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2025-03-21T14:48:29.521620
| 2019-12-24T16:13:36 |
349067
|
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"authors": [
"Iosif Pinelis",
"M. Dus",
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|
Stack Exchange
|
Supremum of $ a_n = a_{n-1}^3 - a_{n-2} $
Let $a_1=0$ and let $ - \ln(2) < a_2 < \ln(2) $
Define
$$ a_n = a_{n-1}^3 - a_{n-2} $$
Then
$$ \sup_{n>2} a_n = a_2 $$
And
$$ \inf_{n>2} a_n = - a_2 $$
How to prove that ?
Could you explain where this problem comes from and why you made such a conjecture ?
It looks like you need the condition $a_2\ge0$.
Why people are closing this question? It seems an interesting example to me
@EmanueleTron If Im correct a Lienard system is a differential equation. So you are saying this can be turned into a differential equation ? How ?
possibly related : https://math.stackexchange.com/questions/3363543/z-n1-v-z-n-5-v-1-z-n-with-v-3-06328648997749
It seems that $a_n = 2 a_{n-1}^5 - a_{n-2}$ satisfies the same property !
... but it does not !
The question is really about the iteration behaviour of the maps $(x,y) \mapsto (y, y^3-x)$ with various starting points. We have a fixed point $(0,0)$ and a $6$-cycle $$(1,0),(0, -1), (-1, -1), (-1, 0), (0, 1), (1, 1)$$
It appears that for something like $-0.797 < x < 0.797$, $(0,x)$ is on an invariant curve. For example, here are $10000$ iterates starting at $(0,0.796)$:
For starting values with slightly larger $x$, the process seems to become unstable.
Thank you Robert. You anticipated my 2 next questions. Me and my mentor ( tommy1729 who gave me this idea ) . The value ln(2) can indeed be stretched to a higher value. A first remark is that tommy noted that the computation or estimate of this number strongly depends on the numerical precision. With 8 digits or so it is hard to get above ln(2) ; you get the illusion of divergence. A closed form for this number is ofcourse desired ; the 1st followup question. Secondly is there convergence or divergence at that number ? Or stated differently : do all valid x lie on an open or a closed interval?
Merry Xmas from me and tommy. I wonder how it is possible to get a score of +6 but also 3 close votes :/ Does not make sense to me.
For $b:=a_2$, the GCD of the polynomials $a_{10}=a_{10}(b)$ and $a_9(b)+b$ is
$$b \left(b^2+1\right) \left(b^{78}-b^{76}+b^{74}-b^{72}-8 b^{70}+8 b^{68}-8 b^{66}+8 b^{64}+28 b^{62}-28
b^{60}+28 b^{58}-28 b^{56}-59 b^{54}+59 b^{52}-59 b^{50}+59 b^{48}+85 b^{46}-85 b^{44}+85 b^{42}-85
b^{40}-86 b^{38}+86 b^{36}-86 b^{34}+86 b^{32}+61 b^{30}-61 b^{28}+61 b^{26}-60 b^{24}-30 b^{22}+30
b^{20}-30 b^{18}+27 b^{16}+9 b^{14}-9 b^{12}+9 b^{10}-6 b^8-3 b^6+3 b^4-3 b^2+1\right).
$$
The only root of this GCD in $(0,1)$ is $b_*\approx0.637295\in(0,\ln2)$.
So, for $a_2=b_*$ we have $a_{10}=0=a_1$ and $-a_9=a_{11}=a_2$, so that the sequence $(a_n)$ is of period $9$; we also have $\max_n a_n=a_2$ and $\min_n a_n=-a_2$.
Thank you for your answer. There are probably infinitely many cycles. Marry Xmas.
I encountered a similar problem with the iteration $u_n=\sqrt{u_{n-1}^2+1}-u_{n-2}$, where there is fundamentally a $9$-cycle in the shape of a maple leaf
(replace the +1 by 0 to see this). I asked an expert 20 years ago who told me that using the
KAM (Kolmogorov-Arnold-Moser) theorem, one could prove that the "curve" drawn
above by R. Israel is not a curve at all, but a very thin strip of width 0.07
or something. Don't ask me for the proof, I have no idea.
Very interesting. Im not an expert with KAM, I only heard of it. The type of recursions that give these kind of behaviours like in the OP or the one you give are conjectured to have a pattern by my mentor.
However more investigation is needed even for the precise statement of the conjectures.
There might be a way without KAM. I will let you know if I can find such a way ( I and some friends have ideas that might work ).
Thank you for your contribution !! +1
|
2025-03-21T14:48:29.522139
| 2019-12-24T18:52:47 |
349072
|
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}
|
Stack Exchange
|
Maschler's bargaining set-an incomplete step in a proof
I have a problem with the concept of the bargaining set which is given below in some detail.
Let $N=\{1,\ldots,n\}$ be a set of players and $v:2^N\to \mathbb{R}$
a superadditive game (meaning $S,T \subseteq N ,\,S \cap T=\emptyset \Longrightarrow v(S)+v(T)\le v(S \cup T).$) in coalitional function form.
For $y\in\mathbb{R}^C$ and $S\subseteq N$ write $y(S)$ for $\sum_{i\in C}y_i.$ So $y(\emptyset)=0.$
Write ${\cal S_{kl}}=\{T \subseteq N\,|\,k \in T,\, l \notin T\}.$
Let $x\in \mathbb{R}^D$ be an individually rational imputation with $x(D)=v(D)$.
What is the relationship between the following two formulas-I'd like them to be equivalent $\forall y\in \mathbb{R}^C$:
$(\exists D \in {\cal S_{lk}})\big(D \cap C=\emptyset \;\wedge\; x(D)\leq v(D)\big) \quad\vee$
$\Big((\forall i \in C)(y_i > x_i)\quad\wedge$
$(\forall D \in {\cal S_{lk}})\big(x(D \setminus C)+y(D \cap C) > v(D)\big)$
$\ \Longrightarrow \ y(C)>v(C)\Big).$
and
$(\exists D \in {\cal S_{lk}})\big(D \cap C=\emptyset \;\wedge\; x(D)\leq v(D)\big) \quad\vee$
$\Big((\forall i \in C)(y_i \geq x_i)\quad\wedge$
$(\forall D \in {\cal S_{lk}})\big(x(D \setminus C)+y(D \cap C) \geq v(D)\big)$
$\ \Longrightarrow \ y(C)\geq v(C)\Big).$
We may assume that the first part before $\vee$ which is the same in both
is not satisfied.
The only difference is by replacing some $>$ with $\geq.$
The problem arises when $D\cap C=\emptyset,$ otherwise they are equivalent by increasing coordinates of $y$ and slight considertaions.
Perhaps here may be used the superadditivity of $v$?
I believe that the second is equivalent to this
$\Big(k,l\in N \;\wedge\; k \neq l \;\wedge\; C \in {\cal S_{kl}} \quad \Longrightarrow$
$(\exists D \in {\cal S_{lk}})\big(D \cap C=\emptyset \;\wedge\; x(D)\leq v(D)\big)\quad\vee$
$ \min\big\{y(C)\,|\, x(D \setminus C) +y(D \cap C) \geq v(D),\,
D \in {\cal S_{lk}},\, y_i \geq x_i,\, i \in C \big\} \geq v(C)\Big),$
which is my aim to obtain from the first displayed formula with strict
inequalities which are problematic to obtain these weak inequalities and $\min$. The entire thing here is to obtain weak linear inequalities connected by Boolean connectives that determine the Maschler bargaining set which is a quite interesting concept in cooperative game theory. The formulas are supposed to describe the bargaining set $\cal M$, which is such that there is no justified objection of a player $k$ to $l$. And it is of the form
finite union of polytopes in $\mathbb{R}^N.$
|
2025-03-21T14:48:29.522319
| 2019-12-24T19:25:05 |
349074
|
{
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|
Stack Exchange
|
An absolute value eigenvalue question
Denote by $\circ$ the Kronecker product, let $|\cdot|$ denote the matrix/vector of absolute values, and let $e$ be the vector of all ones. Comparison is entrywise, i.e., $y \ge x$ is equivalent to $y_i \ge x_i$ for all $I$.
Problem 1: Let $A$ be a real $n\times n$ matrix, $n>1$, and assume
$$
(A \circ A)e = (n-1)e.
$$ Then there exists $x \neq 0$ with $|Ax| \ge |x|$.
A weaker formulation (Problem 2):
Assume
$$
|Ae| = n e.
$$
Then there exists $x \neq 0$ with $|Ax| \ge |x|$.
The first problem is sharp in the sense that assuming $(A \circ A)e = (n-1)e$, then there exists no $x \neq 0$ with $|Ax| > |x|$.
Equality, i.e., there exists $x \ne 0$ with $|Ax| \ge |x|$, is true for the matrix with zero on, $+1$ above and $-1$ below the diagonal, i.e. for $A=(a_{ij})$ with
$$
a_{ij}=
\begin{cases}
+1 & i<j\\
0 & i=j\\
-1 & i>j\\
\end{cases}
$$
as you stated it, $e$ would be a solution to Problem 2, right?
|
2025-03-21T14:48:29.522424
| 2019-12-24T19:57:06 |
349076
|
{
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"authors": [
"Benjamin Steinberg",
"Julian Quast",
"Martin Brandenburg",
"Profinite Questioner",
"Tim Campion",
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|
Stack Exchange
|
Is there a free profinite abelian group on a profinite set?
Let $\mathit{Profinite}_{\mathrm{Ab}}$ be the category of profinite abelian groups, and let $\mathit{Profinite}_{\mathrm{Set}}$ be the category of profinite sets. Does the forgetful functor
$$\mathit{Profinite}_{\mathrm{Ab}} \to \mathit{Profinite}_{Sets}$$
admit a left adjoint?
I am a beginner to this kind of question; I do not even know if both the domain and codomain categories admit all small colimits.
Changed using "mathit", which corrects "$Profinite$" to "$\mathit{Profinite}$".
Just using the universal properties, one sees that the profinite completion $G_I$ of the free abelian group $\mathrm{Z}^{(I)}$ is naturally the free profinite abelian group over the set $I$, in the sense that every map $I\to H$ for $H$ a profinite abelian group uniquely extends to a continuous homomorphism $G_I\to H$. (I see this is not the asked question, this is just a remark.)
To adapt to the question, I guess that $I$ being profinite, one can start from $\mathbf{Z}^{(I)}$ and call a congruence subgroup, a subgroup $L$ of finite index such that the composed map $I\to \mathbf{Z}^{(I)}/L$ factors through some finite quotient of the profinite set $I$. Then the pro-congruence completion of $\mathbf{Z}^{(I)}$ (which is a quotient of the profinite completion) ought to be the required free profinite abelian group over the profinite set $I$.
I think for some profinite set $S = \varprojlim S_i$, you can take $\varprojlim_i \widehat{\mathbb Z}[S_i]$.
Yes, the free functor (i.e. the left adjoint to the "forgetful" functor) exists.
Let $Ab^{fin}$ be the category of finite abelian groups and $Set^{fin}$ the category of finite sets. Because each of these categories are essentially small and have finite limits, the categories of pro-objects $Pro(Ab^{fin})$ and $Pro(Set^{fin})$ are complete and cocomplete. In fact, they are opposite to locally finitely presentable categories -- if $C$ is essentially small with finite limits then $Pro(C) \simeq Fun^{lex}(C,Set)^{op}$ where $Fun^{lex}$ denotes the category of finite-limit-preserving functors. So equivalently, we're asking whether the forgetful functor $Fun^{lex}(Ab^{fin}, Set) \to Fun^{lex}(Set^{fin}, Set)$ has a right adjoint.
The forgetful functor $U_!: Fun^{lex}(Ab^{fin}, Set) \to Fun^{lex}(Set^{fin}, Set)$ is by definition the left Kan extension $Lan_{y_{Ab^{fin}}} (y_{Set^{fin}}\circ U^{op})$ where $y_C: C^{op} \to Fun^{lex}(C,Set)$ is the co-Yoneda embedding and $U: Ab^{fin} \to Set^{fin}$ is the forgetful functor. If we look at the full presheaf categories $Fun(C,Set)$ rather than $Fun^{lex}(C,Set)$, the right adjoint to this exists and is given by the functor $U^\ast: Fun(Set^{fin},Set) \to Fun(Ab^{fin}, Set)$ given by precomposition by the forgetful functor $U: Ab^{fin} \to Set^{fin}$.
Since $U: Ab^{fin} \to Set^{fin}$ preserves finite limits, $U^\ast$ restricts to a functor $Fun^{lex}(Set^{fin}, Set) \to Fun^{lex}(Ab^{fin}, Set)$, and because $Fun^{lex}$ is a full subcategory of $Fun$, the same calculation as at the link above shows that this functor is the adjoint we are searching for.
Why does Fun^{lex} admit all small colimits? I understand it admits filtered colimits (which commute with finite limits in Set) but how do I compute a finite colimit in Fun^{lex}?
@ProfiniteQuestioner It is a reflective category of the presheaf category.
I am missing something basic here. Do you conclude the existence of the left adjoint (to the inclusion of Fun^{lex} into Fun) by (i) constructing it (if so, how?), or by (ii) proving that Fun^{lex} admits all small colimits (if so, how?) and invoking the adjoint functor theorem, or (iii) exhibiting Fun^{lex} of a localization of Fun along certain morphisms (if so, how?). I am sorry these are such basic questions.
Using the profinite completion of the free abelian group as suggested by @YCor, seems a simpler solution.
@ProfiniteQuestioner Where are you learning about profinite groups and profinite sets? Do they not cover such facts? My reference is Adamek and Rosicky's Locally Presentable and Accessible Categories, though that's not the most direct reference
@BenjaminSteinberg Perhaps, but the description given above is exceedingly simple: the "free" functor, regarded as a functor $Fun^{lex}(Set^{fin}, Set)^{op} \to Fun^{lex}(Ab^{fin}, Set)^{op}$, is just precomposition with the forgetful functor $U: Ab^{fin} \to Set^{fin}$.
|
2025-03-21T14:48:29.522745
| 2019-12-24T20:55:29 |
349082
|
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"Jeremy Rickard",
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}
|
Stack Exchange
|
MAGMA-question concerning dual modules of bimodules
Let $G$ be a finite group and let $H_1,H_2\leq G$.
Let char$(k)=p>0$, $k$ a field, large enough.
Let $T$ be a $(kH_1, kH_2)$-bimodule given in MAGMA.
Moreover, let $T$ be finitely generated projective as left $kH_1$-module.
Then, $T$ has a left dual $\widetilde{T}:=\text{Hom}_{kH_1}(T,kH_1)$.
I'd like to ask the following question:
Is there a way to access / construct this dual module $\widetilde{T}$ as $(kH_2,kH_1)$-bimodule with MAGMA?
Thanks in advance for the help.
What you call the left dual is isomorphic as a bimodule to the vector space dual $\text{Hom}_k (T,k)$. Does that help?
|
2025-03-21T14:48:29.522817
| 2019-12-24T21:42:36 |
349087
|
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"site": "mathoverflow.net",
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"url": "https://mathoverflow.net/questions/349087"
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|
Stack Exchange
|
Inequalities $\pi(x^a+y^b)^\alpha\leq \pi(x^c)^\beta+\pi(y^d)^\gamma$ involving the prime-counting function, where the constants are very close to $1$
Let $\pi(x)$ be the prime-counting function, I'm curious about if a suitable variant of the second Hardy–Littlewood conjecture (this corresponding Wikipedia)
$$\pi(x^a+y^b)^\alpha\leq \pi(x^c)^\beta+\pi(y^d)^\gamma\tag{1}$$
can be proved, where the constants $0<a,b,c,d\leq 1$ and the constants $0< \alpha,\beta,\gamma\leq 1$ are very close to our upper limit $1$, for all real numbers $x<y$ with $L<x$ for a suitable choice of a constant $L$.
Question. Is it possible to prove any statement of the type $(1)$ under the cited requirements, for constants $0<a,b,c,d\leq 1$ and constants $0< \alpha,\beta,\gamma\leq 1$ all these (all together/
simultaneously) very close to $1$, for all real numbers $x<y$ for a suitable $L<x$? Many thanks.
I don't know if this type of proposals $(1)$ are in the literature, or are essentially the same original second Hardy–Littlewood conjecture, when we require that those constants are very close to $1$.
If there is relevant literature answer my question as a reference request and I try to search and read those statements from the literature.
References:
[1] G. H. Hardy and J. E. Littlewood, Some problems of ‘Partitio numerorum’ III: On the expression of a number as a sum of primes, Acta Math. (44): 1–70 (1923).
I'm waiting to know if this question is potentitally interesting, feel free to criticize it, it is very useful to me with the purpose to avoid ask bad questions. Many thanks for help and the patience of all users.
I've undeleted this post, that I've deleted two years ago. Can you provide feedback a about if the question is potentially interesting? Many thanks.
|
2025-03-21T14:48:29.522956
| 2019-12-24T22:56:47 |
349090
|
{
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"Anthony Quas",
"https://mathoverflow.net/users/11054"
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"url": "https://mathoverflow.net/questions/349090"
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|
Stack Exchange
|
Name for a probability density ''symmetrized'' by a permutation group?
Let $p$ be a probability density function over random variable $X$, and $G$ a compact permutation group over the outcomes of $X$. For each $g\in G$, let $p_g$ indicate the probability density function of the random variable $g(X)$. Now, imagine that one "symmetrizes" $p$ by mixing together the densities induced by each $g\in G$,
$$
\tilde{p}(x)=\int_G p_g(x)\; d\mu
$$
where $\mu$ indicates the Haar measure of $G$. Is there a name/reference describing such an operation?
conditional expectation over the $G$-invariant sigma-algebra?
|
2025-03-21T14:48:29.523032
| 2019-12-25T01:24:48 |
349091
|
{
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"Andreas Blass",
"darij grinberg",
"https://mathoverflow.net/users/2530",
"https://mathoverflow.net/users/44191",
"https://mathoverflow.net/users/6794",
"user44191"
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"url": "https://mathoverflow.net/questions/349091"
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|
Stack Exchange
|
What is this category of circles?
I am working on a problem in Analysis and somehow I arrived to the following category that I suspect is isomorphic to the category of finite dimensional Hilbert spaces. Sorry if my description sounds a bit vague, I know very little about Category Theory.
Objects are finite collections of oriented circles in the plane. Each circle has an identity map and an associated inversion map in the sense of classical planar geometry.
Morphisms are a bit more difficult to specify:
Suppose that we have two systems, each of them consisting of just one circle, say $C_{p,r}$ with center $p$ and radius $r$, and $C_{q,s}$ with center $q$ and radius $s$. The circles $C_{p,r}$ and $C_{q,s}$ have associated reflections $R_{p,r}$ and $R_{q,s}$, which are maps defined everywhere on the plane but on $p$ and $q$, respectively. One can show that, for every point $x$ different from $p$ and $q$, we have:
$$R_{p,r}\circ R_{q,s}(x) = R_{q,s}\circ R_{p,r}(x).$$
This allows us to define a morphism $T_{\{(p,r),(q,s)\}}$ between $C_{p,r}$ and $C_{q,s}$ equal to either of the expressions above, but it does not seem possible to set a source and a target for it, i.e., is it an arrow from $C_{p,r}$ to $C_{q,s}$ or vice--versa?
However, note the following: let $\mathcal{C}$ be a "reference circle'' with its own reflection $\mathcal{R}$. Then it seems that the category can be localized or something similar. The maps
$$T_{\{(p,r),(q,s)\}}$$
and
$$T_{\{(p,r),(q,s)\}}\circ\mathcal{R}$$
are now different and we can say that one of them takes the reference circle $\mathcal{C}$ to $C_{p,r}$ and the other one takes $C_{q,s}$ to $\mathcal{C}$. I am not so sure about this last statement.
In this category there is also a dagger $^\dagger$ and an inner product $\langle\cdot,\cdot\rangle_\mathcal{C}$, both induced by the choice of $\mathcal{C}$. The notation $\langle\cdot,\cdot\rangle_\mathcal{C}$ reminds me of an inner product on the tangent space.
Your morphisms can have sources and targets - have two morphisms: one with source $C_{p, r}$ and target $C_{q, s}$ and one the other way. The category of sets with bijections as morphisms is a perfectly fine category, and still has sources and targets, because each bijection (in a sense) appears twice. However, it looks to me like the category, if anything, has only one object - the plane - and morphisms that are symmetries of the plane (generated by inversions).
Are you saying any two inversions commute? Because that's not true.
In addition to the error pointed out by @darijgrinberg, the question says nothing about morphisms between objects of more than one circle, gives no information about composing morphisms, and gives no idea why the OP thinks this should have anything to do with the category of finite-dimensional Hilbert spaces. I'll vote to close as unclear.
|
2025-03-21T14:48:29.523250
| 2019-12-25T12:32:11 |
349102
|
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|
Stack Exchange
|
Rankin-Selberg convolution and product of degrees as of Christmas 2019
Almost 5 years ago (time flies), I asked in Rankin-Selberg convolution and product of degrees whether the Rankin-Selberg convolution of two automorphic representations of respectively $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ and
$\operatorname{GL}_{n'}(\mathbb{A}_{\mathbb{Q}})$ gave rise to an automorphic representation of $\operatorname{GL}_{n.n'}(\mathbb{A}_{\mathbb{Q}})$. Paul Garrett answered it by giving the known cases where this was proven at that time.
Have there been breakthroughs so far getting us any closer to such a general result?
Edit August 12th 2022: what about now? I would be especially interested in the case where the degree of the Rankin-Selberg convolution L-function is a prime power, that is $d_{F^{\otimes k}}=(d_{F})^{k}$ with $d_{F}\in\mathbb{P}$. A proof of this equality would entail that the following weakening of Goldbach's conjecture: "every composite even integer is the sum of 2 prime powers" implies that an L-function of composite even degree is a twist of a non primitive L-function.
I don't think so. It's a hard problem.
I think the general solution of this problem lies a few dozen Fields medals away at least, to paraphrase Jack Schwartz's famous remark about artificial intelligence. (This would also apply to the prime-power-degree special case you're hoping for, which doesn't actually seem to have any reason to be easier than the general case.)
Newton and Thorne proved that if $\pi$ is a cuspidal automorphic representation of $\mathrm{GL}_2(\mathbb{A}_{\mathbb{Q}})$ corresponding with a holomorphic cuspidal newform of even integral weight $k\geq 2$, squarefree level, and trivial central character, then for each $n\geq 1$, the $n$-th symmetric power lift $\mathrm{Sym}^n \pi$ is a cuspidal automorphic representation of $\mathrm{GL}_{n+1}(\mathbb{A}_{\mathbb{Q}})$. We have the standard identity
$\mathrm{Sym}^n \pi\otimes\mathrm{Sym}^n\pi = \boxplus_{j=0}^n \mathrm{Sym}^{2j}\pi$,
so by Newton--Thorne, this is an isobaric automorphic representation of $\mathrm{GL}_{(n+1)^2}(\mathbb{A}_{\mathbb{Q}})$. This is not yet known to hold for $\pi$ corresponding with Hecke--Maass forms. Thus this is a thin set of examples, but I find it noteworthy nonetheless. This showed up on the arxiv the day before you posted your question.
Thank you very much for your answer. Are there known important properties, even conjectural, that are stable under RS convolution ?
|
2025-03-21T14:48:29.523541
| 2019-12-25T13:55:05 |
349106
|
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|
Stack Exchange
|
finite subgroups of discrete arithmetic groups
Let K be a totally real multi-quadratic fields and let $\mathcal{O}$ be its ring of integers. I would like to compute the orders of the finite subgroups of the discrete group $\mathrm{SL}_{2}(\mathcal{O})$. These orders are helpful for computing Euler characteristics.
I would like to know whether there is any methods to do the computations and where to find them.
Thanks for all.
the finite subgroups are all abelian
|
2025-03-21T14:48:29.523617
| 2019-12-25T14:15:18 |
349107
|
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|
Stack Exchange
|
Equivariant cohomology algebra of toric variety
Let $X$ be a complex projective and smooth toric variety of complex dimension $n$. It is acted by the real torus $T=(S^1)^n$.
Is it true that the $T$-equivariant cohomology $H^*_T(X,\mathbb{Z})$ of $X$ is isomorphic as a graded algebra to $$H^*(X,\mathbb{Z})\otimes H^*_T(pt,\mathbb{Z})\simeq H^*(X,\mathbb{Z})\otimes \mathbb{Z}[x_1,\dots,x_n],$$
where $x_i$ have degree 2? (to prove such an isomoprhism with coefficients in a field would be fine too.)
Remark. I think I have a proof of such an isomorphism (with $\mathbb{Q}$-coefficients) in the category of graded $H_T^*(pt, \mathbb{Q})$-modules rather than graded algebras (which seems to be well known).
Isn't this false for $\Bbb P^1$? There we have $H^_{S^1}(\Bbb P^1) = H^(\Bbb{CP^\infty} \vee \Bbb{CP}^\infty)$ (as that wedge is the Borel construction on $\Bbb P^1$) which has cohomology $\Bbb Z[x,y]/(xy)$. In particular, it has no nilpotent elements, unlike your tensor product.
As graded $H^\ast_T(pt,\mathbb Q)$-modules it is just the fact, that the action of $T$ on $X$ is Hamiltonian with respect to the invariant Kähler structure. This implies that the action is equivariant formal, which means your isomorphism is an isomorphism of $H^\ast_T(pt,\mathbb Q)$-modules.
@MikeMiller: Why not make your comment into an answer? I'd be interested to know how you see the homotopy equivalence you mentioned.
As per Mark Grant's suggestion, here are some additional details for my comment.
We may write $\Bbb P^1$ with its circle action as the adjunction space $$* \cup_{S^1 \times 0} S^1 \times [0,1] \cup_{S^1 \times 1} *,$$
with the circle acting in the obvious way on $S^1 \times [0,1]$. Passing to the Borel construction, we find that $$\Bbb P^1 \times_{S^1} ES^1 = BS^1 \cup_{ES^1 \times 0} ES^1 \times [0,1] \cup_{ES^1 \times 1} BS^1.$$
The picture is that we are connecting two copies of $BS^1$ by a contractible bit.
To make this precise, collapse the closed contractible subspace $ES^1 \times 1/2$ to a point. As this is contractible, the projection map is a homotopy equivalence; thus $\Bbb P_{hS^1}$ is homotopy equivalent to the wedge of two mapping cones $$\text{Cone}(ES^1 \to BS^1) \vee \text{Cone}(ES^1 \to BS^1).$$ Because $ES^1$ is contractible, each of these cones are homotopy equivalent to $BS^1$ itself.
Thus we see that $$\Bbb{CP}^\infty \vee \Bbb{CP}^\infty \simeq \Bbb P^1_{hS^1};$$ further the projection map $\Bbb P^1_{hS^1} \to \Bbb{CP}^\infty$ sends the two wedge summands identically onto $\Bbb{CP}^\infty$.
Thus $$H^*_{S^1}(\Bbb P^1;\Bbb Z) \cong \Bbb Z[x,y]/(xy),$$ with action of $u \in H^2_{S^1}(pt)$ given by $u \cdot x = x^2, u \cdot y = y^2.$ This is isomorphic as a graded module to $$H^*(\Bbb P^1;\Bbb Z) \otimes H^*_{S^1}(pt;\Bbb Z),$$ but not as an algebra: the equivariant cohomology has no nilpotent elements, whereas the tensor-product algebra does, given by the generator of $H^2(\Bbb P^1) \otimes H^0_{S^1}(pt)$.
A previous version of this answer just wrote down a map $$\Bbb{CP}^\infty \vee [0,1] \vee \Bbb{CP}^\infty \hookrightarrow \Bbb P^1_{hS^1}$$ which Mayer-Vietoris and van Kampen show is an iso on $\pi_1$ and homology, so is a homotopy equivalence by Hurewicz and Whitehead. This seemed cleaner, though.
|
2025-03-21T14:48:29.523828
| 2019-12-25T15:22:10 |
349109
|
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|
Stack Exchange
|
Non-isomorphic curves with isomorphic Jacobians
There are many examples of pairs of non-isomorphic curves of genus 2 or 3 whose Jacobians are isomorphic as (unpolarized) Abelian varieties, see e.g. this post and its answers. This is relatively easy because almost all principally polarized Abelian varieties of dimension 2 and 3 are Jacobians. Ciliberto and Van der Geer have constructed families of examples in genus 4 (in Classication of Algebraic Varieties, Contemp. Math. 162, AMS (1994)). Does anyone know examples in genus $\geq 5$?
|
2025-03-21T14:48:29.523896
| 2019-12-25T15:47:15 |
349110
|
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|
Stack Exchange
|
Real part of entire function property
Is there any characterization of the set of entire functions $f(z) $ such that $\Re(f(z)) \geq \Re(\overline{f(\bar{z})})$ for all $z\in \mathbb{C}^{+} $?
($\Re$ stands for the real part)
Edit: I was thinking in writing $f(z) $ in the form
$$f(z) =f(x+iy) =u(x, y) +iv(x, y) $$
and
$$\overline{f(\bar{z})}=\overline{f(x-iy)} =u(x, - y) - iv(x, - y) $$
so, it is like finding all such continuous functions $u$ for which
$$u(x, y) \geq u(x, - y), \;\;\forall\; y>0, x\in\mathbb {R} $$
Your last inequality implies that $v(x,y):=u(x,y)-u(x,-y)\geq 0,\; y>0$,
and similarly $v(x,y)\leq 0,\; y<0$. Since $v$ is harmonic, this easily implies that $v(x,y)=cy$
for some real constant $c$. Now $w(x,y):=u(x,y)-cy/2$ will satisfy $w(x,y)=w(x,-y)$
This implies $(\partial w/\partial y)(0,y)=0$ and by Cauchy-Riemann,
the conjugate function $w^*$ to $w$ is constant on the real line.
Since $(u+iu^*)(z)=w+iw^*(z)+ciz/2$,
we obtain a characterization: these are exactly those entire functions whose imaginary part is of the form $a+bx$ on the real line, where $a$ and $b$ are constants, $b$ is real. (And your original inequality is
in fact always equality). Or in other words, the general form of such $f$ is $f=g+a+biz$ where $g$ is real on the real line.
Remark. I used the fact that a harmonic function which is positive for $y>0$ and
negative for $y<0$ must be $cy$. This is easy to prove, but is also a special case of a general theorem describing meromorphic functions with
positive real part in the upper half-plane and negative real part in the lower half-plane, see for example Levin, Distribution of zeros of entire functions, Ch. VI, sect. 1, Them 1.
thank you, this is true, and contains more. The function $f(z) =-iz=-ix+y$ satisfy the condition above but it is not real for real $z$.
@Guest: thanks, I corrected.
the content $b$ depends on the choice of the function $g$, is this wright?
@Guest: I made one more small correction, hopefully the last one. Now $b$ does not depend on he choice of $g$, only on $f$ itself. If you impose the condition that $a$ is pure imaginary, (include real part of $a$ into $g$) then the representation becomes canonical: $g$, $a$ and $b$ depend only of $f$.
Ok. I only have a question about the first line when you defined $v$: I know that $v(x, y)=\Im(f(z)) =\frac{f(z) - \overline {f(z)}} {2i}$. So, is your $v$ is the same as the imaginary part of $f$? According to what you wrote it is like having $v(x, y)=\frac{f(z) - \overline {f(\bar{z} )}} {2i}$!!!
@Guest: a priori this is not clear. I DEFINED $v=\Re(f(z))-\Re(f(\bar{z})$. Only in the end, as a consequence of the proof it turns out that this is the imaginary part.
|
2025-03-21T14:48:29.524086
| 2019-12-25T15:52:41 |
349112
|
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|
Stack Exchange
|
Hadamard lemma without integration
Let $I$ be the ideal of smooth germs vanishing at zero. Let $I^{k+1}$ be the ideal generated by $(k+1)$-fold product of such germs. Write $F_k$ for the ideal of $k$-flat germs at zero.
By the product rule $I^{k+1}\subset F_k$. The converse inclusion is usually called Hadamard's lemma, and asserts for $k=1$ that a smooth germ $f$ may be written $f(x) = f(0) + \sum_{i=1}^n x_i g_i(x)$ with smooth $g_i$.
The only proof I know uses integration.
Question. Is it possible to prove Hadamard's lemma (the equality $I^{k+1}= F_k$) without integration?
|
2025-03-21T14:48:29.524153
| 2019-12-25T16:41:05 |
349116
|
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|
Stack Exchange
|
Is there a Calabi-Yau threefold with $h^{1,1}=1$ and $h^{1,2}=0$?
Is there a Calabi-Yau threefold with $h^{1,1}=1$ and $h^{1,2}=0$?
As of three years ago no such CY threefold was known to the authors of https://arxiv.org/pdf/1602.06303.pdf.
|
2025-03-21T14:48:29.524199
| 2019-12-25T19:14:09 |
349122
|
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|
Stack Exchange
|
Are the L-functions of a normalized newform and the corresponding cuspidal representation equal?
Let $f \in S_k(\Gamma_0(N))$ be a normalized newform with Fourier expansion
$$f(z) = \sum\limits_{n=1}^{\infty} a_n e^{2\pi i z n}$$
and $a_1 = 1$. Then $f$ is an eigenfunction of all Hecke operators $T_p$, not just those with $(p,N) = 1$, and for the normalized L-function
$$L^{S_{\infty}}(f,s) = \sum\limits_{n=1}^{\infty} \frac{n^{(1-k)/2}a_n}{n^s}$$
one can add in an archimedean factor to make a completed L-function $L(f,s)$ which is an Euler product
$$L(f,s) = \prod\limits_{p \leq \infty} L_p(f,s)$$
and, for a suitable "contragredient" $g \in S_k(\Gamma_0(N))$ and epsilon root number $\epsilon(f,s)$, satisfies the functional equation
$$L(f,s) = \epsilon(f,s) L(g,1-s).$$
See for example Kowalski's article "Classical Automorphic Forms" in the book "An Introduction to the Langlands Program."
Now, $f$ can be identified in various ways with a cuspidal automorphic form on $\operatorname{GL}_2(\mathbb Q) \backslash \operatorname{GL}_2(\mathbb A_{\mathbb Q})$. There is a cuspidal automorphic representation $\pi = \otimes \pi_p$, unique up to isomorphism, which contains $f$. The normalized newform $f$ is conversely uniquely determined by $\pi$. This is one way to state the "Multiplicity One" theorem.
Also, $\pi$ itself has an L-function $L(\pi,s) = \prod\limits_{p \leq \infty} L(\pi_p,s)$ satisfying its own functional equation $L(\pi,s) = \epsilon(\pi,s) L(\tilde{\pi},1-s)$.
At the primes $p$ not dividing $N$, the representation $\pi_p$ is spherical, and the L-function $L(\pi_p,s)$ arises from the local spherical Hecke algebra of $\operatorname{GL}_2(\mathbb Q_p)$. Here things can be normalized so that
$$L(\pi_p,s) = L_p(f,s). \tag{1} $$
What about the primes dividing $N$? Here $L_p(f,s)$ arises from a Hecke operator $T_p$, but the representation $\pi_p$ could be supercuspidal, or it could have an Iwahori-fixed vector but be non-spherical. With the proper normalizations, can we have equation (1) holding for all primes $p$ (and consequently $L(\pi,s) = L(f,s)$)? Note that when $\pi_p$ is supercuspidal or is induced from a ramified character, $L(\pi_p,s) = 1$.
I expect it's possible to normalize things so that the local factors of $L(f,s)$ and $L(\pi,s)$ agree at all places, but I have never seen any reference which does this.
You know the exact functional equation satisfied by $L(f,s)$ from the Fricke involution, you want $L(s,\pi)$ to satisfy a similar one, which implies they are equal (since changing finitely primes adds some Euler factors in the functional equation), morally being left invariant under $GL_2(\Bbb{Q})$ is really the functional equation.
$L(\pi,s)$ agrees with $L(f,s)$ if $f\in\pi$ is a newform, and this is even true for $\mathrm{GL}_n$. Of course, things are complicated by the fact that there are many ways to define $L(\pi,s)$ and $L(f,s)$. I found the following papers very useful to check various consistencies: Schmidt and Kondo-Yasuda.
Here is a summary based on Kondo-Yasuda. Let $G:=\mathrm{GL}_n$. Let $(\pi,V_\pi)$ be a cuspidal automorphic representation of $G$ over $\mathbb{Q}$ of unitary central character $\omega$. Let $K_p(c)$ denote the subgroup of elements of $G(\mathbb{Z}_p)$ whose bottom row is congruent to $\begin{pmatrix}0 & \cdots & 0 & 1\end{pmatrix}$ modulo $c$. Let $K(c):=\prod_p K_p(c)$. The conductor $c_\pi$ is the smallest $c$ such that $V_\pi^{K(c)}\neq\{0\}$. An element of $V_\pi^{K(c_\pi)}$ is called a global newform: it is an eigenfunction of the convolution by any element of $C_c(K_p(c_\pi)\backslash G(\mathbb{Q}_p)/K_p(c_\pi))$. Let $c_\omega$ be the conductor of $\omega$, and consider the corresponding primitive Dirichlet character modulo $c_\omega$:
$$\chi_\omega(m):=\omega(1,\underbrace{1,\dotsc,1}_{v\mid c_\omega},\underbrace{m,m,\dotsc}_{v\nmid c_\omega})=
\omega(1,\underbrace{m^{-1},\dotsc,m^{-1}}_{v\mid c_\omega},\underbrace{1,1,\dotsc}_{v\nmid c_\omega}),$$
for $m>0$ and $(m,c_\omega)=1$. Note that $c_\omega\mid c_\pi$. Let $\chi_\pi$ denote the Dirichlet character modulo $c_\pi$ induced by $\chi_\omega$.
For $k\in\{1,\dotsc,n-1\}$, we define the $k$-th Hecke operator at $p$ as the characteristic function of
$$H_k(p):=K_p(c_\pi)\,\mathrm{diag}(\underbrace{p,\dotsc,p}_\text{$k$ entries},\underbrace{1,\dotsc, 1}_\text{$n-k$ entries})\,K_p(c_\pi).$$
Accordingly, let $\lambda_{\pi,k}(p)$ be the $k$-th normalized Hecke eigenvalue at $p$:
$$\int_{H_k(p)}f(xg)\,dg=\lambda_{\pi,k}(p)p^{\frac{k(n-k)}{2}}\mathrm{vol}(K_p)f(x),\qquad f\in V_\pi^{K(c_\pi)}.$$
Further, let $\lambda_{\pi,0}(p):=1$ and $\lambda_{\pi,n}(p):=\chi_\pi(p)$.
Theorem (Tamagawa 1963, Satake 1963, Kondo-Yasuda 2010).
$$L(\pi,s)=\prod_p\left(\sum_{k=0}^n(-1)^k\lambda_{\pi,k}(p)p^{-ks}\right)^{-1}.$$
Thanks very much for your answer. I assume the archimedean equality $L_{\infty}(f,s) = L(\pi_{\infty},s)$ is easy to check ($\pi_{\infty}$ for a newform should be like a discrete series representation)
@D_S: Yes. As a rule of thumb, as long as the $L$-function "looks good", i.e., it has an Euler product decomposition and a functional equation of the usual shape, "it is good". If the definition is not OK, it will show up as a problem in the analytic properties.
For $L(\pi_\infty,s)$ I recommend Tate's article "Number theoretic background" in Volume 2 of Borel-Casselman (eds.): Automorphic forms, representations, and $L$-functions.
|
2025-03-21T14:48:29.524519
| 2019-12-25T19:41:30 |
349123
|
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|
Stack Exchange
|
Are hypergraphs $H=(V,E)$ with $|E|=|V|$ $2$-colorable?
A hypergraph $H=(V,E)$ consists of a set $V$ and a collection of subsets $E \subseteq {\cal P}(V)$. A coloring is a map $c: V\to \kappa$, where $\kappa \neq \emptyset$ is a cardinal, such that for every $e\in E$ with $|e|\geq 2$ the restriction $c|_e$ is non-constant.
Question. Is every hypergraph $H=(V,E)$ with $|V|\geq \omega$ and $|E| = |V|$ and $|e| = |V|$ for all $e\in E$ $2$-colorable?
Motivation of question. If we take $V= \omega$ and $E$ to be the collection of computable subsets of $\omega$, then the resulting hypergraph is $2$-colorable - and there are even "balanced" colorings of $\omega$, also referred to as computationally random bitstreams.
Surely an infinite complete graph is a counterexample...
Surely all two subsets of a three set will help form a counterexample. Gerhard "Maybe The Subsets Are Bigger?" Paseman, 2019.12.25.
Sorry - forgot to add the condition that all members of $E$ have cardinality $|V|$
Nice reformulation, thanks @MattF.!
This is essentially done by the Bernstein set construction: if one has $\kappa$ many sets each of size $\kappa$, then order them into ordinal $\kappa$ and recursively choose 2 points from each, so that all these points are distinct. That is, we have $x_\alpha,y_\alpha\in A_\alpha$ with all $x_\alpha,y_\alpha$ distinct. At the end, color each $x_\alpha$ red, each $y_\alpha$ green.
Beautiful, Péter!
|
2025-03-21T14:48:29.524659
| 2019-12-25T19:57:16 |
349126
|
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|
Stack Exchange
|
Is the Hilbert space-filling curve bijective over computable numbers?
The Hilbert curve is a continuous space-filling curve that maps $I$ to $I^n$ where $I$ denotes the unit interval [1]. Like all other space-filling curves, it is not one-to-one. I am wondering if the Hilbert curve, or a common variant thereof, becomes a continuous bijective map if we restrict its domain/range to $\mathbb{CN} \mapsto \mathbb{CN}^n$ where $\mathbb{CN} \subset I$ denotes the set of computable numbers that lie within the unit interval.
At first glance, this seems like a true statement because we have the following two algorithms [2]:
A forward algorithm that takes any number $t \in \mathbb{CN}$ as input, iterates over its binary digits, and sequentially produces the binary digits of all numbers in a corresponding (unique) tuple $T \in \mathbb{CN}^n$.
A reverse algorithm that takes any tuple $T \in \mathbb{CN}^n$ as input, (simultaneously) iterates over the binary digits of $T$'s elements, and sequentially produces the binary digits of a (unique) number $t \in \mathbb{CN}$.
However, I'm not sure if this simple reasoning neglects to account for any of the finer points in computable analysis, especially with respect to the curve's continuity.
[1] As Pietro Majer points out, the original Hilbert curve maps $I$ to $I^2$. In this question, I am referring to its standard n-dimensional extension that uses Gray codes to construct its finite approximations.
[2] The algorithms I mention above is described here. As far as I can tell, these algorithms operate on the standard n-dimensional Hilbert curve (and utilize Gray codes to construct outputs from given inputs).
Thanks in advance for any comments, ideas and pointers to the relevant literature.
Injectivity depends on the curve of course - maybe the original curve maps two computable points to the same element. I think the "right" question here is whether some Hilbert curve which is computable is injective on the computable points.
@NoahSchweber; You are right, being specific matters here. Even though I was talking about the specific curve whose construction was as given in the Wikipedia link, this was not clear. I will add a reference to a specific construction to clarify the question.
I believe the standard curve is not injective on computable points - intuitively, we can find many non-injectivities pretty easily, and this suggests that those non-injectivities are computable.
How can we construct a non-injective example? It seems to me that the standard algorithm based on Gray codes produces unique output bits for each input bit sequence. What am I missing?
For example, in this Hilbert curve: https://en.wikipedia.org/wiki/Hilbert_curve the midpoint $(1/2,1/2)$is the image of three different points, all computable.
The Hilbert curve is the very specific square filling curve $I\mapsto I^2$ defined via a dyadic subdivision, but what should be the definition of a "Hilbert curve" in $I^n$ , i.e a filling cube curve in dimension $n\ge3$? Hilbert does not make any mention to such a generalization, btw. There are of course many constructions of $n$-cube filling curves (already described in the Peano's paper) but it is not clear if you refer to a precise natural generalization of Hilbert construction.
@PietroMajer: I mean the n-dimensional generalization that uses Gray codes. I added this to the question body.
@GeraldEdgar: If we list finite binary approximations of $(1/2, 1/2)$ as a sequence of tuples, we get: $((0.1, 0.1), (0.10, 0.10), (0.100, 0.100), ...)$. Feeding this to the reverse algorithm, we generate the binary digits $(0.1, 0.10, 0.100, ...)$. It is actually easy to see that all further zeroes in the input sequence result in zeroes in the output sequence. Therefore, I deduce that $(1/2, 1/2)$ maps (uniquely) to $1/2$. Where am I going wrong?
Note that if we agree to represent a point of $I^3$ (or in general $I^n$) with a unique expansion, i.e. just merging the digits of the three coordinates: $x_1,y_1,z_1,x_2,y_2,z_2,x_3,\dots$ , then the Peano curve $I\to I^3$ comes from an extremely simple involutory bijection on the ternary strings ${0,1,2}^\mathbb{N} \to {0,1,2}^\mathbb{N}$; thus at least on the representing digits your conjecture it is true. Of course, passing to the quotient, one gets a non-injective curve $I\to I^3$. Points with multiple fiber are easily characterized.
Where am I going wrong? The number $1/2$ has more than one binary expansion. Not only $0.10000\cdots$ but also $0.01111\cdots$.
@GeraldEdgar: OK, the inverse image of $(1/2, 1/2)$ indeed converges to $1/6$ if we use the infinite expansion $0.01111\cdots$. But isn't this a special issue that only manifests for dyadic rationals? If we agree to use finite expansions for dyadic rationals (which is a subset of computable numbers), don't we end up with a one-to-one map?
@MehmetOzanKabak Yes, the issue "just" comes from dyadic rationals. No, we can't resolve this by any kind of agreement - because we cannot compute any specific binary expansion (or Gray code) from a real number. (Technically we even ought to be using signed digit expansions rather than binary...)
Related: https://mathoverflow.net/q/303278/15002
@Arno: I don't mean to be stubborn. I'm trying to understand the root issue as best as I can, but I fail. If we slightly modify the definition of the space-filling function to use finite binary expansions for dyadic rational values, why wouldn't we end up with a one-to-one map?
@MehmetOzanKabak We would end up with a bijection between the computable points, but the map is no longer computable, and it is no longer a space-filling curve.
@Arno: OK, I think I understand the root issue now. There is no way to "fix" forward and reverse maps simultaneously while preserving the space-filling property, regardless whether we operate in $I$ or $\mathbb{CN}$.
The Hilbert curve, due to its fractal nature, is mapping certain subintervals of the unit interval to certain squares in the unit square. On any given resolution, we have a bijection between the subintervals we consider on that scale, and the squares we consider at that scale. The non-injectivity of the actual map then comes from the fact that the intervals intersect in at most one point, whereas the squares can have an entire line as intersection.
Given any fractal space filling curve with these properties, we can compute two distinct points in the interval that get mapped to the same point in the unit square: Pick two non-adjacent intervals that get mapped to adjacent squares, and decide that our target points will be in these. On the next scale, there have to be matching subintervals/squares, and so on. The intersections of the intervals/squares yield points which are computable relative to the original space filling curve.
We thus see that a computable Hilbert curve is a never a bijection on the computable points.
However, given the curve and a point in the square, we can follow the bijections between squares and intervals back to compute some point in its preimage. Hence, the inverse of a computable Hilbert curve is not a function, but it is computable as a multivalued function.
I would like to check if I understand your argument correctly. If we find a point $(x, y)$ that always lies on the boundary of multiple squares regardless of the scale, and if at least one such square maps to an interval that is disjoint from the rest; then this point should be a non-injective point. Do I understand you correctly? If so, consider the two-dimensional curve and the point $(x, y) = (0.5, 0.25)$. If I'm not missing something, this point satisfies our condition. Do you think this point is a non-injective point?
Let's go for (0.5,0.75) instead. This point belongs to the top-left square and to the top-right square on the first level. The top-left square is the image of [0;0.25] and the top-right square is the image of [0.75;1]. So (0.5,0.75) has a preimage in [0;0.25] and one in [0.75;1], so the curve definitely not injective there. Moreover, we can compute a preimage of (0.5,0.75) in each of the two intervals, so restricting to computable points doesn't change anything.
|
2025-03-21T14:48:29.525140
| 2019-12-25T20:14:17 |
349128
|
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|
Stack Exchange
|
A density problem
Let $\langle\cdot,\cdot\rangle$ be the usual scalar product in ${\bf R}^n$ ($n\geq 2$) and let $B$ be the closed unit ball of ${\bf R}^n$.
Denote by $C^0(B,B)$ the space of all continuous functions from $B$ into $B$ endowed with the usual sup-metric. Denote by $\Lambda$
the set of all $f\in C^0(B,B)$ for which there exists some continuous function $\alpha_f:B\to {\bf R}$ such that the set
$$\{(x,y)\in B\times {\bf R}^n : \langle f(x)-x,y\rangle=\alpha_f(x)\}$$
is disconnected. Is it true that $\Lambda$ is dense in $C^0(B,B)$ ?
This question is aimed to provide a completely new proof of Brouwer's fixed point theorem. So, a possible positive answer should be independent of any result based on Brouwer's theorem. To prove the required density, it would be reasonable to use the Baire category theorem.
|
2025-03-21T14:48:29.525347
| 2019-12-26T05:26:20 |
349136
|
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|
Stack Exchange
|
Uniform upper bound on contraction coefficient w.r.t total-variation metric, of a certain set of block-diagonal Markov kernels
Disclaimer. This is related to another question I've asked on the TCS site https://cstheory.stackexchange.com/q/46097/44644. I'm new to information theory (and other relevant fields). It's even possible that I'm not using the appropriate language / terminology to describe my problem. Please any kind of insight, clarification, solution would be very much appreciated.
Keywords: data-processing inequalities; Markov kernel; ergodicity; contractive Markov kernels; Dobrushin coefficient
I $-$ Setup
Let $X=(X,d)$ be a Polish space equipped with the Borel sigma-algebra. Let $\mathcal P(X)$ the the space of all probability distributions on $X$ and let $\mathcal K(X,X)$ be the space of all Markov kernels $K:X \rightarrow \mathcal P(X)$ on $X$.
Now, for $\varepsilon > 0$, $\delta \in [0, 1)$, and some fixed $\mu \in \mathcal P(X)$, define
$$
\mathcal K_{\varepsilon,\delta} := \{K \in \mathcal K(X,X) \mid \mathbb P_{\tilde{x} \sim K(\cdot|x)}(d(\tilde{x},x) > \varepsilon) \le \delta\;\text{for }\mu\text{-a.e }x \in X\}.
$$
For simplicity (and if it helps to simplify things), "...$\text{for }\mu\text{-a.e }x \in X$" may be replaced with "...for all $x \in X$".
I'm interested in obtaining upper-bounds on the quantity $L(\mathcal K_{\varepsilon,\delta},\mu)$ defined by
$$
L(\mathcal K_{\varepsilon,\delta},\mu):=\inf_{K \in \mathcal K_{\varepsilon,\delta}}L(K,\mu),
$$
where
$$
L(K,\mu):= \sup_{\nu \in \mathcal P(X),\; TV(\mu,\nu) > 0}\frac{TV(\mu K,\nu K)}{TV(\mu,\nu)}.
$$
Likewise, I'd like to upper-bound $L(\mathcal K_{\varepsilon,\delta})$ defined by
$$
L(\mathcal K_{\varepsilon,\delta}) := \sup_{\mu}L(\mathcal K_{\varepsilon,\delta},\mu)
$$
Thus $L(\mathcal K_{\varepsilon,\delta})$ is a kind of uniform Lipschitz constant for the kernels in $\mathcal K_{\varepsilon,\delta}$ w.r.t to the total-variation metric on $\mathcal P(X)$. One could consider a modified scenario replacing TV with some other "distance" like relative entropy, Wasserstein, etc. Not sure that would simplify the analysis though...
I.1 $-$ Main focus
For concreteness, we may restrict to the cases where $\mu$ has some "measure-concentration properties" (I'm not yet sure what form this should take...) and
$X$ is $\mathbb R^n$ or $[0, 1]^n$ equipped with and $\ell_p$-norm;
$X$ is the Hamming cube $\{0,1\}^n$;
etc.
II $-$ Questions
Question 2.1 What are good upper-bounds for $L(\mathcal K_{\varepsilon,\delta},\mu)$ and $L(\mathcal K_{\varepsilon,\delta})$ ?
Of course such a "good" upper-bound must somehow depend on the problem parameters $\varepsilon,\delta$ explicitly.
II.2 $-$ A rough bound via Dobrushin coefficients
In view of obtain a (presumably very rough) bound, define the Dobrushin coefficient $D(K) \in [0, 1]$ of a Markov kernel $K$ by
$$
D(K) := \max_{x,x'}TV(K(\cdot|x),K(\cdot|x')) = \frac{1}{2}\max_{A,x,x'}|K(A|x)-K(A|x')|,
$$
where the supremum is taken over all measurable $A \subset X$ and distint points $x,x' \in X$.
By, the (Dobrushin's) data-processing inequality, we have the bound $L(K,\mu) \le D(K)$, from where we get the (perhaps very loose) bound
$$
L(\mathcal K_{\varepsilon,\delta}) \le L(\mathcal K_{\varepsilon,\delta,\mu}) \le \inf_{K \in \mathcal K_{\varepsilon,\delta}} D(K).
$$
However, evening computing a good upper bound on $\inf_{K \in \mathcal K_{\varepsilon,\delta}} D(K)$ seems daunting.
Question 2.2 What is a good upper-bound for $\inf_{K \in \mathcal K_{\varepsilon,\delta}}D(K)$?
II.3 $-$ The case $\delta=0$
We now turn to the important particular case when $\delta = 0$. Consider the subset of deterministic Markov kernels
$$
\mathcal K_\varepsilon := \{K_f \mid f \in \mathcal F_{\varepsilon}\},
$$
$\mathcal F_\varepsilon$ is the set of measurable functions $f:X \rightarrow X$ such that $d(f(x),x) \le \varepsilon$ for all $x \in X$.
$K_f$ is the Markov kernel on $X$ defined by setting
$K_f(A|x) := 1_{f^{-1}(A)}(x)$, for every $x \in X$ and measurable $A \subseteq X$.
Somethings to note:
$\nu K_f = f_{\#}\nu$ for every measurable function $f:X \rightarrow X$ and probability distribution $\nu \in \mathcal P(X)$. For such Markov kernels, we therefore have the following formula
$$
L(K_f,\mu) = \sup_{\nu}\frac{TV(f_{\#}\mu,f_{\#}\nu)}{TV(\mu,\nu)}.
$$
$\mathcal K_\varepsilon \subseteq \mathcal K_{\varepsilon,0}$; thus $L(\mathcal K_{\varepsilon,0},\mu) \le L(\mathcal K_{\varepsilon},\mu)$ and $L(\mathcal K_{\varepsilon,0}) \le L(\mathcal K_{\varepsilon})$.
$\mathcal K_{\varepsilon}$ is precisely the set of Markov kernels considered in my TCS post here https://cstheory.stackexchange.com/q/46097/44644.
Question 2.3. What are good upper-bounds for $L(\mathcal K_{\varepsilon},\mu)$, $L(\mathcal K_{\varepsilon})$, and $\inf_{K \in \mathcal K_\varepsilon} D(K)$?
III $-$ Maybe-be-useful references
The paper https://arxiv.org/pdf/1411.3575.pdf seems to a relevant reference for my problems, but I'm still reading it and learning IT at the same time.
Consider $K(x) = \delta_x$...
@MartinHairer Please could you be more explicit. Except I'm missing something, this suggestion cannot work (i.e cannot produce any informative bound) as it disregards all problem parameters: $\varepsilon$, $\delta$, etc.
@MartinHairer If fact, if $K(x) = \delta_x$, then $TV(\mu K,\nu K) \equiv TV(\mu,\nu)$, which is tautological. No ?
This is my point. Unless I misread your definitions, the trivial example belongs to all the $\mathcal{K}_{\varepsilon,\delta}$, so you can't improve on the trivial bound $1$ unless you restrict yourself to smaller classes of kernels.
@MartinHairer You probably misread the definitions. For example, $L(\mathcal K_{\varepsilon,\delta},\mu) := \inf_{K \in \mathcal K_{\varepsilon,\delta}} L(K,\mu)$. It's an inf on the kernels, not a sup. So the fact that diagonal kernels are "bad" (i.e there all have $L(K,\mu)=1$) is not a problem; $\mathcal K_{\varepsilon,\delta}$ contains other non-trivial kernels when $\varepsilon > 0$. Makes sense ?
I was indeed too quick and thought it was a sup...
@MartinHairer OK, great. Please don't hesitate if you have any insights on the problem. Thanks in advance :)
Although you mention the Hamming cube among your examples, my impression is that you are still more interested in the situation when the arising measure spaces are not atomic (discrete). I am afraid there is not much one can do in the continuous situation. The point is that although you are imposing conditions requiring "locality" of the transition probabilities (you require that for any point $x$ the corresponding transition probability $\pi_x$ have the property that $\pi_x B(x,\epsilon)>1-\delta$), these conditions (quite appropriate when talking about weak convergence) do not help at all in what concerns the possible singularity of the transition probabilities, and it is this singularity that is responsible for the total variation distance.
A very clear illustration of this phenomena is provided precisely by the deterministic case you mention at the end of your question. If the measure $f(\mu)$ (which you denote $f_\#\mu$) is not finitely supported, then $L(K_f,\mu)=1$ (take an $f(\mu)$-small subset of $X$, and let $\nu$ be the normalized restriction of $\mu$ to its preimage).
Returning to the general case, you would have to assume that $\mu$-almost all (if you work with the measures $\nu$ absolutely continuous with respect to $\mu$) or all (if you work with all measures $\nu$) transition probabilities $\pi_x$ are absolutely continuous with respect to a common measure $\lambda$ and impose further conditions on the total variation distances $\|\pi_x-\lambda\|$.
Thanks for the detailed response; upvoted! The singularities you talk about would go away if I consider Wasserstein metric instead of TV in my questions, right ? Back to TV, I've tried to compute bounds for Gibbs kernels $K_\varphi(x'|x) = e^{-2\varphi(x,'x)/\lambda}d\gamma$ for base measure $\gamma \in \mathcal P(X)$, a bandwidth parameter $\lambda$ and potentials $\varphi: X\times X \rightarrow \mathbb R$ which are $R$-Lipschitz w.r.t first argument, and I obtain useful bounds of the form $L(K,\mu) \le D(K) \le (e^{2RD/\lambda}-1) / (e^{2RD/\lambda} + 1) < 1$, where $D=diam(X)$.
My motivation for considering the kernels in $\mathcal K_{\varepsilon,\delta}$ and later $K_{\varepsilon}$ was to create kernels which don't move too far per step. The Gibbs distribution seems to be a "mollified" version of this idea. I wonder if there are other ways to do this.
I mean the usual singularity of measures (so that two probability measures are singular if the total variation of their diference is 2) which indeed does not affect the transportation distance. Of course, in your Gibbs setup there won't be any singular transition probabilities, and the Markov operator can well be contracting in the total variation metric.
OK, thanks. It's possible my quick-and-dirty bound for Gibbs kernels is not very optimal. Q: Is there any concrete reference on TV-contraction of Gibbs kernels ? Thanks in advance.
If I understand correctly you just use the fact that for these transition probabilities $|\pi_x-\pi_y|$ is bounded away from 2 - which is essentially Doeblin's condition.
|
2025-03-21T14:48:29.525875
| 2019-12-26T09:17:47 |
349141
|
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|
Stack Exchange
|
Algebra of Hecke operators on $M_k(\mathrm{SL}_2\mathbb{Z})$ is an integral domain?
Let $M_k(\mathrm{SL}_2\mathbb{Z})$ be the space of modular forms of (integer) weight for the full modular group. Let $\mathbf{H}$ denote the Algebra generated by the Hecke operators $T_n$. Is $\mathbf{H}$ an integral domain? Specifically, let $T_mT_n=0,$ can we conclude either $T_m=0$ or $T_n=0$ on $M_k(\mathrm{SL}_2\mathbb{Z})$?
If you replace $M_k$ with the subspace $S_k$ of cusp forms, then the answer should be "yes", as a consequence of Maeda's conjecture (see https://arxiv.org/abs/1207.3480).
You mean the (commutative, normal for the Petersson inner product thus diagonalizable) complex algebra $\Bbb{T(C)}$ of endomorphisms of the complex vector space $M_k(SL_2(Z))$ ($k$ even) generated by the identity and the $T_n$.
If $\dim(M_k(SL_2(Z))=1$ then it is $= \Bbb{C}$, otherwise it is not an integral domain, as it contains $E_k$ plus a cusp eigenform $f$, take some $p$ such that $\sigma_{k-1}(p)\ne a_p(f)$ then $T_p E_k= \sigma_{k-1}(p)E_k, T_pf=a_p(f)f$ ie. the minimal polynomial of $T_p $ is $(X-\sigma_{k-1}(p))(X-a_p(f))g(X)$ so that $(T_p-a_p(f)) (T_p-\sigma_{k-1}(p))g(T_p)=0$.
From that $E_4^3-E_6^2=1728\Delta$ has only one simple zero at $i\infty$ we get the $\Bbb{C}$-basis of modular forms with rational coefficients $$ M_k(SL_2(Z))=\sum_{4a+6b=k} \Bbb{C}E_4^a E_6^b$$ Since $T_n E_4^a E_6^b$ has rational coefficients too this implies the matrix of $T_n$ in this basis has rational entries so that the minimal polynomial of $T_n$ is in $\Bbb{Q}[X]$ and hence $\Bbb{T(Q)},\Bbb{T(Z)}$ are not integral domains neither.
On the other hand $T_nT_m\ne 0$ because $T_nT_m E_k = \sigma_{k-1}(n)\sigma_{k-1}(m)E_k$.
Maeda's conjecture is saying $S_k(SL_2(Z))$ is generated by the Galois orbit of a single eigenform $f$, in which case $\Bbb{T(Q)}|_{S_k(SL_2(Z))}$ is isomorphic to $\Bbb{Q}(\{a_n(f)\})$ which is an integral domain. $\Bbb{T(C)}|_{S_k(SL_2(Z))}$ is an integral domain iff $\dim(S_k(SL_2(Z)))=1$.
|
2025-03-21T14:48:29.526022
| 2019-12-26T10:28:17 |
349145
|
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|
Stack Exchange
|
Normalized Laplacian matrix versus walk Laplacian matrix (or normalized adjacency matrix versus walk adjacency matrix)
In graphs, found that two different normalization matrices exist for Laplacian and adiacency matrix. I will ask about the adjacency matrix (for the Laplacian matrix the questions are the same). The first normalization matrix of the adjacency matrix is known as walk adiacency matrix, and is defined as
$$N_{walk}=D^{-1}A$$
where $A$ is th adjacency matrix and $D$ is the degree matrix. The sum of each row of $N_{walk}$ is $1$, so I see the mean of the word "normalized" used in $N_{walk}$.
The second (and, strangely for me, most common version) instead is:
$$N_{norm}=D^{-1/2}A D^{-1/2}$$
so, for me, the following questions arise:
For each row, the sum of $N_{norm}$ is still $1$? I don't think, so, why is it know as a "normalized" matrix? What is normalized in this matrix if the sums is not $1$?
Why $N_{norm}$ is "better" than $N_{walk}$?
Is there any intuitive explaination for $N_{norm}$? while I still can see $N_{walk}$ as a real "normalized" version of $A$, what is the meaning, in an intuitive way, of dividing each entry $a_{ij}$ for $\sqrt{d_i}\sqrt{d_j}$? Where it comes from?
$N_{norm}$ is symmetric, while the other matrix is not.
Ok, but why “normalised”?
Not sure if that is the official answer, but one sense in which it is 'normalized' is that its spectral radius is 1.
For the Laplacian, the second normalization gives a matrix with all ones on the diagonal (if no isolated vertices). So that may be why it is referred to as the "normalized Laplacian". The name for the adjacency matrix may just be borrowed from this since I think that normalized Laplacians are much more studied than normalized adjacency matrices (I do not recall seeing the latter notion before). Also, as Dima pointed out, the second type of normalization preserves the symmetry of the matrix. In the case of the Laplacian, positive semidefiniteness is also preserved, which is probably desirable.
|
2025-03-21T14:48:29.526202
| 2019-12-26T10:37:45 |
349147
|
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|
Stack Exchange
|
Examples of non-cubulated hyperbolic groups
What is known regarding which hyperbolic groups are cubulated?
I take it the usual definition of cubulated is acting properly and cocompactly on a CAT(0)-cube complex.
My impression is that not all of them are, but I didn't manage to find references with a counterexample.
Are there known ways to create non-cubulated hyperbolic groups?
Are there famous examples of non-cubulated groups?
A weaker notion, "PW" is just acting properly on a CAT(0) cube complex (possibly infinite-dimensional). It's much weaker than cubulable, still not satisfied by all hyperbolic groups (see AGenevois's answer), while satisfied by many groups.
The Ollivier-Wise Rips machine produces hyperbolic groups with an infinite Property T subgroup and arbitrary f.p. quotient (hence possibly without Property T). These hyperbolic groups don't have Property PW, hence are not cubulable.
If a group $G$ satisfies Kazhdan's property (T), then any action of $G$ on a CAT(0) cube complex has a global fixed point. See Niblo and Roller's article Groups acting on cubes and Kazhdan's Property (T). Examples of hyperbolic groups which satisfy this property include:
Uniform lattices in quaternionic hyperbolic spaces.
Random groups in Gromov's model for some density.
Also, by a theorem due to Gromov, any hyperbolic group admits a quotient which is hyperbolic and has property (T).
Do you know if uniform lattices in quaternionic hyperbolic spaces have torsion or are torsion free?
I am not quite familiar with lattices in quaternionic hyperbolic spaces, but there are reflections in such spaces, so I would say that there exist torsion-free lattices and lattices with torsion.
@YanivShakhar: They're linear, and hence virtually torsion free. Probably there are explicit examples of quaterionionic hyperbolic lattices with torsion, but if you want to construct a hyperbolic group with torsion and property (T), just take any non-trivial element $g$ and kill a sufficiently high power $g^n$.
As @AGenevois says in his answer, the standard examples of non-cubulated hyperbolic groups are those with Kazhdan's property (T), such as quaternionic hyperbolic lattices.
Complex hyperbolic lattices (in dimension >2) provide a more delicate class of examples. On the one hand, they are not cubulable, by a theorem of Delzant--Py. On the other hand, they are known to have the Haagerup property (I think this is proved in the book by Bekka--de la Harpe--Valette), so they also don't have property (T).
As far as I know, they are the only class of examples of hyperbolic groups known to be Haagerup but non-cubulable. I'd be interested to hear of others.
That they're Haagerup is a 1974 result of Faraut-Harzallah.
You can get more examples by taking free products of uniform complex-hyperbolic lattices in $PU(n,1)$, $n\ge 2$. What is unknown, I think, is the existence of 2-dimensional hyperbolic groups which have the Haagerup property but do not admit cubulations.
@Misha — sure, but I wouldn’t class those as genuinely “different” examples. I agree that 2-dimensional examples would be particularly nice to have.
|
2025-03-21T14:48:29.526415
| 2019-12-26T11:14:53 |
349151
|
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|
Stack Exchange
|
On the cellularity of the $G_\delta$-topology
Given a topological space $X$, let $X_\delta$ be the topology on $X$ generated by the $G_\delta$ subsets of $X$. Let $c(X)$ be the cellularity of $X$, that is, the supremum of cardinalities of families of pairwise disjoint non-empty open subsets $X$. Thus $X$ is ccc if and only if $c(X)=\aleph_0$.
In 1972 Juhász proved:
Let $X$ be a compact Hausdorff space. Then $c(X_\delta) \leq 2^{c(X)}$.
Compactness can actually be replaced with pseudocompactness+regularity in Juhász's result.
Some form of compactness is essential in Juhász's theorem. For every cardinal $\kappa$ there is a (regular) ccc space $X$ such that $c(X_\delta) > \kappa$; some examples of such a space can be found in this other Mathoverflow question of mine. However, none of those examples is Lindelof, as they all contain large closed discrete subsets. This motivates the following question:
QUESTION: Let $\kappa$ be any cardinal. Is there a Lindelof regular ccc space such that $c(X_\delta) > \kappa$?
To find a consistent example of a Lindelof ccc regular space such that $c(X_\delta) > 2^{\aleph_0}$ it suffices to take a small modification of Gorelic's example of a Lindelof space with points $G_\delta$ and cardinality larger than the continuum (see the remark at the beginning of page 606 of Gorelic's paper).
A related Mathoverflow question is: Covering compact Hausdorff spaces with closed $G_\delta$ sets
|
2025-03-21T14:48:29.526537
| 2019-12-26T11:55:03 |
349153
|
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"Sam",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349153"
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|
Stack Exchange
|
Algorithm to generate free unlabelled trees uniformly at random
I am implementing an algorithm to generate free unlabelled trees uniformly at random (uar). For this I found this paper by Herbert S. Wilf (The uniform selection of trees. 1981. In Journal of Algorithms). This paper defines the procedure Free which generates free unlabelled trees by calling one of two smaller procedures (either Bicenter or Forest) with a certain probability each. These two procedures, in turn, are based on the ranrut procedure, which generates rooted unlabelled trees uar. You can find the reference in Combinatorial Algorithms For Computers and Calculators. Albert Nijenhuis and Herbert S. Wilf. 2nd Edition. Academic Press. (I'll try to make this post as self-contained as I can).
First, I'm quite convinced to have managed to implement the ranrut procedure so that each rooted unlabelled tree is generated uar. However, I'm having some problems with the algorithm to generate free unlabelled trees (here) and I was hoping that you could help me. In this post I would like to ask two questions.
I have one problem with language. Specifically, with one particular word, which I do not understand in the context of the paper. The word is adjoined and appears in the Forest procedure, as seen highlighted in this image:
The image is an extract of Wilf's paper that shows the full procedure Forest. Now, the word adjoined is used in the sentence
Then exit with $j$ copies of $T'$ adjoined to $\mathcal{F}'$.
Should adjoined be understood here as the disjoint union of graphs? Namely, should I perform the operation $\mathcal{F}' \oplus T'_1 \oplus \cdots T'_j$, where $\oplus$ denotes disjoint union of graphs ? If not, how should I interpret it?
Wilf says that, in order to generate uar a forest of $m$ trees, we first choose two integers $(j,d)$ with probability (as shown in the above image)
$$
Prob(j,d) = \frac{d\cdot \alpha(m - jd,q) \cdot a_d}{m \cdot \alpha(m,q)}.
$$
For this we need the values $a_n$ (the number of rooted unlabelled trees of $n$ vertices), and $\alpha(m,q)$ (the number of rooted forests of $m$ vertices whose trees have at most $q$ vertices each).
I have written an algorithm to correctly calculate $a_n$. I know it is correct because I can compare the values obtained with its integer sequence.
I would like to know if I'm calculating $\alpha(m,q)$ correctly. However, I haven't been able to find the integer sequences for $\alpha(m,q)$. Do you know where I can find it?
I hope to have made my doubts understandable and the questions clear.
Thanks to you all for your time.
Edit
This post was motivated because my implementation of this algorithm did not generate trees uniformly at random. In my desperation I came here for help. Although the questions above still stand (because the implementation of the above might still be wrong), one of the "bugs" was caused by an error in Wilf's paper. My advisor pointed me to this document (Graph theory package for Giac/Xcas - Reference Manual, September 2018) where this error is mentioned and corrected (page 38, footnote 1.1). For the sake of self-containment, I tell you here what the error is and how it has been corrected. The Free procedure that Wilf defines in this paper, says that in order to generate an $n$-vertex free tree u.a.r. you have to generate a bicentroidal tree with probability $p$, or a random forest of $n-1$ vertices with probability $1-p$ (the roots of this forest are connected to a new vertex). Wilf says that
$$p ={1 + a_{n/2} \choose 2}/a_n$$
where $a_n$ is the number of rooted unlabelled trees of $n$ vertices. As pointed out in the cited document in this edit, the denominator is wrong. Instead of $a_n$, it should be $f_n$, the number of unlabelled free trees on $n$ vertices.
Based on the images, yes, I say you have the right idea. In your disjoint union, make sure the T'_i are all isomorphic to T'. Gerhard "Follow The Recipe Very Closely" Paseman, 2019.12.26.
Okay. That is exactly what I was doing but I wasn't sure at a 100%. Now it only remains the other question of the $\alpha(m,q)$ :)
I have edited the original post with a correction of a small error in the original paper. However, I would appreciate to have my other questions answered, since they refer to parts of my code that still need debugging.
I wonder if you actually need this algorithm, or you just need one fast in practice. Walk at random on $K_n$ until every vertex is visited (about $n\log n$ steps on average). Mark the edge along which each vertex was visited for the first time. That's a uniform random free tree. It's very fast: 6000 trees/sec on 1000 vertices for me.
I like the idea of using a simpler algorithm. This would alleviate my job a lot more. However, I think you are describing a procedure for generating u.a.r. labelled free trees (as described by Aldous here). But I want to generate u.a.r. unlabelled free trees.
I hope I didn't misunderstand your proposal. If I didn't, the only question that remains unanswered in the original post is to find the integer sequences of $\alpha(m,q)$.
Why don't you compare your algorithm and your calculation of $\alpha(m,q)$ with the source code on which the cited documentation is based? In particular, this function: https://github.com/marohnicluka/giac/blob/master/graphe.cc#L7149
Yes, thank you. I compared my algorithm and the computation of the $\alpha(m,q)$ in the link you gave and the numbers coincide, up to $n=~400$.
I know this is not what the "answers" are for, but I see no other way of reaching you, unfortunately. I am currently working on implementing Wilf's algorithm myself and I would like to contact you about some problems I am having. Is there any way I can reach you? Kind regards.
@Sam, I've just read your comment (I am very sorry! three years later...). I don't think I ever got a notification of your comment. If you are still struggling, go to my profile in MO, and you'll find a link to my webpage, which contains all necessary contact info. If you just want to see the implementation in C++, see this and this files.
Thanks to the effort of several people, I got the answers I was looking for. I write this so that other people in the future can have a direct answer to the questions above without having to read the comments.
The first question was whether the word adjoined (in the Forest procedure) referred to disjoint union of graphs. This comment written by Gerhard Paseman confirmed this. Thanks to this comment and careful reading of the Free procedure, any interpretation of the word adjoined other than disjoint union should be ruled out.
The second question was about the calculation of the numbers $\alpha(m,q)$. Timothy Budd added a comment where I was pointed to an implementation of this algorithm where I could find the computation of $\alpha(m,q)$. Budd advised me to compare the result of that algorithm with mine. This proved useful, and the results of my implementation and those of the code to which I was pointed coincide.
Thanks to everyone who helped in finding the answers to my questions, even to those who did not answer them directly. I really appreciate your help.
|
2025-03-21T14:48:29.527012
| 2019-12-26T11:58:11 |
349154
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349154"
}
|
Stack Exchange
|
What is the Teichmuller metric on the Teichmuller space of a closed surface of genus 1?
Howard Masur's research asserts that if $S_g$ is a closed surface of genus $g\geq2$, then the Teichmuller space $T(S_g)$ does not have nonpositive curvature. His proof relies on the existence of similar Strebel rays. However, similar Strebel rays does not exist in $T(S_g)$ if $g=1$. On the other hand, $T(S_1)=\mathbb H$ does admit a hyperbolic metric.
Questions:
Does the Teichmuller metric and the hyperbolic metric for $g=1$ coincide?
If not, does it has nonpositive curvature?
Is the metric well-defined for $g=1$?
@WillSawin Shouldn't it be? Are you saying it could vanish everywhere? Indeed I haven't seen any reference on this subject. Can you provide one or give a short explanation?
The answer is yes, up to renormalization.
Precisely, the bijection $\mathbb{H}^2\to Teich(\mathbb{T})$ you refer to induces an isometry from $(\mathbb{H}^2,d_{\mathbb{H}^2})\to (Teich(\mathbb{T}),2d_{Teich})$.
This is exactly Theorem 11.20 in Farb and Margalit's book A primer on mapping class groups, available here.
|
2025-03-21T14:48:29.527121
| 2019-12-26T13:11:17 |
349156
|
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|
Stack Exchange
|
Independence depth of linearly dependent random variables
Suppose, $\Xi$ is a collection of random variables. We call $\Xi$ $k$-independent, iff any $k$ distinct elements of $\Xi$ are mutually independent. For example, $2$-independence is pairwise independence and $|\Xi|$-independence is the full mutual independence of random variables from $\Xi$.
Let's define independence depth of $\Xi$ as the maximal number $k$, such that $\Xi$ is $k$-independent.
Suppose $X_1, ... , X_n$ are non-constant real random variables, such that $X_1 + ... + X_n = 0$. What is the largest possible independence depth of $\{X_1, ... , X_n\}$?
I only know the following fact:
Independence depth of $\{X_1, ... , X_n\}$ is strictly less than $n - 1$.
Suppose $\{X_1, ... , X_{n}\}$ is $(n-1)$-independent. Then suppose $Y = -X_n$. Thus we have $Y = \sum{i = 1}^{n - 1} X_k$.
Let's define $\chi_X$ as a characteristic function of a random variable $X$. Then we have $\forall x, y \in \mathbb{R}, k \leq n -1 $:
$$\chi_{X_k}(x)\Pi_{i = 1}^{n-1} \chi_{X_i}(y) = \chi_{X_k}(x)\chi_{Y}(y) = Ee^{i(xX_k + yY)} = Ee^{i((x + y)X_k + \sum_{i = 1}^{k - 1}yX_i + \sum_{i = k+1}^{n-1} yX_i)} = (\Pi_{i = 1}^{k - 1}\chi_{X_i}(y))\chi_{X_k}(\Pi_{i = k + 1}^{n-1}\chi_{X_i}(y))$$
From that and the facts, that characteristic functions are continuous and $\chi_{X_1}(0) = ... = \chi_{X_{n-1}}(0) = 1$ it follows, that $\exists \epsilon > 0$, such that $\forall x \in \mathbb{R}, |y| < \epsilon, k < n - 1$ we have $\chi_{X_k}(x + y) = \chi_{X_k}(x)\chi_{X_k}(Y)$. From that and the fact, that $\mathbb{R}$ is an Archimedean field, we can conclude, that $\forall x, y \in \mathbb{R}, k < n - 1$ we have $\chi_{X_k}(x + y) = \chi_{X_k}(x)\chi_{X_k}(Y)$. And we know, that all non-zero functions with this property are of the form $x \mapsto e^{cx}$. Thus we can conclude, that $\forall k < n - 1$ we have $\chi_{X_k}(x) = e^{ic_kx}$ and thus $X_k = c_k$ almost surely. Thus all $X_k$ and $Y$ (as the sum of them) are constants.
It is a follow-up of this question: https://mathoverflow.net/q/349010/110691
If the rv's have expectations, then $k>1$ is impossible, by the argument I already gave in your first question: take $E(\ldots|X_1)$ to see that $X_1$ is constant.
Of course, this suggests that the full question has the same answer, with just some technical difficulties added.
Here's a different argument.
Pick $t>0$ such that $P[|X_i|>t]\leq 1/n$ for all $i.$ The event $|X_i|>tn$ is a subset of the union of the events $|X_j|>t$ for $j\neq i,$ so
$$P[|X_i|>tn] \leq \sum_{j\neq i} P[|X_j|>t \text{ and }|X_i|>tn] \leq \frac{n-1}{n}P[|X_i|>tn].$$
Since each $X_i$ is essentially bounded you can use the argument in Christian Remling's comment, or $0=\operatorname{Var}(\sum X_i)=\sum \operatorname{Var}(X_i)>0.$
The largest possible independence depth of $\{X_1,\dots,X_n\}$ is $1$. That is, for any natural $n\ge2$, there are no pairwise independent random variables (r.v.'s) $X_1,\dots,X_n$ such that (i)
$X_1+\dots+X_n=0$ almost surely (a.s.) and (ii) for all real $c_1,\dots,c_n$ and all $i\in[n]:=\{1,\dots,n\}$ we have $P(X_i=c_i)\ne1$.
Indeed, suppose the contrary: that $n\ge2$, $X_1,\dots,X_n$ are pairwise independent r.v.'s such that
$X_1+\dots+X_n=0$ a.s., and for all real $c_1,\dots,c_n$ and all $i\in[n]$ we have $P(X_i=c_i)\ne1$.
Let $Z=(Z_1,\dots,Z_n):=X-Y$, where $X:=(X_1,\dots,X_n)$ and $Y=(Y_1,\dots,Y_n)$ is an independent copy of $X$. Then $Z_1,\dots,Z_n$ are symmetric pairwise independent r.v.'s such that
$Z_1+\dots+Z_n=0$ a.s., and for all $i\in[n]$ we have $P(Z_i=0)\ne1$.
Take now any real $a>0$ and introduce
$$W_i:=W_{i,a}:=Z_i\,I\{|Z_i|\le a\},
$$
where $I$ denotes the indicator. Then the $W_i$'s are bounded symmetric pairwise independent r.v.'s, whence
\begin{equation}
E\Big(\sum_{i\in[n]}W_i\Big)^2=\sum_{i\in[n]}EW_i^2. \tag{1}
\end{equation}
On the other hand, recalling the condition $Z_1+\dots+Z_n=0$ a.s., introducing the random set $\mathcal J_a:=\{j\in[n]\colon |Z_j|>a\}$, and finally letting $a\to\infty$, we have
\begin{align*}
E\Big(\sum_{i\in[n]}W_i\Big)^2
&=\sum_{J\subseteq[n]}E\Big(\sum_{i\notin J}W_i\Big)^2\,I\{\mathcal J_a=J\}\\
&=\sum_{J\ne\emptyset}E\Big(\sum_{i\notin J}W_i\Big)^2\,I\{\mathcal J_a=J\}\\
&\le n\sum_{J\ne\emptyset}\sum_{i\notin J}EW_i^2\,I\{\mathcal J_a=J\}\\
&= n\sum_{J\ne\emptyset}\sum_{i\notin J}EZ_i^2\,I\{|Z_i|\le a\}\,I\{\mathcal J_a=J\}\\
&\ll\max_{i\ne j}EZ_i^2\,I\{|Z_i|\le a\}\,I\{|Z_j|>a\} \\
&=\max_{i\ne j}EZ_i^2\,I\{|Z_i|\le a\}\,P(|Z_j|>a) \\
&=\max_{i\ne j}EW_i^2\,P(|Z_j|>a) \\
&=o\Big(\max_{i\in[n]}EW_i^2\Big)
=o\Big(\sum_{i\in[n]}EW_i^2\Big),
\end{align*}
which contradicts (1), as desired. (In the above multi-line display, $\sum_{J\ne\emptyset}$ denotes the summation over all non-empty $J\subseteq[n]$, $A\ll B$ means $A\le CB$ for some real $C>0$ not depending on $a$, and $\max_{i\ne j}$ denotes the maximum over all distinct $i$ and $j$ in $[n]$.)
|
2025-03-21T14:48:29.527854
| 2019-12-26T15:47:46 |
349164
|
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|
Stack Exchange
|
Metric projection on CAT(0) tangent cone
Let $(Y,d)$ be a complete and separable CAT(0) space, fix $y \in Y$. Then, consider the tanget cone $(T_yY,d_y)$ at $y$, i.e. the metric cone over the space of directions, and denote by $0_y$ the 'tip' of such cone. It is well known that $(T_yY,d_y)$ is CAT(0) space and, for any $C\subset T_yY$ convex and closed, the CAT(0) condition grants the existence of a metric projection $P_C$ assigning to each vector $v \in T_yY$, its projection $v^C= $argmin$_Cd_y(v,\cdot)$.
Define an 'inner product' (cleary in absence of an underlying linear structure) imposing $2\langle v,w\rangle_y := \vert v\vert_y^2 +\vert w\vert_y^2 - d^2_y(v,w)$ for any $v,w \in T_yY$ (here $\vert \cdot\vert_y$ = $d_y(\cdot,0_y)$).
I am interested in the properties of the metric projection $P_H$ in this framework for a specific $H \subset T_yY$. Suppose H is flat region passing through $0_y$ and isometric to $\mathbb{R}^2$ inside $T_yY$. Let $h \in H$, $v \in T_yY$ and consider $v^H$, the metric projection of $v$ onto $H$.
My question: is there any relation between the quantities $\langle h,v\rangle_y$ and $\langle h,v^H\rangle_y$?
The reference I am looking into is 'A course in metric geometry - Burago Burago Ivanov' where it treats a splitting theorem for Hadamard spaces.
|
2025-03-21T14:48:29.527983
| 2019-12-26T15:49:55 |
349165
|
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|
Stack Exchange
|
A vector bundle analogy of the Nash embedding theorem
Let $E$ be a Riemannian vector bundle over a manifold. Can $E$ be considered as a subbundle of trivial bundle $M\times \mathbb{R}^n$ for some $n$ such that the metric of each fiber is the restriction of Euclidean metric of $\mathbb{R}^n$?
Should "trivial subbundle" be "subbundle of a trivial bundle"?
After applying Steven Landsburg's correction, this is a definition chase using the fact that there is a classifying map $f: M \to BO(k)$, where $BO(k)$ is the Grassmannian of $k$-planes in $\Bbb R^\infty$, and the fact that $f$ factors through some finite-dimensional Grassmannian $\text{Gr}(n,k)$. It is not analagous to Nash's theorem (which is much more difficult).
Even more is true: if the vector bundle has a metric connection, then by a result of Narasimhan and Ramanan, there is a classifying map from $M$ to a sufficiently high-dimensional Grassmannian that classifies both metric and connection on $E$. This seems to me the correct "linearisation" of the Nash embedding theorem.
@StevenLandsburg sorry for my delay. Thanks for your correction. I revise it.
|
2025-03-21T14:48:29.528101
| 2019-12-31T17:21:17 |
349455
|
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"url": "https://mathoverflow.net/questions/349455"
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|
Stack Exchange
|
Number of orthants intersected by a convex hull
I'm trying to figure out the following problem:
Let $x_1,\ldots,x_k\in\mathbb{R}^n$ be some points for some $k<n$. Let $\mbox{conv} (x_1,\ldots,x_k)$ be their convex hull. I'm looking for a tight (possibly with an example) upper bound for the number of orthants that such convex hull can intersect with (depending on $n$ and $k$).
I've figured that as the convex hull of two points may intersect with up to $n$ orthants, by induction I can bound the number of intersections of $k$ points with different orthants to be $O(n^{k-1})$. I want to prove that this bound is tight or find a tighter one, hopefully with a concrete example.
Any ideas? Thanks!
Consider the $k-1$ dimensional simplex given by $\alpha_1+\alpha_2+\cdots \alpha_k=1, \alpha_i\geq 0$. The equations $e_i\cdot (\sum_{j=1}^k \alpha_j x_j)=0$ for $1\le i\le n$ describe $n$ hyperplanes that cut our simplex into several regions. Here $\cdot$ is the dot product and $e_i\in \mathbb R^n$ is the $i$-th coordinate vector.
It is easily seen that the number of orthants intersecting $\text{Conv}(x_1,x_2,\dots,x_k)$ is the same as the number of regions that our simplex was divided in. Therefore the question is actually equivalent to asking: "What is the highest number of regions that $\mathbb R^{k-1}$ can be divided into by $n$ hyperplanes?" The answer is given by
$$1+n+\binom{n}{2}+\cdots+\binom{n}{k-1}$$
and can be proved by induction on both $k$ and $n$ (this fact is classical and has appeared on MO before, for example here). In particular the $O(n^{k-1})$ bound is tight.
So it is indeed $O(n^{k-1})$ and a polynomial in $n$. Given this, If we drop the requirement $n \gt k$ and observe that the answer is $2^n$ for $n \lt k$ , that determines the answer uniquely.
@AaronMeyerowitz yes! I prefer not to define it in separate cases, since the polynomial evaluated at $n<k$, does give the correct value $2^n$. Really the inductive proof makes everything clear, since it shows that the answer satisfies $a_{n,k}=a_{n-1, k}+a_{n-1,k-1}$ for all $n,k\geq 1$.
It is true that the linked question has the stated result as one of the answers given. However it doesn’t answer the question asked.
Yes, the linked question is much harder and still doesn't have a satisfactory answer.
|
2025-03-21T14:48:29.528574
| 2019-12-31T23:57:25 |
349461
|
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"Henry",
"Hollis Williams",
"Igor Rivin",
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"JRN",
"Jeff Harvey",
"Jim Conant",
"John Coleman",
"Kenneth Derus",
"Lennart Meier",
"Manfred Weis",
"Martin Brandenburg",
"Martin Peters",
"Martin Sleziak",
"Michael Hardy",
"N. Virgo",
"Nik Weaver",
"Nikolaj-K",
"Piyush Grover",
"R Hahn",
"Rebecca J. Stones",
"Roland Bacher",
"Sam Hopkins",
"Solveit",
"Timothy Chow",
"Todd Trimble",
"Tom Copeland",
"Tri",
"Wlod AA",
"Z. M",
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Stack Exchange
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Mathematicians with both “very abstract” and “very applied” achievements
Gödel had a cosmological model. Hamel, primarily a mechanician, gave any vector space a basis. Plücker, best known for line geometry, spent years on magnetism. What other mathematicians had so distant interests that one wouldn’t guess one from the other?
(Best if the two interests are not endpoints of a continuum, as may have been the case of past universalists like Euler or Cauchy. For this reason, maybe best restrict to post-1850 or so?)
The point of asking is that it seems not so rare, but you don’t normally learn it other than by chance.
Edit: Now CW, works best with “one mathematician per answer” (and details of actual achievement, e.g. “war work on radar” may have been creative for some but maybe not all who did it).$\,\!$
Almost all known mathematicians are also known for their applied achievements, especially the older mathematicians. For instance, topologist Karol Borsuk, when living under the WWII German occupation, had developed an entertaining game from sales of which he derived his income which helped him to survive in those harsh times. In later years, algebraic topologists, the first students of S.Novikov, namely Vitia Bukhshtaber and Sasha Mishchenko, worked intensively on applied projects. The list of examples is endless.
The only mention of Gauss is in passing, in an answer mentioning several other people. But Gauss did important applied work in geodesy and physics. After all, his name was given to the unit for magnetic flux density. However, I don't feel competent to write an answer about him.
@FaheemMitha, Gauss (1777–1855), of course! However, he is just barely post-1850.
A narrower question would be about mathematicians who hold patents (inventions).
@WlodAA I missed the 1850 cutoff. Though it seems a bit arbitrary.
I asked exactly the same question elsewhere and got shut down with almost no answers. https://mathoverflow.net/questions/345636/examples-of-mathematicians-who-excelled-in-pure-and-applied-mathematics
@Tom, that' MO-life/game, it comes with the territory.
@Wold AA, haha, no, that's the way some very highly-opinionated, demonstrative people get their jollies when dealing with outsiders, as you find in most facets of life. Don't condone it by normalizing it.
Possible duplicate of Examples of Mathematicians who excelled in Pure and Applied Mathematics. (I apologize for answering this question and not the earlier one. I was not aware of the earlier one.)
@TomCopeland, your and my statements coexist peacefully in perfect harmony.
Einstein invented a refrigerator. Although Einstein is probably not a "mathematician"?
It may be a stretch to call Gödel's cosmological model applied, even relative to other cosmology.
@Nikolaj-K You have a point :-) I guess I’m really on about the distance, i.e. interests so far apart that they fly in the face of common prejudice / pigeonholing. So in his case I was ready to be surprised by almost anything even remotely applied.
@Wlod AA, reading about Perelman in "Poincare's Prize" by Szabo and the revisionist account "Grothendieck : The Myth of a Break" by Claude Lowbry, maybe disillusionment stemming from gatekeepers is more serious and prevalent than I would like to think. (Rota's advice: Start your own journal.)
@Joel Reyes Noche: My earlier question got closed although it was basically the same question for reasons no-one cared to explain to me, whereas this later version got 58 upvotes.
@HollisWilliams, it seems that sentiments have changed since your question was asked.
John von Neumann was the first person to come to my mind.
He published over 150 papers in his life: about 60 in pure mathematics, 60 in applied mathematics, 20 in physics, and the remainder on special mathematical subjects or non-mathematical ones.
[…]
In a short list of facts about his life he submitted to the National Academy of Sciences, he stated, "The part of my work I consider most essential is that on quantum mechanics, which developed in Göttingen in 1926, and subsequently in Berlin in 1927–1929. Also, my work on various forms of operator theory, Berlin 1930 and Princeton 1935–1939; on the ergodic theorem, Princeton, 1931–1932."
Not to forget his computer architecture with a single central processing unit
And at the opposite extreme, his contributions to set theory --- you can't get more pure than that. The modern definition of an ordinal as the set of all smaller ordinals is due to him. (I was told the paper where he did this was written when he was in high school.)
In principle, it was John vin Neumann who once and forever, up to details, defined life, and, following Ulam's additional ideas, constructed first artificial living "organism". His notion of life was somewhat narrow which made sense as the crucial starting point.
Also the first person to define the concept of a computer virus, now ubiquitous.
I realise that Wlod was referring to the computer virus when he talked about the construction of an artificial living organism.
Alan Turing is one mathematicians with both “very abstract” and “very applied” achievements: https://en.wikipedia.org/wiki/Alan_Turing
His applied work is diverse: He broke nazi codes and he worked on the chemical basis of morphogenesis, predicting oscillating chemical reactions before they were observed.
@MichaelHardy, "He [Turing] broke nazi codes" -- Turing did no such thing, but, yes, he was a part of a huge team working on breaking Enigma (German encrypting machine). It has started with Polish mathematicians. British mathematicians/chessplayers were later in charge. They enrolled in a future topologist (homotopy), young -- at the time -- Peter Hilton. etc.
Would it be possible to expand this answer? E.g. giving examples of his abstract and applied achievements.
Jean Leray started in fluid mechanics. His work might not be applied enough for your question, but when he was made prisoner during WWII he concealed his knowledge about fluid mechanics thinking he could be asked to provide expertise to the enemy. Instead he developed sheaf theory and spectral sequences while a prisoner.
Interesting! Thank you.
So in a sense, sheaf theory is "pacifistic" mathematics. I like that.
David Mumford did foundational work in algebraic geometry, but then later in his career did a lot of work in the applied areas of vision (especially computer vision) and pattern theory.
Did any of the computer vision work pan out (no pun intended)?
@Tri: not an expert so can't say for sure, but my impression is that yes his work in vision was also fundamental, e.g., see https://en.wikipedia.org/wiki/Mumford%E2%80%93Shah_functional which apparently is actually used in modern image programs.
Has everyone forgotten John Nash? I believe his contributions to game theory are among the most prominent examples of mathematical ideas which are widely used in other fields.
On the pure end, the De Giorgi-Nash(-Moser) theorem is a landmark of elliptic and parabolic PDE, which establishes the uniform control of solutions of linear PDE with no assumptions on the smoothness of the coefficients - this is widely used in applications to the nonlinear case. Nash also resolved the isometric embedding problem in differential geometry with a very clever analytic approach. The main part of his proof established a particular instance of what is now known as the Nash-Moser implicit function theorem. The statement is somewhat innocuous (following Richard Hamilton's formulation, it extends the implicit function theorem from Banach spaces to 'tame Frechet spaces') but Nash's proof was very daring. Nash also resolved the isometric embedding problem in a different way, by looking for a low-regularity solution. The 'impossible'-looking thing about this paper is that he shows that n(n+1)/2 simultaneous PDE can be satisfied by finding only n+2 different functions, the key being that this is impossible if the n+2 functions are even as much as twice-differentiable. Nash's proof is remarkably direct.
(I remember, when he got the Nobel Prize in 1994, sitting at a lunch table with a group of mathematicians in different fields, each of whom knew of Nash from his work in their subject, all trying to figure out whether "their" Nash was the guy getting the prize.)
"On the pure end ... do I need to list examples?" Yes, please!
@Nathaniel The De Giorgi-Nash(-Moser) theorem is a landmark of elliptic and parabolic PDE, which establishes the uniform control of solutions of linear PDE with no assumptions on the smoothness of the coefficients - this is widely used in applications to the nonlinear case.
@Nathaniel Nash also resolved the isometric embedding problem in differential geometry with a very clever analytic approach. The main part of his proof established a particular instance of what is now known as the Nash-Moser implicit function theorem. The statement is somewhat innocuous (following Richard Hamilton's formulation, it extends the implicit function theorem from Banach spaces to 'tame Frechet spaces') but Nash's proof was very daring - I think that to most analysts it 'looks' like it has no chance of working.
@Nathaniel and Nash also resolved the isometric embedding problem in a different way, by looking for a low-regularity solution. The 'impossible'-looking thing about this paper is that he shows that n(n+1)/2 simultaneous PDE can be satisfied by finding only n+2 different functions, the key being that this is impossible if the n+2 functions are even as much as twice-differentiable. Nash's proof is remarkably direct.
@slcvtq thank you, that's very useful information. I felt it would improve the answer greatly, so I've edited it in.
@Nathaniel He also has a significant paper where he shows short-time existence for solutions of a fluid problem, although I'm unfamiliar with any of the details. In algebraic geometry, he showed that any manifold type can be realized as a real affine algebraic variety. I'm not sure how significant this result is, although it's well-known.
@Nathaniel: your edits really improve the answer, thank you.
Stanisław Ulam is best known for his work on the Manhattan Project, but his contributions to pure mathematics include pioneering work on measurable cardinals and formulating the reconstruction conjecture in graph theory.
Abraham Robinson made contributions to mathematical logic (model theory, nonstandard analysis) and aerodynamics (airplane wing design).
In December of 1942 Robinson wrote to his supervisors in Jerusalem that he had decided to participate in the general struggle against Fascism and apply his knowledge in applied mathematics to this end. He remarked that there was no effort for him to turn to applied problems. Robinson addressed the problem of comparison between single-engine and twin-engine planes for which he suggested an analog of the variational method by Ludwig Prandtl. He also worked on the problem of structural fatigue and collapse of a flying boat.
In 1944 Robinson married Renée Kopel, a fashion photographer. Abby lived with Renée up to his terminal day.
Robinson was a member of the group studying the German V-2 missiles as well as of a mission of the British Intelligence Objectives Subcommission which concerned intelligence gathering about the aerodynamical research in Germany. In 1946 Robinson returned to Jerusalem to pass examinations for the Master degree. The results were as follows: “physics good, mathematics excellent.” In this short period Abby worked together with Theodore Motzkin.
In 1946 the Royal College of Aeronautics was founded in Cranfield near London. Robinson was offered the position of a Senior Lecturer with salary 700 pounds per year. It is worth mentioning that Robinson was the only member of the teaching staff who learned how to pilot a plane. In Cranfield Abby became a coauthor of delta-wing theory for supersonic flights, and in 1947 he learned Russian in order to read the Soviet scientific periodicals.
To gain the PhD degree, Robinson joined the Birkbeck College which was intended for mature working students and provided instructions mainly in the evening or on weekends. Abby's supervisor in the college was Paul Dienes, a Hungarian specialized mainly in function theory. Dienes instigated Abby's interest in summation methods (which resulted lately in Abby's work with Richard Cooke who also taught in the Birkbeck College). Dienes was a broad-minded scientist with interests in algebra and foundations. In 1938 he published the book Logic of Algebra, the topic was close to Abby's train of thoughts. In this background Robinson returned to logic and presented and maintained the PhD thesis “On the Metamathematics of Algebra” in 1947.
In 1951 Robinson moved to Canada where he worked at the Department of Applied Mathematics of Toronto University. He delivered lectures on differential equations, fluid mechanics, and aerodynamics. He also supervises postgraduate students in applied mathematics. Abby worked on similarity analysis and wrote “Foundations of Dimensional Analysis” which was published only after his death in 1974.
Robinson was the theorist of delta-wing, but his Farnborough research in the area was highly classified. In Toronto Robinson wrote his magna opus in aerodynamics, Wing Theory, which was based on the courses he delivered in Cranfield as well as on his research in Canada. Robinson invited as a coauthor John Laurmann, his former student in Cranfield. The book addressed airfoil design of wings under subsonic and supersonic speeds in steady and unsteady flow conditions. James Lighthill, the creator of aeroacoustics and one of the most prominent mechanists of the twentieth century, appraised most of the book as “an admirable compendium of the mathematical theories of the aerodynamics of airfoils and wings.” Robinson performed some impressive studies of aircraft icing and waves in elastic media, but in the mid-1950s his interest in applied topics had been fading. Robinson continued lecturing on applied mathematics, but arranged a seminar of logic for a small group of curious students.
My impression was that Robinson's achievements in applied math are much less influential than his pure math work. His work on applied math is barely mentioned in the obituary which I read of him, although that might just be because of lack of knowledge on the part of the author.
Some things Kolmogorov did were fairly "pure", but he is one of the eponyms of the Kolmogorov–Smirnov test.
And also contributed to the theory of turbulence in fluid mechanics.
Actually, Kolmogorov is one of the finest examples! An extraordinary span of crucial results in pure mathematics but also for applications.
The modern approach to the theory of probability, based on the set theory and measure theory is due to Andrey Kolmogorov, he is the father of the modern probability theory.
G. H. Hardy is well-known for his work in number theory, but also for the so called Hardy–Weinberg principle in genomics.
An interesting article on this collaboration: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2475721/
Steve Vickers works on topos theory and pointfree topology, but also
He was responsible for the adaptation of the 4K ZX80 ROM into the 8K ROM used in the ZX81 and also wrote the ZX81 manual. He then wrote most of the ZX Spectrum ROM, and assisted with the user documentation.
Vladimir Arnold's work ranges from the very abstract (cohomology ring of the colored braid group, symplectic topology, Maslov Index, real algebraic geometry, invariants of plane curves,...) to the applied (stability of the solar system & Arnold diffusion, Cat map, singularity theory,...) and very applied (gömböc, book on Huygens & Barrow, Newton & Hooke,...). Admittedly, his work forms a continuum, but of wide breadth.
Several known for “pure” work have rather applied contributions to geometrical optics:
Carathéodory (1937) Geometrische Optik
Chaundy (1919) The aberrations of a symmetrical optical system
Whittaker (1907) The theory of optical instruments
Hausdorff (1896) Infinitesimale Abbildungen der Optik
Hensel (1888) Theorie der unendlich dünnen Strahlenbündel
Kummer (1861) Über atmosphärische Strahlenbrechung
Weierstrass (1856) Zur Dioptrik
Sturm (1845) Mémoire sur la théorie de la vision
Listing (1845) Beitrag zur physiologischen Optik
Gauss (1843) Dioptrische Untersuchungen
Liouville (1842) Démonstration d’un théorème de M. Biot sur les réfractions astronomiques
Möbius (1830) Kurze Darstellung der Haupt-Eigenschaften eines Systems von Linsengläsern
Monge (1798) Mémoire sur le phénomène d’Optique, connu sous le nom de Mirage
Gauss is well-known to astronomers. I once talked to an astronomer who was surprised to find that I considered Gauss to be a mathematician.
Israel Gelfand ... was a prominent Soviet mathematician. He made significant contributions to many branches of mathematics, including group theory, representation theory and functional analysis ...
The Gelfand–Tsetlin (also spelled Zetlin) basis is a widely used tool in theoretical physics and the result of Gelfand's work on the representation theory of the unitary group and Lie groups in general.
Gelfand also published works on biology and medicine. For a long time he took an interest in cell biology and organized a research seminar on the subject.
But interestingly, he apparently “hotly denied being a mathematical biologist” and spoke of “the unreasonable ineffectiveness of mathematics in the biological sciences” (p. 29).
That is interesting!
The first person who comes to mind is Hilbert, although his work on quantum mechanics might not qualify as „very applied“. But Wiener and von Neumann certainly fit the bill. Many of the prominent pure mathematicians in the UK (and presumably elsewhere) were involved in war work and did some very applied stuff there—a prominent example would be Robert Rankin. Gröbner in Austria is an analogous case.
Added as an edit: the fact that many pure mathematicians have worked as code breakers is mentioned elsewhere but of course Bletchley Park is a rich source of examples. Turing‘s contribution there is common knowledge but he was one of several, including, notably, William Tutte and Peter Hilton.
Hilbert also came up with the Einstein-Hilbert action before Einstein ...
Einstein never came up with the Einstein-Hilbert action at all, he derived the Einstein equations using a geometric argument and never originally thought of obtaining them by varying an action.
John von Neumann started in axomatic set theory. By the end he was on the (U.S.) Atomic Energy Commission.
I am surprised that no one has mentioned Terence Tao yet. Speaking of his very abstract achievements in math to the audience of this site is like carrying coals to Newcastle. But he has some cool applied achievements as well. Such as compressed sensing, helping physicists model neutrino oscillations and the Polymath project.
Terence Tao is obviously an excellent mathematician, but he didn't model neutrino oscillations. He explained some math behind work of physicists who were modeling neutrino oscillations.
He did nothing to model neutrino oscillations, but he did explain some of the linear algebra which neutrino physicists were seeing in their work.
What about James Simons, of the Chern–Simons form, and working in topology and with manifolds.
He is perhaps more known for his more applied works, making tons of money from the stock market and funding things related to mathematics (including Quanta Magazine, arxiv, conferences, etc).
An interesting biography of Simons was recently published: https://www.goodreads.com/en/book/show/43889703-the-man-who-solved-the-market
I have a friend to whom De Rham was famous for completely different reasons — as author of a “corner-cutting algorithm” used in Computer Aided Geometric Design (of car bodies).
Littlewood is best-known for his pure math research, but during the first World War he worked on ballistics. I suspect there were others who put aside pure math for more applied topics during the wars.
Shing-Tung Yau, well-known to pure mathematicians for the proof of the Calabi conjecture, the Donaldson-Uhlenbeck-Yau theorem, the positive mass theorem in general relativity, and differential Harnack inequalities and various gradient estimates for PDE, also has interesting papers in applied topics such as
"Genus zero surface conformal mapping and its application to brain surface mapping" (with X. Gu, Y. Wang, T.F. Chan, and P.M. Thompson)
"GPU-assisted high resolution, real-time 3-D shape measurement" (with S. Zhang and D. Royer)
"Geometric understanding of deep learning" (with N. Lei, Z. Luo, and X. Gu)
I am not sure why this answer has been downvoted. Yau is an author of many more papers on imaging, discrete Ricci flow, graph theory, etc. than those listed, and some of these papers are more legitimately applied works than some of the contributions of other mathematicians mentioned here and upvoted. To put a randomly chosen example: how is https://www.sciencedirect.com/science/article/pii/S0167839618301249 not an applied paper?
Interesting example, thanks for sharing. Seems that Yau really does have more of an interest in applied mathematics than I thought.
Pafnuty Liebovich Chebyshev contributed to probability and number theory, among other matters - but also worked on steam engine "linkage" design. I also remember being told that he had developed a theory of tensile strength of textile strings - but I can't find evidence of that for now.
Andrew Gleason solved Hilbert's Fifth Problem, contributed to the foundations of quantum mechanics by proving Gleason's Theorem and was a serious cryptographer during and after WWII.
I am surprised the name of Bernhard Riemann did not come up already. He founded a few fields of mathematics and it is a bit funny to justify his presence in this list so I'll be short. On the abstract side, his work in number theory. On the applied side, his work on equations of mathematical physics, including hyperbolic equations and propagation of shocks.
How about Arne Beurling? Complex and harmonic analysis on the "very abstract" side, breaking nazi codes on the "very applied". See https://en.wikipedia.org/wiki/Arne_Beurling and https://bookstore.ams.org/SWCRY/
Gunnar Carlsson
In pure math, he works in homotopy theory, having resolved the Segal Conjecture, as well as in manifold topology, with cases of Borel and Novikov conjectures, and also in algebraic K-theory.
In applied math, he is one of the founders of the field of topological data analysis and among the first (if not the first) to develop persistent homology. With Gurjeet Singh he created the Mapper algorithm based on the Reeb graph. That has been used in many, many settings, for example finding a new genetic marker for breast cancer. He co-founded the company Ayasdi, which continues to develop topological (and topology-inspired) tools along often in conjunction with machine learning, with clientele which ranges from academics to government to industry in an impressive array of areas.
James Glimm did his Ph.D. in C $^*$-algebras, where he made long-lasting contributions. He switched fields soon after, and (as Wikipedia says) he has been noted for contributions to C*-algebras, quantum field theory, partial differential equations, fluid dynamics, scientific computing, and the modeling of petroleum reservoirs.
James Munkres
Very abstract: Obstructions to the smoothing of piecewise-differentiable homeomorphisms (doi:10.1090/S0002-9904-1959-10345-1)
Very applied: Munkres assignment (a.k.a. Hungarian) algorithm for Linear Assignment problems
Claude Elwood Shannon. Do I need to say more?!
Shannon has single-handedly both introduced a new mathematical theory, Information Theory and was the author of the first and fundamental results on his Information Theory. Other theories were introduced by a crowd where one or two people stand out and get extra credit. In the case of Shannon's Information Theory, there was nothing like this.
Information Theory has exciting chapters:
data compression;
error-correcting codes;
encryption
Each of these three theories assists the respective technologies.
Shannon is known also for very practical financial activities like making fun of the casino business, where he induced certain casino modifications, etc.
Shannon's 1940 PhD Thesis at MIT was entitled An Algebra for Theoretical Genetics, and I've heard that its content kept on getting rediscovered in the 60s by multiple researchers.
Kolmogorov had a very high opinion on Shannon. Kolmogorov applied Shannon's ideas (Shannon's entropy function) to solve an old and outstanding problem of classification of Bernoulli shifts (he got half of it, which was a wonderful achievement).
Answers should be self-contained, so yeah, you should say more...
But this one is! C'mon, ...
Yeah, ...no. Although you're onto a nice in-joke here, but that would better apply to Kolmogorov rather than Shannon...
I mostly know Shannon for his paper on communication theory, which certainly had very pure spinoffs in the work of others, but is itself in the "very applied" category. What is the 'pure' work he is known for?
Shannon's information theory had some forerunners. I have some referenes and history in my answer at https://stats.stackexchange.com/a/463828/11887
On the "applied" side of the ledger, Shannon's masters thesis is famous for laying out how digital circuits work. You can find it if you Google "The most significant master's thesis of the 20th century."
On the "pure" side of the ledger, I was reflecting on @N.Virgo 's comment; to my mind, Shannon's perspective (entropy, channel capacity, concentration of measure in high dimensions, etc) seems to so alien compared to the earlier history of (more applied) electrical engineering that his work seems remarkably pure. But perhaps that's a matter of taste.
Maybe the most succinct argument for the breadth of Shannon's interests is in his (now-declassified) 1945 paper on early cryptography. It's divided into three sections, with the 2nd and 3rd being relevant here: (I) Mathematical Structure of Secrecy Systems, (II) Theoretical Secrecy, and (III) Practical Secrecy.
Noga Alon has hundreds of contributions in combinatorics, but also co-authored the foundational paper on streaming algorithms
that has been cited more than 1800 times according to Google Scholar:
Alon, Noga, Yossi Matias, and Mario Szegedy. "The space complexity of approximating the frequency moments." Journal of Computer and system sciences 58, no. 1 (1999): 137-147.
Sam Karlin and Rick Durrett are among the leading probabilists who also contributed to Mathematical biology.
Olga_Ladyzhenskaya
made fundamental contributions to the Theory of PDE and to fluid mechanics.
that foundational paper looks very theoretical
Ronald Fisher is considered an establishing figure in the field of modern statistics. A story I've heard goes that his work in statistics was mentioned to a biologist, who responded "He worked in statistics? I knew he was a huge influencer in biology but had no idea he did statistics."
I believe that his work on design of (agricultural) experiments was a motivation for some of his combinatorial work (on block designs, etc.).
On the internet I came across someone who didn't know that James Clerk Maxwell was known for anything besides the invention of color photography.
With Fisher it can be hard to distinguish between the pure and the applied. Fisher's fiducial methods, introduced in order to solve the Behrens–Fisher problem, is beset with perplexing theoretical questions and with questions in the philosophy of scientific induction, and at the same time the Behrens–Fisher problem is so extremely "applied" that statistics textbooks for people studying biology or psychology or economics, who have no understanding of theoretical mathematics even at the secondary-school level, present an approximate numerical method for dealing with the problem.
@MichaelHardy: With Fisher it can be hard to distinguish between the pure and the applied. Indeed. In fact, as a statistician, I think of many of the "applied" examples in other answers to this question as "pure" examples!
See on rereading Fisher by L Savage. Quote: In addition to Fisher's illustrious career as a statistician he had one almost as illustrious as a population geneticist, so that quite apart from his work in statistics he was a famous, creative, and controversial geneticist. Even today, I occasionally meet geneticists who ask me whether it is
true that the great geneticist R. A. Fisher was also an important statistician.
Fisher held two chairs in genetics, ( ...) , but was never a professor of statistics. ...
Another quote from L Savage: Indeed, my recent reading reveals Fisher as much more of a mathematician than I had previously recognized. I had been misled by his own attitude toward mathematicians, especially by his lack of comprehension of, and contempt for, modern abstract tendencies in mathematics (...). Seeing Fisher ignorant of those parts of mathematics in which I was best trained, I long suspected that his mastery of other parts had been exaggerated, but it now seems to me that statistics has never been served by a mathematician stronger in certain directions than Fisher was.
Norbert Wiener -- well known for his profound mathematics but also as the father of cybernetics.
"Wiener is considered the originator of cybernetics, a formalization of the notion of feedback, with implications for engineering, systems control, computer science, biology, neuroscience, philosophy, and the organization of society.
Norbert Wiener is credited as being one of the first to theorize that all intelligent behavior was the result of feedback mechanisms, that could possibly be simulated by machines and was an important early step towards the development of modern artificial intelligence."
Also:
"During World War II, his work on the automatic aiming and firing of anti-aircraft guns caused Wiener to investigate information theory independently of Claude Shannon and to invent the Wiener filter. (To him is due the now standard practice of modeling an information source as a random process—in other words, as a variety of noise.) His anti-aircraft work eventually led him to formulate cybernetics."
Also ... $\to \infty$.
See: https://en.wikipedia.org/wiki/Norbert_Wiener
Wiener’s Cybernetics sounds rather abstract to me...?
I am not sure what "Robert Solovay's checksum utility" is, but it sounds very applied, and is mentioned in hundreds of LaTeX input files.
He is also one of the giants of set theory, perhaps best known for his model of set-theory in which every set of reals is Lebesgue measurable.
This is the same Solovay as the Solovay-Strassen primality test, is it not?
@ErickWong Yes. And also the same as Solovay as the Solovay-Kitaev theorem. That theorem describes how uniformly a finite;y generated subgroup fills up $SU(2)$. That sounds very pure, but it has major implications in quantum computing (taking the subgroup's generators to be quantum operations for which good error-correction is available).
Although the Lebesgue-measure model is surely Solovay's best-known set-theoretic achievement, I (and I suspect also some other set theorists) consider his and Dana Scott's Boolean-valued models at least equally important. At a first-year grad student in spring of 1967, I audited a class by Tony Martin on independence results. Most of the course was based on Cohen's book, but at the very end Tony briefly described Boolean-valued models. I immediately thought "Oh, so that's what this semester has really been about!"
@AndreasBlass Agreed. It did not occur to me to mention Boolean-valued models as his most important result because (at least for me, and perhaps for my generation) they look so natural (dare I say obvious?).
Raoul Bott. The Bott-Duffin theorem, which is essentially the result of Bott's doctoral thesis (in electrical engineering; the director was Richard Duffin), gives a constructive proof that a positive-real function is the impedance of a transformerless network. This is a basic result in electrical engineering and control theory. Bott's many accomplishments as a topologist are well known.
Surprisingly to me, Garrett Birkhoff also did some very applied mathematics (Wikipedia says "During and after World War II, Birkhoff's interests gravitated towards what he called "engineering" mathematics."). I have his book Hydrodynamics in front on me, and it has plots of experimentally-determined results, photos of objects plunging through water, but also sections on group theory and a Lie algebras. Of course, he also coauthored Algebra with Mac Lane, and is well-known for lattice-theoretic work. He also worked on computational mathematics and numerical linear algebra.
Thierry Coquand works mainly in formal topology, constructive algebra, and foundations, but he was one of the early authors (and namesake!) of the Coq proof assistant. In fact, most people probably know Coquand only as the creator of Coq but not as a logician/algebraist.
I think that they are quite the same thing. The theory behind Coq is quite foundational and logic.
@Z.M The theory, sure; a lot of type theory work was driven by Coq. But according to my understanding, Coquand was also involved in a lot of the "dirty" implementation work.
Eduard Stiefel went from characteristic classes and Lie group representations and topology, to (early) numerical programming and computation of orbits for NASA.
Apologies, this duplicates an answer at @Tom’s question which I hadn’t noticed.
Albert Einstein, in addition, to be a colossus in Physics, had patents (inventions), and he had a significant contribution to Differential Geometry (and, on the top of it, also to tensor analysis, including Einstein notation).
One of his patents was for a refrigerator design! His "very abstract" may have been less abstract than a lot of the people mentioned in other answers, but his "very applied" was a lot more applied than, say, compressed sensing.
Indeed there is a recent and interesting book on the subject (József Illy, 2012).
There is Ernst Zermelo, who is well-known for his work in logic, but who was also a pioneer in optimisation and what is now called control theory.
Zermelo also started as assistant to Max Planck, publishing papers on thermodynamics in a famous polemic with L. Boltzmann 1895–1896.
@FrancoisZiegler Indeed, and this is easily accessible in his Collected Works. See also Heinz-Dieter Ebbinghaus´ biography of Zermelo.
I am slightly surprised that Henri Poincaré is not already on this list. Perhaps it is because almost all of his work could be considered applied mathematics. But his contributions to the foundations of algebraic topology were extremely important, and seem "pure" to me.
Poincaré is also the pioneer of automorphic forms. I don't know whether this is pure.
Poincar'e was also a mining engineer.
Perhaps Helmut Wielandt might be mentioned. As well as his work on finite group theory, he has a famous theorem on doubly stochastic matrices, and another elegant proof (albeit of a previously known theorem) that the equation $AB -BA = I$ can't hold in any (real or complex) normed algebra.
Michel Demazure worked on group schemes as a member of Bourbaki. But he is also known for his work in computer vision for recovering the 3d geometry of a scene by comparing the positions of known points in two still photos of the scene.
John McCleary will give a talk at the JMM in a couple of weeks on "Hassler Whitney and Fire Control in WWII." Whitney "was assigned to work on fire control, the mathematics of aiming weapons for accuracy." Here's the abstract.
Dana Scott's achievements include work in pure set theory and also work in computer science. He proved that there are no measurable cardinals in Gödel's constructible universe and (with Solovay) developed the Boolean-valued-model view of forcing. He also introduced Scott domains (though not with that name) for denotational semantics of programming languages.
Misha Gromov has written on the formalization of genetic and biomolecular structures and the thinking process. Some articles from his website:
Mathematical slices of molecular biology
Functional labels and syntactic entropy on DNA strings and proteins
Pattern formation in Biology, Vision and Dynamics
Crystals, proteins, stability and isoperimetry
Ergostructures, Ergodic and the Universal Learning Problem: Chapters 1,2
Structures, Learning and Ergosystems Chapters 1-4, 6
Memorandum Ergo
Math Currents in the Brain
Learning & Understanding, chapter 1,2
Great Circle of Mysteries: Mathematics, the World, the Mind
Arguably, the greatest ever mathematical logician, Emil Leon Post, was among the main founders of Computer Science (Informatics/Informatique).
The fate was cruel to him, in more than one way, hence no wonder that he is vastly unappreciated.
Emil Post had significant contributions to algebra too. But let's concentrate on mathematical logic.
People don't appreciate the Emil Post's theorem about the elementary logic: tautologies = theorems. It may seem trivial but there are hardly any textbooks which include a complete(!!) proof. There is an objective reason why this theorem is not trivial. Indeed, a minor modification of the axioms of Boolean algebra leads to systems which are very hard to tell from actual Boolean algebras. On occasions, it takes intensive computer programs to decide the issue.
Emil Post had developed formalization independently of David Hilbert (there are trade-offs between the approaches by these two mathematicians).
Emil Post has proved the incompleteness theorem years before Kurt Gödel (again, there were trade-offs between the two).
Emil Post has developed the theory of algorithms independently of Alan Turing; occasionally, people talk about Post-Turing machines.
What work are you considering applied?
As I have been reading Rota's Indiscrete thoughts lately I had in mind the following mathematician, Jacob T. Schwartz.
Citing from the book,
If a twentieth century version of Emerson's Representative Men were ever to be written, Jack Schwartz would be the subject of one of the chapters. The achievements in the exact sciences of the period that runs from roughly 1930 to 1990 may well remain unmatched in any foreseeable future. Jack Schwartz' name will be remembered as a beacon of this age. No one among the living has left as broad and deep a mark in as many areas of pure and applied mathematics, in computer science, economics, physics, as well as in fields which ignorance prevents me from naming.
Let me mention that this is the Schwartz from Dunford--Schwartz. Some widely cited articles in other fields include 'On the Existence and Synthesis of Multifinger Positive
Grips' (Robotics), 'Probabilistic algorithms for verification of polynomial identities', 'Ultracomputers', 'Affine invariant model-based object recognition', 'On meaning', 'Programming with sets: An introduction to SETL',...
Ronald Graham spent his career at Bell Labs working on applied problems such as scheduling theory, but is also known for his work in Ramsey theory. In that context Graham's number held the record for many years as the largest natural number ever used in a serious mathematical proof. Such numbers are not really part of applied mathematics.
Graham's number is not a good example; it was a number he made up during an interview with Martin Gardner, and was not used in a serious mathematical proof. See here https://en.wikipedia.org/wiki/Graham%27s_number#Publication or for that matter here https://mathoverflow.net/a/118650/2926
@ToddTrimble Whether or not Graham's number deserves its hype, it is still illustrative of the fact that Graham did both pure and applied mathematics.
Sadly, the late Ron Graham. See also https://mathoverflow.net/questions/365170/ron-l-graham-s-lesser-known-significant-contributions
Stephen Smale, who is mainly known for his contributions to topology and topological dynamics, also did important work in mathematical economics.
Smale, Steve, Global analysis and economics, Handbook of mathematical economics, Vol. 1, 331-370 (1981). ZBL0477.90014.
It seems like the field of economics that Smale contributed to, general equilibrium theory, is just as non-applied in its nature as his work in topology. His more recent work in biology and learning might be more relevant: "Emergent behavior in flocks" and "On the mathematical foundations of learning" both by Cucker and Smale, and "Learning theory estimates via integral operators and their approximations" and "Estimating the approximation error in learning theorry" and "Shannon sampling and function reconstruction from point values" all by Ding-Xuan Zhou and Smale
Richard Arenstorf worked in number theory and in orbital mechanics. He is best remembered for the Arenstorf orbit used by the Apollo program.
His pure and applied work weren’t that far apart. Both involved a lot of classical analysis. He struck me as sort of a 19th century mathematician, even though he was born in 1929.
Just came across a page of 25+ of Mark Goresky’s Engineering publications.
Mikhail L. Zeitlin, or Gel’fand-Zeitlin basis fame (1950), later switched to “game theory, the theory of automata, computer science, physiology, and mathematical methods of biology”.
The big bird Yuri Manin
Manin is known for his work in algebraic geometry.
He is also father of quantum computing together with Richard Feynman.
Frank Garside (he doesn't have a wikipedia page but he has this https://en.wikipedia.org/wiki/Garside_element) was responsible for solving the conjugacy problem in the braid group, and then became the mayor of Oxford.
George Reid was Senior Wrangler and then an algebraist before becoming Mayor of Cambridge in 1990-91. I am not sure this type of example is what the question was looking for
Richard von Mises did seminal work on the philosophical foundations of probability in terms of long-run frequencies starting in the 1930s. This led to a series of attempts to revise his approach to fix problems, culminating in concepts of algorithmic randomness based on computability from Chaitin, Kolmogorov, Martin-Löf, etc.
Von Mises was also an engineer who is known for contributions to aerodynamics and solid mechanics.
Jerrold Marsden made major contribution to symplectic geometry but also was a key contributor to problems in celestial mechanics and numerical methods.
I think you mean symplectic geometry — which originally was celestial mechanics.
The person with the most citations with "mathematics" on Google Scholar is Eric Lander. He started as a representation theorist, and then moved into molecular biology and genetics.
Several answers (on Beurling, Gleason, Gröbner, Littlewood, Rankin, Robinson, Turing, Ulam, Whitney) suggest that applied work was often classified. I also heard about Vieta being his King’s cryptographer and Monge’s first work being classified.
Two quotes to illustrate that sometimes this means “interesting achievement”:
(Notices AMS 63, p. 505): Above are excerpts from two Nash letters that the National Security Agency (NSA) declassified and made public in 2012. In these extraordinary letters sent to the agency in 1955, Nash anticipated ideas that now pervade modern cryptography and that led to the new field of complexity theory. (In the obituary for Nash that appears in this issue of the Notices, page 492, John Milnor devotes a paragraph to these letters.)
and sometimes apparently not:
(Mac Lane 1976, p. 138): faute de mieux, finds himself in New York as Director of the Applied Mathematics Group of Columbia University, instructed to hire many fresh mathematical brains to help with the research side of the war effort. One of his first acts was to hire Samuel Eilenberg—as well as Irving Kaplansky, George Mackey, Donald Ling, and many others. During the day we all worked hard at airborne fire control... [The] report (more exactly, part 2 on “Aerial Gunnery Problems,” as cited in the bibliography) was initially classified confidential and hence buried in the Government Archives. By now it is declassified, but hardly interesting.
I’d be curious about more examples of declassified work that turned out significant (and would happily move this answer there if someone asks — I just hesitate to spam the site with another big-list question myself).
It would hardly be spam it is such an interesting question! Better that someone with ample reputation ask it, too.
I’ll let someone else do it, if they feel the subquestion is worth splitting off.
Quite a bit of mathematical work/papers (or non-mathematical work by mathematicians) done for DOD contractors were never classified.
As I note in another answer, Hassler Whitney also worked on fire control during the war.
Bill Tutte is another mathematician who did classified work during WWII.
A similar story to that of Leray is David Gilbarg. He originally did his PhD on algebraic number theory with Emil Artin, but then switched to more applied topics because of the Second World War, becoming more well-known for work on PDE theory and fluid dynamics.
William Tutte. He is well known for his contributions to graph and matroid theory, including pioneering the enumeration of planar graphs, and introducing the so called Tutte polynomial. He is less well known for his work on deciphering German codes during World War II, similar to Turing. According to Wikipedia "During the Second World War, he made a brilliant and fundamental advance in cryptanalysis of the Lorenz cipher, a major Nazi German cipher system which was used for top-secret communications within the Wehrmacht High Command. "
Tutte was mentioned (in passing) in an earlier answer, https://mathoverflow.net/questions/349461/mathematicians-with-both-very-abstract-and-very-applied-achievements/349466#349466
@GerryMyerson thanks for pointing that out.
While his other family members outdid Andrzej Trybulec in topology and geometry, he added to these specializations also his creation of Mizar -- the computer proof-checker.
Eugene Dynkin, of probability (Dynkin’s Lemma, among many other things) and Lie algebra (Dynkin Diagrams, in fact according to Wikipedia the whole positive root formalism is worked or by him) fame.
Dan Quillen.
Abstract achievements: his Fields Medal winning work on algebraic $K$-theory, plus inventing model categories, homotopical algebra, and an axiomatic approach to abstract homotopy theory.
Applied achievements: his 1964 PhD thesis was about partial differential equations, and was used in many subsequent applied papers on systems of PDEs, including an Annals paper by Goldschmidt. Later, in the 1980s, he did work in Riemannian geometry and functional analysis and he invented the notion of 'superconnection' in differential geometry and analysis. This work has been applied to work on Heat kernels and Dirac operators, Deformation quantization, elliptic operators, index theory, and topological quantum field theory.
Just as with Grothendieck - just because his work had been USED by applied people, does not make it applied.
I respectfully disagree. Both Grothendieck and Quillen published highly cited papers in applied mathematics. Just because I emphasized further applications by later authors does not mean their papers were not already applied. I leave it to experts to decide (via their votes) if these contributions were "very applied" or "pretty abstract" as you wrote on the Grothendieck answer.
@DavidWhite, I have to agree with Igor. Both Quillen and Goldschmidt’s work on overdetermined systems of PDE are highly abstract and are quite distant from applied math. All of the citations are from equally abstract pure math papers. I do not know of any applied math papers citing their work. It is in fact difficult to find any nontrivial interesting examples, other than already well known ones, of the type of systems they study.
Thanks for this additional context. It was fun for me, as a pure mathematician, to learn a bit about the applied contributions of Quillen and Grothendieck. I'm not surprised that those contributions are of a more theoretical nature, and happy to hear it from experts.
Frank Ramsey wrote two papers in economics --- one on optimal taxation and one on optimal savings --- that remain the foundation of both much theoretical work and of much practical policy-making. His achievements in pure mathematics probably don't need to be reviewed here.
Georg Kreisel did hydrodynamics during and after WWII. Among other things, he determined that the floating harbors used in the D-Day invasion would be stable in heavy seas.
What about "very abstract"?
@David White Proof theory and the metamathematics of constructivity. (F.R.S in 1966 for that.)
Not sure whether this counts: Martin Hairer has developed a musical soft-ware (Amadeus) which is still used, it seems.
Martin Hairer is mentioned here, too: Prominent non-mathematical work of mathematicians.
Alexander Grothendieck
Abstract achievements: his Fields Medal winning work on derived functors, plus a whole new approach to algebraic geometry that has shaped generations of mathematicians after (see EGA, SGA, FGA)
Applied achievements: his PhD thesis was in functional analysis, and his early papers focused on "the theory of nuclear spaces as foundational for Schwartz distributions" (among other topics). This work has applications to stochastic PDEs, elliptic PDEs, probability theory and mathematical statistics, physics, engineering, and kernel functions (hence, data analysis and machine learning). Following citations on Google Scholar is a fun way to see citations to Grothendieck's work from a wide variety of application areas. Grothendieck also published a paper (cited 60 times) on solution spaces to a general class of PDEs.
Grothendieck did pretty abstract functional analysis, so I don't see how it counts.
I have to concur that this seems a bit of stretch. I very much doubt, given Grothendieck's political leanings, that he would have intentionally worked on anything applied.
Yes I think this is too big of a stretch.
Now Clausen–Scholze continued Grothendieck's work on functional analysis. Is that also applied?
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349471
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Stack Exchange
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A variant on characteristic $p$ de Rham cohomology
I was thinking about de Rham cohomology in characteristic $p$, and in particular the recent question about Poincare residues, and I came up with the following construction.
Let $k$ be a perfect field of characteristic $p$ and let $A$ be a regular $k$-algebra. Let $\Omega^j$ be the Kahler $j$-forms, let $Z^j$ be the closed $j$-forms, $B^j$ the exact $j$-forms and $H^j = Z^j/B^j$. The inverse Cartier operator is the unique isomorphism $C^{-1} : \Omega^j \to H^j$ satisfying
$$C^{-1}(\alpha \wedge \beta) = C^{-1}(\alpha) \wedge C^{-1}(\beta) \quad C^{-1}(f) = f^p \quad C^{-1}(df) = f^{p-1} df$$
for $f \in A$. (It is easy to see that there is at most one such map, a nice exercise to see that it is well defined, and not at all clear that it is an isomorphism.)
The inverse operator is an isomorphism $Z^j/B^j \to \Omega^j$, which we can also consider as a surjection $Z^j \to \Omega^j$. By abuse of notation, I'll write $C$ for the surjection $Z^j \to \Omega^j$ as well. We thus have two maps $Z^j \to \Omega^j$: The surjection $C$, and the obvious inclusion.
Define a differential form $\alpha \in \Omega^j$ to be forever closed if, for all $i$, we have $C^i(\alpha) \in Z^j$. Note that we must have $C^{i-1}(\alpha) \in Z^j$ for it to make sense to define $C^i(\alpha)$, so this condition spells out as "we impose that $\alpha$ is closed, and therefore $C(\alpha)$ is defined, and we impose that $C(\alpha)$ is closed, and therefore $C^2(\alpha)$ is defined, etcetera."
Define a forever closed form $\alpha$ to be "eventually exact" if $C^k(\alpha)$ is $0$ for $k$ sufficiently large. Note that exact forms are eventually exact, since the exact forms are the kernel of $C$. Define the eventual cohomology, $EH^j$, to be the forever closed forms modulo the eventually exact forms.
It looks like $EH^{\bullet}$ is always finite dimensional, and forms a graded ring. It does not appear that the dimension of $EH^j$ gives topological betti numbers -- it appears to give something like the multiplicity of the highest weight part of the cohomology.
Is this some object people have studied before?
Old version had a statement about Meyer-Vietores which I no longer think is right. If $X$ is affine and $X = U \cup V$ with $U = { f \neq 0 }$ and $V = { g \neq 0 }$, then I thought before that any forever closed form $\gamma$ on $U \cap V$ was of the form $\alpha + \beta$ for $\alpha$ defined on $U$, $\beta$ defined on $V$ and both forever closed. But all I can really show is that, for any $N$, I can find $\alpha$ and $\beta$ with $\alpha+\beta=\gamma$ and $\alpha$ and $\beta$ each $N$-fold closed.
To be able to compute the iterates of the Cartier operator it is convenient to understand how $C$ interacts with the de Rham differential:
Cartier isomorphism induces an isomorphism of complexes $$(\Omega^{*}_A,d_{dR})\simeq (H^{*}(\Omega^{\bullet}_A),\beta)$$ where $\beta$ is the Bockstein differential provided by the distinguished triangle $$\Omega^{\bullet}_A\to R\Gamma_{cris}(A/W_2(k))\to \Omega^{\bullet}_A$$ It shows that for a closed form $\alpha$ the image $C(\alpha)$ is closed iff the class $[\alpha]$ is annihilated by the Bockstein homomorphism which in turn is equivalent to the liftability of $\alpha$ to class in $H^i_{cris}(A/W_2(k))$. Passing to cohomology in the above isomorphism, composing it with the Cartier isomorphism and iterating this procedure $(i-1)$ times we get an isomorphism $$(\Omega^{*}_A,d_{dR})\simeq (E_i^{(1-i)*,*},\beta_i)$$ of the de Rham complex with the complex appearing on the $i$-th page of the Bockstein spectral sequence associated to the crystalline cohomology complex.
These facts can be seen easily from the following description of the Cartier isomorphism: choose a lift $\tilde{A}$ of $A$ to a complete formally smooth algebra over $W(k)$ equipped with a lift $\widetilde{Fr}$ of the Frobenius endomorphism of $A$(the existence of such lift follows from the vanishing of the relevant obstruction groups which is implied by smoothness of $A$ over $k$). The Cartier operator applied to a form $\omega\in \Omega^i_A$ is then given by $C(\omega)=\overline{\frac{\widetilde{Fr}^*(\tilde{\omega})}{p^i}}$ where $\tilde{\omega}$ is any lift of $\omega$ to a form on $\tilde{A}$ and $\overline{\cdot}$ denotes the reduction.
By tracing through the construction of the Bockstein differentials we get the following
Lemma. For a closed form $\alpha$ the $i$-th iteration of the Cartier operator is defined and gives a closed form if and only if
$[\alpha]\in H^j(\Omega_A^{\bullet})$ lifts to a class
$\widetilde{[\alpha]}$ in the crystalline cohomology of $A$ over
$W_{i+1}(k)$. The $(i+1)$-th iteration of the Cartier operator is zero
if and only if the class $p^i\widetilde{[\alpha]}\in
H^j_{cris}(A/W_{i+1}(k))$ vanishes.
Combining these conditions for all $i$ we get that a form is forever closed iff its class is in the image of the map $H^j_{cris}(A/W(k))\to H^i_{dR}(A/k)$ and it is eventually exact iff the class is in the image of $H^j_{cris}(A/W(k))[p^{\infty}]\to H^i_{dR}(A/k)$.
Crystalline cohomology of $A$ coincides with the cohomology of the $p$-adically completed de Rham complex of any lift of $A$ to $W(k)$ and for the purposes of computing the above invariants we can replace $H^j_{cris}(A/W(k))$ by the (non-complete) de Rham cohomology $H^j_{dR}(\widetilde{A}/W(k))$ where $\widetilde{A}$ is any lift. The quotient $H^j_{dR}(\widetilde{A})/H^j_{dR}(\widetilde{A}){[p^{\infty}]}$ is a $W(k)$-lattice in the finite-dimensional vector space $H^j_{dR}(\widetilde{A}[1/p]/W(k)[1/p])$(it is finite-dimensional e.g. by comparison with singular cohomology).
It indeed seems to follow that $EH^j$ is a finite-dimensional vector space over $k$ with dimension at most the $j$-th rational Betti number of any lift of $A$.
Thanks! Am I reading correctly to say that $H_{dR}(\tilde{A})/H_{dR}(\tilde{A})[p^{\infty}]$ lives in $H_{dR}(\tilde{A}[1/p]/W(k)[1/p])$ as those de Rham classes of $\tilde{A}[1/p]$ which can be represented by differential forms in $\Omega^{\bullet}(\tilde{A})$ (i.e. without denominators)? And then $EH$ is the tensor product of this with $W(k)/p W(k) \cong k$?
Wait, I'm suspicious of your removing the $p$-adic completion. Let $A$ be the coordinate ring of a supersingular elliptic curve with $1$ point removed and let $\tilde{A}$ be an elliptic curve mins a point over $\mathbb{Z}_p$ deforming $A$. I believe that every $1$-form is eventually effective, so $EH=0$. But let $\omega$ be the invariant differential form. I am pretty sure that $p^k \omega$ is not exact for any $k$. I suspect the resolution is that there is an element in the $p$-adic completion of $\tilde{A}$ whose differential is $\omega$, but that we need to work with this completion.
I am now less confident in the last sentence. I took $y^2 = x^3+x$. The invariant form is $\omega = \tfrac{dy}{3x^2+1} = \tfrac{dx}{2y}$, which reduces to just $dy$ in characteristic $3$. So $\omega$ is exact modulo $3$, but is not exact on the $3$-adic lift. This seems to me to be a problem for your formula $H_{dR}(\tilde{A})/H_{dR}(\tilde{A})[3^{\infty}]$ (or for my understanding of it) because it seems to me that $\omega$ is a nonzero class in $H_{dR}(\tilde{A})/H_{dR}(\tilde{A})[3^{\infty}]$ . I thought that I could work with the $3$-adic completion of $\tilde{A}$ instead, (continued)
but I computed $100$ coeffcients of the power series $\sum p_n x^n$ for which $d \sum p_n x^n y = \omega$ and the $p_n$ did not approach zero $3$-adically, or even stay in $\mathbb{Z}3$; I found that $v_3(p_n)$ seemed to stabilize at $-2$. Probably I just misunderstand the notation $H{dR}(\tilde{A})$. Thanks for any help!
@DES-SupportsMonicaAndTransfolk I don't think that the reduction of the module $H_{dR}(\tilde{A})/H_{dR}(\tilde{A})[p^{\infty}]$ is necessarily equal to $EH$. Rather, the reduction surjects onto $EH$ but there can be a non-torsion class in the de Rham cohomology over $W(k)$ that is congruent to a torsion class and it happens in your example: $dy/(3x^2+1)$ is equal to $dy+3\frac{-x^2}{x^2+1}dy$(this is just saying that $\omega$ reduces to an exact form, as you're saying).
So the quotient of the de Rham cohomology by the torsion is a rank two module(because the cohomology of a punctured elliptic curve in characteristics zero is two-dimensional) that surjects onto one-dimensional $EH$ in characteristic $p$. I agree that the cohomology of complete elliptic curve does not contribute to the $EH$ but there is a log-form $dy/y$ invariant under Cartier operator that makes $EH$ one-dimensional, I think.
Thanks! I think my remaining complaints/confusions are focused on the word $W(k)$-lattice in the sentence "The quotient $H^j_{dR}(\widetilde{A})/H^j_{dR}(\widetilde{A}){[p^{\infty}]}$ is a $W(k)$-lattice..." I thought $W(k)$-lattice meant finitely generated. But it looks like, in this example, the quotient in question is a non-finitely generated $W(k)$-submodule. If you agree, then I think I have no more confusion.
@DES-SupportsMonicaAndTransfolk Oh, I don't have a proof but I was hoping that $H_{dR}(\tilde{A})$ doesn't contain infinitely $p$-divisible elements in which case the quotient by the torsion is a $p$-adically separated submodule of a finite-dimensional vector space $H_{dR}(\tilde{A}[1/p]/W(k)[1/p])$ so it is finitely generated. How do you see that the quotient is non-finitely generated in this example?
@DES-SupportsMonicaAndTransfolk My first comment wasn't quite to the point: if we denote $M=H^j_{dR}(\tilde{A}/W(k))/H^j_{dR}(\tilde{A}/W(k))[p^{\infty}]$ then there is a surjection $M/p\to EH^j$ and the element $dx/y$ is killed by this map just because its classe is divisible by $3$ in the cohomology of $\tilde{A}$ but, more interestingly, $-x^2/(3x^2+1)dy$(this is an element whose class times $3$ is equal to the class of $dx/(2y)$) reduces to a non-zero class $-x^2dy$ modulo $p$ but $-x^2/(3x^2+1)dy$ is congruent to $-x^2dy$, the latter representing a torsion class because $-3x^2dy=d(-x^3y)$
... is exact. So $x^2/(3x^2+1)dy$ dies in the eventual homology even though this class is non-torsion in the de Rham cohomology over $W(k)$.
Note that $d(x^n y) = n x^{n-1} y dx + x^n dy = (n x^{n-1} 2y^2 + x^n (3x^2+1)) \omega$. We have $n x^{n-1} 2y^2 + x^n (3x^2+1) = (2n+3) x^{n+2} + (2n+1) x^n$. So $x^n \omega$ is cohomologous to $- \tfrac{2n+3}{2n+1} x^{n+2} \omega$ and thus $\omega$ is cohomologous to $\pm \left( \prod_{j=0}^{k-1} \tfrac{4j+3}{4j+1} \right) x^{2k} \omega$. Taking $k = 3^{2m+1}$ for large $m$ seems to give that $\omega$ is infinitely $3$-divisible in $H^1_{DR}(\tilde{A}[1/3])$.
I don't trust my ability to get indexing right in the previous comment, so work through a few cases yourself if you want to get all the details right.
That seems right, thanks! In fact, for any torsion-free $\mathbb{Z}_p$-module $M$ with a finite-dimensional $M[1/p]$ the dimension of the image of the map $M\to M/p$ over $\mathbb{F}_p$ is equal to the $\mathbb{Z}p$-rank of the completion of $M$ so the above discrepancy in dimensions can only happen in the presence of divisible elements. I wonder, however, whether this divisible part of $H{dR}$ is related to some well-known cohomological invariants.
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2025-03-21T14:48:29.533291
| 2020-01-01T06:23:59 |
349480
|
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}
|
Stack Exchange
|
Exponential maps on a level set
Thanks in advance for any help or a link for me to read up.
Consider a known differentiable function $f(x_1,\ldots,x_d)$, where $x_i$ are real numbers and $(x_1,\ldots,x_d) \in\Bbb R^d$ i.e. are points in the $d$ dimensional Euclidean space. Given the zero level-set $D = \{(x_1,\ldots,x_d)\in\Bbb R^d\, |\, f(x_1,\ldots, x_d) = 0\}$, let's assume that $D$ forms a $d-1$-dimensional manifold and that $D$ is not empty.
Starting from a point on $D$, $p \in D$, how do I compute the exponential map? At every such point $p$, what I know are:
normal vectors $\mathbf n(p)$, computed from $\frac{\mathrm{d}f}{\mathrm{d}x}$ and then normalize.
tangent space $T_p(D)$, computed using $\mathbf v(p)\cdot\mathbf n(p) = 0$ for all $\mathbf v\in T_p(D)$.
I can sample a random tangent vector $\mathbf v \in T_p(D)$ and try to construct the exponential map from here.
In sum, my question is: how to compute the exponential map and trace out a geodesic in $D$?
Thank you.
The answer is essentially the same as for any hypersurface. Probably the easiest approach is to use the implicit function theorem to write $D$ locally as a graph $x_d = h(x_1, \dots, x_{d-1})$ and use the characterization of geodesics on a graph. Basically, a parameterized curve is a constant speed geodesic if and only if its acceleration, as a curve in Euclidean space, is everywhere normal to the hypersurface.
|
2025-03-21T14:48:29.533427
| 2020-01-01T07:05:51 |
349482
|
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|
Stack Exchange
|
gradient curve $\gamma$ defined on $(-T,0]$, can't be extended from $\gamma(-T)$?
Let $f$ be a semi-concave function on an Alexandrov space $X$.
Denote $\gamma_p(t)$ the $f$-gradient curve with $\gamma_p(0)=p$, i.e.
$$
\gamma^+_p(t)=\nabla_{\gamma_p(t)}f.
$$
If $X$ is a Riemannian manifold, $\nabla_{\gamma_p(t)}f=\nabla f(\gamma_p(t))$.
For any $p\in X$ and sufficiently small $t\geqslant 0$, there is $p'\in X$ such that $\gamma_{p'}(t)=p$. In other words, there exists a maximal $T>0$, $\gamma_p(t)$ can be defined on $(-T,0]$ with $\gamma_p(0)=p$.
Let $q=\lim_{t\to -T}\alpha_p(t)$. Then $q$ may be a critical point of $f$, i.e. $\nabla_q f=0$. Or $|\nabla_q f|>0$, but if we connect gradient curve $\gamma_{q}(t)$ defined on $(-T',0]$ ($\gamma_q(0)=q$) with gradient curve $\gamma_p(t)$ at $q$, this is not a gradient curve since it's possible that
$$
\nabla_q f\neq \gamma_p^+(-T).
$$
Can one show some examples for this case?
The answer depends on your definition of semi-concave functions. If you only require them to be semi-concave on geodesics then an obvious example is given by $X$ equal to the closed unit ball in $\mathbb R^n$ and $f=d(\cdot, \partial X)$. Gradient curves starting on the boundary can not be extended to the left.
If you require the function to be semi-concave on the double of $X$ then such example is impossible and gradient curves can always be extended to the left at points where $\nabla_pf\ne 0$. This follows by a standard homology argument. I'll leave you to fill the details but briefly it goes like this. Suppose we have a point $p\in X^n$ such that $\nabla_pf\ne 0$ but the gradient curve starting at $p$ can not be extended to the left. Suppose $p\notin\partial X$. Then $H_n(X,X\backslash \{p\},\mathbb Z_2)\cong H_{n-1}(\Sigma_pX,\mathbb Z_2)\ne 0$. On the other hand since the gradient curve starting at $p$ does not extend to the left we can use the gradient flow $\phi_t$ of $f$ to push any relative cycle in $C_n(X,X\backslash \{p\},\mathbb Z_2)$ into $X\backslash \{p\}$ which implies that it's equal to zero in homology which gives a contradiction. If $p\in\partial X$ then the same argument applies to the flow of the double of $f$ on the double of $X$ which has no boundary.
Thank you! I know that for any $p$ with $\nabla_p f\neq 0$, there exist $q$ such that the gradient flow starting at $q$ go through $p$. I'm sorry that may question was not clear, so I update it right now. Please read it again.
Sorry, I misunderstood the question and I am still not sure what you are asking. Are you asking for an example when the gradient curve starting at $q$ doesn't pass through $p$? That's not possible because forward gradient flow is Lipschitz.
Thank you! I want to find an example, that, for any $t>-T$, $\nabla_{\gamma(t)} f=\gamma^+(t)$, but at the limit point $\gamma(-T)$, the equality fails.
This is impossible because the limit of gradient curves is a gradient curve. So for $t_i\to -T+$ and $q_i=\gamma(t_i)$ the gradient curves starting at $q_i$ will converge to the gradient curve starting at $q=\gamma(-T)$. This means that the curve $\gamma$ on $[-T,0] is actually a gradient curve.
Thank you! But if the gradient curve $\gamma$ can be defined on $[-T,0)$, we know that there is $q'$ such that the $f$-gradient flow starting at $q'$ goes through $\gamma(-T)$. This contradicts to the maximal of $T$. On the other hand, I check the proof of "the limit of gradient curves is a gradient curve" in Petrunin's paper "Semiconcave...". If gradient curves $\alpha_n(t)\to \alpha(t)$, we can only get that $\alpha^+(t)=\nabla_{\alpha(t)}f$ for almost every $t$. So I am puzzled now.
There would only be such $q'\ne q$ if $\nabla_qf\ne 0$. Which shows that if $T$ is finite and $q=\gamma(-T)$ then $\nabla_qf= 0$. So there is no contradiction. The same thing happens here as for smooth functions on smooth manifolds. Backward gradient flow maximally extended either goes on forever or ends at a critical point.
|
2025-03-21T14:48:29.533953
| 2020-01-01T09:03:33 |
349484
|
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"David Loeffler",
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"sort": "votes",
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|
Stack Exchange
|
What is known about the cohomological dimension of algebraic number fields?
What is the cohomological dimension of algebraic number fields like $\Bbb{Q}$, $\Bbb{Q}[i]$, $\Bbb{Q}[\sqrt{3}]$ or similar? I'm interested in computing the cohomological dimension of $\Bbb{A}^1_k$ with $k$ a finite extension of $\Bbb{Q}$.
By definition, an algebraic number field is a finite extension
of the field of rational numbers $\Bbb Q$.
An algebraic number field $K$ is called totally imaginary
if it has no embeddings into $\Bbb R$.
For example, the field $\Bbb Q(i)$ is totally imaginary,
while $\Bbb Q$ and $\Bbb Q(\sqrt{3})$ are not.
We fix an algebraic closure $\overline K$ of $K$, and we write $\Gamma_K={\rm Gal}(\overline K/K)$ for the absolute Galois group of $K$.
Let $p$ be a prime number. By definition, the $p$-cohomological dimension ${\rm cd}_p(K)$ is the supremum of the degrees of nonzero cohomology over all finite $\Gamma_K$-modules $M$ whose cardinality is a power of $p$.
The cohomological dimension ${\rm cd}(K)$ is the supremum of ${\rm cd}_p(K)$
over all prime numbers $p$.
Theorem 1 (well-known) The cohomological dimension ${\rm cd}(K)$ of an algebraic number field $K$ is $2$ if $K$ is totally imaginary, and $\infty$ otherwise.
Following a suggestion of David Loeffler, I state and prove two other theorems, from which Theorem 1 follows.
Theorem 2. The 2-cohomological dimension ${\rm cd}_2(K)$ of an algebraic number field $K$ is $2$ if $K$ is totally imaginary, and $\infty$ otherwise.
Proof. Assume that $K$ is not totally imaginary. Let $\Omega_{\Bbb R}$ denote the set of real embeddings of $K$; then $\Omega_{\Bbb R}$ is nonempty.
We take $M=\mu_2$. For any odd natural number $n\ge 3$ we have
$$ H^{n}(K,\mu_2)\cong\bigoplus_{v\in\Omega_{\Bbb R}} H^n(K_v,\mu_2)\cong
\bigoplus_{v\in\Omega_{\Bbb R}}
H^1(\Gamma_{\Bbb R},\mu_2)
\cong\bigoplus_{v\in\Omega_{\Bbb R}}\Bbb Z/2\Bbb Z\neq 0.$$
(For the first isomorphism, see J. S. Milne, Arithmetic Duality Theorems, Theorem I.4.10(c). For the second isomorphism, see Atiyah and Wall, Cohomology of Groups, IV.8, Theorem 5, in: Cassels and Fröhlich (eds.), Algebraic Number Theory.)
Thus ${\rm cd}_2(K)=\infty$, as required.
Now assume that $K$ is totally imaginary. Then for any $n\ge 3$ and any finite $\Gamma_K$-module $M$ (in particular, for any $\Gamma_K$-module $M$ whose cardinality is a power of 2)
we have
$$H^n(K,M)\cong\bigoplus_{v\in\Omega_{\Bbb R}}H^n(K_v,M)=0,$$
because $\Omega_{\Bbb R}=\varnothing$. (For the isomorphism, again see Milne, Theorem I.4.10(c).) Thus ${\rm cd}_2(K)\le 2$.
Let $\Omega_f$ denote the set of finite places of $K$. We take $M=\mu_2$. We have
$$H^2(K, \mu_2)\cong\left\{(a_v)\in\bigoplus_{v\in\Omega_f}\Bbb Z/2\Bbb Z\
\mid\ \sum_{v\in \Omega_f} a_v=0\right\}, $$
and hence, $H^2(K,\mu_2)\ne 0$.
Thus ${\rm cd}_2(K)=2$, as required.
Theorem 3. For any odd prime $p$,
the $p$-cohomological dimension ${\rm cd}_p(K)$ of any algebraic number field $K$ equals $2$.
Proof. For any $n\ge 3$ and any finite $\Gamma_K$-module $M$ whose cardinality is a power of $p$, we have
$$H^n(K,M)\cong\bigoplus_{v\in\Omega_{\Bbb R}}H^n(K_v,M)\cong
\bigoplus_{v\in\Omega_{\Bbb R}} H^1(\Gamma_{\Bbb R},M)=0,$$
where $H^1(\Gamma_{\Bbb R},M)=0$ because the group $\Gamma_{\Bbb R}$
is of order 2 while $M$ is of odd cardinality (this is an easy exercise; see also Atiyah and Wall, IV.6, Corollary 1 of Proposition 8). Thus ${\rm cd}_p(K)\le 2$.
Let $\Omega_f$ denote the set of finite places of $K$. We take $M=\mu_p$. We have
$$H^2(K, \mu_p)\cong\left\{(a_v)\in\bigoplus_{v\in\Omega_f}\Bbb Z/p\Bbb Z\
\mid\ \sum_{v\in \Omega_f} a_v=0\right\}, $$
and hence, $H^2(K,\mu_p)\ne 0$.
Thus ${\rm cd}_p(K)=2$, as required.
Maybe it's worth remarking that the infinite cohomological dimension is a local problem at the prime 2 only. For $K$ any number field and $p$ an odd prime, the $p$-cohomological dimension (i.e. the sup of the degrees of nonzero cohomology among all finite $Gal(\overline{K} / K)$-modules whose cardinality is a power of $p$) is always 2.
@DavidLoeffler: Thank you! I will edit my answer.
|
2025-03-21T14:48:29.534189
| 2020-01-01T10:44:45 |
349486
|
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|
Stack Exchange
|
Conjugacy classes of the mapping class group
I am not sure if this is a well known problem, but I was not able to find anything online that answered my question.
Is it known how to tell whether two elements of the mapping class group of a surface are conjugate?
I'd be curious whether considering conjugation in $\mathrm{MCG}^+$ (group of oriented self-diffeotopies) is equivalent to conjugation (of elements of $\mathrm{MCG}^+$) in $\mathrm{MCG}^\pm$ (the group of all self-diffeotopies, which is also $\mathrm{Out}(\Gamma_g)$ for a closed surface of genus $g$). For instance this fails in the $2$-torus, since a non-trivial unipotent matrix is conjugate to its inverse in $\mathrm{GL}_2(\mathbf{Z})$, but not in $\mathrm{SL}_2(\mathbf{Z})$.
An exponential-time solution to the conjugacy problem in the mapping class group was given by Jing Tao, in:
Tao, Jing(1-OK)
Linearly bounded conjugator property for mapping class groups. (English summary)
Geom. Funct. Anal. 23 (2013), no. 1, 415–466.
Tao's main contribution is to prove that two conjugate periodic mapping classes have a conjugator of linear length; the pseudo-Anosov case had already been done by Mazur--Minsky.
To actually determine conjugacy in practice, I think the state of the art in this area is the recent papers of Bell--Webb, who show how to compute distance in the curve graph and Nielsen--Thurston type in practice. Bell has even implemented some of their algorithms (see here). See also the parallel results of Birman--Margalit--Menasco.
Added (4 March 2020): Mark Bell pointed out to me that his software can indeed effectively solve the conjugacy problem in specific examples. Flipper decides conjugacy of pseudo-Anosovs, while Curver decides conjugacy of periodic automorphisms.
|
2025-03-21T14:48:29.534345
| 2020-01-01T11:13:20 |
349489
|
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"Gerhard Paseman",
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|
Stack Exchange
|
Calculating radii allowing for circular placement of polygonal linkage's joints
Given a planar polygonal linkage defined by a sequence of $n$ hinge joints $(j_0,\,\cdots,\,j_{n-1},j_n = j_0)$ with links of fixed lengths $\lbrace\|j_{k+1}-j_k\|=d_k\ |\ 0\le k\lt n\rbrace$ between adjacent joints,
how can the set of radii be calculated that allow for placing all joints of the linkage on a circle with such a radius, while preserving the lengths of its links?
In view of the answers I want to stress that all joint shall lie on the same circle and, that the linkage is cyclically closed, i.e. its joints and links resemble a connected 2-regular graph.
There is already a question that is related to polygonal linkages: Is the area of a polygonal linkage maximized by having all vertices on a circle? which however was not concerned with determining the radius of the circle on which all joints lie when the area enclosed by the linkage is maximal.
Being able to calculate the smallest and the largest of those radii would be sufficient for my purposes, which I will state in a later edit.
In reply to the answers I'd like to add the following findings:
a sharp lower bound for the radius on which all joints of a linkage can simultaneously lie, can be found by differentiating the radius of the circumcircle of a triangle w.r.t. one of its side-lengths:
$$\frac{\partial\left(\frac{x\,y\,z}{\sqrt{2(x^2y^2+y^2z^2+z^2x^2)-(x^4+y^4+z^4)}}\right)}{\partial x}=0\iff x=\sqrt{z^2-y^2} $$
which indicates that the circumcircle of a pair of adjacent links is minimal if the longer of the two is the hypotenuse of a right triangle and that the lower bound for entire linkage equals the maximum of all those minima.
if one divides the link-lengths by the by the sought radius, the resulting values are of the form $2\sin\left(\frac{\varphi_{ij}}{2}\right)$ and function for determining the radius can set up via the trigonometric addition theorems; that function can in turn be converted to rational one by utilizig $$\sin\left(\frac{\varphi_{ij}}{2}\right)=\frac{2t}{1+t^2}\quad \text{and}\quad \cos\left(\frac{\varphi_{ij}}{2}\right)=\frac{1-t^2}{1+t^2}$$
This problem has an algebraic-number solution (set up a system of quadratic equalities between the squared link lengths and the squared distances between hinge points) but with unsolvable Galois groups, so there is not going to be a nice closed formula solution. See: Varfolomeev, V.V.: Galois groups of the Heron–Sabitov polynomials for inscribed pentagons. Mat. Sb. 195 (2004) 3–16 Translation in Sb. Math. 195: 149–162, 2004.
Let the radius be $r$. Of course we need $r \ge d_k/2$ for all $k$. If $\theta_k$ is the angular coordinate of joint $j_k$, with say $\theta_0 = 0$, we have
$$ \theta_{k+1} - \theta_k = \pm 2 \arcsin(d_k/(2r))$$
and we need $\theta_{n} = 2 \pi m$ for some integer $m$. By symmetry we may assume the first $\pm$ is $+$, so there are potentially $2^{n-1}$ choices there.
I suspect that for moderately large $n$ it may be best to handle this numerically, though in principle we could use a system of polynomials.
It should be noted that the set of radii can be empty, suggesting the challenge of maximizing the number of hinge points lying on a circular arc. (Are the joints supposed to lie in cyclic order, or does one double back as Robert seems to assume?) Gerhard "Starting The Arc Of 2020" Paseman, 2020.01.01.
@GerhardPaseman the joints shall all lie on a single circle and the question amounts to for which radii that is possible; I see that I have to edit my question to make that point more explicit.
@Manfred, indeed I am mistaken about the set of radii being empty; my bicycle chain argument suggests otherwise. However, I am using the fact that the linkage is like a bicycle chain. Gerhard "In A Mechanically Topological Sense" Paseman, 2020.01.02.
Some simple reductions.
For n=2 and 3 (rigid linkages) , there is a unique answer. Scaling the problem so that the longest link has length one, this results in radii between $1/2$ and $1/s$ where $s^2=3$. Depending on parity, these are also the absolute minimum values of radii possible.
If one is wiling to settle for a near optimum, one can do a "bicycle chain" argument (assuming a really accommodating chain and appropriate pipe, but hey, such assumptions are part of the benefits package of a theoretician), place the chain in a pipe of appropriate parity (or hopeful diameter) and start placing the chain in the pipe so that many links are flexed to touch the inner diameter of the pipe. One may get many joints on the inner surface, and one may be able to manage all but one joint this way on the pipe. The reason for this argument is that one can attempt to predict the discrepancy, and then adjust the pipe toward a more accommodating diameter. In particular, if one can do this while keeping one joint outside of the pipe after each intermediate placement, one can by continuity estimate an exact radius, which for large n should not differ much from the trial radius.
For lengths all the same, one can do stellar patterns by tracing a pattern on a regular ngon (for k relatively prime to n, place link j on point jk mod n). The desired radii should be easily calculated and are probably in the literature (for some interesting value of 'literature'). This gives for this special case a starting set of radii, and one wonders if between two such radii there is an intermediate radius that has a (most likely nonregular) link placement.
A correction to Robert's post: The set of radii can be infinite for some cases with even n: just use an arc. I suspect the set of radii is finite for odd n.
Gerhard "Building A Circle Of Ideas" Paseman, 2020.01.01.
Here is a nice way to see the infinite set of radii for even n=2m: take an open ended linkage of m links, and place it on a large arc. Now reflect a copy across the (perpendicular bisector of the) arc, making a closed linkage of n links on that arc. Gerhard "Building Linked Circle Of Ideas" Paseman, 2020.01.01.
Gerhard, thanks for your answer but it appears to me that you have overlooked the condition that the linkage is a cyclic one and thus the smallest non-trivial case consists of $n=4$ links. It is the cyclicity what makes the problem hard to solve explicitly.
@Manfred, I don't think so. If (when drawn on the plane) the links do not intersect, then one has a different problem. For n=4, assume the links are the sides of a rectangle. I maintain that the set of radii is infinite. If my picture is wrong for this case, please edit to include what cases you are considering for n=4. Gerhard "Trying To Frame Correct Question" Paseman, 2020.01.02.
Ok, I now see what I had in mind, but didn't mention; you are of course right, if no restriction is made on the turning angles; my silent condition is however that all turns are to the same side, e.g. to the left when traversing the polygon. Sorry for that omission.
|
2025-03-21T14:48:29.534955
| 2020-01-01T11:43:09 |
349491
|
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|
Stack Exchange
|
Encoding a random variable with mutual information constraints
For random variables X and Y, is there any one-bit variable $Z=f(Y)$, such that $I(X;Z)\geq I(X;Y)/B$ where $B$ is the number of bits to represent $Y$?
If X and Y are independent, finding Z is trivial. I think you assume that X and Y are not independent, OK? Also, if X is identical to Y, or X=g(Y), the answer is clear!
We exclude such trivial cases.
Your definition of B is ambiguous. For instance, if Y can have 3 different values, is B=2 or B=ln(3) or B=H(Y) ?
No. The problem of choosing $Z$ with a rate constraint to maximize $I(X;Z)$ is called the information bottleneck problem and is characterized by the solution to an integral equation.
There is no limit to the complexity of how the information about $X$ appears in $Y$.
Consider that after observing $Y=y$ then $X$ is characterized by the conditional distribution $P_{X|Y=y}$. To preserve information about $X$ then $Z$ must describe which of these conditional distributions appears. But there is no limit to the amount of complexity in this weighted collection of conditional distributions $(P_{Y=y}\cdot P_{X|Y=y})_y$.
For a specific example, consider $(X,Y)$ with the joint-distribution:
\begin{equation}
P_{X,Y} = \frac{1}{6}
\begin{bmatrix}
0,1,1\\
1,0,1\\
1,1,0
\end{bmatrix}.
\end{equation}
Then one can check any nontrivial $Z$ as you described gives the same $H(Z),\ H(X,Z)$. One can also check $I(X;Z)=\log_2 3-\frac{4}{3} < I(X;Y)/H(Y)=1-\frac{1}{\log_2 3}$.
|
2025-03-21T14:48:29.535099
| 2020-01-01T12:42:00 |
349494
|
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|
Stack Exchange
|
Defining chain complexes for cellular spaces with local coefficients
Let $X$ be a nice finite cellular complex (a regular CW complex or a simplicial one), equipped with a local system $\mathcal{F}$ of free rank 1 modules over some Noetherian commutative ring $R$. What might be the most natural way to define the chain complex associated with this object?
In the majority of textbooks I saw authors follow the original recipe by Steenrod, and start by choosing arbitrarily a reference point (or a leading vertex in the case of simplicial complexes) in each cell, and then twisting the boundary operator by the module authomorphism corresponding to the homotopy class of paths connecting the respective reference points (and contained within the closure of the cell considered).
I have an impression, that at least for finite cellular complexes it is possible to define the chain complex with local coefficients in a more natural way. Let us think of $\mathcal{F}$ as of a locally constant sheaf of rank 1 free modules over $R$. In the stratification of $X$ associated with the cellular structure, let $S_d$ stand for the stratum consisting of the interiors of $d$-dimensional cells. Topologically, $S_d$ is a disjoint union of a finite number of $d$-disks. Let us define $d$-chains as global sections of the inverse image of $\mathcal{F}$ with respect to the inclusion $S_d \hookrightarrow X$ (the resulting module is naturally the product of copies of $R$, but since there are only finitely many of them, we can think of it as of the direct sum). Now, for any couple of incident cells (a $d$-cell and one of its $(d-1)$-faces), there is a natural isomorphism between the modules of global sections of pullbacks of $\mathcal{F}$ to them. To get the twisted differential it suffices to use this isomorphism (instead of explicit basises in $C_d$) in the definition of the boundary operator.
Even if this construction is limited to finite cellular complexes only, this way of getting rid of arbitrariness still seem to me to be too easy. Is there any flaw in this reasoning?
PS: The condition of finiteness of the cellular complex is indeed necessary, as can be seen from an example of a real line considered as a cellular complex consisting of 1-D cells $]n, n-1[$ and 0-D cells $\{n\}$ for $n \in \mathbb{Z}$. The pullback of $\mathcal{F}$ to the union of 1-D cells is a product of countly many copies of $R$, and thus contains a non-zero constant global section, which is a non-zero cycle in the considered construction. Maybe this can be fixed by treating the local system $\mathcal{F}$ as a cosheaf?
There is a discussion somewhat along these lines in my algebraic topology book (available online) in Section 3.H, pages 327-336.
Thank you, Allen. I remember this part of your book, and both of the proposed approaches (that of tensoring the chain modules of the universal cover over the group ring of the fundamental group and the more geometric one, with singular simplexes living in the group bundle). Indeed, with either of these approaches one can avoid arbitrary choices.
However, my question was about the right to use the sheaf machinery, which seems already to contain a good deal of the construction of the chain complex with twisted coefficients. The question originated from some research in mathematical physics, where I would like to be able to say "this global section of $\mathcal{F}$ over cells of the top dimension is a cycle". Since nobody seems following this road, I am afraid there are some gotchas I do not see.
There is a section 12,4 on local coeficients and local systems in our book Nonabelian Algebraic Topology (groupoids.org.uk/nonab-a-t.html} but it will take a while to get used to the machinery. Many of the basic ideas go back to A.L. Blakers, and to JHC Whitehead, but are put in a new form. For example there is a tight relation between crossed complexes and the more classical chain complexes with operators.
It seems that in the case when the space $X$ comes equipped with with a cellular structure, the right way to represent local systems is through the "locally constant constructble sheaves". If the cellular complex is regular, a constructible sheaf $\mathcal{F}$ is completely characterized by its sections $\mathcal{F}(\sigma)$ over the interior of each cell $\sigma$ and the restriction homomorphisms $\rho_{\sigma \tau}:\mathcal{F}(\tau) \to \mathcal{F}(\sigma)$ for each incidence of a cell $\sigma$ and its face $\tau$ (that is, a constructible sheaf is a functor from the face-relation poset to the target category, see e.g. https://arxiv.org/pdf/1303.3255.pdf section 4.2.2). In the case of a non-regular cellular complex (e.g. a semi-simplicial one) there might be several homomorphisms for a couple of incident cells. If $\mathcal{F}$ is locally constant, all homomorphisms $\rho_{\sigma \tau}$ are isomorphisms, and one can constuct a locally constant constructible cosheaf with the same data (by using $\rho^{-1}_{\sigma \tau}$ as extension homomorphisms).
The key fact is that if the space $X$ is compact, the ordinary cellular cohomologies and homologies with local coefficients are equivalent to the cohomologies of the locally constant constructible sheaf and the homologies of the corresponding cosheaf, respectively (I am not quite sure about what happens when $X$ is not compact, but it seems that in this case one should use cohomologies with local support and Borel-Moore homologies). The recipe for the construction of the (co)chain complex is exactly the same as proposed in the question (see https://repository.upenn.edu/cgi/viewcontent.cgi?article=4722&context=edissertations section 2.2.2).
This mostly answers my question, hope this might help somebody.
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2025-03-21T14:48:29.535536
| 2020-01-01T13:33:52 |
349496
|
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|
Stack Exchange
|
Vanishing of first co-homology with coefficients modular representations of small dimension
Is the following true:
For any $n$ there exists $p_0$ s.t. for any finite group $G$ of Lie type of rank $\leq n$ and characteristic $p\geq p_0$ and any (irreducible) $\mathbb F_p$ representation $V$ of $G$ of dimension $\leq n$ we have
$$H^1(G,V)=0$$
I think the answer to your question is yes for example with $p_0 = n+3$.
See the following paper:
"Small Representations Are Completely Reducible", Robert M. Guralnick, J. Algebra (1999).
Theorem A in this paper says the following:
Let $k$ be a field of positive characteristic $p$. Let $G$ be a finite group containing no nontrivial normal $p$-subgroup. Let $V$ be a $kG$-module such that $G$ acts faithfully on $V$.
(a) If $\dim V \leq p − 2$, then $V$ is completely reducible;
(b) If $\dim V \leq p − 3$, then $H^1(G, V) = 0$.
|
2025-03-21T14:48:29.535657
| 2020-01-01T16:35:38 |
349508
|
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|
Stack Exchange
|
Curvature of Lie groups
Can a Lie group be of negative sectional curvature?
In fact every Lie group contains an entire curve which means that a Lie group does not admit a complete metric with negative sectional curvature
This question makes sense only if you say more precisely how you define a metric on the Lie group: what kind of metric, satisfying which conditions?
I'm talking about the Hermitian metric (defined by a positive (1,1)form )
I agree that this is still unclear — but would guess that Milnor’s “Special Example 1.7” gives a positive ($=$ negatively curved...) answer to what is probably intended. Does it?
The problem for me is: how a complex Lie group on the one hand admits a Riemannian metric with negative sectional curvature and on the other hand it contains copies of \ C therefore it contains entire curves (image of \ C by holomorphic function ) and we know that the presence of an entire curve implies the inexistence of a metric with negative sectional curvature
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2025-03-21T14:48:29.535762
| 2020-01-01T17:09:21 |
349512
|
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Stack Exchange
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What number of colorings can guarantee that for every k-element subset there exists a coloring assigns different colors for elements from this subset?
Let $M(n, k)$ be a minimal number $m$ such that there exists set $C$ ($|C|=m$) of colorings of n-element set $[n]$ with $k$ colors such that for every $k$-element subset $K$ of $[n]$ there exists coloring $c\in C$ such that $c$ assigns different colors for all elements of $K$.
Is there any research about this object?
Is there any closed-form formula for $M(n, k)$? May be for small $k$?
It's obvious that $M(n, 2)=\lceil\log_2(n)\rceil$ (simple divide and conquer idea), but $M(n, 3)$ is not so obvious for me.
These are known as Perfect Hash Families. The notation $PHF(N;n,v,k)$ denotes a set $\mathcal H$ (hash functions) with $|\mathcal H|=N$, of $v$-colorings of the base $[n]$, such that for any $X\subset [n]$ with $|X|=k$, there exists at least one hash function which assigns it different colors. Let's denote by $f(n,v,k)$ the smallest $N$ for which a $PHF(N;n,v,k)$ exists. In this notation, you are interested in $f(n,k,k)$.
M. L. Fredman and J. Komlós prove in the paper "On the Size of Separating Systems and Families of Perfect Hash Functions" the following bounds:
$$\frac{v^{k-1}}{(v)_{k-1}}\cdot\frac{\log n}{\log(v-k+2)}\le f(n,v,k)\le \frac{k\log n}{\log\left(1-\frac{(v)_k}{v^k}\right)}$$
where $(v)_k=v(v-1)\cdots (v-k+1)$ is the falling factorial. This bound was improved by J. Körner and K. Marton in "New Bounds for Perfect Hashing via Information Theory" and there is a large literature making sharper bounds for particular families of parameters. Here is a large table of known parameters of realizable perfect hash families maintained by Ryan Dougherty.
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2025-03-21T14:48:29.535911
| 2020-01-01T17:18:03 |
349513
|
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|
Stack Exchange
|
Canonical étale path between a point and its ''nearby'' point
Consider the punctored line $X=\Bbb{A}^1_k\setminus \{s_1,\ldots,s_n\}$ over some field $k$. A(n étale) path in $X$ between two geometric points $x$ and $y$ is, by definition, an isomorphism between the fiber functors $\operatorname{Fib}_x$ and $\operatorname{Fib}_y$.
Here the functor $\operatorname{Fib}_x:\mathbf{FÉt}(X)\rightarrow \mathbf{Set}$ is the functor sending a finite étale cover $E$ of $X$ to the set underling the fiber $E_x$.
A path between $x$ and $y$ induce an isomorphism $\pi^{ét}_1(X,x)\simeq \pi^{ét}_1(X,y)$.
The spectrum of $\mathcal{O}^{sh}_{X,x}$ ($sh$=strict henselization) is a small (étale) neighborhood around $x$ and has essentially two geometric points, $x$ and the (separable closure of the) generic point $\eta$.
I'm wondering if there is a canonical way of choosing a path between $x$ and $\eta$, i.e. I'm wondering if there is a canonical way of choosing an isomorphism
$$\pi_1^{ét}(X,x)\simeq \pi^{ét}_1(X,\eta)$$
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2025-03-21T14:48:29.536017
| 2020-01-01T17:52:11 |
349519
|
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|
Stack Exchange
|
State of the art on strong full exceptional collections
I am trying to understand on which kinds of varieties we can find strong full exceptional collections of sheaves. (Classical examples are of course projective spaces (Beilinson) and Grassmann varieties and quadrics (Kapranov).)
As far as I understand, the "strongness" of an exceptional collection implies that the derived category of coherent sheaves is equivalent to the derived category of modules over an associative algebra.
If one is content with working with DG algebras one can just focus on exceptional collections, while allowing higher Ext groups between the sheaves in the collection. However, I would like to know what currently is known about varieties with strong exceptional collections.
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2025-03-21T14:48:29.536101
| 2020-01-01T19:59:37 |
349533
|
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|
Stack Exchange
|
Freedom problem in hyperbolic groups
I will start with a general algorithmic question:
Question. (A faithfulness decision problem.) Suppose that $G, H$ are finitely-presented groups with decidable word problem. Is injectivity decidable for homomorphisms $f: G\to H$ (defined by the images of generators of $G$)?
[A very minor side issue: I am not even sure if this decision problem has an established name. For some reason, Max Dehn did not consider it.]
The non-injectivity problem is trivially semidecidable.
More specifically, I am interested in decidability of faithfulness for homomorphisms from (finitely generated, nonabelian) free groups to (nonelementary) hyperbolic groups $H$. The decision problem then becomes:
Freedom problem. Do given elements $g_1,...,g_n\in H$ generate a rank $n\ge 2$ free subgroup?
Things that I know:
If $H$ is locally quasiconvex (or, more generally, all finite rank free subgroups of $H$ are undistorted), then the freedom problem is decidable. The same applies to the injectivity problem for homomorphisms from general finitely-presented groups $G\to H$ to locally quasiconvex hyperbolic groups. The injectivity problem is decidable if $H$ is the fundamental group of a compact hyperbolic 3-manifold.
It is natural to expect that the decidability of the freedom problem in hyperbolic groups $H$ is closely related to distortion of free subgroups in $H$. One can conjecture that if every f.g. free subgroup in $H$ is recursively distorted then the freedom problem is decidable. One can also conjecture that hardness of the freedom problem correlates to the degree of distortion of free subgroups. (For instance, for "hyperbolic hydra groups" of Brady-Dison-Riley, the distortions of some free subgroups are given by Ackerman functions.) I do not know how to approach any of these conjectures.
So, my question is if anything else is known about these problems.
One more thing: There is a "dual" problem about surjectivity of group homomorphisms. I am not asking about this one. (It is more related to the subgroup membership problem and more is known about it.)
This question seems to be about a special case of your general question. I seem to remember asking Martin Bridson about the decidability of the freeness (freedom?) of subgroups of hyperbolic groups, and I believe that he thought that it was unknown but likely to be undecidable.
BTW if $H$ is hyperbolic and $G$ is a f.p. group with non-solvable word problem, the question of injectivity for homomorphisms $G\to H$ is quite easy to solve :)
@YCor: Yes, of course; ditto for infinite simple groups, groups containing $Z^2$, etc., which is why I assumed that $G$ is free, so we have an ample supply of homomorphisms.
You might like to compare with Question 9.4 of https://arxiv.org/abs/1003.5117 . Bridson and I asked if the finite presentation problem is (uniformly) solvable over hyperbolic groups. If the freedom problem is unsolvable in some hyperbolic group, then the answer to our question is (as expected) "no".
As regards the first question (with $H$ not hyperbolic), it has a negative answer even with $G$ free:
This question has an answer saying that there exists a f.p. $H$ group with solvable word problem in which injectivity of homomorphisms $\mathbf{Z}\to H$ is undecidable. (Hence injectivity of homomorphisms $F_n\to H\ast F_{n-1}$ is undecidable too.)
Nevertheless for $H$ hyperbolic, it is decidable whether an element has infinite order (because the conjugacy problem is solvable and there are finitely many finite order conjugacy classes, so it is enough to test conjugacy with each representative).
(Please don't accept this answer, because the freeness problem in hyperbolic groups is more interesting.)
Ah, I see. Thank you!
Can't you test finiteness in a hyperbolic group without using the conjugacy problem? Since there are only finitely many finite conjugacy classes of finite order there is a fixed number $n$ so that every element of finite order has order dividing $n$. To check finiteness you just raise an element to the power $n$ and use the word problem to check if you get $1$.
@BenjaminSteinberg (You mean test finite order.) Indeed, you're right. It looks like a simpler approach.
|
2025-03-21T14:48:29.536408
| 2020-01-01T21:37:43 |
349539
|
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|
Stack Exchange
|
Compatiblity of completion and fibre products. (Formal completion and formal groups)
Let $S$ be a scheme (not necessarily locally noetherian), $X$ a smooth separated group scheme over $S$, and $\hat{X}$ be the formal completion along with the identity section.
Then does the group scheme-structure on $X$ induce the formal group scheme-structure on $\hat{X}$?
i.e., do the multiplication $m : X \times_S X \to X$ and the inversion $X \to X$ factor through $\hat{X}$, and these satisfy the group law?
And is the correspondence $X \to \hat{X}$ a functor from the category of nice group schemes to the category of the formal groups?
If this is true, then an abelian scheme induces a formal group of $\dim$-variables.
(The multiplication is the image of $T$ under $A[[T]] \to A[[X, Y]]$, the inversion is the image of $T$ under $A[[T]] \to A[[T]]$. )
I have shown that a morpshim of schemes induce a unique morphism on their formal completion:
$S$ a scheme, $X, Y$ $S$-schemes given with sections (assume its sheaf of ideals are finitely presented), $f : X \to Y$ an $S$ morphism which is compatible with their given sections.
Then there exists only one morphism $\hat{f} : \hat{X} \to \hat{Y}$ which is compatible with $f$.
So it seems that if we can show the compatibility of completion with fibre products (i.e., $ (X \times_S Y)^{\hat{}} \cong \hat{X} \hat{\times}_S \hat{Y}$, including the canonical map to $X \times_S Y$), the highlighted statement is true.
For, since we can take $\hat{X} \hat{\times}_S \hat{X}$ as the completion of $X \times_S X$, there exists a unique map $m : \hat{X} \hat{\times}_S \hat{X} \to \hat{X} $, compatible with the multiplication on $X$.
And similar to the inversion.
The group law follows from the uniqueness of the completion of morphism.
And if this is true, then a morphism between nice schemes induces a morphism of formal groups, and hence we have a functor from the category of "nice" group scheme to the category of formal groups.
Is my opinion right?
And please show the compatibility of completion with fibre products.
P.S. I have posted a similar question.
The answer says that using the functor of points we can show that $\hat{X} \hat{\times} \hat{X} \to X \times X \to X$ factors through $\hat{X}$.
But I can't understand it.
Thank you very much!
Notice that most of what you have written is summarized by the Lemma "Any finite product preserving functor between categories with finite products induces a functor between the group objects".
|
2025-03-21T14:48:29.536622
| 2020-01-01T22:46:18 |
349540
|
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|
Stack Exchange
|
How to obtain the rational solution of a linear system efficiently?
Suppose that I have a linear system $AX=b$ with $A\in\mathbb{Z}^{n\times m}$ and $b\in\mathbb{Z}^{n}$. Assume that $AX=b$ has exactly one rational solution. Then how can I obtain this solution efficiently?
I know LU decomposition is a choice for this purpose. But it seems that the denominators may expand in the process of computing a LU decomposition, which I hope to avoid if the rational solution has small denominators.
If I already have a numerical solution of $AX=b$, does it help?
It is not necessary that $m=n$ though there is exactly one solution. If the rational solution has no large denominators, then I hope no large denominators appear in the computation.
Maybe this is relevant: https://cs6505.wordpress.com/2018/10/02/linear-equations-i/. One can solve in $O(n^\omega L)$ time an $n\times n$ system with integer entries which take at most $L$ bits to represent.
I've voted to reopen. It is a perfectly legitimate question, concerned with the well known phenomenon of "intermediate expression swell" in computer algebra. Such linear algebra problems may easily arise in applied mathematical research, where practitioners are not expected to be proficient in linear algebra over rings (like the integers). Hence this question could be useful to them.
The following answer is only theoretical. Suppose you first reduce your matrix to Smith normal form $D = S A T$, where $S$ and $T$ are integral matrices with integral inverses (the determinants of $S$ and $T$ are each either $+1$ or $-1$) and $D$ is diagonal, whose diagonal elements are the so-called elementary divisors of $A$. Then you reduce your problem to solving $DY=c$, where $c = S b$ and $X = T Y$. Since by $c$ is still integral and $D$ is now diagonal, the denominators in $Y$ will be no larger than the largest entry in $D$ (the largest elementary divisor of $A$). Then, since $X$ is obtained from $Y$ by multiplication with another integral matrix, the denominators in $X$ will be no larger than the lcm of the denominators in $Y$.
Unfortunately, I don't think that reduction to Smith normal form is a cheap operation, so whether it is worth going through that step in your problem will depend on other factors.
EDIT: Actually, the Smith normal form is overkill for your problem. The Hermite normal form is basically the integer version of the LU decomposition that you already had in mind: $U = S A$, where $U$ is integral upper triangular, and $S$ is integral with integral inverse. The new system you need to solve is $U X = S b$, which can be done by back substitution, where at each step you only divide by the corresponding diagonal element of $U$. Hence, the final denominators in $X$ will be no larger than the lcm of the diagonal elements of $U$.
The linked Wikipedia article on the Hermite normal form mentions that modern algorithms for its computation have polynomially bounded time and space complexity (respectively, in terms of the matrix size and logarithmic size of the matrix entries). If you (or your computer algebra system) use such an algorithm, it might be your best bet for provably avoiding intermediate expression swell. The practicality of this method for your problem of course would still be subject to experimentation.
EDIT 2: I've also found a very well written survey of several algorithms for solving rational linear systems, with practical timing comparisons in the case of sparse systems.
Cook, William; Steffy, Daniel E., Solving very sparse rational systems of equations, ACM Trans. Math. Softw. 37, No. 4, Paper No. 39, 21 p. (2011). ZBL1365.65121. (author's copy)
Part of the hypothesis of OP is that the system has exactly one rational solution. Is there anything to be gained through use of this hypothesis?
@GerryMyerson I don't see how, at least in this factorization approach.
This is standard and can be done using $p$-adic lifting, see for example the following article:
J.D. Dixon, Exact solution of linear equations using $P$-adic expansions, Numer. Math. 40 (1982) pp. 137–141, doi:10.1007/BF01459082.
This also exploits that there is a unique solution.
See also https://doi.org/10.1016/j.jsc.2003.07.004
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2025-03-21T14:48:29.536976
| 2020-01-01T23:57:33 |
349545
|
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|
Stack Exchange
|
Packings with block size equal to $6$?
In design theory the following is the defintion of a packing :
Definition : A $(v,k)$-packing is a pair $(V, \mathcal{B})$ of a finite set $V$ of cardinality $\vert V \vert = v$ and a finite set $\mathcal{B}$ of $k$-subsets of $V$ is a of order $v$ and block size $k$ such that any pair of elements of $V$ appears in at most one element of $\mathcal{B}$. The elements of $S$ are the points, and those of $\mathcal{B}$ are the blocks.
One of the interesting problems related to packings is finding a $(v,k)$-packing with the maximal number of blocks. Let $D(v,k)$ denote the number of blocks in a maximal packing. This is a well-studied problem. I am interested only in a particular case $k=6$ but I could not find any papers studying it?
Question: What is the minimum number $v_0$ for which $D(v_0,6)\geq 100$ ? is there any online database for packings (like the one available for coverings)?
It know that $66 \geq v_0\geq 56$, the first bound is because of the existence of a block design $(66,6,1)$ and the second bound follows from the Johnson bound $$D(v,k) \leq \left\lfloor \frac v k \left\lfloor \frac {v-1}{k-1}\right\rfloor \right\rfloor$$.
Edit: if I understand well the relation between packings and constant weight codes then what I am looking for is the smallest $v_0$ such that $A(v_0,10,6)\geq 100$ in this website there is a lot of bounds on those numbers, in particular, $A(60,10,6) \geq 104$ from $TD(6,10)–TD(6,2)$ (with $96$ blocks) and adding six $6$-lines in the groups and two $6$-lines in the hole. But I don't know how to construct the packing from this information?
This implies that $v_0 \leq 60$ (only four values remaining, probably an open problem)
There should be some tight relation between packing and coverings. Something like a packing for n, k leads by complementation to a covering for n,n-k. Check out Handbook of Combinatorial Designs. Also check out La Jolla repository. Gerhard "Little Fuzzy On Packing Thinking" Paseman, 2020.01.01.
I think that taking the complement of a each block in a packing produces another packing (the same goes for coverings) . Their sizes are very close. but different. we have $D(v,k)\leq f(k,v)\leq C(v,k)$ for some known function $f$. where $C(v,k)$ is the minimal size of a covering
@Gerhard, it took me time, but I did not manage to find your article on la Jolla Repository, do you have a link ? Thank you !
The website has been rearranged. The older version had a link to a paper by many authors (I believe Greg Kuperberg was one), and my sometimes faulty memory tells me that packing was related to covering in that paper (this may be a wrong assertion). At this point, you should wait awhile until someone with a better memory responds here, and then contact Dan Gordon if you don't get satisfaction here. Gerhard "Sorry To Raise Hope Prematurely" Paseman, 2020.01.01.
The CRC Handbook of Combinatorial Designs, 2nd ed., (2007) has seven pages on packings. Skimming through those pages (email me for a copy) says almost nothing about $k=6$ and I know a lot of work went into $k=5$, so I suspect your question is pretty open? The CRC section does reference an old survey by W. H. Mills and R. C. Mullin, "Coverings and Packings" in Contemporary Design Theory, Wiley, 1992.
Perhaps, Gerhard means this article https://arxiv.org/abs/math/9502238
@KenW.Smith, I already checked the book thanks. Yeah, there are a lot of articles focused on the case $k=5$. But, I was asking if there is some new advances since the publication of the book (as today we have more computer power than years ago). Otherwise, as you said, it should still be open.
@ Max Alekseyev, thanks for the article. I will read it and see if I can use the same algorithm for packings.
I have discovered some new information, but I did not understand the construction, I did not quite get how to "add six 6-lines in the groups" if somebody can help?
|
2025-03-21T14:48:29.537386
| 2020-01-02T00:14:02 |
349547
|
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}
|
Stack Exchange
|
Division of space by hyper-planes
It is a well known and lovely result that the maximum number of regions that $\mathbb R^{k}$ (with $k$ positive) can be divided into by $n$ hyperplanes is given by
$$1+n+\binom{n}{2}+\cdots+\binom{n}{k}.
$$ and occurs when they are in general position. It is clear that the minimum with distinct hyperplanes is $n+1$ (when they are parallel ) and the next smallest is $2n$ when all but one are parallel and also when each pair has the same (nonempty ) intersection.
For which $n,k,m$ is it possible to divide $\mathbb R^{k}$ into exactly $m$ regions using $n$ distinct hyperplanes?
Fir purposes of induction it might be better to drop the requirements that the hyperplanes be distinct.
Do you forbid repeated hyperplanes?
Also, these are affine hyperplanes, right?
Yes, affine hyperplanes. I’m willing to allow repeated hyperplanes although then the answer is the union of the possibilities for $1\leq j\leq k$ distinct hyperplanes.
Let's denote by $S_{k,n}$ the set of possible integers $m$, such that $\mathbb R^k$ can be divided into $m$ regions by $n$ hyperplanes. If we denote by $S^{P}_{k,n}$ the set defined similarly but for the projective space $\mathbb {RP}^k$. We have that $S_{k,n}=S^P_{k,n+1}$ since every affine arrangement can be lifted to a projective arrangement with the same number of regions by adding the hyperplane at infinity. Similarly any projective arrangement gives rise to affine arrangements with the same number of regions by deleting one of the hyperplanes. The following result solves the problem for $k=2$:
Theorem: (N. Manturov, "Classification of arrangements by the number of their cells") We have $m\in S^P_{2,n}$ if and only if there exists some integer $0\le r\le n-2$ such that
$$(n-r)(r+1)+\binom{r}{2}-\min\bigg\{ n-r, \binom{r}{2}\bigg\}\le m\le (n-r)(r+1)+\binom{r}{2}.$$
The proof in that paper involves quite a bit of casework and it seems like it wouldn't easily generalize to higher dimensions, however I find one particular aspect very interesting. The way it is shown that all numbers satisfying these inequalities do work is by exhibiting a very simple family of arrangements $\mathbb B^c_{a,b}$ which are obtained as the union of a pencil of $a$ lines (all passing through the same point), a simple arrangement of $b$ lines (no three lines concurrent), such that $c$ of the intersection points of the simple arrangement lie on the first pencil. This leads us to a (possibly too optimistic?) conjecture:
Conjecture: If an arrangement of $n$-hyperplanes in $\mathbb {RP}^k$ has $m$ regions, then there exists an arrangement of $n$-hyperplanes, obtained as the union of $n$ generalized pencils (one for each type), which also has $m$ regions.
By a generalized pencil, I mean taking two disjoint projective plane planes of dimension $r_1,r_2$ in a $(r_1+r_2+1)$-dimensional projective space, picking a generic arrangement of hyperplanes in the first space, and taking the span of each hyperplane with the second projective space. We say that the pencil has type $(r_1,r_2)$. If this conjecture were to be true, there would be some hope of a proof strategy along the lines of starting with an arbitrary arrangement and somehow deforming it into such a form while preserving the number of regions.
Even if you all you did was tightly bound the change in the number of regions, that would do a lot to limit the search space. Gerhard "Sometimes Satisfied With Second Best" Paseman, 2020.01.03.
@GerhardPaseman If the proposed conjecture were true it wouldn't just limit the search space, it would completely answer the question, since one could write down explicitly the intervals whose union gives $S_{n,k}$. I guess my point was that dealing with the numbers themselves (exactly or bounding them) gets really hard really fast, but maybe there is some hope if we translate it to a geometric statement.
I guess it the theorem should have the subscripts in the other order $S^P_{2,n}.$
Yes, sorry. It's fixed.
The motivation of the question was the observation that the least number of lines in the plane that give five regions is $n=4$ parallel lines. Surprisingly, as long as $m \gt 5$ and $m \leq \binom{n}2+n+1$, the maximum number of regions possible using $n$ lines, there is a division of the plane into $m$ regions using no more than $n$ lines.
I suspected this based on this easy observation (and some computation):
If the plane is divides by $t$ parallel classes of lines with sizes $a_1,\cdots,a_t$, no three lines concurrent, then the plane is split into $\frac{n^2-s}2+n+1$ regions where $n=\sum a_i$ and $s=\sum a_i^2.$
The nice construction by Manturuv does capture some small numbers of regions which this does not, but those attained are enough to establish the result claimed above up to $n=50$ except for $m=5$ and $m=17.$ Naively running over all partitions of $n$ becomes time consuming.
So for $k=2$ then set of attainable number of regions is the union of intervals. Once $n-r \leq \binom{r}2$ we have an interval starting at $$(n-r)(r+1)+\binom{r}{2}-(n-r)=nr-\binom{r+1}2$$ with the previous interval ending at $$(n-r+1)r+\binom{r-1}2=nr-\binom{r+1}2+1.$$ Since this happens near $r=\sqrt{2n},$ we can certainly attain any number of regions $n\sqrt{2n} \leq m \leq n\frac{n+1}2+1$ using exactly $n$ lines.
|
2025-03-21T14:48:29.537756
| 2020-01-02T01:28:08 |
349554
|
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|
Stack Exchange
|
The $r$-dimensional volume of the Minkowski sum of $n$ ($n\geq r$) line sets
Let $n$ line sets be $\mathcal{S}_i=\{a\mathbf{h}_i:0 \le a \le 1\}$, for $1 \le i \le n$, where $\{\mathbf{h}_1,\cdots,\mathbf{h}_n\}$ is a vector group of rank $r$ in the $r$-dimensional Euclidean space. Define the Minkowski sum of two sets as $\mathcal{S}_1+\mathcal{S}_2=\{\mathbf{s}_1+\mathbf{s}_2:\mathbf{s}_1\in\mathcal{S}_1,~\mathbf{s}_2\in\mathcal{S}_2\}$. How to compute the $r$-dimensional volume of $\mathcal{S}_1+\cdots+\mathcal{S}_n$?
Let $M$ be the matrix whose rows are the vectors $\boldsymbol{h_i}$. Then the $r$-dimensional volume of $\mathcal{S}=\mathcal{S}_1+\cdots+\mathcal{S}_n$ is equal to the sum of the absolute values of the $r\times r$ minors of $M$. I don't know who originally showed this, but one can show that $\mathcal{S}$ can be tiled with $r$-dimensional parallelotopes whose volumes are the $r\times r$ minors. This follows for instance from the proof of Lemma 2.1 here.
Thanks for your explanation. This problem can be further extended as a new problem I asked several minutes ago.
The keyword you are looking for is "zonotope", which is defined to be the Minkowski sum of line segments. An early reference for zonotope is: P. McMullen, “On zonotopes”, Transactions of the American Mathematical Society, Vol. 159, 1971.
Following your notation, the $r$-dimensional volume of the zonotope $\mathcal{S}_{1} + ... + \mathcal{S}_{n}$ is equal to
$$\displaystyle\sum_{1\leq i_{1} < i_{2} ... < i_{r}\leq n} \big\vert{\rm{det}}\left(\mathbf{h}_{i_{1}},\mathbf{h}_{i_{2}},...,\mathbf{h}_{i_{r}}\right)\big\vert.$$
For reference, see eqn. (57) in "Combinatorial Properties of Associated Zonotopes" by G.C. Shephard, Canadian Journal of Mathematics, 1974. In that paper, there is an extra factor $2^{r}$ in front of the above expression since the line segments there are defined by $\{a\mathbf{h}_{i} : -1\leq a \leq 1\}$ instead of the OP's convention: $0\leq a \leq 1$. In the very end of this paper, Shephard credits McMullen for drawing attention to this formula. The same formula also appears as Exercise 7.19 in G.M. Ziegler, Lectures on Polytopes, Vol. 152, Springer, 2012; screenshot below:
Sorry for essentially repeating Richard's answer, his answer came before I finished typing.
I don't have access at the moment to the paper of Shephard, but presumably he is looking at line segments from $-\boldsymbol{v_i}$ to $\boldsymbol{v_i}$, not $0$ to $\boldsymbol{v_i}$. That is why there is an extra factor of $2^d$. If you take $\boldsymbol{h_i}$ to be the $i$th unit coordinate vector in Ryan Chen's question, then the zonotope is a unit cube of volume 1, not $2^r$. The matrix $M$ is the $r\times r$ identity matrix.
@RichardStanley: You are right, edited.
|
2025-03-21T14:48:29.537971
| 2020-01-02T02:41:54 |
349555
|
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|
Stack Exchange
|
Wirtinger calculus on general holomorphic vector bundles
For any holomorphic vector bundle $\mathbb{C}^n \rightarrow E \rightarrow M$ on a complex manifold, one can define a Dolbeault operator $\overline{\partial}_E$ which obeys the Leibniz rule and has the property $\overline{\partial}_E\circ\overline{\partial}_E = 0$. Given a local frame $(z^1,\dots,z^n)$ with $z^j = x^j + \mathrm{i}y^j$, is this enough to define a Wirtinger calculus which behaves as if $E$ were the tangent bundle? Namely, for a function $f: M \rightarrow \mathbb{C}$ can $\overline{\partial}_E f$ be then locally expressed as $\sum_j \left(\frac{\partial f}{\partial x^j} + \mathrm{i}\frac{\partial f}{\partial y^j}\right)\mathrm{d}\overline{z}^j$, and is there an analogous operator $\partial_E$ such that $\partial_E +\overline{\partial}_E$ is the complex deRham differential?
If you equip $E$ with a hermitian metric, then there is a unique connection $\nabla$ which is compatible with the metric and for which $\nabla^{0,1} = \bar{\partial}_E$. This is called the Chern connection and can be written as $\nabla = \nabla^{1,0} + \nabla^{0,1} = \partial_E + \bar{\partial}_E$. Is this the kind of thing you're looking for?
Yes, thank you! But do we know anything about the operator $\partial_E$, for instance, an expression explicit or in terms of the Dolbeault operator of the underlying manifold? Are $\partial_E$ and $\overline{\partial}_E$ dependent on the hermitian metric?
The operator $\partial_E$ depends on the hermitian metric, but $\bar{\partial}_E$ doesn't.
Also, the operator $\bar{\partial}_E$ acts on sections of $E$, so $\bar{\partial}_Ef$ is not defined.
|
2025-03-21T14:48:29.538100
| 2020-01-02T10:06:55 |
349574
|
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|
Stack Exchange
|
Set of numbers with unique results from sums
I need to create a set of numbers where any amount of them can be added together and each result will always give a unique answer, so we always know that the result was created from adding exactly these numbers and no other combination in the set can work. I'm not sure how many numbers I will need in the set at this point, so am just looking for a general rule. If someone can give me a name for the concept or something which I can then research myself I'm happy to do this, just struggling to word the problem properly to find an answer myself. Thanks in advance for any help!
"Dissociated" is one term in use. Powers of 2 are perhaps the simplest construction.
The term would be "subset sum distinct sets". An initial reference to start with is "A construction for sets of integers with distinct subset sums" by Tom Bohman.
|
2025-03-21T14:48:29.538198
| 2019-12-26T19:46:07 |
349172
|
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|
Stack Exchange
|
$G1$ interpolating curves with symmetric slopes in ends of segments
given a set $\lbrace p_i| 1\le i \le n\rbrace =\lbrace(x_1,y_1),\,\cdots,\,(x_n,y_n)\rbrace$ of points , which method can be recommended to calculate a sequence of angles $\left(\varphi_1,\,\cdots,\,\varphi_n\right)$ with
$$\left(\cos(\varphi_i),\,sin(\varphi_i)\right)^T\left(x_{i+1}-x_i,\,y_{i+1}-y_i\right)=-\left(\cos(\varphi_{i+1}),\,sin(\varphi_{i+1})\right)^T\left(x_{i+1}-x_i,\,y_{i+1}-y_i\right)$$
for cases $i=1,\ i=n-1$ or $\left((p_i-p_{i-1})\times(p_{i+1}-p_i)\right)\cdot\left((p_{i+1}-p_i)\times(p_{i+2}-p_{i+1})\right)\ge 0$
and
$$\left(\cos(\varphi_i),\,sin(\varphi_i)\right)^T\left(x_{i+1}-x_i,\,y_{i+1}-y_i\right)=\left(\cos(\varphi_{i+1}),\,sin(\varphi_{i+1})\right)^T\left(x_{i+1}-x_i,\,y_{i+1}-y_i\right)$$
otherwise,
that minimizes $\max\left|\frac{\left(\cos(\varphi,\,\sin(\varphi)\right)^T\left(x_{i+1}-x_i,\,y_{i+1}-y_i\right)}{\|p_{i+1}-p_i\|}\right|$
I have the feeling that the problem should have a simple geometric meaning but discerning it from the formulation above is not at all easy. Am I mistaken? If not, can you let us know what it is?
@fedja your feeling is right; the prototypical case is $G1$ interpolation with circular segments or with quadratic Wilson-Fowler splines. The benefit would be not having issues defining slopes in the first and last point. My hope is that the interpolating curves will tend to have less varying curvature between the nodes.
|
2025-03-21T14:48:29.538310
| 2019-12-26T20:27:40 |
349175
|
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|
Stack Exchange
|
Characteristic polynomial of the line graph (originally dual graph)
I am quite sure I have seen somewhere the connection between the characteristic polynomial of a (finite undirected) graph and its dual. I am not able to find it currently. Could you please refer me to the result?
(eigenvalues would be enough)
By dual, I mean the graph where edges become vertices and adjacent if they intersect in a vertex.
By characteristic polynomial I mean that of the adjacency matrix.
I removed my answer since now the question is more specific and needs a new answer.
The term you are looking for is not "dual graph" (which is usually a term reserved for planar graphs) but "line graph": https://en.wikipedia.org/wiki/Line_graph
For a formula for the adjacency matrix eigenvalues of a line graph in terms of the signless Laplacian eigenvalues of the original graph, see section 1.4.5 of https://homepages.cwi.nl/~aeb/math/ipm/ipm.pdf.
@sam thanks. You may post this as a answer.
The term for the graph construction you are talking about is "line graph" (see the Wikipedia article). The eigenvalues of the adjacency matrix of the line graph $L(\Gamma)$ are closely related to the signless Laplacian eigenvalues of the original graph $\Gamma$, as explained for instance in section 1.4.5 of these notes (see also Chris Godsil and Gordon Royle's textbook on algebraic graph theory). An important spectral feature of line graphs is that their (adjacency matrix) eigenvalues are at least $-2$.
|
2025-03-21T14:48:29.538432
| 2019-12-26T22:57:27 |
349182
|
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|
Stack Exchange
|
Identifying the two points of a subspace homeomorphic to a Sierpinski space
Let $X$ be a $\Delta$-generated space having a subset $A=\{a,b\}$ such that the relative topology is the Sierpinski topology with for example $\{a\}$ closed and $\{b\}$ open (the Sierpinsky space is a $\Delta$-generated space). I consider the quotient set map $X\to X / (a=b)$ with $X / (a=b)$ equipped with the final topology.
I want to prove that $X\to X / (a=b)$ is a homotopy equivalence. I do
not know if I am clumsy or if the statement is just wrong.
The idea I am exploring is to start from the proof that the Sierpinski space is contractible. Let $H:[0,1] \times A \to A$ be the set map defined by $H(0,a)=a$, $H(0,b)=b$, $H(1,a)=H(1,b)=a$ and $H(u,a)=H(u,b)=b$ for $u\in ]0,1[$. Then $$H^{-1}(b) = \big(]0,1[ \times \{a,b\}\big) \cup \big(\{0\} \times \{b\}\big).$$ Since $]0,1[ \times \{b\} \subset ]0,1[ \times \{a,b\}$, we deduce that $$H^{-1}(b) = \big(]0,1[ \times \{a,b\}\big) \cup \big([0,1[ \times \{b\}\big)$$ which is open. Therefore $H$ is continuous and it is a homotopy between the identity of $A$ and the constant map $a$. The statement above is therefore true when $X=A$.
Then I extend $H$ to a set map $H:[0,1]\times X \to X$ by setting $$\forall x\in X\backslash A, \forall u\in [0,1], H(u,x)=x.$$ One has
$$H^{-1}(b) = \big(]0,1[ \times \{a,b\}\big) \cup \big([0,1[ \times \{b\}\big)\\H^{-1}(a) = \big(\{1\} \times \{a,b\}\big) \cup \big(\{0\} \times \{a\}\big)
\\
H^{-1}(A) = [0,1] \times A
\\
\forall x\in X\backslash A, H^{-1}(x) = [0,1]\times \{x\} .
$$
I need to prove that $H:[0,1]\times X \to X$ is continuous. If $U$ is an open of $X$, there are three mutually exclusive cases:
$U\cap A =\varnothing$. Then $H^{-1}(U)=[0,1]\times U$ which is an open of $[0,1] \times X$.
$A\subset U$. Then $H^{-1}(U)=[0,1]\times U$ as well which is an open of $[0,1] \times X$.
$b\in U$ and $a \notin U$. Then $$H^{-1}(U)=H^{-1}(U\backslash \{b\}) \cup H^{-1}(b).$$ We obtain $$H^{-1}(U)=\big([0,1]\times (U\backslash \{b\})\big) \cup \big(]0,1[ \times \{a,b\}\big) \cup \big([0,1[ \times \{b\}\big).$$
Since $[0,1[\times (U\backslash \{b\}) \subset [0,1]\times (U\backslash \{b\})$, we obtain $$H^{-1}(U)=\big(\{1\}\times (U\backslash \{b\})\big) \cup \big(]0,1[ \times \{a,b\}\big) \cup \big([0,1[ \times U\big).$$ Since $b\in U$ by hypothesis, we obtain $$H^{-1}(U)=\big(\{1\}\times (U\backslash \{b\})\big) \cup \big(]0,1[ \times \{a\}\big) \cup \big([0,1[ \times U\big).$$
I cannot prove that $H^{-1}(U)$ is open...
|
2025-03-21T14:48:29.538572
| 2019-12-26T23:59:11 |
349186
|
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|
Stack Exchange
|
Large "computably un-simplifiable" computable well-orderings
Question
Suppose $A,X$ are computable well-orderings. Say that $A$ is $X$-unsimplifiable if there is no computable well-ordering $B$ whose ordertype is strictly less than that of $A$ but such that Duplicator has a computable winning strategy in the Ehrenfeucht-Fraisse game on $A$ and $B$ of length $X$.
(See below for the definition of these games.)
My question is:
Is there any computable well-ordering $X$ such that there are $X$-unsimplifiable computable well-orderings of ordertype arbitrarily large below $\omega_1^{CK}$?
If we drop the requirement that the winning strategies for Duplicator be computable, the answer is of course negative (this is an easy generalization about the usual results on elementary equivalence of ordinals). But that argument is extremely non-effective - basically, we have to guess at which intervals in $A$ are "large," and that can't be done computably.
I suspect that the answer is yes, and indeed that $X$-unsimplifiable $A$s exist for every $X$ of sufficiently large ordertype (say, $\ge\omega^2$). However, I don't see how to prove it; in particular, when I try to build such an $A$ for a given $X$ I keep winding up with an ill-founded linear order.
Details of the game $EF_X(A,B)$
Players Challenger and Duplicator alternately pick individual elements of $A$ or $B$ - for simplicity, we assume they have disjoint domains - with Challenger going first and on each turn Duplicator picking from whichever of the two structures Challenger did not pick from. Additionally, on each of their moves Challenger plays an element of $X$ smaller than all elements of $X$ played up to that point; basically, $X$ functions as a "clock" which eventually runs out (since $X$ is a well-ordering).
Once Challenger has no legal moves, the game ends. The winner is determined as follows: a pair of tuples $\overline{a}\in A$ and $\overline{b}\in B$ have been built by Challenger and Duplicator at this point, and Duplicator wins iff the structures $(A;\overline{a})$ and $(B;\overline{b})$ satisfy the same atomic sentences.
I think $n$ is $1+\log_2 n$-unsimplifiable. (Thinking of $k$-unsimplifiable as "revealed in time $k$" basically.) If the analogy holds then I guess $X\ge\omega$ is enough...
Or need a weird sense in which $2^{\omega^2}=\omega_1^{CK}$
$n$ is $1+\log_2 n$-unsimplifiable for finite $n$.
Fix any order $B$, $|A|>|B|$.
Ch picks $a_1\in A$ near the middle of $A$. Then Du picks some $b_1\in B$.
Either (i) $|A_{<a_1}|>|B_{<b_1}|$ or (ii) $|A_{>a_1}|>|B_{>b_1}|$.
Ch picks $a_2<a_1$, near the middle of $A_{<a_1}$, in case (i), and $a_2>a_1$, near the middle of $A_{>a_2}$, in case (ii). Continuing, we maintain the property
that
$|A_{<a_k}|>|B_{<b_k}|$ or $|A_{>a_k}|>|B_{>b_k}|$.
So Ch always goes where Du has little room left. Eventually Du runs out of room.
Ch picks the greatest element of $X$ not yet picked.
|
2025-03-21T14:48:29.538785
| 2019-12-27T01:12:58 |
349189
|
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|
Stack Exchange
|
Prove $\frac{\text{Area}_1}{c_1^2}+\frac{\text{Area}_2}{c_2^2}\neq \frac{\text{Area}_3}{c_3^2}$ for all primitive Pythagorean triples
A while ago I asked this question on MSE here. After placing a bounty it got quite a bit of attention but unfortunately it has yet to be resolved. After getting some advice from MO Meta I have decided to post the question here (note this is the same as area just multiplied by two and $a_n,b_n > 0$).
Does,
$$\frac{a_1b_1}{c_1^2}+\frac{a_2b_2}{c_2^2} = \frac{a_3b_3}{c^2_3}$$
for any three different primitive Pythagorean triples $(a_n,b_n,c_n)$?
My personal belief, for what it's worth, is that this does not occur and I am actively trying to disprove it. However, I would appreciate a counter example just as much.
Some results so far:
User @mathlove on MSE has found the following necessary condition,
The following is a necessary condition for $c_i.$
It is necessary that for every prime $p$, $$ \nu_p(c_1)
\leq \nu_p(c_2)+\nu_p(c_3)$$ $$\nu_p(c_2)\le \nu_p(c_3)+\nu_p(c_1)$$
$$\nu_p(c_3)\le \nu_p(c_1)+\nu_p(c_2)$$ where $\nu_p(c_i)$ is the
exponent of $p$ in the prime factorization of $c_i$.
(Proof can be found here)
In order to search for these values I created an exhaustive algorithm to search for these triples with help from here. This is what I've found,
For $c^2 < 10^{14}$
$$\frac{a_1b_1}{c_1^2}+\frac{a_2b_2}{c_2^2} \neq \frac{a_3b_3}{c^2_3}$$
and,
$$\frac{1}{c_1^2}+\frac{1}{c_2^2} \neq \frac{1}{c^2_3}$$
Note that the trouble finding these triples appears to come from dividing by the square of the hypotenuse as there are many solutions to $a_1b_1 + a_2b_2 = a_3b_3$. It seemed at first that these solutions may just be extremely unlikely to occur (ratios lining up perfectly) hence why nothing was found but it looks like there is a little more to this. A bug in my initial code made me accidentally search for solutions to this,
$$\frac{a_1b_1}{c_1}+\frac{a_2b_2}{c_2} = \frac{a_3b_3}{c_3}$$
Which yielded these very interesting values for $c < 10^7$,
$$\frac{3*4}{5} + \frac{20*21}{29} = \frac{17*144}{145}$$
$$\frac{20*21}{29} + \frac{119*120}{169} = \frac{99*4900}{4901}$$
$$\frac{119*120}{169} + \frac{696*697}{985} = \frac{577*166464}{166465}$$
$$\frac{696*697}{985} + \frac{4059* 4060}{5741} = \frac{3363*5654884}{5654885}$$
This pattern has a well defined structure to it. Notice the recursive nature where one of the terms always comes from the sum of the previous. Additionally the LHS numerators are both one apart and on the RHS the $b_3$ and $c_3$ are also a distance one from each other.
Background and motivation
A resolution one way or another to the original question could help to resolve a couple of (presumably not so important) but pesky open problems in number theory. I am preparing a website that I will link to eventually that will give the full background. However, it is too lengthy for this post and in accordance with advice from META MO I will omit it to keep this as brief as possible. Additionally I am not a research level mathematician so please forgive any unintended ignorance when responding to comments.
Edit for bounty
Joe Silverman and Constantin-Nicolae Beli have already given some good insight into the problem, I am putting a bounty on this with the hope that it will get more attention. I don't have much reputation so doing anything, even just commenting would go a long way for me. Looking at the problem as it stands I see one main problem and a subproblem that may help answer the main problem.
Main problem
Prove the original statement or find a counter example.
Subproblem
Is the solution set mentioned by Constantin-Nicolae Beli the only solutions to $\frac{a_1b_1}{c_1}+\frac{a_2b_2}{c_2} = \frac{a_3b_3}{c_3}$ and would that also be the same solution set for
when the hypotenuse is squared?
Important update
Going back and looking at the background of where this comes from. I found that what I am asking for is equivalent to this,
$$\frac{\left(c_{1}-x_{1}\right)\left(c_{1}+x_{1}\right)}{c_{1}^2}+\frac{\left(c_{2}-x_{2}\right)\left(c_{2}+x_{2}\right)}{c_{2}^2}=\frac{\left(c_{3}-x_{3}\right)\left(c_{3}+x_{3}\right)}{c_{3}^2}$$
Where $c_n$ is the associated hypotenuse of the primitive triple and $x_n$ is an integer solution to the circle,
$$x^2+y^2=2c^2$$
I wanted to mention this connection as it is related to the solution set mentioned by Constantin-Nicolae Beli.
It may be worth noting that any version where some of the $a_n, b_n < 0$ can be rearranged so all are positive - so it's equivalent to ask if there are any nontrivial answers (where "trivial" means one of the $a_n, b_n = 0$)
For subproblem 1, the answer seems like it should trivially be "no", just by parity considerations. $c$ is always odd in a primitive Pythagorean triple.
So you are looking for solutions
$$
\bigl( [a_1,b_1,c_1],[a_2,b_2,c_2],[a_3,b_3,c_3]\bigr) \in (\mathbb P^2)^3(\mathbb Q)
$$
to the equations
$$
a_1^2+b_1^2=c_1^2,\quad
a_2^2+b_2^2=c_2^2,\quad
a_3^2+b_3^2=c_3^2,\quad
a_1b_1c_2^2c_3^2+a_2b_2c_1^2c_3^2+a_3b_3c_1^2c_2^2=0.
$$
These equations define a surface $S$ in $(\mathbb P^2)^3$. In general, we don't have very many tools for determining the rational points on algebraic surfaces, but a first question, which you should be able to answer, is to determine the geometry of $S$. For example, it has a singularity at $([0,0,1],[0,0,1],[0,0,1])$. You should find the others, resolve the singularities, and determine the Kodaira dimension of $S$. That will at least suggest what the set of solutions might look like. (A program such as Magma might be able to do this computation for you.)
Another approach is to use some of the obvious symmetries to map to a simpler surface. For example, let $\pi:S\to S$ be the map that swaps $[a_1,b_1,c_1]$ and $[a_2,b_2,c_2]$, i.e., $\pi$ is the restriction to $S$ of the automorphism of $(\mathbb P^2)^3$ that swaps the first two factors. Compute the quotient variety $S/\pi$, see if you can determine its rational points, and then study which ones (if any) lift to $S$. Similarly for the automorphism that cyclically permutes the three factors of $(\mathbb P^2)^3$, and for the full $\mathcal S_3$ permutation group.
It may be easier to analyze the surface $\frac{(m_1^2 - n_1^2)m_1 n_1}{(m_1^2 + n_1^2)^2} + \frac{(m_2^2 - n_2^2)m_2 n_2}{(m_2^2 + n_2^2)^2} = \frac{(m_3^2 - n_3^2)m_3 n_3}{(m_3^2 + n_3)^2}$ as a subset of $(\mathbb{P}^1)^3$ (gotten using the usual parametrization of Pythagorean triples).
Thank you so much for pointing this out. I was a bit afraid of this and maybe I should have specified my background a bit more. I just got done with real analysis and am currently self studying abstract algebra so I'm afraid this is a bit out of my scope but I'll do my best. I'm going to try to pick up some algebraic geometry from the springer series in the next couple months or so. Hopefully I can pursue this route a bit further when I get there. Do you have any thoughts on the recursive nature of the solutions when c is not squared? It's pretty puzzling to me...
@PMaynard What better way to learn some algebraic/Diophantine geometry than to have an example like this in mind. There's a fantastic book called "Ideals, varieties, algorithms" by Cox, Little and O'Shea which might be a good place to start if you're interested in concrete examples like this (rather than certain other Springer books called, say, "Algebraic geometry").
Not really an answer, but a suggestion. Did you try to solve the "wrong" problem, where at the denominators you have $c_i$, not $c_i^2$? It seems quite interesting.
If we restrict ourselves to solutions of the type you found, then a Pythagorean triple with $b=a+1$ satisfies $a^2+(a+1)^2=c^2$, which writes as $(2a+1)^2-2c^2=-1$. The fundamental unit of ${\mathbb Z}[\sqrt 2]$ is $1+\sqrt 2$, of norm $-1$. Hence if $(x_n,y_n)$ are the solutions of the Pell equation $x^2-2y^2=\pm 1$, with $x_n+y_n\sqrt 2=(1+\sqrt 2)^n$, then $x_n^2-2y_n^2=(-1)^n$. So we have Pythagorean triplets $(a,a+1,c)$ if we take $2a+1=x_n$ and $c=y_n$, with $n$ odd. Such a triplet looks like $(\frac{x_n-1}2,\frac{x_n+1}2,y_n)$.
If we want a triplet with $c=b+1$ then we have the equation $a^2+b^2=(b+1)^2$ and we get $(a,b,c)=(a,\frac{a^2-1}2,\frac{a^2+1}2)$, with $a$ odd.
Then one can verify that $(a_1,b_1,c_1)=(\frac{x_{2k-1}-1}2,\frac{x_{2k-1}+1}2,y_{2k-1})$, $(a_2,b_2,c_2)=(\frac{x_{2k+1}-1}2,\frac{x_{2k+1}+1}2,y_{2k+1})$ and $(a_3,b_3,c_3)=(x_{2k},\frac{x_{2k}^2-1}2,\frac{x_{2k}^2+1}2)$, with $k\geq 2$, satisfy the equation $\frac{a_1b_1}{c_1}+\frac{a_2b_2}{c_2}=\frac{a_3b_3}{c_3}$. This checks straightforwardly if we write $x_n=\frac{\alpha^n+\beta^n}2$ and $y_n=\frac{\alpha^n-\beta^n}{2\sqrt 2}$, where $\alpha =1+\sqrt 2$ and $\beta =1-\sqrt 2=-\alpha^{-1}$.
A legitimate and nice problem you may think of is whether these are the only solutions or not. It also has some geometric meaning since $\frac{ab}c$ is the height perpendicular on $c$ in the triangle.
Very interesting. Unfortunately I haven't had time to study this so thank you for the insight. I want to believe that these are the only solutions because these are the only ones up to $10^7$, but then again there could always be more until shown otherwise. Honestly, there is just so much here to study so hopefully I can spend some more time on this as well. I was aware of the ratio being equivalent to the height. It is frustrating for $c^2$ because I can't apply any geometric meaning to that yet. What I'd also really like to know is if these solutions have any bearing on the $c^2$ case?
The pell equation above shows up in where this problem comes from. When I post the full background later this may help. So I'm wondering if maybe this form could also be a necessary condition for $c^2$ as well (although we still don't know if these are the only solutions). If this is the case it could make the problem easier to tackle.
I have updated the question to add some context that might be related to your findings. This is where the Pell equation shows up.
|
2025-03-21T14:48:29.539495
| 2019-12-27T03:11:11 |
349191
|
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|
Stack Exchange
|
Is there a good approximation for this Gaussian-like integration?
Is there an analytic solution or approximation for the following Gaussian-like integration? $\frac{1}{\eta^{2n}} \frac{1}{\sqrt{2 \pi}} \int_{-\eta}^{+\eta} e^{-x^2/2} x^{2n} dx$? The numerical plot suggests that it initially decrease faster, but reach a steady decrease of $(2n)^{-1.06}$ numerically when $2n > 100$ for all $\eta$.
The integral is asympotically $Θ(1/n)$, so the $1.06$ is actually $1$. Are you asking for the coefficient in the $Θ$?
Write $x^{2n} = e^{2n \ln x} $. When $n$ becomes large, the integrand is strongly peaked around the maximum of the exponent, i.e., at $x=\sqrt{2n} $. If that's within the range of integration, saddle point approximation should be good.
@MichaelEngelhardt I understand the question as $n→+\infty$ for fixed $η$, so $x=\sqrt{2n}$ is outside the range of integration.
@LeechLattice Ok, then the dominant contribution comes from the endpoints, $x=\pm \eta $.
An exact result (in terms of the incomplete Gamma function) and the large-$n$ asymptotics are as follows:
$$\frac{1}{\eta^{2n}} \frac{1}{\sqrt{2 \pi}} \int_{-\eta}^{+\eta} e^{-x^2/2} x^{2n} dx=\pi^{-1/2}2^{n} \eta^{-2 n} \left[\Gamma \left(n+\tfrac{1}{2}\right)-\Gamma \left(n+\tfrac{1}{2},\tfrac{1}{2}\eta^2\right)\right]$$
$$\rightarrow \frac{e^{-\eta^2/2} \eta}{\sqrt{2 \pi } n},\;\;\text{for}\;\;n\gg 1.$$
The convergence to the large-$n$ result is shown in the plot for $\eta=5$ (blue is the integral, gold the large-$n$ value):
Thanks! Can you send me your address so that I can acknowledge you properly? It still looks like that the trend is less than 1/n
Can you give me a link to the 2nd gamma function which has two arguments? I can only find the definition for the first one at https://en.wikipedia.org/wiki/Gamma_function
https://en.wikipedia.org/wiki/Incomplete_gamma_function ; no need for an acknowledgment, but for completeness, the proper way to refer to MO postings is explained here: https://meta.mathoverflow.net/a/4352/11260
See paragraph 3.8 in https://people.sc.fsu.edu/~%20jburkardt/presentations/truncated_normal.pdf and https://people.smp.uq.edu.au/YoniNazarathy/teaching_projects/studentWork/EricOrjebin_TruncatedNormalMoments.pdf where recursive Formulas are given for the moments of a truncated normal distribution.
For the truncated normal distribution, is there a formal publication? I am using truncated normal distribution in my paper, so that I'd like to reference it.
For example https://www.amazon.com/Handbook-Normal-Distribution-Second-Statistics/dp/0824793420
Mathematica 12.0 answers
AsymptoticIntegrate[Exp[-x^2/2]*x^(2*n), {x, -\[Eta], \[Eta]}, {n, Infinity, 1},
Assumptions -> n>0&&n\[Element] Integers && \[Eta] > 0]/ Sqrt[2*Pi]/\[Eta]^(2*n)//Simplify
$$\frac{e^{-\frac{\eta ^2}{2}} \eta ^{-2 n} \left(\eta ^2\right)^{n+\frac{1}{2}}}{\sqrt{2 \pi } n} $$
Thanks you so much! Can you send me your address so that I can acknowledge you properly?
The answer seems can be simplified further as $ \frac{e^{-\frac{\eta^2}{2}} \eta}{\sqrt{2 \pi} \frac{1}{n}$
Yes, the command of Mathematica Simplify[%,
Assumptions -> n > 0 && n \ [Element] Integers && \ [Eta] > 0] performs $$\frac{e^{-\frac{\eta ^2}{2}} \eta }{\sqrt{2 \pi } n} .$$
|
2025-03-21T14:48:29.539986
| 2019-12-27T04:23:37 |
349192
|
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|
Stack Exchange
|
Random walks: How many times does the largest component change?
My understanding is that for an unbiased random walk (starting at the origin) on $\mathbb R$ with $N$ steps that the expected number of sign changes is $O(\sqrt N)$. For a biased walk I believe the expected number is $O(1)$. In either case it is $O(\sqrt N)$.
Now suppose we have a random walk $(X_n,Y_n)$ on $\mathbb R^2$ and are interested in the number of times the largest component changes. Formally define $M(n) = 1$ if $X_n \ge Y_n$ and $M(n) = 2$ if $X_n < Y_n$. We want to bound $\mathbb E|\{n \le N: M(n) \ne M(n+1)\}|$.
To get a bound we only need to consider the random walk $X_n-Y_n$ on $\mathbb R$. By the previous result the sign changes $O(\sqrt N)$ times on expectation. But sign changes correspond to the largest component changing and we're done.
For dimensions $d>2$ this trick no longer works. In that case is anything known about the expected number of times the largest component changes? Are there any known order bounds depending on $d$ and $N$? I suspect $O(\log(d) \sqrt N)$ bounds might be possible.
I am trying to find answers online but I can't seem to even find a reference for the $O(1)$ result I mentioned.
An $O(\sqrt{N})$ bound seems clear since you can look at the 2-dimensional random walks obtained by specifying two directions in advance and then only keeping the steps along those.
For biased r/w, I guess the question is whether more than one component of the drift vector takes the maximal value. If not, say the $n$th component takes the maximal value. Now each time $X_i=X_n$, there is a probability $p_i$ that this never occurs again. So the total number of switches is bounded above by a sum of $d-1$ geometric random variables, so that for all $N$, the expected number of switches up to time $N$ is bounded above as required.
The question of how the number of switches scales with $d$ is interesting...
For a large class of unbiased random walks, the expected number of switches is of order $\Theta (\sqrt{N \log d})$.
It is closely related to the growth of regret in online learning, see, e.g., [1]; it is fine if the increments are biased as long as all components have the same mean.
The exact answer depends on the step distribution of the random walk. For concreteness, suppose $X_n=(X_n(i) : 1 \le i \le d)$ is a random walk in $\mathbb R^d$ with $d$ independent components, and
each component has $\pm 1$ increments of the same mean $\mu \in (-1,1)$. Denote $M_n:= \max_{j\le d} X_n(j)$. Let $J_n$ denote the index of a maximal component at time $n$.
(Precisely, let $J_0=1$. Given an integer $n \ge 1$, suppose that $J_{n-1}$ has already been defined. If $X_{n}(J_{n-1})=M_n$ then take $J_n:=J_{n-1}$; otherwise,
set $J_n$ to be the minimal $j$ such that $X_{n}(j)=M_{n}$.) Finally, let
$S_n:=\sum_{k=1}^n {\mathbf 1}_{J_k \ne J_{k-1}}$ be the number of times the maximal component switches by time $n$. Observe that for $n \ge 1$,
$$ M_{n+1}-M_n =X_{n+1}(J_n)-X_n(J_n)+ 2 \cdot{\mathbf 1}_{J_{n+1} \ne J_n} \, . $$
Therefore $M_n-n\mu-2S_n$ is a martingale for $n \ge 0$, so for all $N>0$,
$${\mathbb E} M_N-N\mu- 2 {\mathbb E}S_N=0 \,. \quad (*)$$
The multivariate central limit theorem, and the standard asymptotics for the maximum of $d$ Gaussians (see, e.g., Solution 18.7, page 348 in [2]), imply that as $N \to \infty$,
$${\mathbb E}\Bigl[\frac{M_N-N\mu}{\sigma\sqrt{N}}\Bigr] \to \sqrt{2\log d} \, , $$
where $\sigma^2=1-\mu^2$ is the variance of the increments. By (*), as $N \to \infty$,
$${\mathbb E}\Bigl[\frac{S_N}{\sigma\sqrt{N}}\Bigr] \to \sqrt{(\log d)/2} \, . $$
The analysis above can be extended to the case where
each independent component has increments of the same mean $\mu$ and variance bounded above and below by positive constants. (In that case $M_n-n\mu-cS_n$ will be a super- or sub-martingale depending on the value of $c>0$.) If the increments of different components have different means, then one can restrict attention just to those components where the increments have a maximal mean.
[1] Towards Optimal Algorithms for Prediction with Expert Advice.
Nick Gravin, Yuval Peres, and Balasubramanian Sivan.
Proceedings of the Twenty-Seventh Annual ACM-SIAM Symposium on Discrete Algorithms, 2016, 528-547
[2] Karlin, Anna R., and Yuval Peres. Game theory, alive. Vol. 101. American Mathematical Soc., 2017.
|
2025-03-21T14:48:29.540283
| 2019-12-27T07:52:34 |
349196
|
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|
Stack Exchange
|
Extension of superharmonic functions
Let $V$ be a bounded open set in $\mathbb{R}^{n}$ with $n\geq2$ and $W$ be an open neighborhood of the boundary $\partial V$ of $V$. If $u$ is superharmonic on $W$, is there a way to extend $u$ to a function $\tilde{u}$ that is superharmonic on $\complement V\cup W$ and is equal to $u$ at least on $\overline{V}\cap W$? ($\complement$ means the complement in $\mathbb{R}^{n}$ and $\overline{V}$ is the closure of $V$)
See Theorem 2.18 in https://www.amazon.com/Subharmonic-Functions-Mathematical-Society-Monographs/dp/0123348013 to this end.
In general, this is not possible. Consider the case $n=2$ take the unit disk for $V$,
and some ring, for example $1/2<|z|<2$ for $W$. Function $u(z)=\log|z|$ is harmonic in
$W$ but cannot be extended from any neighborhood of the unit circle to the closure
of the unit disk as a superharmonic function. The obstacle is clear:
$$\int_{|z|=1} \frac{\partial u}{\partial n} ds=\pi>0,$$
where $\partial/\partial n$ is the differentiation along the outer normal.
While for superharmonic functions in the disk this integral is always $\leq 0$,
which follows from the Green formula
$$\int\int_V\Delta u dxdy=\int_{\partial V}\frac{\partial u}{\partial n} ds.$$
Another reason why such an extention may be impossible is that your function
can blow up to $-\infty$ at a boundary point of $W$ which is interior for $V$. For example, with the same $U,W$ an extension of $\log|z-1/2|-2\log|z|$ to the unit disk is evidently impossible,
though the first obstacle does not exist for this function.
A correct extension theorem of the type that you propose would be something
like this: if
$$\int_{\partial V}\frac{\partial u}{\partial n}ds<0$$
then $u$ extends from SOME neighborhood $W_1\subset W$ of $\partial V$ to $V$ but
in general $W_1\cap V$ will be smaller than $W\cap V$.
This is just a preliminary statement requiring some smoothness assumptions on $u$ and $V$, because in general
neither $\partial u/\partial n$ nor $ds$ nor $\partial/\partial n$
are defined.
|
2025-03-21T14:48:29.540434
| 2019-12-27T10:27:51 |
349202
|
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"url": "https://mathoverflow.net/questions/349202"
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|
Stack Exchange
|
Only discrete topology gives trivial topos?
Given a Grothendieck site $\mathsf{(C,\tau)}$, if the associated Grothendieck topos $\mathsf{Shv(C,\tau)}$ is trivial, i.e. $\mathsf{Shv(C,\tau)}$ consists of the terminal sheaf $*$ only, can we conclude that the topology $\tau$ is the discrete topology—the finest topology on $\mathsf{C}$, namely for every object $X$, every sieve is a covering sieve?
Note that the converse is true by applying the sheaf condition to the empty covering sieve of $X$. And to show the above to be true (if it is), one only needs to show the empty sieve of $X$ is a covering for every $X$.
Even more generally, does a strict refinement of a topology give strictly less sheaves?
Remark: It's not hard to show on the other extreme that, every presheaf on $\mathsf{C}$ is a sheaf iff the topology $\tau$ is the chaotic topology (or indiscrete topology), i.e. the only covering sieve of any $X$ is $h_X$.
The correspondence between Grothendieck topology and localization of a topos is bijective (see SGA 4), so if two topologies have the same sheaves then they are equal.
Could you be more precise?
I'm affraid I don't see how to be more precise than "If two topologies have the same sheaves then they are equal". Are you asking for a reference ?
Right, e.g. where in SGA4; other reference would be also OK.
Just to record that I find a reference here: https://stacks.math.columbia.edu/tag/00ZP and so indeed a strict refinement of a topology gives strictly less sheaves.
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2025-03-21T14:48:29.540560
| 2019-12-27T10:56:30 |
349205
|
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|
Stack Exchange
|
Reciprocal expansion of modified Bessel function
I am reading Sherstyukov and Sumin - Reciprocal expansion of modified Bessel function in simple fractions and obtaining general summation relationships containing its zeros. The authors say they are using Krein's series, but I never heard about this series. The condition is that the zeros of the function must be simple. It resembles the Mittag-Leffler theorem (?). What happens if the zeros are not simple? Is it correct anyway? I see other series like
$$\frac{1}{J_0(x)}=\sum _{k=1}^{\infty } -\frac{2 (x-1) (x+1) j_{0,k}}{\left(j_{0,k}-1\right) \left(j_{0,k}+1\right) \left(x-j_{0,k}\right) \left(j_{0,k}+x\right) J_1\left(j_{0,k}\right)}-\frac{-x-1}{2 J_0(1)}-\frac{x-1}{2 J_0(1)}
$$
that get it better the above paper.
Welcome to MO. As it stands, the question is hard to understand, since you keep referring us to the quoted paper. The question should be self-contained, so that the reader can understand it without going to the paper.
It also is not clear, at least to me, what the question here is. Is it a question about the Bessel function? Or a question about Krein's series? You are more likely to get a useful answer if you ask a clear, well-defined question.
What does "that get it better the above paper" mean?
I really do not if it class of transformation it is useful but the following converge very quick
$$\frac{1}{I_0(x){}^2}=\sum _{k=1}^{\infty } -\frac{2 (x-1) (x+1) \left(-i \left(x^2+1\right) \left(j_{0,k}\right){}^3 I_0\left(i j_{0,k}\right)-i x^2 j_{0,k} I_0\left(i j_{0,k}\right)+2 x^2 I_1\left(i j_{0,k}\right)-i \left(j_{0,k}\right){}^5 I_0\left(i j_{0,k}\right)-2 \left(j_{0,k}\right){}^4 I_1\left(i j_{0,k}\right)\right)}{\left(\left(j_{0,k}\right){}^2+1\right){}^2 \left(\left(j_{0,k}\right){}^2+x^2\right){}^2 I_1\left(i j_{0,k}\right){}^3}-\frac{-x-1}{2 I_0(1){}^2}-\frac{x-1}{2 I_0(1){}^2}$$
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2025-03-21T14:48:29.540702
| 2019-12-27T11:59:54 |
349206
|
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"loup blanc"
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|
Stack Exchange
|
Diagonal Lyapunov equation with rank 1
Given the discrete-time Lyapunov equation (1):
$$
A^T P A - P = bb^T
$$
such that $P$ shall be diagonal and positive definite and $b$ is a column vector. How to characterize $A$ and $b$, where there are diagonal solutions $P \succ 0$? More precisely,
$$
S = \{A \in \mathbb{R}^{n\times n}, b \in \mathbb{R}^{n} \mid \exists \textrm{ diagonal } P \succ 0 \textrm{ for } (1) \}
$$
How to characterize $S$?
@RodrigodeAzevedo Can you please be more specific? About the characterization?
Do you have access to Mathematica? This question can be solved via quantifier elimination and boils down to finding the parameters for which a linear program is feasible. The question needs further editing, too. In a Lyapunov equation, $A$ and $b$ are given and the goal is to find $P \succ 0$. Here, the goal is to find the sets in which $A$ and $b$ live such that the Lyapunov equation is solvable. By the way, set $S$ should include the set in which $b$ lives, too.
@RodrigodeAzevedo Thanks for the comments. Unfortunately, I don't have access to Mathematica. I try to see whether there might be another way to get it. I agree with your comments, however, I feel that the question reflects well your statement. Maybe add 'Solvability of' in the title?
If you make $P = \mbox{diag} (x_1, \dots, x_n)$, you should be able to write the linear program in ${\rm x} \geq 0_n$.
You can get a necessary condition via Gaussian elimination.
@RodrigodeAzevedo How? That is not at all evident to me. Note that $A\otimes A - I$ need not be invertible.
@FedericoPoloni Assuming the matrices are $n \times n$, don't we have $\binom{n+1}{2}$ linear equations in $x_1, \dots, x_n \geq 0$? I am not vectorizing.
Yes, that seems the correct number of equations, but you also need to eliminate the nonlinear $b$ variables to get a condition on $A$ only. And in any case it is not that easy to get a condition via Gaussian elimination, since you need 'branches' to check if pivots are zeros or not.
@FedericoPoloni Why would the condition be on $A$ only? Essentially, one has a linear system with non-negativity constraints. Using Gaussian elimination, one should be able to conclude if the linear system is even feasible. If it is, then one can use quantifier elimination. If it is not, then the linear program is also infeasible. Using quantifier elimination in, say, Mathematica, may produce a quantifier-free formula large enough to fill some 100 A4 pages, or more.
@RodrigodeAzevedo Hmm, looks like I was looking at an earlier edit of the question when the condition was asked on $A$, not on the pair $(A, b)$. OK, I take back the comment on eliminating $b$.
It seems hopeless to me. The considered equation
$A^TPA-P=bb^T$ can be rewritten $\phi(P)=bb^T$, where $\phi=(A^T\otimes A^T-I_{n^2})$ -if we stack the matrices row by row into vectors-.
If $spectrum(A)=(\lambda_i)_i$, then $spectrum(\phi)=(\lambda_i\lambda_j-1)_{i,j}$. Thus, in general , there is a unique solution
$P=(\phi)^{-1}(bb^T)$. On the other hand, since if $P$ is a solution, then $P^T$ too, $\phi^{-1}$ is an automorphism of the symmetric matrices.
Condition 1. The obtained symmetric $P$ is diagonal; there are $n(n-1)/2$ equations (dependent or not) linking the entries of $A=[a_{i,j}],b$.
For example, when $n=2$ and $b=[88,-72]^T$, the condition is
When $n=3$, writing conditions takes up a lot of space!
Condition 2. $P>0$. That certainly works if the $(|\lambda_i|)_i$ are $>1$.
Is P missing in the definition of $\phi$?
No, $\phi$ is an endomorphism of $M_n$. cf. Kronecker product.
I make a different attempt to answer the question. $A,b \in S$ if there exists $c$ and $d$ such that
$$
\begin{bmatrix}
A & b \\ c & d
\end{bmatrix}
\begin{bmatrix}
P & 0 \\ 0 & 1
\end{bmatrix}
\begin{bmatrix}
A^T & c^T \\ b^T & d
\end{bmatrix}
=
\begin{bmatrix}
P & 0 \\ 0 & 1
\end{bmatrix}
$$
This is equivalent to
$$
\begin{bmatrix}
P^{-1/2} & 0 \\ 0 & 1
\end{bmatrix}
\begin{bmatrix}
A & b \\ c & d
\end{bmatrix}
\begin{bmatrix}
P^{1/2} & 0 \\ 0 & 1
\end{bmatrix}
$$
being orthonormal. Thus $A$ is a submatrix of an (diagonally conjugated) orthonormal matrix. By Theorem 2.1 in Fiedler,1996 $A$ has at least $n-1$ singular values equal to 1 and the remaining one is less than 1. Diagonal conjugation may change this of course.
|
2025-03-21T14:48:29.540988
| 2019-12-27T14:48:25 |
349209
|
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|
Stack Exchange
|
On a continuous function as a substitute of the prime-counting function in the second Hardy–Littlewood conjecture satisfying certain asymptotics
It it well-known that the prime-counting function $\pi(x)$ satisfies the prime number theorem and that were in the literature two related conjectures to this arithmetic function, these are: the Riemann hypothesis, I mean in this ocassion the equivalence due to von Koch ([1]), and the second Hardy–Littlewood conjecture (see the corresponding Wikipedia or [2])
I would like to know if it is in the literature a continuous function $f(t)$ on some interval $(L,\infty)$, for some constant $L>0$, satisfying the following conditions $$\pi(x+y)\leq \pi(x)+f(y)\tag{C1}$$
for all positive real numbers $x$ and $y$ such that $x\leq y$ where $L<x$. And also that satisfies
$$|f(x)-\pi(x)|=O\left(\sqrt{x}\log x\right)\tag{C2}$$
as $x\to\infty$, and also satisfying the asymptotic equivalence $$\pi(x)\sim f(x)\tag{C3}$$
as $x\to\infty$.
Question*. Do you know any continuous function $f(x)$, on our interval $(L,\infty)$ being $L$ a positive constant, satisfying previous conditions $\text{(C1)}$, $\text{(C2)}$ and $\text{(C3)}$? Alternatively, is it possible to prove the existence of such continuous function? If it is in the literature please refer it answering as a reference request and I try to search and read the statement from the literature. Many thanks.
If isn't in the literature I'm asking about what work can be done to
elucidate the problem in my Question, that is at least feedback about the existence of such function: I don't know if it's easy to prove that such a continuous function (and its corresponding interval $(L,\infty)$) exists. If you can to prove the existence and you want to add more remarks about the function(s) $f(x)$ feel free to do it.
References:
[1] Helge von Koch , Sur la distribution des nombres premiers, Acta Mathematica volume 24, Article number: 159 (1901).
[2] G. H. Hardy and J. E. Littlewood, Some problems of ‘Partitio numerorum’ III: On the expression of a number as a sum of primes, Acta Math. (44): 1–70 (1923).
*An hour ago I've asked a similar question without a good mathematical content (thanks for the user in comments), I've updated it in a new post.
@Wojowu many thanks for your feedback in my (previous recent) deleted post (now I understand that from the asymptotic of the logarithmic integral and the function $\frac{x}{\log x}$, and that $\frac{y}{\log y}$ is a continuous and strictly increasing function the question in my deleted post has not a good mathematical content). Feel free to add your valuable comments again.
Many thanks for the upvoter.
While I was searching literature I've found the paper by professor Matt Visser, Primes and the Lambert $W$ function, Cornell University Library (November 2013); the reference that I know is arXiv:1311.2324v2
|
2025-03-21T14:48:29.541202
| 2019-12-27T15:39:50 |
349210
|
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|
Stack Exchange
|
Can "description" of models revive formalism?
A model of a theory is a structure (e.g. an interpretation) that satisfies the sentences of that theory. Wikipedia
Let $A$ be a set of sentences in some language that has only one extra-logical primitive symbol $R$.
Let's say that theory $T$ describes a model $A$ if and only if
$T$ proves the existence of a pair $\langle M, [R] \rangle $, where $ [R] $ is a set of all ordered pairs $\langle a,b \rangle $ where $a,b \in M $ and $ a \ R \ b$.
if we relativise all sentences of $A$ to $M$ (i.e. bound all of their quantifiers $\in M$), and replace each atomic sentence $a \ R \ b$ in them by $\exists r \in [R] \ (r=\langle a,b \rangle)$.
then all the resulting sentences hold in $T$.
We'd say by then that $T$ describes [not prove] that $A$ is satisfied in $\langle M, [R]\rangle$.
However a theory might describe a model of all its axioms, in the exact above sense, but might still not be able to express it in a single sentence!
To clarify: we can have a theory $T$ which prove the existence of sets $M$ and $[R]$ obeying all conditions depicted above, and such that $A$ is the set of all axioms of $T$ itself (even if $A$ is finite); and at the same time $T$ cannot prove the single sentence:
$\forall \phi [(T\vdash \phi) \to \langle M,[R]\rangle \models \phi)]$
and so cannot prove the single sentence:
$\exists M \exists [R] (\langle M,[R]\rangle \models T)$.
So even though $T$ is describing a model of itself, still $T$ cannot prove the sentence "there exists a model of $T$", and so doesn't prove its own consistency.
This mean that in principle we can extend theory $T$ by enlarging its language, adding axioms and inference rules, in such a manner that we have more expressive theories $T_1, ..,T_n$, where each $T_{i+1}$ extends $T_i$, and at the same time each $T_i$ can describe a model of $T_{i+1}$ in the above spoken sense. We can even have $T$ describing a model of $T_n$, i.e. without necessarily proving the sentence "there exists a model of $T$". All of this without implying that $T$ is inconsistent, nor needing to weaken $T$ below expressing basic arithmetic facts.
In some semantic sense description of a model of a theory [in the above particular sense], is next to saying that it is consistent, actually it is next to saying that it is possibly true of something! And constructively speaking that what matters.
Although this in no way touches Godel's incompleteness theorems, but it in some sense develops a parallel approach available to formalism through "describing" models by going higher and higher up in description steps of more complex structures, without affecting the consistency of the base theory!
The only theory that I know of it being able to describe a model of its finitely many axioms in the above sense is New foundations "NF"
Question: Are there other known examples of such theories?
Question 2: had there been formalist approaches along that line?
OK I'll try. Thanks
I made the changes. That's the best I can do. I hope its clear now.
In what sense are interpretations structures? The latter are (semantic) objects, the former essentially live at the level of syntax.
I copied the wikipedia here. I'll give the reference.
@AndrésE.Caicedo I understand the referred wikipedia article as those structures constituting an interpretation of the sentences of the relevant theory, i.e. in a semantic sense. Much as showing a particular horse give an interpretation to the sentence that uniquely refer to it.
Awful, that sentence in the wikipedia article is terribly misleading.
Anyway, ZF and ZFC are examples. If we want the theory to be finitely axiomatizable, NBG is an example.
@AndrésE.Caicedo which pair of classes NBG (or ZF\ZFC) defines that it describe as a model of it in the sense written above?
This is a standard textbook exercise. There are several answers here and in math.stackexchange with details. If no one else provides a link, I'll see if I can find one of them. Otherwise I'll post a sketch.
@AndrésE.Caicedo notice that the pair of classes I've asked for must be OBJECTS IN the universe of discourse of ZF\ZFC (i.e., must be sets) or NBG (i.e., must be classes that are quantified upon); that is to say: not merely extensions of definable predicates in these theories.
Yes, of course.
The answer here provides a decent sketch for the case of ZFC.
@AndrésE.Caicedo, Noah actually said that he didn't answer that for the finite case?
|
2025-03-21T14:48:29.541667
| 2019-12-27T16:26:50 |
349212
|
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|
Stack Exchange
|
Functional approach vs jet approach to Lagrangian field theory
Context: I am a PhD student in theoretical physics with higher-than-average education on differential geometry. I am trying to understand Lagrangian and Hamiltonian field theories and related concepts like Noether's theorem etc. in a mathematically rigorous way since the standard physics literature is sorely lacking for someone who values precision and generality, in my opinion.
I am currently studying various text by Anderson, Olver, Krupka, Sardanashvili etc. on the variational bicomplex and on the formulation of Lagrangian systems on jet bundles. I do not rule the formalism yet, but made significant steps towards understanding.
On the other hand, most physics literature employs the functional formalism, where rather than calculus on variations taking place on finite dimensional jet bundles (or the "mildly infinite dimensional" $\infty$-jet bundle), it takes place on the suitably chosen (and usually not actually explicitly chosen) infinite dimensional space of smooth sections (of the given configuration bundle).
Even relatively precise physics authors like Wald, DeWitt or Witten (lots of 'W's here) seems to prefer this approach (I am referring to various papers on the so-called "covariant phase space formulation", which is a functional and infinite dimensional but manifestly "covariant" approach to Hamiltonian dynamics, which also seems to be a focus of DeWitts "The Global Approach to Quantum Field Theory", which is a book I'd like to read through but I find it impenetrable yet).
I find it difficult to arrive at a common ground between the functional formalism and the jet-based formalism. I also do not know if the functional approach had been developed to any modern standard of mathematical rigour, or the variational bicomplex-based approach has been developed precisely to avoid the usual infinite dimensional troubles.
Example:
Here is an image from Anderson's "The Variational Bicomplex", which shows the so-called augmented variational bicomplex. Here $I$ is the so-called interior Euler operator, which seems to be a substitute for integration by parts in he functional approach.
Later on, Anderson proves that the vertical columns are locally exact, and the augmented horizontal rows (sorry for picture linking, xypic doesn't seem to be working here, don't know how to draw complices)
are locally exact as well. In fact for the homotopy operator $\mathcal H^1:\mathcal F^1\rightarrow\Omega^{n,0}$ that reconstructs Lagrangians from "source forms" (equations of motion) he gives (for source form $\Delta=P_a[x,y]\theta^a\wedge\mathrm d^nx$) $$ \mathcal H^1(\Delta)=\int_0^1 P_a[x,tu]u^a\mathrm dt\ \mathrm d^nx. $$
On the other hand, if we use the functional formalism in an unrigorous manner, the functional derivative $$ S\mapsto\frac{\delta S[\phi]}{\delta \phi^a(x)} $$ behaves like the infinite dimensional analogue of the ordinary partial derivative, so using the local form of the homotopy operator for the de Rham complex (which for the lowest degree is $f:=H(\omega)=\int_0^1\omega_\mu(tx)x^\mu\mathrm dt$) and extending it "functionally", one can arrive at the fact that if an "equation of motion" $E_a(x)[\phi]$ satisfies $\frac{\delta E_a(x)}{\delta\phi^b(y)}-\frac{\delta E_b(y)}{\delta\phi^a(x)}=0$, then $E_a(x)[\phi]$ will be the functional derivative of the action functional $$ S[\phi]=\int_0^1\mathrm dt\int\mathrm d^nx\ E_a(x)[t\phi]\phi^a(x). $$
I have (re)discovered this formula on my own by simply abusing the finite dimensional analogy and was actually surprised that this works, but it does agree (up to evaluation on a secton and integration) with the homotopy formula given in Anderson.
This makes me think that the "variation" $\delta$ can be considered to be a kind of exterior derivative on the formal infinite dimensional space $\mathcal F$ of all (suitable) field configurations, and the Lagrangian inverse problem can be stated in terms of the de Rham cohomology of this infinite dimensional field space.
This approach however fails to take into account boundary terms, since it works only if integration by parts can be performed with reckless abandon and all resulting boundary terms can be thrown away. This can be also seen that if we consider the variational bicomplex above, the $\delta$ variation in the functional formalism corresponds to the $\mathrm d_V$ vertical differential, but in the augmented horizontal complex, the $\delta_V=I\circ\mathrm d_V$ appears, which has the effect of performing integrations by parts, and the first variation formula is actually $$ \mathrm d_V L=E(L)-\mathrm d_H\Theta, $$ where the boundary term appears explicitly in the form of the horizontally exact term.
The functional formalism on the other hand requires integrals everywhere and boundary terms to be thrown aside for $\delta$ to behave as an exterior derivative. Moreover, integrals of different dimensionalities (eg. integrals over spacetime and integrals over a hypersurface etc.) tend to appear sometimes in the functional formalism, which can only be treated using the same concept of functional derivative if various delta functions are introduced, which makes me think that de Rham currents (I am mostly unfamiliar with this area of mathematics) are also involved here.
Question: I would like to ask for references to papers/and or textbooks that develop the functional formalism in a general and mathematically precise manner (if any such exist) and also (hopefully) that compare meaningfully the functional formalism to the jet-based formalism.
It sounds like you want a calculus on infinite dimensional manifolds (of smooth functions), with applications to variational calculus. Pedro's answer points to his excellent work as an example of where this is done. But you should also be aware that mathematicians have not in general been neglecting infinite dimensional calculus. Textbooks on Banach manifolds appeared already in the '60s (Dieudonné, Lang). The Nash-Moser implicit function theorem gave an impetus to Fréchet manifolds. The Kriegl-Michor monograph surveys many advances and generalizations within the subject.
Integrals of horizontal differential forms defined on jet bundles give you a particular way to define functions ("functionals") on infinite dimensional manifolds of field configurations. Then you can apply to them your favorite rigorous infinite dimensional calculus. It is then just a matter of building up a dictionary between the jet and functional formalisms. But from your question, you seem to already know it.
I do not know if it is good form for MO to cite one's own papers when answering a question, but I will take the chance. This matter is addressed in quite a bit of detail in my joint paper with Romeo Brunetti and Klaus Fredenhagen,
R. Brunetti, K. Fredenhagen, P. L. Ribeiro, Algebraic Structure of Classical Field Theory: Kinematics and Linearized Dynamics for Real Scalar Fields, Commun. Math. Phys. 368 (2019) 519-584, arXiv:1209.2148 [math-ph].
There we discuss only scalar fields, but the discussion remains essentially unchanged for sections of fiber bundles or even fibered manifolds. In what follows, I will assume the former.
As a rule, the functional formalism is more general, for it is clear that given
a smooth $d$-form $\omega$ on the total space of the $k$-th order jet bundle $J^k\pi$ of the fiber bundle $\pi:E\rightarrow M$ with $D$-dimensional typical fiber $Q$ over the smooth $d$-dimensional (space-time) manifold $M$ (we assume all finite-dimensional manifolds here to be smooth, Hausdorff, paracompact and connected) - think of $\omega$ as a "Lagrangian density" - we have that $$F_K(\varphi)=\int_K (j^k\varphi)^*\omega\ ,\quad\varphi\in\Gamma(\pi)$$ is a functional on the space $\Gamma(\pi)=\{\varphi\in\mathscr{C}^\infty(M,E)\ |\ \pi\circ\varphi=\mathrm{id}_M\}$ of smooth sections of $\pi$ for each compact region $K\subset M$ (i.e. $K$ has nonvoid interior) and each $k=0,1,...\infty$. The case $k=\infty$ can be handled just like for finite $k$ since the total space of the infinite-order jet bundle $J^\infty\pi$ is the countable projective limit of the finite-dimensional manifolds $J^k\pi$ and therefore is a metrizable Fréchet manifold, see e.g. the aforementioned paper or the book by Andreas Kriegl and Peter Michor, The Convenient Setting of Global Analysis (AMS, 1997) for a discussion of the manifold structure of $J^\infty\pi$. For a better handling of the boundary terms which appear when performing functional derivatives (see below), it is convenient to replace $K$ with the multiplication of $(j^k\varphi)^*\omega$ by some $f\in\mathscr{C}^\infty_c(M)$ and then integrate over the whole of $M$, thus yielding the functional $$F(\varphi)=\int_M f(j^k\varphi)^*\omega\ ,\quad\varphi\in\Gamma(\pi)\ .$$
Speaking of which, $\Gamma(\pi)$ has an infinite-dimensional manifold structure
modelled on the locally convex vector spaces $\Gamma_c(\varphi^*V\pi)$ of smooth sections with compact support of the pullback of the vertical bundle $V\pi=\ker T\pi$ of $\pi$ by each $\varphi\in\Gamma(\pi)$ (these locally convex vector spaces are all canonically topologically isomorphic to each other) - the corresponding (topological) manifold structure is the so-called Whitney topology on $\Gamma(\pi)$. There is in addition a natural smooth structure on $\Gamma(\pi)$ modelled on that of $\Gamma_c(\varphi^*V\pi)$ - it can be proven that smooth curves $\gamma:\mathbb{R}\rightarrow\Gamma(\pi)$ are necessarily of the form $\gamma\in\mathscr{C}^\infty(\mathbb{R}\times M,E)$ with $\gamma(t,\cdot)=\gamma(t)=\gamma_t\in\Gamma(\pi)$ (i.e. $\pi\circ\gamma_t=\mathrm{id}_M$) for all $t\in\mathbb{R}$ and for all $a<b\in\mathbb{R}$ there is $K\subset M$ compact such that $\gamma(t,p)$ is constant in $t\in[a,b]$ for all $p\not\in K$. This entails that $\gamma'_t\in\Gamma_c(\gamma_t^*V\pi)$ for all $t\in\mathbb{R}$ since we are differentiating along a single fiber of $\pi$ when differentiating with respect to the curve parameter $t$. Given that notion of smooth curves, smooth maps are just the ones that map smooth curves to smooth curves (see A. Kriegl, P. Michor, loc. cit. for many more details). Given the specific notion of smooth curves and smooth maps that $\Gamma(\pi)$ has, it is easy to see why the "Poincaré-lemma-type" argument used by Anderson to solve the inverse problem of calculus of variations can be recast in functional form with essentially no change as you did, for the core of the argument still remains essentially finite-dimensional.
In order to see where the boundary terms come from, consider the second functional $F$ above. The variational derivative consists only in taking the derivative of $F(\gamma_t)$ with respect to $t$ at $t=0$, where $\gamma:\mathbb{R}\rightarrow\Gamma(\pi)$ is a smooth curve on $\Gamma(\pi)$ so that $\gamma_0=\varphi$, $\gamma'_0=\delta\varphi\in\Gamma_c(\varphi^*V\pi)$ and applying the chain rule (which, by the way, does hold in this setting). One then applies the standard variational formula $j^k\delta\varphi=\delta(j^k\varphi)$ (recall that $j^k\varphi$ only takes into account base = horizontal derivatives of sections, whereas $\delta\varphi$ is just a fiber = vertical derivative. The desired commutativity comes from local triviality of $\pi$ or, more generally, the implicit function theorem in the case of arbitrary fibered manifolds) together with integration by parts - the result is the Euler-Lagrange derivative of $\omega$ plus a sum of terms proportional to derivatives of positive order of the cutoff function $f$. This latter sum yields the boundary terms in the (distributional) limit when $f$ becomes the characteristic function of a compact region $K$ with smooth boundary $\partial K$ - in this limit, $F$ becomes $F_K$ defined above.
The functional formalism is genuinely more general than the jet bundle formalism also because it can handle a large class of nonlocal functionals. Allowing these is seen to yield better algebraic properties (closure under "pointwise" = "field-wise" products, etc.) than just considering local ones, which is convenient for field quantization at a later stage, among other things. Moreover, there is a very simple and elegant characterization of local functionals within the functional formalism which does not mention jet bundles anywhere - see e.g. Proposition 2.2, pp. 535-539 of the above paper.
This is meant as a long comment to the very good answer by Pedro Ribeiro.
There is a nice analog of the variational bicomplex in the functional framework. Namely, the space of differential forms on $M \times \Gamma(E)$ (where $E$ is a fiber bundle over $M$) comes with a natural bigrading $\Omega^{p, q}(M \times \Gamma(E))$ induced by the product structure, i.e. the dual of the decomposition $T_{m, \phi} (M \times \Gamma(E)) = T_m M \times T_\phi \Gamma(E)$ of the tangent space. Moreover, the jet map
$$
j^k: M \times \Gamma(E) \to J^k E, \qquad (m, \phi) \mapsto j^k_m \phi
$$
yields a morphism from the variational bicomplex $\Omega^{p, q}(J^k E)$ to the bicomplex $\Omega^{p, q}(M \times \Gamma(E))$ with the exterior differential. Personally, I find the functional bicomplex easier to understand than the variational one; and as remarked by Pedro the functional framework is more flexible as it also handles non-local Lagrangians. On the other hand, the jet bundle approach has advantages for simulation, because you stay in the finite-dimensional setting which makes it easier to discretize while preserving the (symplectic) geometry.
This is very interesting, could you provide some references for this approach?
This observation goes back to Zuckerman: ACTION PRINCIPLES AND GLOBAL GEOMETRY. In my master thesis (Chapter 5), I reformulated a few basic concepts of symplectic geometry in the functional framework for field theories.
Oh yeah, I have the Zuckerman paper, it's just very difficult to read as its not typeset properly. I might retype it in LaTeX if I get the freetime to do it. And thanks for the link to your thesis, I will go through it some time.
|
2025-03-21T14:48:29.542576
| 2019-12-27T16:29:21 |
349213
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/349213"
}
|
Stack Exchange
|
Moduli stacks of ($\mu$-)semistable sheaves
$X$: smooth projective surface/$\mathbb{C}$
$H$: ample divisor
Then, is the moduli stack of (Gieseker)semistable sheaves for $H$ with a fixed Chern class dense in the moduli stack of $\mu$-semistable sheaves for $H$ with the Chern class?
(I am especially interested in the case $\rho(X)=1$ & rank $2$ semistable sheaves.)
Thank you !
|
2025-03-21T14:48:29.542643
| 2019-12-27T16:51:14 |
349216
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
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"Binai",
"Mohan",
"Sasha",
"abx",
"https://mathoverflow.net/users/40297",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/349216"
}
|
Stack Exchange
|
K3 Surfaces : Derivations and automorphisms
The literature about K3 surfaces is extensive. Let us consider the fact that the zero locus of any smooth homogeneous degree 4 equation is a K3 surface.
Let $R$ the quotient ring of a homogeneous polynomial in $4$ variables and degree $4$.
I am not being succesfull to find anything related to the study of derivations and locally nilpotent derivations on these sort of ring. What could be a reference for
a description of the derivations on $R$?
a description of the locally nilpotent derivations on $R$?
a description of the automorphism groups of $R$?
any condition for the group of automorphisms of $R$ to be finite/finitely generated?
What do you call a derivation on a surface? If you mean a tangent vector field, they do not exist ($\neq 0$) on a K3.
@abx: "do not exist" means "=0". But the question is, indeed, quite unclear.
It was really bad written and it was missing an essential part of the data. Sorry for my mistake in not double checking it. I tried to clarify now...
is just given by the Euler sequence. They are just derivations of the polynomial ring (which are obvious) which takes the equation of the hypersurface to a multiple of the same equation.
@Mohan It is pretty clear that the derivations should have this property related to the polynomial ideal that we are taking the quotient, but I want to know a characterization (if it exists).
I can try to say a little about your first question, on derivations of $R$, in characteristic 0. As Mohan said, the derivations of $R$ are just derivations of the polynomial ring taking the defining equation to itself, but I'd add that it's a bit subtle what the collection of such derivations looks like. It's clear we have derivations of every positive degree $k$, for example by first applying the Euler operator and then multiplying by a polynomial of degree $k$. So, one natural question you can ask is whether there's a derivation of $R$ of negative degree. Note that such a derivation is locally nilpotent (though not all locally nilpotent derivations have negative degree).
The general context for your question is that we're given a ring $R=S(X,L):=\bigoplus H^0(X,L^m)$, which is naturally the section ring of a (smooth) projective variety with a chosen ample line bundle. By a theorem of J.M. Wahl ("A cohomological characterization of $\mathbb P^n$") if $R$ admits a derivation of negative degree, then $X$ must in fact be $\mathbb P^n$. Thus, in your example there will be no derivations of negative degree. (This crucially uses the characteristic 0 hypothesis.)
In fact, one can rule out differential operators of negative degree of any order on your $R$, not just derivations: one can show that the existence of a differential operator of order $m$ and degree $-e$ gives rise to an element of $H^0(X,(\mathrm{Sym}^m T_X)(-e))$, and in turn to an ample subsheaf of $\mathrm{Sym}^m T_X$. By powerful results of Miyaoka (Corollary 8.6 of "Deformations of a Morphism along a Foliation and Applications") this in turn forces $X$ to be uniruled (again, in characteristic 0!), which your K3 surface is not.
It is pretty clear that the derivations should have this property related to the polynomial ideal that we are taking the quotient, but I want to know a description (if it exists), i.e, what "the collection of such derivations looks like".
Ah, for sure that characteristic 0 is a crucial step. I would be happy with answers in this case!
|
2025-03-21T14:48:29.542882
| 2019-12-27T17:00:34 |
349217
|
{
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"Anthony Quas",
"Deane Yang",
"R. N. Marley",
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"https://mathoverflow.net/users/137336",
"https://mathoverflow.net/users/613"
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"url": "https://mathoverflow.net/questions/349217"
}
|
Stack Exchange
|
Help to understand a limit $\varepsilon\rightarrow 0$ computation on a fluid mechanic paper
In Córdoba and Gancedo - Contour dynamics of incompressible 3-D fluids in a porous medium with different densities (page 4) I read that if
$$ v (x_1,x_2,x_3,t)=-\frac{\rho_2-\rho_1}{4\pi} \operatorname{PV}\int_{\mathbb{R}^2} \frac{(y_1,y_2,\nabla f(x-y,t)\cdot y)}{[\lvert y\rvert^2 + (x_3 - f(x-y,t)^2)]^{3/2}}\ dy$$
and for $\varepsilon $ positive and $x=(x_1,x_2)$, we define
$$ v^1(x_1,x_2,f(x,t),t) =\lim_{\varepsilon \rightarrow 0} v\bigl(x_1-\varepsilon \partial_{x_1} f(x,t) , x_2 - \varepsilon \partial_{x_2} f (x,t), f(x,t) + \varepsilon , t\bigr),$$
then they say
\begin{align*}
v^1(x_1,x_2,f(x,t),t) ={} & v(x_1,x_2,f(x,t),t) \\
& {}+ \frac{\rho_2-\rho_1}{2}\frac{\partial_{x_1} f(x,t)(1,0,\partial_{x_1}f(x,t))}{1+(\partial_{x_1}f(x,t))^2 +(\partial_{x_2}f(x,t))^2 } \\
& {}+ \frac{\rho_2-\rho_1}{2}\frac{\partial_{x_2} f(x,t)(0,1,\partial_{x_2}f(x,t))}{1+(\partial_{x_1}f(x,t))^2 +(\partial_{x_2}f(x,t))^2 }.
\end{align*}
I don't understand how they get that expression for $v^1$. Please, any help or idea is welcome.
yeah, you are right.
I have made an edit. Thank you.
You might try a more descriptive title. Your title suggests (before one clicks on it) that someone is looking for help with their calculus homework.
I second @AnthonyQuas's suggestion.
I've also edited it. Than for your suggestion.
|
2025-03-21T14:48:29.542995
| 2019-12-27T17:53:41 |
349221
|
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"Gjergji Zaimi",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349221"
}
|
Stack Exchange
|
The (possibly negative) integers arising from sum of elements in any column of the character table
Let $G$ be a finite group and $A$ be the character table for the irreducible complex representations. The sum of elements in any row of the character table is a positive integer as its equals to the multiplicity of irreducible representation corresponding to that row inside $V$, where $V$ is the representation on the group algebra induced by the conjugacy action of $G$.
But how to understand the sum of elements in any column of the character table in terms of representation theory? It's an integer by Galois theory but might not be positive (e.g $G=M_{11}$). In the case of Weyl groups, is there a simple formula? At least, can we determine its sign from $G$?
There is also the possibility of a Stembridge $q=-1$ phenomenon (aka cyclic sieving): maybe the absolute value of a column sum counts the fixed points of some set with involution. We can see this for the sum of all characters evaluated at the identity, since in some cases (when all representations are real) this counts the number of elements of the group that are involutions. Also, this picture wouldn't necessarily help with an interpretation for the sign.
If $G$ is a finite group whose complex irreducible representations are all realizable over $\mathbb{R}$, (eg $G = S_{n}$), then for any $x \in G$, the sum of the entries of the column corresponding to (the conjugacy class of ) $x$ is precisely the number of square roots of $x$ in $G$, so is, in particular, a non-negative integer (and is strictly positive if $x$ has odd order).
|
2025-03-21T14:48:29.543134
| 2019-12-27T18:00:17 |
349222
|
{
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"Phil Tosteson",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349222"
}
|
Stack Exchange
|
Characterization of degeneracy of spectral sequence of a fiber bundle at the second term
Let $f\colon E\to B$ be a fiber bundle of compact manifolds with fiber $F$. Assume that the push-forward $Rf_*(\underline{\mathbb{F}})$ in the derived category of the constant sheaf with coefficients in a field $\mathbb{F}$ is isomoprhic (in the derived category) to the direct sum of its cohomology sheaves (with appropriate shifts). As far as I understand, this implies that the spectral sequence of this fibration degenerates at the second term.
My question is whether the converse is true: namely if the spectral sequence degenerates at the second term, does it imply that the push-forward of the constant sheaf is isomoprhic in the derived category to the direct sum of its cohomology sheaves?
I don't think the converse is true, but you might be interested in Deligne's first paper http://www.numdam.org/item/PMIHES_1968__35__107_0/
|
2025-03-21T14:48:29.543226
| 2019-12-27T18:06:02 |
349223
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Hans",
"Mike Shulman",
"Smiley1000",
"Todd Trimble",
"https://mathoverflow.net/users/128639",
"https://mathoverflow.net/users/155881",
"https://mathoverflow.net/users/2926",
"https://mathoverflow.net/users/49"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349223"
}
|
Stack Exchange
|
A formula for a right adjoint in terms of a left
For those familiar with (covariant) Galois connections, you may have noticed that they can be viewed as categorical adjunctions. A Galois connection is a pair of maps between posets $X$ and $Y$
$$ f_{\bullet}: X \rightleftharpoons Y: f^{\bullet}$$
such that $f_{\bullet} x \preceq y$ if and only if $x \preceq f^{\bullet} y$.
If we view $X$ and $Y$ as small categories, then order-preserving maps are simply functors between such categories as
$$Hom(f_\bullet x, y) \cong Hom(x, f^\bullet y)$$
is equivalent to the condition above (naturality is trivial).
In the case of posets, there is an explicit formula to calculate the right adjoint from the left adjoint if one exists (sim. left adjoint from right adjoint).
You can check that
$$f^\bullet (y) = \bigvee f_{\bullet}^{-1}(\downarrow y)$$
is a right adjoint (if $f_\bullet$ preserves joins I believe??).
Now here is my question: Are there other areas of mathematics or other examples where an explicit formula for a right adjoint in terms of the left adjoint appears?
References most welcome. Thanks!
(No, it's $f_\bullet$ preserves sups.) Do you know the adjoint functor theorem, which generalizes this formula? https://ncatlab.org/nlab/show/adjoint+functor+theorem
Thanks @ToddTrimble I have read the ncatlab article you sent before but I'm afraid I got lost in some of the technical details. Do you or anyone else know a more readable reference for the Adjoint Functor Theorem?
More readable, huh. Well, I first learned it from Categories for the Working Mathematician; you could try there. It's probably in a bunch of places, but I am not intimately familiar with the more recent introductory textbooks. But (since I had a certain amount to do with the writing of the nLab article): which technical details lost you, or where did you first find yourself getting lost?
Just so that it is recorded here as an answer, here's the formula from the "naive" adjoint functor theorem that directly generalizes the one given in the post for posets:
$$f^\bullet(y) = \lim_{(x,\alpha)\in (f_\bullet \downarrow y)} x$$
where the limit is over the comma category $(f_\bullet \downarrow y)$, whose objects are pairs of $x\in X$ with $\alpha:f_\bullet(x)\to y$.
The subtlety in adjoint functor theorems comes from the fact that this comma category is of the same "size" as the categories $X$ and $Y$, while non-poset categories rarely have limits of the same size as themselves. Thus, the adjoint functor theorems used in practice impose various technical conditions enabling this limit to be replaced by a smaller one; see e.g. the nlab article for details.
Thank you . If you wouldn't mind, can you give any intuition as to how the comma category construction comes into play here?
The same way that $f_\bullet^{-1}(\downarrow y)$ appears in the poset formula. The object $f^\bullet y\in X$ comes with a counit map $f_\bullet f^\bullet y \to y$ and is universal among such objects equipped with such maps, so to construct it we take the limit over all objects $x\in X$ equipped with such a map $f_\bullet x \to y$.
Shouldn't it be the colimit instead of the limit?
It is common for the structure of objects in a category to be corepresentable, and so the formula
$$Hom(f_\bullet x, y) \cong Hom(x, f^\bullet y)$$
is a formula for the right adjoint.
As an example of how this works, in the category of groups, you have the fact that
$$ |G| \cong \hom(\mathbb{Z}, G) $$
where $|G|$ means the underlying set, and that the group operation is given by the natural transformation
$$ \hom(\mathbb{Z}, G) \times \hom(\mathbb{Z}, G) \cong \hom(F_2, G) \to \hom(\mathbb{Z}, G) $$
induced by the map $\mathbb{Z} \to F_2$ sending $1 \mapsto xy$, where $F_2$ is the free group on two elements $x$ and $y$. (it is the coproduct of $\mathbb{Z}$ with itself, thus the first isomorphism)
So, if $f^\bullet$ is a group-valued functor, you can compute it as follows:
The underlying set of $f^\bullet X$ is given by $\hom(f_\bullet \mathbb{Z}, X)$
The group operation is $\hom(f_\bullet \mathbb{Z}, X) \times \hom(f_\bullet \mathbb{Z}, X) \cong \hom(f_\bullet F_2, X) \to \hom(f_\bullet \mathbb{Z}, X)$
and this information suffices to determine the group; the identity and inverse can be given directly by similar formulas if desired.
One classical example I am thinking of is the suspension-loopspace adjuction. However, I think it would be impossible to say that the suspension of a space is some formula given by the loopspace of a space. But it is so for posets! I think the obstruction is what @Mike Shulman is talking about concerning small limits.
I like your answer @Anon, do you know of any more examples in other categories?
|
2025-03-21T14:48:29.543678
| 2019-12-27T19:43:22 |
349228
|
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|
Stack Exchange
|
Open convex hull of a closed set
Let $X$ be a closed set in a Euclidean space of finite dimension and suppose that its convex hull $H$ is open. I can prove that, in this case, $H$ is a Cartesian product of a line with an open convex set of dimension lower by one. (See https://mathstodon.xyz/@11011110/103381278180283137 for proof sketch.) It's also not hard to realize any such product as an open convex hull of a closed set. Does this fact appear in the literature anywhere?
|
2025-03-21T14:48:29.543737
| 2019-12-27T21:45:48 |
349233
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349233"
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|
Stack Exchange
|
Initial value problem with heterogeneous initial values
In all the references I checked the standard initial value problem for an ODE is stated as:
\begin{equation}
\begin{cases}
y'=F(y,t)\\
y(t_0)=y_0
\end{cases}
\end{equation}
for some $F:\mathbb{R}^{n+1}\to\mathbb{R}^n$, and under hypotheses on $F$ one can conclude existence, uniqueness, etc.
My question is: is there a standard way to address the problem:
\begin{equation}
\begin{cases}
y'=F(y,t)\\
y_i(t_i)=y_{0,i}, \quad i=1,\ldots, n
\end{cases}
\end{equation}
where the "initial condition" is not given for the same $t_0$ for all the components of $y$, but at different $t_is$? where by address I mean, to begin with, check existence and uniqueness conditions, e.g. do some version of Peano, Osgood and Caratheodory theorems hold? any reference would be appreciated.
I asked the same question here https://math.stackexchange.com/questions/3485504/system-of-initial-value-problems-with-initial-conditions-at-different-t-0
and now I am trying MO.
My thoughts:
For example, using the integral formulation of the problem, in the case $n=2$ for simplicity, we have:
\begin{align}
y_1(t)&=y_1(t_{01})+\int_{t_{01}}^t F_1(y_1(s),y_2(s),s)d s\\
y_2(t)&=y_2(t_{02})+\int_{t_{02}}^t F_2(y_1(s),y_2(s),s)d s
\end{align}
Now if $F_1$ and $F_2$ are Lipschitz with constant $L<1/(2|t_{01}-t_{02}|)$ on the interval $I=[t_{01}-t_{02}]$ then on $I$ there exist a unique solution via the standard argument by the contraction mapping theorem. In fact define a norm on $C^1(I, \mathbb{R}^2)$ by $\lVert y-z\rVert=\max_t{|y_1(t)-z_1(t)|}+\max_t{|y_2(t)-z_2(t)|}$. Then, if $T$ is the integral operator associated with the IVP:
\begin{equation}
\lVert Ty-Tz\rVert\le\max_t\int_{t01}^t|F_1(y(s))-F_1(z(s)|d s+\max_t{\int_{t02}^t|F_2(y(s))-F_2(z(s)|d s}
\end{equation}
\begin{equation}
\le2\max_t L\int_I(|y_1(s))-z_1(s)|+|y_2(s))-z_2(s)|)d s
\end{equation}
if $t \in [t_{01},t_{02}]$
Is this the standard way to proceed? Is there any more general/better way?
|
2025-03-21T14:48:29.543880
| 2019-12-27T22:15:15 |
349235
|
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|
Stack Exchange
|
Limit of quotient stacks
Let $k$ be a field (we can set it to be either perfect or algebraically closed if necessary), let $G$ be a (split) reductive group over $k$. Let $(X_i)$ be a filtered projective system of finite type $k$-schemes with affine transition maps and set $X = \lim_i X_i$ the projective limit of this system. We also assume $G$ acts on $X$ and the $X_i$ and that all the transition maps are $G$-equivariant.
Consider $[X/G]$ the quotient stack of $X$ by $G$ and all the quotient stacks $[X_i/G]$ we then have an inverse system in the $(2)$-category of algebraic stacks. Moreover we have a natural map $[X/G] \rightarrow \lim_i [X_i/G]$.
By the theory of Alper (see https://arxiv.org/abs/1005.2398) we can expect this map to be an adequate homeomorphism (loc cit) which is not far from being an isomorphism.
Question 1 : Is this map an isomorphism of algebraic stacks ?
The following paper suggests it is true (https://arxiv.org/abs/1307.4669) in the first line of the proof of 1.4.2.b.
Question 2 : Same as 1 but I only assume $k$ to be a noetherian ring ?
This is true for formal higher categorical reasons. Namely, taking homotopy quotients of sheaves of sets by a sheaf of groups (yielding sheaves of groupoids) commutes with homotopy limits indexed by categories with simply connected nerve. This is a special case of a more general commutativity result in $\infty$-topoi, which is tersely explained at https://ncatlab.org/nlab/show/commutativity+of+limits+and+colimits#coproducts_commute_with_connected_limits.
|
2025-03-21T14:48:29.544052
| 2019-12-27T22:44:43 |
349237
|
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|
Stack Exchange
|
Uncountability of the real numbers from LLPO without countable choice
Does there exist a proof of the uncountability of the real numbers that uses analytic LLPO (the statement that any real number $x$ satisfies either $x \leq 0$ or $x \geq 0$) but avoids Excluded Middle and the Axiom of Choice (including dependent and countable choice)?
You may use Unique Choice.
I find it difficult to apply LLPO because of the overlapping cases and the unavailability of Countable Choice.
Note that $x\le 0$ or $0\le x$ holds for all (countably many) rational $x$. So you will need further properties of the reals to prove uncoutability.
Isn't the usual proof perfectly constructive, without using any classicality axioms?
@MikeShulman Yes, but it also uses countable choice, which I'm not allowing
in order to understand your question i seem to need your definition/construction of the real numbers, since this alone is not trivial in the absence of countable choice.
@Frank'aWaaldijk I'm defining the real numbers using Dedekind cuts
@Frank'aWaaldijk In fact, you can even use the Cauchy real numbers. The problem appears to be difficult regardless
Hmm, it's doable if $\forall x \in \mathbb{R} . x \geq 0 \lor \lnot (x \geq 0)$ but I don't see immediately how we'd do it just with analytic LLPO. Interesting question! Are you trying to show that the reals are uncountable without choice, by any chance?
@AndrejBauer Yes, and this is a stepping stone towards that
My guess would be that LLPO is not going to help with uncountability of reals, at least not the strong form "for every sequence of reals there is a real that avoids it". This is so because LLPO does not allow you to construct a discontinuous map (in the Johstone topological topos LLPO holds but there are no discontinuous maps there). As soon as you can construct maps that "jump", you can employ Levy-type proofs of uncountability, see this note.
Also, are you aware of Ingo Blechsmidt's recap at https://groups.google.com/forum/#!topic/constructivenews/jSvzqu1LUis?
@AndrejBauer Thank you. I haven't seen that thread before
@tj_ The point is that LLPO is a non-constructive principle because there's no effective procedure for deciding which of $x \leq 0$ or $x \geq 0$ is true. So by working in a restricted logic which doesn't have Excluded Middle or Countable Choice, but does have access to the non-constructive principle LLPO, can we still prove the real numbers uncountable?
I would try to port Bas Spitters' example (Appendix A of this note) to Johstone's topological topos. If you can do that, you'll show that LLPO cannot prove the strong form of uncountability. If you cannot do it, it migth give you a clue on how to prove it.
The answer is in spirit: yes. One can prove uncountability of the reals without using countable choice by using a modified version of LLPO which I call fLLPO (functional LLPO).
Still, fLLPO is also not essentially necessary in my opinion. In my view, constructive uncountability of the reals relies on countable choice only indirectly.
Generally speaking, countable choice is mostly needed in situations where we prefer to keep our objects slightly generic, to help our 'bookkeeping' remain light.
But if we define the reals numbers a bit more precise, as is common practice for instance in intuitionistic mathematics INT, then uncountability follows directly (without choice and without fLLPO).
To see this, take the following
Definition of the explicit reals:
A sequence $x=[l_0,r_0],[l_1,r_1],\ldots$ of binary rational intervals is an explicit real number iff for all $n\in\mathbb{N}$ we have
$r_n, l_n\in 2^{-n}\cdot\mathbb{Z}$
$r_n-l_n=2^{-n+1}$
$[l_{n+1},r_{n+1}]\subset[l_n,r_n]$
This definition can be a bit cumbersome, for instance when multiplying explicit real numbers, because it forces one to specify any real number with a strict convergence rate.
And, given a Cauchy sequence of rationals $c$ which has no explicit modulus of convergence, one needs countable choice to see that $c$ is Cauchy-equivalent to an explicit real number. This is in my opinion the indirect reason why countable choice is 'necessary' to prove uncountability of the real numbers (given by the Cauchy-sequence definition).
The other way round is choice-unproblematic: any explicit real $x$ determines a unique Cauchy-sequence of rationals, just take the left end points $(l_n)_{n\in\mathbb{N}}$.
Now the explicit real numbers are easily seen to be uncountable without using choice, through a simple diagonal argument.
Proposition (without choice): the explicit reals are uncountable.
proof
Let $x_0, x_1,\ldots$ be a sequence of explicit reals. Then we can construct an explicit real $y=[s_0,t_0],[s_1,t_1],\ldots$ such that $y\# x_n$ for all $n\in\mathbb{N}$.
To see this, firstly we put $s_0=0, t_0=2$.
Next, suppose that for given $n$, the left and right end points $s_{2i+1}, t_{2i+1}, s_{2i+2}, t_{2i+2}$ have been defined for all $i< n$, such that moreover $[s_{2i+2}, t_{2i+2}]\ \#\ x_i$.
We now consider the left and right end points of $(x_n)_{2n+2}$, which is the $2n\!+\!2$ -th binary interval of the sequence $x_n$. Call those end points $u_{2n+2}, v_{2n+2}$ respectively, then $v_{2n+2}-u_{2n+2}=2^{-2n-1}$ since $x_n$ is an explicit real.
Since $t_{2n}-s_{2n}=2^{-2n+1}$, there are a unique smallest binary rational $q\in 2^{-2n-2}\cdot\mathbb{Z}$ and a subsequent unique smallest binary rational $r\in 2^{-2n-1}\cdot\mathbb{Z}$ such that
(i) $[q,q+2^{-2n-1}]\subset [s_{2n}, t_{2n}]$
(ii) $[q,q+2^{-2n-1}]\ \# \ [u_{2n+2}, v_{2n+2}]$
(iii) $[q,q+2^{-2n-1}]\subset [r,r+2^{-2n}]\subset [s_{2n}, t_{2n}]$
We can now define $s_{2n+1}=r, t_{2n+1}=r+2^{-2n}$ and $s_{2n+2}=q, t_{2n+2}=q+2^{-2n-1}$.
We leave it to the reader to verify that $y=[s_0,t_0],[s_1,t_1],\ldots$ is a well-defined explicit real number such that $y\# x_n$ for all $n\in\mathbb{N}$.
In slightly more general phrasing we therefore find:
Corollary (without choice): The set of real numbers $\mathbb{R}$ contains the uncountable subset $E$ of explicit reals.
But we would like to see that this subset $E$ is in fact all of $\mathbb{R}$, in the sense that any real number is equivalent to an explicit real.
For this we can use a functional version of LLPO called fLLPO, and for clarity we need to look at the representation of real numbers through Cauchy-sequences. We will call the set of Cauchy-sequences $\mathbb{R}_{\rm basis}$.
fLLPO (functional Lesser Limited Principle of Omniscience)
There is a function $f_{\geq 0}:\mathbb{R}_{\rm basis} \rightarrow \{0,1\}$ such that for all $x\in \mathbb{R}_{\rm basis}$ we have $f_{\geq 0} (x) = 0 \rightarrow x\leq 0$ and $f_{\geq 0} (x) =1 \rightarrow x \geq 0$
Notice that according to Bishop, if we assume LLPO then such a function should exist by the very meaning of $\forall\exists$.
Proposition (using fLLPO):
Every real number is equivalent to an explicit real number.
proof (short version, I will fill in later)
For a given $x\in \mathbb{R}_{\rm basis}$ we can apply the function $f_{\geq 0}$ step-by-step to determine uniquely an explicit real $y$ equivalent to $x$.
Without assuming countable choice, are the explicit reals a field?
You modified both the definition of reals and LLPO, so I wonder how much your answer is related to the original question.
@Andrej: I did not modify the definition of the reals, and if you cannot see the relation between LLPO and fLLPO... well. Do you have a real question here?
The OP clearly stated that they use the Dedekind reals, whereas you are using rapidly converging Cauchy sequences of rationals and Cauchy reals. The distinction between LLPO and fLLPO is precisely the absence of propositional truncation in the latter.
$E$ is not a subset of reals because the real number $0$ is represented by many different explicit reals. Are you using an implied Bishop-style equality relation on $E$, and similarly for $\mathbb{R}$? In this case there is no problem whatsoever with uncountability of reals, because the Cauchy sequences never get quotiented and a simple diagonal argument, much like the one you spelled out, simply works.
@AndrejBauer your last remark is patently untrue. If you think that a simple diagonal argument will show that the Cauchy reals are uncountable, then please write this argument down in an answer of your own... good luck. I repeat there is nothing wrong with my definitions, nor is there anything wrong with my answer. I asked the OP in the comments if the definition of the reals was limited to a specific one and the OP replied: "In fact, you can even use the Cauchy real numbers. The problem appears to be difficult regardless".
@jkabrg Yes the explicit reals form a field, but the proof calls for tedious bookkeeping, as I mentioned in the answer.
sorry, I should have added that the explicit reals form a field also in the absence of countable choice, but the proof then requires etc.
Andrej's point is that a Cauchy real number is by definition an equivalence class of Cauchy sequences, rather than a single Cauchy sequence on its own. As a result, your "explicit reals" are not a subset of the real numbers
Also, showing that the real numbers contain an uncountable subset is easy: The irrational numbers are always uncountable. And even then, so what? In Recursive foundations, the natural numbers have an uncountable subset
Yes but my answer shows a lot more, namely that the uncountable subset contains a representative of all the real equivalence classes. That is precisely the point of using fLLPO.
I've had enough now. I give a very nice answer to an intriguing question which no one has managed to answer in 2 months time, and my efforts are met with hostile and poor reading, as well as mathematical errors. This is Mathoverflow, supposedly for professionals, and I have no inclination at all to spell out the obvious.
You are correct, I apologize for being so impolite and rash. I was thinking of Cauchy sequences with explicit moduli (essentially the ones you call explicit). Although, if one is going to work without countable choice, one simply shouldn't use Cauchy sequences with $\forall\exists$ definition, as those won't even form a Cauchy-complete field (suitably quotiented). So my only remark left is whether you have a notion of equality of sequences (a la Bishop) that you didn't mention. Otherwise, how is $E$ a subset of reals?
That is a very nice reply, thank you Andrej! Please allow me some time to address the equivalence issue that you mention, as I have a pressing matter outside mathematics.
I wrote down what I think equality of sequences should be in my post, and I think there's no problem at all. It's just the obvious thing.
This is not an answer but rather an alternative proof of @FrankaWaaldijk's observation that explicit Cauchy sequences are uncountable. The only difference is that we use explicitly given moduli of convergence, rather than a fixed rate of convergence, and that we explicitly impose an equality relation (which I think is implied in @FrankaWaaldijk's answer).
We work in Bishop-style constructive mathematics but without countable choice. In particular, a set is a collection with an imposed equality relation, and a function is an operation that respects the imposed equalities on the domain and the codomain.
Definition: An (explicit) Cauchy sequence is a sequence of rational number $q : \mathbb{N} \to \mathbb{Q}$ together with a strictly increasing function $\mu : \mathbb{N} \to \mathbb{N}$, called modulus, such that $\forall k, m, n \in \mathbb{N} \,.\, |q_{\mu(k) + m} - q_{\mu(k) + n}| < 2^{-k}$.
Two Cauchy sequences $(q, \mu)$ and $(q', \mu')$ are considered equal when $|q_{\mu(i)} - q'_{\mu'(j)}| \leq 2^{-i-j}$ for all $i, j \in \mathbb{N}$.
Lemma: There is an operation which takes as input an interval $[\ell, r]$ with rational endpoints and a Cauchy sequence $(q, \mu)$, and outputs an interval $[\ell', r']$ such that $|r' - \ell'| = |r - \ell|/3$ and $[\ell', r'] \subseteq [\ell, r]$, and at most finitely many terms of $q$ are inside $[\ell', r']$.
Proof. We describe the procedure for computing $r'$ and $\ell'$. Let $k$ be the least number such that $2^{-k} < |r - \ell|/12$. Now, if $q_{\mu(k)} > (\ell + r)/ 2$ then let $[\ell', r'] = [\ell, (2 \ell + r)/3]$, otherwise let $[\ell', r'] = [(\ell + 2 r)/3, r]$. In the first case, we have for all $n$
$$
q_{\mu(k)+n} > q_{\mu(k)} - 2^{-k} > (\ell + r)/2 - |r - \ell|/12 > (\ell + 2 r)/3.
$$
Thus, at most the first $\mu(k)$ terms of $q$ are contained in $[\ell', r']$. The other case is treated similarly. $\Box$
Theorem: The set of explicit Cauchy sequences is uncountable.
Proof.
Let $C$ be the set of all rational Cauchy sequences and $a : \mathbb{N} \to C$ a sequence of Cauchy sequences. We construct a sequence $s$ which avoids every sequence enumerated by $a$.
First, define a sequence of nested intervals $[\ell_{-1},r_{-1}] \supseteq [\ell_0, r_0] \supseteq \cdots$ by taking $[\ell_{-1}, r_{-1}] = [0, 1]$ and letting $[\ell_{k+1}, r_{k+1}]$ be the interval obtained from the Lemma applied to $[\ell_k, r_k]$ and $a(k)$. Now define the sequence $s : \mathbb{N} \to \mathbb{Q}$ by $s_k = (\ell_k + r_k)/2$. By the Lemma, $s_k$ differs from $a(k)_k$ and therefore $s$ is not enumerated by $a$. $\Box$
Discussion
Authors often require a fixed modulus of convergence for all sequences. For example, Bishop requires $|q_m - q_n| < 1/m + 1/n$ (if memory serves me right), and another popular choice is $|q_m - q_n| \leq 2^{-min(m, n)}$. Which particular modulus we prefer, if any, is not too important because we can always find a subsequence of a given sequence that has any desired modulus:
Lemma: Suppose $(q, \mu)$ is a Cauchy sequence and $\nu : \mathbb{N} \to \mathbb{N}$ a strictly increasing function. Then there is a reindexing function $\iota : \mathbb{N} \to \mathbb{N}$ such that $(q, \mu)$ is equal to $((q \circ \iota), \nu)$.
Proof. Exercise. $\Box$
Of course, one would want to show the uncountability of the set of Cauchy sequences without explicit moduli. That is, suppose we say that $q : \mathbb{N} \to \mathbb{Q}$ is standard Cauchy when $$\forall k \in \mathbb{N} . \exists \mu \in \mathbb{N} . \forall m, n \in \mathbb{N} \,.\, |q_{\mu+m} - q_{\mu+n}| < 2^{-\mu}. \tag{1}$$
Can we show uncountability of standard Cauchy sequences? Well, one seemingly needs to choose a suitable $\mu$ at each step of diagonalization.
However, as a mitigating factor I want to point out that (1) is not a workable definition of Cauchy sequences, unless we do have countable choice, because we cannot even show that a Cauchy sequence of Cauchy sequences converges.
@jkabrg asked whether the explicit Cauchy sequences (with the given equality) form a complete field. I did not verify all the details, but I think the answer is positive so long as one sticks to the principle that without countable choice $\forall\exists$ statements should be replaced by moduli. That is, an explicit Cauchy sequence of explicit Cauchy sequences converges to an explicit Cauchy sequence.
I personally am not a big fan of using setoids, or sets with imposed equality relations. The reason is as follows. Suppose we want to work in a "naturally occurring" constructive mathematical environment $\mathcal{E}$ (a topos, a higher-order fibration, a realizability model, a type theory). Then we do not get to pick equality relations because $\mathcal{E}$ already has a built in notion of equality. If $\mathcal{E}$ also has quotients then we may get the desired equality on a set by quotienting it. Setoids and Bishop's sets-with-equality are not even object of $\mathcal{E}$, but rather objects of an exact completion of $\mathcal{E}_{\mathrm{ex}}$. They live in the wrong place! To put it differently, even though Bishop succeeds in being agnostic about excluded middle, he fails to be ecumenical about equality because he commits to a certain kind of semantic models known as exact completions. These include realizability models but exclude sheaves on a space.
Surely Bishop would require $1/m+1/n$, not $1/m - 1/n$?
Absolutely, thanks and fixed.
Is it obvious that the function constructed in the theorem respects the imposed notion of equality (i.e., if I input different sequences of Cauchy sequences representing the same sequence of reals, I get the same real out)?
@JamesHanson: if you are referring to the construction of $s$ from $a$, then note that I never claimed it to be equality preserving. In fact it is not, nor does it need to be for the purposes of proving the theorem (which only requires existence of $s$). I hope I was careful enough to distinguish operations from functions (equality preserving operations).
I see. Thank you.
|
2025-03-21T14:48:29.545026
| 2019-12-27T23:28:05 |
349239
|
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|
Stack Exchange
|
Groups for which all projections of $C^*_{\text{red}}G$ belong to $\mathbb{C}G$
Revision: According to comment of Wojowu we give a complete revise for this post.
A group $G$ is a pr-group if all projections of $C^*_{\text{red}} G$ are contained in its dense subalgebra $\mathbb{C}G$.
What is a counterexample of this situation?
Is there a characterisation of pr-groups?
Can't you take for any $G$ the augmentation map $\mathbb C[G]\to\mathbb C$ and compose it with inclusion $\mathbb C\hookrightarrow\mathbb C[G]$?
@Wojowu yes my question is stupid for finite group since the group algebra is the reduced algebra. But this property can be defined for arbitrary group
I should say " Let G be a group with non trivial torsion"
@Wojowu or I could say "a pr-group is a group for which all projections of $C^*_{red} G$ belongs to $\mathbb{C}G$"
Which idempotents are counted as trivial? Only $0$ and $1$, or others?
@LSpice yes these are counted as trivial. On the other hand for every torsion $g$ the element $1/n\sum_0^n \lambda^i g^i$ is a non trivial idempotent.
Where $g^n=1$ and $\lambda$ is the nth root of unity.
Could you edit your question?
@YCor Yes I completely changed it. Thanks for your suggestion.
@Wojowu according to your comment I changed my question. Thanks for your atention.
@LSpice I modified the question. Please see this new version.
If the group is torsion-free, pr-ness is clearly implied by the Kadison–Kaplansky conjecture, so providing a counterexample in the torsion-free case seems to be difficult.
However, every group containing the free product $\Gamma=\mathbb Z/n \ast \mathbb Z$ is non-pr. Indeed, let $g$ denote the generator of $\mathbb Z/n$ and $t$ the generator of $\mathbb Z$, and let $p = \frac1n\sum_{k=0}^{n-1} g^n$ be the projection coming from the torsion part.
The group C*-algebra $C^*_r(\mathbb Z)$ has lots of unitaries which have infinitely many nonzero Fourier coefficients: for instance, $u = \exp(i\cdot(t+t^{-1}))$ (it corresponds via Fourier transform to the function $\exp(2i\cdot\cos\theta)$ on the circle). Let's write its Fourier expansion as $u = \sum_{k\in\mathbb Z} u_kt^k$.
Now, $upu^*$ is clearly a projection in $C^*_r(\Gamma)$. Its Fourier coefficient at $t^k g t^{-\ell}$ is equal to $u_k\overline{u_\ell}/n$, so infinitely many Fourier coefficients are non-zero, and hence in can't belong to $\mathbb C\Gamma$.
I believe, one can relax the freeness condition to something significantly milder (intuitively, one requires an element $t$ of infinite order whose conjugation action on $g$ produces sufficiently many independent elements), but I haven't thought much further.
|
2025-03-21T14:48:29.545491
| 2019-12-28T00:08:58 |
349240
|
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|
Stack Exchange
|
indecomposable modules of gentle algebras
Let $A = \mathcal{k}Q/I$ be a gentle algebra (where $\mathcal{k}$ is algebraically closed). In the paper Auslander-Reiten Sequences with Few Middle Terms and Application to String Algebras, Butler and Ringel show that string and band modules classify the indecomposable modules of $A$ (pages 157–161). To flesh out the details a little more, for each string $c$ of $Q$ they produce a string module $M(c)$. And for each cyclic string $b$ they produce a family of band modules $M(b,x,n)$ where $x \in \mathcal{k}^*$ and $n \geq 1$.
I am trying to compare this to the classification of indecomposable representations of the $2$-Kronecker quiver. But as an example I don't see where the indecomposable representation
$$\mathcal{k}\overset{0}{\underset{1}{\rightrightarrows}} \mathcal{k}$$
appears in Butler and Ringel's classification. What am I missing?
If the $2$-Kronecker quiver is
$$ \overset 1\circ\overset{\alpha}{\underset{\beta}{\rightrightarrows}} \overset 2\circ ,$$
then the representation
$$\mathcal{k}\overset{0}{\underset{1}{\rightrightarrows}} \mathcal{k}$$
corresponds to the string module $M(\beta)$.
|
2025-03-21T14:48:29.545599
| 2019-12-28T05:49:15 |
349242
|
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|
Stack Exchange
|
If $p^2 - q^2$ is a perfect square where $p$ and $q$ are primes $> 5000$ then is one of its prime factors always greater than $17$?
Is it true that if $p^2 - q^2$ is a perfect square where $p$ and $q$ are primes $> 5000$ then it has a prime factor greater than $17$?
Note: This question was asked in MSE but did not receive an answer. Hence posting in MO.
What is the question here?
looking at MSE post, this appears to be a conjecture. This has to be spelled out, IMHO.
This is not an answer, but just an observation. If $p^2-q^2=a^2$, then $(a,q,p)$ form a Pythagorean triple. It follows that there are $c$ and $d$ such that $p=c^2+d^2$; $a=2cd$ and $q=c^2-d^2$. Since $q$ is supposed to be prime, $c-d$ must be 1, so $c=(q+1)/2$ and $d=(q-1)/2$. It follows that $p=(q^2+1)/2$ and $a=(q^2-1)/2$. Hence you’re asking that if $q$ is prime over 5000 and is such that $(q^2+1)/2$ is prime, must one of $q-1$ and $q+1$ have a factor greater than 17.
This seems very likely: there are something like $n^7$ numbers up to $e^n$ with all factors 17 or less, so it seems likely that there will only be finitely many pairs of such numbers differing by 2.
Definitely there are finitely many such numbers: if $q-1$ and $q+1$ are 17-smooth, they may be written as $ax^3$ and $by^3$ for certain $a,b$ taken from a finite set, and $y/x$ would be too good rational approximation of $(a/b)^{1/3}$. Instead of Roth theorem you could you Faltings or something else.
The title asks whether all the prime factors exceed $17$, but the body asks whether one prime factor exceeds $17$. Please edit for consistency.
@GerryMyerson Thanks for spotting the typpo. Fixed.
I'm voting to close this question because it is now attracting useful answers on MSE which in my view is a more fitting place for recreational number theory
@GeoffRobinson Thanks. Can you please fix the typpo for readability?
@NilotpalKantiSinha : I thought I had deleted the comment, I have now, because I found that the observation was made by someone else on MSE.
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2025-03-21T14:48:29.545891
| 2019-12-28T06:27:12 |
349243
|
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|
Stack Exchange
|
A definable gender problem related to Sacks reals
Let $L$ be the ground model, and $a\in2^\omega$ be a Sacks-generic real over $L$. Note that any real $x\in S=(2^\omega\cap L[a])\setminus L$ is still Sacks-generic over $L$. Now assume that $\mathsf E$ is an OD (ordinal-definable) equivalence relation in $L[a]$ on the set $S$, with exactly two equivalence classes, say $M$ and $F$. (Two genders of the Sacks reals.) Are $M$ and $F$ necessarily OD themselves?
My idea of a counterexample is as follows. Let $F$ be the set of all continuous 1-1 maps $2^\omega$ onto $2^\omega$, coded in $L$. Then $F$ is a group under the superposition. Moreover if $x,y\in S$ then it is known that $y=f(x)$ for some $f\in F$. Now if $H\subseteq F$ is a subgroup coded in $L$ then the relation:
$x \mathrel{\mathsf E_H} y$ iff $y=f(x)$ for some $f\in H$
is an OD equivalence on $S$, and there is no immediate idea as how to OD-define the $\mathrel{\mathsf E_H}$-class of $a$ (w/o a reference to $a$). Now the goal is to define $H$ such that $\mathrel{\mathsf E_H}$ has exactly two equivalence classes on $S$ in $L[a]$. The principal non-commutativity of $F$ looks to be a huge obstacle though.
[Added Feb 4 at 5:50]
And finally it is established, there are two distinct but OD-indiscernible populations of Sacks-generic reals over $L$, arXiv . In fact this is an unpublished result of Solovay dated back to 2002. A similar result holds for $E_0$-large forcing, via a known ``canonization'' theorem, but the problem is open for other popular forcing notions.
That's a really nice question!
In my paper http://jdh.hamkins.org/ehrenfeuchts-lemma-in-set-theory/, we prove a negative answer in the extension with two Sacks reals.
Yes thm 4.6 in Ehrenfeucht’s Lemma in Set Theory, in fact the result can be improved so that the indiscernible sets of reals are countable and disjoint and the union is lightface $\varPi^1_2$ as in Golshani etc. MLQ 2017, 63, 1-2, 19-31. Those examples are designed to yield exactly that, the plain Sacks model is a different thing.
The question is now answered in this recent paper, where it is shown that in the extension $L(s)$ of the constructible universe $L$ by a single Sacks real $s$, there is a definable pair $\{a,b\}$ of objects (where $a$ and $b$ are subsets of real numbers) such that neither $a$ nor $b$ is ordinal definable.
This shows that $a$ and $b$ are indiscernible in $L(s)$ in the following strong sense: for any formula of set theory $\phi(x,y,\alpha)$, where $\alpha$ is an ordinal parameter, $\phi(a,b,\alpha) \leftrightarrow\phi(b,a,\alpha)$ holds in $L(s)$.
|
2025-03-21T14:48:29.546120
| 2019-12-28T06:30:30 |
349244
|
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|
Stack Exchange
|
Representations are determined by characters : Groups and Lie algebras
I know that any finite-dimensional complex representation of a finite group $G$ is determined by its characters. This is immediate, in view of the complete reducibility of this category modules.
My question is, do we need complete reducibility when we work in a category of modules over complex finite-dimensional semisimple Lie algebras in order to objects are characterized by their characters?
Thank you.
Assume a module $M$ is determined up to isomorphism by its character $\chi$. Write $\chi = \chi_1 + \cdots + \chi_n$ as a sum of irreducibles. Let $M_i$ be a simple module affording $\chi_i$. The direct sum $M_1 \oplus \cdots \oplus M_n$ affords $\chi$ and by assumption is isomorphic to $M$. Hence $M$ is a direct sum of simple modules, so semisimple. Complete reducibility holds.
@JayTaylor please explain, why is it always possible to write a character (of a module over a Lie algebra) as a sum of characters of irreducibles? $\chi = \chi_1 + \cdots + \chi_n$?
Characters are additive in short exact sequences, so the character can't tell the difference between a direct sum and a nontrivial extension. In particular, the character of any module can be written as a sum of characters of simple modules (namely those of its Jordan-Hölder composition series).
@BertramArnold Such series exists for infinite-dimensional modules as well? because I remember for arbitrary highest weight modules of Kac-Moody algebras such series doesn't exist. Kindly explain to me more. thanks.
For finite dimensional algebras over algebraically closed fields, simple modules are characterized up to isomorphism by their characters. Probably algebraically closed is not needed.
@BenjaminSteinberg. Thanks. But I am interested in general Kac moody setup where algebras are infinite-dimensional mostly.
Doc, I am not sure what your question is, but the answer is yes. Whatever definition of character you are using, any two extensions of $M$ by $N$ will have the same character. Thus, a non-trivial extension has the character as $M\oplus N$. Bingo: non-isomorphic modules will have the same characters...
Thanks. can you please tell me what is the meaning of extensions?
@GA316 In this context, I believe it means a short exact sequence of the form $0\to M\to M'\to N\to 0$ With a split short exact sequence, $M'\cong M\oplus N$, but if you do not have complete reducibility, then there will exist non-split short exact sequences.
@Aaron Exactically, Aaron explained it well. A character is always a homomorphism from the Grothenideck group to some other group. Since $M^\prime$ and $M\oplus N$ give the same element of the Grothendieck group, they cannot be distinguished by a character...
Thank you. Can you suggest some references regarding these?
What are "these"?
A good reference is Humphrey's book "Representations of semisimple Lie algebras in the BGG category $\mathcal{O}$", see chapter 3 for extensions.
|
2025-03-21T14:48:29.546346
| 2019-12-28T07:32:04 |
349245
|
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|
Stack Exchange
|
Is there a reasonable notion of spectral theorem on a pre-Hilbert space?
I'm trying to understand how bad things could possibly get without Cauchy completeness as a criterion for Hilbert spaces in quantum mechanics. Obviously, doing calculus on a pre-Hilbert space would be complicated but could there still be some reasonable version of the spectral theorem? Why or why not? Some elaboration and perhaps even some references would be appreciated.
One example to consider is the Laplacian $\Delta$ on the pre-Hilbert space $E = C_c^\infty(\Omega)$, where $\Omega$ is some nice domain (e.g. a ball). Then $\Delta$ is symmetric, everywhere defined, and negative definite, so you would hope any "reasonable" version of the spectral theorem would apply to it. Now you would also hope any "reasonable" version of the spectral theorem would let you define the semigroup $e^{t\Delta}$ on $E$, but this is impossible since solutions of the heat equation do not stay compactly supported.
Here is a simple example that shows that the idea of spectral theory on pre-Hilbert spaces in the sense you are asking is hopeless. Consider the pre-Hilbert space consisting of the restrictions of all complex polynomials to $[0,1]$, as a dense subspace of $L^2[0,1]$. Then let $A$ be the operator of multiplication by $x$. The spectral projections of this operator are characteristic functions; none of them except $0$ and $1$ are polynomials.
There is a very nice setup which is suitable for precise mathematical understanding of quantum mechanics including the “delta-function-like eigenvectors”: that of a rigged Hilbert space. It is a Hilbert space $H$ together with a fixed dense continuous inclusion of a locally convex (often assumed nuclear) topological vector space $\Phi\hookrightarrow H$. An example to think of is the inclusion of the Schwartz space $\mathcal S(\mathbb R^n)$ into $L^2(\mathbb R^n)$.
And indeed, there is a very satisfactory spectral theory of selfadjoint operators on rigged Hilbert spaces which gives, for instance, the precise meaning to the statement “the delta functions $\delta_x$, $x\in[0,1]$ form a complete system of generalised eigenvectors for the operator of multiplication by $x$ on $L^2([0,1],\mathrm{Leb})$”.
A systematic treatment can be found in the classical source
I. M. Gelfand and N. J. Vilenkin. Generalized Functions, vol. 4: Some Applications of Harmonic Analysis. Rigged Hilbert Spaces. Academic Press, New York, 1964.
Applications to some classical problems of quantum mechanics can be found in the Ph.D. thesis R. de la Madrid, Quantum Mechanics in Rigged Hilbert Space Language (2001).
Serious question, is there any substantive advantage to working with rigged Hilbert spaces other than being able to rigorously interpret the physicists' delta functions? My feeling has always been that the "right" way to make delta functions rigorous is via spectral theory, but that could just be ignorance on my part.
@NikWeaver, Some aspects of self-adjoint extensions of (restrictions of) self-adjoint operators (e.g., Friedrichs extensions in the semi-bounded case) are nicely explained in terms of $H^1\to L^2 \to H^{-1}$ and such, in my opinion.
@paulgarrett: thank you, that's a good example.
@paulgarrett do you know a good reference that discusses self-adjoint extensions in the framework of Rigged Hilbert spaces?
It should be mentioned that Amiel Feinstein, who translated that book into English discovered an error in Gelfand-Vilkenin's proof, involving the treatment of sets of measure zero, which was then corrected by the following paper of G. G. Gould: https://londmathsoc.onlinelibrary.wiley.com/doi/10.1112/jlms/s1-43.1.745
I'll also add another oft-unmentioned caveat. People coming from the physics literature often expect that "a complete system of eigenvectors" means that for each spectral value there must exist an eigenvector. This is not the case, as can be seen in the multiplication operator $\frac{1}{1+n}$ on $\ell^2$, which has a complete set of eigenvectors but for which there is no eigenvector, not even in the generalized sense, corresponding to the spectral value $0$.
|
2025-03-21T14:48:29.546657
| 2019-12-28T09:03:34 |
349247
|
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|
Stack Exchange
|
Reading list for Equivariant Cohomology
I was applying equivaraint cohomology, in particular in the symplectic setting, for some time, but I feel like I am missing some nice books/course notes/articles. Could you advise me some literature, in particular where basic examples are worked out?
I think that the following notes are quite good, but I would like something more detailed and containing more examples/exercises: Anderson - Introduction to equivariant cohomology in algebraic geometry.
We has a go at some kind of nonabeiian equivariant cohomolgy in "Spaces of maps into classifying spaces for equivariant crossed complexes"
R Brown, M Golasiński, T Porter, A Tonks - Indagationes Mathematicae, 1997 - Elsevier , but I am not sure if it can help in the algebraic geometry stuation.
Loring W. Tu has given a series of lectures on equivariant cohomology, which are available here. These are being made into a book, which is expected to be published in March 3, 2020.
I recommend Equivariant de Rham cohomology: theory and applications which in my opinion is a well-made introduction in equivariant cohomology. Moreover the article of Tymoczko "An introduction to equivariant cohomology and homology, following Goresky, Kottwitz, and MacPherson" in Snowbird lectures in algebraic geometry contains a lot of computations and examples of equivariant cohomology of GKM spaces.
Thanks a lot Panagiotis! I see that the article of Tymoczko is also available on arxiv https://arxiv.org/abs/math/0503369, that's nice
I would like to point out that the term "equivariant cohomology'' is ambiguous. To those unfamiliar with modern algebraic topology, it means Borel cohomology, the cohomology theory that is the subject of this question. That was introduced by Borel in the late 1950's or early 1960's. However, in 1966, Bredon introduced a more general version of equivariant cohomology (I know the year since I was at a talk he gave that year at IAS). Intuitively, it is reasonable to think of orbits $G/H$ as like equivariant points, since their orbit spaces are points. Bredon cohomology starts with a contravariant functor, $M$ say, from the homotopy category of orbits to abelian groups. Then, on $G$-spaces, an ordinary cohomology theory $H^*_G(X;M)$ with coefficients in $M$ is a homotopy functor to graded abelian groups that satisfies the obvious equivariant analogues of the Eilenberg-Steenrod axioms, but with dimension axiom stating that
$$H^*(G/H;M) = H^0(G/H;M) =M(G/H).$$
These are the ordinary cohomology theories of modern algebraic topology. Let $A$ be an abelian group. One example of a coefficient system is the constant coefficient system at $A$, denoted $\underline{A}$. It has $\underline{A}(G/H) = A$ for all $H\subset G$. Let $EG$ be a free contractible $G$-space and define the homotopy orbit $X_{hG}$ of a $G$-space $X$ to be the orbit space $(EG\times X)/G$, where $G$ acts diagonally on the product. Then the Borel cohomology of $X$ is defined to be $H^*(X_{hG};A)$. However, an easy comparison of definitions or axioms shows that Borel cohomology is just the special case $H^*_G(EG\times X;\underline{A})$ of Bredon cohomology. For that reason, no algebraic topologist today would consider writing a book just about Borel cohomology. Unfortunately, there are no textbooks about Bredon cohomology either as far as I know. However, one already somewhat outdated book, not a textbook but a 1995 status report on a subject, ``Equivariant homotopy and cohomology theory''
http://www.math.uchicago.edu/~may/BOOKS/alaska.pdf
starts out with an exposition of equivariant cohomology, including how Borel and Bredon cohomology each apply to the proof of classical P.A. Smith theory, which dates from the late 1930's.
For the desired symplectic emphasis, I’d warmly recommend these (of which Panagiotis’ ref. cites #2–5):
Atiyah, Michael F.; Bott, Raoul, The moment map and equivariant cohomology, Topology 23, 1-28 (1984). ZBL0521.58025.
Berline, Nicole; Getzler, Ezra; Vergne, Michèle, Heat kernels and Dirac operators. Berlin etc.: Springer-Verlag. vii, 369 p. (1992). ZBL0744.58001.
Guillemin, Victor W.; Sternberg, Shlomo, Supersymmetry and equivariant de Rham theory. With reprint of two seminal notes by Henri Cartan, Berlin: Springer. xxiii, 228 p. (1999). ZBL0934.55007.
Audin, Michèle, Torus actions on symplectic manifolds, 2nd edition. Basel: Birkhäuser. viii, 325 p. (2004). ZBL1062.57040.
Meinrenken, Eckhard, Equivariant Cohomology and the Cartan Model, Encyclopedia of mathematical physics, Vol. 2, pp. 242-250 (2006). ZBL1170.00001.
Vergne, Michèle, Applications of Equivariant Cohomology, Proc. Int. Cong. Math. Zürich, Vol. 1, pp. 635-664 (2007). MR2008m:53194.
Do you also have a suggestion regarding the order in which one should look at them?
6, 5, 4, 3, 2. (1: anytime.)
Thanks.. I am now reading 5
Some additional resources, which are more on the algebraic side than the symplectic side:
1) Bill Fulton's Eilenberg lectures on Equivariant Cohomology in Algebraic Geometry, available at David Anderson's website. My understanding is that the plan is for these notes to be compiled into a book at some point.
2) Michel Brion's Equivariant Cohomology and Equivariant Intersection Theory.
A draft of the ECAG book is now available on my website: https://people.math.osu.edu/anderson.2804/ecag/index.html
As mentioned in the comments, Loring Tu has since written a book on equivariant cohomology which assumes little background and is quite clear. I think it prepares one quite well for reading The Moment Map and Equivariant Geometry by Atiyah and Bott.
Here is Tu's book: https://press.princeton.edu/books/hardcover/9780691191744/introductory-lectures-on-equivariant-cohomology.
|
2025-03-21T14:48:29.547041
| 2019-12-28T09:22:31 |
349248
|
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|
Stack Exchange
|
Extending fermat's last theorem
Question: Is the following claim true?
$$\sum_{q=0}^{u}(n+qd)^m\ne a^m \ \ \ \ \forall n,u,d,a\in\mathbb{N}, \ \ \ \ m\in \mathbb{Z}_{\ge 4} $$
I am check upto $[1\le n \le 60,1\le u \le 10,1\le d\le 20,1\le a\le 100,4\le m\le 10]$ without finding a counter example.
PARI/GP
for(n=1,60,for(u=1,10,for(d=1,20,for(m=4,10,for(a=1,100,if(sum(q=0,u,(n+q*d)^(m))==a^m,print([n,u,d,m,a])))))))
Moreover if we put $u=1$ and $m\ge 3$ have no counter example by fermat's last theorem
Below are lots of counter examples for $ m = 3 $ mention but not found for $ m> 3 $
General solution for m=3
? for(n=1,60,for(u=1,10,for(d=1,20,for(m=3,3,for(a=1,100,if(sum(q=0,u,(n+q*d)^(m))==a^m,print([n,u,d,m,a])))))))
[3, 2, 1, 3, 6]
[6, 2, 2, 3, 12]
[9, 2, 3, 3, 18]
[11, 3, 1, 3, 20]
[12, 2, 4, 3, 24]
[15, 2, 5, 3, 30]
[18, 2, 6, 3, 36]
[21, 2, 7, 3, 42]
[22, 3, 2, 3, 40]
[24, 2, 8, 3, 48]
[27, 2, 9, 3, 54]
[30, 2, 10, 3, 60]
[31, 5, 2, 3, 66]
[33, 2, 11, 3, 66]
[33, 3, 3, 3, 60]
[36, 2, 12, 3, 72]
[39, 2, 13, 3, 78]
[42, 2, 14, 3, 84]
[44, 3, 4, 3, 80]
[45, 2, 15, 3, 90]
[48, 2, 16, 3, 96]
[55, 3, 5, 3, 100]
Formula
$$\sum_{q=0}^{u}(n+qd)^{m}=\sum_{i=0}^{m} \binom{u+1}{i+1}\sum_{j=i}^{m}\binom{m}{j}n^{m-j}d^j\sum_{k=0}^{i}(i-k)^j(-1)^k\binom{i}k $$
Where $n,d\in \mathbb{R}$ and $u,m\in \mathbb{Z^*}$ and $0^0=1$
Proof : https://math.stackexchange.com/q/3476906/647719
Note: This question was posted in MSE(17/12/19) and got some upvotes but no answer hence posting in MO
Related posts
https://math.stackexchange.com/q/3473570/647719
|
2025-03-21T14:48:29.547141
| 2019-12-28T09:23:02 |
349249
|
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|
Stack Exchange
|
On intrinsic volumes
Let $\Gamma$ be a convex polytope in $\mathbb R^n$. The $k$-th intrinsic volume of $\Gamma$ is the number
$$
\text{v}_k(\Gamma)=\sum_{\Delta\in{\mathcal B}(\Gamma,k)}\text{vol}_k(\Delta)\psi_\Gamma(\Delta)\,,
$$
where $\mathcal B(\Gamma,k)$ is the set of $k$-dimensional faces of $\Gamma$, $\text{vol}_k(\Delta)$ the Lebesgue measure of $\Delta$ and $\psi_\Gamma(\Delta)$ the outer angle of $\Delta$.
Now consider the expression
$$
\text{v}_k^\varphi(\Gamma)=\sum_{\Delta\in{\mathcal B}(\Gamma,k)}\varphi(E_\Delta)\text{vol}_k(\Delta)\psi_\Gamma(\Delta)\,,
$$
where $\varphi$ is a real function defined on the Grassmannian manifold of $k$-dimensional $\mathbb R$-linear subspaces of $\mathbb R^n$ and $E_\Delta$ is the linear subspace parallel to the face $\Delta$.
For any $\varphi$, the functional $\text{v}^\varphi_k$ defines a $k$-homogeneous, translation invariant weak valuation on the space of convex polytopes of $\mathbb R^n$. If $\varphi$ is continuous and orthogonally invariant, $\text{v}^\varphi_k$ yields a continuous, $k$-homogeneous, rigid motion invariant valuation on convex bodies and so $\text{v}_k^\varphi$ becomes a multiple of the $\text{v}_k$.
As far as I know the functional $\text{v}_k^\varphi$ has been introduced by Kazarnovskii in the paper entitled On the zeros of exponential sums. (Soviet Math. Doklady 23.2 (1981), pp. 347–351), however it might have been defined before by someone else.
Can anybody provide further information about the valuation $\text{v}_k^\varphi$?
a remark is that these valuations are obtained by integrating universal differential forms on the normal bundle of the convex set. Namely, they are linear images of the normal cycle. The universal differential forms can be obtained by tweaking the ones giving usual intrinsic volumes by phi.
Thank you for tour comment. This weighted version of the intrinsic volume admits a corresponding mixed form generalizing mixed volume. I wonder if the mixed version verifies an Alexandroff-Fenchel inequality!
You may find relevant the following recent paper by Thomas Wannerer https://arxiv.org/pdf/1907.11606.pdf. In Thm 1.2 he characterizes those functions $\phi$ on the Grassmannian for which the valuation you wrote extends by continuity in Hausdorff metric to all convex compact sets.
Thanks a lot for pointing me to this paper.
|
2025-03-21T14:48:29.547312
| 2019-12-28T09:36:06 |
349251
|
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|
Stack Exchange
|
$X$ with $H^*(X)=$affine Verma module?
Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra over $\mathbf{C}$, and $\widehat{\mathfrak{g}}_\kappa$ the associated affine Lie algebra. It is the central extension of the loop algebra $\mathfrak{g}((t))$ using the invariant bilinear form $\kappa$ on $\mathfrak{g}$. Write $V_\kappa(\mathfrak{g})=\text{Ind}^{\widehat{\mathfrak{g}}_\kappa}_{\mathfrak{g}[[t]]\oplus \mathbf{C}c}\mathbf{C}$ for the Verma module of $\widehat{\mathfrak{g}}_\kappa$, where on $\mathbf{C}$, $\mathfrak{g}[[t]]$ acts by $0$ and the central element $c$ acts by $1$.
Is there a space (topological space, stack, etc.) $X$ whose cohomology is $H^*(X)=V_\kappa(\mathfrak{g})$?
For instance, if $Y$ is a smooth algebraic surface then $X=\coprod_{n\ge 0}\text{Hilb}^nY$ has cohomology of the form $V_\kappa(\mathfrak{h})$, where the vector space $\mathfrak{h}=H^*(Y)$ is viewed as a commutative Lie algebra with form $\kappa(\alpha,\beta)=\int_Y\alpha\wedge\beta$. I am curious if there is a construction for non-commutative Lie algebras.
If the natural thing is not to consider cohomology but some other functor from spaces to vector spaces, that is fine too (though ideally it would generalise the above example).
@Ebz The grading should be whatever it is in the Grojnowski-Nakajima construction, which I can't remember offhand, an element in $H^\star(X^{[n]})$ probably has grading $\star+f(n)$ for some function $f(n)$ (probably linear or quadratic).
@Ebz I think it's hard enough to come up with a natural example with $H^*(X)=V_\kappa(\mathfrak{g})$ as vector spaces, I am happy to worry about the extra structure later.
|
2025-03-21T14:48:29.547443
| 2019-12-28T11:23:42 |
349259
|
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|
Stack Exchange
|
Is the result of Schmidt conditional to RH
From this page:
https://en.wikipedia.org/wiki/Chebyshev_function#Asymptotics_and_bounds
A theorem due to Erhard Schmidt states that, for some explicit positive constant $K$, there are infinitely many natural numbers $x$ such that $$ψ(x)>x+K√x$$
Since the original paper is in German. I read this paper: https://projecteuclid.org/download/pdf_1/euclid.acta/1485887467
My question is: Is the result of Schmidt conditional to Riemann Hypothesis.
Even if Schmidt's proof uses RH, it is relatively easy to prove much stronger lower bounds assuming RH is false.
@Wojowu: I know that. I am asking on that result if it is related to RH or its negation.
1. It is known unconditionally that, as $x$ tends to infinity,
$$\psi(x)-x=\Omega_{\pm}(x^{1/2}).\tag{$1$}$$
This is Corollary 15.4 in Montgomery-Vaughan: Multiplicative number theory I.
2. In fact Hardy and Littlewood proved the stronger result
$$\psi(x)-x=\Omega_{\pm}(x^{1/2}\log\log\log x).\tag{$2$}$$
This is Theorem 15.11 in Montgomery-Vaughan: Multiplicative number theory I.
3. Schmidt (1903) proved the analogue of $(1)$ for the function
$$f(x):=\sum_{k=1}^\infty\frac{1}{k}\pi(x^\frac{1}{k}).$$
His proof is essentially the same as of the above quoted Corollary: if the Riemann Hypothesis is false, then one has a better result, while if the Riemann Hypothesis is true, then one has a precise form of the stated result with an implied constant given in terms of the lowest lying zero of $\zeta(s)$. So Schmidt's result is unconditional as well, but it differs slightly from the statement attributed to him in the Wikipedia article.
4. Hardy and Littlewood (1916) attribute $(1)$ to Schmidt, and they quote it as Theorem 2.241. Precisely, they say that "This is substantially the well-known result of Schmidt". The stronger statement $(2)$ is Theorem 5.8 in their paper.
P.S. As Greg Martin kindly pointed out, $(2)$ is really due to Littlewood (1914).
But still the result appeared in Wikipedia is true and unconditional as I read in the linked paper. Page 139, Theorem 2.241.
@Helena: I agree, and this is what item 1 in my post is about. I also added items 2 and 4 in my post.
Remark: the result (2) is actually due to Littlewood alone (even though its full proof first appeared in a paper of Hardy and Littlewood).
@GregMartin: Thank you! See the "P.S." section in my post.
It is elementary, take $K=1/30 < 1/|\rho_0|$ where $\rho_0\approx 1/2+i14.13$ is the first zero, if $\psi(x)-x\le Kx^{1/2}$ for $x$ large enough, then $x-\psi(x)+K x^{1/2}+C\ge 0$ for all $x$, where $C$ is a suitable real constant. Let $$F(s):=\int_1^\infty (x-\psi(x)+Kx^{1/2}+C) x^{-s-1}dx= \frac1{s-1}+\frac{\zeta'(s)}{s\zeta(s)}+\frac{K}{s-1/2}+\frac{C}{s}.$$
By the non-negativity of the integrand, it has a singularity at its abscissa of convergence $\sigma$. But the RHS is analytic on $(1/2,\infty)$, thus $\sigma=1/2$. And by the non-negativity again we have, as $\Re(s) \to 1/2$, $$|F(s)|\le F(\Re(s))\sim \frac{K}{\Re(s)-1/2}.$$
This contradicts that $$F(s)\sim\frac{\zeta'(s)}{s\zeta(s)}\sim\frac{1/\rho_0}{s-\rho_0}\qquad\text{as $s\to \rho_0$}.$$
Showing that $K$ arbitrary large works should be much more difficult
|
2025-03-21T14:48:29.547789
| 2019-12-28T14:26:38 |
349264
|
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|
Stack Exchange
|
What is the max number of the shortest path in a complete graph？
Give a complete graph $K_n$(the edge of the graph has positive weight), define the max number of the shortest paths between two vertices to be the number of paths that have the same length and the length is the shortest among any other paths.
I come across two questions.
define x to be the number of the shortest path between two vertices a and b. a and b is considered to be a vertex pair. How to set the edge weight to reach the max sum of the number of shortest path when considering K vertex pairs, notice that K vertex pairs have 2*K different vertices？
consider all vertex pairs which consist $C_n^2$ vertex pairs,notice that one vertex can belong to one or many vertex pairs. How to set the edge weight to reach the max sum of the number of shortest path？
I am not sure what you are asking: In a complete graph, all vertices are connected, by an edge, so there is always a unique shortest path of length 1.
The first line of the question reads "the edge of the graph has positive weight". This means that edges can have all sorts of length (any real number). It is standard to define the length of a path as the sum of the weight of the edges. The only thing which is not clear is whether positive means $>0$ or $\geq 0$.
The edge weight is always larger than 0,and the number of shortest path from a to b means between a and b there may be 4 path which are 4;1,1,1,1;2,3;1,4 then the shortest distance between a and b is obviously 4,and 4 and 1,1,1,1 these two paths both have length 4. These numbers reprrsent the edge weight of the path, now I can change the edge weight that I can change the path 2,3 into the path2,2,so the number of my shortest paths increase by 1. The problem is how to choose the edge weight of the graph.
|
2025-03-21T14:48:29.547942
| 2019-12-28T14:39:52 |
349265
|
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|
Stack Exchange
|
Iterated removal of singleton Pythagorean triples
Consider the set of all Pythagorean triples (positive integers $a, b, c$ such that $a^2 + b^2 = c^2$, not necessarily coprime). Then for each integer that appears in exactly one triple, remove that triple from the set and repeat this process an infinite number of times.
For example, 3 is a member of only one triple (3, 4, 5), so that triple is removed. Afterwards, 5 is a member of only one triple (5, 12, 13), so that triple can also be removed, and so on.
Call a triple stable if it is never removed by this process, and a positive integer stable if it is a member of at least one stable triple.
For example, 15 is stable because taking each elementwise product of the triples (3, 4, 5) and (5, 12, 13) results in a "magic square" where each row and column is a Pythagorean triple. Since each element in the square is a member of two different triples including only other elements of the square, every row, column, and element of such a square is stable.
$\begin{matrix}
15 & 36 & 39\\
20 & 48 & 52\\
25 & 60 & 65
\end{matrix}$
My questions are
Is every stable integer connected to one or more cycles of stable triples as described above? Or are there stable integers which are stable merely because they are the root of an infinitely ascending tree of Pythagorean triples that can't be removed in a countable number of steps?
and
Is there a simple algorithm to determine whether an integer is stable?
|
2025-03-21T14:48:29.548064
| 2019-12-28T15:08:57 |
349267
|
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|
Stack Exchange
|
Angle of analyticity of semigroup
Is there any known parabolic PDEs in the literature where the angle of analyticity of the associated semigroup is $<\pi/2$ ?
For example, the angle of heat semigroup in $L^2$ is exactly $=\pi/2$. I'm wondering if there is a known example where the angle is e.g., $\pi/4$ or other value $< \pi/2$.
Yes, such examples are known. For some references, see for instance Secion 7.2.6 in "W. Arendt: Semigroups and Evolution Equations: Functional Calculus, Regularity and Kernel Estimates (2004)".
@JochenGlueck Thank you. The example given is about the realization of the semigroup in $L^p$. Do you know any example in Hilbert space, say $L^2$ for example?
Good question... Are you interested in real coefficients only, or also in PDEs with complex coefficients?
Yes I'm interested in real coefficients, but if necessary no problem with complex ones.
With complex coefficients it's quite easy: just consider the PDE $\dot u = e^{i\frac{\pi}{4}} \frac{\partial^2}{\partial x^2} u$ on, say, $L^2(\mathbb{R})$ (though one might argue whether this is really a "parabolic" PDE). I don't know an example with real coefficients at the moment, but I would suspect that such examples exist.
Too long for a comment. Actually I do not think that it is written anywhere but these kind of counterexamples are usually provided by the Ornstein-Uhlenbeck semigroup, generated by $\Delta+Bx \cdot \nabla$, where $B$ is a matrix. Assuming that all eigenvalues of $B$ have negative real parts, then an invariant measure $\mu$ exists (and is given by a Gaussian density). It turns out that the angle of analiticity in $L^p$ of the invariant measure can be computed exactly and can be smaller than $\pi/2$, even for $p=2$. This can be found in a paper by Chill, Fasangova, Pallara and myself.
Chill, R.; Fašangová, E.; Metafune, G.; Pallara, D., The sector of analyticity of the Ornstein-Uhlenbeck semigroup on (L^p) spaces with respect to invariant measure, J. Lond. Math. Soc., II. Ser. 71, No. 3, 703-722 (2005). ZBL1123.35030.
To obtain similar examples in unweighted spaces one has only (but patiently) to compensate the weight thus obtaining an operator with a linear drift and a quadratic potential. This works however only from dimension $2$ on; in the one dimensional case the operator is always self-adjont but still the angle of analitycity is different from $\pi/2$ in $L^p$, for $p$ different from $2$. There is also a paper by E. Priola and myself dealing with non-analytic Markov semigroups where one finds other examples.
Metafune, G.; Priola, E., Some classes of non-analytic Markov semigroups, J. Math. Anal. Appl. 294, No. 2, 596-613 (2004). ZBL1067.47055.
It would be nice to have a direct approach, avoiding the detour.
There are examples but you need singularities or unbounded coefficients; the uniformly parabolic case with regular coefficients is indeed a perturbation of the laplacian. In 1D, if you perturb the harmonic oscillator $D^2-x^2$ by a linear drift $bxD$, the angle of analyticity depends on $b$.
|
2025-03-21T14:48:29.548280
| 2019-12-28T15:56:09 |
349272
|
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|
Stack Exchange
|
an identity between two elliptic integrals
I would like a direct change of variable proof of the identity
$$\int_0^{\arctan\frac{\sqrt{2}}{\sqrt{\sqrt{3}}}} \frac{1}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2\phi }}d\phi=\int_0^{\arctan\frac{1}{\sqrt{\sqrt{3}}}}\frac{1}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^22\phi }}d\phi\,.$$
I need it as part of a paper on Legendre's proof of the "third singular modulus."
your $d\phi$ are not where they should be.
@DimaPasechnik thank you, now corrected.
Instead of writing $\sqrt{\sqrt3}$, you might use either $3^{1/4}$ or $\sqrt[4]{3}$, LaTeX for the latter is \sqrt[4]{3}.
Isn't math.se a right forum for such questions?
Mathematica reduces it to $$F\left(2 \cot ^{-1}\left(\sqrt[4]{3}\right)|\frac{1}{4} \left(\sqrt{3}+2\right)\right)=2 F\left(\tan ^{-1}\left(\frac{\sqrt{2}}{\sqrt[4]{3}}\right)|\frac{1}{4} \left(\sqrt{3}+2\right)\right)$$ with http://reference.wolfram.com/language/ref/EllipticF.html
@user64494 Once again, not everything for which a CAS provides an answer (but no explanation!) deserves your "isn't MSE a right forum for such questions". Note that the OP desires a proof, presumably seeking understanding
@user64494 I agree with Yemon's response. Even worse, such comments serve to "shame" the OP for even asking here, when in fact the question is perfectly on-topic, for the reasons Yemon gave. Please reflect carefully on this.
@ToddTrimble I don't know either the motivation or intention of the person who commented above, but it's pretty clear that there is a lot of expertise on MSE for computing every definite integral under the sun, certainly a lot more than has ever been demonstrated on MO. There's actually a case to be made for some definite integral questions that MSE is a better place because there are more experts there. I would ask you to reflect on whether any suggestion that MSE might be better is automatically "shaming" rather than actually directing the OP to a more useful site.
@ElectricPenguin Thanks. Interesting point. I do question your assertion that there is more skill at Mathematics for solving such problems, but next time I see a case where this community finds a definite integral question on-topic and is defeated by it, I'll think about it. But based on a long history, it seems user64494 says the same thing just about every time he sees a definite integral problem for which Mathematica gives an answer (and, tbh, user's formulation sounds a bit rhetorical to me, as if asking "shouldn't you know better?", thus prompting my comment).
@ToddTrimble, no problem! I agree with you that the way it was phrased in this particular example was non-optimal and I certainly don't know about past histories.
Two edits in the bounty offer: $theta$ should be replaced by $\theta$ and $\arctan(1/\sqrt(3)$ should be replaced by $\arctan(1/\sqrt(\sqrt(3))$
I think $\theta$ should be equated to $2\arctan 3^{-1/4}$, see the answer box where I have tried to work this out in some detail.
@Carlo Beenakker Yes, you are right. I miscopied the identity...$\theta$ should be equated to $2\arctan 3^{-\frac{1}{4}}$. Thank you.
Since the bountied question has changed substantially, now asking for the application of an identity in Legendre's Traite des fonctions elliptiques, I am starting a new answer. Legendre defines
\begin{align}
&F(\phi,k)=\int_0^{\phi}\frac{d\phi'}{\sqrt{1-k^2\sin^2\phi'}},\\
&\sin\phi=\frac{\sin(\theta/2)}{\sqrt{\tfrac{1}{2}+\tfrac{1}{2}\Delta(\theta)}},\;\;\Delta(\theta)=\sqrt{1-k^2\sin^2\theta},
\end{align}
and then derives the identity
$$F(\phi,k)=\tfrac{1}{2}F(\theta,k).$$
Now we apply this to $k^2=\frac{2+\sqrt{3}}{4}$, $\theta=2 \arctan 3^{-1/4}$ and find
$$\sin\phi=\frac{2 \sin (\theta/2)}{\sqrt{\sqrt{4-\left(\sqrt{3}+2\right) \sin ^2\theta}+2}}=\sqrt{3}-1,$$
and thus $\phi=\arcsin(\sqrt{3}-1)=\arctan\left(3^{-1/4}\sqrt{2}\right)$. Hence, Legendre's identity gives
$$F\left(\arctan\left(3^{-1/4}\sqrt{2}\right),\frac{2+\sqrt{3}}{4}\right)=\frac{1}{2}F\left(2 \arctan 3^{-1/4},\frac{2+\sqrt{3}}{4}\right)$$
or equivalently
$$\int_0^{\arctan\left(3^{-1/4}\sqrt{2}\right)}\frac{d\phi'}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2 \phi'}}=\int_0^{\arctan 3^{-1/4}}\frac{d\phi'}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2 2\phi'}},$$
which is the identity in the OP.
Not yet an answer, but a bit too long for a comment. The Legendre normal form of these elliptic integrals might be a first step, at least by introducing simpler integration limits:
\begin{align}
&I_1=\int_0^{\arctan\frac{\sqrt{2}}{\sqrt{\sqrt{3}}}} \frac{d\phi}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2\phi }}=\int_0^{\sqrt{3}-1}\frac{dt}{\sqrt{(1-t^2)(1-\frac{2+\sqrt{3}}{4}t^2)}}, \\
&I_2=\int_0^{\arctan\frac{1}{\sqrt{\sqrt{3}}}}\frac{d\phi}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^22\phi }}=\frac{1}{2}\int_0^{3^{1/4}(\sqrt{3}-1)}\frac{dt}{\sqrt{(1-t^2)(1-\frac{2+\sqrt{3}}{4}t^2)}}.
\end{align}
|
2025-03-21T14:48:29.548602
| 2019-12-28T17:55:17 |
349274
|
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|
Stack Exchange
|
Limit of alternated row and column normalizations
Let $E_0$ be a matrix with non-negative entries.
Given $E_n$, we apply the following two operations in sequence to produce $E_{n+1}$.
A. Divide every entry by the sum of all entries in its column (to make the matrix column-stochastic).
B. Divide every entry by the sum of all entries in its row (to make the matrix row-stochastic).
For example:
$E_0=\begin{pmatrix}
\frac{2}{5} & \frac{1}{5} & \frac{2}{5} & 0 & 0\\
\frac{1}{5} & 0 & \frac{7}{10} & \frac{1}{10} & 0\\
0 & 0 & 0 & \frac{3}{10} & \frac{7}{10}
\end{pmatrix}\overset{A}{\rightarrow}\begin{pmatrix}
\frac{2}{3} & 1 & \frac{4}{11} & 0 & 0\\
\frac{1}{3} & 0 & \frac{7}{11} & \frac{1}{4} & 0\\
0 & 0 & 0 & \frac{3}{4} & 1
\end{pmatrix}\overset{B}{\rightarrow}\begin{pmatrix}
\frac{22}{67} & \frac{33}{67} & \frac{12}{67} & 0 & 0\\
\frac{44}{161} & 0 & \frac{12}{23} & \frac{33}{161} & 0\\
0 & 0 & 0 & \frac{3}{7} & \frac{4}{7}
\end{pmatrix}=E_1$
What is the limit of $E_n$ as $n \to \infty$?
Additional remarks.
In my problem, the matrix has $c\in \{1,2,\dots,5\}$ rows and $r=5$ columns (note that the two letters are reversed, but in the original context of this problem these letters $r$ and $c$ do not actually stand for rows and columns). So $E_0$ can be $1\times 5$, $2\times 5$, ... or $5\times 5$.
We denote with $(e_n)_{ij}$ the entries of $E_{n}$; hence $(e_n)_{ij}\in[0;1]$ and $\forall i \sum_{j=1}^{r}(e_n)_{ij}=1$ for $n>0$.
I managed to express $(e_{n+1})_{ij}$ as a function of $(e_{n})_{ij}$ :
$$(e_{n+1})_{ij}=\frac{\frac{(e_{n})_{ij}}{\sum_{k=1}^{c}(e_n)_{kj}}}{\sum_{l=1}^{r}\frac{(e_n)_{il}}{\sum_{k=1}^{c}(e_n)_{kl}}}$$
What I can't seem to find now is an expression $(e_{n})_{ij}$ as a function of $(e_{0})_{ij}$, to be able to calculate $\underset{n \to +\infty }{lim}(e_n)_{ij}$
I wrote code to compute this iteration; when I ran it with the previous example $E_0$, I found out that:
$E_0=\begin{pmatrix}
\frac{2}{5} & \frac{1}{5} & \frac{2}{5} & 0 & 0\\
\frac{1}{5} & 0 & \frac{7}{10} & \frac{1}{10} & 0\\
0 & 0 & 0 & \frac{3}{10} & \frac{7}{10}
\end{pmatrix}\overset{n \rightarrow+\infty}{\rightarrow}E_n=\begin{pmatrix}
\frac{7}{25} & \frac{3}{5} & \frac{3}{25} & 0 & 0\\
\frac{8}{25} & 0 & \frac{12}{25} & \frac{1}{5} & 0\\
0 & 0 & 0 & \frac{2}{5} & \frac{3}{5}
\end{pmatrix}$
Not only do the row sums equal $1$, but the column sums equal $\frac{3}{5}$: it seems that in this process column sums converge to $\frac{c}{r}$.
I'm not a mathematician so I was looking for a simple inductive proof. I tried to express $E_2$ (and so on) as a function of $E_0$, but it quickly gets overwhelming, starting from $E_2$...
Does 'transition' just mean "permutation of a set of matrices"? How are your transitions defined when the relevant sums are 0?
A column is never equal to 0. I even barely never have 0 as coefficient, but I added some in my example to make it easier to understand and manipulate.
@LSpice: It is reasonable to assume the OP takes the numbers to be $\ge 0$. If a row or column is identically $0$ then the corresponding transition is defined to leave it as it is. No, "transition" means either the operation $t$ or $t'$ applied to some given matrix.
I actually does $t$ $t'$ a infinite number of times, and the output finally never changes (even thought I never really get to it).
As with many problems of this type, finding the "limiting outcome" is much better done by finding 1) invariants of the transition, and 2) commonalities of fixed points. You have done the latter; there are enough of the former and latter to solve this in general.
When $E_0$ is square (i.e., $r = c$) this procedure is called Sinkhorn iteration or the Sinkhorn-Knopp algorithm (see this Wikipedia page). You can find a wealth of results by Googling those terms, the most well-known of which is that if $E_0$ has strictly positive entries (and again, is square) then the limit of $E_n$ indeed exists and is doubly stochastic.
The paragraph below applies to a different problem, where row normalization is split out from column normalization, so I have an $"E_{n+1/2}"$ which will be sometimes different from both $E_n$ and $E_{n+1}$. (If the starting matrix "looks like" an order k square stochastic matrix, then it will be invariant under both normalizations.)
Note that some cases will not converge to a single limit. Given a c by r matrix of all ones, normalizing by column (of r entries) results in all entries being 1/r, while normalizing by rows gives all entries being 1/c, so the sequence will fluctuate between these two. Except when the entries are all zero, I would expect a similar oscillation with any other starting nonzero binary matrix. You might be able to establish oscillation for matrices with more distinct values.
Getting back to the posted problem, the transformations have an invariance under permutations of rows, similarly of columns. Thus if the input looks like an upper two by two diagonal block matrix and a lower two by three nonzero matrix, the upper block may converge on a stochastic two by two block, while the lower will be influenced (if it converges at all) by a ratio of 3/2. So the block structure will influence the results, and the ratio c/r might not apply.
Gerhard "Goes This Way And That" Paseman, 2019.12.28.
It turns out I am wrong; some binary matrices (and non binary too) stabilize: those with a nonzero upper k by k minor which are order k stochastic matrices, and zero elsewhere, are invariant under the transform. The oscillation occurs when the nonzero part is nonsquare. Gerhard "Is Shaping Up The Argument" Paseman, 2019.12.28.
Possibly the question changed between your post and now, but it looks like the original poster was considering "normalize by columns then normalize by rows" to be a single operation, so your example converges to the matrix with all entries = 1/r.
@Bill, indeed, I will edit to clarify my perception. Gerhard "Should Read Post More Carefully" Paseman, 2019.12.28.
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2025-03-21T14:48:29.548988
| 2019-12-28T18:06:09 |
349275
|
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Stack Exchange
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Important combinatorial and algebraic interpretations of the coefficients in the polynomial $[n]!_q = (1+q)(1+q+q^2) \ldots (1+q+\cdots + q^{n-1})$
What are some important combinatorial and algebraic interpretations of the coefficients in the polynomial
$$[n]!_q = (1+q)(1+q+q^2) \ldots (1+q+\cdots + q^{n-1})?$$
As motivation, I will give three interpretations, ask for a fourth, and raise a related question about the unimodality. I would be particularly interested in answers using the RSK-correspondence or subspaces of $\mathbb{F}_q^n$.
Given a permutation $\sigma \in \mathrm{Sym}_n$, let $\mathrm{inv}(\sigma)$ denote the number of inversions of $\sigma$; that is, pairs $(x,y)$ with $x < y$ and $\sigma(x) > \sigma(y)$. Then $[n]!_q = \sum_{\sigma \in \mathrm{Sym}_n} q^{\mathrm{inv}(\sigma)}$.
An element $x \in \{1,\ldots, n-1\}$ is a descent of $\sigma \in \mathrm{Sym}_n$ if $\sigma(x) > \sigma(x+1)$. The major index $\mathrm{maj}(\sigma)$ is the sum of the descents of $\sigma$. Then $[n]!_q = \sum_{\sigma \in \mathrm{Sym}_n} q^{\mathrm{maj}(\sigma)}$. I think this is due to MacMahon.
In the 'inside-out' version of the Fisher–Yates shuffle on an $n$-card deck, at step $j-1$, card $j-1$ from the top is swapped with one of cards in positions $0, 1, \ldots, j-1$ from the top, chosen uniformly at random. These choices are enumerated by $1 + q + \cdots + q^{j-1}$. After $n$ steps (starting with $j=1$), each permutation has equal probability. (This is essentially coset enumeration in the symmetric group by the chain $\mathrm{Sym}_1 \le \mathrm{Sym}_2 \le \ldots \le \mathrm{Sym}_n$.) Hence $[n]!_q$ enumerates permutations according to the sum of the positions chosen at each stage.
Does the normal Fisher–Yates shuffle have a similar combinatorial interpretation? Is there a more natural interpretation of the $q$-power, still using the inside-out Fisher–Yates shuffle?
Finally, (1) makes it easy to see that $[n]!_q$ is symmetric, i.e. the coefficients of $q^m$ and $q^{\binom{n}{2}-m}$ are the same: use the Coxeter involution, thinking of $[n]_q!$ as the Poincaré series of the Coxeter group $\mathrm{Sym}_n$. This can also be seen in a similar way from (2). But it does not seem to be obvious from (3).
Which interpretation is the best way to show that the coefficients in $[n]!_q$ are unimodal, i.e. first increasing then decreasing?
Statistics on the symmetric group with the same distribution as inversion number (and major index, etc.) are called “Mahonian,” which might be a helpful keyword for you.
Shouldn't the statement that "$[n]_q!$ is symmetric" (I guess you mean $[n]!_q$ there, for consistency with the rest of the question) be that the coefficient of $q^m$ is the same as that of $q^{\binom n 2 - m}$, not of $q^{n - m}$?
@LSpice: yes, thank you for the correction.
This answer concerns a geometric/Lie-theoretic interpretation of $[n]!_q$.
$[n]!_q$ gives the number of points in the full flag variety of full flags of subspaces in an $n$-dimensional vector space $\mathbb{F}_q^n$ over the finite field $\mathbb{F}_q$.
Recall that the full flag variety (over any field) has a natural stratification, the Bruhat stratification. Due to the above point-counting remark, it follows that the coefficient of $q^i$ in $[n]!_q$ is the number of $i$-dimensional cells in the Bruhat stratification.
There is also a way to deduce the unimodality of the coefficients from this geometric perspective. Namely, the partial order on the Bruhat cells whereby $C \leq C'$ if $C$ is contained in the closure of $C'$ is called the Bruhat order or strong order. The strong order can be viewed as an order on the symmetric group because the Bruhat cells are naturally labeled by permutations. Strong order is graded, and the rank sizes are precisely the coefficients of $[n]!_q$ (i.e., the number of permutations with given inversion number). Richard Stanley showed in the "Weyl groups..." paper cited below that in this situation (when you have a complex projective variety with a cellular decomposition satisfying certain conditions), the poset in question is necessarily graded, rank-symmetric, rank-unimodal, and strongly Sperner, which in particular implies the unimodality of the coefficients of $[n]!_q$. His proof employed the hard Lefschetz theorem and so can hardly be called elementary, but it is conceptual.
Stanley, Richard P., Weyl groups, the hard Lefschetz theorem, and the Sperner property, SIAM J. Algebraic Discrete Methods 1, 168-184 (1980). ZBL0502.05004.
I have trouble parsing this sentence. Does it mean the same thing as "It gives the number of full flags of subspaces …"? If so, then this seems to be explaining the significance of the number $[n]!_q$ itself, not of the coefficients when it is viewed as a polynomial in $q$.
Yes, this is an interpretation of the number $[n]!_q$, not the coefficients of the polynomial per se. To get an interpretation of the coefficients you can use the Bruhat stratification. I will edit to include this.
Since a Bruhat cell of dimension $i$ corresponds to a symmetric-group element of length $i$, and since the length of a symmetric-group element is its number of inversions, then this is the same as interpretation (1) in the post, right?
@LSpice: that’s right but this is a more geometric way of looking at it, and still fundamental imo.
I also added something about unimodality from the Bruhat stratification perspective.
I think that the symmetry can be seen conceptually by the fact that conjugation by the long element takes an element of length $\ell$ to one of length $\frac1 2\lvert\mathsf A_{n - 1}\rvert - \ell$ (in the language of inversions, that it swaps inversions and non-inversions).
I would say “Poincaré duality” (in the context of the cited paper of Stanley’s) implies symmetry.
(In the geometric setting, I guess that this is the observation that (given a Borus $(B^+, S)$), $i$-dimensional cells in the Bruhat decomposition wrt $B^+$ are naturally in bijection with ($\frac1 2\lvert\mathsf A_{n - 1}\rvert - i$)-dimensional cells in the Bruhat decomposition wrt the $S$-opposite Borel $B^-$.)
If you're just interested in unimodality, and not any of the other properties, it's a very quick proof from Hard Lefschetz, right?
@WillSawin: I don't think there's a way to do it that's much faster than what Stanley does, but what Stanley does is pretty quick (takes just a few pages).
In FindStat, the database of combinatorial statistics, the major index of a permutation is http://www.findstat.org/StatisticsDatabase/St000004/. By clicking "search for distribution" there, you will find other statistics that (FindStat thinks) are equidistributed with major index. I get 41 results. As mentioned in a comment above, these statistics are called "Mahonian" in honor of MacMahon.
This answer concerns an interpretation of $[n]!_q$ coming from invariant theory.
There is a natural action of $S_n$ on $V=\mathbb{C}^n$, and hence on polynomial functions on $V$. This action preserves degree. The coinvariant algebra is the quotient of the polynomial functions on $V$ by (the ideal generated by) the $S_n$-invariant functions of positive degree. This (finite-dimensional!) algebra inherits an $S_n$ representation. As an ungraded $S_n$ module it is isomorphic to the regular representation (this was first proved by Chevalley). But it also inherits a grading, giving a natural grading to the regular representation of the symmetric group. Various fundamental results in invariant theory (the Chevalley-Shephard-Todd theorem, Molien's theorem, etc.) imply that the Hilbert series of the coinvariant algebra is $[n]!_q$. Hence the coefficient of $q^i$ in $[n]!_q$ is the dimension of the coinvariant algebra in degree $i$.
There is a connection here to RSK, and since the question asked about RSK, let me explain it.
Let me use $R_n$ to denote the coinvariant algebra. Since as an ungraded algebra, $R_n$ is the regular representation of $S_n$, we have $R_n \simeq \bigoplus_{\lambda \vdash n} V_{\lambda}^{f^{\lambda}}$, where $V_{\lambda}$ is the $S_n$-irrep indexed by the partition $\lambda$, and $f^{\lambda} := \mathrm{dim}(V_{\lambda})$. Recall that also $f^{\lambda} = \#$ Standard Young Tableaux (SYTs) of shape $\lambda$, and that we have the famous hook-length formula $f^{\lambda} = \frac{n!}{\prod_{u\in \lambda}h(u)}$, where $h(u)$ is the hook-length of the box $u\in \lambda$.
But we again get more by considering the grading on $R_n$. Namely, for a partition $\lambda\vdash n$, define the fake degree polynomial $f^{\lambda}(q)$ by
$$f^{\lambda}(q) := \sum_{i \geq 0} \# \textrm{ (copies of $V_{\lambda}$ in the degree $i$ component of $R_n$)} \cdot q^i.$$
Note that $f^{\lambda}(q)$ is a $q$-analog of $f^{\lambda}$. Stanley (see Section 4 of the paper cited below and Enumerative Combinatorics, Vol. 2, 7.21.5) proved a $q$-analog of the hook-length formula: $f^{\lambda}(q) = q^{b(\lambda)} \frac{[n]!_q}{\prod_{u\in \lambda}[h(u)]}$, where $b(\lambda) := 0\lambda_1 + 1\lambda_2 + 2\lambda_3 + \cdots$. Furthermore, Lusztig gave a statistical formula for $f^{\lambda}(q)$. Namely, for an SYT $T$, a descent of $T$ is an entry $i$ for which $i+1$ is in a strictly lower row, and the major index of $T$, denoted $\mathrm{maj}(T)$, is a sum over all descents $i$ of $T$ of the value $i$. Then in unpublished work (again see Stanley's paper cited below), Lusztig showed that $f^{\lambda}(q) = \sum_{\substack{T \textrm{ a SYT}, \\ \mathrm{sh}(T) =\lambda}}q^{\mathrm{maj}(T)}$.
Now note that the decomposition $R_n \simeq \bigoplus_{\lambda \vdash n} V_{\lambda}^{f^{\lambda}}$ implies $[n]!_q = \sum_{\lambda\vdash n}f^{\lambda} \cdot f^{\lambda}(q)$, and hence due to Lusztig's result that the coefficient of $q^i$ in $[n]!_q$ is the sum over all SYT $T$ with $n$ boxes and with $\mathrm{maj}(T)=i$ of $f^{\mathrm{sh}(T)}$.
Here's where the connection to RSK comes in. Recall that the usual Robinson-Schensted (no Knuth!) algorithm is a bijection between permutations $w \in S_n$ and pairs $(P,Q)$ of n-boxed SYTs of the same shape. Usually $P$ is called the insertion tableau and $Q$ the recording tableau. It is not hard to see from the usual "insertion and bumping" description of R-S that the descents of $w$ are exactly the same as the descents of its recording tableau $Q$. In particular, their major indices are the same. Thus R-S proves bijectively that
$$ \sum_{w \in S_n} q^{\mathrm{maj}(w)} = \sum_{\substack{T \textrm{ a SYT} \\ \textrm{with $n$ boxes}}} f^{\mathrm{sh}(T)} \cdot q^{\mathrm{maj}(T)},$$
where the RHS is the interpretation of $[n]!_q$ we just mentioned, and the LHS is the well-known one due to MacMahon mentioned in the original question.
Stanley, Richard P., Invariants of finite groups and their applications to combinatorics, Bull. Am. Math. Soc., New Ser. 1, 475-511 (1979). ZBL0497.20002.
Please could you add some details so that this becomes an explicit interpretation of the coefficients of the polynomial?
@MarkWildon: edited.
@MarkWildon: I added some info about how this coinvariant algebra stuff is somewhat connected to RSK.
Very nice: I like the bijective proof of the identity using the major index of the recording tableaux. I've already upvoted you, but your answer deserves more.
For a simple but noncombinatorial proof of the unimodality of the coefficients of $[n]!_q$, see
George E. Andrews,
A theorem on reciprocal polynomials with applications to permutations and compositions. Amer. Math. Monthly 82 (1975), no. 8, 830–833.
For an even simpler proof (in my opinion, at least), see Enumerative Combinatorics, vol. 1, second ed., Exercise 1.50(c).
There is a nice generalization to Catalan objects as follows.
Acyclic orientations of unit-interval graphs
Note that for the complete graph on $n$ vertices,
the number of acyclic orientations weighted by ascents (edges oriented from smaller to larger label) is given by $[n]_q!$.
Now, there are a Catalan number of unit interval graphs, which can be indexed by area sequences. The complete graph has area sequence $(0,1,2,\dotsc,n-1)$.
The weighted sum over acyclic orientations of an unit interval graph with area sequence $(a_1,\dotsc,a_n)$ is given by the product $[a_1+1]_q\dotsb [a_n+1]_q$.
Summing all these polynomials over all $(n+1)^{-1}\binom{2n}{n}$ area sequences of length $n$ gives the $n$th Touchard-Riordan polynomial, which is defined as the sum over all perfect matchings of $2n$ vertices on a circle, weighted by $q$ to the number of crossings.
Perfect matchings
In particular, $[n]_q!$ is the sum over all perfect matchings on $2n$ vertices, where vertices $1,2,\dotsc,n$ are matched with $n+1,n+2,\dotsc,2n$.
Rook placements on Ferrers board
Let $a$ be an area sequence.
A Ferrers board with row lengths given by
$a_i+n-i$ for $i=1,\dotsc,n$ allows for $[a_1+1]_q\dotsb [a_n+1]_q$ different ways to place $n$ non-attacking rooks,
and the weight is now given by rook inversions.
I have some more info and references here.
Thank you for a very nice answer. It deserves more than my up-vote.
After posting the question, I remembered an interpretation using formal characters of representations of $\mathrm{SL}_2(\mathbb{C})$ that gives the shortest proof I know that $[n]!_q$ is unimodal.
Write $1+q+\cdots + q^{j-1}$ as $q^{(j-1)/2}(q^{-(j-1)/2} + q^{-(j-3)/2} + \cdots + q^{(j-1)/2})$ and substitute $q= Q^2$ to get $Q^{j-1}(Q^{-(j-1)} + Q^{-(j-3)} + \cdots + Q^{j-1})$. Hence
$$[n]!_q = Q^{\binom{n}{2}}( Q^{-1} + Q) ( Q^{-2} + 1 + Q^2) \ldots (Q^{-(n-1)} + Q^{-(n-3)} + \cdots + Q^{n-1}).$$
Up to the power of $Q$, the right-hand side is the formal character of the tensor product of the (unique) irreducible representations of $\mathrm{SL}_2(\mathbb{C})$ of dimensions $2, 3, \ldots, n$. Since any formal character is a sum of the characters of irreducible representations, and each irreducible representation has a unimodal character, this proves that $[n!]_q$ is unimodal in $Q$, and hence in $q$.
Each irreducible character is unimodal, but the sum may be non-unimodal. For example $V=V(2)+V(3)+V(3)$. Here dim$V_1=2$, dim$V_0=1=$ dim $V_2$. The even and odd parts of sum of unimodal characters would be unimodal.
The polynomials have nonnegative, symmetric, unimodal coefficients. So the the product of polynomials also has this property.
This is the same as the answer of Ira Gessel.
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2025-03-21T14:48:29.549966
| 2019-12-28T19:30:33 |
349283
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|
Stack Exchange
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Software and ideas for workshops and conferences with long-distance participants
Conferences and workshops are often great - getting together and being together is an ideal setting for doing research and learning things. However, there are various reasons to encourage the possibility of having conferences and workshop through "cyber space". The cost of travel, in terms of time, money, $CO_2$, is large.
Q1: My first question is about software that makes it possible to run lectures with long-distance participation. What are the features and what are additional features we would like to see.
Q2 My second question is about what could be good ideas for running a long-distance workshops or conferences. (This question is mainly experience-based rather than opinion-based.) Hybrid setting with conventional conference that allows some participation from cyber space? Completely cyber conference where each participant stays in his office/study.?
I have some experience with the TCS+ seminars which are completely in cyberspace and participants from the audience can ask questions. They work rather well. (But only a few participants per talk.) I am not familiar of similar things in mathematics.
Very related question: Tools for long-distance collaboration (This is about tools for long distance collaboration, in general. Not specific to seminars/workshops/conferences and also asked nine years ago.)
I have much more in opinions than in experience to offer. It seems that you want to "preserve the conference experience" in a software medium. If this is so, you should details those parts you want preserved. If instead you are reimagining what a conference could look like, and are asking for such imaginings, then rewrite the question to emphasize that aspect. Either way, I suspect you intend a different question from what is written. Gerhard "Isn't Sure What's Really Wanted" Paseman, 2019.12.28.
In pure mathematics there's the ECHT which is fully electronic (although some talks have also a local audience)
I have no ideas/experience, thus I cann't provide an answer, but I think that your project is very good and important. I've evoked in the past that similar things, workshops, conferences or a journal, could be worked in sites as this MathOverflow, as natural extensions.
In addition to the homotopy type theory seminar already mentioned, there is also the electronic Computational Homotopy Theory Seminar:
https://s.wayne.edu/isaksen/echt/
Every week, the audience consists of a physically present group, and others who use Zoom to connect. It's pretty seamless to click the link and join (no need for the organizer to manage who has access, etc). The organizer has the ability to mute all microphones, and then audience members can unmute themselves to ask a question. The organizer can presumably also kick people out (e.g., if someone got the link and was trying to be disruptive). Maybe you could reach out to Dan Isaksen for advice, as this seminar is very well organized and successful.
A group of homotopy theorists also published a paper about "less climate-impactful conferencing." This was after they ran a conference that had two hubs instead of one. You could imagine generalizing their model to regional hubs. Hope this helps!
Re: "LESS CLIMATE-IMPACTFUL CONFERENCING". Our goal is to illustrate the feasibility of multi-hub conferences and to encourage others to follow our example by simply organising a bunch of them and showing that they work well: the first one (summer 2018) was this one: https://www.mpim-bonn.mpg.de/HAMP the second one (summer 2020) risks unfortunately being cancelled because of the current pandemic https://www.mpim-bonn.mpg.de/grt2020 the third one (summer 2021) is in advanced stages of planning, but no website exists yet. The fourth one (summer 2022) already has a core group of three organisers.
The HoTTEST seminar comes to mind, intended for 60 min talks and 30 min discussion. You can check the link:
https://www.uwo.ca/math/faculty/kapulkin/seminars/hottest.html
for details on the software they use and further information. They also mention a couple of other seminars done electronically as well.
The Northeastern Combinatorics Network has been running a Virtual Combinatorics Colloquium since March 2018 using Zoom and encouraging local viewing parties. Not exactly a conference, but perhaps a relevant existing program.
https://sites.google.com/view/northeastcombinatoricsnetwork/virtual-combinatorics-colloquium
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2025-03-21T14:48:29.550301
| 2019-12-29T04:17:58 |
349300
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Stack Exchange
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Estimating a "transition" matrix from data
I have observation vectors $X_1$, $X_2$, $X_3$, and so on.
I assume that there is a matrix A such that $AX_n = X_{n+1}$.
How do I calculate A?
I'm sure this is a standard problem but I don't know what to search for in Google...
Perhaps vector autoregression (VAR) models are what you are looking for.
That answers my question. Thank you!
Vector Autoregression (VAR) is possibly what you are looking for. Usually there is a zero-mean error term so that your expression holds in expectation.
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2025-03-21T14:48:29.550369
| 2019-12-29T06:08:54 |
349301
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Stack Exchange
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Is there a Poincare residue in characteristic $p$?
The Poincare residue I mean is there one here:
https://en.wikipedia.org/wiki/Poincar%C3%A9_residue
Basically, I would like a nice way to use a meromorphic $n$-form on $\mathbf{P}^n_{\mathbf{F}_p}$ to get an $(n-1)$-form on the hypersurface given by the pole.
I suspect that one exists, but phrased in some fancy language. I'm not sure what to look for. A reference (and some decoding of it) would be very welcome.
I've been thinking about this, and I want to record a few thoughts. Let $k$ be a field of characteristic $p$, let $X$ be a smooth $n$-dimensional variety, let $D$ be a Cartier divisor and let $U = X \setminus D$.
We cannot hope to have a natural map from $H_{DR}^n(U)$ to $H^{n-1}_{DR}(D)$ which looks anything like the residue map (also known as the Gysin map). Take $p$ odd. Take $X$ to be the affine plane with coordinates $(x,y)$, and let $D$ be $\{ y=0 \}$.
Let $\alpha$ be the $2$-form $x^{2p-1} y^{-p-1} dx \wedge dy$ and consider the automorphism $\phi(x,y) = (x+y,y)$ of $X$. This preserves the divisor $D$ and acts trivially on $D$, so $\alpha$ and $\phi^{\ast}(\alpha)$ should have the same residue. In other words, $\phi^{\ast} \alpha - \alpha$ should have residue $0$. Now, $\phi^{\ast} \alpha - \alpha = \sum_{j=0}^{2p-2} \binom{2p-1}{j} x^j y^{p-2-j} dx \wedge dy$. If we compute residues naively, the residue of $\sum_{j=0}^{2p-2} \binom{2p-1}{j} x^j y^{p-2-j} dx \wedge dy$ should be $\binom{2p-1}{p-1} x^{p-1} dx$. Also, $\binom{2p-1}{p-1} \equiv 2 \neq 0 \bmod p$ by Lucas' theorem. But $x^{p-1} dx$ is not exact in characteristic $p$. So working naively can't give us a residue which is well defined in $H^{\ast}_{dR}$. Moreover, it doesn't make sense to fix this by defining $\binom{2p-j}{p-j} x^j y^{p-2-j} dx dy$ to have a nonzero residue for other values of $j$, because $x^j y^{p-2-j} dx dy$ is exact for all $0 \leq j \leq 2p-2$ except $p-1$.
There is a very deep thing to do. We can lift $X$, $D$ and $U$ up to flat schemes over some dvr of mixed characteristic (for example, if $k = \mathbb{F}_p$, we could take $p$-adic lifts) and take the de Rham cohomology of these lifts. There is a ton of very hard literature on this sort of idea, starting with the research of Monsky and Washnitzer. Indeed, there is a Gysin sequence in Monsky-Washnitzer cohomology: See
Monsky, P., Formal cohomology. II: The cohomology sequence of a pair, Ann. Math. (2) 88, 218-238 (1968). ZBL0162.52601.
I don't feel confident to summarize this paper.
I went looking for something more elementary to do and came up with an interesting idea: Although $x^{p-1} dx$ isn't exact, it is in a sense "almost exact". The exact forms are the kernel of the Cartier operator, and $x^{p-1} dx$ is in the kernel of the square of the Cartier operator. Define $EH^n$ to be $n$-forms modulo forms which are killed by some power of the Cartier operator. (This is a definition for top dimensional forms only; see my recent question for what I think the more general definition should be.) I think I can build a Gysin map $EH^n(X) \to EH^{n-1}(D)$. But I'm going to wait a bit to see if someone answers my other question before I write more.
Okay, let me spell out this idea in a bit more detail.
First of all, let's recall how residue works when $\omega$ only has a simple pole along $D$. First, choose an open set $X'$ on which $D$ is principal, with generator $t$, and on which there is a vector field $\vec{v}$ with $\langle \vec{v}, dt \rangle = 1$. Set $U' = X' \cap U$ and $D' = X \cap D$. If $\omega$ has only a simple pole on $D'$, then $t \omega$ extends to $X'$. Contracting $t \omega$ against $\vec{v}$ gives an $(n-1)$-form, which we can then restrict to $D'$. The final result is independent of the choices of $t$ and $\vec{v}$, and is the residue of $\omega$ to $D'$. We can cover $X$ by open sets $X'$ as above and compute the residue on each such set, and since the result is independent of our choices, we get a well defined residue on $D$. Nothing here uses characteristic $0$ (and we even get a specific differential form for our residue, not a cohomology class.)
Now, suppose that $\omega$ has a pole of order $N$, and let $\mathcal{C}$ be the Cartier operator. Then $\mathcal{C}(\omega)$ has a pole of order at most $1+(N-1)/p$. Applying the Cartier operator $k$ times for $k$ large enough that $p^k \geq N$, we get a differential form with a pole of order $\leq 1$. We can take the residue $\mathrm{Res}(\mathcal{C}^k \omega)$ of that form. But then we should apply the "inverse Cartier operator" $k$-times to this residue. The Cartier operator from top dimensional forms to top dimensional forms is surjective, but has a kernel, so what this really means is to find some $n-1$ form $\alpha$ on $D$ with $\mathcal{C}^k(\alpha) = \mathrm{Res}(\mathcal{C}^k \omega)$. So $\alpha$ is only defined modulo the kernel of $\mathcal{C}^k$. In other words, this residue is a class in $EH^{n-1}(D)$ in the sense I describe above. This is a map $\Omega^n(U) \to EH^{n-1}(D)$.
It is also not hard to show that this map passes down to a map $EH^n(U) \to EH^{n-1}(D)$.
I don't know if this is helpful, but I think it is the best you can do.
|
2025-03-21T14:48:29.550735
| 2019-12-29T08:17:44 |
349304
|
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"Gerry Myerson",
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|
Stack Exchange
|
Does there exist such a sequence $B$ when $p>5$?
Let $A = (a_1, a_2, \ldots, a_n)$ be the sequence of odd primes are less than or equal to a prime number $p$.
Let $C$ be the infinite ascending sequence of composite numbers that their factors are all in $A$.
Let $B$ be a sequence of $n$ consecutive numbers of $C$ such that $b_n - b_{n-1} = a_2 - a_1, \ b_{n-1} - b_{n-2} = a_3 - a_2, \ldots, \ b_2 - b_1 = a_n - a_{n-1}$.
For example, when $p=5,\ A=3,5, \ C= 9,15,25,27,45,75,81,..., \ B=25,27$.
Does there exist such a sequence $B$ when $p>5$?
Alternative question:
If there exist such a sequence $B$ when $p>5$, then $b_1$ must be greater than $3p$?
ref: see the informative comments of the same question on MSE.
Have you made any progress for any $p$ other than $p=5$? Is there any $p$ for which you can prove impossibility? any $p$ for which you have found such $B$?
@Gerry Myerson I guess there's no such $B$ when $p>5$. See the informative comments of the same question on MSE,that's all: https://math.stackexchange.com/questions/3486259/does-there-exist-such-a-sequence-b-when-p5
When you crosspost, please mention so and include a link when you make the post (not just in response to later questioning).
@LSpice Thanks, I see that and updated the content.
|
2025-03-21T14:48:29.551186
| 2019-12-29T09:17:30 |
349305
|
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|
Stack Exchange
|
Integral cohomology of compact Lie groups and their classifying spaces
Let $G$ be a compact Lie group and let $BG$ be its classifying space. Let $\gamma\colon \Sigma G \to BG$ be the adjoint map for a natural weak equivalence $G \xrightarrow{\sim} \Omega BG$. Taking rational cohomology, it is well-known, that the map $\gamma$ induces an isomorphism between indecomposable elements $QH^*(BG,\mathbb{Q})$ and primitive elements $\mathrm{Prim}\, H^{*-1}(G,\mathbb{Q})$. But I want to know what happens integrally.
To be more precise, the integral cohomology ring modulo torsion $H^*(G,\mathbb{Z})/\mathrm{Tor}$ is still a Hopf algebra isomorphic to an exterior algebra (see Mimura, Toda - Topology of Lie groups, I and II, Chapter VII, Theorem 1.22). Is the homomorphism induced by $\gamma$ still surjective onto primitive elements? If not, what one can say about orders of cokernels?
|
2025-03-21T14:48:29.551287
| 2019-12-29T09:36:48 |
349306
|
{
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"authors": [
"Alexandre Eremenko",
"Dave L Renfro",
"M. Winter",
"hirak chatterjee",
"https://mathoverflow.net/users/108884",
"https://mathoverflow.net/users/130874",
"https://mathoverflow.net/users/15780",
"https://mathoverflow.net/users/25510"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/349306"
}
|
Stack Exchange
|
Surface Area of a Superellipsoid
We know the surface area of a supersphere (x^n+y^n+z^n=1) can be easily obtained using the gamma function (here is the discussion). But what happens when we consider a superellipsoid ($x^m +y^n+z^p=1; m \neq n\neq p $), or more generally for the implicit function $(x/a)^m +(y/b)^n+(z/c)^p=1; a \neq b\neq c; m \neq n\neq p $
@Matt F.: Depends of what is meant by "closed-form" formula. Any formula will contain some integral. But Gamma function (for the sphere) is also an integral, is not it?
@AlexandreEremenko I would argue that the Gamma function in the formulas for spheres is just there to give a common expression for both, odd and even dimensions. You do not need it for, say, even dimensions. Here you can give a closed form using only factorials.
Some of the references I give in this 2 June 2007 sci.math post may be of interest.
The surface area of an $n$-dimensional ellipsoid is expressed in terms of hyperelliptic integrals, see SURFACE AREA AND CAPACITY OF ELLIPSOIDS IN n
DIMENSIONS by
Garry J. Tee
In dimension 3, they are elliptic integrals, and the result is due to Legendre. The paper also mentions approximate formulas.
For $n\neq m\neq p$ one can write the integrals but they have no standard name: Surface area of an $\ell_p$ unit ball?
I have tried to solve with the article you have suggested, can you find out how to employ the integral for different exponents?
|
2025-03-21T14:48:29.551417
| 2019-12-29T09:45:31 |
349308
|
{
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"Student",
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|
Stack Exchange
|
Uniform boundedness for Strichartz constants in Tao's book
This is a question from Tao's book "Nonlinear Dispersive
Equations". We define the space $S^0(I\times R^d)$ as the closure of the Schwartz functions under the norm
$$\|u\|_{S^0}=\sup_{(q,r)\text{ admissable}}\|u\|_{L_t^q L_x^r}, $$
where $(q,r)$ is an admissible Strichartz pair satisfying $\frac{2}{q}+\frac{d}{r}=\frac{d}{2}$. Denote by $N^0$ the dual space of $S^0$. Now consider the Schrödinger equation
$$i\partial_t u+\Delta u=F $$
with initial datum $u_0$. It is claimed that using Duhamel and Strichartz we have
$$\|u\|_{S^0}\lesssim_{d}\|u_0\|_{L_x^2}+\|F\|_{N^0} $$
((3.26) in the book). What makes me confused is the index $d$, which says that the estimate constant should only depend on the space dimension. I don't see this directly from the Strichartz estimates stated in Tao's book. By those Strichartz estiamtes in the book ((2.24) to (2.26)), the estimate constants should also depend on the precise choice of the admissable pairs but not only on the space dimension. I observe that by the footnote "23" (p. 134), it is cautioned that some issues may occur in the case $d=2$, since the set of admissable pairs is not compact. So my guess would be that the Strichartz constants should depend continuously on the pair $(q,r)$, and the uniform boundedness follows. In fact, the set for $(q,r)$ is also unbounded, but I quess it is meant for the set of the pairs $(\frac{1}{q},\frac{1}{r})$, where compactness holds due to the relation $\frac{2}{q}+\frac{d}{r}=\frac{d}{2}$. However, I don't see the continuous dependence of the Strichartz estimates on the pairs from the proof given in the book.
Thank you for your help in advance!
Standard interpolation theorems (e.g., Riesz-Thorin) will give uniform bounds on the constants in the Strichartz estimates in terms of the constants of the endpoint Strichartz estimates (and in fact these interpolation theorems show that the best constants do indeed vary continuously, although uniform boundedness suffices for applications). As noted, when $d=2$ the endpoint is missing and one needs to modify norms such as $S^0$ either by restricting to a compact set of exponents, or inserting an appropriate weight that depends on the Strichartz constants.
@TerryTao Thank you very much Prof. Tao for your quick reply. The idea of interpolation perfectly explains how it works.
|
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