$ such that for all $s \in (t - \delta, t] \cap N^c$, $|f(t) - f(s)| < \epsilon$.
Is there a well-known class corresponding to these functions? Is it the case that all functions satisfy this property? This class is certainly not the a.e. left-continuous functions. I am not exactly sure if this is the class of functions that are equal to a left-continuous function a.e.
|
2025-03-21T14:48:29.744651
| 2020-01-30T22:34:56 |
351575
|
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|
Stack Exchange
|
What does second order set theory give us that is new?
There is a natural analogy between the theories PA and ZFC. See the linked question by Gro-Tsen here.
Peano arithmetic (PA) is a first order approximation to the natural numbers. As is well known, there are true facts about the natural numbers that we cannot prove in PA, but we can prove them in stronger systems that seem to be entailed by philosophical assumptions about the natural numbers.
Informally, my question is: Does this aspect of the analogy extend to ZFC?
In a trivial way, the answer to my question is yes. Since we (supposedly) believe the axioms of ZFC, we accept the statement Con(ZFC) that asserts ZFC is consistent, and also Con(ZFC+Con(ZFC)), and so forth. My question then is whether or not there are non-trivial examples of higher order considerations giving us theorems about sets that are not provable in ZFC, just as the Paris-Harrington theorem is true but not provable in PA.
To attempt motivating an example, consider the following: The "axiom of union" in ZFC is a first order approximation of the philosophical assertion that putting collections together results in a new collection. (The classic paradoxes show that we have to be careful what we mean by a collection, but let's ignore that issue for now.) In particular, if we have some collection $S$ of subsets of the natural numbers, then we have no problem asserting the existence of $\bigcup S$ as a subset of $\mathbb{N}$.
If we have a first order, recursively enumerable theory $T$, extending ZFC, we can define a truth predicate $P_n$ for the formulas cut off at level $\Sigma_n$. Philosophically, we thus expect the existence of a full truth predicate---not one definable in any first order way in $T$ (that would be impossible by Tarski's theorem), but one that "exists" nonetheless. Can we leverage this fact, or other facts like it, to make philosophical advances in understanding "true" set theory?
I suspect the truth predicate is outside the arithmetical hierarchy as well. However, we need an expert or someone with more than a cell phone to research this properly. Definability of truth might be a good search phrase. Gerhard "Not Ready For Big Screen" Paseman, 2020.01.30.
Do the usual arguments pro- or anti-large-cardinals fit into this? Or are you looking only for "unobjectionable" claims?
@NoahSchweber If you have an argument that by considering the usual axioms of ZFC from a second order perspective, large cardinals are affected, that would be very interesting.
Reflection principles ("for any property of the universe of all sets we can find a set with the same property") are second order, and often seen as "entailed by philosophical assumptions" about sets. Cantor asserts something to this effect, but one has to be careful with "properties" to avoid inconsistency. You may like Maddy, Believing the Axioms.
@Conifold Thanks for those two links. I did like Maddy's paper! My question may be viewed somewhat orthogonal to that paper though. Regardless of whether or not the axioms we have are an accident of history, and ignoring whatever motivating factors we have for accepting them, if we expect them to mirror operations that hold in some "true" set theory then what else can we say about that universe? (i.e., a universe with second order union, second order separation, etc...)
Does this mean something provable in second order ZFC? See also Williams, The Structure of Models of Second-order Set Theories.
|
2025-03-21T14:48:29.745034
| 2020-01-31T00:21:34 |
351581
|
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|
Stack Exchange
|
References for topological quantum field theory
I have been studying TQFTs and I am mainly reading
Lurie's proof of the cobordism hypothesis: https://www.math.ias.edu/~lurie/papers/cobordism.pdf
Walker's notes: https://canyon23.net/math/tc.pdf
However, neither of them are completely rigorous. As Lurie mentions on his homepage, his paper is an exposition. Has this proof been completely rigorously done anywhere? Also, I was very much enjoying Walker's notes, but they aren't complete. Are there any references that introduce TQFTs and explain how the definition of Extended TQFTs as $\left(\infty,n\right)$-functors comes up?
"Has this proof been completely rigourously done anywhere?": No, definitely not. Calaque and Scheimbauer in their paper do give a rigorous definition of the symmetric monoidal (∞,n)-category of bordisms. But there is nothing beyond.
Ayala-Francis write the proof assuming a result on factorization homology: https://arxiv.org/abs/1705.02240
@ThiKu, I found it quite hard to read the Ayala-Francis paper. What do I need to read before it? I.e., What are the prereqs?
I have no idea. I just wanted to hint you to that paper, which seems to be the state of the art.
|
2025-03-21T14:48:29.745152
| 2020-01-31T01:00:00 |
351583
|
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|
Stack Exchange
|
A parsimonious large cardinal axiom
The ordering of large cardinals by consistency strength is well known.
I was wondering what one can say regarding an ordering by direct implication.
In particular, I am looking for is a parsimonious axiom of the form "There exists an XYZ cardinal" which 'picks up' as many other large cardinals as possible by directly implying that they exist.
So the best I can think of is:
Extendible cardinal $\implies$ Supercompact cardinal $\implies$ Measurable cardinal $\implies$ Loads of other large cardinals (There are also many other in between these categories that are directly implied, I believe)
But is it possible to go further ?
Honestly, I don't understand what you are asking. A table with all known large cardinals and for any two of them a decision on whether or not the existence of one kind of cardinal implies the existence of the other?
I0 cardinals?
|
2025-03-21T14:48:29.745239
| 2020-01-31T01:23:46 |
351584
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351584"
}
|
Stack Exchange
|
Theory of integration for functions from $\mathbb{Z}_{p}$ to $\mathbb{Z}_{q}$ for distinct primes $p,q$
Let $p$ and $q$ be prime numbers. When $p=q$, Mahler's Theorem gives a complete description of $C\left(\mathbb{Z}_{p};\mathbb{Z}_{p}\right)$, the space of continuous functions from $\mathbb{Z}_{p}$ to $\mathbb{Z}_{p}$. I'm wondering (possibly in vain) if there might be a comparable classification of $C\left(\mathbb{Z}_{p};\mathbb{Z}_{q}\right)$ when $p$ and $q$ are distinct.
I ask only because I've been doing $p$-adic harmonic analysis, but have found myself having to brave the wilds of $L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}_{q}\right)$, the space of all $f:\mathbb{Z}_{p}\rightarrow\mathbb{C}_{q}$ so that:$$\sup_{\mathfrak{z}\in\mathbb{Z}_{p}}\left|f\left(\mathfrak{z}\right)\right|_{q}<\infty$$
Pontryagin duality lets me do Fourier analysis on $L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}\right)$; for $p=q$, on the other hand, I can use things like the volkenborn integral, or the amice transform / mazur-mellin transform—$p$-adic distributions, in general. The problem is, without a structure theorem like Mahler's for the $p\neq q$ case, though I can define “integration” on $L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}_{q}\right)$ by elements of its dual space (continuous functionals $\varphi:L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}_{q}\right)\rightarrow\mathbb{C}_{q})$, I don't see a way to do useful computations for the specific, non-abstract functions that I'm trying to fourier analyze.
So, I guess what I'm really asking is: how do you take the "integral" or "fourier transform" of such a function?
Any thoughts? Reference recommendations? Etc.?
${\mathbb Z}_p$ and ${\mathbb Z}_q$ are homeomorphic; hence so are $C({\mathbb Z}_p,{\mathbb Z}_p)$ and $C({\mathbb Z}_p,{\mathbb Z}_q)$.
Okay then... what is the formula for the homeomorphism? Because it sure ain't the identity map. xD
@MCS This answer may be useful.
Every compact, totally disconnected metric space without isolated points is homeomorphic to the Cantor set.
|
2025-03-21T14:48:29.745393
| 2020-01-31T03:48:30 |
351587
|
{
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"authors": [
"Maksym Voznyy",
"MyNinthAccount",
"https://mathoverflow.net/users/140496",
"https://mathoverflow.net/users/95511"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351587"
}
|
Stack Exchange
|
Resolved: Two more generators needed for a Z/6 elliptic curve
We are searching for the rank 8 elliptic curves with the torsion subgroup Z/6 using newly discovered families similar to Kihara's (Kihara's family is described in https://arxiv.org/pdf/1503.03667.pdf).
Today we came across a curve
$[0,8169768624655967629114128598,0,-451787550647310420612086468536366715869054405951830599,0]$
Both Magma Calculator (http://magma.maths.usyd.edu.au/calc/) and mwrank with $-b12$ return 6 generators for this curve.
Magma V2.20-10 (STUDENT) runs out of memory running the following code:
SetClassGroupBounds("GRH");
E := EllipticCurve([0,8169768624655967629114128598,0,-451787550647310420612086468536366715869054405951830599,0]);
MordellWeilShaInformation(E);
Sagemath returns $8$ for the upper bound of analytic rank, even for max_Delta=$3.3$ (we are still testing for higher max_Delta):
E = EllipticCurve([0,8169768624655967629114128598,0,-451787550647310420612086468536366715869054405951830599,0])
E.analytic_rank_upper_bound(max_Delta=3.3,root_number="compute")
Is there a way to find two more generators?
A similar question for the $6$ <= Rank(E) <= $7$ situation was successfully resolved by Jeremy Rouse (One more generator needed for a Z/6 elliptic curve) but our software chokes when we are trying to follow his instructions.
Max
Doesn't Magma's TwoPowerIsogenyDescentRankBound (Fisher's code) prove at step 5 (just beyond 4-descent, but not yet 8-descent) the rank is at most 6 in about 20 seconds and 120 MB of memory?
Thank you, MyNinthAccount! Resolved.
Just to give a more complete answer:
SetVerbose("cbrank",1);
E := EllipticCurve([0,8169768624655967629114128598,0,\
-451787550647310420612086468536366715869054405951830599,0]);
TwoPowerIsogenyDescentRankBound(E);
/---------------------------------------------------\
| SUMMARY TABLE Step No : 6 5 4 3 2 1 |
|---------------------------------------------------|
| dim_2 ( E'(Q)/phi E(Q) ) <= 4 5 5 5 5 |
| dim_2 ( E(Q)/phi'E'(Q) ) <= 4 5 5 5 5 |
| Therefore rank E(Q) <= 6 |
\---------------------------------------------------/
Total time: 22.980 seconds, Total memory usage: 96.16MB
|
2025-03-21T14:48:29.745539
| 2020-01-31T05:15:37 |
351590
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Martin Brandenburg",
"Neil Strickland",
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"https://mathoverflow.net/users/2841",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351590"
}
|
Stack Exchange
|
Classification of the functors on the category of cyclic groups
Let $\mathsf{Grp}$ be the category of groups and let $\mathsf{Cyc}$ be the subcategory of cyclic groups.
As seen in the posts here and there (and their answers), a functor $F: \mathsf{Cyc} \to \mathsf{Cyc}$ is a very structured/restrictive notion, we are then lead to wonder whether there exists such a functor which is non-equivalent to the identity or the trivial functor, or if there is a such functor with $F(C_1) \not \simeq C_1$. As pointed out by Martin Brandenburg and Jeremy Rickard, $C_1$ is a retract of $F(C_1)$, so that $F(C_1)$ must be a retract of $F^2(C_1)$, and more generally, $F^n(C_1)$ is a retract of $F^{n+1}(C_1)$, which means that $F^{n+1}(C_1)$ is isomorphic to a semidirect product $F^n(C_1) \ltimes N_n$; now $F^{n+1}(C_1)$ is a cyclic group, so the semidirect product is in fact a direct product and moreover $gcd(|F^n(C_1)|,|N_n|) = 1$.
Question: What are the functors on the categroy of cyclic groups?
Remark: $Aut(-)$ is not such a functor because $Aut(C_8) \simeq C_2 \times C_2$ (and $Aut^2(C_8) \simeq S_3$).
In his answer, Neil Strickland provides examples of functors $F$ with $F(C_1) \not \simeq C_1$ and with $F^2(C_1) \not \simeq F(C_1)$, but with $F^3(C_1) \simeq F^2(C_1)$.
Bonus question: Is there a functor $F: \mathsf{Cyc} \to \mathsf{Cyc}$ such that $F^{n+1}(C_1) \not \simeq F^n(C_1)$ for all $n$?
Remark: If so, the sequence $(F^n(C_1))_n$ cannot be periodic (for $n$ large enough), because then (as shown above) $F^{n+1}(C_1) \simeq F^{n}(C_1) \times N_n$ with $|N_n|>1$ for all $n$.
I think that there are too many functors to classify them.
The first step is to describe the morphisms in $\mathsf{Cyc}$ by generators and relations and thereby to give a description of functors into any category. But it will be quite complicated and not easy to simplify even for simple target categories.
Another remark: $\mathsf{Cyc}$ has at least two additional structures, for example it is $\mathbb{Z}$-linear (aka preadditive) and symmetric monoidal. It is much easier to classify the functors which preserves one of both of these structures.
@MartinBrandenburg Is there a non-constant functor $F$ with $F(C_1) \neq C_1$?
@MartinBrandenburg: Which bifunctor provides a symmetric monoidal structure?
The first question is answered by Neil below, the second: the tensor product of abelian groups. We have $C_n \otimes C_m = C_{\mathrm{gcd}(n,m)}$ for $n,m \geq 0$.
I'll use additive notation, and I'll assume that you are only considering finite cyclic groups. Let $\mathbf{Cyc}_p$ be the category of cyclic $p$-groups. Given $i,j\geq 0$ we can define $Q(p;i,j)\colon\mathbf{Cyc}\to\mathbf{Cyc}_p$ by $Q(p;i,j)(A)=\{a\in p^iA\colon p^ja=0\}$. We can also define a constant functor $C(p;i)\colon\mathbf{Cyc}\to\mathbf{Cyc}_p$ by $C(p;i)(A)=\mathbb{Z}/p^i$. Now suppose we have a collection of functors $F_p$, one for each prime $p$, each of the form $Q(p;i,j)$ or $C(p;i)$, and that only finitely many of the functors $F_p$ are constant. Then the group $F(A)=\prod_pF_p(A)$ is cyclic for all $A$, so we get a functor $F\colon\mathbf{Cyc}\to\mathbf{Cyc}$. I don't know if that gives all possible functors, but it certainly gives a reasonably rich supply of them.
As a very specific example, the functor $F(A)=(A/2A)\times(\mathbb{Z}/3)$ is non-constant with $F(0)\neq 0$.
UPDATE: Here's a more exotic example. If $X$ is a based set of size $1$ or $3$, there is a unique group structure for which the basepoint is the identity. If $A$ is cyclic of order $1$ or $7$ then we can impose an equivalence relation with $a\sim a^2\sim a^4$ for all $a$, and then $A/\sim$ has size $1$ or $3$ with basepoint $0$ and so has a group structure. This construction gives a functor on the category of groups of order $1$ or $7$, and we can compose with $A\mapsto A/7$ to get a functor $\mathbf{Cyc}\to\mathbf{Cyc}_3$. I am sure that there are many variations on this theme.
From the other questions I assume that $C_0 = \mathbb{Z}$ belongs to $\mathsf{Cyc}$.
Is there a functor $F$ with $F^2(C_1) \neq F(C_1)$?
If $G$ is the functor in my update, and $FA=\mathbb{Z}/7\times GA$ then $F0=\mathbb{Z}/7$ and $F^20=\mathbb{Z}/21$.
|
2025-03-21T14:48:29.745914
| 2020-01-31T05:18:00 |
351591
|
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"fedja",
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"url": "https://mathoverflow.net/questions/351591"
}
|
Stack Exchange
|
Subspace proximity
Given two rank-$k$ projection matrices $\Pi_1, \Pi_2 \in \mathbb{R}^{n \times n}$ (i.e. $\Pi_1 = \Pi_1^2$ and $\Pi_2 = \Pi_2^2$), what is the best function $c_{n,r}$ such that $\|\Pi_1 - \Pi_2 \|^2_F \leq c \| \Pi_1 (\Pi_1 - \Pi_2) \Pi_1 \|^2_F$?
When $\Pi_1$ and $\Pi_2$ span orthogonal subspaces, it is easy to see that $c =2$. For random projector matrices, $c_{n,r}\leq2$ still seems to hold computationally. For small perturbations of random projectors, $c_{n,r}$ appears to be much larger than $n$.
There seems to be a misprint somewhere: if you take projections to two very close lines in $\mathbb R^2$, the constant blows up as written, doesn't it?
|
2025-03-21T14:48:29.746000
| 2020-01-31T06:18:30 |
351593
|
{
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"authors": [
"Brauer Suzuki",
"Derek Holt",
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"https://mathoverflow.net/users/332108",
"https://mathoverflow.net/users/34538",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351593"
}
|
Stack Exchange
|
On the iterated automorphism groups of the cyclic groups
Let $C_n$ be the cyclic group of order $n$. Its automorphism group $Aut(C_n)$ is a group of order $\varphi(n)$ isomorphic to $(\mathbb{Z}/n\mathbb{Z})^{\times}$ the multiplicative group of integer modulo $n$. This last group is abelian but not always cyclic, the first non-cyclic being $Aut(C_8) \simeq C_2 \times C_2$. Moreover the iterated automorphism groups $Aut^m(C_n)$ are not always abelian, as $Aut^2(C_8) \simeq S_3$.
By watching the table here for the group structure of $(\mathbb{Z}/n\mathbb{Z})^{\times}$, we cannot expect an easy classification for the set of groups $Aut(C_n)$, but (Q1) what about the following set? $$\{Aut^m(C_n) \ | \ n \ge 1, m \ge 0 \}$$
If it is also non-easy, (Q2) what about the set of groups $Aut^m(C_n)$ which are isomorphic to $Aut^{m+1}(C_n)$? For $n \le 15$, it is exactly $\{C_1,S_3,D_8 \}$ as shown by the following table:
$$\scriptsize{ \begin{array}{c|c}
G &C_1&C_2&C_3&C_4&C_5&C_6&C_7&C_8&C_9&C_{10}&C_{11}&C_{12}&C_{13}&C_{14}&C_{15} \newline
\hline
Aut(G) &-&C_1&C_2&C_2&C_4&C_2&C_6&C_2^2&C_6&C_4&C_{10}&C_2^2&C_{12}&C_6&C_2 \times C_4 \newline
\hline
Aut^2(G) &-&-&C_1&C_1&C_2&C_1&C_2&S_3&C_2&C_2&C_4&S_3&C_2^2&C_2&D_8 \newline
\hline
Aut^3(G) &-&-&-&-&C_1&-&C_1&-&C_1&C_1&C_2&-&S_3&C_1&- \newline
\hline
Aut^4(G) &- &- &-&-&-&-&-&-&-&-&C_1&-&-&-&-
\end{array} }$$
Let $\alpha(n)$ be the smallest $m \ge 0$ such that $Aut^m(C_n) \simeq Aut^{m+1}(C_n)$. Then:
$$\scriptsize{ \begin{array}{c|c}
n &1&2&3&4&5&6&7&8&9&10&11&12&13&14&15 \newline
\hline
\alpha(n) &0&1&2&2&3&2&3&2&3&3&4&2&2&3&2
\end{array} }$$
Surprisingly, for $n<15$ (but not for $n=15$), $\alpha(n)$ is exactly the number of iterations of the Carmichael lambda function $\lambda$ needed to reach $1$ from $n$ (OEIS A185816). (Q3) Why?
A specific OEIS sequence has just been created (A331921) providing the value of $\alpha(n)$ for $n<32$; in this case, $\alpha(n) \le 5$ and $\{\mathrm{Aut}^5(C_n) \ | \ n < 32 \} = \{C_1,S_3,D_8,D_{12},\mathrm{PGL}(2,7) \}$. A usual laptop cannot compute $\alpha(n)$ for $n \ge 32$ (you are welcome to contribute to this sequence), we just know that $\alpha(32) \ge 8$. Here is the sequence $|Aut^n(C_{32})|$ for $n \le 8$:
$$\scriptsize{ \begin{array}{c|c}
n &0&1&2&3&4&5&6&7&8 \newline
\hline
|Aut^n(C_{32})| & 2^5&2^4&2^4&2^6&2^{7}3&2^{9}3&2^{11}3&2^{16}3&2^{41}3^3 5^1 7 \newline
\hline
\text{IdGroup}(Aut^n(C_{32})) & [32,1]&[16,5]&[16,11]&[64,138]&[384,17948]&[1536,?]&[6144,?]&?&?
\end{array} }$$
We can also consider the sequence of $n$ such that $\exists m \ge 0$ with $Aut^m(C_n) \simeq C_1$:
$1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 14, 18, 19, 22, 23, 27, 38, 46, 47, 54, 81, \dots$ (A117729).
It seems to be a (strict) subsequence of A179401, so (Q4) why $Aut^m(C_n) \simeq C_1$ $\Rightarrow$ $\varphi^2(n) = \lambda^2(n)$?
It is an open problem whether the sequence $Aut^n(G)$ stabilizes (see here). For a given finite group $G$ there are three possibilities:
(1): $\exists m \ge 0 $ such that $Aut^{m+1}(G) \simeq Aut^{m}(G)$,
(2): not (1) and $\exists m \ge 0$ such that $\exists r>0$ with $Aut^{m+r}(G) \simeq Aut^{m}(G)$,
(3): not (1) and not (2), i.e., $\forall m \ge 0$ and $\forall r>0$ then $Aut^{m+r}(G) \not \simeq Aut^{m}(G)$.
The case (1) means that the sequence $(Aut^m(G))_m$ is constant for $m$ large enough, (2) means that it is periodic non-constant for $m$ large enough, and (3) means never periodic. The existence of finite groups in cases (2) or (3) is open. (Q5) Can we rule out the cases (2) and (3) for the cyclic groups?
If not, let redefine $\alpha(n)$ as the smallest $m \ge 0$ such that $\exists r>0$ with $Aut^{m}(C_n) \simeq Aut^{m+r}(C_n)$ if $C_n$ is in cases (1) or (2), and $\infty$ in the case (3).
In your table for $C_{32}$ you can add that the last two automorphism groups have orders $2^{16}\cdot 3$ and $2^{41}\cdot 3^3\cdot 5\cdot 7$.
@BrauerSuzuki: Nice, how did you compute them? Did you identify the group structures? Can you go beyond?
It is crucial to convert the automorphism group to a more efficient structure like PcGroup (for solvable groups) or PermutationGroup, at each iteration (can be done in GAP or MAGMA). Beyond that, I run out of memory.
It seems to me that the structure of ${\rm Aut}^{m}(C_{n})$ also depends heavily on the prime factorization of $n$, and I don't really see any reason to expect the answer to Q1 to be any more tractable than determining the structure of ${\rm Aut}(C_{n})$.
For example (just to illustrate) , if we choose a prime $p$, greater than $3$, and then we take a pair of primes $q_{1}$ and $q_{2}$ so that each $q_{i}-1$ is divisible by $p$, but not by $p^{2}$, and we set $n = q_{1}q_{2}$, then ${\rm Aut}^{2}(C_{n})$ has a direct factor isomorphic to
${\rm GL}(2,p)$, and is not solvable. Furthermore the involvement of ${\rm PSL}(2,p)$ in ${\rm Aut}^{m}(C_{n})$ persists for $m >2$.
I just tried some computations with $n=11 \times 31$, and the orders of the series of automorphism groups seem to be increasing rapidly, starting $300$, $5760$, $46080$, $566231040$.
Q1 asks for a classification of the set ${Aut^m(C_n) \ | \ n>0,m \ge 0 }$ (up to equivalence). This set could be easy to describe "negatively", in the sense that it could be all the finite groups except a tractable list.
@DerekHolt Is it known whether, for a given finite group $G$, $Aut^n(G)$ is constant for $n$ large enough?
@GeoffRobinson: and what about the existence of a finite group $G$ for which ${\rm Aut}^{n}(G)$ does not stabilize? Any known cyclic group example? Derek's example $C_{341}$ is a candidate, but it is known? Is there a finite group $G$ whose sequence ${\rm Aut}^{n}(G)$ is (for $n$ large enough) periodic non-constant?
There are several related posts on this issue, such as this one and this one. But I am not aware of any known example in which the orders of the groups in the chain of automorphism groups of a finite group have been proved to be unbounded.
Do you think we can translate Q5 into a purely number theoretic problem?
|
2025-03-21T14:48:29.746318
| 2020-01-31T07:04:43 |
351595
|
{
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"Carlo Beenakker",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351595"
}
|
Stack Exchange
|
How do you refer to the feasible set of solutions to a mixed-integer program?
I frequently want to refer to the feasible set of solutions to a mixed integer programming instance. Is there a name for a subset of $\mathbb{R}^n\times\{0,1\}^m$ of the form $\{(x,a)| Ax + Ba\leq b\}$, where $x\in \mathbb{R}^n$ and $a\in\{0,1\}^m$?
the "feasible region" has been used as an alternative to "feasible set"
Sure, but I'm hoping for maybe something like "Mixed-integer polytope" used in some literature somewhere. My complaint is essentially that I just want a short, concise handle to refer to the feasible region of a MIP
|
2025-03-21T14:48:29.746388
| 2020-01-31T08:45:32 |
351596
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/351596"
}
|
Stack Exchange
|
History and relationship on g-K modules and spherical functions
I am learning Harmonic analysis on real reductive Lie groups. It seems to me that there are two kinds of treatment. One is through $(\mathfrak g, K)$ modules (e.g., Wallach, Real reductive groups), the other is through spherical functions (e.g., Gangolli and Varadarajan, Harmonic analysis of spherical functions on real reductive groups). Both of the books say that the theory is developed by Harsh-Chandra (and of course with others). I would like to understand the history and the relationship between these two approaches. Could any experts explain this here or suggest some references?
|
2025-03-21T14:48:29.746472
| 2020-01-31T09:33:30 |
351597
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Iosif Pinelis",
"Mateusz Kwaśnicki",
"Nate Eldredge",
"Yoav Kallus",
"https://mathoverflow.net/users/108637",
"https://mathoverflow.net/users/20186",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/56931",
"r_faszanatas"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/351597"
}
|
Stack Exchange
|
Does convergence in law imply convergence in convex distance?
For two random variables $X$ and $Y$ taking values in $\mathbb{R}^m$, the convex distance $d_c$ is defined as
$$d_c(X,Y) = \sup_{h} \lvert \operatorname{E}(h(X)) - \operatorname{E}(h(Y)) \rvert,$$
where the supremum is taken over all indicator functions of measurable convex subsets of $\mathbb{R}^m$.
For $m=1$, it is easy to see that $d_c$ coincides with the Kolmogorov distance whenever $X$ and $Y$ are continuous, i.e. we have that $d_c(X,Y)= \sup_{x \in \mathbb{R}} \lvert F_X(x) -F_Y(x) \rvert$, where $F_X$ and $F_Y$ denote the cumulative distribution functions of $X$ and $Y$, respectively.
In particular, if $m=1$, we have that if a sequence $(X_n)$ of continuous real-valued random variables converges to another continuous random variable $Y$ in distribution, then $d_c(X_n,Y) \to 0$ as $n \to \infty$. Does this implication continue to hold if $m \geq 2$?
I could neither find a reference, nor a proof of this myself.
EDIT
If continuity of the random variables is defined as having non-atomic distributions, the implication is false (see the counterexample by Iosif Pinelis below).
If continuity is defined as having CDFs which are absolutely continuous with respect to the Lebesgue measure, the question is still open.
This doesn't seem right, even for $m=1$. Convergence in Kolmogorov distance is stronger than convergence in distribution, not weaker. Take for instance deterministic $X_n = 1/n$ and $Y=0$. And they don't converge in convex distance either, by taking $h = 1_{{0}}$ or $1_{[-1,0]}$. Is there a mistake, or assumptions missing?
@NateEldredge Thany you much for your comment, I somehow forgot to add the crucial assumption that the random variables are continuous.
So, in the second-to-last paragraph, you want $Y$ to be continuous as well as $X_n$?
@NateEldredge Yes, corrected this as well, now the question should be well-posed.
How do you define a continuous random vector in $\mathbb R^m$?
@IosifPinelis like in $\mathbb{R}$: its cumulative distribution function is continuous. Are there other (non-equivalent) ways to define continuity?
The other way is absolute continuity with respect to the Lebesgue measure. As illustrated by Iosif's answer, this is not equivalent.
@r_faszanatas: Well, there are many ways, but this one is particularly flawed: according to it, the uniform distribution on an interval is continuous if and only if the interval is neither horizontal nor vertical, which seems rather artificial.
@MateuszKwaśnicki this is something I should know by heart, thanks a lot for bringing it up in a friendly way. I completely agree that my definition is not a good one.
The answer is no. For instance, let $Y$ be uniformly distributed on the unit sphere in $\mathbb R^m$ (not the ball, but the sphere) and, for each natural $n$, let $X_n:=(1+1/n)Y$. Then the distribution of each of the random vectors $Y,X_1,X_2,\dots$ is non-atomic and hence, in particular, the values of this distribution on any singleton set are $0$; also, the cumulative distribution function of this distribution is continuous.
Moreover, $X_n\to Y$ in distribution, but $d_c(X_n,Y)=1$ for all $n$.
Thanks a lot for this counterexample. Given the comments above: Can you add to your answer that continuity is defined the way I commented (cumulative distribution function is just continuous)? Also, it would be nice to settle the question for the better definition of absolute continuity with respect to the Lebesgue measure.
@r_faszanatas : A better definition of the continuity of a random vector is that its distribution be non-atomic (which in the one-dimensional case coincides with the usual definition of the continuity of a random variable). The distributions of $X_n$ and $Y$ in the counterexample are indeed non-atomic.
As always, I learned a lot from this question, thanks again. So now the question is, if the implication holds in case at least the limiting random variable has a CDF which is absolutely continuous with respect to the Lebesgue measure.
@r_faszanatas : I think the answer is affirmative if $Y$ is absolutely continuous. To keep things in good order, though, I think this additional question should be posted separately.
just asked this as another question, see https://mathoverflow.net/questions/351870/does-convergence-in-law-to-absolutely-continuous-limit-imply-convergence-in-conv
|
2025-03-21T14:48:29.746732
| 2020-01-31T09:56:58 |
351601
|
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|
Stack Exchange
|
Complexity of pseudo-inverse of random matrix
Assume that $\mathbf{A}_{M\times N}$ is a sparse complex matrix. Then, what is the complexity of computation of its pseudo inverse, i.e.,
$$\mathbf{A}^{\mathrm{H}}(\mathbf{A}\mathbf{A}^{\mathrm{H}})^{-1}.$$
For a usual $\mathbf{A}$ (not sparse), the complexity of computation is
$$\alpha_1 N M^{\nu-1}+\alpha_2 M^{\nu}+\alpha_3 M^2 N^{\nu-2},$$
where $2.37 \leq\nu\leq 3$.
Assume that $\mathbf{A}$ is sparse in a way that $\mathbf{A}\mathbf{A}^{\mathrm{H}}$ is also sparse. Is it better for the sparse case?
What random distribution on the sparse matrices do you use?
It is not important. For example, consider the entries have bernouli distribution with $p<<1$.
|
2025-03-21T14:48:29.746814
| 2020-01-31T10:34:44 |
351604
|
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|
Stack Exchange
|
Intuition for the McGerty-Nevins compactification of quiver varieties
In Section 4 of the paper Kirwan surjectivity for quiver varieties (Inventiones Math. 2018) McGerty and Nevins define a compactification of the moduli space of representations
of the preprojective algebra associated to a quiver as the moduli space of representations of another quiver (the "tripling" of the original quiver) with certain relations. This compactification plays a key role in this paper and is defined by an explicit procedure obtaining the vertices and edges for the "tripled" quiver starting with the vertices and edges of the original quiver. My question is: what is the intuition behind the definition of such a compactification? Are there other constructions of (modular) compactifications of quiver varieties?
If I recall correctly I believe the intuition is roughly this: think of representations of a quiver as an analog of coherent sheaves on a smooth projective curve, modules for its preprojective algebra (having to do with doubling the quiver) as sheaves on the cotangent bundle to the curve, and modules for this tripled version as sheaves on the compactified cotangent bundle (projectivization of cotangent direct sum the trivial line bundle).
@DavidBen-Zvi Thank you very much! Sorry, I have a related question: in https://arxiv.org/abs/0712.4160 Mirkovic and Vybornov construct compactifications of quiver varieties by embedding them into the affine Grassmannian. Does their compactification also have a modular desription in terms of the quiver, in particular is it related to the McGerty-Nevins compactification?
@DavidBen-Zvi Sorry, in addition to this Mirkovic and Vybornov mention Wilson's embedding of the Calogero-Moser space into the adelic Grassmannian which also provides a compactification of the Calogero-Moser space (a particular case of a quiver variety) -- is Wilson's compactification related to the McGerty-Nevins compactification? Can this be deduced from your work with Nevins?
I don't know the answers to these questions, but I would expect these compactifications are the same. An example is Hilbert schemes or moduli of torsion-free sheaves on the cotangent to a curve (say on $A^2$) or its noncommutative version (which is what gives CM spaces) - you compactify it by thinking of framed t-f sheaves on a compactification of the cotangent, and it doesn't matter which (eg $P^2$ vs $P^1\times P^1$) since your sheaf is trivialized at $\infty$ anyway..
(From Hilbert scheme POV you just want to make sure points are not allowed to run off to $\infty$ in a family - if you do you connect $Hilb_n$ together for all $n$, which is what happens in the adelic Grassmannian.)
@DavidBen-Zvi Thank you very much! Sorry to bother you again, but is it known if the compactification of the Calogero-Moser space in your comment as framed torsion-free sheaves on the compactification of the noncommutative cotangent bundle to the curve is the same as the closure of the Calogero-Moser space in the adelic Grassmannian under Wilson's embedding? Maybe the fact that they are the same follows from your work with T. Nevins?
@DavidBen-Zvi Sorry, I was also wondering if in your work with T. Nevins (or elsewhere) there is an algebro-geometric description of Wilson's embedding of the Calogero-Moser space in the adelic Grassmannian? For example, in Section 3 of your paper with T. Nevins, as far as I understand, both the Calogero-Moser space and the Sato Grassmannian (in which the adelic Grassmannian lies) are described as certain moduli spaces of D-bundles on a curve -- does this provide an algebro-geometric description of Wilson's embedding?
Yes you can understand Wilson's embedding this way -- the adelic Grassmannian itself is the moduli problem for D-bundles which are unframed (ie torsion-free sheaves on the NC cotangent bundle, not its compactification) which are trivialized generically along the curve (see eg our paper https://arxiv.org/abs/0807.4992), and ON THE LEVEL OF POINTS there's a natural embedding of the CM spaces here (ie the framing gives you a canonical trivialziation on the locus of the curve where the D-bundle is locally free).
Whether you consider this an algebro-geometric identification of the closures is a different question - IIRC this embedding doesn't make sense in families, ie you don't get a generic trivialization along the curve direction from the CM space, but I could be misremembering. In any case you have to be careful since the adelic Grassmannian is a funny object algebro-geometrically (it's the pushforward to the curve of an ind-scheme over the Ran prestack of the curve).
|
2025-03-21T14:48:29.747105
| 2020-01-31T10:57:20 |
351605
|
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|
Stack Exchange
|
$L^{\infty}$ as colimit
I recently read a result (in Jarchow's book) that any ultrabornological space can be expressed as a colimit (in the category LCS) of Banach spaces. My question is the following.
Let $\mu$ be a finite measure on $\mathbb{R}$. Is there an uncountable set of $p_i \in [0,\infty)$ and finite Borel measures $\mu_i$ on $\mathbb{R}$ for which one can construct an inductive system $\{L^{p_i}_{\mu_i}(\mathbb{R})\}$ of locally convex spaces for which
$$
\injlim L^{p_i}_{\mu_i}(\mathbb{R}) \cong L^{\infty}_{\mu}(\mathbb{R})?
$$
Edit:
Alternatively, I look for a system such that
$
\injlim L^{p_i}_{\mu_i}(\mathbb{R})
$
is dense in $L^{\infty}_{\mu}(\mathbb{R})$ when its topology is relaxed from the final topology (which need not be metrizable) to the (relative) Banach topology of $L^{\infty}_{\mu}(\mathbb{R})$.
The answer to this question is YES -- but it is useless!
In fact, a theorem of Valdivia (which you can find, e.g., as Theorem 6.5.8 in the book Barrelled Locally Convex Spaces of Bonet and Perez-Carreras) states that given any infinite-dimensional separable Banach space $F$, one can represent any ultrabornological locally convex space $E$ (this means, that $E$ has some representation as an inductive limit of Banach spaces) which does not carry the finest locally convex topology as an inductive limit of Banach spaces isomorphic to $F$ such that, in addition,the linking maps between the steps are nuclear operators.
This theorem is rather useless because it is too good: Every ultrabornological space has such a representation and thus you cannot deduce any nice properties from it.
Oh, so these representaitons are never "explicit"; I'm assuming...
"The theorem is so good that it is useless." Got to remember that one.
@AIM_BLB Just a rewording of the penultimate line of Jochen's answer.
@JochenWengenroth I have been thinking about this answer for some time (as you may notice) and is there a characterization/(can more be said about) of LF spaces which are the limit of finite-dimensional Fréchet spaces? (Possibly after the topology is relaxed in the limit).
Then every linear map into any locally convex space is continuous because this is so for finite dimensional spaces. I think that there is only one such LF-space.
When I read your question, I assumed that you were asking whether there is a way to do this in the context of the underlying structure rather in the spirit of the answer above which addresses the situation where the spectrum consists of spaces which are merely isomorphic to the appropriate ones. Here are some suggestions (rather than an answer).
For clarity, I will work in the context of a space $\Omega$ with a $\sigma$–algebra $\cal A$ (rather than the reals) and the $L^1$-case. The family of positive finite measures forms a lattice and so the corresponding family $\{L^1(\mu)\}$ can be regarded both as an inductive and a projective spectrum in a natural way. So we can take inductive and projective limits in various senses.
Your query refers to the former but I am afraid that it seems to me that the results are not of interest (a purely subjective opinion, of course). One can take projective limits in three senses—that of Banach spaces (with linear contractions as morphisms), strict topologies, and locally convex spaces. Only the first two are of interest in the context of your query.
The first case yields the Banach space of bounded, measurable functions, the second one the same space with a weaker, but still complete, lc topology. If you are interested in duality theory, I would suggest that the latter space is the interesting one—its dual is the space of $\sigma$-additive measures on $\cal A$.
do you happen to have a reference to this type of literature?
I am afraid not. I’m afraid that this is a deeply unfashionable area, so one has to work these things out for oneself with no hope of publication.
It's a very short answer but this may be of interest to you:
Davis, Henry W.; Murray, F. J.; Weber, J. K. Jr., Families of $L_p$-spaces with inductive and projective topologies, Pac. J. Math. 34, 619-638 (1970). ZBL0187.05802.
https://projecteuclid.org/download/pdf_1/euclid.pjm/1102971942
|
2025-03-21T14:48:29.747415
| 2020-01-31T11:10:56 |
351606
|
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|
Stack Exchange
|
Any way to prove Monotone Continuity?
This question is cross-posted on math.stackexchange.
Let $A \succeq B$ be defined as $A$ is at least as probable as $B$. $\succeq$ obeys the following axioms. For any $A, B, C$,
Non-negativity: $A \succeq \emptyset$
Additivity: If $A\cap C=B\cap C=\emptyset$, then $A\succeq B$ if and only if $A\cup C \succeq B\cup C$
Transitivity: If $A \succeq B$ and $B\succeq C$, then $A\succeq C$
Is there any way I can prove Monotone Continuity from these axioms?
Monotone Continuity: For $A_1,A_2,A_3,...$ and $B$, if
(i) $A_i \subseteq A_{i+1}$,
(ii) $A=\bigcup_i A_i$,
(iii) $A\cap B=\emptyset$, and
(iv) $B\succeq A_i$, then $B\succeq A$.
HI, and welcome to the MathOverflow. I noticed you cross posted a similar question on the sister site MathSE: this is not forbidden, but you have to state it in the body of the question (I did the same as you on my first post here).
More than saying this, there is a principle that you should not simultaneously post on both sites (even with cross-references) so as to avoid duplication of effort. If you feel a question is between the level of the two sites, post on MSE first, and if there is no answer in a couple of days, try at MO.
The answer is no. E.g., let $X:=\{1,2,\dots\}$ with $P(A):=\sum_{x\in A}2^{-x}$ for all $A\in2^X$. For any $A$ and $B$ in $2^X$, let $A \succeq B$ be defined as ($P(A)>P(B)$ or $A=B$). Then the relation $\succeq$ satisfies all your axioms.
However, it does not have the monotone continuity property. E.g., let $B:=\{1\}$ and $A_i:=\{2,\dots,i+1\}$ for $i=1,2,\dots$. Then $A_i \subseteq A_{i+1}$ and $B\succeq A_i$ for all $i$, and $A\cap B=\emptyset$ for $A:=\bigcup_i A_i$. Yet, $B\not\succeq A$.
|
2025-03-21T14:48:29.747558
| 2020-01-31T11:27:39 |
351608
|
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|
Stack Exchange
|
Partitioning the set of Pauli words into abelian pieces
Let $\sigma_x,\sigma_y,\sigma_z$ be the Pauli matrices. A Pauli word of length $n$ is defined as the tensor product $\otimes_{i=1}^n\sigma_i$ of operators $\sigma_1,\dots,\sigma_n\in\{\mathbf 1,\sigma_x,\sigma_y,\sigma_z\}$. The set $PW_n$ of Pauli words of length $n$ has cardinality $4^n$. A subset $A\subset PW_n$ is called abelian if $u\circ v=v\circ u$ for any $u,v\in A$.
Question. What is the number of partitions of the set $PW_n\setminus \{\otimes_{i=1}^n\mathbf 1\}$ into $2^n+1$ abelian subsets $A_{1},\dots,A_{2^n+1}$ of cardinality $|A_i|=2^n-1$?
This problem was posed in December 2019 by a Francis Drake (probably, a pseudo) on page 41 of Volume 3 of the Lviv Scottish Book.
I think that your project is very good, I would like to see more notebooks containing problems from other groups/departments of mathematics in this website. Yours is a very good idea. Isn't required a response to this comment.
|
2025-03-21T14:48:29.747654
| 2020-01-31T13:46:29 |
351610
|
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|
Stack Exchange
|
Hadwiger number in vertex collapse in a bipartite graph
If $G=(V,E)$ is a finite graph, let the Hadwiger number $\eta(G)$ equal the largest integer $n$ such that the complete graph $K_n$ is a minor of $G$.
Is there a bipartite graph $G$ on more than $3$ vertices such that whenever $2$ non-adjacent vertices are collapsed, then the Hadwiger number increases?
|
2025-03-21T14:48:29.747859
| 2020-01-31T14:04:32 |
351614
|
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"Bertram Arnold",
"Dan Petersen",
"Fernando Muro",
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|
Stack Exchange
|
Are exterior algebras intrinsically formal as associative dg algebras?
(Cross-posted from mathematics stackexchange.)
Fix a finite dimensional vector space $V$ over a field of characteristic zero, and let $R=Sym(V[1])$ be the free graded commutative algebra generated by $V$ in cohomological degree $-1$, but thought of as a formal associative dg algebra (we forget that it is commutative).
Then it is easy to see that (edit: any $A_{\infty}$-deformation of) $R$ is almost formal, meaning that in a minimal $A_{\infty}$-model, the higher products eventually vanish. Indeed, $R$ is concentrated in non-positive cohomological degrees, so all products of an $A_{\infty}$-structure $\mu_{n} : R^{\otimes n} \rightarrow R[2-n]$ have to vanish for $n>>0$.
Question 1: Is $R$ intrinsically formal, so that for any deformation, $\mu_{n}$ can be assumed to be zero for $n>2$?
Question 2: What if $R=Sym(V[-1])$, so that $R$ is concentrated cohomologically in non-negative degrees? (I'm thinking of the cohomology ring of a torus, but as an associative algebra, rather than a commutative algebra.) Can there be non-trivial $A_{\infty}$-structures on $R$, or is $R$ intrinsically formal?
You say that you consider $R$ as a formal associative DG-algebra. This means that you endow it with the trivial differential, hence it is itself a minimal model. There are no higher $\mu_n$'s.
Thanks. I mean to ask whether an exterior algebra has non-trivial $A_{\infty}$-deformations, or whether it is `intrinsic formality'. I'll clarify.
It seems that the answer to Q2 is positive, at least in characteristic zero. If $R$ has cohomology $Sym(V[-1])$, then I think you can show it has an augmentation $R \rightarrow k$, and that the Koszul dual $R^{!}=RHom_{R}(k,k)$ is quasi-isomorphic to $Sym(V^{*})$. Taking the Koszul dual again, we get a quasi-isomorphism $R \simeq Sym(V[-1])$, as desired. Is there a more elementary argument? I still don't know about Q1, or about positive characteristic.
The Koszul dual $R^{!}=RHom_{R}(k,k)$ is not quite $Sym(V^{*})$, but rather the completion thereof. So more naturally to work with the dual coalgebra $Sym(V)$. But I still think the argument is valid.
I deleted my answer to the question - as pointed out by Bertram Arnold the theorem I quoted doesn't apply to the situation here.
No problem. Nonetheless, the paper of Saleh looks interesting, and the argument simple: https://arxiv.org/pdf/1609.02540.pdf. For the record, let me give an elementary example about the difference between Comm and Ass. Let =(), in degree 0. Just a polynomial algebra. It's rigid as a commutative algebra, since $H^{1}(T_{})=0$, but it certainly admits deformations as an associative algebra.
Here are examples of nontrivial $A_\infty$-structures extending the product on $\operatorname{Sym}(\mathbb R^3[-1])$ and $\operatorname{Sym}(\mathbb R^5[-1])$, respectively. Below the fold, I have kept my original answer, which got the main ideas right but almost all degrees wrong.
Fix coordinates $\xi_1,\xi_2,\xi_3$ on $\mathbb R^3[-1]$ and set
$$
\mu_3(f_1,f_2,f_3) = \xi_2\xi_3 (\partial_{\xi_1}f_1)(\partial_{\xi_1}f_2)(\partial_{\xi_1}f_3)\ .
$$
This defines a $A_\infty$-structure: Every term in $[\mu_3,\mu_3](f_1,\dots,f_5) = \sum \mu_3(\dots,\mu_3(\dots),\dots)$ contains $\xi_1^2 = 0$ and hence vanishes; similarly, $[\mu_3,\mu_2](f_1,\dots,f_4)$ vanishes if one of the $f_i$ is a multiple of $\xi_2$ or $\xi_3$, and it also vanishes if one of them is the identity since $[-,-]$ preserves normalized Hochschild cochains. Thus the only expression we have to check is
\begin{align*}
[\mu_2,\mu_3](\xi_1,\xi_1,\xi_1,\xi_1) &= \color{red}{-}\xi_1\mu_3(\xi_1,\xi_1,\xi_1) -\mu_3(\xi_1^2,\xi_1,\xi_1) \\
&\ + \mu_3(\xi_1,\xi_1^2,\xi_1) - \mu_3(\xi_1,\xi_1,\xi_1^2) \\
&\ + \mu_3(\xi_1,\xi_1,\xi_1)\xi_1\\
&= -\xi_1\xi_2\xi_3 + \xi_2\xi_3\xi_1 = 0
\end{align*}
(The red sign comes from the Koszul sign rule.)
This $A_\infty$-algebra has a nontrivial Massey product $0\notin \langle\xi_1,\xi_1,\xi_1\rangle = \xi_2\xi_3 + \xi_1A + A\xi_1$ and hence is not formal.
Essentially the same argument applies to the $A_\infty$-structure on $\mathbb R^5[-1]$ defined by the only nontrivial bracket
$$
\mu_3(f_1,f_2,f_3) = \xi_2\xi_3\xi_4\xi_5 (\partial_{\xi_1}f_1)(\partial_{\xi_1}f_2)(\partial_{\xi_1}f_3)\ .
$$
Recall that we can associate to any (cohomologically) graded algebra $(A,\mu_A)$ the Hochschild cochains
$$
C^n(A) = \prod_{p+q = n}\operatorname{Hom}^p(A^{\otimes q},A)
$$
which come equipped with a differential defined by precomposing with the multiplication of cyclic neighbours in the tensor product. As the totalization of a double complex, it also comes with a canonical filtration, namely
$C^n_{\ge m}(A) = \prod_{{\substack{p+q=n\\q\ge m}}} \operatorname{Hom}^p(A^{\otimes q},A)$. The skew-symmetrization of the pre-Lie structure $f\circ g(c_1,\dots,x_{m+n-1}) = \sum_{i=1}^m \pm f(x_1,\dots,g(x_i,\dots,x_{i+n-1}),\dots,x_{m+n-1})$ defines a filtered shifted dgla structure on $C^*(A)$, and $A_\infty$-structures with trivial differential $(A,0,\mu_A,\mu_3,\dots)$ correspond to Maurer-Cartan elements $(\mu_3,\dots)$ in $C^2_{\ge 3}(A)$.
Now consider any such element $\mu$; its component of lowest filtration defines a degree $2$ cohomology class in the associated graded, since $\mathrm d\mu = -\frac{1}{2}[\mu,\mu]$ lies in higher filtration (as $[C^p_{\ge m}(A),C^q_{\ge n}(A)]\subset C^{p+q-1}_{\ge m+n-1}(A)$). If this cohomology class is $0$, i.e. $\mu = \mathrm d\eta$ up to terms of higher filtration, we can act by the gauge symmetry $\mu\mapsto e^{-\eta}\circ\mu = \mu - \mathrm d\eta\, + (\text{higher filtration})$ to obtain an equivalent Maurer-Cartan element which lies in higher filtration. Since the Hochschild cochains are complete with respect to the canonical filtration, the limit of this process is well-defined, i.e. if all obstructions vanish we obtain that our Maurer-Cartan element is gauge equivalent to $0$.
If $A = \operatorname{Sym}(V)$ is the commutative algebra on a graded vector space $V$, we can compute $C^*(A)$ with its filtration explicitly: We have $C^*(A) \simeq \operatorname{Ext}_{A\otimes A^{op}}(A,A)$, and since the multiplication map $A\otimes A^{op}\to A$ is $\operatorname{Sym}$ of the codiagonal $V\oplus V\to V$, we obtain a Koszul resolution $A\simeq (\operatorname{Sym}(V\oplus V\oplus V[1]),\mathrm d)$. Explicitly, we choose a basis $x_i$ of $V$; then the resolution is generated by elements $x_i,x_i'$ in degree $|x_i|$ and $y_i$ in degree $|x_i|-1$, with $\mathrm dy_i = x_i' - x_i''$. (This is where we use characteristic $0$.) It follows that $C^*(\operatorname{Sym}(V))\simeq \widehat{\operatorname{Sym}}(V\oplus V^*[-1])$, where the filtration is given by the degree in the $V[-1]$-variables, and the hat indicates that we take the completion with respect to this filtration.
Now let us specialize to the two cases in your question, namely $V$ is concentrated in degree $-1$ or $+1$. In the second case, $C^*(A)$ is concentrated in nonnegative degrees, and $HH^2(A) := H^2(C^*(A))\cong \operatorname{Sym}^* V^*\otimes \Lambda^2 V$, with the relevant part in $\operatorname{Sym}$-filtration $\ge 3$. In the first case, $V\oplus V^*[-1]$ lives in degrees $-1$ and $2$, and the deformation group is given by $HH^2_{\ge 3}(A) = \prod_{p\ge 3} \Lambda^{2p-2}V\otimes \operatorname{Sym}^p V^*$ and thus also does not vanish. Indeed, Kontsevich's formality theorem shows that Maurer-Cartan elements of $\hbar C^*_{\ge 3}(A)[[\hbar]]$ are in bijection with those of its cohomology, i.e. an infinitesimal deformation can be extended to a formal deformation iff it satisfies the Maurer-Cartan equation in in the cohomology. In some cases, the formal infinite series involve only finitely many terms and therefore give rise to actual deformations, as is the case in the two examples above.
Thanks! I like this argument. I haven't seen before the extra filtration on Hochschild cochains, and its relation to formality. Could you provide some pointers to the literature where I can see other examples of this kind of argument?
I suppose the filtration restriction on $HH^{}(A)$, to get minimal $A_{\infty}$-structures rather than non-minimal and curved gadgets, should be related to the Quillen sequence $A \rightarrow {\rm Der}^{\ast}(A) \rightarrow HH^{}(A)[1]$. In Ultimately, it should be expressible in terms of ${\rm Der}^{\ast}(A)$ rather than cochains, no? And in this graded commutative case, I think the map $A \rightarrow Der^{}(A)$ is zero, so that probably $Der^{}(A)$ is just embedding to contain the filtered piece you are describing.
Note that there seems to be a dual $$ missing somewhere. I think actually $HH^{}(Sym(V))=Sym(V) \otimes Sym(V^{}[-1])$. Note that is certainly true if $V$ is in degree $0$, by HKR. If correct, then when $V$ is of pure degree $1$, $V^{}[-1]$ is of pure degree $0$, and so I would get $HH^{2}(A)=\bigwedge^{2}V$. Or did I mess up degrees? But it kind of makes sense. I guess there should be non-homogeneous deformations of $A$ by "super Clifford algebras".
You are right, I forgot to dualize in the HKR theorem. Note that the Hochschild class you write down lies in filtration $0$, so it corresponds to a curved deformation. But if $V$ is concentrated in degree $-1$, we get lots of interesting classes in higher filtration. I'll update my answer.
I agree with this answer. But let me add that there is a subtlety involved in making sense of this type of argument. To be precise in the second paragraph we are taking the transfinite composition of infinitely many gauge transformations and for this to be meaningful we also need to be sure that the sequence of gauges can be made to converge to zero with respect to the filtration. So we should prove in general that obstructions vanish in the cohomology of $C^{2}_{\geq n}$, rather than that their images in $HH^2$ vanish.
Right. These curved deformations are what I guess give something like "super Clifford algebras".
Thanks for the correction. One more: $HH^{2}(A) \simeq \bigwedge^{2}(V) \otimes Sym(V^{*}[-1])$, in the $+1$ case. Doesn't affect the conclusion, however.
Something still doesn't seem right, though. Consider the Quillen sequence $R \rightarrow {\rm Der}(R) \rightarrow C^{}(R)$, where ${\rm Der}(R)$ is derived bimodule derivations. Then ${\rm Der}(R)$ controls deformations of $R$ as a dga, and Bertram's calculation shows that almost all of those "curved" deformations in the $+1$ case actually come from $H^{1}({\rm Der}(R))=\bigwedge^{2}(V) \otimes Sym^{\geq 1}(V^{}[-1])$, so there should be a huge number of uncurved deformations. Some confusion about filtration degrees?
You are, again, completely right, and I messed up the filtrations; the space of first-order deformations is indeed given by $\operatorname{Sym}^{\ge 3} V\otimes \Lambda^2 V$. I think I have found two explicit non-formal $A_\infty$-structures, which you can find in my answer.
Thanks. I'll think more about this. In comments below question, I thought I had an argument for intrinsic formality in this case, so I'm surprised. Maybe what is wrong with my argument is that the spectral sequence computes the associated graded of the Koszul dual, not the Koszul dual, and so a priori one could get a filtered deformation of polynomials for the Koszul dual.
An aside: deformations of $Sym(V[-1])$ to a non-minimal $A_{\infty}$-algebra arise in nature. Interpret $Sym(V[-1])$ as Lie algebra cochains for the abelian Lie algebra $V^{}$, and now deform the bracket on $V^{}$ to something non-trivial. That's dual to deforming the differential in $Sym(V[-1])$.
|
2025-03-21T14:48:29.748521
| 2020-01-31T14:05:59 |
351615
|
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|
Stack Exchange
|
Is $\mathfrak p=\omega_1$ equivalent to the existence of a Hausdorff gap without infinite pseudointersection?
Given two set $A,B$ we write $A\subset^* B$ if the complement $A\setminus B$ is infinite.
A Hausdorff gap is a transfinite family $\langle A_\alpha,B_\alpha\rangle_{\alpha\in\omega_1}$ of infinite subsets of $\omega$ satisfying the following two properties:
$\bullet$ $A_\alpha\subset^* A_\beta\subset^* B_\beta\subset^* B_\alpha$ for any countable ordinals $\alpha\le\beta$;
$\bullet$ for any infinite subset $I\subset \omega$ there exists a countable ordinal $\alpha$ such that $A_\alpha\not\subset^* I$ or $I\not\subset^* B_\alpha$.
A in infinite subset $I\subset\omega$ is called a pseudointersection of a Hausdorff gap $\langle A_\alpha,B_\alpha\rangle_{\alpha\in\omega_1}$ if $I\subset^* (B_\alpha\setminus A_\alpha)$ for any $\alpha\in\omega_1$.
The definition of the small uncountable cardinal $\mathfrak p$ implies that under $\mathfrak p>\omega_1$ each Hausdorff gap has an infinite pseudointersection.
What does happen under $\mathfrak p=\omega_1$?
Problem. Is $\mathfrak p=\omega_1$ equivalent to the existence of a Hausdorff gap without infinite pseudointersection?
Yes. This is a result due to Nyikos and Vaughan from 1983, appearing the paper
Nyikos, Peter J.; Vaughan, Jerry E., On first countable, countably compact spaces. I: ((\omega_ 1,\omega^*_ 1))-gaps, Trans. Am. Math. Soc. 279, 463-469 (1983). ZBL0542.54004.
Hausdorff gaps without a pseudo-intersection are called tight gaps, and Theorem 1.2 of the paper shows that these exist if and only if $\mathfrak{p}=\omega_1$.
Thank you so much. Your answer was very helpful!
|
2025-03-21T14:48:29.748634
| 2020-01-31T14:07:33 |
351616
|
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|
Stack Exchange
|
Is this matrix invertible?
Suppose that $P$ is a finite partition of the unit $d$-dimensional hypercube composed of $m$ hyperrectangles $R_1,...,R_m$
Denote $u_1,...,u_m$ the middle-point of each hyperrectangle : $u_i = \frac{b_i-a_i}{2}$ for $R_i = \times_{j=1}^{d} [a_{i,j},b_{i,j}]$
For a given dimension $j \in 1,...,d$, define the matrix $E^{(j)}$ of size $m\times m$ by :
$$E_{k,l}^{(j)} = \frac{\lambda([0,u_{k,j}] \cap R_{l,j}])}{\lambda(R_{l,j})}$$
where $u_{k,j}$ denotes the $j$th coordinate of $u_k$ and $R_{l,j} = [a_{l,j},b_{l,j}]$ denotes the projection of $R_l$ onto the dimension $j$.
Questions :
1° What can you say about the space $ S^{(j)} = \left\{ x \in \mathbb{R}^m : E^{(j)}x = u_{.,j}\right\}$ ?
2° Are the matrices $E^{(j)}$ invertible ?
3° What about $S = \cap_{j=1}^m S^{(j)}$ ?
I started by trying to understand the structure of the matrices $E^{(j)}$. I understand that their diagonals are $1/2$, that all their values are between 0 and 1, but I can't say much about the off-diagonal terms. I guess that a majority of values will be either 0 or 1.
How can I get e.g. information about the kernel of such matrices ? This will give me a base of the affine subspace.
|
2025-03-21T14:48:29.748747
| 2020-01-31T14:18:58 |
351617
|
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|
Stack Exchange
|
Does minimal degree $n$ imply a $K_n$ minor
Is it true that any finite graph has a $K_n$ minor, where $n$ is a minimal vertex degree?
Not if the graph is infinite (e.g. $n=3$ and the infinite binary tree).
No.
The edge-graph of the icosahedron is regular of degree five, but does not have a $K_5$ minor because it is planar (Kuratowski's theorem).
This is a great (counter)example.
Also the smallest if I am not mistaken. No $K_4$-minor means the graph has a vertex of degree 2.
More generally, it is a classic result (independently due to Kostochka and Thomason) that minimum degree $(\alpha+o(1))n \sqrt{\log n}$ suffices to force a $K_n$ minor, where $\alpha$ is an explicit constant. Conversely, there are random graphs with minimum degree $\Omega(n\sqrt{\log n})$ that do not contain a $K_n$ minor. See here to access the paper by Thomason.
Update. Alon, Krivelevich, and Sudakov have recently given a new short proof of the Kostochka-Thomason result.
Where can I read about this in detail?
I added a link to the corresponding paper by Thomason. Note that Thomason's result is stated for average degree, but every graph with average degree $d$ contains a subgraph of minimum degree $\frac{d}{2}$.
|
2025-03-21T14:48:29.748861
| 2020-01-31T14:35:37 |
351619
|
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|
Stack Exchange
|
Can power of a prime number be approximated by product of powers of two adjacent prime numbers?
With prime numbers $a < b < c$ and no primes exist in ranges $(a, b)$ and $(b, c)$, is it possible that there exists positive integers $x$, $y$, $z$ such that $|a^x c^z-b^y|=2$?
Do you intend to rule out the case a=2? Gerhard "Two Is Also A Prime" Paseman, 2020.01.31.
@GerhardPaseman If $a=2$, $|a^x c^z - b^y|$ is odd, so ...
So the one case where you can prove there are such terms close to each other is ruled out. Gerhard "Distance One Seems Not Wanted" Paseman, 2020.01.31.
Write a as b-alpha, c as b+beta, and delta as alpha times beta. Modulo delta, the sum is b^y(b^{x+z-y)-1) + (zbeta - xalpha)b^{y-1}. Since alpha and beta are both even, there are lots of times when this implies the difference is divisible by four. Gerhard "Not Seeing Many Solutions Here" Paseman, 2020.01.31.
|
2025-03-21T14:48:29.749206
| 2020-01-31T15:09:33 |
351626
|
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|
Stack Exchange
|
Dunford-Pettis theorem
Let $\mathcal{F}$ be a bounded set in $L^{1}(\Omega)$. Then $\mathcal{F}$ has compact closure in the weak topology $\sigma(L^{1},L^{\infty})$ if and only if $\mathcal{F}$ is equi-integrable, that is,
\begin{cases}
\forall\epsilon>0\ \exists\delta>0\text{ such that}\\
\int_{A}|f|<\epsilon\ \forall A\subset\Omega,\text{measrable with }|A|<\delta,\ \forall f\in\mathcal{F}
\end{cases}
and
\begin{cases}
\forall\epsilon>0\ \exists\omega\subset\Omega,\text{ measurable with }|\omega|<\infty\text{ such that}\\
\int_{\Omega\backslash\omega}|f|<\epsilon\ \forall f\in\mathcal{F}.
\end{cases}
I would like to known if there is a difference between the bounded and unbounded case?
Lemma$^*$: Let $\mu_{d}$ be a (diffuse) measure in $\mathcal{M}_{d}(\Omega)$ and let $(v_{n})$ be a sequence of functions in $W^{1,p}_{0}(\Omega)$, bounded in $L^{\infty}(\Omega)$, and converging to a function $v$ $\text{cap}_{p}-$q.e., then $(v_{n})$ converges to $v$ $\mu_{d}-$a.e. and
$$\underset{n\rightarrow+\infty}{\operatorname{lim}}\int_{\Omega}v_{n}d\mu_{d}=\int_{\Omega}v d\mu_{d}.$$
What precisely do you mean by "bounded case" and by "unbounded case"? By the way, your second sentence is not correct.
I means that $\Omega$ is a bounded/unbounded subset, this is the definition given in [H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Theorem 4.30 (Dunford–Pettis)]
In the ``unbounded'' case equiintegrability does not imply relative weak compactness; consider $f_n=$ the indicator function of $[n, n+1]$ in $L_1(\mathbb{R})$.
|
2025-03-21T14:48:29.749351
| 2020-01-31T15:43:28 |
351632
|
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|
Stack Exchange
|
The Hubbard-Stratonovich transformation
is there a way to extend the Hubbard-Stratonovich transformation
$$e^{\frac{1}{2}Ks^2}=\left(\frac{K}{2\pi}\right)^{1/2}\int_{–\infty}^\infty e^{–Kx^2+Ksx}dx$$
to the case $e^{\frac{1}{2}Ks^p}$ for $p \in \mathbb{N}$?
To the downvoters: This is an excellent and deep question. +1
@HyyFly: It would be nice to elaborate a bit on your question and in particular give some motivation. What applications do you have in mind for possible generalizations of the Hubbard-Stratonovich transformation? The latter is just the physics parlance for the formula for the Fourier transform of a Gaussian, used in reverse. In any case, if you want to entice people to write useful and thoughtful answers, how you write your question must show some effort.
As to the question itself:
The problem of figuring out when "simple functions" like the exponential of a quadratic has a simple Fourier transform was addressed in the article "When is the Fourier transform of an elementary function elementary?" by Etingof, Kazhdan and Polishchuk in Selecta Math. 2002.
If all you want is a way of breaking quantum field theory vertices with $p$ legs into trivalent vertices, then one can do that by iterating the Hubbard-Stratonovich transformation (aka intermediate field representation). See, for instance, the article "Loop vertex expansion for $\Phi^{2k}$ theory in zero dimension" by Rivasseau and Wang in J. Math. Phys. 2010 and the follow up "Note on the intermediate field representation of $\phi^{2k}$ theory in zero dimension" by Lionni and Rivasseau.
For a QFT where you want to split fermion multilinears into a bilinear and auxiliary field there is also https://arxiv.org/abs/1706.06494
As Abdelmalek Abdesselam mentions you need to iterate such transformations. Let me sketch one way this can work, (in a formal perturbative sense):
Starting from $\exp(s^p)$, you can consider the integral over two new variables $x,y$
$$
\int dx \int dy \exp(s^{p-2} y + x(y-s^2))
$$
Clearly doing the $x,y$ integrations together will set $y=s^2$ (and there will also be some numerical prefactor you can adjust) so you obtain the original term $\exp(s^p)$.
However the log of the integrand here is now of order $p-1$ rather than $p$, so you can do another such transformation involving $s,x,y$ and more new variables to further reduce the total order until you hit $p=3$, where the procedure terminates.
There is a mathematical physics interpretation here, if you are interested: if we view $\exp(s^p)$ as the (exponential of an) action of some physical system, and canonically associate to that an L-infinity algebra, what this procedure is doing is producing a strictification of said algebra: namely an L-infinity algebra quasi-isomorphic (read: physically equivalent) to the original one, but all of whose higher (than bilinear) brackets vanish. The end result looks like a Chern-Simons type theory.
|
2025-03-21T14:48:29.749564
| 2020-01-31T16:26:33 |
351635
|
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|
Stack Exchange
|
On the center-stable manifold theorem for sets
Suppose I have a dynamical system $f:S \to S$ where $S \subset \mathbb{R}^n$ and $S$ is compact and $f$ is twice differentiable. Assume there exists a function $V$ such that $V(f(x)) < V(x)$ unless $x$ is a fixed point of $f$. This implies convergence of $\lim_{k \to \infty}f(x_k)$ with $x_k = f(x_{k-1})$ and moreover you can argue about convergence to the set of fixed points for any initial condition (not point-wise convergence, the set of limit points for a specific initial condition is connected and closed). My question is can I argue that the set of initial conditions so that the the dynamical system converges to a set of fixed points with the Jacobian of these fixed points having an eigenvalue of magnitude greater than one is of measure zero? Essentially, can I argue that the whole component lies on the center-stable manifold? The answer is yes when we have point-wise convergence, i.e., $\lim x_k$ exists.
|
2025-03-21T14:48:29.749697
| 2020-01-31T17:43:43 |
351640
|
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|
Stack Exchange
|
Extent of “unscientific”, and of wrong, papers in research mathematics
This question is cross-posted from academia.stackexchange.com where it got closed with the advice of posting it on MO.
Kevin Buzzard's slides (PDF version) at a recent conference have really unsettled me.
In it, he mentions several examples in what one would imagine as very rigorous areas (e.g., algebraic geometry) where the top journals like Annals and Inventiones have published and never retracted papers which are now known to be wrong. He also mentions papers relying on unpublished results taken on trust that those who announced them indeed have a proof.
He writes about his own work:
[...] maybe some of my work in the p-adic Langlands philosophy relies
on stuff that is wrong. Or maybe, perhaps less drastically, on stuff
which is actually correct, but for which humanity does not actually
have a complete proof. If our research is not reproducible, is it
science? If my work in pure mathematics is neither useful nor 100
percent guaranteed to be correct, it is surely a waste of time.
He says that as a result, he switched to formalizing proofs completely, with e.g. Lean, which guarantees correctness, and thus reusability forever.
Just how widespread is the issue? Are most areas safe, or contaminated? For example, is there some way to track the not-retracted-but-wrong papers?
The answer I accepted on academia.stackexchange before the closure gives a useful general purpose method, but I'd really appreciate more detailed area-specific answers. For example, what fraction of your own papers do you expect to rely on a statement "for which humanity does not actually have a complete proof" ?
If my work in pure mathematics is neither useful nor 100 percent guaranteed to be correct, it is surely a waste of time. There would appear to be a gap in this argument, unless the author has a general proof that "not useful" plus "not 100% guaranteed to be correct" implies "waste of time", in which case there are myriad special cases outside of mathematics, many of them quite surprising and unsettling.
Just to be clear, this depends on precisely what you mean by "wrong" doesn't it? You have not defined this term yet. I imagine that there are many ways to be wrong, not all of the same kind. For example, does a trivial typographical error count? Grammatical errors? etc. Complete proof? What is that?
@Somos : I mean it as a substantial error making the proof incomplete as such.The paper in Annals mentioned by Buzzard being an example : the opposite statement seems to have been proved later on.
There has been substantial discussions on this site on the meaning (and ambiguity) of the word "wrong"; if somebody can track these discussions and link them it would be useful.
Possible duplicate: https://mathoverflow.net/questions/291890/what-percentage-of-published-mathematics-papers-are-correct/
Perhaps relevant: https://en.wikipedia.org/wiki/List_of_incomplete_proofs.
@ Sam Hopkins : thanks for this reference, but my question is precisely about papers which have not had an errata issued yet should have one.
what fraction of your own papers do you expect to rely on a statement "for which humanity does not actually have a complete proof" Zero, but the catch is that I insisted on claiming the proof of the reduction of A to B (B being published by other people and universally accepted as valid) instead of claiming A itself at least twice (not that my co-authors were very happy about it, but I find it a good practice if one has trouble going through the whole proof tree of B yourself). On the other hand, I'm guilty of publishing at least 2 papers in which some gap was later found (and closed).
A complete proof from what axioms? fedja just described the perfectly acceptable practice of working in a high-level system where you just take some B on faith, a.k.a. make it part of your axioms. (Likewise people in Paris used to speak of “axiom Valiron”: everything in this book is true.) Of course, the risk in doing this is, your resulting axiom system might be inconsistent. But then so might ZFC, right? (1/2)
So I see “relying on stuff for which humanity does not actually have a complete proof” (a contradiction might be found) as hugely different from “relying on stuff that is wrong” (a contradiction has been found). (2/2)
The actual answers will probably suffice for you, but I still recommend that you look at Thurston’s response to the Jaffe-Quinn paper that was mentioned in one of the answers. You may find it helpful in alleviating some of the underlying concerns you seem to have about mathematics as a whole.
"Are most areas safe, or contaminated?"
Most areas are fine. Probably all important areas are fine. Mathematics is fine. The important stuff is 99.99999% likely to be fine because it has been carefully checked. The experts know what is wrong, and the experts are checking the important stuff. The system works. The system has worked for centuries and continues to work.
My talk is an intentionally highly biased viewpoint to get people talking. It was in a talk in a maths department so I was kind of trolling mathematicians. I think that formal proof verification systems have the potential to offer a lot to mathematicians and I am very happy to get people talking about them using any means necessary. On the other hand when I am talking to the formal proofs people I put on my mathematician's hat and emphasize the paragraph above, saying that we have a human mathematical community which knows what it is doing better than any computer and this is why it would be a complete waste of time formalising a proof of Fermat's Last Theorem -- we all know it's true anyway because Wiles and Taylor proved it and since then we generalised the key ideas out of the park.
It is true that there are holes in some proofs. There are plenty of false lemmas in papers. But mathematics is robust in this extraordinary way. More than once in my life I have said to the author of a paper "this proof doesn't work" and their response is "oh I have 3 other proofs, one is bound to work" -- and they're right. Working out what is true is the hard, fun, and interesting part. Mathematicians know well that conjectures are important. But writing down details of an argument is a lot more boring than being imaginative and figuring out how the mathematical world works, and humans generally do a poorer job of this than they could. I am concerned that this will impede progress in the future when computers start to learn to read maths papers (this will happen, I guess, at some point, goodness knows when).
Another thing which I did not stress at all in the Pittsburgh talk but should definitely be mentioned is that although formal proof verification systems are far better when it comes to reliability of proofs, they have a bunch of other problems instead. Formal proofs need to be maintained, it takes gigantic libraries even to do the most basic things (check out Lean's definition of a manifold, for example), different systems are incompatible and systems die out. Furthermore, formal proof verification systems currently have essentially nothing to offer the working mathematician who understands the principles behind their area and knows why the major results in it are true. These are all counterpoints which I didn't talk about at all.
In the future we will find a happy medium, where computers can be used to help humans do mathematics. I am hoping that Tom Hales' Formal Abstracts project will one day start to offer mathematicians something which they actually want (e.g. good search for proofs, or some kind of useful database which actually helps us in practice).
But until then I think we should remember that there's a distinction between "results for which humanity hasn't written down the proof very well, but the experts know how to fill in all of the holes" and "important results which humanity believes and are not actually proved".
I guess one thing that worries me is that perhaps there are areas which are currently fashionable, have holes in, and they will become less fashionable, the experts will leave the area and slowly die out, and then all of a sudden someone will discover a hole which nobody currently alive knows how to fill, even though it might have been the case that experts could once do it.
It is nice to have a reply from the actual author here.
My perspective on what you say in your fifth paragraph is that writing formal proofs is a lot like programming, so if mathematicians start doing it more they will get to "enjoy" all the nuisances that occur in programming.
I enjoyed the talk slides and have been lurking around some of your other online channels for this kind of stuff, but when I see "It was in a talk in a maths department so I was kind of trolling mathematicians" I can't help thinking of "Rex Kramer, Danger Seeker" (look it up on YouTube as I'd rather not link in case other readers of MO take it the wrong way)
Thank you very much for this answer, I had not understood this was "trolling", I'm quite relieved !
For those interested, there's a nice older question about correcting published errors https://mathoverflow.net/questions/31337/how-do-i-fix-someones-published-error?rq=1 I guess the wiki is still lacking.
The only reason that you need a huge library before you can define a manifold is that mathematicians set things up that way. You need three years of undergraduate math before you can teach gemetric topology. Formalization just reflects this fact. If we want to do better, then we need to organize math in better ways. Example: homotopy type theory lets you define a circle directly in 5 lines. (But there is a price to pay.)
Do you really think that only one in ten million important research papers is flawed? That is delusional.
@TonyK: what are you talking about? There obviously aren't 10000000 "important stuff" research papers, so I think we need to read Kevin's "99.99999%" as hyperbole.
@AndrejBauer: this is MathOverflow. Where numbers are numbers.
@TonyK: Oh please.
@AndrejBauer: OK, I think you must agree that Kevin Buzzard is saying that the chance that there exists an important research paper with a serious error is vanishingly small. And that in itself is delusional in my opinion.
You're probably right. But you know what the real problem is? It's not that there are probably faulty important papers out there, but rather that we have no idea how many or which ones.
"... experts will leave the area and slowly die out, and then all of a sudden someone will discover a hole which nobody currently alive knows how to fill, even though it might have been the case that experts could once do it." A natural solution to this problem is to incentivize grad students and advanced undergrads to writing papers filling in these gaps, e.g. for the journal of "Trivial Remarks" or "Not Research Level Questions" (or perhaps less self-depracting venues). MO rep or similar doesn't create this incentive because it doesn't count towards publications.
The problem that I have with contemporary research mathematics is that the proofs for important results are usually very complicated and involved. Actually, so complicated and involved that a generic research mathematician working in that area still has a certain chance of not spotting any crucial mistake or gap in the argument. This chance is actually so high, that people usually still do not believe in the argument if he / she says that 'he / she went through the proof and assured him- or herself that it is correct'. Only if an 'elder' says this, people actually start believing the argument.
As Kevin Buzzard himself admits in his answer, he somewhat exaggerated his point for effect.
However, I'd submit that if you were unsettled by his talk, then that's a good thing. I don't think that the proper reaction is to look for reassurance that mathematics really is fine, or that the problems are restricted to some easily quarantined corner.
Rather, I think the proper reaction is to strive for a more accurate view of the true state of the mathematical literature, and refuse to settle for comforting myths that aren't based on reality. Some of the literature is rock-solid and can stand on its own, much more of it is rock-solid provided you have access to the relevant experts, and some of it is gappy but we don't really care. On the other hand, some small percentage of it is gappy or wrong and we do care, but social norms within the mathematical community have caused us to downplay the problems. This last category is important. It is a small percentage, but from a scholarly point of view it is a serious problem, and we should all be aware of it and willing to acknowledge it. If, every time someone brings it up, we try to get them to shut up by repeating some "propaganda" that makes us feel good about mathematics, then we are not solving the problem but perpetuating it.
Some related concerns were raised over 25 years ago by Jaffe and Quinn in their article on Theoretical Mathematics. This generated considerable discussion at the time. Let me quote the first paragraph of Atiyah's response.
I find myself agreeing with much of the detail of the Jaffe–Quinn argument, especially the importance of distinguishing between results based on rigorous proofs and those which have a heuristic basis. Overall, however, I rebel against their general tone and attitude which appears too authoritarian.
My takeaway from this is that Jaffe and Quinn made many valid points, but because this is a sensitive subject, dealing with societal norms, we have to be very careful how we approach it. Given the way that the mathematical community currently works, saying that someone's work has gaps and/or mistakes is often taken to be a personal insult. I think that if, as a community, we were more honest about the fact that proofs are not simply right or wrong, complete or incomplete, but that there is a continuum between the two extremes, then we might be able to patch problems that arise more efficiently, because we wouldn't have to walk on eggshells.
I have had some push-back against the rather gung-ho approach I have been taking to highlight the problems I perceive with the way mathematics is currently being done, and I think Tim's post above does a very good job of highlighting the same issues but in a more balanced way. Thanks. Some of the issues I highlight in my talk are huge gaps in the literature which the experts have absolutely no doubt can be filled. I personally am uncomfortable about this but not everyone is. However there are other gaps, he said, treading on eggshells, where there might be more of a problem.
@KevinBuzzard : I do like the way you point out that everyone eventually dies and that therefore gaps which can be filled only by a tiny number of experts are at risk of becoming permanent gaps. I share your feeling that these sorts of gaps are worrisome. Maybe something can be done about them without ruffling too many feathers, because there's no implicit accusation of an incomplete argument, just a call that the experts perform "community service."
The classification of finite simple groups could serve as a model. Although you express concern that even the current plan might not be completed successfully, at least it's a huge improvement over the situation as it stood in, say, 1985. Credit should be given to the experts who acknowledged the need for a proper writeup and who dedicated an enormous amount of time and effort to the project.
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2025-03-21T14:48:29.750894
| 2020-01-31T18:03:25 |
351641
|
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|
Stack Exchange
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Vanishing of Hochschild homology of a category
Let $A$ be a dg- or $A_{\infty}$-category (with $\mathbb{Z}$-graded Hom sets, over a field of characteristic $0$). Let $HH_*(A)$ be the Hochschild homology of $A$.
Suppose that $HH_n(A)=0$ for all $n \in \mathbb{Z}$. Does this imply that $A$ is the zero category?
If not, then what assumptions can I add to $A$ to make this true (e.g. smoothness, properness, ect)?
Remark: I have heard the heuristic that Hochschild homology can be viewed as "differential forms" on the "spectrum" of the non-commutative category $A$ (this heuristic is presumably motivated by the Hochschild-Kostant-Rosenberg theorem). Hence it's natural to expect that $A$ vanishes if it admits no non-zero differential forms.
Note also that if $A$ is a (possibly non-commutative) $k$-algebra, then one can check that $HH_0(A) =0$ iff $A=0$.
This precise question was phrased as the vanishing conjecture in Hochschild homology and semiorthogonal decompositions. But we now know that there exist so called (quasi)phantom categories, which give counterexamples. These are categories with vanishing Hochschild homology, and vanishing (resp. torsion) Grothendieck group. As they are admissible subcategories of derived categories of smooth projective varieties, they have all nice properties you could want for their dg enhancements. An overview of some constructions:
Gorchinskiy, Sergey; Orlov, Dmitri, Geometric phantom categories, Publ. Math., Inst. Hautes Étud. Sci. 117, 329-349 (2013). ZBL1285.14018.
Böhning, Christian; Graf von Bothmer, Hans-Christian; Katzarkov, Ludmil; Sosna, Pawel, Determinantal Barlow surfaces and phantom categories, J. Eur. Math. Soc. (JEMS) 17, No. 7, 1569-1592 (2015). ZBL1323.14014.
Galkin, Sergey; Katzarkov, Ludmil; Mellit, Anton; Shinder, Evgeny, Derived categories of Keum’s fake projective planes, Adv. Math. 278, 238-253 (2015). ZBL1327.14081.
Galkin, Sergey; Shinder, Evgeny, Exceptional collections of line bundles on the Beauville surface, Adv. Math. 244, 1033-1050 (2013). ZBL1408.14068.
Böhning, Christian; Graf von Bothmer, Hans-Christian; Sosna, Pawel, On the derived category of the classical Godeaux surface, Adv. Math. 243, 203-231 (2013). ZBL1299.14015.
Alexeev, Valery; Orlov, Dmitri, Derived categories of Burniat surfaces and exceptional collections, Math. Ann. 357, No. 2, 743-759 (2013). ZBL1282.14030.
What is interesting is that Hochschild cohomology can detect their non-vanishing, and all kinds of interesting behavior regarding deformation theory arises, see
Kuznetsov, Alexander, Height of exceptional collections and Hochschild cohomology of quasiphantom categories, J. Reine Angew. Math. 708, 213-243 (2015). ZBL1331.14024.
Based on your comment regarding Hochschild COhomology detecting nonvanishing in the above examples, am I correct to infer that there are no known examples of non-zero categories with HH_(A)=HH^(A)=0? Are there any conjectures regarding whether such examples should exist?
If a category is non-zero then the zeroth Hochschild cohomology will be nonzero, as it contains the identity morphism between the diagonals.
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2025-03-21T14:48:29.751116
| 2020-01-31T18:05:54 |
351642
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|
Stack Exchange
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construct a bijective map between subsets of binary sequence
Consider the binary sequence $\{0,1\}^N$ where $N$ is an even integer (for simplicity). Let $M_k := \{\beta\in \{0,1\}^N \rvert \sum_{j=1}^N \beta_j = k\}$ (i.e., $M_k$ is the set that contains all binary sequences with length $N$ and with exactly $k$ 1's). The question is: for general $k \le N/2$, is it always possible to construct a bijective map $\phi: M_k\rightarrow M_{N-k}$ such that $\phi(\beta)_j \ge \beta_j$? (Here, $\phi(\beta)_j$ is the $j^{th}$ coordinate of the binary sequence $\phi(\beta)$.)
Obviously, both sets $M_k$ and $M_{N-k}$ have the same number of elements. When $k = 1$, the construction of $\phi$ is simple. However, for general $k$ and $N$, I am not sure how to systematically construct such a function $\phi$. Any hint or reference would also help a lot. Thank you.
The answer is "yes, this is always possible." The bijection on ranks $k$ and $N-k$ induced from a symmetric chain decomposition of the Boolean lattice gives you what you're asking for. For basics on this topic, see these slides.
Thank you very much, Sam. This looks like exactly what I need.
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2025-03-21T14:48:29.751241
| 2020-01-31T18:23:02 |
351646
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|
Stack Exchange
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The isometry group of a product of two Riemannian manifolds
Under what conditions is the isometry group of a product of two Riemannian manifolds the product of the isometry groups of each one of the components?
One counterexample is a product of two isometric manifolds, since the isometry contains the involution which cannot be a product. I was wondering if there are some general criterion.
Of course, $\ M^a\times M^b\ $ will have a bunch of additional isometries (where $M^n := M\times\ldots\times M$; -- $n$-fold product). Possibly, there are no other non-trivial examples.
The argument here is due to Mahan Mj:
If $A, B$ are compact and not isometric, then the isometry group is the product.
Here is the point: Let $X=A \times B$ be the product manifold with the product metric. Let $f: X \to X$ be an isometry. Then each $f(A \times \{b\})$ is totally geodesic for every b.
so that $X$ fibers over $B$ with fibers $f(A).$ Further there is a section of the bundle map through every point. These sections are isometric to $B.$ So unless the factors are interchanged (when $A$ and $B$ are isometric), the isometry must be a bundle map. covering some isometry of the base (you can prove this from the existence of sections).
It follows that if $A, B$ are not isometric, the isometry group is just the product. else there is a quotient map of $Isom(X)$ onto $\mathbb{Z}/2\mathbb{Z}$ with kernel $Isom(A) \times Isom (A)$ (when $A, B$ are isometric).
The argument fails when $A, B$ are not compact. Notice that the isometry group of the plane is far larger than the isometry group of the line squared.
I think you might be assuming that $A$ and $B$ are indecomposable when saying "unless the factors are interchanged (when $A$ and $B$ are isometric)", as pointed out by Wlod AA's comment above.
I think, I don't understand the argument. Where does it use that $A$ and $B$ are compact?
I was trying to understand Igor Rivin's answer and finally wrote a post on Thuses (a very nice young platform for math blogging, which I enthusiastically recommend). Actually, perhaps this is more or less the same proof, but with less things along the lines, at least, one can see explicitly, where irreducibility and compactness are used.
I tried commenting on Thuses, but it won't let me register. You define a map $\phi_b : A \to A$ by $\phi_b(a) = \pi_A(A'_b)$ which does not seem to be an element of $A$. Did you mean $\phi_b(a) = \pi_A(f(a,b))$?
You may find this answer of mine of some interest. There, I explain a special case when one can prove that $I(M \times N) \cong I(M) \times I(N)$ with simple differential geometric arguments involving sectional curvatures. And then (in the edit) I sketch a proof of a much more general result using some holonomy techniques taken from Kobayashi & Nomizu. In a nutshell, this result says that if you have a product of some irreducible Riemannian manifolds and at most one flat Riemannian manifold, then the isometry group of the product is the product of the isometry groups of the factors plus permutations of isometric factors (no compactness or even completeness assumption required). One can also try to drop the irreducibility assumption, but then to say something interesting about the isometry group and how it relates to those of the factors, one needs to add the assumptions of completeness and simply connectedness (at least I don't see how to proceed otherwise). Here is the resulting statement:
Proposition. Let $M = M_1 \times \ldots \times M_k$ be a Riemannian product of complete simply connected Riemannian manifolds. We have an obvious embedding of Lie groups $i \colon I(M_1) \times \ldots \times I(M_k) \hookrightarrow I(M)$. The following are equivalent:
$i$ is a local isomorphism, i. e. it gives an isomorphism on the identity components of the above groups: $I^0(M_1) \times \ldots \times I^0(M_k) \simeq I^0(M)$;
$\dim I(M) = \dim I(M_1) + \cdots + \dim I(M_k)$;
At most one of $M_i$'s has a nontrivial Euclidean de Rham factor.
Moreover, if these conditions are satisfied, then $i$ is an isomorphism if and only if there does not exist $i \ne j$ such that $M_i$ and $M_j$ share isometric de Rham factors. (If all $M_i$'s are irreducible, it means that no two of them are isometric.)
In order to prove this, just de Rham decompose each factor, stack all Euclidean subfactors together, and apply the second proposition from my answer linked above.
There is also the following perspective: assume that $M$ and $N$ are homogenous and denote by $H_M \trianglelefteq I(M)$ (resp. $H_N \trianglelefteq I(N)$) the isotropy group (a stabiliser of a point). Then $I(M) \times I(N)$ acts on $M\times N$ transitively and thus $I(M \times N) / I(M) \times I(N)$ maps to $H_{M\times N}$. In fact, it further projects to $H_{M \times N} / H_M \times H_N$, and I think this is an isomorphism. What you are actually proving is that if $M$ and $N$ have different curvature, $H_{M \times N} =H_N \times H_M$, right?
(Is it important that $M$ is of non-negative curvature and $N$ is flat? I'd suspect, the same must hold whenever $M$ and $N$ have curvature of different signs)
(upd: of course $H_M \subset I(M)$ and $H_N \subset I(N)$ are not normal, unless $M$ itself is a Lie group
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2025-03-21T14:48:29.751636
| 2020-01-31T18:33:12 |
351648
|
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Stack Exchange
|
Using irreducible characters of the orthogonal group as basis for homogeneous symmetric polynomials
The irreducible characters of the orthogonal group $O(2N)$ are given by
$$ o_\lambda(x_1,x_1^{-1},...x_N,x_N^{-1})=\frac{\det(x_j^{N+\lambda_i-i}+x_j^{-(N+\lambda_i-i)})}{\det(x_j^{N-i}+x_j^{-(N-i)})}$$
I was playing with them as basis for the space of homogeneous symmetric polynomials. I wanted to write the function
$$p_3=\sum_{i=1}^N (x_i^3+x_i^{-3})$$
as a linear combination of $o_\lambda$'s.
I started with $N=2$ and in that case I found that
$$ p_3=o_{(3)}-2o_{(2,1)}+o_{(1)}.$$
However, I then tried with $N=4$ and in that case I found that
$$ p_3=o_{(3)}-o_{(2,1)}+o_{(1,1,1)}.$$
I was expecting the coefficients to be the same, i.e. to be independent of $N$. (Independence of $N$ indeed holds when using characters of the unitary group, in which case the coefficients are the characters of the permutation group.)
Have I made some mistake or are the coefficients indeed dependent on $N$?
The coefficients do depend on $N$. A way to get around this and deal with "universal characters" was found by Koike and Terada (Young-diagrammatic methods for the representation theory of the classical groups of type $B_n$, $C_n$, $D_n$).
(I have looked more carefully at this theory of "universal characters" mentioned by Stanley and am updating this answer according to what I learned. All of this was contained in the answer by Stanley, I am just unpacking it for the sake of those like me and the OP who may be confused.)
The following material is in the paper Young-diagrammatic methods for the representation theory of the classical groups of type $B_n$, $C_n$, $D_n$ by Koike and Terada, also in works by King (such as Modification Rules and Products of Irreducible Representations of the Unitary, Orthogonal, and Symplectic Groups, Journal of Mathematical Physics 12, 1588, 1971) and a very readable 1938 book by Murnaghan (The Theory of Group Representations).
$\DeclareMathOperator\Or{O}\DeclareMathOperator\U{U}$So functions $o_\lambda$ correspond to irreducible characters of $\Or(2N)$ only when $\ell(\lambda)\le N$. The set of such actual characters forms a basis for the space of symmetric functions on variables $\{x_1,x_1^{-1},\dotsc,x_N,x_N^{-1}\}$. However, the coefficients in the expansion of power sums in general depend on $N$.
On the other hand, partitions with $\ell(\lambda)>N$ do not define irreducible representations, so $o_\lambda$ is not an actual character. In that case they are replaced by another symmetric function, $o_\lambda\to o_{\widetilde{\lambda}}$ with a modified partition $\widetilde{\lambda}$.
When these functions are used, the large-$N$ expansions continue to hold for small $N$. In that sense they are 'universal'.
When a Schur function $s_\mu$ is decomposed in terms of $o_\lambda$, which corresponds to the branching rule of $\U(2N)\supset \Or(2N)$, one has $\ell(\lambda)\le \ell(\mu)$, so if $\ell(\mu)\le N$ only actual characters appear. This is not true if $N<\ell(\mu)\le 2N$, and then universal characters must also be used. This is not discussed in the classic book "Representation theory" by Fulton and Harris, for example, where only the case $\ell(\mu)\le N$ appears.
The modified partition $\widetilde{\lambda}$ is defined as follows. For $O(2N)$, let $m=2\ell(\lambda)-2N$. Then remove from the Young diagram of $\lambda$ a total of $m$ adjacent boxes, starting from the bottom of the first column and keeping always at the boundary of the diagram. In this way the changes to be implemented are, in sequence, $\lambda'_1\to\lambda'_2-1$, then $\lambda'_2\to\lambda'_3-1$, etc. until the procedures stops at some column $c$. If $m$ is too large and there are not enough boxes to accommodate this procedure, or if the remaining diagram is not a partition, then $o_\lambda=0$; otherwise $o_\lambda=(-1)^{c-1}o_{\widetilde{\lambda}}$.
For example, the universal decomposition for $p_4$ is
$$p_4=o_4-o_{31}+o_{211}-o_{1111}+o_0.$$
If $\lambda=(1,1,1,1)$ and $N=6$, we must remove $8-6=2$ boxes, in which case we get $c=1$ and $\widetilde{\lambda}=(1,1)$. Hence, for $O(6)$ we have $o_{1,1,1,1}=o_{1,1}$ and the universal relation reduces to $p_4=o_4-o_{31}+o_{211}-o_{11}+o_0$. When $N=4$, we must remove $8-4=4$ boxes from $\lambda=(1,1,1,1)$. We end up with no boxes at all, so $o_{1,1,1,1}=o_\emptyset$ for $O(4)$. We must remove $6-4=2$ boxes from $\lambda=(2,1,1)$, arriving at $o_{2,1,1}=o_2$. The relation reduces to $p_4=o_4-o_{31}+o_{2}$. Finally, take $N=2$. We cannot remove $8-2=6$ boxes from $(1,1,1,1)$, so $o_{1,1,1,1}=0$ for this group; when we remove $4-2=2$ boxes from $(3,1)$, the result is not the diagram of a partition, so $o_{3,1}=0$ for this group; removing $6-2=4$ boxes from $(2,1,1)$ leads to the empty partition, and the removing procedure ends in the second column so $c=2$, hence $o_{2,1,1}=-o_\emptyset$ for this group. Thereby the relation reduces to $p_4=o_4$.
Unfortunately, the modification rule is incorrectly stated in the recent "The Random Matrix Theory of the Classical Compact Groups" (Cambridge University Press, 2019), by Elizabeth Meckes.
By the "character of a classical group," I mean a certain symmetric function that encodes the dimensions of the weight spaces as coefficients of monomials. It is these coefficients that depend on $N$ for $O(N)$ or $\mathrm{Sp}(2N)$. That is because the symmetric functions are being evaluated at $x_1,\dots,x_N,x_1^{-1},\dots,x_N^{-1}$ or $x_1,\dots,x_N,x_1^{-1},\dots,x_N^{-1},1$. For instance, the constant term of $e_2(x_1,\dots,x_N,x_1^{-1},\dots,x_N^{-1})$ is $N$, where $e_2$ is an elementary symmetric function.
@RichardStanley I don't understand what you mean. When we write $p$'s in terms of $s$'s, the coefficients don't depend on $N$. In other words the funcional relation is the same for any number of variables. The question was whether the same is true when $p$'s are written in terms of $o$'s.
@RichardStanley and of course this touches on the problem of the right expression for the $o$'s. The OP presents a formula which is incorrect according to the source I mentioned.
@RichardStanley By the way, according to my calculations, $e_2=o_2+o_{11}+1$ for any $N$.
I should have made my answer just a comment, since as Marcel notes it does not address the actual question.
I think it is worth pointing out that this sort of dependence on $N$ also happens for $GL_N$ if you consider algebraic, and not just polynomial representations. For example, if we compute the character of the adjoint representation we see it has constant term $N$.
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