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2025-03-21T14:48:31.639728
| 2020-07-29T11:13:19 |
366849
|
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|
Stack Exchange
|
Kontsevich's A-infty cohomology classes of the moduli space of curves
In his paper "Feynman diagrams and low-dimensional topology," Kontsevich attaches to each $A_\infty$ algebra a cohomology class (with complex coefficients) on the moduli space of smooth, complex curves of genus g with n marked points.
He does this by noting that the moduli space of ribbon graphs with metric (where each vertex has degree at least 3) of genus g and n boundary components, is isomorphic to the moduli space of curves above. He uses the given $A_\infty$ algebra to produce a partition function $Z(\Gamma)$ for a ribbon graph $\Gamma$.
Finally, he says that $\sum_\Gamma{Z(\Gamma).\Gamma}$ is a closed cochain, and that it is closed "due to the higher associativity conditions on the algebra."
Would someone be able to explain why 1. the cochain is closed; 2. why the sum is finite; 3. why this gives you a cochain rather than a chain?
I'm tried several of the references and tried to work it out myself but I can't see what Kontsevich is saying here. (I am aware that to answer this question, one probably needs to already be aware of Kontsevich's idea - and I'm hoping someone here is.)
In particular, I've looked at Harer's "The cohomology of the moduli of curves" to understand the ribbon graph moduli isomorphism as Kontsevich mentions this as an "exposition" of the result (to not much success). I've looked at Kontsevich's "Notes on $A_\infty$-algebras, $A_\infty$-categories, and noncommutative geometry," but this is even more incomprehensible to me (since he vastly generalizes the above construction). I've also looked at various papers referenced in Kontsevich's paper.
|
2025-03-21T14:48:31.639864
| 2020-07-29T12:20:01 |
366853
|
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|
Stack Exchange
|
Free resolutions of universal enveloping algebras for simple, finite dimensional Lie algebras
I'm currently studying Anick's resolution on the context of universal enveloping algebras for certain Lie algebras, namely some of the smallest cases: $A_1,A_2,A_3,B_2,G_2$, and so on.
What are some references on other free resolutions that are convenient for universal enveloping algebras for simple, finite dimensional Lie algebras? Free resolutions of associative algebras in general (such as Anick's resolution) are also welcome. I'm aware that the bar resolution "induces" Anick's resolution.
|
2025-03-21T14:48:31.639926
| 2020-07-29T12:23:14 |
366855
|
{
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"authors": [
"Carlo Beenakker",
"MSMalekan",
"Manuel Norman",
"https://mathoverflow.net/users/11260",
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"site": "mathoverflow.net",
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|
Stack Exchange
|
Latest progress on Tarski numbers
Two questions, the first: What is the smallest non negative integer that we do not know yet is the Tarski number of a group?
The second question is the same as in the title: What is the latest progress on Tarski numbers?
For the second question, I think you may find the following papers interesting: M. Ershov, G. Golan, M. Sapir. The Tarski numbers of groups, Advances in Mathematics, Vol. 284, 2015, 21-53; A. Rejali, A. Yousofzadeh. Configuration of groups and paradoxical decompositions. Bull. Belg. Math. Soc. Simon Stevin 18 (2011), no. 1, 157-172; A. Yousofzadeh, A constructive way to compute the Tarski number of a group, J.
Algebra. Appl, 17, no. 1 (2018), 1850139
@CarloBeenakker That is, we know that there is a group with Tarski number 5. Could you please provide a reference?
Q1: The state of the art as reported in 2014, see arXiv:1406.2097, is that the only numbers which are known to be Tarski numbers of some groups are 4,5,6. Tarski numbers $<4$ are forbidden, which suggests that 7 is the answer to the question "What is the smallest non negative integer that we do not know yet is the Tarski number of a group?".
Q2: Recent progress, in addition to the references by Manuel Norman: In arXiv:1603.04212 it is shown that the Tarski number of Burnside groups is in the range $[6,14]$.
The last arXiv link is blank.
broken link fixed, thanks.
|
2025-03-21T14:48:31.640063
| 2020-07-29T13:00:23 |
366856
|
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"lambda"
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|
Stack Exchange
|
Why is a dynamical system not a dynamic system?
This is a research question in the history of math, I suppose.
As a non-native english speaker I became used to mathematical expressions like 'dynamical' and 'tangential'. When using them in daily conversation as substitutes for 'dynamic' and 'tangent' I got frowned upon by native english speakers who claimed to have never heard of these words before.
Some references suggest
https://www.merriam-webster.com/dictionary/dynamical
https://www.merriam-webster.com/dictionary/tangential
that indeed they can almost mean the same as 'dynamic' and 'tangent' but for some reason nobody seems to use these words that way.
So whereas in English:
A dynamic person-a dynamical system
the adjective is different
it is in French
une personne dynamique- un système dynamique
and in German
Eine dynamische Person - ein dynamisches System
the adjective is the same.
I am wondering when (and maybe also why) these expressions started deviating in English.
"Dynamic system" is also used, although less common.
https://link.springer.com/chapter/10.1007/978-3-662-04868-9_2
https://www.amazon.com/Introduction-Dynamic-Systems-David-Luenberger/dp/0471025941
"Tangential" is a perfectly fine word and I'm surprised people suggested otherwise. For example, I would say "your remarks are tangential to the main issue" to suggest that someone was getting off track. Saying "tangent" in that sentence sounds weird to me. As for dynamics, my inclination is that "dynamical" puts the emphasis on "being in the subject of dynamics" while "dynamic" means its usual thing (changing a lot, etc.). There is a similar phenomenon in "arithmetic" vs "arithmetical".
While we're discussing the language, why "a vector space" and "a moduli space" ('vector' and 'moduli' as nouns) in English but "un espace vectoriel" and "un espace modulaire" ('vectoriel' and 'modulaire' as adjectives) in French?
By the way, since, as you mention, this is in some way a history question, if it gets closed here then you might try HSM.
@LSpice: I guess this has to do with the fact that adjectives in English come before the noun, and thus such a position for the related noun makes it a determinative, so makes it semantically equivalent to an adjective.
@LSpice: English is perfectly happy with noun-noun compounds separated by space. "Dinner table", "meal ticket", "brick wall", "chicken soup". In grammar this is called attributive nouns. English is just much happier using nouns as adjectives compared to French which demands the preposition (compare, "mur en briques", "table à dinner" etc.)
@SylvainJULIEN, right, I certainly agree that that's how it's functioning; I'm just curious why, for example, English doesn't call them "vectorial spaces" and "modular spaces", or why French doesn't call them … whatever the analogue would be in French ("un vecteur espace"?).
@LSpice: I've definitely used "un vecteur espace" at least once when lecturing in French.... The (Swiss) students found it hilarious, along with all my Americanisms.
@WillieWong, re https://mathoverflow.net/questions/366856/why-is-a-dynamical-system-not-a-dynamic-system#comment926793_366856 —nothing like (trying) to speak another language to make one more understanding of others' imperfect practice in one's own language. :-)
The difference between "historic" and "historical" is somewhat analogous. But not exactly. Both "historic event" and "historical event" are used but have different meanings. As for "dynamic system" and "dynamical system", I can't think of when I would use the former.
Here are the two entries from Anthony Lo Bello's Origins of Mathematical Words (John Hopkins, 2013) which is very informative, entertaining, and perhaps curmudgeonly. In his parlance, following the lexicographer Samuel Johnson, a "low word" is one with an "irregular combination" of roots that has "little or no etymological legitimacy."
dynamical The Greek noun [dunamis] means power. The corresponding Greek adjective is [dunamikos], pertaining to power. The correct English adjective is therefore dynamic. To superimpose the vestige -al of the Latin adjectival ending -alis upon the stem of a Greek adjective is often the product of ignorance and produces a low word. In other cases, the addition of the Latin suffix to the Greek adjective is due to the fact that a different meaning is intended from that of the Greek adjective; thus, dynamic was an established word, so one spoke of dynamical systems rather than dynamic systems to avoid confusion.
tangential See the entry tangent. The Latin adjectival suffix -alis was added to the stem of the participle tangens, tangentis, which was already an adjective but felt to be a noun, the tangent.
Thing is, they do not mean the same thing. At least, not in theory.
Dynamic the adjective means "exhibiting continual change".
Dynamics the noun means "the study of forces and their relation to motion".
Dynamical the adjective means "relating to the study of dynamics."
A "dynamic" system is a system exhibiting continual change. A "dynamical" system is a system relating to the study of dynamics. (Since OP is Chinese, this is also why DS is 動力系統 and not 不定系統.)
Similarly,
Tangent the adjective means the geometric notion of touching but not intersecting.
Tangent the noun refers first to the geometric construct of the line tangent to a shape, and then also to the idea of "objects that can be split off without making a turn", whence the idea of "going on a tangent" when you derail the discussion with something related but not directly relevant. (You won't be going on a tangent if you change the topic or the direction of discussion abruptly.)
Tangential the adjective refers to the quality of "tangent". Hence you make a "tangential remark" while you "go on a tangent". Hence you look for "tangent lines" while compute "tangential forces". (The force itself is not tangent, but it is directly along the line that is tangent to the object.)
This, of course, gets muddled by the fact that English is perfectly happy with attributive nouns. English being remarkably loosey goosey about grammar for a Western language, understanding in theory why things are the way they are is probably much less useful than accepting their use as a convention. (After all, what is language but a convention to enable communication?)
Homework exercise: discuss transverse versus transversal
Lo Bello (see my answer for bibliographic details) on transversal: Though transversus is already an adjective, Albertus Magnus (1193-1280) superimposed the adjectival suffix -alis upon the stem to create the adjective transversalis, from which the English adjective was derived. Albert wrote the first original commentary on Euclid's Elements of Geometry in the Latin language,
Actually, there's a (somewhat famous) comment of Whitehead on this: "Transversal" is a noun; the adjective is "transverse.": https://en.wikipedia.org/wiki/Transversality_(mathematics)
My "exercise" was actually posed with @PeterDalakov's solution in mind.
Surely the adverb should then be "transversely", but that same wiki page insists on "transversally".
Early uses of "dynamical" go back to the 19th century at least:
On a dynamical theory
of gratings, Lord Rayleigh (1907)
On the dynamical
theory of gases, J.C. Maxwell (1865)
Thermo-dynamical as an adjective was also common, see for example W. Gibbs's Thermodynamical Model (1900).
This is a subtle question of scientific jargon. Indeed, in everyday language, "dynamic" is preferred, and "dynamical" at most is seen as an awkward synonym. My feeling is that "dynamic system" would be the everyday language way of referring to a system that was actually changing in time, moving. Whereas a "dynamical system" in physics, mathematics and at least to some extent beyond is a system that by its nature is capable of exhibiting change in time, i.e., being dynamic; moreover, we make a statement about the reason for that capability, i.e., there is some description of what causes its specific dynamism. One distinguishes between a merely kinematical description and a dynamical description - a kinematical description merely addresses how something moves, whereas a dynamical description addresses why something moves (say, some variational principle).
Indeed, with respect to the latter sentence, there's a clear contrast: A "dynamic description" would be a description that itself changes in time; a "dynamical description" is one that explains the change in time of some other object that is being referred to.
Why are you using "kinematical" instead of "kinematic"? I don't think that in the term "kinematic description", the label "kinematic" is about the description itself having an attribute from within kinematics. If a physical process were explained using thermodynamics, I think the term "thermodynamic explanation" would sound like a reasonable label. At the same time, the terms "electric/electrical" have a usage distinction but there is no term "magnetical" to go along with "magnetic".
@KConrad - Indeed, I don't see a clear distinction between "kinematical" and "kinematic". I probably just used "kinematical" here for symmetry with "dynamical". I didn't think much about it. "Kinematic description" sounds fine. And "magnetical" is hilarious, I have to start using that as a joke. You illustrate perfectly how language isn't something one can press into simple rules. It's phantastomatic.
You mean phantastomatical... :)
|
2025-03-21T14:48:31.640855
| 2020-07-29T13:17:21 |
366858
|
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"Asaf Shachar",
"Mateusz Kwaśnicki",
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|
Stack Exchange
|
Is a function of several variables convex near a local minimum when the derivatives are non-degenerate?
This is a cross-post.
Let $U \subseteq \mathbb R^n$ be an open subset, and let $f:U \to \mathbb R$ be smooth. Suppose that $x \in U$ is a strict local minimum point of $f$.
Let $df^k(x):(\mathbb R^n)^k \to \mathbb R$ be its $k$ "derivative", i.e. the symmetric multilinear map defined by setting
$df^k(x)(e_{i_1},\dots,e_{i_k})=\partial_{i_1} \dots \partial_{i_k}f(x)$.
Assume that $df^j(x) \neq 0$ for some natural $j$. Let $k$ be the minimal such that $df^k(x) \neq 0$. Since $x$ is a local minimum, $k$ must be even.
Suppose now that $df^k(x)$ is non-degenerate, i.e. $df^k(x)(h,\dots,h) \neq 0$ for any non-zero $h \in \mathbb R^n$. (Since $x$ is a minimum, this is equivalent to $df^k(x)$ being positive-definite, i.e. $df^k(x)(h,\dots,h) > 0$ for any non-zero $h \in \mathbb R^n$).
Question: Is $f$ is strictly convex in some neighbourhood of $x$?
In the one-dimensional case, when $f$ is a map $\mathbb R \to \mathbb R$, the answer is positive:
We have $f^k(x)>0$, and the Taylor expansion of $f''$ near $x$ is
$$
f''(y) = {1 \over (k-2)!} f^{(k)}(x)(y - x)^{k-2} + O((y - x)^{k-1}).
$$
Thus, $f''(y)>0$ for $y \ne x$ sufficiently close to $x$, so $f$ is strictly convex around $x$.
Returning back to the high-dimensional case, if $k>2$, we have $\text{Hess}f(x)=df^2(x)=0$, and I guess that we should somehow prove that $\text{Hess}f(y)$ becomes positive-definite for $y$ sufficiently close to $x$.
Perhaps we need to understand the Taylor's expansion of $\text{Hess}f$ around $x$, similarly to the one-dimensional case, but I am not sure how to do that.
Is there a nice way?
Comment:
It is certainly not enough to assume that $df^k(x)$ is non-zero. Indeed, consider $ f(x,y) = x^2 y^2 + x^8 + y^8$.
$f$ has a strict global minimum at $(0,0)$.
$$\det(\text{Hess}f(x,y))=3136 x^6 y^6 + 112 x^8 + 112 y^8 - 12 x^2 y^2,$$ which is negative when $x=y$ is small and nonzero. Thus, $f$ is not convex at a neighbourhood of zero.
Note that $\text{Hess}f(0,0)=0$; The first non-zero derivative at $(0,0)$ is the fourth-order derivative $df^4(0)$. It is degenerate, however, since $df^4(0)(h^1e_1+h^2e_2,h^1e_1+h^2e_2)=4(h^1)^2(h^2)^2$ vanishes when either $h_i$ is zero.
So, non-vanishing of some derivatives does not ensure convexity.
@Mateusz already gave you a counter example, but I think that $d^kf > 0$ is definitely too weak. Morally you want $Hess(f)$ to be positive semi definite in a neighborhood, and this suggests that since you are in the case $Hess(f)(0) = 0$, you want a suitable number of higher derivatives of the matrix valued function $Hess(f)$ to take values in the symmetric positive semidefinite matrices. Assuming $d^3f > 0$ does not control $\partial_x \partial^2_{yy} f$ (which is essentially what Mateusz used in his example).
So something like $d^k Hess(f)$ being positive would probably work. (So if $k = 4$ is the smallest $k$ for which $d^k f \neq 0$ you want $\partial_v \partial_v Hess(f)$ to be a PD matrix.) This can be written as for any $v, w\neq 0$ that $d^kf(v,v,\ldots, v, w,w) > 0$.
Let
$$\begin{aligned} f(x,y) & = x^4 - x^2 y^2 + y^4 \\ & = \tfrac{1}{2} x^4 + \tfrac{1}{2} y^4 + \tfrac{1}{2} (x^2 - y^2)^2 . \end{aligned}$$
Then $f$ is a strictly positive (except at the origin, of course) homogeneous polynomial of degree $4$, and hence $d^j f(\vec 0) = 0$ for $j < 4$ and $d^4 f(\vec 0) > 0$ (indeed: $d^4 f(\vec 0)(\vec h, \vec h, \vec h, \vec h) = 4! f(\vec h) > 0$ whenever $\vec h \ne \vec 0$). On the other hand, $$\partial_{xx} f(0,y) = -2 y^2 < 0$$ whenever $y \ne 0$, and so $f$ is not convex near $0$.
Thanks. Just one question: Why does the fact that $f$ is a homogeneous polynomial of degree $6$ imply that $d^6f$ is non-degenerate? Can we deduce that without explicitly computing $d^6f$?
@AsafShachar: if $f$ is a homogeneous polynomial, it is equal to its taylor polynomial of that degree, and hence if $v = (x_0, y_0)$ up to some numerical constant $d^6f(v,v,\ldots, v)$ is the same as $f(x_0, y_0)$.
@WillieWong and Asaf Schachar: Thanks! I edited this into the answer (and also reduced the degree from 6 to 4).
Let $n=1$, $f(t)=t^2 + |t|^{7/2}\sin(1/|t|)$ for $t\ne0$, $f(0):=0$. Then $f'(0)=0$ and $f''(0)=2>0$, so that $0$ is a strict local minimum of $f$. However, $f''(t)\sim-|t|^{-1/2}\sin(1/|t|)$ as $t\to0$, and so, $f$ is not convex (let alone strictly convex) in any neighborhood of $0$.
Here are the graphs $\{(t,f(t))\colon|t|<0.1\}$ (left) and $\{(t,f''(t))\colon|t|<0.1\}$ (right).
Thanks, this is a nice example. However, I assumed that $f$ is smooth. I think you should keep this answer, to show what can happen without smoothness.
|
2025-03-21T14:48:31.641138
| 2020-07-29T13:26:12 |
366859
|
{
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"Carlo Beenakker",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/366859"
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|
Stack Exchange
|
Relation between the eigenvalues of a block matrix and the eigenvalues of its diagonal blocks
Consider the $(m+n) \times (m+n)$ block matrix
$$M = \begin{bmatrix} A & B \\ C & D \end{bmatrix}$$
I need references where they are talking about the relation between the eigenvalues of $M$ and the eigenvalues of $A$ and $D$. I want to learn under what circumstances such a relation exists. The simplest of such a possibility is $B=0$ or $C=0$. I believe there are other non-trivial cases. Kindly share some references. Thank you.
suppose $A=0=D$; then still $M$ can have any set of eigenvalues (symmetrically arranged around $0$), so how can there be any relationship between the eigenvalues of $M$ and those of $A,D$ ?
@CarloBeenakker, Thank you. In general, there need not be any relations. I want to know any special circumstances under which such a relation holds. Some papers where they talk about similar questions. This is what I am looking for.
To see what you might expect for a relation, consider the case of a $2\times 2$ matrix $M=\begin{pmatrix}a&b\\
c&d\end{pmatrix}$, with eigenvalues $\lambda_\pm=\tfrac{1}{2}(a+d)\pm\sqrt{4bc+(a-d)^2}$. Knowledge of $a$ and $d$ is a constraint on the sum of the eigenvalues (the trace of $M$), but there is no other constraint. I'm pretty sure this carries over to the higher dimensional case: knowledge of the matrices $A$ and $D$ fixes the trace of $M$, and hence the sum of the eigenvalues, but if you are free to vary $B,C$ there is no other constraint.
|
2025-03-21T14:48:31.641267
| 2020-07-29T13:34:46 |
366862
|
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|
Stack Exchange
|
Constructing weakly-dependent process with certain decay rate of dependency coefficients
Let $(X_{t})_{t \in \mathbb{N}}$ be a real-valued stationary stochastic process over probability $(\Omega,\mathcal{F},\mathbb{P})$, such that for $p\geq 2$, $X_{t} \in L_{p}(\mathbb{P})$ and it holds:
\begin{equation}
\label{eq:weak_dep}
|| E[X_{t+k}|\mathcal{F_{t}}] - E[X_{t+k}] ||_{p} \leq M_{p}\psi_{p}(k),
\end{equation}
where $\mathcal{F}_{t} = \sigma(X_{s}: s\leq t)$, $M_{p}:=||X_{t}-E[X_{t}]||_{p}$ , $\psi_{p}({\cdot})$ is a monotonically non-increasing function. Aforementioned condition is a mixingale like projective weak-dependency condition on $(X_{t})_{t\geq 1}$ (introduction in the book "Weak Dependence With Examples and Applications" by Dedecker et.al, more precisely Definition 2.7 on page 19 therein ). I would like to understand how to construct such a process if we are given a polynomial decay rate $\psi_{p}(k) = k^{-\gamma}$, $\gamma \in (0,\infty)$. Has anybody seen an example or could point me some relevant ideas?
|
2025-03-21T14:48:31.641374
| 2020-07-29T15:02:52 |
366871
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/366871"
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|
Stack Exchange
|
Concrete examples of statements not provable in PRA + $\epsilon_0$-induction that are provable in PA?
It is well-known that $\mathbf{PRA}$ plus $\epsilon_0$-induction on bounded formulas cannot prove all $\mathbf{PA}$ theorems (essentially because $I\Sigma_1$ plus $\epsilon_0$-induction on bounded formulas is finitely axiomatizable while the latter isn't). Are there any concrete examples known (preferable "natural")?
(This is an exact copy of my question on MSE, but I expect I won't get any answer there, so I hope I'm allowed to cross-post it here)
What exactly do you mean by “$\epsilon_0$-induction”? The schema of $\epsilon_0$-induction for all formulas most definitely proves the schema of $\omega$-induction for all formulas, i.e., PA. The schema is not finitely axiomatizable unless it is restricted to formulas of complexity $\Sigma_n$, or something. (PRA is also not finitely axiomatizable, by the way.)
@EmilJeřábek Oh, yes, sorry: on bounded formulas. And yes, thanks for pointing that out, I meant: essentially because $I\Sigma_1$ is finitely axiomatizable. I hope it is good now :)
Ok. (The single axiom equivalent to) $I\Sigma_2$ is not provable in $I\Sigma_1+{}$ bounded $\epsilon_0$-induction (as the latter is a consistent $\Pi_3$ theory, while the former implies the uniform $\Sigma_3$-reflection principle). Is that concrete enough?
@Emil That sounds pretty concrete, although I cannot really follow the argument (I'm just starting out in proof theory). I've only come across reflection in the case of Con(PA) is equivalent to $\Pi_1$ reflection.
Another point: it leaves open the question of a "natural" example.
One example is $I\Sigma_2$ (or rather, the conjunction of its finite axiomatization).
Notice that the theory $I\Sigma_1+\epsilon_0$-induction for bounded formulas is axiomatized by $\Pi_3$ sentences: in particular, the $\epsilon_0$-induction schema can be written in prenex form as
$$\forall u\,\forall x\,\exists y\,\forall z\,\bigl[\bigl(\neg\theta(u,y)\land(z\prec y\to\theta(u,z))\bigr)\lor\theta(u,x)\bigr],$$
where $\theta\in\Delta_0$ and $\prec$ is the order representing $\epsilon_0$. By a theorem of Leivant [1], $I\Sigma_2$ is not provable from any set of $\Pi_3$ (or even $\Sigma_4$) sentences consistent with $Q$. More generally, for $n\ge1$, $I\Sigma_n$ is not provable from any set of $\Sigma_{n+2}$ sentences consistent with $Q$.
The reason is that $I\Sigma_n$ proves the uniform $\Sigma_{n+1}$-reflection schema for $Q$:
$$\mathrm\forall x\,\bigl(\mathrm{Pr}_Q(\ulcorner\phi(\dot x)\urcorner)\to\phi(x)\bigr),\qquad\phi\in\Sigma_{n+1},$$
where $\mathrm{Pr}_Q$ denotes the formalized provability predicate for $Q$. This schema is equivalent to the single sentence
$$\def\rfn{\mathrm{RFN}}\rfn_Q(\Sigma_{n+1})=\forall x\,\bigl(\mathrm{Pr}_Q(x)\to\mathrm{Tr}_{n+1}(x)\bigr),$$
where $\mathrm{Tr}_{n+1}$ is the truth definition for $\Sigma_{n+1}$-formulas (defined so that it is vacuously true for non-$\Sigma_{n+1}$-formulas). Moreover, it is easy to see that
$$\rfn_Q(\Sigma_{n+1})\equiv\rfn_Q(\Pi_{n+2}).$$
Now, if we assume for contradiction that $\rfn_Q(\Pi_{n+2})$ is provable from a consistent extension of $Q$ by $\Sigma_{n+2}$ sentences, there is a single $\Sigma_{n+2}$ sentence $\phi$ consistent with $Q$ such that
$$Q+\phi\vdash\rfn_Q(\Pi_{n+2})\vdash\mathrm{Pr}_Q(\ulcorner\neg\phi\urcorner)\to\neg\phi,$$
that is,
$$Q+\phi\vdash\mathrm{Con}_{Q+\phi},$$
contradicting the second incompleteness theorem.
Reference:
[1] Daniel Leivant: The optimality of induction as an axiomatization of arithmetic, Journal of Symbolic Logic 48 (1983), no. 1, pp. 182–184, doi: 10.2307/2273332.
Yes! Thanks! And this only uses the consistency of PRA + $\epsilon_0$-induction, but you need that anyway, otherwise the result is of course false. A few short questions: 1) what do you mean with "uniform"? 2) is it hard to show that $I\Sigma_n$ proves the uniform $I\Sigma_{n+1}$-reflection schema for $Q$? 3) What is the complexity of the truth predicate of up to $n$ quantifiers ($\text{Tr}_n$); I think $\Delta_n$?
I've accepted your answer; but I remind that the question of a natural example is still open. Or are there such that imply $I\Sigma_2$ easily?
(1) What I presented above are called uniform (or global) reflection principles, as opposed to local reflection principles, which are the schemata $\mathrm{Pr}Q(\ulcorner\phi\urcorner)\to\phi$ for sentences $\phi$. (2) It’s not that easy, as it relies on cut elimination. But basically, you convert a proof of a $\Sigma{n+1}$ sentence into a cut-free sequent refutation of its negation $\psi$, which is $\Pi_{n+1}$; assuming for contradiction that $\psi$ is true, you prove by induction that all the sequents in the proof with $\psi$ removed are true. This is $\Pi_n$-induction, ...
... as apart from $\psi$, all the sequents have $\Sigma_n$ antecedents and $\Pi_n$ succedents. (3) The truth definition for $\Sigma_n$ formulas is $\Sigma_n$.
I know about cut elimination, but I don't understand what you mean by "a cut-free sequent refutation of its negation $\psi$".
We are proving inside $I\Sigma_n$ that if a sentence $\phi\in\Sigma_{n+1}$ is provable in $Q$, then $\phi$ is true. $\psi$ is $\neg\phi$, written in a prenex normal form so that $\psi\in\Pi_{n+1}$. A refutation of $\psi$ is a proof of the sequent $\psi\Longrightarrow{}$. Though now that I think about it, it’s not necessary to make the proof into a refutation. We can just take directly a cut-free proof of $\Longrightarrow\phi$, and assuming $\phi$ is false, show by induction that all sequents in the proof with $\phi$ deleted are true.
First, a truth definition for $\Sigma_n$ formulas is $\Sigma_n$, and a truth definition for $\Pi_n$ formulas is $\Pi_n$. You can’t mix the two. Cut-free sequent calculus has the subformula property, respecting the polarity of subformulas, hence in a proof of a $\Sigma_{n+1}$ sentence $\phi$ (with a single leading $\exists$ quantifier), every sequent in the proof has $\phi$ and $\Pi_n$ formulas in the succedent, and $\Sigma_n$ (really, $\Pi_{n-1}$) formulas in the antecedent. A truth definition for such sequents can be written as $\Pi_n$ as the semantics of sequents effectively negates ...
... formulas in the antecedent, except that you can only apply the truth definition to the part of the sequent without $\phi$, as $\phi$ itself has too large complexity. This is the reason why you have to check the truth of $\phi$ before the inductive argument. If it’s true, we are done as that’s what we want to prove, and if it’s false, we have to obtain a contradiction. So, assume it is false, and consider a $\Pi_n$ truth definition for sequents in the proof that fixes the value of $\phi$ as false, and evaluates other formulas in the normal way. Then, using $\Pi_n$ induction (which is ...
... equivalent to $I\Sigma_n$, see e.g. Hájek&Pudlák), prove that all sequents in the proof are valid under this truth definition. In the end, you obtain that the empty sequent (that is, $\Longrightarrow\phi$ with $\phi$ discarded as false) is valid, which is a contradiction.
I think I must be confusing some of your terminology: OK, we have a cut free proof of $\implies \phi$, and we can prove (using our $\Pi_n$ truth definition) that all sequent above this are true (what do you mean by valid?), and we have $\neg\phi$ (for contradiction). How does that lead to a prove of the empty sequent?
Valid just means true under all assignments. The last sequent of the proof is $\Longrightarrow\phi$. Under the truth definitoin that fixes $\phi$ as false, or equivalently, that removes it from the succedent of any sequent before checking its truth, this is the same as the empty sequent $\Longrightarrow$.
By the way, I really really appreciate all your help and comment. Thank you.
Ah, I see. That clarifies it. It's not about obtaining a proof of the empty sequent, rather on the assumption of $\neg\phi$, you obtain that the empty sequent is true, which by definition of the (adjusted) truth predicate is impossible. Right? I'm trying to think why we need $\neg\phi$ to change the truth definition. It seems we can just "hard code" the falsehood of $\phi$ in it no matter what.
If you hardcode $\phi$ as false, you need $\phi$ to be actually false in order to prove soundness of the inference rules for the induction step. In particular, write $\phi=\exists x,\psi(x)$, and consider an $\exists$-right inference from $\Gamma\Longrightarrow\Delta,\psi(t)$ to $\Gamma\Longrightarrow\Delta,\phi$.
Yes, I see. If we know $\phi$ to be false then so must $\psi(t)$, and hence if $\Gamma \implies \Delta, \psi(t)$ is true, one of $\Delta$ must be true, or one of $\Gamma$ be false, but under these conditions $\Gamma \implies \Delta,\phi$ is true, so we remain in good shape.
The final thing I don't understand is why your $\Sigma_{n+1}$ reflection scheme proves $\text{RFN}(\Sigma_{n+1})$ over $I\Sigma_n$? What is the complexity by the way of $I\Sigma_n$? as a single sentence? $\Pi_{n+3}$?
$I\Sigma_n$ is $\Pi_{n+2}$, just count the quantifiers. You would know this if you looked at the Leivant paper in the first place. Concerning the FINAL THING, let $\psi$ be a conjunction of axioms of $I\Delta_0+\mathrm{EXP}$ (which is $\Pi_2$, and enough to verify the properties of $\mathrm{Tr}n$), and given a $Q$-proof of a $\Sigma{n+1}$ sentence $\sigma$ (the “$x$”), construct a proof of $\psi\to\mathrm{Tr}_{n+1}(\ulcorner\sigma\urcorner)$, and apply the reflection principle.
That's okay. Thank-you for all the help. There is no supervisor and MSE usually doesn't answer in this area. I will definitely look at the paper, I think with a bit of patience I should have the level to process it. I'm not a PhD student btw. Again, thanks for your help. I'm sorry to have bothered you so often.
That was a typo, by the way, I meant $I\Sigma_{n+1}$.
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2025-03-21T14:48:31.641972
| 2020-07-29T15:51:07 |
366874
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/366874"
}
|
Stack Exchange
|
Decompose a power of the radius as a sum of squares of forms
I am interested in a decomposition
$$(x_1^2+...+ x_n^2)^d=\sum_{i=1}^m (P_i(x_1,\dots,x_n))^2$$
where $d>1$ is odd, $m\leq n$, $P_i$ are forms of degree $d$.
Of course, the greatest common divisor for all $P_i$ is 1,
to prevent the trivial decomposition with $P_i=x_i(x_1^2+...+ x_n^2)^{(d-1)/2}$.
|
2025-03-21T14:48:31.642035
| 2020-07-29T15:53:42 |
366875
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Dylan Wilson",
"Jeff Strom",
"Matt Feller",
"Nicholas Kuhn",
"Simon Henry",
"Tyrone",
"https://mathoverflow.net/users/102519",
"https://mathoverflow.net/users/132451",
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"https://mathoverflow.net/users/3634",
"https://mathoverflow.net/users/54788",
"https://mathoverflow.net/users/6936"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366875"
}
|
Stack Exchange
|
When is an increasing union a colimit?
Let's consider a diagram $\Phi: \lambda \to \mathcal{T}_*$
$$
X_0 \to X_1 \to \cdots \to X_\xi \to X_{\xi+1} \to \cdots
$$
of pointed spaces,
indexed by some ordinal $\lambda$, in which each $X_\xi$ is a
subspace of a certain space $Y$ and each map $X_\zeta\to X_\xi$ is
the inclusion map.
I have the idea that 'morally' the categorical colimit of the diagram is simply the union $X = \bigcup_{\xi< \lambda} X_\xi$ (with the subspace topology). Certainly this is not actually the case in general -- we need some topological restrictions.
I'm willing to impose really strong topological restrictions.
First of all, I'm happy to assume
all of the maps $X_\zeta \to X_{\xi}$ and $X_\xi \to Y$
are cofibrations. I'd even be pretty happy to have an argument in which each map $X_\xi \to X_{\xi + 1}$ is obtained by attaching a cell, or a cone.
Furthermore, I'm content to work with a category of spaces (such as CGWH) for which cofibrations
are necessarily inclusions of closed subspaces (up to homeomorphism).
In general, of course, there is a comparison map $c: \mathrm{colim}\ \Phi\to X$, and it is
clearly surjective. It would suffice, therefore, to prove that $c$ is a cofibration.
Unfortunately, my diagram-wrangling skills are coming up short. I'd appreciate pointers to a good argument or a useful reference.
Isn't that always the case as soon as the union is equiped with the initial topology ? If the problem is that you don't want to use the initial topology, what topology do you put on the union ?
The union gets the subspace topology.
how about section 3.2 here: http://neil-strickland.staff.shef.ac.uk/courses/homotopy/cgwh.pdf ?
Let me know if this doesn't work, I'm too busy right now to think about it. Take $X=\mathbb{N}$ and give it the cofinite topology. Let $X_n={1,\dots,n}$ for each $n\geq1$. Then $X_n$ is discrete, and $X_{n+1}$ is obtained from $X_n$ by attaching a $0$-cell. Each $X_n\subseteq X$ is closed, and I think it it a cofibration (it has a halo in $X$?). As sets $X=\bigcup X_n$, but as spaces $colim;X_n$ is discrete. Note that everything is compactly generated (although not weak Hausdorff?).
Let $X_n={1, 1/2, \ldots, 1/n, 0}$ as a subspace of $\mathbb{R}$. When we take the colimit, we just get a countable set with the discrete topology, which is different from the subspace topology of the union of the $X_n$'s in $\mathbb{R}$.
It looks like Matt's example shows that there is no interesting set of restrictions that work!
Conversely, I think Lemma 3.7 in Strickland's notes may do the job for me. I'm checking details before suggesting Dylan Wilson upgrade to an answer, though.
@MattFeller I see your point, but I don't think these inclusions can be thought of as cone attachments.
ah, I missed the thing about some ambient space we were taking the subspace topology in... I'm inclined to agree with Matt/Nick. Surely an inclusion between finite sets is a cofibration (you are attaching a cone on the empty set.)
@DylanWilson Not in the category of pointed spaces!
Ok, I think I've got the proof I need. Strickland's Lemma 3.7 requires a compactness condition: for every compact $C$ in $Y$, $C\cap X = C\cap X_\xi$ for some $\xi$. I can prove this condition when each $X_\xi \to X_{\xi+1}$ is obtained by a a cone attachment. I'll post details when I've got them written up.
|
2025-03-21T14:48:31.642296
| 2020-07-29T16:19:17 |
366877
|
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|
Stack Exchange
|
Bloch–Kato–Selmer group of a one-dimensional representation
Let $L/\mathbb{Q}$ be a finite extension and let $V$ be a one-dimensional $L$-linear representation of $G_{\mathbb{Q}}$ which is given by $\chi\rho^*\kappa^n_{\text{cyc}}:G_{\mathbb{Q}}\rightarrow L^\times$, where $\rho$ is a Dirichlet character mod $p$, $\chi$ is a Dirichlet character mod $d$ for some $d$ prime to $p$ (both viewed as galois characters) and $\kappa_{\text{cyc}}$ is the cyclotomic character for the prime $p$. In particular the restriction of $V$ to $G_{\mathbb{Q}_p}$ is de Rham. Further we assume $n\leq0$ and $\chi\rho^*(-1)\cdot(-1)^n=-1$. To be precise, this representation comes from the Artin motive corresponding to $\chi$ twisted by $\rho^*$ and the Tate motive $\mathbb{Q}(n)$.
Let $S=\{\text{primes dividing $d$},p,\infty\}$ as a subset of the places of $\mathbb{Q}$. I want to show (if true) that
the Bloch–Kato–Selmer group
$$
H^1_f(\mathbb{Q},V)=\ker\left(H^1(G_{\mathbb{Q},S},V)\longrightarrow\bigoplus_{v\in S}\frac{H^1(\mathbb{Q}_v,V)}{H_\text f^1(\mathbb{Q}_v,V)}\right),
$$
is zero. Here the $H_\text f^1(\mathbb{Q}_v,V)$ are "local conditions", namely
$$
H_\text f^1(\mathbb{Q}_v,V)=\begin{cases}\ker(H^1(\mathbb{Q}_l,V)\overset{\operatorname{res}}{\longrightarrow}H^1(I_l,V)),& v=l\neq p\\
\ker(H^1(\mathbb{Q}_p,V)\longrightarrow H^1(\mathbb{Q}_p,B_{\text{cris}}\otimes_{\mathbb{Q}_p},V)),&v=p\\
H^1(\mathbb{R},V),&v=\infty,
\end{cases}
$$
where $I_l\subseteq G_{\mathbb{Q}_l}$ is the inertia subgroup.
One can show relatively easily that in this case one has $H^1_\text f(\mathbb{Q}_v,V)=0$ for all $v\in S$ and therefore it reduces to the question of whether the usual localization map
$$
H^1(G_{\mathbb{Q},S},V)\longrightarrow\bigoplus_{v\in S} H^1(\mathbb{Q}_v,V)
$$
is injective. To go further, I assume one has to use that $H^1(G,V)=\projlim_{n\in\mathbb{N}}H^1(G,T/p^nT)\otimes_{\mathcal{O}_L}L$, where $T\subseteq V$ is a $G$-stable $\mathcal{O}_L$-lattice and $G\in\{G_{\mathbb{Q},S},G_{\mathbb{Q}_v}\}$, and then use Pitou–Tate duality. In Neukirch, Schmidt, and Wingberg - Cohomology of number fields, §9.1 they discuss some cases where the localization map is zero, but it is mainly about finite simple $G$-modules and I don't see how I can apply this here. One can also show that $H_\text f^1(\mathbb{Q},V)=0$ if and only if
$$
H_\text f^1(\mathbb{Q},V/T)=\ker\left(H^1(G_{\mathbb{Q},S},V/T)\longrightarrow\bigoplus_{v\in S}\frac{H^1(\mathbb{Q}_v,V/T)}{H_\text f^1(\mathbb{Q}_v,V/T)}\right)
$$
is finite. But I don't know how to prove this either. It would be great if someone could help.
|
2025-03-21T14:48:31.642578
| 2020-07-29T16:49:56 |
366878
|
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|
Stack Exchange
|
Generalization of elementary symmetric polynomials
The elementary symmetric polynomials (ESPs) are defined as -
\begin{align*}
E_{1}^{1}
&= X_1,
\\
E_{1}^{2}
&= X_1 + X_2,
\\
E_{2}^{2}
&= X_1 X_2,
\\
E_{2}^{3}
&= X_1 X_2 + X_1 X_3 + X_2 X_3,
\\
E_{k}^{n}
&= \sum_{1 \leq j_1 < \dotsb < j_k \leq n} X_{j_1} \dotsm X_{j_k}.
\end{align*}
Notice that in the above examples, all the coefficients are 1.
Now we can generalize $E_{k}^{n}$ by the following function
$$ G_{k}^{n} = \sum_{1 \leq j_1 < \dotsb < j_k \leq n} S_{j_1 \cdots j_k} \: X_{j_1} \dotsm X_{j_k}, $$
where $S_{j_1 \cdots j_k}$ is symmetric. If $S_{j_1 \cdots j_k} = 1$, the above equation reduces to the ESP.
Edit: although $S_{j_1 \cdots j_k}$ is symmetric, $G_{k}^{n}$ is not (in the previous versions I incorrectly wrote $G_{k}^{n}$ is a symmetric polynomial).
My questions are -
Does the polynomial $G_{k}^{n}$ have a name?
Do decompositions of $G_{k}^{n}$ exist, like the ESP (by Lee - Power sum decompositions of elementary symmetric polynomials, 2016)?
I apologize if my notations are not standard. I have edited some of them to look familiar.
Edited. Thanks!
The polynomial is still not symmetric, unless all the $S $es are equal.
As @darijgrinberg implies, I think you may be running into the confusion about "how symmetric". A symmetric polynomial is symmetric in all its variables, whereas a symmetric tensor is symmetric only in the indices that are passed to it. (I'm also not sure what you gain by thinking of tensors here.) For example, any degree-1 tensor, say $S_i = i$ in the index notation, is symmetric; but $G^2_1 = X_1 + 2X_2$ is not a symmetric polynomial.
|
2025-03-21T14:48:31.642700
| 2020-07-29T17:02:11 |
366881
|
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"authors": [
"Liviu Nicolaescu",
"Xing Gu",
"https://mathoverflow.net/users/100553",
"https://mathoverflow.net/users/20302"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366881"
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|
Stack Exchange
|
The $\operatorname{spin}^c$ index for manifolds
$\DeclareMathOperator\spin{spin}\DeclareMathOperator\ch{ch}\DeclareMathOperator\ind{ind}$In the paper Čadek, Crabb, and Vanžura - Obstruction theory on 8-manifolds, the authors discussed the "$\spin^c$-index" for a $\spin^c$ manifold $M$ (display (3.1) of the paper):
$$y\in K^0(M)\mapsto \ind(y)=(e^{c/2}\hat{A}(\tau M)\ch(y))[M]\in\mathbb{Z}, $$
where $c$ is the $\spin^c$ class, $\tau M$ is the tangent bundle of $M$, and $\hat{A}$ is the Hirzebruch signature:
$$\hat{A}(\tau M)=1-p_1(\tau M)+\dotsb. $$
I was wondering if there is a more coherent context in which the invariant $\ind(y)$ is discussed.
If if you think of $y$ as defined by a complex vector bundle, then $\mathrm{ind}(y)$ is the index of the $spin^c$ Dirac operator corresponding to the $spin^c$ structure $c$ and twisted by the vector bundle $y$ . This idea goes back to work of Atiyah and Hirzebruch
@Liviu Nicolaescu Thank you!
I suggest consulting section 26 of Hirzebruch's Topological Methods in Algebraic Geometry and the references therein. In particular, it contains the following statement:
Theorem 26.1.1. Let $d$ be an element of $H^2(X, \mathbb{Z})$ whose reduction mod 2 is the Whitney class $w_2(X)$, and $\eta$ a continuous $GL(q, \mathbb{C})$-bundle over $X$. Then
$$\hat{A}(X, \tfrac{1}{2}d, \eta) = \chi^m\left[e^{\frac{1}{2}d}\cdot\operatorname{ch}\eta\cdot\sum_{j=0}^{\infty}\hat{A}_j(p_1, \dots, p_j)\right]$$
is an integer.
References to proofs are given in subsections 26.3-26.5.
|
2025-03-21T14:48:31.642845
| 2020-07-29T17:13:31 |
366884
|
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"authors": [
"Gregory J. Puleo",
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"https://mathoverflow.net/users/6322",
"vidyarthi"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/366884"
}
|
Stack Exchange
|
List chromatic index of a particular graph
Consider the graph $G$ of order $n$ consisting of two disjoint cliques of even order $\frac{n}{2}=p+1$ (where $p$ is odd prime) joined by a bipartite graph (that is, deleting the edges of the two disjoint cliques from $G$ leaves a bipartite graph) of maximum degree $p$. Then, does the graph have list chromatic index $\le 2p+1$? The bipartite graph is also quite specific, in that it has one vertex in each partite set of degree exactly equal to $0,1,2,\dotsc,p$.
My view is that, by Schauz - Proof of the list edge coloring conjecture for complete graphs of prime degree paper, we have that the disjoint cliques are chromatic edge-choosable. In addition, the edges joining the two cliques is a bipartite graph, which is again chromatic edge-choosable by the Galvin's theorem. Thus, it makes me think the above question has a positive answer. By the way, the graph has chromatic index equal to $2p$, that is the graph is of class $1$. Any hints?
@GregoryJ.Puleo thanks! edited the post.
Greedy coloring works here to show $2p$-choosability, I believe, and the hypothesis that $p$ is prime doesn't appear to be necessary. Write the cliques as $A = \{a_1, \ldots, a_{p+1}\}$ and $B = \{b_1, \ldots, b_{p+1}\}$, taking the notation so that $a_i$ has exactly $i-1$ neighbors in $B$ and vice versa.
First color the edges in the bigraph between $A$ and $B$; observe that each such edge is adjacent (in $L(G)$) to at most $2p-1$ previously colored edges when it is processed, thus has a color available. (Alternatively, just use Galvin's theorem for this part; then these edges only need to have lists of size $p$.)
Then color the edges $a_ia_j$ within $A$, ordering the edges so that $i + j$ is non-increasing. Observe that an edge $a_ia_j$ with $i \leq j$ has, within the clique $A$, exactly $p+1-j$ previously-colored adjacent edges at its $a_i$-endpoint and $(p+1)-i-1 = p-i$ previously-colored adjacent edges at its $a_j$-endpoint, for a total of $$2p+1-(i+j)$$ previously-colored adjacent edges within $A$. Furthermore, $a_ia_j$ has exactly $$(i-1) + (j-1) = i+j-2$$ previously-colored adjacent edges going to $B$. Thus, each edge $a_ia_j$ within $A$ is adjacent to exactly $2p-1$ previously-colored edges when it is processed, and therefore has a color available. Coloring $B$ the same way finishes the proof.
great! but I think first coloring the edges of $A$ (or $B$) and then the edges of the bipartite graph and lastly $B$ (or $A$) would also work. But, for this, I would use the paer refereed and the Galvin' stheorem.
by the way, would replacing the bipartite graph with arbitrary bipartite graph have any effect (I dont think so)? If so, then I think we could extend this method to prove edge chromatic choosability for all graphs with maximum degree$\ge\frac{n}{2}$, whre $n$ is the order of the graph
The degree constraint is essential here for arguing that each edge has few enough previously-colored adjacent edges going to $B$. In the extreme case where the bipartite graph was $K_{p+1, p+1}$, the whole graph would just be $K_{2p+2}$, and then no matter how you slice it the last edge you consider will have $2(2p+1) - 1 = 4p+1$ previously-colored adjacent edges.
ok, let us limit the degree of the bipartite graph to a maximum of $p$, then I think it should be possible,right?
I suspect there would still be far too many previously-colored adjacent edges for the late edges within $A$. Note that the last edge considered within $A$ will have $2p-1$ previously-colored adjacent edges just within $A$, and therefore couldn't afford to be incident to any edges going to $B$. I think the only way to relax the hypothesis you stated in the question and have this proof still go through is to allow vertex $a_i$ to have degree at most $i-1$ in $B$, rather than degree exactly $i-1$ (and likewise for $B$-vertices).
|
2025-03-21T14:48:31.643104
| 2020-07-29T17:16:38 |
366885
|
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|
Stack Exchange
|
Von Neumann analysis on a finite difference hyperbolic scheme
I am doing a Von Neumann analysis on a staggered finite difference scheme (for Maxwell's Equations).
The finite difference scheme is:
$$
\mathbf{u}_v|^{n+2}_{i,j} - \mathbf{u}_v|^{n}_{i,j} = - A \frac{\lambda_y}{2} \left[ (\mathbf{u}_v|^{n-1}_{i+1,j+1} - \mathbf{u}_v|^{n-1}_{i+1,j-1}) + (\mathbf{u}_v|^{n-1}_{i-1,j+1} - \mathbf{u}_v|^{n-1}_{i-1,j-1}) \right] + B \frac{\lambda_x}{2} \left[ (\mathbf{u}_v|^{n-1}_{i+1,j+1} - \mathbf{u}_v|^{n-1}_{i-1,j+1}) + (\mathbf{u}_v|^{n-1}_{i+1,j-1} - \mathbf{u}_v|^{n-1}_{i-1,j-1}) \right] \\+ \lambda_y^2 (\mathbf{u}_v|^{n}_{i,j+2} - 2\mathbf{u}_v|^{n}_{i,j} + \mathbf{u}_v|^{n}_{i,j-2}) \\+ \lambda_x^2 (\mathbf{u}_v|^{n}_{i+2,j} - 2\mathbf{u}_v|^{n}_{i,j} + \mathbf{u}_v|^{n}_{i-2,j})
$$
where the matrices $\mathbf{u}_v$, $A$ and $B$ are:
$$
A = \begin{bmatrix}
0 & 1 & 0\\
1 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix},
B = \begin{bmatrix}
0 & 0 & 1\\
0 & 0 & 0 \\
1 & 0 & 0
\end{bmatrix},
\mathbf{u}_v = \begin{bmatrix}
u_1\\
u_2 \\
u_3
\end{bmatrix}
$$
and $\lambda_x = \frac{\Delta t}{\Delta x}, \lambda_y = \frac{\Delta t}{\Delta y}$
The values of $\mathbf{u}_1|^n_{i,j}$ are computed with even values of $i,j,n$, and the values of $\mathbf{u}_2|^n_{i,j}, \mathbf{u}_3|^n_{i,j}$ are computed with the odd values of $i,j,n$. The caveat is that you need to know more values for initial conditions.
By applying the Fourier inversion formula, and assuming for simplicity that $\Delta x = \Delta y$ we get (see Strikwerda, Finite Difference Schemes and Partial Differential Equations 2ed.):
$$
G^3 + 4 \{ \lambda^2[\sin^2(\theta_x) + \sin^2(\theta_y)] - 1 \} G = 4i\lambda[A \sin(\theta_y)\cos(\theta_x) - B \sin(\theta_x)\cos(\theta_y)],
$$
where $G$ is the amplification matrix. I think I have understood properly the concept of Von Neumann analysis, and I want to derive sufficient stability conditions for $\lambda$.
The problem is that there are too many possible cases to consider for each branch of the roots of the polynomial and for each value of $\theta_x$ and $\theta_y$, and though the explicit value of $G$ may be very hard to find, some good estimates for $\lambda$ may be possible.
That doesn't look like a "leap-frog" scheme to me. The values on the right hand side ought to all be at time step $n+1$, but instead they are at steps $n-1$ and $n$.
The scheme is leap-frog because $\mathbf{u}_1|^n$ are computed with the odd values of $n$ and $\mathbf{u}_2|^n, \mathbf{u}_3|^n$ are computed with even values of $n$. I think I could have explained this better.
There, I have edited the question...
I suspect that it is still not correct. And it doesn't behave the way you claim, at least to my understanding. Additionally, what your comment describes would often be referred to as a "staggered" scheme, but not as leapfrog (which means something quite specific).
I understand. Maybe for correctness we can call this simply a staggered scheme.
|
2025-03-21T14:48:31.643545
| 2020-07-29T17:39:17 |
366888
|
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|
Stack Exchange
|
Extending Vigenère method using arbitrary function
Monoalphabetic substitution cipher consists of applying a one letter shift in each letter of plain text. So if $p = p_1 \ldots p_r$ is plain text then encrypted text is $e = q_1 \ldots q_r$ where each $q_i = p_i + k$ where $k$ is the key (the shift). This cipher is vulnerable using freqüency letters in alphabet.
The Vigenère cipher improves that using a multiletter key. Now the key is $k = k_1 \ldots k_s$ and each symbol of encrypted text is $q_i = p_i + k_i$. According to "Introduction to Modern Cryptography" by Katz and Lindell, this is vulnerable by Kasiski's method.
My question is if Kasiski's method or any other method make this vulnerable if we extend the Vigenère cipher to another cipher with any function $f$ which add to $q_i$. That is. $q_i = p_i + f(i)$. [Shift cipher has $f$ constant ($f(i) = k$) and Vigenère has $f$ periodic.]
Do you have references where I could find some information about this
I don't know if I had to post this entry in Crypto because it's about cryptography but I need academic references.
Suppose that I create a list of intgers $(a_1,a_2,a_3,\dots)$, where the $0\le a_i<26$ are chosen randomly, e.g., using quantum effects or micro-temperature changes. Then I share the list with you, and we use it for the cipher you've explained, setting $f(i)=a_i$. The net effect is that we're using a one-time pad, so that's secure. On the other hand, the function $f(i)=0$ is obviously quite insecure! So your question is really more about what functions can be detected via statistical analysis of transcripts, and as such, is probably too broad to have a satisfactory answer.
Can you give me some references about that "The net effect is that we're using a one-time pad, so that's secure"?
For being concrete, what about quadratic function: $f(x) = ax^2 + bx + c$
I don't know a reference, but isn't it obvious that it's a one-time pad? If you replace $\mathbb Z/26\mathbb Z$ with $\mathbb Z/2\mathbb Z$, then you get exactly the usual one-time pad where each bit of the message is XOR'd with the corresponding bit of the random list of bits. As for the concrete example that $f$ is an unknown quadratic polynomial, that's a great question, but comes back down to a Vigenere, since the sequence ${f(i)\bmod26:i\ge1}$ is eventually periodic. It's an example of a dynamical system over $\mathbb Z/26\mathbb Z$. So some version of Kasiski should work.
OK. I have an educated guess: if $f$ is eventually non-periodic, then cipher is not vulnerable to Kasiski's method or frequency analysis. Eg. decimal expansion of $\sqrt{p}$ with base 26 where $p$ is prime.
Sounds reasonable, although since $p$ is your secret key, it would need to be quite large. And probably you'd want to take the part of the base 26 expansion starting after the decimal point, i.e., the expansion of the $\sqrt{p}-\lfloor\sqrt p\rfloor$.
|
2025-03-21T14:48:31.643766
| 2020-07-29T17:39:58 |
366889
|
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"Martin Hairer",
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|
Stack Exchange
|
"Return map" for Brownian motion
Consider a Brownian motion $W$ reflected at the boundary of a domain $D$ in Euclidean space. I want to look at the process obtained by "restricting" it to the boundary.
I was thinking of taking the local time $L(t)$ at the boundary, and then defining my new process as $W(L^{-1}(t))$.
Is this the appropriate way to do it? Does it define a Markov process on the boundary? If so, what is known about this process for, say, a $d$-dimensional round ball (eg on its regularity/ size and frequency of jumps)?
You probably mean $W(L^{-1}(t))$ which would be a reasonable definition and would give you a Markov process on the boundary that locally looks like a Cauchy process. (In the case of $D$ a half-plane it would precisely be a Cauchy process.)
Thank you, that answers it. Fixed the error you mentioned.
This has been studied since 1960s, by Molchanov, Hsu, Kolsrud and others. If you like exact references, you may have a look at page 4 of my recent preprint at arXiv:1912.00072.
|
2025-03-21T14:48:31.643875
| 2020-07-29T17:40:20 |
366891
|
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"Ho Man-Ho",
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|
Stack Exchange
|
Serre-Swan theorem for infinite rank Frechet or Hilbert bundles?
Is there a version of Serre-Swan theorem for infinite rank Frechet or Hilbert bundles? If it existed, it would imply if $E\to X$ and $F\to X$ are infinite rank Frechet or Hilbert bundles over a smooth compact manifold $X$, then $E\cong F$ if and only if $\Gamma(X, E)\cong\Gamma(X, F)$.
(I posted this question about half an hour ago, but don't know why it disappeared. So I am posting it again.)
Edit: I just found that my post did not disappear. I first posted this one on math stackexchange but I thought I did it on mathoverflow. That's why I didn't see my first post here. Well...
(infinite-dimensional) Hilbert bundles over compact spaces are all trivializable because $U(H)$ is contractible, a theorem of Kuiper. So I suppose the answer to your question is yes, for unexciting reasons.
@MikeMiller Yes. The group U(H) is contractible, so the answer is yes. Actually, the structure group of the class of Hilbert bundles I have in mind is actually a subgroup of U(H), which is not contractible.
Which subgroup then?
@ThomasRot Actually the class infinite rank bundles I have in mind comes from the setup of the Atiyah-Singer family index theorem. Let $\pi:X\to B$ be a smooth fiber bundle with typical fiber $Z$. Let $E\to X$ be a finite rank complex vector bundle. The structure group of the infinite rank Frechet bundle $p:\pi_*E\to B$ is $\textrm{Diff}(Z, E)$, consisting of pairs $(f, \phi)$ of maps, where $\phi:Z\to Z$ and $f:E\to E$ is linear on the fibers satisfying $p\circ f=\phi\circ p$.
|
2025-03-21T14:48:31.644027
| 2020-07-29T17:52:19 |
366893
|
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"Chris Ramsey",
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|
Stack Exchange
|
Doubly-stochastic partial-isometric matrices
An $n\times n$ matrix $A$ with nonegative real entries $a_{ij}$ is said to be doubly stochastic if $\sum_{i=1}^na_{ij} = 1$,
for all $j$, and $\sum_{j=1}^na_{ij}=1$, for all $i$.
Much is known [1] about the algebraic structure of the semigroup $\Omega _n$ formed by all doubly stochastic $n\times n$ matrices. For
example, permutation matrices are the only invertible doubly stochastic matrices whose inverse is also doubly stochastic.
On the other hand [3], the idempotent elements in $\Omega _n$ are precisely the direct sums of $k\times k$ matrices of the form
$$
\pmatrix{
1/k & 1/k & \ldots & 1/k \cr
\vdots & \vdots & \ddots & \vdots\cr
1/k & 1/k & \ldots & 1/k \cr}
$$
together with their conjugates by permutation matrices.
Question: Which doubly stochastic matrices are partial isometries
(i.e. satisfy the equation $AA^tA = A$)?
See [2] for the characterization of normal, partial isometric, doubly stochastic matrices.
[1] Farahat, H. K., The semigroup of doubly-stochastic matrices, Proc. Glasg. Math. Assoc. 7, 178-183 (1966). ZBL0156.26001.
[2] Prasada Rao, P. S. S. N. V., On generalized inverses of doubly stochastic matrices, Sankhyā, Ser. A 35, 103-105 (1973). ZBL0301.15005.
[3] Sinkhorn, R., Two results concerning doubly stochastic matrices, Am. Math. Mon. 75, 632-634 (1968). ZBL0162.04205.
@ChrisRamsey But I think the equation in the question describes an isometry. I think the right equation is $(AA^t)^2=AA^t$, right?
@vidyarthi Any doubly stochastic idempotent will be a partial isometry. The OP gives a description of the idempotents and most are not permutations.
@ChrisRamsey see my answer now.
@vidyarthi, here are two results from Halmos' "A Hilbert Space Problem Book", which say that the two characterizations are equivalent: (Problem 127) A bounded linear transformation $U$ is a partial isometry if and only if $U^*U$ is a projection, and (Corollary 3) A bounded linear transformation $U$ is a partial isometry if and only if $U = UU^*U$.
The following is an attempt to validate the conclusion proposed by
@vidyarthi.
Theorem: Every
doubly-stochastic partial-isometric matrix is the product of a
permutation matrix and a doubly-stochastic projection.
Proof:
Given a doubly-stochastic partial-isometric matrix $A$, one has that $A^tA$ and $AA^t$ are doubly-stochastic
projections, so by Theorem 2 in (Sinkhorn, R., Two results concerning doubly stochastic matrices, Am. Math. Mon. 75, 632-634 (1968). ZBL0162.04205) there are permutation matrices $U$ and $V$ such that
$$
U^tA^tAU = P(k_1)\oplus P(k_2)\oplus \cdots \oplus P(k_n)
$$
and
$$
V^tAA^tV = P(l_1)\oplus P(l_2)\oplus \cdots \oplus P(l_m),
$$
where, for any integer $k$,
$$
P(k):=
\pmatrix{
1/k & 1/k & \ldots & 1/k \cr
\vdots & \vdots & \ddots & \vdots\cr
1/k & 1/k & \ldots & 1/k \cr}.
$$
Replacing $A$ with $V^tAU$, we may assume that $U$ and $V$ coincide with the identity matrix and hence $U$ and $V$ will henceforth
be ommitted.
Set $c(k)=(1/\sqrt k,1/\sqrt k,…,1/\sqrt k) ∈ \mathbb R^k$, so that $c(k)$ is a unit vector spanning the range of $P(k)$.
Moreover the range of the
projection $A^tA$ above admits an orthonormal basis formed by the vectors
$$
u_1 = c(k_1)\oplus 0_{k_2}\oplus \cdots \oplus 0_{k_n},
$$
$$
u_2 = 0_{k_1}\oplus c(k_2)\oplus \cdots \oplus 0_{k_n},
$$
$$...$$
$$
u_n = 0_{k_1}\oplus 0_{k_2}\oplus \cdots \oplus c(k_n),
$$
a similar reasoning yielding a basis $\{v_1, v_2, …, v_m\}$ for the range of $AA^t$. The initial and final
projections of a partial isometry share rank, so $n=m$, and we claim that, up to a permutation of indices $i$, one has
that $k_i=l_i$, for all $i$.
Notice that $A$ maps $\hbox{span}\{u_i\}$ isometrically onto $\hbox{span}\{v_i\}$ so, for $i\neq j$, one has that
$Au_i$ and $Au_j$ are orthogonal vectors. However these vectors have nonnegative coordinates so their support (set of
indices for nonzero coordinates) must be disjoint. By the pigeonhole principle each $u_i$ must therefore be mapped
under $A$ to a scalar
multiple of some $v_j$. By positivity and norm preservation these scalars must coincide with 1 so there is a permutation $\sigma $ such that $Au_i=v_{\sigma (i)}$, for all $i$.
Observe that, being doubly-stochastic, $A$ leaves invariant the linear functional $\Sigma$ which sums all of the coordinates of a vector. Noticing that $\Sigma(u_i)=\sqrt{k_i}$, while $\Sigma(v_j)=\sqrt{l_j}$, we deduce that $k_i=l_{\sigma (i)}$.
It is now easy to see that there exists a permutation matrix $W$ such that $Wu_i = v_{\sigma (i)}$. Letting
$$
B=W^tA,
$$
we then have that $Bu_i=u_i$, while $B^tB=A^tA$.
It follows that $B$ is a partial isometry coinciding with the identity operator on its initial space, and hence that $B$ coincides with its initial projection $B^tB$. This leads to
$$
A=WB=WB^tB=WA^tA.
$$
In the last statement, I think you concluded that each block of $A$ is $P(n)$ instead of $A=P(n)$, right?
by the way, I have modified my answer similarly. See if it helps
“... you concluded that each block of is () instead of =(), right?” Yep! This is what I meant by “Dealing with each block separately...”
Here is the desired counterexample: we have the product of $\frac1{3}\begin{pmatrix}1&1&1\1&1&1\1&1&1\end{pmatrix}$ and $\frac1{3}\begin{pmatrix}-1&2&2\2&-1&2\2&2&-1\end{pmatrix}$ to be a stochastic partial isometric matrix $\frac1{3}\begin{pmatrix}1&1&1\1&1&1\1&1&1\end{pmatrix}$. Hence, your argument (as well as mine before) fails. Modified my answer
@vidyarthi, If I understand it right you are claiming that the third matrix in your message is a counter-example. If so I don't think I agree because this matrix is itself a doubly stochastic projection.
No, I meant the second of the matrix has negative entries, and hence is not a permutation matrix, and not stochastic, so your fist statement is thus wrong
@vidyarthi, I understand you have shown one way to decompose the 3rd matrix which does not filfill the conditions in my statement, but I am sure you can find another decomposition which works!
ok, but then I think you should change your answer to just " Every doubly stochastic matrix can be written as a product of a doubly stochastic projection with a permutation matrix" instead of "is"
@vidyarthi, I'd say most people would take "is" as a synonym to "can be written as". For example I believe that it is standard to say that every positive integer "is" the product of finitely many primes, even though often an integer may also be written as a product of numbers that are not prime.
so shall I delete my answer?
In my opinion your argument does not prove the result. On the other hand deleting your answer is entirely up to you.
no, since you refer to me in your answer, so I thought of asking you before deleting
@vidyarthi, If your question is whether deleting your post would adversely affect my answer, the answer is no.
From here, we have that a square matrix is a partial isometry if and only if it is of the form $A=UD=EU$, where $D, E$ are idempotent and $U$ is unitary. Translating this to our case, we have that a doubly stochatic matrix is a partial isometry when it is a product of an orthogonal matrix (scaled by a scalar) with an idempotent matrix (again scaled by appropriate scalar) both of whose row and column sums equals $1$. The rank of the matrix equals that of the unscaled idempotent matrix.
To further elaborate as to why the unscaled matrices $E,D,U$ have the said property, suppose $A=EU$, where $E$ is idempotent and $U$ be unitary (orthogonal), we obtain, for the eigenvector $v=(1\ 1\ 1\ldots\ 1)^t$ of $A$, we get that $Av=v\implies EUv=v=E^2Uv=E(EUv)=Ev$, thereby showing that $v$ is an eigenvector of $E$ with eigenvalue $1$, thereby clearly implying $E$ has row sum of each row equal to $1$. A similar reasoning with the transpose of $A$ shows that $E$ should also have each column sum equal to $1$, since $E^t$ is also idempotent. Now, using that $A=UD$, and the vector $v^t$ as a left eigenvector , we obtain that both $D$ and $U$ also have both row and column sum equal to $1$.
I do not see why $U$, $D$, and $E$ must be doubly stochastic. But the guess is certainly very interesting!
@Ruy yes, they need not be. Added the scaling factors
Now I should say that I do not see why , , and must be scalar multiples of doubly stochastic matrices.
I think I can prove that your conclusion is corrrect, but the proof is a bit envolving. Will try to write it down soon.
I think this is not yet correct. The matrix
$
E={1\over30}\pmatrix{ 25& -5& 10 \cr
-5& 25& 10 \cr
10& 10& 10}
$
is idempotent and has $(1,1,1)$ as a fixed point and yet it is not doubly-stochastic.
@Ruy as I said, the discussion in the second discussion applies to unscaled matrices, that is, the row and column sums are same for the matrix. The matrix you gave as an example is definetely doubly stochastic. The row and column sums are all $1$
Stochastic matrices must have nonnegative entries (which often represent transition probabilities).
@Ruy thanks. edited the post. see now
Let us continue this discussion in chat.
|
2025-03-21T14:48:31.644743
| 2020-07-29T18:21:30 |
366894
|
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|
Stack Exchange
|
A Rademacher ‘root 7’ anti-concentration inequality
Let $r_1,r_2,r_3,\dotsc$ be an IID sequence of Rademacher random variables, so that $\mathbb P(r_n=\pm1)=1/2$, and $a_1,a_2,\dotsc$ be a real sequence with $\sum_na_n^2=1$. For $S=\sum_na_nr_n$, does the following inequality always hold?
$$
\mathbb P\left(\lvert S\rvert\ge1/\sqrt7\right)\ge1/2.\tag{*}\label{star}
$$
Is this a known result, or conjecture?
Stated in another way, for each finite sequence $a_1,a_2,\dotsc,a_N$ with $\sum_na_n^2=1$, if we look at all $2^N$ sums $\pm a_1\pm a_2\pm a_3\pm\dotsb\pm a_N$, do we always have at least as many with absolute value at least $1/\sqrt7$ as with absolute value strictly less than $1/\sqrt7$?
Background
The statement is very similar to a conjecture of Tomaszewski, that at least as many of the $2^N$ sums have absolute value less than or equal to 1 as have value greater than one. This conjecture is equivalent to the concentration inequality $\mathbb P(\lvert S\rvert\le1)\ge1/2$, and is still unproven, other than possibly in a (not currently peer reviewed, as far as I know) paper Keller and Klein - Proof of Tomaszewski's conjecture on randomly signed sums posted on the arXiv recently.
We can also consider more general anti-concentration inequalities, which bound $\mathbb P(\lvert S\rvert \ge x)$ from below. For $x=0$ the lower bound is trivially 1 and, for $x > 1$, it is $0$. For $0 < x < 1$, then, as in a previous answer of mine, the Paley–Zygmund inequality can be used to obtain
$$
\mathbb P(\lvert S\rvert \ge x) \ge (1-x^2)^2/3,
$$
but this is far from optimal. Also, for $x=1$, there does exist a strictly positive anti-concentration bound, as shown in the answers to another question of mine, An $L^0$ Khintchine inequality. It was shown by Oleszkiewicz, in 1996, that a (non-optimal) lower bound of $1/10$ applies when $x=1$ (in On the Stein property of Rademacher sequences) and also conjectured that the optimal bound is $7/32$, but this is still open as far as I am aware.
Running a Monte Carlo simulation to randomly pick the values of $a$ and compute the probability suggests that the optimal lower bound for $\mathbb P(\lvert S\rvert\ge x)$ is piecewise constant in $x$, so that there is only a small set of $x$ values at which the bound changes (specifically, $x=0,1/\sqrt7,1/\sqrt5,1/\sqrt3,2/\sqrt6,1$). As mentioned, for $x=1$ it is an open conjecture, so my question here is regarding the smallest non-trivial value for $x$.
Note that \eqref{star} is best possible, in the sense that it does not hold if either of the inequalities are strict and, consequently, does not hold if either the $1/\sqrt7$ inside the probability or the $1/2$ outside is increased. Considering $a_n=1/\sqrt7$ for $n\le7$, we obtain $\mathbb P(\lvert S\rvert > 1/\sqrt7)=29/64$, and considering $a_n=1/\sqrt2$ for $n\le 2$ gives
$$
\mathbb P(\lvert S\rvert\ge1/\sqrt7)=\mathbb P(\lvert S\rvert > 0)=1/2.
$$
This example also shows that for any $x \gt 0$, the anti-concentration bound can never be better than 1/2, so \eqref{star} is also optimal in this sense, and it would immediately follow that we have optimal inequalities
$$
\mathbb P(\lvert S\rvert\ge x)\ge1/2 \tag{**}\label{starstar}
$$
for all $0 \lt x \le 1/\sqrt7$. In fact, it is not difficult to prove \eqref{starstar} for $x\le0.693/\sqrt7$ as follows: Let $\lVert a\rVert_\infty=\max_n\lvert a_n\rvert$. We split into two cases,
$\lVert a\rVert_\infty\ge x$. In this case, choose $n$ such that $\lvert a_n\rvert\ge x$. Flipping the sign of $r_n$ does not affect the distribution of $S$, but if $\lvert S\rvert < x$ then it changes its value such that $\lvert S\rvert\ge x$, so that $\mathbb P(\lvert S\rvert\ge x)\ge\mathbb P(\lvert S\rvert \lt x)$, giving the result.
$\lVert a\rVert_\infty < x$. In this case, if we let $\Phi(x)$ be the standard normal cumulatiive probability function, the Berry–Esseen theorem gives
$$
\mathbb P(\lvert S\rvert\ge x)\ge 2\Phi(-x)-2C\lVert a\rVert_\infty \gt 2\Phi(-x)-2Cx
$$
for a global constant $C$. We can use $C=0.56$ (as stated in the Wikipedia page, this was proven by Shevstova in 2010, in An improvement of convergence rate estimates in the Lyapunov theorem). Evaluating the right hand side with $x=0.693/\sqrt7$ gives a value greater than 1/2. QED
Finally, I mention that I have confirmed \eqref{star} numerically on a dense grid of values for $a$ and, by bounding the interpolation errors, should in principle lead to a (rather unsatisfactory) proof.
What am I missing? If $a_1=\ldots=a_7=1/\sqrt 7$, isn't the sum at least $1/\sqrt 7$ in absolute value with probability 1?
Yes, and $1\ge1/2$, so that example satisfies the inequality.
Sorry - I thought you were saying the $(1/\sqrt 7,\ldots,1/\sqrt 7)$ example showed both inequalities are sharp, but in fact, it only shows you can't change the $1/\sqrt 7$ in the probability.(while the $1/\sqrt 2$ example shows you can't improve the 1/2).
@Anthony: I edited to clarify
Slightly unrelated but it seems that this (https://mathoverflow.net/questions/187938/lower-bound-for-prx-geq-ex/188087#188087) answer of Fedja avoids using a Barry Essen or Paley Zigmond approach for a similar problem so there might be some useful ideas there
Addressed in Theorem 1.3 in Dvořák and Klein - Probability mass of Rademacher sums beyond one standard deviation (not yet peer reviewed). It describes a computer program that verifies $\Pr[\lvert S\rvert \geq 1/\sqrt{7} - \epsilon] \geq 1/2$, with concrete $\epsilon > 0$. Giving it more time (polynomial in $1/\epsilon$), it empirically seems we may take $\epsilon\to 0$.
Apologizes for the unsatisfactory answer :)
Now there is a complete proof due to Lawrence and Julien (yet to be published): https://arxiv.org/pdf/2306.07811.pdf
|
2025-03-21T14:48:31.645168
| 2020-07-29T18:49:39 |
366896
|
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|
Stack Exchange
|
Resolving singularities in one fell swoop
From what I understand, resolution of singularities (in characteristic 0) is proved and implemented inductively. You repeatedly blow up your variety along subsets of your singular locus in a way that decreases the "severity" of your singular locus.
At the end of "On the problem of resolution of singularities in positive characteristic" (link), Hauser says:
"From Hironaka’s theorem it follows (at least in characteristic zero) that there does exist another ideal structure on the singular locus of a variety so that the induced blowup with this center resolves the singularities in one single stroke. Formidable!"
So in contrast to the usual inductive way of resolving singularities, there is a way of resolving singularities "all at once."
Question: What does Hauser mean by "another ideal structure" on the singular locus?
The following was my initial thought. If $k$ is a field of characteristic 0 and $X$ is a variety over $k$ with singular locus $Z$, we want some ideal sheaf $\mathcal{I}$ such that $\mathcal{O}_X/\mathcal{I}$ has support $Z$, and such that blowing up $X$ at $\mathcal{I}$ is smooth. Actually constructing $\mathcal{I}$ is presumably difficult, since it should compensate for any difficulties that arise from resolving $Z$.
However, blow-ups are determined by their center, so this initial thought isn't right. For example, there are isolated singular points that require multiple blow-ups to be resolved. How could such a singular point ever be resolved by a single blow-up? Perhaps there's a larger center containing the singular locus that one should blow up?
Edit: As Donu points out, if a singular locus can be resolved, it can be resolved by a single blow-up for entirely formal reasons. So the locus cut out by an ideal that resolves our singular locus in one fell swoop should contain, but need not be equal to, our singular locus. This answers the previous paragraph.
The question then becomes whether the singular locus determines the "resolving ideal" in any tractable way.
There is more to this than is at first apparent. Consider a small resolution of an ordinary threefold double point. This is not the blowing up of any ideal sheaf that is supported on the double point for the simple reason that there is no Cartier divisor that is supported on the (codimension 2) exceptional locus. It is the blowing up of an ideal sheaf, but not of an ideal sheaf supported on the singular locus.
The reasons for this are rather formal. Any projective birational morphism of varieties is the blow up of some ideal, see chap II, theorem 7.17, of Hartshorne. So although the statement you quote sounds striking, I don't think it is a terribly useful.
Let me add a few more remarks, even though it won't answer your modified question. Blowing up a complicated ideal or closed subscheme would hard understand geometrically. What makes Hironaka, and various other resolution proofs, useful is not just that the resolution exists, but that it can be achieved by blowing up along a succession of smooth centres.
Thanks for the help! So in general, this ideal has to cut out more than just the singular locus. I guess the question then becomes whether the singular locus determines such an ideal in any tractable way.
Why should that ideal sheaf be supported on the singular scheme as Hauser (rather curiously!) asserts?
Hi Jason, if you're directing the question to me, then I don't see why either. I'm just saying it's the blow of some ideal.
|
2025-03-21T14:48:31.645449
| 2020-07-29T19:32:45 |
366900
|
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|
Stack Exchange
|
When is it true that if $G$ is isomorphic to a spanning subgraph of $H$ and vice versa, then $G$ is isomorphic to $H$?
When is it true that if $G$ is isomorphic to a spanning subgraph of $H$ and $H$ is isomorphic to a spanning subgraph of $G$, then $G$ is isomorphic to $H$?
Clearly this is true if $G$ and $H$ are finite graphs; however, this is not necessarily true for infinite graphs. For instance, let $G$ be an infinite clique together with infinitely many isolated vertices and let $H$ be two disjoint infinite cliques together with infinitely many isolated vertices. We posed this problem in Corsten, DeBiasio, and McKenney - Density of monochromatic infinite subgraphs II (see Problem 2.12), but since this question seems more basic and is only tangential to our results, I figured I would ask here as well.
Addendum 1: After doing some more digging, I found this related post Non-isomorphic graphs with bijective graph homomorphisms in both directions between them which just asked for examples of such graphs $G$ and $H$ where $G$ and $H$ are not isomorphic.
Addendum 2: There was a comment yesterday, which for some reason seems to have been deleted, suggesting the term "co-hopfian graph." I found this paper Cain and Maltcev - Hopfian and co-hopfian subsemigroups and extensions which defines co-hopfian graphs (see the paragraph before Lemma 4.5) as those in which every injective homomorphism from $G$ to $G$ (i.e. injective endomorphism) is an isomorphism. It's unclear to me if this makes a difference in the characterization, but I now believe my question is equivalent to "Which graphs $G$ have the property that every bijective homomorphism from $G$ to $G$ (i.e. bijective endomorphism) is an automorphism." Sorry to overdo it, but my original question has now become three questions:
Which graphs $G$ have the property that every injective endomorphism is an automorphism? (equivalently, when is it true that if $G$ is isomorphic to a subgraph of $H$ and $H$ is isomorphic to a subgraph of $G$, then $G$ is isomorphic to $H$?)
Which graphs $G$ have the property that every bijective endomorphism is an automorphism? (equivalently, when is it true that if $G$ is isomorphic to a spanning subgraph of $H$ and $H$ is isomorphic to a spanning subgraph of $G$, then $G$ is isomorphic to $H$?)
Are the answers to 1 and 2 the same?
It's reasonably easy to find locally finite (but still disconnected) examples via a similar trick. For example if you take $G$ to be the collection of all odd length paths, together with countably many isolated vertices and $H$ to be the collection of all even length paths together with countably many isolated vertices, then they are both isomorphic to a spanning subgraph of the other.
The answers to 1 and 2 are not the same: A ray has injective endomorphisms which are not automorphisms (e.g. map every vertex to its successor), but the only bijective endomorphism is the identity.
This is an extended comment rather than an actual answer.
I think that any answer to your questions 1 and 2 is likely to be rather involved since the properties you ask are sensitive to small local changes in the graph. To illustrate what I mean, consider the following two graphs:
$G_1$ has vertices $u_n, v_n$ for $n \in \mathbb Z$, and edges $u_nu_{n+1}$ and $u_{2n}v_n$ for $n \in \mathbb Z$, and $u_{2n+1}v_n$ for $n \in \mathbb N$.
$G_2$ is obtained from $G_1$ by removing $v_0$.
Although these graphs are almost the same, they differ in terms of the properties you are asking:
$G_1$ has a bijective endomorphism which is not an isomorphism (defined by $u_n \mapsto u_{n+2}$ and $v_n \mapsto v_{n+1}$).
On the other hand, the only injective homomorphism from $G_2$ to itself is the identity: note that $v_0$ is a vertex of degree $2$ whose neighbours also have degree $\leq 2$, thus $v_0$ must be mapped to itself under any injective homomorphism. From there it is not hard to inductively show that the identity is the only injective endomorphism.
|
2025-03-21T14:48:31.645722
| 2020-07-29T20:03:58 |
366902
|
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|
Stack Exchange
|
Explicit resolution of $\Omega^1_C$ for prestable curve $C$
Suppose $C$ is a complex projective curve (or a compact $1$-dimensional connected reduced complex space). If $C$ is smooth, then its module of differentials $\Omega^1_C$ is locally free. If $C$ is a singular prestable curve (i.e. its singularities are at worst ordinary nodes), then $\Omega^1_C$ is coherent, but not a locally free sheaf.
Can we find a canonical finite complex of vector bundles $E^\bullet$ on $C$ resolving $\Omega^1_C$?
It's easy to find such a canonical resolution locally near a node. Let $p\in C$ be a node and assume that the two branches meeting at $C$ are $A,B$. Then, we may embed $C$ locally into $A\times B$ as $A\times\{p\}\cup\{p\}\times B$. This is the zero locus of a tautologically defined section $s$ of the bundle $L =\mathcal O_A(p)\boxtimes\mathcal O_B(p)$. Then, the conormal sequence $0\to L^\vee|_C\to\Omega^1_{A\times B}|_C\to\Omega^1_C\to0$ is a (local) resolution by a two term complex of vector bundles. I don't see any obvious way to globalize this construction though.
My main motivation is to set up a Dolbeault type complex to better understand the vector space $\text{Ext}^1(\Omega^1_C,\mathcal O_C)$ which is the space of first order deformations of $C$. More generally, I also want to use this type of Dolbeault complex to understand the deformations of (possibly nodal) pseudo-holomorphic maps in almost complex manifolds.
Here is a first thought. If it had a finite resolution as you write (canonical or otherwise), if $C$ is irreducible and sheaf of differential forms is torsion free, then it would be locally free by Auslanser-Buchsbaum. This will solve a long standing conjecture known as Berger's conjecture. So, the answer is most likely to be unknown in your case.
Embed the prestable curve $C$ as a Cartier divisor in a surface $S$. Then the resolution is $\mathcal{O}_S(-\underline{C})|_C \to\Omega_S|_C$.
... "surface" --> "smooth surface".
Maybe this is obvious, but is it clear that we can always embed into a smooth surface? Also, this is not canonical, but is there some explicit constructive way to get such an $S$ and an embedding?
I suppose we could choose a 1 (complex) parameter deformation of C which resolves all the nodes simultaneously (xy = 0 --> xy = t for small t) and then the total space of this deformation could serve as $S$.
If $C$ is prestable then its dualizing sheaf is a line bundle; if $C$ is stable this bundle is ample so you can use it to embed in projective space. Now project from a well chosen linear space to obtain an immersion in $\mathbb P^2$, and blow up $\mathbb P^2$ at those singularities which arose from projection (as opposed to those which are part of the abstract pre-stable curve). Now you have an embedding of $C$ in a rational surface. How is that?
Yes, this is pretty explicit. In the unstable case, I think we can just stabilize by adding marked points $p_1,\ldots,p_n$ and then take the ample line bundle $\omega_C(p_1 + \cdots + p_n)$ instead of $\omega_C$.
@MohanSwaminathan yes, that works. Stability is equivalent to ampleness of the log dualizing sheaf.
|
2025-03-21T14:48:31.645976
| 2020-07-29T21:49:06 |
366907
|
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|
Stack Exchange
|
A question about chaining Vinogradov notation
This is not a research question, but I hope it is still legitimate to ask for this platform. Suppose $A(x)$, $B(x)$, $C(x)$, $D(x)$ are positive-valued functions of $x$, and
$A(x) \ll B(x)$ and $ C(x) \ll D(x)$, where $ \ll$ is the Vinogradov symbol. Suppose, further, that $B(x) \le C(x)$. There are two ways of expressing these chain of "inequalities" in one line:
$$ A(x) \ll B(x) \le C(x) \ll D(x) $$
and
$$ A(x) \ll B(x) \ll C(x) \ll D(x) $$
The first has the advantage of being more precise, while at the same time could be misreading, in that the implied constant in $A(x) \ll C(x)$ might not be taken to be $1$. Is there a convention for writing such chains of "inequalities"? If so, which one of the two formulations is preferable?
At the risk of being pedantic, the preferable notation is the one that cannot be misinterpreted by the reader. In this case, I'd use $$A(x)\ll B(x)\quad\text{and}\quad B(x)\le C(x)\quad\text{and}\quad C(x)\ll D(x).$$ While maybe a little more to write, it has the advantage of being unambiguous.
You refer to the implied constant in $A(x) \ll C(x)$, but I think you mean the implied constant in $B(x) \ll C(x)$.
What I meant was that a reader may simply drop $B(x) $ and deduces that $A(x) \le C(s)$, which, of course, does not have to be the case. I understand that this would be highly unlikely, but I am trying to be on the safe side by considering all possibilities.
|
2025-03-21T14:48:31.646124
| 2020-07-30T00:09:29 |
366910
|
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|
Stack Exchange
|
Existence results for Lagrangian solutions to the Incompressible Euler Equation?
It is known that if a function (which we shall call the lagrangian flow, or lagrangian trajectory) $$X:(\mathbb{R}/\mathbb{Z})^3 \times [0,T] \to \mathbb{R}^3$$ with $X \in H^1_t$ (i.e. has weak time derivatives $v=\partial_tX \in L^2((\mathbb{R}/\mathbb{Z})^3\times [0,T])$) minimises the energy lagrangian $$L[X]=\int_{(\mathbb{R}/\mathbb{Z})^3 \times [0,T]} v^2$$ subject to $X(\cdot,0),X(\cdot,T)$ fixed, and $X(\cdot,t)$ being a $\textbf{volume preserving}$ diffeomorphism of $(\mathbb{R}/\mathbb{Z})^3$ for each $t$,
then $u(\cdot,t)=v(\cdot,t) \circ X(\cdot,t)^{-1}$ is a weak solution to the incompressible Euler equation $$\dot{u} + \nabla \cdot (uu) = \nabla p, \nabla \cdot u = 0$$ in a distributional sense.
My question is about existence results once $X(\cdot,0),X(T,\cdot)$ are fixed, and depends on what we mean by diffeomorphisms:
if we require $X(0,t)$ to be $C^{\infty}$ volume preserving diffeomorphisms for each $t$, can we guarantee the existence of a minimiser?
if we require $X(0,t)$ to be $C^1$ volume preserving diffeomorphisms for each $t$, again, can we guarantee the existence of a minimser?
I expect both of the above answers are no or at least unknown, since I've found no mention of them in the literature I've read. The case I'm therefore really interested in is when by 'diffeomorphism' we only require that $$X(\cdot,t) \text{ is bijective, and measure preserving}$$ Correct me if I'm wrong but the above theory still holds even in this loose definition. Can we now, i.e. in the loosest sense, get existence of minimisers? Perhaps with some conditions on $X(\cdot,0)$ and $X(\cdot,T)$? Anything known, perhaps some papers, about this case would be appreciated.
I suggest readong the following three classical references: [Shnirelman (1997). On the nonuniqueness of weak solution of the Euler equation, CPAM, 50(12)] for the nonexistence of minimizers, [Brenier (1989) The least action principle and the related concept of generalized flows for incompressible perfect fluids. Journal AMS 2(2)] for existence of generalized solutions, and finally [Ebin Marsden (1970) Groups of diffeomorphisms and the notion of an incompressible fluid. Ann. of Math. (2) 92] for existence of regular solutions with endpoints close to each other.
|
2025-03-21T14:48:31.646298
| 2020-07-30T03:44:14 |
366912
|
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Stack Exchange
|
Maximum principle for an elliptic like operator
I am trying to prove some monotonicity of a solution of a given pde; after considering a quantity like $ \phi(x) = x \cdot \nabla v(x)$ ($v$ is the solution of a given pde) I arrive at something along the lines of
$$-\Delta \phi(x)+ \phi(x) + 2 \int_0^1 \frac{ \phi(tx)}{t} dt = f(x) \ge 0 \qquad B_1$$ with $ \phi=0$ on $ \partial B_1$ where $B_1$ is the unit ball in $ R^N$. Also assume some smoothness in $v$ and hence this integrand is somewhat behaved. So my question is after making various assumptions on smoothness and.... are we able to prove a maximum principle? ie. can we show $ \phi \ge 0$ in $B_1$?
A first observation is that this nonlocality is kind of singular, not at all the standard form of a convolutional kernel. A second observation is that the answer is liekely no: indeed you already have $\phi(0)=0$, so the strong maximum principle cannot hold. Of course one cannot rule out a very special situation where the weak MP holds but not the strong one. But I doubt it.
that is a good point... i didn't think of that. thanks
(For an actual answer, see the edit below.)
Let $\phi$ be smooth near zero and non-negative. Suppose that the Taylor expansion of $\phi$ at zero is non-trivial, and let $P(x)$ be the leading term. Then $P(x)$ is a non-negative homogeneous polynomial of degree $2 k \geqslant 2$. Then $-\Delta P$ is a homogeneous polynomial of degree $2k - 2$ which is not everywhere positive (it is negative along the line $x_0 \mathbf R$ where $x_0$ is a minimum of $P$ over the unit sphere), while both $\phi$ and the integral term are homogeneous of degree $2k$. Since $-\Delta P(x)$ is the leading term in the expansion of $f(x)$, it follows that $f(x)$ cannot be positive in a neighbourhood of the origin. Therefore, if $f \geqslant 0$, then $\phi$ is necessarily not everywhere positive!
I am not sure what happens if $P(x)$ has zero Taylor expansion near zero, though. I bet the answer is similar, but I fail to see an straightforward argument.
Edit: The following seems to be a complete solution, although I did not check it carefully.
Suppose that $\phi$ is smooth in the unit ball, non-negative, and satisfies the integro-differential equation in question. Consider the symmetrisation $\phi^\star$ of $\phi$:
$$ \phi^\star(x) = \frac{1}{|x|^{N-1}} \int_{\partial B} \phi(|x| u) du = \int_{SO(N)} \phi(O x) dO $$
(both integrals with respect to normalised measures). Then $\phi^\star$ is a rotationally invariant solution of the equation in question, with $f$ replaced by its symmetrisation $f^\star$. We will prove that $\phi^\star$ is identically zero, so that $\phi$ is identically zero, too.
So the problem becomes one-dimensional: if $\phi^\star(x) = \psi(|x|)$ and $f^\star(x) = g(|x|)$, then we have $\psi \ge 0$ on $(0, 1)$, $\psi(0) = \psi'(0) = \psi(1) = 0$ and
$$ -\psi''(r) - \frac{N-1}{r} \psi'(r) + \psi(r) + 2 \int_0^1 \frac{\psi(r t)}{t} \, dt = g(r) \geqslant 0 . $$
Observe that $\psi''(r) + \tfrac{N-1}{r} \psi'(r) = r^{1-N} (r^{N-1} \psi'(r))'$. Thus,
$$ (r^{N-1} \psi'(r))' \leqslant r^{N-1} \psi(r) + 2 \int_0^r r^{N-1} \, \frac{\psi(s)}{s} \, ds . $$
Integrating both sides, we get
$$ r^{N-1} \psi'(r) \leqslant \int_0^r s^{N-1} \psi(s) ds + 2 \int_0^r \frac{r^N - s^N}{N} \, \frac{\psi(s)}{s} \, ds , $$
and integrating both sides again (assuming that $N \ne 2$, which requires as slightly different argument),
$$ \begin{aligned} \psi(r) & \leqslant \int_0^r \frac{s^2 (1 - (s / r)^{N - 2})}{N - 2} \, \frac{\psi(s)}{s} \, ds \\ & + 2 \int_0^r \frac{r^{N + 1} - s^{N + 1} - (N + 1) s^N (r - s)}{N (N + 1)} \, \frac{\psi(s)}{s} \, ds . \end{aligned} $$
In particular,
$$ \psi(r) \leqslant C r^2 \int_0^r \frac{\psi(s)}{s} \, ds $$
for some constant $C$. Gronwall's inequality applied to $\psi(r) / r$ implies that $\psi(r) \le 0$, and hence $\psi(r) = 0$, as claimed.
it will take me some time to digest your answer; thank you very much.
You're welcome. I may have messed something up, please do ask me if you find anything unclear.
|
2025-03-21T14:48:31.646594
| 2020-07-30T04:11:00 |
366913
|
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|
Stack Exchange
|
automorphisms of an étale cover of a curve
The base field is algebraically closed and of chatacteristic zero. If $X$ is a smooth projective curve and $Y\to X$ is an étale covering of $X$ of degree $d$, then what can we say about the automorphism group $\mathrm{Aut}(Y/X)$ of $Y$ over $X$? Is it always nontrivial?
By etale covering do you mean finite etale map? If so, this is just Galois theory. The map corresponds to a finite index index $d$ subgroup $H\le\pi_1(X)$, and the automorphism group is just the automorphism group of the corresponding $\pi_1(X)$-set. IIRC this should just be the quotient of the normalizer of $H$ by $H$. Sometimes it'll be trivial. For example you can fix a surjection $f : \pi_1(X)\rightarrow S_3$ (symmetric group on 3 things), and let $H$ be the preimage of an order 2 subgroup of $S_3$. The resulting cover will have trivial aut group for the same reason that $Q(\sqrt{3})$ does
I mean finite étale maps. I am confused that there may exist a nontrivial finite separable extension $F$ of $K(X)$ with trivial automorphism group $\mathrm{Aut}(F/K(X))$ and $F$ corresponds to a curve $Y$ by $Y=F(X)$.
What is $F(X)$?
I just checked your additional comments and try to make the picture clear. $K$ means the base field, and my comments comes from the correspondences between 1. smooth algebraic curves and its function field $X\leftrightarrow K(X)$, 2. finite etale morphism $Y\to X$ and finite separable field extension $K(Y)/K(X)$, and 3. the automorphism groups $\mathrm{Aut}(Y/X)\leftrightarrow \mathrm{Aut}(K(Y)/K(X))$.
Could explain what is the corresponding $\pi_1(X)$-set? How can say that the automorphism is trivial in your example?
A $\pi_1(X)$-set is a set (finite in our case) equipped with an action by the group $\pi_1(X)$. By the Galois correspondence, if you fix a geometric point $x\in X$, taking fibers gives an equivalence of categories between finite etale covers of $X$ and finite $\pi_1(X,x)$-sets. So any question internal to the category of finite etale covers of $X$ (e.g., existence of automorphisms), can be translated into a combinatorial problem involving sets with group-action.
|
2025-03-21T14:48:31.646763
| 2020-07-30T05:01:02 |
366914
|
{
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"Alapan Das",
"Lucia",
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"url": "https://mathoverflow.net/questions/366914"
}
|
Stack Exchange
|
Looking for a nontrivial upper bound to the function $f(n) = \sum_{d\mid n}\phi(d)/d$
For a natural number $n$, let $f(n) := \sum_{d\mid n}\frac{\phi(d)}{d}$, where $\phi$ is the Euler-phi function. I believe this is a multiplicative function. I'm looking for a nontrivial (sublinear) upper bound for it - ie, a reasonably simple and explicit monotonically increasing function $B(n)$ such that
$f(n)\le B(n)$ for all $n\ge 1$, and moreover
$\lim_{n\rightarrow\infty}\frac{B(n)}{n} = 0$
This function came up in my research. I can see that the worst case scenario happens when $n$ has many distinct prime divisors, though I'm not an analytic number theorist it's not obvious how to get a nontrivial upper bound to me. In any case, this is a simple enough looking function that I assume there must be results on this somewhere (which are presumably better than anything I could come up with given some time).
I'm particularly interested in an upper bound to the function $f$ when it's restricted to numbers of the form $p\pm 1$, where $p$ is a prime, in case this particular case has been studied.
Since $\phi(d)\le d$, clearly $f(n) \le d(n)$ where $d(n)$ denotes the number of divisors of $n$. Pairing a divisor $d$ with $n/d$, and noting that at least one of $d$ or $n/d$ must be $\le \sqrt{n}$, one sees easily that $d(n) \le 2\sqrt{n}$. Of course much more be said (e.g. $d(n) \le C(\epsilon) n^{\epsilon}$ for any $\epsilon >0$), but this may be enough for you. Look up bounds for divisor function in (just about) any elementary number theory book.
This is a multiplicative function; on prime powers, it satisfies $f(p^r) = \sum_{j = 0}^{r} \phi(p^j) p^{-j} = \sum_{j = 0}^{r} (1 - p^{-1}) = (1 - p^{-1}) (r + 1)$ since $\phi(p^j) = p^j (1 - p^{-1})$. So $f(n) = \frac{\phi(n)}{n} \tau(n)$, where $\tau(n) = # {d \mid n}$ denotes the divisor function. Of course $\frac{\phi(n)}{n} \leq 1$, while upper bounds for the divisor function are well-known: https://en.wikipedia.org/wiki/Divisor_function. Note that these are "worst case" bounds, which are at primorials $n = p_1 p_2 \cdots p_m$, which cannot be of the form $p \pm 1$.
If $n=\prod_{i=1}^{r} p_i^{a_i}$, then $f(n)=\prod_{i=1}^{r}(\sigma_i+1)$, where $\sigma_i=(1-\frac{1}{p_i})a_i$.
@Lucia Ah, great! That's an excellent observation. If you'd like to put that as an answer I'd be happy to mark this question answered. Thanks!
Also, $ln(n)+\sum_{i=1}^{N} ln(p_i)= \sum_{i=1}^{N} (\sigma_i+1)ln(p_i)(\frac{p_i}{p_i-1})$. Say, $\sum_{i=1}^{N} ln(p_i)(\frac{p_i}{p_i-1})=\chi$ Hence, $f(n) \leq \left(1+\frac{ln(n)}{\chi}\right)^{\chi}$. Here, $N$ can be chosen as big as possible. $p_i := i$th prime number.
Now, the function $T(x)=(1+\frac{c}{x})^x$ is minimum when $x≈\frac{c}{e-1}$. So, from the bound of $f(n)$, we can choose primes (or the limit $N$, the prime factors of $n$ must be chosen) such that $\chi≈\frac{ln(n)}{e-1}$. Hence, $f(n)≤e^{\frac{ln(n)}{e-1}}=n^{\frac{1}{e-1}}$. So, $\lim \limits_{n \to \infty} \frac{f(n)}{n} ≤ \frac{1}{n^{\frac{e-2}{e-1}}}=0$.
|
2025-03-21T14:48:31.647081
| 2020-07-30T05:14:49 |
366915
|
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|
Stack Exchange
|
Is this inequality of the random matrices correct?
I need to confirm whether the following inequality correct.
Let $\xi_i\in\{\pm 1\}$ be independent random signs, and let
$A_1,\ldots, A_n$ be $m\times m$ Hermitian matrices. Let $\sigma^2 = \|\sum_{i=1}^n Var[\xi_i]A_i^2\|$. Then
$$Pr\bigg(\bigg\|\sum_{i=1}^n\mathbb{E}[\xi_i]A_i-\sum_{i=1}^n\xi_iA_i\bigg\|\geq
t\bigg)\leq2m\exp(-t^2/2\sigma^2).$$
It is said to be cited from the paper "User-Friendly Tail Bounds for Sums of Random
Matrices
". I can only find Theorem 1.5, which is only for normal distribution or equal probability for ${-1,+1}$ .
Is the inequality correct?
Note that the inequality is slightly different from a previous post. The inequality in that post is false accroding to a comment.
|
2025-03-21T14:48:31.647158
| 2020-07-30T05:58:32 |
366916
|
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|
Stack Exchange
|
Hilbert scheme of Lefschetz fibration
Let $\pi:E\to\mathbb{C}$ be a Lefschetz fibration where $E_0$ is the singular fiber over $0\in\mathbb{C}$. In the case when $E$ is a smooth complex surface, Donaldson & Smith proved that, the relative Hilbert scheme ($n$-points on the same fiber) $Hilb^n(E,\pi)$ is still smooth and $\pi^{[n]}:Hilb^n(E,\pi)\to\mathbb{C}$ is a nodal degeneration with the singular fiber over $0\in\mathbb{C}$.
I wonder if there is any extension to $Hilb^2(E,\pi)$ for $E$ of higher dimension (here $n=2$ since $Hilb^2(X)$ is always smooth for smooth $X$ in any dimension). If there is not, could anyone show some counterexample?
|
2025-03-21T14:48:31.647237
| 2020-07-30T06:03:45 |
366917
|
{
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"authors": [
"András Bátkai",
"Brendan McKay",
"Fedor Petrov",
"Josiah Park",
"fedja",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/366917"
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|
Stack Exchange
|
Log-concavity inequality
Let $x,y,$ and $t$ be fixed real numbers, $1<x<y$, $0<t<1$. Does the following inequality hold for some $c$
$$\frac{\log{(tx+(1-t)y)}}{\log^t{x}\log^{(1-t)}{y}}>\frac{\log{(sw+(1-s)z)}}{\log^s{w}\log^{(1-s)}{z}}$$
for $1<x<y\leq w<z$, $|z-w|\leq c|y-x|$, $c>0$, and $0<s<1$?
Intuition says this is true since $\log$ becomes ''flatter'' as its argument grows, so that the ratio on the right hand side should be closer to one when $z$ and $w$ are larger. Also how large can $c$ become?
Edit: I can more or less see how to do this with crude error bounds. I guess I am curious if there are simpler ways of seeing this.
Is there no connection between $s$ and $t$?
@AndrásBátkai Nope. $t$ is just fixed, while $s$ free to vary.
Since $c$ doesn't appear in the inequality and $|z-w|\le c|y-z|$ is true for some $c>0$, it seems that condition is irrelevant. Or did you omit an upper bound on $c$?
@BrendanMcKay It should hold for some fixed $c$ depending on the $x,y$ and $t$. I'm interested in larger $c$. Also, if the $z$ and $w$ become spaced out enough one can find counterexamples to the inequality. Maybe this clarifies things?
Yes, it clarifies that the problem statement needs clarifying. If it is only true for some $c$, you have say "for some $c$" or "there exists $c$ such that".
Are the brackets missed in the numerators?
@FedorPetrov Yes, the convex combinations are within the logarithms.
May $c$ depend on $x, y, t$?
@FedorPetrov Yeah, it should, I expect.
How do you understand the deniminators if $x<1$?
@FedorPetrov I will adjust it to $x>1$ for this reason.
The second derivative of $\log\log x$ is decreasing, so just compare it with $2$ parabolas with the same leading coefficient, in which case the optimal $c$ is just $2\sqrt{t(1-t)}$ if I haven't screwed my algebra.
@fedja Thanks! I think it could be true for larger $c$, but I will check the details.
No, that's the limit: If the differences are small or you are far away from the origin, then the second derivative doesn't change noticeably and $\log\log x$ is indistinguishable from a parabola in the range you are interested in (I assume you want a uniform bound in $x,y,z,w$).
@fedja Yes, absolutely. I had an application in mind where the $z>>x$ which had me thinking otherwise. Thanks again.
|
2025-03-21T14:48:31.647411
| 2020-07-30T06:19:28 |
366918
|
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|
Stack Exchange
|
Cut and Join for Hurwitz number with multiple spin
Let me introduce some background of cut and join equation for spin Hurwitz number with the completed cycle as mentioned in
https://arxiv.org/pdf/1103.3120.pdf
We fix two partition $\mu $ and $\nu$ of a fixed integer $d$, r, s are also fixed such that
$ 2g-2+\ell(\mu)+\ell(\nu)=rs$, where $\ell()$ denote the length of the partition.
A disconnected genus $g$ double Hurwtiz number $h_{g,\mu,\nu}^r$ with completed $(r+1)$ cycle is defined as (Definition 2.5 in the paper)
$$h_{g,\mu,\nu}^r := \frac{1}{\prod_{i=1}^{\ell(\mu)} \mu_i \prod_{i=1}^{\ell(\nu)} \nu_i}\sum_{|\lambda|=d}\chi_{\mu}^{\lambda}\Big(\frac{p_{r+1}}{(r+1)!}\Big)^s\chi_{\nu}^{\lambda} $$
We can now give the geometric meaning of Hurwitz numbers with
completed cycles it the count of the stable map with weight given by the inverse of Automorphism.
Call a stable map $f: C → \mathbb{CP}^1$
an r-covering if (i) it
has a finite number of preimages of $0$ and $\infty$ given by $\mu$ and $\nu$ respectively.
and (ii) all its singular loci
have multiplicity r except possibly the preimages of $0$ and $\infty$
The generating function for this number in the paper is given by eq (31)
$$H_{r+1}(\beta, p_1\ldots, q_1\ldots) := \sum_{s,\ell(\mu), \ell(\nu)}h_{g,\mu,\nu}^{r+1}\beta^s p_{\mu}q_{\mu} $$
The above can be written in a infinite wedge space expression in the following way
$$ H_{r+1}(\beta, p_1\ldots, q_1\ldots)= \langle\exp(\sum \frac{q_m}{h m} \alpha_m)\exp(\beta \mathcal{F}_{r+1}) \exp(\sum \frac{p_m}{h m} \alpha_{-m}) \rangle$$
Now these generating functions are the solution to the cut and join the equation
$$ \frac{\partial}{\partial \beta}H_{r+1}= Q_{r+1}H_{r+1}$$
where $Q_i$ as defined in theorem 5.2 these are the cut and join opertors.
My question is we allow spin say $r_1$ and $r_2$ or finitely many.We can define a similar Hurwitz number with a completed cycle. In literature, there exists a cut and join formula for these numbers?
$$ H_{r_{1}+1, r_2 +1}(\beta, p_1\ldots, q_1\ldots)= \langle\exp(\sum \frac{q_m}{h m} \alpha_m)\exp(\beta_1 \mathcal{F}_{r_1+1}+ \beta_2 \mathcal{F}_{r_2+1}) \exp(\sum \frac{p_m}{h m} \alpha_{-m}) \rangle$$
|
2025-03-21T14:48:31.647569
| 2020-07-30T06:22:11 |
366919
|
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"D.S. Lipham",
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|
Stack Exchange
|
Connectedness of compact metric space
Let $X$ be a compact metric space satisfying the following condition: for any given positive number $\delta>0$, only finitely many components of $X$ have diameter larger than $\delta$.
For a given component $P$ of $X$, let us remove a connected compact subset $C$ of $P$ such that $P\setminus C$ is disconnected. Consider a disjoint partition $P\setminus C=A\cup B$, with $A$ and $B$ at a positive distance. Suppose that $x\in A$ and $y\in B$.
Are then $x$ and $y$ in different quasi-components of $X\setminus C$? or can we find a separation of $X\setminus C=M\cup N$ such that $x\in M$ and $y\in N$?
Can you clarify whether we can assume $x\in A$ and $y\in B$? Or is it only that a partition $A\cup B$ of $P\setminus C$ exists, and $x,y$ might both be in $A$?
sorry for the confuse, actually, I want to express that $x\in A$ and $y\in B$.
(This answers the previous version of the question, for components but not quasicomponents).
I think, it us true in every topological space: if $X$ is a topological space with connected components $P$ and $Q_j$: $X=P\sqcup (\sqcup_j Q_j)$, and you remove a set $C$ from a connected component $P$ so that it becomes disconnected: $P\setminus C=\sqcup A_i$, where $A_i$ are the connected components of $P\setminus C$, then the connected components of $X\setminus C$ are $A_i$'s and $Q_j$'s. The reason is that the connected components are the inclusion-maximal connected subsets, and each connected subset of $X\setminus C$ lies inside a single connected component of $X$.
what are the sets A_i'?
connected components of $P\setminus C$
If P is a segment with length 1, and there is a sequence of segmenst with length 1 which are all parallel to $P$ converging to $P$ , $X$ is the union of these segmetns. Then we remove the midpoint $p$ of $P$, now $P\setminus p$ is disconnected, but $P\setminus p$ is thecomponent of the $X$.
You probably mean that $P\setminus p$ is the component of $X\setminus p$, but it is not the case: the components must be connected, and $P\setminus p$ is not.
yeah, that is the strange point, It's a little counterintuitive, but $P\setminus p$ is exactly the component of $X$.
@YeeNeil I think you might be talking about quasi-components. Your example is example 2 in this wikipedia page: https://en.wikipedia.org/wiki/Locally_connected_space#Quasicomponents
However, connected components really are always connected. They are precisely the connected subsets which are maximal under the inclusion ordering.
Thanks for your comment, then I think my question is that are $x$ and $y$ in different quasi-components of $X\setminus C$, do you have some idea on it?
@Fedor Petrov, I think I confuse the quasi-component with component, now I have correct the question, sorry for that!
Your compactum $X$ is finitely Suslinian (by defnition). As such, every connected subset of $X$ is locally connected (see this paper), and this implies that the components of $X\setminus C$ are the same as the quasi-components of $X\setminus C$. See this paper, Theorem 2.1. @Fedor Petrov already argued that $x$ and $y$ are in different components of $X\setminus C$, therefore they are also in different quasi-components.
Note that $C$ does not have to be connected or compact. It can be any subset of $P$ such that $P\setminus C$ is disconnected.
Yeah, as far as I can see, the idea is that finitely Suslinian implies hereditarily locally connected, and then hereditarily locally connected implies (q=c) according to Theorem 2.1. But I wonder, by the definition in my question, can we say that $X$ is finitely Suslinian, note that we only requirement that the components with diameter than a given positive number $\delta>0$ is finitely, and this of course different with that a sequence a pairwise disjoint subcontinua form a null sequence.
@YeeNeil Your objection is perfectly valid. But I think that a more elementary approach can be taken to show that $x$ and $y$ are in different quasi-components. I will update my answer when I get the chance.
that sounds good, thanks for your works!
|
2025-03-21T14:48:31.647827
| 2020-07-30T06:53:22 |
366920
|
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|
Stack Exchange
|
Containment of Bruhat cells on flag variety
This question was posted at MSE, but it did not receive any answer there.
Let $G$ be a connected semisimple algebraic group over $\mathbb C$, $X$ the flag variety of $G$, $B_0$ a Borel subgroup, $\mathbb O$ a $B_0$-orbit on $X$.
Question: Can we always find a Borel subgroup $B_x$ so that the open $B_x$-orbit $\mathbb O'$ contains $\mathbb O$?
The obvious attempt would be to choose a point $\bar x \in \mathbb O$, and then take $B_x$ to be a Borel in opposite relative position to $B_{\bar x}$. But not all choices work. For example when $\mathbb O$ is the open orbit, choices are pretty restricted, and I'm failing to find a construction that works in general.
Any help would be appreciated.
To make sure I understand: when $\mathbb O$ is open the only possible choice is $B_x=B_0$, correct? So it is "pretty restricted" but not very difficult to make =)
@imakhlin Yes, I think you are right. I meant to say that, if a procedure for choosing $B_x$ exists for any $B_0$-orbit $\mathbb O$, then it would receive more restrictions as $\mathbb O$ becomes bigger, and at the extreme case when $\mathbb O$ is the biggest possible orbit this procedure should give the unique solution $B_x = B_0$.
I'd say that the relevant fact here is as follows. For two Borels $B_1$ and $B_2$ with a common maximal torus $T$ let $x_1$ be the unique $T$-fixed point in the open $B_1$-orbit. Then the $B_2$-orbit $B_2x_1$ lies in the open $B_1$-orbit. For instance, you can see this by proving that
(a) for every $x\in B_2x_1$ the orbit closure $\overline{Tx}$ contains $x_1$ and
(b) the open $B_1$-orbit consists precisely of those points $x$ for which $x_1\in\overline{Tx}$.
(These are fairly basic properties of Schubert decompositions.)
In your setting this means that you can choose any maximal torus $T\subset B_0$, let $x'$ be the unique $T$-fixed point in $\mathbb O$ and consider the unique Borel $B\supset T$ for which the orbit $Bx'$ is open. This $B$ will be your $B_x$. In fact, this $B$ is the opposite (with respect to $T$) Borel of the stabilizer of $x'$. So in terms of your last paragraph you can choose any $\bar x$ but you have to choose the opposite Borel correctly: it must intersect $B_0$ in a maximal torus that fixes $\bar x$.
|
2025-03-21T14:48:31.648008
| 2020-07-30T07:11:43 |
366921
|
{
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|
Stack Exchange
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Weight enumerator classifiers
Let $f(x,y)$ be a polynomial with integer coefficients. What conditions guarantee that this is the weight enumerator of a binary linear code of size $n$ and dimension $k$?
I’m almost certain that the answer to this question is unknown...so instead i’ll settle for anything that is conjectural.
There’s a list of necessary conditions:
$f$ must be homogeneous of degree $n$ with non-negative coefficients.
The $x^n$ coefficient has to be $1$ since the zero vector is the unique weight $0$ vector.
The $y^n$ coefficient has to be $0$ or $1$ since the all $1$’s vector either belongs to the code or doesn’t.
The sum of the coefficients has to be $2^k$ since every vector has a unique weight and so is counted exactly once by some coefficient.
The MacWilliams transform ($g(x,y) = \frac{1}{2^k}f(x+y,x-y)$) has to have all of the above properties but with coefficient sum $2^{n-k}$ since if $f$ corresponds to a code then $g$ would correspond to the dual code.
Are there any more necessary conditions missing?
3a. If the $y^n$ coefficient is $1$ then $f(x,y) = f(y,x)$
(proof: 1's complement).
This answer has grown so split it into sections. So, let me summarize.
First in Section 1 we give a counterexample showing more conditions are needed to guarantee we have the weight enumerator of a linear code. In Section 2 we show that the conditions in the question do classify weight enumerators for constant weight linear codes. Finally in Section 3 we give some additional necessary conditions (but not a complete solution).
Section 1. A "MacWilliams coincidence" example:
Let $f(x,y) = \sum_{j=0}^n a_j x^{n-j}y^j$. One other simple condition not in the question is $a_j \leq \binom{n}{j}$. But we also need some lower bounds of the coefficients $a_j$.
Consider the nonlinear code
\begin{align*}
00000000 & & 11111111 \\
11000000 & & 00111111 \\
10100000 & & 01011111 \\
10010000 & & 01101111 \\
10001000 & & 01110111 \\
10000100 & & 01111011 \\
10000010 & & 01111101 \\
10000001 & & 01111110
\end{align*}
which is binary of length $8$ with $16$ elements.
It's weight enumerator is
$$ f(x,y) = x^8 + 7x^6y^2 + 7x^2y^6 + y^8$$
and
$$ \frac{1}{16}f(x+y, x-y) = f(x,y).$$
This example is in "The MacWilliams identities for nonlinear codes" by MacWilliams, Sloane, and Goethals which can be viewed on Sloane's webpage.
We see $f$ satisfies all the conditions we have listed so far.
However, $f$ cannot be the weight enumerator of any linear code.
Indeed if it were for some linear code then $u+v$ must have weight $2$ for any two codewords $u$ and $v$ of weight $2$.
This means for any two weight $2$ codewords $u$ and $v$ they have exactly one coordinate in common.
We may assume the first two weight $2$ codewords are $11000000$ and $10100000$.
If the next weight $2$ codeword is $10010000$, then $11110000$ would be a codeword.
Alternatively the next weight $2$ codeword could be $01100000$, but then we cannot complete to seven codewords of weight $2$ without forcing a codeword on weight $4$.
In additional to the conditions in the question we need some lower bound on coefficients. We see here that the seven codewords of weight $2$ force the existence of a weight $4$ codeword. (I guess the above shows if $a_2 > 3$, then $a_4 > 0$.)
Section 2. MacWilliams is enough for constant weight:
A constant weight linear is a linear code where all codewords have the same weight (with the exception of the all zero codeword).
It is known that any $k$-dimensional constant weight binary linear code has generator matrix whose columns consist (with possible replication allowed) of all nonzero binary words of length $k$ along with zero columns.
For example, the matrix
$$
\begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 & 1 \\
0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 1
\end{bmatrix}$$
generates $2$-dimension constant weight of weight $6$.
So, the weight generator of a constant weight linear code looks like
$$f(x,y) = x^n + (2^{k}-1)x^{n-m2^{k-1}}y^{m2^{k-1}}$$
for some $k$, $m$, and $n$.
It turns out that this is the only possible way to satisfy our conditions with only $x^n$ and one other monomial.
Assume $f(x,y) = x^n + (2^{k}-1)x^{n-a}y^a$ satisfies MacWilliams in the sense that
$$\frac{1}{2^k}f(x+y,x-y) = \frac{1}{2^k} \left( (x+y)^n + (2^k - 1)(x+y)^{n-a}(x-y)^a\right)$$
has integer coefficients.
This would mean that
$$\frac{1}{2^k} \left( (x+y)^n - (x+y)^{n-a}(x-y)^a\right)$$
has integer coefficients.
The coefficient of $x^{n-1}y$ in
$$(x+y)^n - (x+y)^{n-a}(x-y)^a$$
is $2a$.
Hence, $2^k$ divides $2a$ and so $a = m2^{k-1}$.
Section 3. some (easy) lower bounds:
Here are two fairly simple conditions based off of $a_1$ and $a_{n-1}$ to get lower bounds on the coefficients.
Let $[n] = \{1,2,\dots, n\}$ and let $\binom{A}{j}$ denote all $j$-element subsets of a set $A$. We let $\{e_i : i \in [n]\}$ denote the standard basis. For any $B \subseteq [n]$ set $e_B = \sum_{i \in B} e_i$. Assume $f(x,y)$ is the weight enumerator of a linear binary code $C$. Then there is some $A \subseteq [n]$ with $|A| = a_1$ such that $e_i \in C$ if and only if $i \in A$. It follows that $e_B \in C$ for $B \subseteq A$. So, $e_B$ is a codeword of weight $j$ for each $B \in \binom{A}{j}$. Hence, we get the condition
$$\binom{a_1}{j} \leq a_j$$
for each $1 \leq j \leq n$.
Also, we have $A \subseteq [n]$ with $|A| = a_{n-1}$ such that $e_{[n]} + e_i \in C$ if and only if $i \in A$. Now for any $B \subseteq A$ we have
$$\sum_{i \in B} (e_{[n]} + e_i) = |B|e_{[n]} + e_B$$
which is either $e_B$ or $e_{[n]} + e_B$ depending on the parity of $|B|$.
Hence, we have
$$\binom{a_{n-1}}{2j} \leq a_{2j}$$
and
$$\binom{a_{n-1}}{2j+1} \leq a_{n-2j-1}.$$
Ideally this sort of idea could be continued in a more complicated way with $a_2, a_3, \dots$ for codes with larger minimum distance. But this at least cuts down the candidate space some from the conditions in the original question. There are certainly more conditions. As we saw earlier $a_2 > 3$ implies $a_4 > 0$. It would be nice to find some rule saying $a_2 > m$ implies $a_4 > \phi(m)$ for some $\phi$.
Is what you are doing in its full form basically how the linear programming bound is obtained, I wonder--I know a bit of coding theory but the LP bounds, invariant theory, etc. are not my strong points.
@kodlu I am not sure. I like coding theory, but I am not any sort of expert. I'll have a read a bit about LP bounds. I am having fun weight distribution inverse problem. Thanks for the tip.
|
2025-03-21T14:48:31.648399
| 2020-07-30T09:00:13 |
366926
|
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|
Stack Exchange
|
Grothendieck dessins d'enfants - current surveys or text you can recommend?
I was recommended this forum to be the leading site for algebraic geometry, so I would like to ask you a question about Grothendieck dessins d´enfants. My background is in maps on surfaces (graph embeddings), and I got fascinated about their application to dessins d´enfants.
As an example, I found it completely stunning when I saw the first time that the least genus embedding of $K_{n,n,n}$ (the complete tripartite graph with $3n$ vertices), when viewed as a dessin d'enfants, specifies the Fermat curve. (cf. Jones & Singermann "Maps, Hypermaps, and Triangle Groups" in: Schneps (ed.): The Grothendieck Theory of Dessins d’Enfants. London Mathematical Society Lecture Notes Series 200, Cambridge University Press, Cambridge 1994)
So I would like to look for more areas of potential application of graph embeddings to dessins d'enfants, but don't have a good overview yet over recent developments in this field.
Would you be able to recommend a recent survey of dessins d'enfants, or a good recent text to me?
I am not sure whether I am using the right tags for this, please feel free to adjust!
I don't know to what extent it fits your criteria, but it's probably worth having a look at the book by Ernesto Girondo and Gabino González-Diez, ‘Introduction to Compact Riemann Surfaces and Dessins d'Enfants’ (LMS Student Texts 79 (2012)).
@Gro-Tsen Merci beaucoup! Very grateful for your suggestion! As I am new to this forum, I am not sure how to give you upvote credit for your help? Would be happy to express my thanks
Szamuely's "Galois Groups and Fundamental Groups" sections 4.6-4.9 are a great background to the subject matter, but they don't go into much detail about the dessins themselves.
@PrimeRibeyeDeal Thank you very much for your recommendation! Very interested in that
See Jones & Wolfart, "Dessins d'Enfants on Riemann Surfaces" (Springer, 2016).
@TomHarris Thank you for this suggestion! I will definitely take a look.
|
2025-03-21T14:48:31.648579
| 2020-07-30T09:11:45 |
366927
|
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|
Stack Exchange
|
Intersection modulo 2 theory for infinite dimensional manifolds?
For finite dimensional manifolds, there is a lot of theory about when the number of intersections (modulo $2$) of certain objects are preserved under homotopy. I'll give two quick examples:
Let $f:X \to Y$ be a smooth map from a compact manifold $X$ to a connected manifold $Y$ of the same dimension. Then if both $x,y \in Y$ are regular values of $f$, we know that $f^{-1}(x), f^{-1}(y)$ are both 0-dimensional manifolds, compact, and so finite sets. In fact the sizes of these sets are the same modulo $2$.
Let $X,Y \subset Z$ be two transverse submanifolds of $Z$, one of which is compact. The intersection $X \cap Y$ is as above a finite set of points. In fact its size is preserved (modulo $2$) under homotopy of $X,Y$ to $X',Y'$, provided we end with transverse submanifolds $X',Y'$.
I was wondering if any of this theory generalises to infinite-dimensional manifolds? Take the above two theorems for example, do they have analogues versions for $X,Y,Z$ banach manifolds? What about frechet manifolds? I'm aware compactness will be an issue, but can we generalise nonetheless, for example if we assume the intersection number is finite?
One can speak of transersality of intersections in an infinite dimensional context.
If one goes beyond Hilbert manifolds (e.g. Banach, Frechet) one needs to be a bit careful with the definition of transverse, because one needs to impose splitting conditions. For a submanifold one typically demands that the tangent space $T_xX$ admits a closed complement. Then it it is not enough to demand that $T_xX +T_xY=T_xM$ at every point of intersection, but one needs that $T_xX\cap T_xY$ is closed and complemented. To be able to ignore this issue, let me assume that $Z$ is a Hilbert manifold and $X,Y\subseteq Z$ are submanifolds.
Then if $X$ is closed (as a subset of $Z$), is of finite codimension $m$, and $Y$ is compact of finite dimension $n$, and the intersection is transverse then $X\cap Y$ is a compact submanifold of dimension $n-m$. Moreover, if $f:Y\rightarrow Z$ denotes the inclusion, and $g$ is homotopic to $f$ and also transverse to $X$ than $g(Y)\cap Z$ is cobordant to $f(Y)\cap Z$. In particular, this means that the mod $2$ intersection number is well-defined in this context. If the normal bundle of $X$ is oriented, and $Y$ is also oriented, everything works with orientations as well.
A nice class of mappings are Fredholm mappings. Smale famously proved if $f:M\rightarrow N$ is a smooth Fredholm mapping of index $k$, that $f$ has regular values (without Fredholm this can be false), and that the preimage of a regular value is a manifold of dimension equal to $k$. If $f$ is a proper map, then this manifold is compact. The cobordism class of the regular value is independent of the regular value, and the proper Fredholm cobordism class. This can be used to distinguish proper Fredholm mappings. Together with Alberto Abbondandolo we upgraded this invariant to a full invariant (infinite dimensional framed cobordism classes) in the case $N$ is the Hilbert space. This is in this paper here:
MR4058178 Prelim Abbondandolo, Alberto; Rot, Thomas O.; On the
homotopy classification of proper Fredholm maps into a Hilbert space.
J. Reine Angew. Math. 759 (2020), 161–200. 58B15 (47A53 47H11)
We also discuss the framed cobordism classes of non-positive index in this paper. The index one case, for simply connected Hilbert manifolds, is done in our recent preprint.
https://arxiv.org/abs/2005.03936
This was supposed to be a comment but got too long.
The general result which encompasses both your examples in finite dimensions is the following: If $Y\subseteq Z$ is a submanifold of codimension $k$ and $f:X\to Z$ is a map transverse to $Y$, then $f^{-1}(Y)\subseteq X$ is a submanifold of codimension $k$.
In your second example $f:X\to Z$ is the inclusion, and so for $f^{-1}(Y)=X\cap Y$ to be a finite set you need $\dim(X)+\dim(Y)=\dim(Z)$, i.e. $X$ and $Y$ to have complementary dimensions. If the manifolds are infinite dimensional, this doesn't seem to make sense.
Your first example, where you have a map $f:X\to Z$ and regular values $y,z\in Z$, does generalize to the setting of proper Fredholm maps. MO user Thomas Rot has done some work on this - see these slides of a talk he gave at the Skye conference in 2018. In particular, if the Fredholm index
$$
\dim\ker df_x - \dim\operatorname{coker} df_x
$$ of each differential $df_x:TX_x\to TZ_{f(x)}$ is $k$ for all $x\in X$, then the pre-image of a regular value is a well-defined $k$-dimensional unoriented cobordism class. When $k=0$ this is an integer mod $2$.
Surely there is more to say, perhaps Thomas himself will come along and answer.
That is some nice timing!
@ThomasRot Ha ha! Nice timing indeed! Now, perhaps someone will deposit £1,000,000 into my account...
I can do an upvote...
Not exactly what you are after, but for ($C^1$) Fredholm maps of index 0 there is a modulo 2 mapping degree. This can be generalized to an integer-valued degree only for so-called oriented Fredholm maps (of index 0).
For the case that the image manifold is actually a Banach space, there is also a degree for compact and certain non-compact and actually even multivalue perturbations of such maps.
The integer-valued degree (in the oriented case) was developed by M. Furi and P. Benevieri (and for a different notion of orientation also by V. G. Zvyagin and N. M. Ratiner), while I am not sure who was the first to note the much simpler modulo 2 case.
More details and history are in my monograph M. Väth, Topological Analyis, De Gruyter, Berlin, New York 2012.
The modulo two case goes back to Cacciopoli probably. Smale also defined it.
For the history, one probably has to distinguish the $C^2$ and $C^1$ case: The $C^2$ case is completely analogous to the finite-dimensional case. But the $C^1$ case does not follow by approximation as in finite dimensions but requires a rather different approach. My guess is that it was indeed one of the 4 mentioned authors who did the $C^1$ case first.
|
2025-03-21T14:48:31.649349
| 2020-07-30T09:31:53 |
366928
|
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|
Stack Exchange
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detecting a semi-free module from its bar-resolution
Let $A$ be a DG-algebra over a field (say $k$). A DG-module $M$ over $A$ is said to be semi-free if it admits an exhaustive filtration $0 = M_0 \subset M_1 \subset \ldots \subset M_p = M$ such that the cones of $M_i \rightarrow M_{i+1}$ are finite direct sums of shifts of finite dimensional free $A$-modules (I insist on the finite dimensional hypothesis which I need for my problem).
In case $A$ is proper and $M$ is finite dimensional (perfect?) DG module over $A$ then the bar-resolution of $M$ is an (infinite) semi-free resolution of $M$.
I am wondering if there is a way to detect that $M$ itself is semi-free from its bar resolution? I am not sure what to expect. Perhaps that one can extract a finite resolution from the bar-resolution when $M$ is semi-free?
Since I am looking for positive result, extra-hypotheses can be made on $A$. I just don't want to assume that $A$ is smooth (since in my example it is not).
So, that "finite dimensional" hypothesis you need, what does it mean?
finite dimensional free-modules. I have the feeling some people allow infinite dimensional free modules in the construction of the Bar-resolution.
Finite rank, you probably mean. Your question is not clear because that seems to be your hypothesis on $M$.
sure, that is also my hypothesis on $M$.
then it holds by hypothesis, you don't need the bar resolution.
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2025-03-21T14:48:31.649468
| 2020-07-30T10:12:11 |
366933
|
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|
Stack Exchange
|
Deep applications of the Pettis integral?
In the Notes section of chapter 2 of Diestel and Uhl's Vector Measures they make the comment:
"Presently the Pettis integral has very few applications. But our prediction is that when (and if) the general Pettis integral is understood it will pay off in deep applications." ( Google books link )
That was back in $1977$ and when I did a little searching to find out how well that prediction had fared I found little: wikipedia provides a cursory description and Encyclopedia of Mathematics is arguably better but terser. This question asks why (Dunford-)Pettis integrals are useful, but I would not say it covers deep applications per se.
So: are there deep applications to the Pettis integral, and if so, what are they?
Diestel and Uhl commented on this in Measure theory and its applications:
That's excellent, thank-you!
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2025-03-21T14:48:31.649565
| 2020-07-30T12:59:33 |
366943
|
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|
Stack Exchange
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Connected components of bounded linear operators of $V = (\mathcal C(U(1), \mathbb C) , \lVert \cdot \rVert_\infty)$
This question is related to this one.
Consider the complex Banach space $V=(\mathcal C(U(1), \mathbb C), \Vert \cdot \Vert_\infty)$ where $\mathcal C(U(1), \mathbb C)$ is the space of continuous complex maps defined on the unit circle endowed with the sup norm.
Define $G$ as the operator that maps an element $f \in V$ to the element $G(f)$ such that
$$G(f) : z \in U(1) \mapsto zf(z) \in \mathbb C.$$
Is $G$ in the same connected component that the identity in the subset of bounded linear invertible operators of $V$ denoted by $\mathcal{L}(V)^*$
Note: I moved this question @mathoverflow and deleted the math.stackexchange one.
In fact the general linear group of this Banach space is contractible. There is a proof in this article of Mityagin. A different article of his promises a simpler argument but I couldn't find an online copy.
|
2025-03-21T14:48:31.649676
| 2020-07-30T13:04:33 |
366944
|
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"Alkan",
"Gerry Myerson",
"Mark Wildon",
"Piyush Grover",
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|
Stack Exchange
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Why are attempts to define chaos with discrete states so scarce?
Interestingly, the theory of nested recurrence relations has been correlated with “discrete chaos” by Golomb (1991) and Tanny (1992).
And in literature, there are very few studies that have different approaches on chaos in integer sequences. One curious example that has an interesting definition is by Fraenkel (1994).
Question: Are there other definitions of deterministic chaos for integer sequences that are generated by a recurrence relation? If the answer is no, why are attempts to define chaos with discrete states so scarce?
For usual discrete dynamical systems such as maps, a positive maximal Lyapunov exponent is usually considered as a definition of deterministic chaos.
Interestingly, all of this seems to hinge on a preprint that was apparently never published.
Many thanks for your comment. Best regards.
Isn't all this addressed by symbolic dynamics ? https://en.wikipedia.org/wiki/Symbolic_dynamics
@PiyushGrover: Unless there is some facet of it that I don’t know about, symbolic dynamics discretises states of a continious-state dynamics for analysis.
@PiyushGrover, thanks for your comment, also please consider that simple example question in order to see perspective of question: "How can we define chaos in a recurrence relation $Q(n) = Q(n-Q(n-1)) + Q(n-Q(n-2))$ (Hofstadter's Q) ?"
It appears that the never published preprint is available at https://oeis.org/A005185/a005185_1.pdf
@GerryMyerson, many thanks for your comment. I also believe that this preprint is a nice example to unpublished manuscripts that are very valuable. ( It has also several interesting components such as quasiperiodic solutions.) Best regards.
Has anyone made a serious attempt to do this using finite state automata? I'm thinking of some of the chaotic (as they seem to me) configurations that arise in Conway's Game of Life.
@MarkWildon: I don't know that there is such an attempt but this sounds promising and probably very interesting, at least to me. Best regards.
This is different, but I'm reminded of Terry Tao's enlightening article Structure and randomness in combinatorics, where he explains the motivation and technique for decomposing systems into parts which can be analyzed combinatorically/algebraically and probabilistically.
@PrimeRibeyeDeal: Yes, it is different but extremely helpful article in order to investigate coexistence of regularity and irregularity and their curious decompositions. So thanks for sharing. Best regards.
The biggest problem I see with discrete-state chaos is that you would want it to capture some characteristic features of normal chaos (i.e., in the sense of chaos theory), namely:
Chaos is sensitive to initial conditions, i.e., arbitrarily small changes to the initial condition blow up.
Chaotic dynamics are aperiodic, i.e., they do not repeat.
Chaotic dynamics are bounded.
Mind that there is no final agreement on how to define chaos (as there is no big need for it), so the above are only what most people consider to be necessary conditions.
Anything named chaos that utterly fails to meet these criteria would probably raise false expectations and you are probably better off with another name for whatever interesting property you found.
In this light, I see two, not entirely unrelated problems with extending the concept of chaos to discrete states:
Sensitivity to initial conditions requires some notion of distance that allows for infinitesimally close states (to every state, not just selected ones). Discrete states do not have this. You might cheat around this by letting finite distances grow to infinity, but then you lose boundedness.
If a deterministic dynamics can only have a $n$ discrete states, it will inevitably be periodic. If, on the other hand, you have infinitely many, discrete states, your chaos it is not bounded.
The sequences in question kind-of alleviate this issues by introducing a state-dependent delay into the sequence:
$$ Q_t = Q_{t−Q_{t-1}} + Q_{t-Q_{t-2}}$$
Thus, the actual state of the dynamics is not only defined by its current value of the sequence ($Q_t$), but by its entire past, or at least a slice of it that can be arbitrarily large.
To avoid repetition, you still need unbounded states: Otherwise your state and thus your delay would be limited and you would have periodicity eventually.
However, due to the combinatorial explosion ensuing from having a arbitrarily dimensional state, you can approach infinity much more slowly.
Still, you have unbounded states and thus I would consider it problematic to denote this as chaos.
Finally note that every computer simulation of chaos is inevitably only a discrete approximation, as you only have finitely many floating-point numbers (or whatever number format you use).
For usual discrete dynamical systems such as maps, a positive maximal Lyapunov exponent is usually considered as a definition of deterministic chaos.
Note that a positive Lyapunov exponent does not suffice for chaos. It is only a necessary property.
As Piyush Grover mentioned, symbolic dynamics provides a solid foundation for studying the dynamics with a discrete phase space. Usually one considers just a finite alphabet of symbols, however one may also consider all of $\mathbb{Z}$ as the alphabet. If one computes the topological entropy of the dynamical system to be positive, it may be considered to be chaotic.
A likely obstacle though in the definition of topological entropy is that it assumes that the phase space (or at the least the chaotic attractor) is compact. While there is no “one true” definition of chaos, a system where everything runs off to infinity would not be considered to be chaotic – even if everything runs off to infinity in a slightly different way.
To get around that, perhaps you could take your integer sequences modulo $p$ and show they have positive topological entropy for some and/or all $p \geq 2$.
To get around that, perhaps you could take your integer sequences modulo $p$ – If you do that, what you are assessing with your symbolic apparatus is not the full state of the dynamical system anymore. The compactness aspect of chaos relates to the actual state of the system, not a projection of it. You cannot run the system completely in modulo $p$.
|
2025-03-21T14:48:31.650102
| 2020-07-30T13:14:29 |
366945
|
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"Phobos",
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|
Stack Exchange
|
Conditional entropy - solve example
Given a random variable $X$ that is uniformly distributed on $[-b,b]$ and $Y=g(X)$ with
$$g(x) = \begin{cases} 0, ~~~ x\in [-c,c] \\ x, ~~~ \text{else}\end{cases}$$
Now I want to compute the information dimension $d(X), d(Y)$ and the conditional information dimension $d(X|Y)$ and show that $d(X) = d(X|Y) + d(Y)$ in this case.
The information dimension is defined as
$$
d(X) = \lim_{m\rightarrow \infty} \frac{H(\hat{X}^{(m)})}{m}
$$
with
$$
\hat{X}^{(m)} := \frac{\lfloor2^m X \rfloor}{2^m}
$$
the quantization of $X$.
For a discrete distribution, $d(X) = 0$, and for a continuous one-dimensional distribution, $d(X) = 1$. For a mixed distribution with discrete and continuous components of the form $P_X = d P_X^{(ac)} + (1-d) P_X^{(d)}$, the information dimension is $d(X)=d$.
Now I know, that the random variable X has a continuous component $\Rightarrow d(X) = 1$. The distribution $P_Y$ is a discrete-continuous mixture: $$ P_Y = \begin{cases} \frac{c}{b}, ~~~Y=0\\ \frac{1}{2b},~~~Y \in [-b,-c] \cap [c,b]\\ 0,~~~\text{else} \end{cases}$$
Therefore, $d(Y)=\frac{b-c}{b}$.
Now my question is the following: how do I compute the conditional information dimension?
$$d(X|Y) = \lim_{m \rightarrow \infty} \frac{H(\hat{X}^{(m)}|Y)}{m} = \int_\mathcal{Y} d(X|Y=y)dP_Y(y) = \mathbb{E}_{Y\sim P_Y}(d(X|Y=Y))$$
From the context, it appears that $b\in(0,\infty)$ and $c\in[0,\infty)$ (and, likely, $c\le b$). Anyway, let $c_1:=\min(b,c)$. Note that
With probability $1$, either $Y=0$ or $c_1<|Y|\le b$;
The conditional distribution of $X$ given $Y=0$ is the uniform distribution on the interval $[-c_1,c_1]$ and hence $d(X|Y=0)=1$.
The conditional distribution of $X$ given $Y=y$ with $c_1<|y|\le b$ is the Dirac distribution supported at point $y$ and hence $d(X|Y=y)=0$.
So,
$$d(X|Y)=\int_{\mathbb R}d(X|y)P(Y\in dy)=P(Y=0)=\frac{c_1}b=\frac{\min(b,c)}b.$$
oh makes sense, thank you a lot!
|
2025-03-21T14:48:31.650242
| 2020-07-30T14:46:32 |
366952
|
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}
|
Stack Exchange
|
The Beltrami equation and Neumann series
Let $\mu: \mathbb{C} \to D(0,1).$ A quasiconformal map is a $W^{1}_{2,loc}-$solution to the Beltrami equation $\bar{\partial}f = \mu \partial f$. In this paper, the authors remark that one can formally solve for $f$ in this equation by the Neumann series $$\bar{\partial}f = \mu + \mu T \mu + \mu T \mu T \mu + \cdots$$ where $T$ is the Beurling transform $$Tf(z) = \partial\left(\frac{\bar{\partial}f}{\partial z}\right)^{-1}(z) = -\frac{1}{\pi} p.v \int_{\mathbb{C}}\frac{f(w) \ \mathrm dw}{(z-w)^2}.
$$
I cannot see how, even formally, this is true, and no source I can find shows how to do this. What is the heuristic behind this?
The story of the solution of the Betrami equation using the Beurling transformation in $L^2$ has an elegant and elementary explanation in:
Adrien Douady and X. Buff, Le théorème d’intégrabilité des structures
presque complexes, The Mandelbrot set, theme and variations, London
Math. Soc. Lecture Note Ser., vol. 274, Cambridge Univ. Press,
Cambridge, 2000, pp. 307–324. MR 1765096
I can't see how it becomes precisely what you have quoted, but is similar. If we let $f=(I-\mu L)^{-1}\mu$, where $L$ the Beurling transform, then $(I-\mu L)f=\mu$, so $f=\mu+\mu L f$. If $\varphi$ satisfies $\bar\partial \varphi=f$, then $\bar\partial (z+\varphi)=f=\mu+\mu L f=\mu+\mu\bar\varphi=\mu(\bar z+\bar\varphi)$. So $z+\varphi$ is the solution you are looking for, modulo various details explained clearly in that paper.
|
2025-03-21T14:48:31.650374
| 2020-07-30T15:06:07 |
366953
|
{
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"Guilherme Mazanti",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/366953"
}
|
Stack Exchange
|
Sharp assumption for preserving Lebesgue measurability by left composition
Let $g: [0, 1] \to \mathbb R$ be a Lebesgue-measurable function (in the classical sense: the inverse images of Borel sets are Lebesgue-measurable). It is a classical fact in analysis that $f \circ g$ is Lebesgue-measurable as soon as $f$ is continuous, for instance, or Borel-measurable (the inverse images of Borel sets are Borel), but not necessarily if $f$ is only Lebesgue-measurable. My question is: what is the sharpest assumption that one can put on $f$ guaranteeing that $f \circ g$ is Lebesgue-measurable for every Lebesgue-measurable $g$?
More precisely, consider the class
$$
\begin{aligned}
\mathcal F = \{f: \mathbb R \to \mathbb R \mid {} & g: [0, 1] \to \mathbb R \text{ is Lebesgue-measurable} \implies \\
& f \circ g \text{ is Lebesgue-measurable}\}.
\end{aligned}
$$
$\mathcal F$ contains all Borel-measurable functions, but does it contain other functions? Or is it equal to the set of all Borel-measurable functions?
This question looks quite natural and I imagined there should be some classical result in real analysis providing its answer, but I could not find it in standard textbooks. I initially conjectured that $\mathcal F$ coincides with the set of all Borel-measurable functions. This would mean that, if $f$ is not a Borel-measurable function, then there exists a Lebesgue-measurable $g$ such that $f \circ g$ is not Lebesgue-measurable. The idea would be to pick such an $f$, pick a Borel set $A$ such that $B = f^{-1}(A)$ is not Borel, and try to construct a Lebesgue-measurable $g$ such that $g^{-1}(B)$ is not Lebesgue-measurable, but I cannot see how to construct such a $g$ while keeping its Lebesgue-measurability. Any ideas or any references to this question?
Edit: made the statements about Lebesgue- or Borel-measurability more precise.
The answer is: $\cal F$ is the family of universally measurable functions.
For simplicity, let us consider functions on $[0,1]$ rather than on $\mathbb R$. Let $\cal B$ be the family of Borel sets, $\cal B^\star$ the family of universally measurable sets, and $\cal L$ the family of Lebesgue sets.
Clearly, it is sufficient that $f$ is universally measurable: every $\cal B/\cal L$-measurable function is in fact $\cal B^\star/\cal L$ measurable (for clearly $\cal L^\star = \cal L$); see, for example, the PlanetMath entry.
The above condition turns out to be necessary, too. Indeed: suppose that $f$ is not universally measurable, that is, $A := f^{-1}(B) \notin \cal B^\star$ for some $B \in \cal B$. We will construct a continuous (!) function $g$ such that $g^{-1}(A)$ is not in $\cal L$. Of course, this implies that $f \notin \cal F$.
There is a Borel probability measure $\mu$ such that $A$ is not $\mu$-measurable. Let $\lambda$ be the Lebesgue measure on $[0, 1]$. Considering $\tfrac{1}{2} \mu + \tfrac{1}{2} \lambda$ rather than $\mu$, we may assume that the distribution function $h$ of $\mu$ is strictly increasing. Removing the atoms of $\mu$ and renormalising it, we can make $\mu$ atomless and $h$ is continuous.
We have $\mu(E) = \lambda(h(E))$ for every Borel set $E$. If $h(A)$ were Lebesgue measurable, we would have two Borel sets $F_1$ and $F_2$ such that $F_1 \subseteq h(A) \subseteq F_2$ and $\lambda(F_2 \setminus F_1) = 0$. But then $E_1 = h^{-1}(F_1)$ and $E_2 = h^{-1}(F_2)$ would be Borel sets such that $E_1 \subseteq A \subseteq E_2$ and $$\mu(E_2 \setminus E_1) = \lambda(h(E_2 \setminus E_1)) = \lambda(F_2 \setminus F_1) = 0 ,$$ and consequently $A$ would be $\mu$-measurable.
Now if $g$ is the inverse of $h$, then $g$ is continuous and strictly increasing, and $g^{-1}(A) = h(A)$ is not Lebesgue-measurable.
Thanks for the excellent reply! That's a very nice construction!
Take any $f\in\mathcal F$. Take any real $b$ and any real $a>0$, and let $g(x):=ax+b$ for $x\in[0,1]$. Then the function $g\colon[0,1]\to\mathbb R$ is Borel-measurable and hence the function $h:=f\circ g$ is Borel-measurable. So, for any Borel set $A\subseteq\mathbb R$, the set
$$f^{-1}(A)\cap[b,a+b]=ah^{-1}(A)+b:=\{ax+b\colon x\in h^{-1}(A)\}$$
is Borel, for any interval $[b,a+b]$, which implies that the set $f^{-1}(A)$ is Borel. So, $f$ is Borel-measurable.
Thus, $\mathcal F$ coincides with the set of all Borel-measurable from $\mathbb R$ to $\mathbb R$.
I now realize my question is not necessarily clear enough: I use two notions of measurability, Lebesgue and Borel. Whenever I said only "measurable", I mean Lebesgue-measurable. I will edit the question to make this more precise.
|
2025-03-21T14:48:31.650652
| 2020-07-30T15:12:06 |
366955
|
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"sort": "votes",
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|
Stack Exchange
|
Algebras with a simple preserving duality and finite global dimension
Algebras with a simple preserving duality (an anti-automorphism preserving pointwise a primitive full set of ortohogonal idempotents) and finite global dimension include important classes of algebras such as Schur algebras and blocks of category $\mathcal{O}$, which are also quasi-hereditary.
Question: Is there a (natural) example of an algebra with a simple preserving duality and finite global dimension that is not quasi-hereditary?
|
2025-03-21T14:48:31.650723
| 2020-07-30T15:33:51 |
366957
|
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"authors": [
"Penelope Benenati",
"Steven Stadnicki",
"fedja",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/115803",
"https://mathoverflow.net/users/7092"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366957"
}
|
Stack Exchange
|
Geometric probabilistic problem on triangles on a plane
We are given a triangle $T$ on a plane $P$, with sidelengths $a$, $b$ and $c$, where $c \ge b \ge a > 0$. A straight line $L$ on $P$ is selected uniformly at random from the set of all the horizontal and vertical straight lines cutting $T$. Note that a.s. there is $1$ and only $1$ uncut side of $T$.
Question: What maximum expected length (as a function of $a$, $b$ and $c$) of the uncut side of $T$ over all possible triangles $T$ on $P$, where the expectation is taken over the random selection of $L$?
"What minimum expected length (as a function of a, b and c)" and "over all orientations of T on the plane and all possible choices of a, b and c" are incompatible. I assume the latter is just a typo (otherwise the story is trivial: put $a=b=c=0$), right?
Thank you fedja. I am correcting the question.
Matt, the randomization is only related to choice of the orientation of the straight line $L$ (either horizontal or vertical) and has nothing to do with the triangle $T$. One can think that a fair coin is tossed to decide if $T$ will be cut by a random horizontal or vertical straight line. About the uncut side length: almost surely the number of sides cut by the random straight line $L$ is equal to $2$, which implies that almost surely there will be one and only one uncut side of $T$. I am interesting in its maximum expected length.
Seconding what Matt F is saying — the question 'makes sense' in the sense that it's a quantity one can calculate for any given triangle easily enough, and probably for any orientation of the given triangle with some effort. But understanding why you're trying to maximize this quantity may help with giving more guidance and avoiding XY problems.
StevenStadnicki, Matt, I understand now the question. It is related to my research. I am developing a random algorithm, which would be very complicated to explain. In this algorithm, the use of random straight lines parallel to the cartesian axes reduces its time complexity, and that's why I am focusing on these two types of lines solely.
Matt, I will provide a "numerically calculated expectation for some scalene triangle" as soon as possible.
Do you still choose the vertical option with probability $1/2$ and the horizontal option with probability $1/2$ or the mixture is different now?
I realized that I was wrong in the problem formulation at the beginning. The probability that the horizontal / vertical option is selected depends therefore on the two sizes of the triangle projection on the two axes.
This is also too long for a comment but it shows where the real problem lies.
With the new formulation, the complementary expectation of the sum of two cut sides is just of the form
$$
\frac {\sum_i a_i^2(|\cos\theta_i|+|\sin\theta_i|)}{\max_i(a_i|\cos\theta_i|)+\max_i(a_i|\sin\theta_i|)}
$$
where $a_i$ are the sides and the angles $\theta_i$ that the sides $a_i$ make with one of the axis can be thought of as the angle $\theta$ by which the triangle is rotated plus some fixed offsets. Now, as long as no axis is parallel to one of the sides, both the numerator and the denominator are just linear combinations of $\sin\theta$ and $\cos\theta$, i.e., are proportional to $\cos(\theta+\rm{something})$. Shifting $\theta$ by that something in the denominator, we see that our function is of the form $\alpha\frac{\cos(\theta+\xi)}{\cos\theta}=\alpha(\cos\xi-\sin\xi\tan\theta)$. Since $\tan\theta$ is monotone between singularities, we conclude that the extremal coordinate systems are just the ones with one axis parallel to one of the sides.
The real question is then "to which side?". I suspect that the maximum of the original expectation is attained when one of the axes is parallel to the largest side but I'm not sure yet.
Thank you very much fedja, this sounds principled and very useful.
A bit too long for a comment, here's one suggestion: take $a\lt b\lt c$ and use $(0,0)$ and $(c,0)$ as two points of the triangle. The third point $(x,y)$ (taking $y\gt 0$ WLOG) can be found in whatever usual way you prefer. Now, instead of rotating the triangle, rotate the lines: we can parametrize the pencils of lines as being in the directions $(\cos\theta, \sin\theta)$ and $(-\sin\theta, \cos\theta)$ for $0\leq\theta\leq\frac\pi2$ (by symmetry). Finding the EV in question is a straightforward if annoying calculation, and then maximizing it over $\theta$ should be similarly straightforward. One problem with this approach — and a problem I suspect is innate to the question — is that the answer is likely to be sensitive to specific conditions/regions on $a,b,c$; it's hard to pin down a better one, but this parametrization feels very 'unnatural' for answering questions that are somewhat innately coordinate-centric.
Thank you Steven. I was thinking about something similar. I think the main problem here is to find a very concise way to avoid complicate calculations through an analysis by cases, by finding a very small number of distinct conditions/cases to simplify the calculation.
All the "symmetric" cases should be grouped together to avoid redundant calculations.
|
2025-03-21T14:48:31.651217
| 2020-07-30T16:08:46 |
366960
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/366960"
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|
Stack Exchange
|
Exponential map of moduli space
Let $\mathcal{M}$ be a projective smooth moduli space over $\mathbb{C}$ (the specific example I have in mind is the moduli of curves $\mathcal{M}_g)$. Consider a point $[X]\in \mathcal{M}(\mathbb{C})$. Then via deformation theory in algebraic geometry, we say that the tangent space $T_{[X]}\mathcal{M}$ consists of lifts (or rather isomorphism classes of lifts) of $X$ to $\mathbb{C}[\varepsilon]/(\varepsilon^2)$. This is naturally a $\mathbb{C}$-vector space. My question is in two parts:
Does this construction match with the tangent space of the analytification $\mathcal{M}^{\mathrm{an}}$ at $X$? (This question doesn't need compact, I think)
Since I assume that $\mathcal{M}$ is projective, then $\mathcal{M}^{\mathrm{an}}$ is compact. If we choose a basis of the projective space, we get an induced metric on $\mathcal{M}^{\mathrm{an}}$ (thanks to the comments for pointing me to this gap!). By Hopf-Rinow, the exponential map $T_{[X]}\mathcal{M}^{\mathrm{an}}\rightarrow \mathcal{M}^{\mathrm{an}}$ is well-defined (if we assume that $[X]$ does not lie in the boundary of $\mathcal{M}^{\mathrm{an}}$. If part 1) is true, the tangent space corresponds to infinitesimal lifts of $X$. Can we explicitly say what the image of such a lift is under the exponential map?
If this isn't possible in general, is there perhaps a way to do it for the moduli of curves, moduli of abelian varieties or the Hilbert scheme?
You must specify what metric you put on your $\mathcal{M}$.
Ah, right. If I assume thought that $\mathcal{M}$ is projective, then it should inhert the metric from the projective space, no?
You need to pick a basis for the projective space (modulo unitary transformations) to get a metric.
One class of varieties that has a canonical metric is hermitian locally symmetric spaces, roughly the same as Shimura varieties, such as the moduli of pp abelian varieties. Analogous to the exponential map is an arithmetic phenomenon, the Serre-Tate coordinates, which identify the formal neighborhood of a point with its tangent space.
|
2025-03-21T14:48:31.651386
| 2020-07-30T17:04:56 |
366966
|
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"george",
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|
Stack Exchange
|
Homotopy equivalent Postnikov sections but not homotopy equivalent
Two pointed, connected CW complexes with the same homotopy groups need not be homotopy equivalent (Are there two non-homotopy equivalent spaces with equal homotopy groups?). Moreover, having the same homotopy and homology groups is also not enough (Spaces with same homotopy and homology groups that are not homotopy equivalent?).
Question: Suppose that two pointed, connected CW complexes have homotopy equivalent $n$-Postnikov sections for every $n \geq 1$. Are the spaces homotopy equivalent?
I expect the answer to the question to be negative, but I'm having a hard time finding a counterexample. So far, I know that such a counterexample would have both spaces have the same homology and homotopy groups, and that both spaces would need to have nontrivial homology and homotopy groups in arbitrarily high dimension.
This is a pretty well-known phenomenon, linked with phantom maps.
One of the first existence results was Brayton Gray's paper
Spaces of the same $n$-type, for all $n$, Topology
5 (1966) 241--243
Clarence Wilkerson classified the spaces of the same $n$-type for all $n$ in
Classification of spaces of the same $n$-type for all $n$, Proc. Amer.
Math. Soc.,
60 (1976) 279--285 (1977)
Wow, thanks! I guess this also proves that I failed to do the obvious googling, namely "spaces of the same n-type for all n." Let me add that in the paper by Adams, cited in both of the references, a very simple counter-example is given, and it is also proven that the answer is positive if the spaces have finite homotopy groups.
|
2025-03-21T14:48:31.651505
| 2020-07-30T17:21:52 |
366969
|
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"sort": "votes",
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|
Stack Exchange
|
Mixing times for the exclusion process with rejection
Consider the following Markov chain on $k$-subsets of $\{1,\ldots, L\}$, equivalently, sequences $x\in \{0,1\}^L$ with $k$ 1's.
Let $p_1,\ldots, p_L\in (0,1)$ and $q_i=1-p_i$.
At each step, choose an adjacent pair of positions $(i,i+1)$ at random, and attempt to swap the values of $x_i$ and $x_{i+1}$. The two nontrivial cases are:
$x_i=0$, $x_{i+1}=1$. Accept this swap with probability $\min\{1,\frac{p_{i+1}q_i}{p_iq_{i+1}}\}$; otherwise stay at the same state.
$x_i=1$, $x_{i+1}=0$. Accept this swap with probability $\min\{1,\frac{p_{i}q_{i+1}}{p_{i+1}q_{i}}\}$.
This is a Metropolis-Hastings acceptance ratio, so it is easy to check that the chain is ergodic with stationary distribution
$$\pi (x) \propto \prod_{i:x_i=0} p_i \prod_{i:x_i=1} q_i.
$$
What is the spectral gap and mixing time of this chain? (An upper bound on the mixing time is sufficient.) Ideally, if the $p_i,q_i$ are bounded away from 0, the mixing time would be $\widetilde O(L^3)$. More generally I would expect dependence on some quantity like $\max_{i,j} \max\{ \frac{p_i}{p_j}, \frac{q_i}{q_j}\}$.
More generally, I would like to answer the question when there are more than 2 states: Let the state space be the subset of $\{0,1,\ldots, k-1\}^L$ with fixed counts $|\{i:x_i=j\}|=c_j$. Let $\sum_{j=0}^{k-1} p_{j,i}=1$ (probabilities summing to 1 for each $1\le i\le L$). Then accept a swap between $x_i=a$ and $x_{i+1}=b$ with probability $\min\{1,\frac{p_{a,i+1}p_{b,i}}{p_{a,i}p_{b,i+1}}\}$.
The mixing time in the case where $p_i=q_i=\frac12$ is well understood; this is a symmetric exclusion process with mixing time $O(L^2 \log \min\{k,L-k\})$. (See Theorem 2.4 in Lacoin 2016 for more precise asymptotics. Note the consider a continuous-time Markov process where each pair of neighboring sites gets exchanged with rate one, hence there is a factor of $L$ difference.)
My motivation for this problem comes from analyzing the mixing time of a parallel tempering Markov chain on a multimodal distribution; here $L$ is the number of temperatures and $x_i$ represents the mode that the sample at the $i$th temperature is close to.
|
2025-03-21T14:48:31.651668
| 2020-07-30T17:23:10 |
366970
|
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|
Stack Exchange
|
Potential p-norm on tuples of positive operators
This is a follow-up to this question on p-norms of tuples of operators.
Consider $\left[\begin{matrix} A \\ B \end{matrix}\right] \in B(H)^2_+$, meaning $A,B\geq 0$, and define
$$
\left\|\left[\begin{matrix} A \\ B \end{matrix}\right]\right\|_p = \|(A)^p + (B)^p\|^{1/p}.
$$
The previous question's answer conclusively showed that the generalization of the above is a norm only when $p=2$. However, the proof does not immediately transfer to this more limited situation (at least I can't see how to do it).
In this new case, it is obvious that this satisfies the triangle inequality on the positive cone for $p=1$ and $p=2$.
Q: For which $p$ does $\|\cdot\|_p$ satisfy the triangle inequality on $B(H)^2_+$?
If the answer is "no" for some $p$ then I would be very interested in a specific example as well as a more general result.
Edit: If this is established for any given p then one can create a norm via Haagerup's decomposition ideas.
|
2025-03-21T14:48:31.651768
| 2020-07-30T17:26:46 |
366971
|
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|
Stack Exchange
|
On noncommutative transcendence degrees
The original transcendence degree for (noncommutative) division algebras is the Gelfand-Kirillov transcendence degree, due to I. Gelfand and K. Kirillov ([ Sur les corps li´es aux algèbres envoloppantes des algèbres de Lie, Inst. Hautes Etudes Sci. Publ. Mat. 31 (1966), 5-19. ]) - in fact, it was introduced simultaneously with the now famous Gelfand-Kirillov dimension.
To the best of my knowledge, this is the most studied version of transcendence degrees for division algebras, considered by Borho and Kraft (1976), Lorenz (1984), Zhang (1996).
Many other versions of transcendence degree for division algebras were also considered by Resco, Rosenberg, Schoenfield, Stafford, etc.
In 1998, J. J. Zhang introduced the lower transcendence degree [ On Lower Transcendence Degree. Adv. Math. 139 (1998), no. 2, 157-193. ]. It is conjecturally equal to the original Gelfand-Kirillov transcendence degree, but has better theoretical properties, and is connected with important open problems in ring theory, and also seems suitable as an invariant in noncommutative projective geometry. For a certain wide class of algebras (Zhang calls them LD-stable) it is easely computable: it is just the $GK$ dimension.
Another important notion of transcendence degree is the homological transcendence degree due to A. Yekutieli and J. J. Zhang [ Homological transcendence degree. Proc. London Math. Soc. (3) 93 (2006), no. 1, 105-137. ], motivated by homological considerations discussed by previous authors. This transcendence degree is both theoretically well behaved and at the same time is easely computable for algebras with good homological properties (such as AS regular graded algebras).
Since, so far, there is no uniform completely satisfactory notion of transcendence degree of division algebras, just these potential candidates, what are the pros and cons, and fruitful ramifications in mathematics, of the most used transcendence degrees in the noncommutative setting?
My question is mostly about
Gelfand-Kirillov transcendence degree
Lower transcendence degree
Homological transcendence degree
but any relevant notion of noncommutative transcendence degree is welcome.
|
2025-03-21T14:48:31.651916
| 2020-07-30T17:29:47 |
366972
|
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|
Stack Exchange
|
L functions of Symmetric power of elliptic curves
Let $E$ be an elliptic curve over the raional field with conductor $N$, which corresponds to the eigenform $f(z)=\sum a_nq^n$. Let $L(Sym^2E,s)$ be the L function of the symmetric power of $E$.I am trying to understand a formula in Zagier's paper "Classical and elliptic polylogarithms and special values of L series" on Page 36 which can be used to compute $L(Sym^2E,3)$.
Let $L_2(f,s):=\zeta(s-1)L(Sym^2E,s)=\sum_{n}b_n/n^s$. By Rankin's method we have the functional equation $L^{*}_2(f,s)=L^{*}_2(f,3-s)$, where $L^{*}(f,s)=(2\pi)^{-2s}N^s\Gamma(s)\Gamma(s-1)L_2(f,s)$.
The Gamma factor $\Gamma(s)\Gamma(s-1)$ has a relation with the Bessel function:
$\int_{0}^{\infty}K_v(y)y^s\frac{dy}{y}=2^{s-2}\Gamma(\frac{s+v}{2})\Gamma(\frac{s-v}{2})$ See page 106 of Bump's book Automorphic forms and representations.
Then Zagier gives the following formula
$$L_2(f,3)=C(A-\frac{A^2}{2})+\frac{1}{16}\sum_{n=1}^{\infty}\frac{b_n}{n^3}G_1(4\pi\sqrt{\frac{nA}{N}})+\frac{2^8\pi^6}{N^3}\sum_{n=1}^{\infty}b_nG_2(4\pi\sqrt{\frac{n}{NA}}),$$
Where $$C=2\pi^2N^{-1}L(Sym^2E,2)$$ and
$$G_1(x):=\int_x^{\infty}t^4K_1(t)dt$$ and $$G_2(x):=\int_x^{\infty}t^{-2}K_1(t)dt$$
Zagier mentioned that the formula $L_2(f,3)$ is obtained by splitting up the integral of the Mellin transform of the K-Bessel function into two pieces in the usual way. I would like to know how this can be done by choosing some $A$ in the formula of $L_2(f,s)$
The method for passing from a functional equation to computing (approximating) values is due to Lavrik. See Appendix B of Henri Cohen's second book (Advanced topics in computational number theory). Also useful is Buhler, Schoen, and Top (Cycles, L-functions and triple products of elliptic curves) who do the symmetric cube, and 4.4 of Martin and Watkins (Symmetric powers of elliptic curve L-functions).
@OmniaOperator, 4.4 of Martin and Watkins use Cauchy residue theorem to shift contour of integration and it turns out to converge really slow as it is the case of Dirichlet series. The method in Zagier's paper use K-bessel function, which speed up the convergence to exponential decay.
I think the convergence is essentially the same, taking about $N$ terms. My reading of the M-W result is the Mellin transform dies like $O(\exp(-x^{2/3}))$ and Zagier's dies like $O(\exp(-\sqrt x))$ (he uses a degree 4 product, rather than the direct degree 3 symmetric square, so his convergence is marginally slower). Here $x=n/N$.
|
2025-03-21T14:48:31.652091
| 2020-07-30T18:19:47 |
366975
|
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|
Stack Exchange
|
On thinking of spacetime as a local Scott domain
An observation of Martin and Panangaden links the study of Lorentzian manifolds and the semantics of programming languages via the theory of Scott domains.
Background:
Recall that if $M$ is a time-oriented Lorentzian manifold with no closed causal curves, then there are two important partial orders on $M$, and one additional important relation:
$(M,<)$ is a poset, where $p < q$ if there is a causal path [1] from $p$ to $q$.
$(M, \ll)$ is a poset, where $p \ll q$ if there is a chronological path [1] from $p$ to $q$.
We say that $p$ is horismotic to $q$, and write $p \to q$, if $p < q$ but $p \not \ll q$ [1].
These relations, including various ways of defining one in terms of the other, were first studied in the abstract by Kronheimer and Penrose.
The observation of Martin and Panangaden is that, at least if $M$ is globally hyperbolic, then one way of defining the $\ll$ relation from the $<$ relation (not studied by Kronheimer and Penrose) is precisely the one suggested by the notation from the perspective of domain theory. That is, we have that
$p \ll q$ if and only if $p$ is way below $q$ with respect to the $<$ poset structure [2]. Moreover, $(M,<)$ is a continuous poset [2]. In fact, $(M,<)$ is a local domain [2].
Everything here has a dual statement given by reversing the time orientation, and in fact $M$ is a strongly bicontinuous poset [3].
What I'd like to understand:
What I'd like to understand is just how strong this connection between relativity and domain theory really is. The first thing that stands out to me is that the domains coming from relativity theory don't seem to be "generic" from the point of view of domain theory -- they have various special properties. I'd like to understand what to make of these special properties from the point of view of domain theory. To start with:
Question 1: Is there an intrinsic domain-theoretic reason to study bicontinuous posets [3] rather than just continuous posets? From a domain-theoretic perspective, are there important examples of bicontinuous posets? Are there things one can do and questions one can ask about bicontinuous posets which can't be done with more general continuous posets?
Next, there are various special properties of the horismotic relation in the domains $M$ coming from relativity theory. Most obviously,
(A) there can be no cycles $x_0 \to x_1 \to \dots \to x_n = x_0$.
There are additional special properties coming from the uniqueness of (lightlike) geodesics. The most succinct way to put it is that
(B) if $w,x,y,z$ are 4 distinct points of $M$, and if 5 of the 6 unordered pairs of elements of $\{w,x,y,z\}$ are related (in either direction) by the $\to$ relation, then so is the 6th pair [4].
Kronheimer and Penrose called any spacetime satisfying a certain fragment of (B) regular [4]. Note that the analogous condition for 2 distinct points of $M$ rather than 4 would say that $M$ is linearly ordered by $\to$, so (B) is in some sense a weakening of linearity under $\to$. If $(P,<)$ is an arbitrary poset, then we can still define $x \to y$ if and only if $x < y$ but $x \not \ll y$ (where $\ll$ is the way below relation [2]), and then (A) will hold automatically. But (B) seems to be a curious property from the perspective of order theory or domain theory.
Question 2: Does condition (B) "mean anything" domain-theoretically? Is it in any way a natural domain-theoretic condition to consider? Are there perhaps "more-natural" conditions which imply it or are implied by it? What can one "do" with such conditions domain-theoretically?
I could go on, but maybe I'll close with one final question. In a domain $M$ which is a Lorentzian manifold, it's the case, essentially by definition, that if $p \ll q$, then there is a chronological path [1] from $p$ to $q$. From a purely order-theoretic perspective, this is pretty well encapsulated by saying that there exists a map $\gamma: [0,1] \to M$ such that $\gamma(0) = p$, $\gamma(1) = q$, $s < t \Rightarrow \gamma(s) \ll \gamma(t)$, such that $\gamma$ preserves infs and sups.
Question 3: Do "chronological paths" according to the above definition ever come up in domain theory?
Footnotes:
[1] A smooth path $\gamma: [0,1] \to M$ with $\gamma(0) = p$ and $\gamma(1) = q$ is causal if $\gamma'(t)$ lies in the closed future lightcone for all $t \in [0,1]$ and chronological if $\gamma'(t)$ lies in the open future lightcone for all $t \in [0,1]$. I believe this makes it the case that $p \to q$ if and only if there is a future-directed lightlike geodesic from $p$ to $q$.
[2] If $P$ is a poset and $p,q \in P$, then we say that $p$ is way below $q$ and write $p \ll q$ if, for every directed set $D \subseteq P$ with sup $\bigvee D$, if $q \leq \bigvee D$, then there exists $d \in D$ such that $p \leq d$. A poset is $P$ said to be continuous if, for every $p \in P$, the set $\{q \in P \mid q \ll p\}$ is directed and has sup given by $p$. A poset $P$ is said to be a domain if it is continuous and has sups of directed sets, and a local domain if it is continuous and has sups of bounded directed sets.
[3] A strongly bicontinuous poset $(P,<)$ is a poset such that $P$ and $P^{op} = (P,>)$ are both bicontinuous posets and moreover we have that $p$ is way below $q$ with respect to $<$ if and only if $q$ is way below $p$ with respect to $>$.
[4] More explicitly, in light of (A), (B) reduces to the following conditions for any 4 distinct points $w,x,y,z \in M$:
If $w \to x \to y \to z$ and $w \to y$ and $x \to z$, then $w \to z$. This is a sort of "locality" property for the $\to$ relation, allowing us to patch together lightlike geodesics from smaller pieces.
(regularity in the sense of Kronheimer and Penrose) If $w \to x \to y$ and $w \to y$ and $x \to z$ and $w \to z$, then $y \to z$ or $z \to y$. Dually, if $x \to y \to z$ and $x \to z$ and $w \to y$ and $w \to z$, then $w \to x$ or $x \to w$. This says that a lightlike geodesic may be extended (at either end) in at most one way.
If $w \to x$, $y \to z$, $w \to y$, $x \to z$, and $w \to z$, then $x \to y$ or $y \to x$. This says that if $w \to z$, then the interval between them $\{p \in P \mid w \leq p \leq z\}$ is linearly ordered by $\leq$ (or equivalently by $\to$).
I think the condition (B) I have formulated is actually too strong in general. Better is the fragment called "regularity" by Kronheimer and Penrose. My mistake is in thinking that any points $x,y$ with a null geodesic from $x$ to $y$ are horismotic. This is false, as there might also be a timelike path from $x$ to $y$, especially if there is interesting topology at work.
|
2025-03-21T14:48:31.652517
| 2020-07-30T19:57:54 |
366978
|
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|
Stack Exchange
|
Finding solutions of the differential equation $x\frac{d}{d x}\left(x\frac{d y}{d x}\right)=4(y-x^{2}y+y^{3})$
In my research I have come across the following non-linear differential equation:
$$x\frac{d}{d x}\left(x\frac{d y}{d x}\right)=4(y-x^{2}y+y^{3})$$
I want to find the general solution of this equation for $x\geq0$, but didn't manage to do it.
I have found some facts about the solutions by investigating the asymptotics:
For $x\ll1$, the solutions go like $y=Cx^2$ for some constant $C$, or diverge.
For $x\gg1$, the solutions go like $y=x$, or go to $0$ (probably after oscillating, found by numerics), or diverge.
In particular, I am interested in a one of the solutions. By looking upon the numerics, I believe that there is only one solution that goes like $y=x$ for $x\gg1$, and it does not diverge at $x=0$. I didn't manage to prove this by now.
I am interested in finding this solution analytically, or if impossible, just find appropriate $C$ analytically.
Thanks!
Did you try to change the independent variable $x=e^t$?
Yes, It was helpful in order to investigate the asymptotics for $x\ll1$, but for arbitrary $x$ it gives
$$\frac{d^{2}y}{dt^{2}}=4\left(y-e^{2t}y+y^{3}\right)$$
which I did not find easier to solve.
All I can currently do is to show that at least one solution with the properties you required exists. But since it is just a minimizer of some quite fancy energy functional (actually, even worse than that: it is actually the limit of a sequence of minimizers on finite intervals; its energy on the whole line diverges to $-\infty$), I doubt very much that anything explicit can be done here. Let me know if you are interested in that existence proof.
Looking at the plot of $z=f(t,y)=4y(1-e^{2t}+y^2)$, $y$ can't be asymptotic to $e^t$ for ever. That would either make $y=y(t)$ infinitely steep or make it cross the line $\sqrt{e^{2t}-1}≈e^{t}$ and make $z<0$, which would make $y=y(t)$ oscillatory. It would oscillate between the two positive and negative regions.
Writing in the form $$ y''=4(y-e^{2t}y+y^3)$$ one can guess that the behavior is similar to the solutions of $y''=4(y+y^3)$ for $t \to -\infty$. The behavior at $+\infty$ looks even more difficult, though oscillating. By the way, it is not clear to me if all solutions exist globally or explode in finite time, due to the presence of the non linear term $y^3$.
@fedja I am interested in your existence proof. Nice if you give the details. Thanks
Mathematica's command DSolve cannot solve your ODE in closed form; so, apparently, a closed-form solution does not exist.
However, it will shown here that, in line with your conjecture, for each real $c$ your ODE has solutions $y$ such that $y(x)=(c+o(1))x^2$ as $x\to0+$.
Indeed, your ODE is
\begin{equation*}
x^2 y''+x y'=4(y-x^2y+y^3) \tag{1}
\end{equation*}
for real $x>0$.
Assuming the initial conditions $y(0+)=y'(0+)=0$ and using the Taylor formulas $y'(x)=x\int_0^1 v(sx)\,ds$ and $y(x)=x^2\int_0^1 (1-s)v(sx)\,ds$ with $v:=y''$, we rewrite (1) as
\begin{equation*}
v(x)=\int_0^1 K(x,s)v(sx)\,ds+4x^4\Big(\int_0^1 (1-s)v(sx)\,ds\Big)^3, \tag{2}
\end{equation*}
where
$$K(x,s):=[4(1-x^2)(1-s)-1]s.$$
For any real $c$, setting now
\begin{equation*}
v(x)=c+x w(x),
\end{equation*}
we rewrite (2) as
\begin{equation*}
w(x)=F(w)(x):=-2cx+\int_0^1 K(x,s)w(sx)\,ds+4x^3\Big(\int_0^1 (1-s)(c+xw(sx))\,ds\Big)^3. \tag{3}
\end{equation*}
Take now any real $m>0$ and $h>0$, and let
\begin{equation*}
W_{m,h}:=\{w\in C[0,h]\colon\|w\|\le m\},
\end{equation*}
where $\|w\|:=\max_{x\in[0,h]}|w(x)|$.
Since $|K(x,s)|$ is convex in $x^2$, for $x\in[0,1]$ we have
\begin{equation*}
\int_0^1 |K(x,s)|\,ds\le\max\Big(\int_0^1 |K(0,s)|\,ds,\int_0^1 |K(1,s)|\,ds\Big)
=\max(19/48,1/2)=1/2.
\end{equation*}
So, for any $w\in W_{m,h}$
\begin{equation*}
\|F(w)\|\le2|c|h+m/2+h^3(|c|+hm)^3/2\le m \tag{4}
\end{equation*}
if $h>0$ is small enough -- which will be assumed in what follows. So, $F$ maps $W_{m,h}$ into $W_{m,h}$.
Moreover, if for some $w$ and $u$ in $W_{m,h}$ we have $\|w-u\|\le t$ for some real $t\ge0$, then similarly to (4) we get
\begin{equation*}
\|F(w)-F(u)\|\le t/2+3h^3(|c|+hm)^2ht/2\le\tfrac23\,t
\end{equation*}
if $h>0$ is small enough. So, $F$ is a contractive map of $W_{m,h}$ into $W_{m,h}$, and hence $F$ has a fixed point. That is, equation (3) has a solution $w\in W_{m,h}$.
Thus, ODE (1) has a solution $y$ with $y''(x)=c+x w(x)$ for $x\in[0,h]$ and $w\in W_{m,h}$, and with $y(0+)=y'(0+)=0$. So, $y(x)=cx^2/2+O(x^3)$ as $x\to0+$.
$\Box$
wouldn't the small-$x$ behavior already follow upon linearization of the ODE? the linearized ODE has solution $c J_2(2x)=cx^2/2+{\cal O}(x^3)$.
Thank you losif, nice to see the proof of the conjecture!
Nevertheless, it does not give new details about the eqution, details that I still hope we can find.
@CarloBeenakker : The linearization needs to be justified. The answer here may be viewed, in part, as such a justification. (However, the method -- which is rather standard -- should work, not just for this ODE, but in a general setting, without even an implicit association with linearization. In view of the Schauder fixed point theorem, one does not even need the right-hand-side function $F$ to be contractive.)
Is this too naive? To obtain the large-$x$ behavior, I take a series expansion of $y(x)$ in powers of $1/x$,
$$y(x)=c x + a_0 + a_1/x + a_2/x^2+ a_3/x^3 +\cdots,$$
substitute into
$$R(x)=x\frac{d}{d x}\left(x\frac{d y}{d x}\right)-4(y-x^{2}y+y^{3}),$$
expand $R(x)$ in powers of $1/x$ and demand that the leading order terms vanish. It follows that $c=1$, $a_{n}=0$ for any even $n$, while for odd $n$ I find
$$a_1=-\frac{3}{8},\;\;a_3=-\frac{9}{128},a_5=-\frac{99}{1024},\;\;a_7=-\frac{11637}{32768},\;\;a_9=-\frac{627669}{262144},$$
so coefficients of the form $a_{2n+1}=-b_n/2^{2n+1}$ with integer $b_n$.
I believe that this is indeed true for solutions that does not go to $0$ or diverge at infinity. This is the way I found the asymptotics for $x\gg1$ I wrote about - but this is not the only solution, there are vanishing and diverging solutions too.
Nevertheless, I do not see how to use it to prove that the solution that does not vanish or diverge at infinity is unique, and how to find the constant $C$ of the $x\ll1%$ asymptotics of this solution, or show it exists.
Good question. It could happen that the convergence radius of the series in $1/x$ is zero ... so one would need a statement about the asymptotics of the $a_n $ to settle that.
@MichaelEngelhardt : I'd guess the radius of convergence of the series expansion is $0$ in this case. Of course, we could use just an asymptotic expansion for $y(x)$, provided that we have the matching asymptotic expansions for $y'(x)$ and $y''(x)$. But of course, we'd need to show that a solution $y$ admitting such asymptotic expansions exists.
|
2025-03-21T14:48:31.652968
| 2020-07-30T20:40:08 |
366980
|
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|
Stack Exchange
|
An explicit (maybe algebraic) isometric embedding of the double torus with constant curvature K = -1
The following question is related to this previous question, Canonical immersion of the double torus:
Is there any known explicit (maybe algebraic) isometric embedding of a genus 2 surface endowed with a metric of constant curvature K = -1 into a Hilbert space of finite dimension?
Isometric embedding into where?
Into a Hilbert space of finite dimension.
Or, to be more concrete, into $\mathbb{R}^n$ for some $n > 0$?
Into an R^n with a fixed inner product.
I’m pretty sure none is known, and it’s an open question whether any such “explicit” embedding or evening just immersion exists.
|
2025-03-21T14:48:31.653042
| 2020-07-30T20:48:46 |
366981
|
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|
Stack Exchange
|
Restriction of holomorphic functions on $G$-invariant subspace
Let $X$ be a complex manifold with a holomorphic action of a complex reductive group $G$. Let $Y \subset X$ be a $G$-invariant reduced complex analytic subspace. Is the restriction
$$
\mathcal{O}_X^G \to \mathcal{O}_Y^G
$$
surjective (on stalks)? Here $\mathcal{O}_X^G$ is the sheaf of $G$-invariant holomorphic functions on $X$ and similarly for $Y$.
After writing an answer in terms of global sections, then realising the original question was in terms of stalks, I realised that I don't understand what "the sheaf of $G$-invariant holomorphic functions on $X$" means: what is $\mathscr O_X^G(U)$, if $U$ is a non-$G$-stable open subset of $X$? I wonder if you might mean the sheaf of holomorphic functions on the quotient $X/G$, which is defined by putting $\mathscr O_{X/G}(\overline U_X) = \mathscr O_X(U_X)^G$ for any open subset $\overline U_X$ of $X/G$ with pullback $U_X \subseteq X$. I'll answer the question that way, but, since the technique I describe is very robust, hopefully I can adapt it to whatever the correct interpretation is.
Fix a point $\overline y \in Y/G$, and a germ $f$ at $\overline y$ (that is, an equivalence class of $G$-invariant holomorphic functions on $Y$ defined on the pullback of an open neighbourhood in $Y/G$ of $\overline y$).
Fix an open subset $\overline U_X$ of $X/G$ containing $\overline y$ that is small enough that $f$ is defined on $\overline U_X \cap Y/G$ and that, if we write $U_X$ for the pullback to $X$ of $\overline U_X$, then $f$ extends to a holomorphic function on $U_X$. Then consider the space $\mathscr V$ of functions on $X$ whose restriction to $Y$ is proportional to $f$. This is a $G$-representation, hence also a $\mathfrak g$-representation.
Let $K$ be the compact form of $G$. Since $K$ is compact, the representation $\mathscr V$ has a non-$0$, $K$-fixed vector $F$. The span of $F$ is annihilated by $\mathfrak k$, hence by $\mathfrak g = \mathfrak k \otimes_{\mathbb R} \mathbb C$; so $F$ itself is $G$-fixed.
Thanks a lot for your answer. Yes, that's the correct interpretation; sorry for not being very precise. On the last point: is it true that every complex representation (possibly infinite-dimensional) of a compact Lie group has a non-zero fixed-point? I wasn't aware of that. Any reference or easy argument?
hmm... that can't be right since that's equivalent to a 1-dimensional subrepresentation so the irreducible representations have no non-zero fixed points. I don't understand your argument then. What am I missing?
No, not every representation has a non-$0$ fixed point, but it does have a canonical projection to its fixed space (i.e., its isotrivial component). In this concrete case, since we are restricting to a $G$-fixed function on $Y$, the canonical form of the restriction means we don't change the restriction. Since the restriction is non-$0$ (else extendibility is easy!), so too is the $K$-fixed function that extends it.
(Sorry, the canonical form of the projection, by averaging over $K$-transforms, means we don't change the restriction.)
|
2025-03-21T14:48:31.653382
| 2020-07-30T22:19:16 |
366983
|
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|
Stack Exchange
|
Known dense subset of Schwartz-like space and $C_c^{\infty}$?
After reading this question, which asked for some examples of commonly used (proper) dense subsets of $C_0^{\infty}(\mathbb{R}^n)$ with the $L^p$-norm I wonder. What are some "well-known" examples of countable subsets of $X\subseteq C_c^{0} \cap \mathscr{S}(\mathbb{R}^n)$ (the Schwartz space) for which
$X$ is dense in $C_c^{0}$ for the topology generated by the sup-norm $\sup_{x \in \mathbb{R}^n} \|f(x)\|$
$X$ is dense in $\mathscr{S}(\mathbb{R}^n)$ for the topology generated by the semi-norms
$
p_{\alpha,\beta,K}(f):=\sup_{x\in K} \|x\|^\alpha \left\| (D^{\beta} f)(x) \right\|
.
$
$\sup_{x\in\mathbb{R}^n}|D^{\beta}f(x)|$ are not the seminorms you should use for $C_c^{\infty}$.
@AbdelmalekAbdesselam thank you, I refined the question.
The completion of your first space is the Banach algeba of continuous functions which vanish at infinity. There is a version of Stone-Weierstraß for this algebra which makes it relatively easy to construct families with the desired property. The only problem is to get a single such function to start with and there is standard way to construct a bell shaped one. One can then take its dilations and translates, and the algebra generated to get an example (taking care to keep the result countable by means of the usual ploy of only using rational coefficients and parameters).
Yes I was going to suggest this too using $\exp(-1/(1-|x|^2)I_{Ball_{\mathbb{R}^n}(0,1)}$. Since it also lies in the Schwartz-space (since its smooth and rapidly-decreasing) but I'm not sure if it is dense in the latter space.
@AnnieLeKatsu Yes but how to show this? And to user131781 using Annie's example we would be taking shifts and linear combinations of this no?
I don't know about "well known" or canonical answers to this question, but it is easy to construct an $X$ that works as follows.
Using the definition of Hermite polynomials given by
$$
H_n(x)=(-1)^n e^{x^2}\left(\frac{d}{dx}\right)^n e^{-x^2}\ ,
$$
we define the one-dimensional Hermite functions
by
$$
h_n(x)=\pi^{-\frac{1}{4}} 2^{-\frac{n}{2}} n!^{-\frac{1}{2}} e^{-\frac{x^2}{2}} H_n(x)
$$
and then the $d$-dimensional Hermite functions by
$$
h_{\alpha}(x_1,\ldots,x_d)=h_{\alpha_1}(x_1)\cdots h_{\alpha_d}(x_d)
$$
for every multiindex $\alpha$.
These functions form an orthonormal basis for the Hilbert space $L^2(\mathbb{R}^d)$ as well as as an unconditional Schauder basis of Schwartz space $\mathscr{S}(\mathbb{R}^d)$.
Clearly, finite linear combinations of the $h_{\alpha}$ with rational coefficients is a countable dense subset of $\mathscr{S}(\mathbb{R}^d)$. To simultaneously satisfy the other condition one can pick a smooth cutoff function $\rho:\mathbb{R}^d\rightarrow [0,1]$, constant equal to $1$ on the ball $B(0,1)$ and equal to zero outside the ball $B(0,2)$. Now take previous linear combinations and multiply them by $x\mapsto \rho(\frac{1}{k}x)$, for $k=1,2,\ldots$ This will give a set $X$ that fulfills the two requirements.
Could you add a reference on density of Hermite functions in the Schwartz space?
https://aip.scitation.org/doi/abs/10.1063/1.1665472 although a better presentation is in the book https://bookstore.ams.org/simon-1
|
2025-03-21T14:48:31.653613
| 2020-07-31T00:08:06 |
366987
|
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|
Stack Exchange
|
A new notion of probability coupling
Let $X$ and $Y$ be two discrete random variables distributed according to $\mu$ and $\nu$, respectively. Consider the following optimization problems
$$\inf_{\pi\in \Pi(\mu, \nu)}\Pr(X\neq Y),$$
where $\Pi(\mu, \nu)$ denotes the set of all couplings of $\mu$ and $\nu$, i.e., all joint distributions $P_{XY}$ whose marginals are given by $\mu$ and $\nu$.
It is very well-known that the above minimization is given by total variation distance:
$$\inf_{\pi\in \Pi(\mu, \nu)}\Pr(X\neq Y) = d_{TV}(\mu, \nu).$$
My question is s follows. Now consider $X\sim \mu$ and $Y_1$ and $Y_2$ are two independent samples of $\nu$. Now, what is
$$\inf_{\pi\in \Pi(\mu, \nu)}\Pr(X\neq Y_1, X\neq Y_2)?$$
Has it appeared before? Any connections with known probability distance measures?
Did you mean $\pi(X \ne Y)$ rather than $\operatorname{Pr}(X \ne Y)$? Also, what is the joint distribution of $(X,Y_1,Y_2)$ once $\pi$ is chosen? (If $\mu=\nu$ and $\pi$ is the diagonal coupling, and both $(X,Y_1)$ and $(X,Y_2)$ are assumed to be distributed according to $\pi$, then we have $X=Y_1$ a.s. and $X=Y_2$ a.s., so it is problematic to require that $Y_1$ and $Y_2$ are independent.)
Yes, I mean $\pi(X\neq Y)$. The joint distribution is given by $\pi(x, y_1, y_2) = \nu(y_1)\nu(y_2)P_{X|Y_1Y_2}(x|y_1,y_2)$ for some conditional distribution $P_{X|Y_1Y_2}$. Then the first marginal should be $\mu$, the second and third should be $\nu$.
If $\mu = \nu$, then the above infiumum should be zero.
But look: the infimum is over couplings $\pi$ of $\mu$ and $\nu$, while under the infimum a triple coupling $\bar\pi$ of $\mu$, $\nu$ and $\nu$ is used. My point is that for a general $\pi$ it is not possible to construct any $\bar\pi$ such that $\pi(A \times B) = \bar\pi(A \times B \times \mathcal Y) = \bar\pi(A \times \mathcal Y \times B)$ and $\bar\pi(\mathcal X \times B_1 \times B_2) = \nu(B_1)\nu(B_2)$. (Here of course $\mathcal X$ is the set of all possible outcomes of $X$, and $\mathcal Y$ the set of all outcomes of $Y$.)
I see your point! Maybe the way it's written is not clear!
The question is this: given two marginals $\mu$ and $\nu$, what is minimum value of $\Pr(X\neq Y_1, X\neq Y_2)$ where $P_{X.Y_1, Y_2}(x, y_1, y_2) = \nu(y_1)\nu(y_2)P_{X|Y_1Y_2}(x|y_1,y_2)$?
Note that this quantity can vanish even if $\mu \neq \nu$, for example when $\mu$ is the law of $\min{Y_1,Y_2}$, so it wouldn't be a good notion of 'distance'...
It's a very nice observation! But I expected it...I think it'll be a certain type of $f$-divergence [which is not strictly convex at 1, and thus it being equal to zero does not necessarily imply $\mu = \nu$].
|
2025-03-21T14:48:31.653825
| 2020-07-31T01:33:30 |
367989
|
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|
Stack Exchange
|
Interpretations of modal logic where $\Box$ means "valid"
Consider the propositional modal language in one propositional letter, $p$.
Recall that a pointed Kripke frame is a Kripke frame $(W,R)$ with a designated world $w_0\in W$, and a sentence is valid in a pointed Kripke frame iff it is true at $w_0$ for every interpretation of the propositional letters as subsets of $W$.
I'm wondering if it's possible to find a finite model in which $\Box$ means "valid". More precisely, is it possible to find a finite transitive reflexive pointed Kripke model $(W,R,w_0, [[\cdot]])$ such that
$w_0 \Vdash \Box A$ if and only if $A$ is valid in $(W,R,w_0)$?
Certainly it can be done in an infinite frame. For instance, over the infinite tree that has omega many daughters at any node, you can make each sentence satisfiable in the frame true at one of the daughters of the base node. And this even works with infinitely many propositional letters.
(For context: I got thinking about this question after coming back to this earlier question about logical interpretations of $\Box$.)
What does $w \Vdash B$ stand for? I am reading your question as "Can we have a finite Kripke model $(W, R, w_0)$ such that, for all $A$, $w_0 \Vdash \Box A$ iff $w_0 \Vdash A$?" but I don't think that's what you are asking. (If it is, consider the trival model.)
@AndrejBauer Sorry there was a typo in an earlier version of the question (I wrote "frame" instead of "model"). Does it make sense now: $w\Vdash B$ just means $B$ is true at $w$ in the pointed Kripke model $(W,R,w_0,[[.]])$ ?
This is impossible since a reflexive frame has the infinite sequence of related points $wRwRwR….$ for each point $w$.
Provability Logic (modal logic GL) corresponds to finite, transitive, irreflexive frames. GL interprets $\Box$ as an encoding of a proof in a language that can formulate its own provability predicate. GL is characterized by $\Box (\Box P \to P) \to \Box P$, which takes reflexivity off the table immediately.
$\def\R{\mathrel R}$No, this is not possible.
Recall that the depth of a point $x$ in a transitive frame $(W,R)$ is the maximal length $d$ of a strictly increasing chain starting at $x$, i.e., $x_1,\dots,x_d$ such that $x_d=x$ and $x_{i+1}\R x_i$, $x_i\not\R x_{i+1}$.
There are formulas in one variable that are satisfiable only in frames of depth $\ge d$ (cf. Thm. 12.21 in Chagrov&Zakharyaschev, Modal logic). Moreover, it is possible to define them in such a way that when satisfied in a model of depth exactly $d$, they force a particular value for $p$ in all points in the root cluster; we will obtain a contradiction from this. An explicit construction of such formulas follows below.
Consider the formulas
$$\begin{align}
\theta_1(p)&=\Box p,\\
\theta_{i+1}(p)&=p^{i+1}\land\Diamond\theta_i(p)\land\Box\Bigl(p^{i+1}\lor\bigvee_{j\le i}\theta_j(p)\Bigr),
\end{align}$$
where
$$p^i=\begin{cases}\phantom{\neg}p&\text{if $i$ is odd,}\\\neg p&\text{otherwise.}\end{cases}$$
We will use the property that these formulas are pairwise contradictory; moreover, the following formulas are valid:
$$\theta_j\to\Box\neg\theta_i,\qquad j<i.\tag{$*$}$$
We can prove this by induction on $i$. For $i=1$, there is nothing to prove. Assuming it holds for $i$, we show it for $i+1$ as follows. Let $j\le i$, and assume for contradiction that $x\R y$ are such that $x\models\theta_j$ and $y\models\theta_{i+1}$. If $j<i$, $y\models\Diamond\theta_i$ contradicts the induction hypothesis. If $j=i$, we have $y\models\neg p^i$. This directly contradicts the definiton of $\theta_1$ for $i=1$; otherwise, the definition of $\theta_i$ gives $y\models\bigvee_{j<i}\theta_j$, which together with $y\models\Diamond\theta_i$ contradicts the induction hypothesis again. This finishes the proof of $(*)$.
Now, assume for contradiction that $(W,R,w_0,V)$ is as in the question, and let $d$ be the depth of $w_0$. The formula $\theta_d(p)$ is satisfiable in $(W,R,w_0)$ by the valuation that makes $p$ true in points of odd depth, and false in points of even depth. That is, $\neg\theta_d(p)$ is not valid in the pointed frame $(W,R,w_0)$, thus by assumption, $w_0\not\models\Box\neg\theta_d(p)$, i.e., there is $x_d$ such that
$$w_0\R x_d\models\theta_d(p).$$
Unwinding the definition, we find a chain $x_d\R x_{d-1}\R\dots\R x_1$ such that $x_i\models\theta_i(p)$. This implies $x_i\not\R x_{i+1}$, as $\theta_i\to\Box\neg\theta_{i+1}$ is valid by $(*)$. Thus, the chain $x_d,\dots,x_1$ is strictly increasing. Since $w_0$ does not have depth $\ge d+1$, we must have $x_d\R w_0$. This implies $w_0\models p^d\lor\bigvee_{j\le d-1}\theta_j(p)$. Using $(*)$, we cannot have $w_0\models\bigvee_{j\le d-1}\theta_j(p)$ as $w_0\R x_d\models\theta_d(p)$, thus we obtain
$$w_0\models p^d.$$
However, since $\neg\theta_d(p)$ is not valid in the frame, $\neg\theta_d(\neg p)$ is not valid there either. Then the same argument as above with $p$ and $\neg p$ swapped gives
$$w_0\models\neg p^d.$$
This is a contradiction.
I formulated the argument above for reflexive transitive frames as requested, but it can be easily adapted to arbitrary finite pointed Kripke frames $(W,R,w_0)$: we take for $d$ the depth of $w_0$ under the transitive closure of $R$, and replace all instances of $\Box$ inside the $\theta_i$ formulas by the defined modality
$$\Box^{\le n}\phi=\bigwedge_{i=0}^n\underbrace{\Box\dots\Box}_i\phi,$$
where $n=|W|$. Note that $\Box^{\le n}$ is the box modality corresponding to the transitive reflexive closure of $R$.
By the way, the situation for intuitionistic logic is quite different. For example, the intuitionistic version of the required property holds for the unique 5-element rooted submodel of the Rieger–Nishimura ladder, more or less because its underlying frame does not map p-morphically onto the unique 4-element rooted subframe of the RN ladder.
Thank you! Yes, I was able to find the model of the intuitionistic version, which is partly what inspired the question.
|
2025-03-21T14:48:31.654885
| 2020-07-31T04:25:35 |
367995
|
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|
Stack Exchange
|
Ask for some reference about isoperimetric constant on Voronoi diagrams?
Given a Poisson point process $\mathcal{P}$ in $\mathbb{R}^2$, the $\textbf{Voronoi cells}$ of a point $p\in \mathcal{P}$ is defined by
$$V(x):=\{y\in \mathbb{R}^2: \|x-y\|=\min_{x'\in \mathcal{P}}\|x'-y\|\}
$$
where $\|\cdot\|$ is the $\ell^2$ norm.
The Delaunay triangulation $DT(\mathcal{P})$ is the dual graph of the Voronoi diagram.(See figure as below)
Given a square $Q$ with side length $s(Q)=n$ and $\hat{Q}$ with side length $s(\hat{Q})=n/25$ such that each cell of $Q$ that is divided into $25^2$ parts by cells of $\hat{Q}$. Assume that there exists at least one point in $\hat{Q}$ with high probability. If we define the $\Gamma:=DT(\mathcal{P})\cap Q-A$ where $A\subset DT(\mathcal{P})\cap Q$ and $\partial_E(A_1, A_2)=\{e=\{x,y\}: x\in A_1, y\in A_2\}$.
How to prove that isoperimetric inequality on Delaunay triangulation $DT(\mathcal{P})$ within a square $Q$,
$$\mathbb{P}(\frac{|\partial_E(A, \Gamma)|}{|A|}\geq \frac{c}{n})\ge 1-e^{-cn}$$
for any $A\subset DT(\mathcal{P})\cap Q$ and $|A|\le 0.5| DT(\mathcal{P})\cap Q|$?
Is there any reference for this problem? Thanks.
|
2025-03-21T14:48:31.654991
| 2020-07-31T05:46:47 |
367997
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/367997"
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|
Stack Exchange
|
A question on $a_i(n) = a_i(\pi(n)) + a_i(n-\pi(n))$ with $a_i(n) = 1$ for $n \le i$
Let $a_i(n) = a_i(\pi(n)) + a_i(n-\pi(n))$ with $a_i(n) = 1$ for $n \le i$ where $\pi(n)$ is the prime-counting function.
By definition, it is obvious that $a_1(n) = n$ and $a_2(n)$ is https://oeis.org/A316434.
Question. Does $\lim_{n\to\infty}\frac{a_i(n)}n$ exist for any fixed value of $i \ge 2$ ?
(Below plot shows initial values of $\frac{a_i(n)}n$ for $2 \le i \le 7$.)
Thanks.
|
2025-03-21T14:48:31.655057
| 2020-07-31T05:49:01 |
367998
|
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|
Stack Exchange
|
pursuit-evasion based on Schroeder's upper bound for graphs of genus $g$
I am following Schroeder's work on pursuit-evasion games on graphs (often called "cops and robbers"). In his 2001 publication ("The copnumber of a graph is bounded by $\lfloor 3/2 {\ \rm genus}(G)+3\rfloor$". In: Categorical perspectives (Kent, OH, 1998). Trends in Mathematics, pp. 243-263. Birkhäuser, Boston 2001) he derived an upper bound for the cop number $c(G)$ that depends on the genus $g$ of the surface on which the graph $G$ can be embedded: $c(G)\leq \lfloor 3g/2 +3\rfloor$.
My most recent reference for this result is "Topological directions in Cops and Robbers" from 2018, Anthony Bonato and Bojan Mohar, arXiv:1709.09050v2 .
This gives $c(G)\leq 4$ if $G$ can be embedded on a torus. Now, I have worked extensively to come up with an example of a graph $G$ that actually hits this bound, i.e. I have searched for $G$ with $c(G)=4$, but with no success. So I am starting to see strong evidence for the conjecture $c(G)\leq 3$ if $G$ can be embedded on a torus. Question: Is someone aware of a more recent reference for this conjecture? It appears lower than any other bound I have seen in the literature so far (N.B. I would also be interested in references beyond torus embeddings)
What you conjecture has been conjectured (more or less explicitly) a few times before. In the paper by Bonato and Mohar that you reference, it is dubbed the Andreae-Schroeder conjecture.
I recently proved that it is true, i.e. the cop-number of toroidal graphs is at most 3, see this ArXiv preprint. See also this preprint, where a general bound $c(G) \leq \frac{4g}{3} + \frac {10}3$ is proved.
I am very impressed by your result, just took a first look. Your second reference with the general bound is amazing: extremely close to the original Schroeder $g+3$ conjecture!!
|
2025-03-21T14:48:31.655495
| 2020-07-31T06:33:54 |
368000
|
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|
Stack Exchange
|
Are there $f, g$ such that $\int_{S^{1}} |f'|^{2}+|g'|^{2}d\theta-2\int_{S^{1}}f'g<0,$ where $f'=\frac{\partial f}{\partial \theta}$
Let $f,g$ are the functions of $S^{1}$. Are there $f, g$ such that
$$\int_{S^{1}} |f'|^{2}+|g'|^{2}d\theta-2\int_{S^{1}}f'g<0,$$
where $f'=\frac{\partial f}{\partial \theta}$?
$f(\theta)=\theta$ is not well defined on $S^{1}.$
OK, comment deleted.
No, there are no such functions. Indeed, if $h$ is periodic and has mean zero, then $\|h\|_2 \le \|h'\|_2$. Take $f,g$ as above and write $g=g_1+c$ with $g_1$ having mean zero. Then
$$
\|f'\|_2^2+\|g'\|_2^2<2\int_{S^1}f'g=2 \int_{S^1} f'g_1 \le 2\|f'\|_2\|g_1'\|_2=2\|f'\|_2\|g'\|_2 \le \|f'\|_2^2+\|g'\|_2^2.
$$
Thank you! For general constant $C$, are there functions $f,g$ such that $$\int_{S^{1}} |f'|^{2}+|g'|^{2}d\theta-C\int_{S^{1}}f'g<0$$?
Of course, yes. Take $f(t)=-\cos t, g(t)=\sin t$, then you get equality with $C=2$.
|
2025-03-21T14:48:31.655717
| 2020-07-31T06:36:45 |
368001
|
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"sort": "votes",
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|
Stack Exchange
|
Extending a representation of a free group to an extension of a mapping torus
Given a free group on $n$ generators, $F_n$, $\phi$ an automorphism of $F_n$, and a non-trivial representation $\rho: F_n \rightarrow \operatorname{Homeo}_+(\mathbb{R})$, are necessary and sufficient conditions known, in terms of $\rho, \phi$, and $n$, for extending the representation $\rho$ to $\tilde{\rho}: F_n \rtimes_\phi \mathbb{Z} \rightarrow \operatorname{Homeo}_+(\mathbb{R})$?
A first necessary condition is that for every $g\in F_n$, the maps $\rho(g)$ and $\rho(\phi(g))$ must be conjugate in $\mathrm{Homeo}_+(\mathbb{R})$. (clearly what you ask is the stronger condition that the conjugating element can be chosen so that it is always the same). Note that this is sufficient in the case $n=1$.
|
2025-03-21T14:48:31.655798
| 2020-07-31T06:42:17 |
368003
|
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"url": "https://mathoverflow.net/questions/368003"
}
|
Stack Exchange
|
Banach embedding of finite dimensional spaces
Recall that: let $0<r<s<2$, then $\ell_r$ uniformly contains a subspace isomorphic to $\ell_s^m$, $m\ge 1$ (see [JS]).
I am wondering whether are any result for the case when $r>s>2$?
[Johnson, William B.; Schechtman, Gideon Embedding $l_p^m$ into $l_1^m$, Acta Math. 149 (1982), 71--85.][JS]
For $2<r<\infty$, if $\ell_s^n$ embeds uniformly into $\ell_r$ for all $n$, then either $s=r$ or $s=2$. This is basically the localization to finite dimensions of the classical dichotomy theorem of Kadec and Pelczynski. The book of Albiac and Kalton is a good source for this.
Thank you so so so much!!! I had a look at the reference you mentioned. It is a beautiful theory.
|
2025-03-21T14:48:31.655877
| 2020-07-31T07:32:18 |
368005
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368005"
}
|
Stack Exchange
|
Asymptotic behavior of a matrix equation and its eigenvalues
We have a matrix valued function $A:\mathbb{R}_+\to \mathbb{R}^{m\times m}$. It is known that $A(\lambda)$ is a positive definite matrix for all $\lambda\in\mathbb{R}_+$ Denoting $\rho_i(A(\lambda))$ the $i^{th}$ eigenvalue of the matrix $A(\lambda)$, we know that asymptotically each eigenvalue of the matrix $A(\lambda)$ is of the form $$\rho_i(A(\lambda)) = \beta_i\lambda + \alpha_{i0} + \frac{\alpha_{i1}}{\lambda} + \frac{\alpha_{i2}}{\lambda^2} + ...$$ where $\beta_i>0,\alpha_{i0},\alpha_{i1},\alpha_{i2,...\in\mathbb{R}}$ are constants independent of $\lambda$.
Now there is a $m\times 1$ matrix $L$ whose elements are some contant reals(independent of $\lambda)$. The matrix $m\times 1$ $c(\lambda)$ is given as $$c(\lambda) = A(\lambda)L$$
Can we show that each element of the matrix $c$, that is say $c_j(\lambda)$ is of the form $$c_j(\lambda) = \gamma_j\lambda + \kappa_{j0} + \frac{\kappa_{j1}}{\lambda} + \frac{\kappa_{j2}}{\lambda^2}+...$$ where $\gamma_j,\kappa_{j0},\kappa_{j1},\kappa_{j2},...\in \mathbb{R}$ are constants independent of $\lambda$
PS : At the end of the day, I am interested in the asymptotic behavior as $\lambda\to\infty$
My attempt
As $A(\lambda)$ is pd, the eigenvectors of this matrix form a basis for $\mathbb{R}^m$. So expanding the constant matrix $L$ along this basis we have $$L = a_1e_1(\lambda) + a_2e_2(\lambda) + ...+a_me_m(\lambda)$$ where $e_1(\lambda),e_2(\lambda),..e_m(\lambda)$ are the eigenvectors of the matrix $A(\lambda)$ and $a_1,a_2,...a_m \in \mathbb{R}$ are some constants.
Substituting this in equation $c = A(\lambda)L$ we get $$c = \sum_{i=1}^ma_1\rho_i(A(\lambda))e_i(\lambda)$$ So it follows from here that $$c_j(\lambda) = \gamma_j\lambda + \kappa_{j0} + \frac{\kappa_{j1}}{\lambda} + \frac{\kappa_{j2}}{\lambda^2}+...$$
unless the eigenvectors $e_i(\lambda)$ behave strange. All we know is $\|e_i(\lambda)\|_2 = 1$. I am not sure if this is enough to complete the proof or there need to be any more assumptions.
cross post : https://math.stackexchange.com/q/3775350/2987
Of course, without assumptions on the behavior of the eigenvectors, your desired conclusion will not hold. E.g., for $t:=\lambda$, let
$$A(t):=\left(
\begin{array}{cc}
2+\cos t & \sin t \\
\sin t & 2-\cos t \\
\end{array}
\right),\quad L:=\left(
\begin{array}{c}
1 \\
0 \\
\end{array}
\right).$$
The eigenvalues of $A(t)$ are $3$ and $1$ for all $t$, whereas
$$AL=\left(
\begin{array}{c}
2+\cos t \\
\sin t \\
\end{array}
\right),$$
and $\sin t$ cannot be represented as $at+b+c/t+d/t^2+\cdots$ for any constants $a,b,c,d,\dots$.
@losifPinelis : Thanks for the example. Looks like the eigenvectors should converge elementwise.
|
2025-03-21T14:48:31.656056
| 2020-07-31T08:02:37 |
368006
|
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|
Stack Exchange
|
Ultrabornological representation for the space of uniformly continuous functions?
Let $\{\omega_i\}_{i\in I}$ be a non-empty set of increasing (not necessarily strictly) continuous functions preserving $0$. Then, for each $i \in I$ define the space
$$
C_{\omega_i}(\mathbb{R}^n,\mathbb{R}^d):=
\left\{
f \in (\mathbb{R}^n,\mathbb{R}^d):\,
\|f\|_{\omega_i,\infty}<\infty
\right\} \mbox{ where }
\|f\|_{\omega_i,\infty}:= \sup_{x \in \mathbb{R}^n} (\omega_i(\|x\|)+1)^{-1}\|f(x)\|.
$$
Then this is a Banach space since the map $f \mapsto f (\omega_i(\|x\|)+1)$ is clearly an isometry with $C_0(\mathbb{R}^n,\mathbb{R}^d)$ for the sup-norm, and the maps between $C_{\omega_i}(\mathbb{R}^n,\mathbb{R}^d)$ and $C_{\omega_j}(\mathbb{R}^n,\mathbb{R}^d)$ can be defined similarly by rescaling analogously. This makes $I$ into a poset with
$$
i\leq j \mbox{ if and only if } \sup_{x \in \mathbb{R}^n}\omega_i(\|x\|)
\leq \sup_{x \in \mathbb{R}^n}\omega_j(\|x\|).
$$
Thus, we can define the LCS colimit of this co-cone $\left\{C_{\omega_i}(\mathbb{R}^n,\mathbb{R}^d)\right\}$.
Now for the question, if $\{\omega_i\}_i$ is taken to be the collection of all monotonically increasing and continuous functions identifying $0$ (i.e.: $\omega(0)=0$) then does $\operatorname{co-lim}_i C_{\omega_i}(\mathbb{R}^n,\mathbb{R}^d)$ contain all uniformly continuous functions?
Note: Here the colimit is in the LCS sense.
Maybe, I miss something, but the answer seems to be easy: If $f:\mathbb R^n\to\mathbb R^d$ is continuous with $f(0)=0$ you can define the weight function $$\omega(r)=\sup\{\|f(x)\|: \|x\|\le r\}$$ which is obviously increasing with $\omega(0)=0$. Moreover, it is continuous at $r\ge 0$ because of the uniform continuity of $f$ on the compact set $\{x\in\mathbb R^n:\|x\|\le r+1\}$. Then $f\in C_\omega(\mathbb R^n, \mathbb R^d)$ holds. The extra assumption $f(0)=0$ can be removed by applying this to $\tilde f(x)=f(x)-f(0)$.
This shows that $C(\mathbb R^n, \mathbb R^d)$ is the co-limit of $\{C_\omega(\mathbb R^n, \mathbb R^d)\}$ in the category of vector spaces. Whether this holds in LCS is a different question.
You mean in the category of topological vector spaces? In that case $C(\mathbb{R}^n,\mathbb{R}^d)$ has compact-convergence topology or some (non-metrizable) final topology?
Also, are these examples where, $I$ is say countable and fully ordered (besides the space of functions of moderate growth?)
You wrote: Does the co-limit contain all uniformly continuous functions? -- This sounds very much like an algebraic question (where the co-limit is the union).
Ah, I see the issue; my error. But I'm mostly looking in the LCS category (I made an edit to help clarify), I'm sorry abou the confusion
Note that a uniformly continuous function on a normed space has a linear growth, $|f(x)|\le a|x|+b$. So you don't need such a wide class of weights ${\omega_i}$. 0n the other hand, adding $1$ in the denominator in the definition of the norms means you do not care to much about being uniformly continuous or just continuous.
|
2025-03-21T14:48:31.656256
| 2020-07-31T09:03:32 |
368009
|
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|
Stack Exchange
|
About an estimate of an oblique derivative problem of Laplace's equation
Suppose $n\geq 2$, set $B=B_r(0)\subset \mathbb{R}^n$, $B^+=\{x\in B|x_n>0\}$, $H=\{x\in B|x_n=0\}$, Let $u\in C^\infty(B^+)\cap C^1(B^+\cup H)$ be a solution of the following oblique derivative problem:
$$\begin{cases}
\Delta u=0 & \text{in} & B^+,\\
au+\beta\cdot\nabla u=0 & \text{on} & H,
\end{cases}$$
where the constant $a\leq 0$, $\beta$ is a constant vector with the $n$-th component $\beta_n>0$. Then $u$ can be uniquely extended to a harmoic function on $B'=B_{\lambda r}(0)$, where $\lambda=\beta_n/|\beta|$. In fact, let $v=au+\beta\cdot\nabla u$ on $B^+\cup H$, and $v(x',-x_n)=-v(x',x_n)$, then $v$ is harmonic on $B$; for any $x_0\in H$, let
$$u(x_0+t\beta)=e^{-at}(u(x_0)+\int_0^t v(x_0+s\beta)e^{as}ds)$$
whenever $x_0+t\beta\in B$, then it's easy to check $u$ is smooth and harmonic in $B'$, and coincides with the previous $u$ in $B'\cap \{x_n\geq 0\}$.
The question is: Does the following estimate necessarily hold, where $C$ is a positive constant independent of $u$?
$$\sup_{B'}|u|\leq C\sup_{B^+}|u|$$
I guessed it's true but could not find the answer. If it's indeed true, please give a sketch of proof, if not, I also want to see the reason.
Erm... The points $x_0+t\beta: x_0\in H, t<0$ do not necessarily cover $B_-$, so your extension is not defined in the whole $B$, as written.
No chance. Consider $n=2$, $\beta=(1,1)$. Then $H=[-1,1]$, right? Now just draw the picture. What do you see? I see a slanted strip that covers some part of the unit disk, but far from the whole unit disk.
@fedja Well, I modified the question a little, which is enough for my purposes.
|
2025-03-21T14:48:31.656382
| 2020-07-31T09:23:56 |
368011
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368011"
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|
Stack Exchange
|
Number of paths in the Bruhat order in the symmetric group
Let $\mathbb{S}_m$ the symmetric group on $m$ letters. Let $v\in\mathbb{S}_m$, and consider paths in the Bruhat order like this: $1\lessdot v_1\lessdot\cdots\lessdot v$, where $\lessdot$ means the covering relation in the (strong) Bruhat order. Let $N_v$ be the number of such paths.
It is intuitively clear that $N_v\leq\ell(v)!$ (for a proof, I found just now the reference), and further that the difference $\ell(v)!-N_v$ is even. Can you prove the latter fact?
Remark. What I said should be true for every finite Coxeter group but I am mostly interested in the symmetric group for now.
For the benefit of people finding this question later: Depending on what definition you have seen of the Bruhat order, it might not be "intuitively clear that $N_v\le\ell(v)!$". But this becomes clear when you know the subword characterization of Bruhat order. See, for example, Björner and Brenti's book "Combinatorics of Coxeter Groups", Theorem 2.2.2.
$\ell(v)!$ is of course even if $\ell(v)>1$, so the statement is really that $N_v$ is even for $\ell(v)>1$. We find a fixed-point free involution on the set of such Bruhat paths. Suppose that $v_2,v_3,\ldots$ are fixed. By the diamond property of Bruhat order there are exactly two possibilities for $v_1$. This gives the involution we want (in fact many of them).
In fact, this argument gives that for any Coxeter group (or even any Eulerian poset) $N_v$ is divisible by $2^{\lfloor \ell(v)/2\rfloor}$.
|
2025-03-21T14:48:31.656504
| 2020-07-31T10:31:25 |
368015
|
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"url": "https://mathoverflow.net/questions/368015"
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|
Stack Exchange
|
Properties of matrix $X=\left[\frac{1}{1-\bar\alpha_i \alpha_j}\right]_{ij}$
Let $\{\alpha_i\}_{i=1}^n$ be complex numbers such that $|\alpha_i|<1$, and consider the following $n\times n$ structured matrix
$$
X=\left[\frac{1}{1-\bar\alpha_i \alpha_j}\right]_{ij}.
$$
Such matrix arises in the solution of particular Stein matrix equations (e.g., see p. 11 of Bhatia, "Positive definite matrices", Princeton University Press, 2007), and seems to be somehow related to the class of Cauchy matrices. $X$ seems to have a interesting structure (and, I guess, it may feature some "nice" properties), however I couldn't find any additional information on this matrix on the web. So my question:
Does matrix $X$ have a name? Are there any known properties of $X$ (such as formulas for the determinant, inverse, etc.)?
P.S. A property that follows from the connection with Stein equations is that $X$ is always positive semidefinite.
This matrix is also related to the Nevanlinna-Pick Theorem. Namely, if $z_i, \lambda_i \in \mathbb D, 1\leq i\leq n$ then $$\left[\begin{matrix} \frac{1- \overline{z_j}z_i}{1-\overline{\lambda_j}\lambda_i}\end{matrix}\right]_{i,j=1}^n \geq 0$$
if and only if there is a holomorphic function $\varphi : \mathbb D \rightarrow \overline{\mathbb D}$ such that $\varphi(\lambda_i) = z_i, 1\leq i\leq n$.
In this theory $K(\lambda_j,\lambda_i) = (1-\overline{\lambda_j}\lambda_i)^{-1}$ is called the Szego kernel. So one could refer to your matrix as the Szego kernel matrix.
Assuming all the $\alpha_j$ are nonzero, the matrices $X$ are Cauchy-like matrices, since you can rewrite them as
$$
X_{ij} = \frac{\alpha_j^{-1}}{\alpha_j^{-1}-\bar{\alpha}_i}
$$
so there are analogous formulas for their determinant and inverse. In particular, $XA^{-1}$ is a Cauchy matrix, where $A = diag(\alpha_i)$, so these formulas follow directly from those for Cauchy matrices.
A variant of this equivalence that does not require the invertibility of $A$ and better exploits symmetry/Hermitianity is the following. Let $B = (I+A)(A-I)^{-1}$ (note that $\alpha_i\neq 1$, otherwise there would be a zero denominator in $X_{ii}$, so $A-I$ is invertible). Then,
$$
B^*X+XB = -2(A- I)^{-*}E(A- I)^{-1}
$$
expands to
$$
( I+A^*)X(A- I) + (A^*- I)X( I + A) = -2E
$$
which reduces to
$$
2A^*XA - 2X = -2E,
$$
which is the Stein equation that you used to define $X$.
So $X$ solves a Lyapunov equation with a diagonal $B$, and hence one can write the more symmetric formula
$$X_{ij} = \frac{-2(\alpha_i-1)^{-*}(\alpha_j-1)^{-1}}{\bar{\beta}_i + \beta_j},$$
where $\beta_i = \frac{\alpha_i + 1}{\alpha_i-1}$ are the diagonal entries of $B$. Note that $\Re\beta_i < 0$ iff $|\alpha_i| < 1$.
Alternatively, the Hermitian matrix $(A^*-I)^{-1}X(A-I)^{-1}$ is a Cauchy matrix wrt the two sequences $\bar{\beta}_i$ and $-\beta_j$.
This trick to convert between Lyapunov and Stein equations is classical (bilinear transform, or Cayley transform).
These matrices are a special case of a much broader class of positive semidefinite (psd) matrices constructed as follows. Let $(\alpha_{i,j})_{i,j=1}^n$ be any psd matrix satisfying $|\alpha_{i,j}| < 1$ for all $i,j \in \{1,\ldots,n\}$. Then the matrix $((1-\alpha_{i,j})^{-c})_{i,j=1}^n$ is positive semidefinite for all $c \geq 0$. This follows immediately from the geometric series formula since an entrywise power of a psd matrix is psd. The matrix you mention is constructed in this way from the matrix $(\alpha_1,\ldots,\alpha_n)^*(\alpha_1,\ldots,\alpha_n)$ which is transparently psd.
|
2025-03-21T14:48:31.656729
| 2020-07-31T11:26:46 |
368016
|
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|
Stack Exchange
|
Computing the derived $I$ invariants $D(G\operatorname{Rep})\to D((G/I)\operatorname{Rep})$
In Scholze's paper The Langlands–Kottwitz approach for the modular curve (published version) about $\operatorname{GL}_2$ when he wants to find the relation between the semisimple trace and the usual trace he claims that if $I = I_K\subset G_K$ is the inertia group of a local number field $K$ with residue field $k$ and $V$ is a representation of $G_k$ such that $I_k$ acts through a finite cyclic quotient then $R_I(V)$, $R_{I}:D(\operatorname{Rep}_{\overline{\mathbb Q_l}}(G_K))\to D(\operatorname{Rep}_{\overline{\mathbb Q_l}}(G_k))$ is represented by $$\dotsb\to 0\to V^I\to V^I(1)\to \ 0 \to\dotsb.$$
I think essentially we can reduce the problem to $\mathbb F_l$ coefficients and we can assume $I=\langle s\rangle$ is a finite cyclic group and $G_K=I\rtimes
\hat{\mathbb{Z}}$ but I can't compute the derived functor in this case. I think that if $w:D(\operatorname{Rep}_{\mathbb F_l}(\hat{\mathbb{Z}}))\to D(\mathbb F_l)$ is the forgetful functor then $R_I\circ w$ is represented by $$0\to V[I]\xrightarrow{s-1} V[I]
\xrightarrow{N}V[I]\to \dotsb$$
but then $R_I$ can't be represented by the complex in the first paragraph.
Any help is appreciated.
By "Scholze's paper about $\operatorname{GL}_2$" do you mean Scholze - The local Langlands correspondence for $\operatorname{GL}_n$ over $p$-adic fields?
No I mean "the langlands kottwitz method for the modular curve"
OK, I have edited in the link.
|
2025-03-21T14:48:31.656851
| 2020-07-31T11:47:18 |
368017
|
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"url": "https://mathoverflow.net/questions/368017"
}
|
Stack Exchange
|
Conjecture of van der Holst and Pendavingh related to bound for Colin de Verdière invariant
In their 2009 paper (“On a graph property generalizing planarity
and flatness”. In: Combinatorica 29.3 (May 2009), pp. 337–361. issn: 1439-6912.
doi: 10.1007/s00493-009-2219-6.), van der Holst and Pendavingh defined a new minor monotone
graph invariant $\sigma(G)$ for a graph $G$: the minimal integer $k$ such that every CW-complex whose 1-skeleton is $G$ admits a certain even mapping into $\mathbb R^k$.
They were able to prove $\mu(G)\leq\sigma(G)+2$, where $\mu(G)$ is the Colin de Verdière number of $G$ which is minor monotone as well (Colin de Verdière. “Sur un nouvel invariant des graphes et un critère de planaritè”. In: Journal of Combinatorial Theory, Series B 50.1 (1990), pp. 11–21.
issn: 0095-8956. doi: 10.1016/0095-8956(90)90093-F.)
My main interest is in the conjecture of van der Holst and Pendavingh in that paper. They conjectured that actually $\mu(G)\leq\sigma(G)$ might hold. Question: What is known about the status of this conjecture? (I have difficulties tracing it as their new invariant $\sigma(G)$ does not seem to have a commonly agreed name yet).
Kaluza and Tancer have actually proved $\mu(G)\leq\sigma(G)$ in 2019: See their proof in the preprint "Even maps, the Colin de Verdière number, and representations of
graphs" on arxiv. Here is the link https://arxiv.org/pdf/1907.05055.pdf
You are right, the invariant $\sigma(G)$ of Holst and Pendavingh does not seem to have an established name yet.
I would like to add one important aspect: it was known that $\mu(G)$ and $\sigma(G)$ can deviate by a large amount for larger values $k$. Now we have the proof of the improved (sharper) bound $\mu(G)\leq\sigma(G)$, but even though this is an improvement, Kaluza and Tancer also showed that a large gap exists already for small values of $k$: They showed there is a graph $G$ such that $\mu(G)\leq7$ and $\sigma(G)\geq8$ ("Even maps, the Colin de Verdière number, and representations of graphs" on arxiv. Here is the link https://arxiv.org/pdf/1907.05055.pdf).
Now, a suspension of $G$ (adding a new vertex to $G$ and connecting it to all vertices of $G$) increases both $\mu(G)$ and $\sigma(G)$ by exactly one (unless $G$ is the complement of $K_2$). Therefore $\mu(G)\leq7$ and $\sigma(G)\geq8$ implies that for every $k \in\mathbb N$, $k \geq 7$, there must exist a graph $G_k$ with $\mu(G)\leq k$ and $\sigma(G)\geq k+1$, i.e. strict inequality for all large values of $k$. Finally, the authors also show that the gap between $\mu(G)$ and $\sigma(G)$ is asymptotically large.
Op-t´Bevers thank you, this is useful and in a certain sense speaks in favour of keeping with the Colin de Verdière invariant. However, what makes the new Holst Pendavingh invariant very attractive is their "strongly topological" definition with even functions, so they look like the natural generalization of Hanani-Tutte
@soerenssen soerenssen you are right about the strongly topological definition. But I have seen Colin de Verdière characterization of embeddings in projective plane and torus. I have not yet seen such characterization for the new invariant (just flat embeddings in ${\mathbb R}^k$)
|
2025-03-21T14:48:31.657054
| 2020-07-31T11:58:28 |
368018
|
{
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"LSpice",
"Zeno cosini",
"https://mathoverflow.net/users/139875",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368018"
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|
Stack Exchange
|
Why $h_t$ maps into $\mathbb{R}^{\nu}$?
I am studying geometric measure theory (Herbert Federer - Geometric measure theory) and I have a question about class $r$ homotopies. Here's the definition, from p. 363, Section 4.1.9:
Suppose $U$ is an open subset of $\mathbb{R}^{n}$, $V$ is an open subset of $\mathbb{R}^{v}$ $f$, and $g$ are functions mapping $U$ into $V$. A homotopy of class $r$ from $f$ to $g$ is a map
$$
h: A \times U \rightarrow V
$$
of class $r$ such that $A$ is an open subinterval of $\mathbb{R}$, $0 \in A$, $1 \in A$,
$$
h(0, x)=f(x) \text { and } h(1, x)=g(x) \text { for } x \in U.
$$
Whenever $t \in A$ we define
$$
h_{t}: U \rightarrow V, \quad h_{t}(x)=h(t, x) \text { for } x \in U,
$$
hence $h_{0}=f$ and $h_{1}=g$; in case $r \geq 1$ we also define
$$
h_{t}: U \rightarrow \mathbb{R}^{v}, \quad h_{t}(x)=\langle(1,0), D h(t, x)\rangle \text { for } x \in U,
$$
hence $\langle(v, w), D h(t, x)\rangle=v h_{t}(x)+\left\langle w, D h_{t}(x)\right\rangle$ for $t \in A$, $x \in U$, $v \in \mathbb{R}$, $w \in \mathbb{R}^{n}$.
Now my question is about the last 3 lines:
in case $r \geq 1$ we also define
$$
h_{t}: U \rightarrow \mathbb{R}^{v}, \quad h_{t}(x)=\langle(1,0), D h(t, x)\rangle \text { for } x \in U,
$$
hence $\langle(v, w), D h(t, x)\rangle=v h_{t}(x)+\left\langle w, D h_{t}(x)\right\rangle$ for $t \in A, x \in U, v \in \mathbb{R}, w \in \mathbb{R}^{n}$.
Actually, we know that $\langle v, Df(a)\rangle = v \cdot \nabla f(a)$. Also, in the last line it is mentioned that $w \in \mathbb{R}^n$, hence $0 = (0, \dotsc,0) \in \mathbb{R}^n$. Now, we have:
$$h_t(x)= \langle (1, 0, \dotsc, 0), Dh(t,x) \rangle$$ which is equal to
\begin{align} h_t(x) &= (1, 0, \dotsc, 0) \cdot \nabla h(t,x)\\ &= (1, 0 , \dotsc, 0) \cdot \left(\frac{\partial h}{\partial t}, \frac{\partial h}{\partial x_1}, \dotsc, \frac{\partial h}{\partial x_n}\right) \\
&= \frac{\partial h}{\partial t}.\end{align}
Now, my question is that why it is said that $h_t: U \rightarrow \mathbb{R}^{\nu}$? I mean, now we should see that $\frac{\partial h}{\partial t} \in \mathbb{R}^{\nu}$, but I don't get it really! If this is just the differential of $h$, it must lie in $\mathbb{R}^{\nu-1}$!
The partial derivative at $t \in A$ of a map $h : A \times U \to V \subseteq \mathbb R^\nu$ is a map $U \to \mathbb R^\nu$; what is the question? (What I don't get is what it means to define $h_t$ twice, on the same domain, in two completely different ways ….)
Perhaps the point is that the notation is unclear, and looks like an inner product, whereas I think the significance is that $Dh(t, x)$ is a map $\mathbb R \times \mathbb R^n \to \mathbb R^\nu$, and all we are doing is applying it to the vector $(1, 0) \in \mathbb R \times \mathbb R^n$. (One could think of this as matrix-vector multiplication.)
Thank you so much @LSpice. Well, as I saw in the book, the definition of the $\langle -,-\rangle$ is exactly as I wrote and it'll be the inner product of the vector $v=(0,1) \in \mathbb{R} \times \mathbb{R}^n$ and $\nabla h(t,x)$. What I don't understand is that why this inner product lies in $\mathbb{R}^\nu$! I mean, in the definition, it is said that $h$ lies in $\mathbb{R}^\nu$. So, as $h_t = \frac{\partial h}{\partial t}$, shouldn't it lie in $\mathbb{R}^{\nu -1}$?
I'm not sure why you would expect the partial derivative of an $\mathbb R^\nu$-valued function to be $\mathbb R^{\nu - 1}$-valued; partial derivatives in some sense "reduce input variables", not "reduce output variables". Anyway, the 'definition' of $\langle{-}, {-}\rangle$ in terms of the gradient that you mention I guess is the one on p. 209, where it is specified only for $Y = \mathbb R$, so that evaluation is the inner product. For more general $Y$, one can only think of evaluation.
Incidentally, your title asks why $h$ maps into $\mathbb R^\nu$, but that's given; I think your question seems to be rather why $h_t$ maps into $\mathbb R^\nu$.
Yes, you're right @LSpice .
I edited the title, thanks for that. About the definition of that product so, how can we define it? I didn't find any definitions and since I'm really new to this topic, I am really confused.
I'm not sure what you mean by "the product". Do you mean the symbol $\langle{-}, {-}\rangle$? If so, then I have said what it means: evaluation of a linear map. If you want to think in coördinates, it is $(\partial h_1/\partial t, \dotsc, \partial h_\nu/\partial t) = (\nabla h_1\cdot(1, 0), \dotsc, \nabla h_\nu\cdot(1, 0))$, where $h_i \colon A \times U \to \mathbb R$ are the component functions.
Got it. Really appreciate it.
|
2025-03-21T14:48:31.657345
| 2020-07-31T12:07:27 |
368021
|
{
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"authors": [
"David Ketcheson",
"Gro-Tsen",
"Hans-Peter Stricker",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368021"
}
|
Stack Exchange
|
Approximated solutions of SEIR models
Numerical solutions of the SEIR equations (describing the spreading of an epidemic disease) – or variations thereof –
$\dot{S} = - N$
$\dot{E} = + N - E/\lambda$
$\dot{I} = + E/\lambda - I/\delta$
$\dot{R} = + I/\delta$
with
$N = \beta I S / M$ = number of newly infected individuals
$\beta = $ infection rate
$\lambda = $ latency period
$\delta = $ duration of infectiosity
$M = S + E + I + R = $ size of the population
yield characteristic and almost symmetric peaks for the function $I(t)$ of numbers of infectious individuals. So $I(t)$ can – by a rough guess – be approximated by a Gauss curve
$$\widetilde{I}(t) = I_0\ \operatorname{exp}\Big({-\big((t-t_0)/\sigma\big)^2}\Big)$$
with $I_0$ the maximal value of $I(t)$, $I(t_0) = I_0$, and $\sigma$ such that $\widetilde{I}(0) = I(0) = 1$, i.e.
$$\sigma = t_0\ /\ \sqrt{\text{ln} I_0}$$
For different values of $\delta$, the reproduction number $R_0 = \beta\cdot\delta$, and a fixed value $\lambda = 2$ we find:
It turns out that an exponent $\sqrt{2}$ instead of $2$ yields better results, i.e.
$$\widetilde{I}(t) = I_0\ \operatorname{exp}\Big({-\big(|t-t_0|/\sigma\big)^{\sqrt{2}}}\Big)$$
My question is fourfold:
Why is a Gauss-like curve a good approximation at all? That means: Why is $I(t)$ so symmetric?
By which considerations could one come up with the exponent $\approx \sqrt{2}$?
By which considerations can the asymmetry of the numerical solution $I(t)$ be understood which becomes apparent when comparing it with the symmetric approximation $\tilde{I}(t)$?
Has anyone an idea how $I_0$ and $t_0$ look like as functions of $\beta,\lambda,\delta,M$?
Just to give another view on the tables above, find here all curves overlayed:
If we drop the ‘E’ and stick to SIR as the simplest case, there is a variant of SIR where recovery, instead of taking place following an exponential process, occurs in constant time: see here. This variant, which can serve as an approximation to the classical case, admits an exact closed-form solution, and has a symmetric peak, so this might partially answer your question. See here for further discussion (and here for meta-discussion).
@Gro-Tsen: Thanks for the comment and the links which I already found helpful. But please help me: What does the meta-discussion have to do with my question? Does the second-to-last link (which surely has to do with my question) refer to the article that was rejected by arXiv?
@Gro-Tsen: By the way: I can provide the same pictures as above for an SEIR model with transition from exposed to infectious and from infectious to recovered in constant times $\lambda =$ latency period and $\delta=$ duration of infectiousness (which I find less unrealistic, too).They are comparably well approximated by Gauss-like functions, so considering resp. neglecting E doesn't change so much - maybe only that there is no closed-form solution.
The meta-discussion is only mentioned to explain why the link doesn't point to the arXiv, and yes, it's the note in question (the one on hal.archives-ouvertes.fr) which was rejected thence.
@Gro-Tsen: Did you make any progress with arXiv? Is your paper still unpublished there?
You can often get an amazingly good fit with $I_0F(t-t_0)^{-2}$ where $F(x)=(ae^{bx}+be^{-ax})/(a+b)$ with positive $a,b$, but I do not have nice approximate algebraic expressions for the parameters, just some stupid equations to solve numerically. I'll think of it a bit.
@fedja: Is your function $F(x)$ known for giving good fits in many contexts? How can these be characterized? Do you have a reference? Does $F(x)$ have a name I can google for?
@Hans-PeterStricker I posted the (partial set of) the corresponding equations. Try them and let me know what you think. Sorry for not answering earlier: had some other fish to fry.
You don't mention your initial conditions or whether your observations hold for arbitrary initial conditions.
@DavidKetcheson: I did mention - maybe not prominently enough - that I assume $\tilde{I}(t=0) = I(t=0) = 1$.
But what about S, E, and R?
@DavidKetcheson: You are right. I implicitely assumed: $S(0) = M-1$, $E(0) = 0$, $R(0) = 0$. (Better would have been to start with $E(0)=1$ and $I(0) = 0$.)
Is your function F(x) known for giving good fits in many contexts? How can these be characterized?
This is too long for a comment but I'd like you to check if the fit is to your satisfaction before I elaborate. I prefer to write everything in the numerator, so my equations will be
$$
\dot S=-\beta IS, \dot E=\beta IS-\lambda E, \dot I=\lambda E-\delta I\,.
$$
Suppose that $I_0$ is the maximum of $I$ attained at the moment $0$ (just shift otherwise).
Then the equations I'm using (I hope I'm copyng them right) are
$$
2a^2(\beta I_0+\lambda+\delta-\mu)=\lambda\delta \beta I_0
\\
6a^2=(\lambda+\delta-\mu)(\beta I_0-\mu)
$$
Once you have solved those for $a,\mu>0$ (assume that $I_0$ is known for the moment and you just want a fitting curve rather than an independent derivation for everything), let $a_\pm=\sqrt{a^2+\frac{\mu^2}4}\mp \frac\mu 2$ (so $a_->a_+$), define
$$
F_{a,\mu}(t)=I_0\left(\frac{a_-\exp(a_+t)+a_+\exp(-a_-t)}{a_-+a_+}\right)^{-2}
$$
and compare it to $I(t)$. If you like the fit, we can discuss where all that nonsense came from and how to write the full system where $I_0$ will be solved for, not given. If not, I'll stop here, so let me know what you think.
The equations are algebraic of third degree, so, unless you are a big fan of Cardano's formulae, you'll have to solve them numerically. That's not hard (almost any decent iteration scheme works). The approximation is pretty good in most cases, IMHO, but it has its limitations so one can find regimes where it breaks though those are usually rather extreme. Enjoy! :-)
Two pictures, as promised. The black curve is the true trajectory, the red one is the computed trajectory (note that the height of the peak is also computed: I finally found a good third equation, so I played it honestly and didn't try to tweak the parameters beyond what my linearized equations gave directly), the green line is the best symmetric approximation you can hope for (half sum of the true trajectory and its reflection around the peak). I believe that the red line is better even without any tweaking and that the precision with which the maximum is determined is also fairly decent, but you can judge by yourself :-).
Thanks a lot for sharing the equations. Before I start evaluating them, may I send a simplified version to you (by mail from [email protected]), just for cross-checking (because you wrote that you're not quite sure if you copied them right)? Furthermore: Do you have a couple of plots demonstrating the goodness of fit of your approximation, ideally compared to the goodness of fit of my much simpler approximation?
@Hans-PeterStricker Yes, I have run quite a few plots. I just thought that that the surest way to evaluate a suggestion is to try it (programming takes 20 minutes to one hour) but, if you want, I'll be happy to share some pictures. Just wait a bit more until I get some free time. I double checked the equations, they seem to agree with what I have. Note that they are just linearizations of something, so if this fit is not good for some particular curve, it usually doesn't mean that the good fit does not exist, merely that you are out of the linear approximation range.
@Hans-PeterStricker OK, posted a couple of pictures. The second one is rather extreme: $R\approx 28$. Also, this is certainly not the curve of the proposed kind that fits the data best, just what the most naiive computation produces on the first run. The fit may be quite improved if you tweak it manually, but I preferred to leave it as it was;-)
Thanks so very much. Did you try or do you plan to publish this?
From an article that user @Gro-Tsen refers to I learned - and give here as a partial answer - that for the case of a vanishing latency period $\lambda = 0$, i.e. for the classical SIR model
$\dot{S} = -N$
$\dot{I} = +N - I/\delta$
$\dot{R} = +I/\delta$
there is a closed formula for $I_{max}$ (i.e. the maximal value of $I(t)$) as a function of $\beta$, $\delta$, and $M$ namely
$$ I_{max} = \frac{R_0 - \log R_0 - 1}{R_0} \cdot M$$
with $R_0 = \beta\cdot\delta$. This is quite nice.
This result is classical and widely known (to mathematical epidemiologists). For details and a number of other related results, see e.g. this review paper.
@DavidKetcheson: Is there also a classical result for the time $t_{max}$ when the maximum $I_{max}$ is reached?
I think the formula given in this answer is not quite correct (it is nearly correct, only when $R(0)$ and $I(0)$ are very small). The correct version is in the reference in my comment.
@DavidKetcheson: You are right, I just saw it. Thanks for the hint.
For $t_{max}$, I'm not aware of an exact formula. For initial conditions like those you are using, you can derive a rough estimate of $t_{max}\approx\ln((\sigma-1)/(\sigma I(0)))/(\beta - \gamma)$ where $\sigma=\beta \delta$ (and working in units where $M=1$).
@DavidKetcheson: Your comment on $t_{max}$ came some minutes too late: I just posted a question on this. But in your comment you already give an answer. Maybe you would like to post it as such.
@DavidKetcheson: Can you give me a short hint how to derive the rough estimate?
|
2025-03-21T14:48:31.658091
| 2020-07-31T12:09:01 |
368023
|
{
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"Bipolar Minds",
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}
|
Stack Exchange
|
Alternating $n$-homomorphism on abelian group is skew of $n$-cocycle
Let $A$ be a finitely generated abelian group. Let $c$ be a 2-cocycle on $A$, where $A$ acts trivially on $\mathbb{C}^\times$. It is well-known that the skew-map
$$ c(a_1,a_2) \longmapsto \frac{c(a_1,a_2)}{c(a_2,a_1)}$$
induces a surjective homomorphism $H^2(A,\mathbb{C}^\times) \to \bigwedge^2(A,\mathbb{C}^\times)$, where $\bigwedge^2(A,\mathbb{C}^\times)$ is the group of alternating bihomomorphisms on $G$. If I remember correctly, this is even an isomorphism, but I couldn't find the statement anymore.
Now, let $d \in Z^3(A,\mathbb{C}^\times)$ be a 3-cocycle on $A$. We can now define a higher skew-map
$$d(a_1,a_2,a_3) \longmapsto \prod_{\sigma \in S_3} d(a_{\sigma(1)},a_{\sigma(2)},a_{\sigma(3)})^{\operatorname{sgn}(\sigma)}.$$
Again, this induces a homomorphism $H^3(A,\mathbb{C}^\times) \to \bigwedge^3(A,\mathbb{C}^\times)$, where $\bigwedge^3(A,\mathbb{C}^\times)$ is now the group of alternating trihomomorphisms. By alternating trihomomorphisms I mean homomorphisms $t:A \otimes A \otimes A \to \mathbb{C}^\times$, s.t. $t(a,a,b)=1=t(a,b,b)$ for all $a,b \in A$. I didn't check, but I believe this is possible for higher $n$-cocycles as well with the obvious definition of a higher skew map.
I believe this is all known, but I couldn't find it anywhere. In particular, I would like to know if the induced homomorphisms are again surjective (or even bijective).
Do you also want $t(a, b, a) = 1$ for all $t$?
@LSpice I think it follows from the other two conditions and $t$ being a homomorphism
@LSpice I got $t(a,b,a)=\frac{t(a,a+b,a+b)}{t(a,a,a+b)t(a,b,b)}$ but probably its the same as yours
No, yours is right and mine (which I have now deleted) is wrong; I was incorrectly treating it as a homomorphism $A \times A \times A \to \mathbb C^\times$.
|
2025-03-21T14:48:31.658215
| 2020-07-31T12:18:11 |
368024
|
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"მამუკა ჯიბლაძე"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368024"
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|
Stack Exchange
|
Primality test for $N=2^a3^b+1$
Can you prove or disprove the following claim:
Let $N=2^a3^b+1$ , $a>0 , b>0$ . If there exists an integer $c$ such that $$c^{(N-1)/3}-c^{(N-1)/6} \equiv -1 \pmod{N}$$ then $N$ is a prime.
You can run this test here. I have verified this claim for all composite $N$ up to $2^{100} \cdot 3^{100}+1$ with $2 \le c \le 100$ , and for all prime $N$ from this list.
The "only if" part is clear - by the same trick as in the answer by @Fedor it follows from $(c^{(N-1)/3}-c^{(N-1)/6}+1)(c^{(N-1)/6}+1)=c^{(N-1)/2}+1$, since for prime $N$ there are $(N-1)/2$ residues modulo $N$ with $c^{(N−1)/2}+1\equiv0\mod N$ but only $(N-1)/6$ residues with $c^{(N−1)/6}+1\equiv0\mod N$.
Yes. Obviously this $c$ and $N$ are coprime. We get $c^{(N-1)/2}+1=(c^{(N-1)/6}+1)(c^{(N-1)/3}-c^{(N-1)/6}+1)$ is divisible by $N$. Therefore $c^{N-1}-1$ is divisible by $N$, and $N-1$ is divisible by $k:={\rm {ord}}(c)$, where ${\rm ord}(x)$ denotes the multiplicative order of $x$ modulo $N$. But $(N-1)/2$ is not divisible by $k$, since $c^{(N-1)/2}\equiv -1\pmod N$. Assume that $(N-1)/3$ is divisible by $k$; then $c^{(N-1)/3}\equiv 1 \pmod N$, $c^{(N-1)/6}\equiv c^{(N-1)/3}+1\equiv 2 \pmod N$ and $$1\equiv c^{(N-1)/3}= (c^{(N-1)/6})^2\equiv 4\pmod N,$$
a contradiction. So neither $(N-1)/2$ nor $(N-1)/3$ is divisible by $k$; thus, $k=N-1$. But $k$ must divide $\varphi(N)$, so $N-1\leqslant \varphi(N)$ and $N$ is prime.
Yes, thank you, fixed
@მამუკაჯიბლაძე true, so many misprints for 8 lines, sorry. Fixed.
|
2025-03-21T14:48:31.658346
| 2020-07-31T13:17:00 |
368030
|
{
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|
Stack Exchange
|
Regarding subspace generated by the polynomial multiples of outer functions
Let $\mathbb{D}$ and $\mathbb{T}$ denote the open unit disk and unit circle in $\mathbb{C}$ respectively. We write $Hol(\mathbb{D})$ for the space of all holomorphic functions on $\mathbb{D}.$ The Hardy spaces on $\mathbb{D}$ are defined as:
$$H^{p}:= \left\{ f\in Hol\left( \mathbb{D}\right) :\sup _{r < 1}\int ^{2\pi }_{0}\left| f\left( re^{i\theta}\right) \right| ^{p}d\theta < \infty \right\} \;\;\;\;(0<p<\infty), $$
$$H^{\infty }:= \left\{ f\in Hol\left( \mathbb{D}\right) :\sup_{z\in D}\left| f\left( z\right) \right| < \infty \right\}.$$
A function $g\in H^p(\mathbb{D})$ is outer if there exists a function $G:\mathbb{T}\longrightarrow [0,\infty)$ with $G\in L^1(\mathbb{T})$ such that
$$g\left( z\right) =\alpha \text{exp}\left( \int ^{2\pi }_{0}\dfrac {e^{i\theta }+z}{e^{i\theta }-z}G\left( e^{i\theta }\right) \dfrac {d\theta }{2\pi }\right) \qquad(z\in \mathbb{D})$$ and $|\alpha|=1$.
Let $\mathscr{P}[h]$ denote the closed subspace generated by the functions $z^n h(z),\; n=0,1,2,....$, ie $\mathscr{P}[h]$ consists of all $H^p$ functions that can be approximated by polynomial multiples of $h$. Note that $\mathscr{P}[1]=H^p$, since polynomials are dense in $H^p$.
I wanted to ask:
I know that if $h$ is not outer, then $\mathscr{P}[h]\neq H^p $.
Will this also imply that $1\notin \mathscr{P}[h]$ with dist$(1,\mathscr{P}[h])>0.$? Will this distance be bounded below by some +real number(irrespective of given outer function h)?
The answer to (1) depends on what you mean by a "subspace generated by (...)". If you mean "closed subspace generated by (...)", as suggested by the remaining part of the quoted sentence, then yes, $\mathscr P[h]$ is closed. :-)
You have basically answered your own question already (and the answer is, yes, $1\notin P[h]$ if $h$ is not outer). This follows because $P[h]$ is invariant under multiplication by $z$, so if we had $1\in P[h]$, then this space would contain all polynomials. (And then the distance is positive automatically, as $P[h]$ is closed.)
By the way, https://math.stackexchange.com/ is a better site to ask such questions.
@ChristianRemling I still have a part of the question remaining which asks if the distance is bounded below by any positive real number...
The answer to your final question is no if $p < \infty$ and yes if $p = \infty$.
Let $a > 0$ and let $h_a$ be a singular inner function given by
$$h_a(z) = \exp(-a(1-z)/(1+z)). $$
Note that $|h_a(z)| \leqslant 1$ and $h_a(z) \to 1$ as $a \to 0^+$. Therefore, by the dominated convergence theorem, $h_a$ converges to $1$ in $H^p$ for every $p < \infty$ (at least when $p \geqslant 1$; I bet the same is true when $p < 1$, but I have never worked with these spaces and so I am not entirely sure). In particular, the distance between $\mathscr P[h_a]$ and $1$ converges to zero as $a \to 0^+$.
On the other hand, the distance in $H^\infty$ between $1$ and any function $f$ which is not outer is at least $1$, because either $f$ has a zero in the unit disk or some non-tangential boundary limit of $f$ is equal to $0$. Thus, the distance between $\mathscr{P}[h]$ and $1$ is either $0$ or $1$.
The $H^\infty$ norm is equal to the supremum norm inside the unit disk, so if $f$ takes values arbitrarily close to $0$ in the unit disk, then $|f - 1|_{H^\infty}$ is at least $1$.
Huh, your comment just disappeared...
Will there be any outer function for which this distance will be some k>0?
There's nothing special about constant $1$, I think, just approximate any $H^p$ function by a polynomial $P_n$, and then this polynomial by an inner function $h_{a_n}$ multiplied by $P_n$.
|
2025-03-21T14:48:31.658614
| 2020-07-31T13:23:40 |
368031
|
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|
Stack Exchange
|
Irreducible representation of $S_n$: contained in tensor powers of the standard representation?
Let $S_n$ be the permutation group and $V = \operatorname{Fun}(X,\mathbb{k})$ functions from $X=\{1,\dotsc,n\}$ to some field $\mathbb{k}$. How can I prove that every irreducible representation of $S_n$ occurs in $V^{\otimes m}$ for integer $m$ large enough?
This is a standard fact when $n!\neq 0$ in $\mathbb{k}$.
This is a standard fact that holds for any faithful representation of any finite group.
how about using schur functions here?
@lambda, is it true for any faithful representation in any characteristic? I don't know that it isn't, but I thought that there might be some problems in case the characteristic of $\mathbb k$ divides $n!$.
Use the orthogonality relation and show that the inner product of any irreducible character of $S_n$ with character of $V^{\otimes m}$ does not vanish for large $m$.
Not to keep beating the same (possibly wrong, to mix metaphors) horse, but, @DianbinBao, isn't that a characteristic-$0$ argument?
@LSpice, yes, it only works for characteristic 0 case.
@LSpice I believe that if the word "occurs" is interpreted correctly for the non-semisimple setting (i.e. a composition factor, not necessarily a summand) then it is true all characteristics, but I guess I did glance past the fact that the OP wanted this for arbitrary fields.
See https://www.ams.org/journals/proc/1962-013-05/S0002-9939-1962-0141710-X/S0002-9939-1962-0141710-X.pdf
@LSpice, yes, it works for characteristic 0 case. As the OP mentioned $n!\neq 0$ in $\mathbb{K}$. The inner product makes sense and the argument still works by Schur's lemma.
Does "occurs" mean as a subfactor, as a subrepresentation, as a summand, ... ?
@DianbinBao, re your comment, I think that the post mentions that this is a known fact when $n! \ne 0$, but does not actually assume that $n! \ne 0$.
@BenjaminSteinberg, wow, yet another illustration of the beauty that's to be found in (that other) Steinberg's works! As @lambda and @DavidESpeyer mention, the sense of 'constituent' there is 'subquotient'. Is there an easy example of a representation that can't be realised as a subobject of a tensor power $V^{\otimes m}$?
(Just to have the name here, @BenjaminSteinberg's reference is to Robert Steinberg - Complete sets of representations of algebras.)
@LSpice, for finite groups every irreducible appears as subrepresentation of the regular representation because the group algebra is Frobenius. R. Steinberg proves the regular representation appears inside of a direct sum of the tensor powers and so you get them as subobjects. But Steinberg also proves his theorem for finite semigroups and in that case you do not get them all as subobjects.
@LSpice for instance of you take the monoid of all maps on n letters then the trivial representation is a quotient of the natural representation on $\mathbb C^n$ but not a subobject of any Tensor power.
There is a nice proof in Alperin's book Local Representation Theory that works in any characteristic and for any faithful module: see Theorem 1 on page 45.
Turning my comment into an answer, Robert Steinberg proved in Complete sets of representations of algebras that if you have a faithful representation of a finite semigroup $S$, then every irreducible representation of $S$ appears as a composition factor of a tensor power of that representation. In fact, he shows that the semigroup algebra $KS$ acts faithfully on the direct sum of the tensor powers, which is a stronger statement since finite semigroups are not usually completely reducible. My favorite proof of this is Passman’s variant of a proof of Rieffel in Elementary bialgebra properties of group rings and enveloping rings: An introduction to Hopf algebras.
If $G$ is a finite group, then $KG$ is a Frobenius algebra and so every irreducible representation is a subrepresentation of the regular representation. Thus in the case of a finite group, every irreducible representation will be a subobject of a tensor power.
For semigroups, this is not true. There are many examples. The easiest is the monoid $T_n$ of all maps on $n$ letters. For $n\geq 2$, the natural representation on $\mathbb C^n$ is faithful and has the trivial representation as a quotient but not as a subrepresentation. The same remains true after taking tensor powers.
It seems to me that the regular representation of $S_n$ occurs, specifically, in $V^{\otimes n}$ where $V$ is functions on an $n$-element set, as the subspace of functions on the $n$th power of an $n$ element set which vanish on all $n$-tuples that are not permutations. So we can take $m=n$ here.
That is true in this case since that module is just the action of the group on functions n to n. The general case can require n the size of S.
Benjamin Steinberg answered the question, but I wanted to unwind his idea into an explicit formula.
Let $V$ be the representation with basis $e_1,\dotsc, e_n$, where a permutation $\sigma$ acts by sending $e_i$ to $e_{\sigma(i)}$.
Let $W$ be an irreducible representation of $S_n$.
Fix a linear form $l$ on $W$. We can map $W$ to $V^{\otimes n}$ by sending $w \in W$ to
$$f(w) = \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dotsm\otimes e_{ \sigma(n) }.$$
Then we have
\begin{align*}
\sigma' (f(w)) & {}= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot \sigma'( e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dotsm e_{ \sigma(n) } ) \\
& {}= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot e_{ \sigma'(\sigma(1))} \otimes e_{ \sigma'(\sigma(2))} \otimes \dotsm \otimes e_{ \sigma'(\sigma(n))} \\
& {}= \sum_{ \sigma \in S_n} l ( (\sigma' \circ \sigma)^{-1}( \sigma'(w)))\cdot e_{ \sigma'(\sigma(1))} \otimes e_{ \sigma'(\sigma(2))} \otimes \dotsm \otimes e_{ \sigma'(\sigma(n))} \\
& {}= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (\sigma' (w) ))\cdot e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dotsm\otimes e_{ \sigma(n) } = f(\sigma' (w))
\end{align*}
using the change of variables $\sigma \mapsto \sigma' \circ \sigma$ in the last line. So $f$ is a homomorphism, and because $W$ is irreducible, and $f$ is nontrivial (as long as $l$ is a nontrivial linear form, since all the terms in the sum give different basis vectors in the tensor product), this map is the inclusion of a subrepresentation, as desired.
One can get embeddings into $V^{\otimes m}$ for higher values of $m$ by just putting repetitions in the sequence of basis vectors being tensored, or for $n-1$ by removing the last term, but one can't go lower than $n-1$, because of the sign representation (except maybe in characteristic $2$, I guess).
Sorry for the silly question—why does the sign representation prevent going lower than $n - 1$?
@LSpice Suppose there is a copy of the sign representation in the tensor product. Take a generator (or the generator, since it's unique up to scaling). Write it in a basis of tensors of basis elements. Find one tensor of basis elements which has a nonzero coefficient. Since only $n-2$ basis elements can appear, two don't. Swap those and derive a contradiction.
|
2025-03-21T14:48:31.659049
| 2020-07-31T14:17:30 |
368034
|
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|
Stack Exchange
|
Implementing Bemporad optimization algorithm
I need to reproduce the results of a relatively old paper of a paper and one of the steps is to solve an IQP problem, the algorithm suggested is this one. Which I don't know it's the best way to do it since it's quite old (I have understanding of continuous optimization, but when it comes to discrete I know bits and pieces). Also I can't really find a C/C++ implementation of any IQP solver, but this one to me doesn't look particularly difficult to implement.
Section 3.2. Reports the algorithm which I'll write down here for reference:
Take the original MIQP, relax all integrality constraints, mark the relaxed QP with its number of guaranteed switches, i.e.−1. Set $f_{opt}=1$,$k_c=−1$,$x_{opt}=[1;...;1]$ and
initialize with the relaxed QPthe list of problems to be solved.
If the list of problems is empty, terminate and output $f_{opt},x_{opt}$.
If there are problems on the list marked by $k_c$, select one of them, remove it from the list, and solve it.If the QP is feasible, denote its cost by $f*$ and its solution by $x*$. Go to step 5. If the QP is infeasible go to 2.
If there are no problems on the list marked by $k_c$,increase $k_c$ by 1 and go to 2.
Fathoming by worse cost: If $f*\geq f_{opt}$,then go to 2
Integer feasibility : If $f* < f_{opt}$ and $x*$ satisfies the integrality constraints, then set $f_{opt}=f*$ and $x_{opt}=x*$ .Go to 2
Feasibility but not integer feasibility Separate the problem. Mark the sub problems by the number of guaranteed switches in the fixed integer variables. Add the subproblems to the list of problems. Goto 3.
It's not 100% clear to me how I form a subproblem to be solved however (I guess this is step 7).
Can you help?
From what I see in figure 2. I think the exploration is exhaustive (with the Outside first strategy). But a subtree isn't explored if it is not feasable.
I think this is the essence of the algorithm, unless I'm missing something.
Professor Bemporad's codes can be downloaded from http://cse.lab.imtlucca.it/~bemporad/hybrid/toolbox/
It's matlab, I'd need C++. I guess it might be easy enough to convert though. It doesn't change the nature of my question anyway, which is how it works.
Are all your variables binary?
Yes, they're. (By the way I just realized that "Outside strategy" is probably what I'm missing, to sure what data structure I would need to achieve that exploration.)
Links can become broken. Please include the full reference.
http://or.stackexchange.com may be a better home for this question.
OK, because all your variables are binary, you can linearize the problem by introducing, for $i<j$, variables $y_{i,j}\ge 0$ to represent $x_i x_j$, together with linear constraints $y_{i,j} \le x_i$, $y_{i,j} \le x_j$, and $y_{i,j} \ge x_i+x_j-1$. Now use an ILP solver.
@RobPratt Maybe you can give a full answer or a reference as I don't understand the transformation from non linear to linear (I mean I got a rough idea, but probably more details would be useful as a reference).
Because all your variables are binary, you can linearize the problem by introducing, for $i<j$, variables $y_{i,j} \ge 0$ to represent the product $x_i x_j$, together with linear constraints
\begin{align}
y_{i,j} &\le x_i\\
y_{i,j} &\le x_j\\
y_{i,j} &\ge x_i+x_j−1
\end{align}
A derivation via conjunctive normal form is given here.
Now use an ILP solver.
|
2025-03-21T14:48:31.659328
| 2020-07-31T15:38:08 |
368040
|
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|
Stack Exchange
|
An integral of composite function of triangle functions
I expected the following formula to hold:
$\int^{2n\pi}_0\cos(\sin t+t/n)dt=0$,
for ${}^\forall n\in\mathbb{N},\ n\geq2$
But I can't prove it.
Could you please tell me.
I'm not quite sure why people are voting to close this. Is it a well-known result? Is it obvious? I think the poster is saying that he found that this seems to be true numerically, and is asking for a proof or reference. I checked it numerically up to $n=10$.
I voted to close because I would have expected an explanation of why it might be expected to hold. More generally the question lacks context. Also it's evidently true for even $n$ by comparing $t$ and $t+n\pi$.
This is a question for MSE.
You can rewrite the integral as
$$
\int_0^{2\pi} \left(\sum_{j=0}^{n-1}\cos\Big(\sin t+\tfrac tn+2\pi \tfrac jn\Big)\right)\,dt.
$$
But $\sum_{j=0}^{n-1}\cos\big(a+2\pi \tfrac jn\big)=0$ for all $a$.
In particular, the equality holds if $\sin t$ is replaced by any $2\pi$-periodic function.
Nice. And for $n=1$ you sum is $\cos(a)$, not 0, which explains why the integral is non-zero for $n=1$. I guess there will be a difference of opinion as to whether this belongs on MO or MSE, but I will say it's a clever trick that I enjoyed seeing.
Thank you very much.
You are smart.
|
2025-03-21T14:48:31.659566
| 2020-07-31T16:37:30 |
368041
|
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|
Stack Exchange
|
Sylow subgroups of abelian profinite groups
If $G$ is a finite abelian group, then we have a decomposition
$$G\cong \prod_{p} G(p)$$
where $G(p)$ is the $p$-Sylow subgroup of $G$. This product makes sense as for all but finitely many primes $p$, we have $G_p=\{0\}$. This is proven by showing that the cardinality of $G$ and $\prod_{p} G(p)$ agree. If we now assume that $P$ is a profinite abelian group, there still exists the notion of a $p$-Sylow subgroup $P(p)$ which is now a pro-$p$-group. I'm curious if there exists an isomorphism
$$P\cong \prod_{p} P(p).$$
One can verify this by computing indices as 'supernatural numbers', in the sense of Serre's "Galois cohomology"; but probably easier is to consider the maps $P \to P(p)$ given by $g \mapsto \lim_{N \to \infty} g^{p^N - 1}$, where $N \to \infty$ in the divisibility order. (I might also consider calling your profinite group something other than $P$, which looks like a pro-$p$ group ….)
Oh, by the way there is a very natural proof for finite groups, which consists in proving that the natural map $\prod_p G(p)\to G$ is a group isomorphism (rather than computing cardinalities).
This is Proposition 2.3.8 of Ribes and Zaleskii - Profinite groups (second edition). (I originally gave references specifically for the finer structure of profinite Abelian groups, but assuming finite generation, in Section 4.3 of the same book.)
I'm particularly interested in the non-finitely generated case. Does anyone know more about that case?
It turns out those Section 4.3 references were mainly about the finer structure of pro-$p$ groups. I modified now to cite Proposition 2.3.8, which is about pronilpotent groups in general (no finite generation hypotheses). Sorry for the confusion!
|
2025-03-21T14:48:31.659711
| 2020-07-31T17:52:02 |
368046
|
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|
Stack Exchange
|
Given multiple posets, what is the probability that a randomly selected (uniform dist) subposet of their product has a max under the product order?
Given multiple totally ordered posets, how do I find the probability that a randomly selected (with uniform distribution) subposet of their product has a maximum under the product order?
I have some vague intuition about the shape of subsets of powers of the unit interval that suggests that it is a relatively small probability, but nothing concrete.
I have no real idea on how to approach the problem after taking the power set of the product. I don't know a general way to separate members of the power set that have maximums under the product ordering fro. Those that don't without manually checking.
Suggestions for relevant texts are appreciated as well.
I take it these are finite posets? So you’re asking what is the probability that a uniformly random subposet of a product of chains (say of sizes $n_1,\ldots,n_k$) has a maximum? Reasonable question, but you might want to edit the post and especially the title (which vaguely says “multiple posets”) to reflect this.
I don't see why they should be finite. You can have maximums in posets of infinite cardinality.
But how do you define a uniform distribution on subposets otherwise?
I have no idea. If that assumption is necessary, then so be it. But I'd prefer not to make that assumption if possible. The example in my head is [0,1]×[0,1]. Can we not define a uniform distribution for picking out subsets of [0,1]×[0,1] with a given property?
There is no “uniform” distribution on any infinite sets.
Ok. Sorry for dragging this out, but I'm not super familiar with measures and probability, so I might be using the wrong words. I'll try to explain differently. Say we have the product of 3 chains. We take the power set of this product and put each element in a hat so that all elements are equally likely to be chosen. I'm trying to find out how to find the probability that the element pulled out of the hat has a maximum using the product order. If that's only possible with finite chains, ok. If I'm using words wrong, ok. Just let me know how to edit the title and question and I will. Thanks
If you’re considering finite chains, then what you just wrote is a totally reasonable question in combinatorial probability. I would be shocked if it had a nice explicit answer, but you could reasonably ask for bounds on this probability.
Note that a special case of your question (when the chains are 2 element chains) is: choose a collection of subsets of some fixed finite set uniformly at random among all such collections: what’s the probability that one set in this collection contains all the others? That might be a question that has been studied. As you suggested, surely it is quite low probability, but people in probability do often like to study ‘rare’ events.
|
2025-03-21T14:48:31.659927
| 2020-07-31T17:55:52 |
368047
|
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|
Stack Exchange
|
Survey of recent developments of the Gelfand-Kirillov dimension
It is almost two decades since the now classical books by McConnell and Robinson's
[ Noncommutative Noetherian rings. With the cooperation of L. W. Small. Revised edition. Graduate Studies in Mathematics, 30. American Mathematical Society, Providence, RI, 2001 ],
and Krause and Lenagan's
[ Growth of algebras and Gelfand-Kirillov dimension. Revised edition. Graduate Studies in Mathematics, 22. American Mathematical Society, Providence, RI, 2000. ],
which are were (and still are in my opinion), the standard references on almost everything related to the Gelfand-Kirillov dimension, appeared.
Time has passed, and a lot of new work on this dimensional invariant has been done.
I am looking for references, surveys and pherhaps lecture notes on the Gelfand-Kirillov dimension which covers relevant developments regarding this invariant in the last 20 years.
Regarding its computational aspects, one has for instance
J. Bueso, J. Gomés-Torrecillas, A. Verschoren, [ Algorithmic methods in non-commutative algebra.
Applications to quantum groups. Mathematical Modelling: Theory and Applications, 17. Kluwer Academic Publishers, Dordrecht, 2003 ],
but it does not cover all aspects of recent developments.
This list is certainly far from being complete, but it contains some important results obtained in the last 20 years.
The following thesis discusses some recent results obtained by Bell (see Section 5):
Michelle Roshan Marie Ashburner (2008). A Survey of the Classification of Division Algebras over Fields. Master Thesis, University of Waterloo
This is a survey on GK dimension of graded PI-algebras:
L. Centrone, On some recent results about the graded Gelfand-Kirillov dimension of graded
PI-algebras, Serdica Math. J. 38(1-3) (2012), 43-68.
Centrone also wrote other papers where he proved some interesting results on GK dimension. For instance, he wrote:
L. Centrone, The graded Gelfand-Kirillov dimension of verbally prime algebras, Linear Multilinear Algebra 59(12) (2011), 1433-1450.
and
L. Centrone, A note on graded Gelfand-Kirillov dimension of graded algebras, J. Algebra
Appl. 10(5) (2011), 865-889.
For some results on Hopf algebras with finite GK dimension, see:
Zhang, G. (2013). Hopf algebras of finite Gelfand-Kirillov dimension. PhD Thesis, University of Washington
To conclude, GK dimension has been recently extended to algebras over commutative domains by Zhang and Bell. Now, GK dimensions can be studied on many new structures. In the following paper, GK is studied for skew PBW extensions
Reyes, A.: Gelfand–Kirillov dimension of skew PBW extensions. Rev. Col. Mat. 47(1), 95–111 (2013)
while in this one it has been studied for rings:
Lezama, O., Venegas, H. Gelfand–Kirillov dimension for rings. São Paulo J. Math. Sci. 14, 207–222 (2020).
I'm not aware of any survey discussing all these new developments.
This is indeed a very good list of references. Thank you very much!
Complementing Manuel Norman's excelent answer, recently I've found a very nice survey about the Gelfand-Kirillov dimension, from 2015, by Jason Bell, called Growth Functions.
This survey discusses many important results involving the Gelfand-Kirillov dimension. Among the more recent ones we have:
Agata Smoktunowicz result (see paper here) that there are no graded domains of GK dimension strictly between $2$ and $3$. This was conjectured by Artin and Stafford, who obtained the weaker result that there are no graded domains of GK dimension strictly between $2$ and $\frac{11}{5}$. Smoktunowicz result is at the same time an extension of Bergman's Gap theorem and, in the context of noncommutative projective algebraic geometry, a proof of the (extremely desirable) fact that the are no noncommutative projective schemes with dimension between $2$ and $3$.
A result by Jason Bell (see the paper here) that shows that if a complex domain of quadratic growth is either PI (and hence close to being commutative) or primitive (and hence as far from being commutative as possible.
A result by Lenagan and Smoktunowicz (see the paper here) that shows that there exists algebras with finite Gelfand-Kirillov dimension for which Kurosh Problem has a negative solution.
I finish this answer with what I discovered to be a challanging open problem in the area: for left and right Noetherian domains, is the Gelfand-Kirillov necessarily an integer? There is a discussion of a similar problem in this MO post.
|
2025-03-21T14:48:31.660220
| 2020-07-31T19:22:07 |
368053
|
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|
Stack Exchange
|
Higher order generalization of Cauchy-Schwarz?
Is there a generalization of the Cauchy-Schwarz inequality along the following lines? Let $V$ be an inner product space (for simplicity of notation, let us work over the real numbers). Let $v_1, \ldots, v_n$ be in $V$. Let $G$ denote the Gram matrix of the $v_i$, namely, $G$ consists of all possible $(v_i, v_j)$, as $i,j = 1, \ldots, n$, where $(-,-)$ is the inner product in $V$. The usual Cauchy-Schwarz inequality, with $n=2$, can be written as follows, to get rid of square roots:
$$ \det(G) = (v_1,v_1)(v_2,v_2) - (v_1,v_2)^2 \geq 0, $$
with equality iff $v_1$ and $v_2$ both belong to some $1$-dimensional subspace of $V$. So in this case, for $n=2$, the LHS is a homogeneous polynomial in $G$ of degree $2$, and equality is achieved iff $v_1$ and $v_2$ both belong to some $1$-dimensional subspace.
For the general $n$ case, is there a higher degree homogeneous polynomial in $G$ which is non-negative for any $v_1, \ldots, v_n$ in $V$, and which vanishes iff the $v_i$, for $i = 1,\ldots, n$ all lie in some $1$-dimensional subspace of $V$?
(I suspect there may be such a polynomial of degree $2 \lfloor \frac{n(n+1)}{4} \rfloor$. So for instance, if $n=2$, the expected degree is $2$. If $n=3$, the expected degree is $6$, and so on.)
Indeed the Gramian is positive semi-definite, so its determinant is always nonnegative, and is positive just when the vectors are linearly independent. See https://en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant
@KevinCasto, yes but, the determinant of the Gramian vanishes iff the vectors are linearly dependent. What I would like is though, a polynomial which vanishes iff the vectors lie in the same $1$-dimensional subspace.
Actually, there's an octic polynomial: $$Q(v_1,v_2,\ldots,v_n) = \sum_{1\le i<j\le n} ((v_i,v_i)(v_j,v_j)-(v_i,v_j)^2)^2.$$
@RobertBryant, ah yes true! The famous sum of squares trick, when working over $\mathbb{R}$. Thank you. How can one obtain all such polynomials? Can one use some form of the positivstellensatz perhaps?
@RobertBryant, could you please write it as an answer? The answer turned out to be simple (and I should have thought about it), but it is guiding me in the right direction (for the problem I am interested in, which inspired this post). Is the post too trivial? Should I delete it?
Yes, because the OP stated that the ground field is $\mathbb{R}$, one can simply take the octic polynomial
$$
Q(v_1,v_2,\ldots,v_n) = \sum_{1\le i < j\le n} \bigl((v_i,v_i)(v_j,v_j)-(v_i,v_j)^2\bigr)^2,
$$
which will do the trick.
Come to think about it, are the squares necessary? We already know that the expressions inside parentheses are nonnegative.
@Malkoun: Yes, you are absolutely correct. I didn't think of that, but, yes, simply the sum of the terms would be enough. Of course, this implies that this construction would work for any ordered field.
If one removes the square, then this (now quartic) polynomial also has a geometric interpretation: it is the trace of the exterior square $\bigwedge^2(T^* T)$ of the square $T^* T$ of the linear transformation $T: {\bf R}^n \to V$ that maps the standard basis to $v_1,\dots,v_n$. (It is also the square of the Frobenius norm of $\bigwedge^2 T$.)
I guess another way to express what Terry is saying is that the identity $$ (v,v)(w,w) = (v,w)^2 + |v\wedge w|^2 = (v,w)^2+(v{\wedge}w,v{\wedge}w)$$ (with the natural inner product on $\Lambda^2(V)$) already shows that $(v,v)(w,w) - (v,w)^2$ is a sum of squares anyway, so the natural quartic polynomial would be $$P(v_1,\ldots,v_n) = \sum_{1\le i < j\le n} |v_i\wedge v_j|^2.$$ In particular, it is expressed as a sum of squares.
|
2025-03-21T14:48:31.660480
| 2020-07-31T19:35:27 |
368054
|
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}
|
Stack Exchange
|
Absolute Galois group of Q and stratification of moduli space of curves
This is slightly related, but distinct from, a question I asked earlier.
The moduli space of ribbon graphs with metric (with all vertices having degree at least 3) is isomorphic to the moduli space of (decorated) smooth complex curves of genus g with n marked points. (cf. Kontsevich's "Feynman diagrams and low-dimensional topology).
But this moduli space of ribbon graphs with metric is naturally stratified by the underlying combinatorial type of ribbon graph (i.e "it breaks up into pieces" where each piece corresponds to exactly one underlying ribbon graph). By Grothendieck's dessin correspondence (and extensions thereof), isomorphism classes of ribbon graphs are in one-to-one correspondence with isomorphism classes of algebraic curves defined over $\overline{\mathbb{Q}}$ (cf. https://arxiv.org/abs/math-ph/9811024).
This means that the moduli space of curves can be partitioned into pieces where each piece corresponds to a unique $\overline{\mathbb{Q}}$ curve.
How well-studied is this?
Given a $\overline{\mathbb{Q}}$ curve, can we explicitly say which other complex curves correspond to the same piece of the partition?
Can this partition be given a group/ring/etc structure? Does the obvious $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ action (by acting on the coefficients defining the curve) on the entire partition have any interesting features?
Does the isomorphism of moduli spaces preserve strata? (Kontsevich says that the moduli of curves is stratified by "euclidean spaces of some dimension modulo some finite groups" - I don't know what he means precisely, but can the $\overline{\mathbb{Q}}$ stratification correspond to another nice stratification on the moduli of curves)?
There's no 1-to-1 correspondence. Different ribbon graphs can give you the same curve.
Really? In particular, the paper linked above says that there is a bijection on the (first sentence) of the second page. This is repeated in the review of this paper on Mathscinet as well. ... (Perhaps I should have said "connected" ribbon graphs instead.)
"Grothendieck discovered that there is a natural bijection between the set of
isomorphism classes of connected ribbon graphs and the set of isomorphism classes of of Belyi maps." Belyi maps are not algebraic curves defined over $\overline{\mathbb Q}$. They are algebraic curves defined over $\overline{\mathbb Q}$, plus extra data.
Ahh, I see. Is "algebraic curves defined over $\overline{\mathbb{Q}}$ are in bijection with Riemann surfaces admitting a Belyi map, but such Riemann surfaces may admit more than one Belyi map, so picking the Belyi map is picking extra data" what is happening here?
Yes, that's exactly it.
One point that still confuses me - a point in the moduli space of metric ribbon graphs corresponds uniquely to a point in the above moduli space of curves, but if two distinct ribbon graphs give the same curve over $\overline{\mathbb{Q}}$, equipping all edges of both with length 1, they would correspond to the same point in the moduli space of curves. What am I missing? (Is the decoration on the moduli space of curves what is making the difference?)
Yes, the decoration is quite significant here.
|
2025-03-21T14:48:31.660698
| 2020-07-31T19:50:11 |
368056
|
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|
Stack Exchange
|
Open covers by ind-affine ind-schemes
Many apologies if this is totally standard! I couldn't find it in the literature.
Background definitions:
A presheaf $X: \textbf{Aff}^\text{op} \to \textbf{Set}$ is an ind-scheme if it is a filtered colimit of schemes, where we regard a scheme as a presheaf via the Yoneda embedding (and then restrict to affine schemes), and the colimit is taken in the category $\text{Psh}(\textbf{Aff})$ of presheaves (i.e. the colimit is taken "pointwise"). One can impose the condition that all of the transition maps are closed immersions, and then call the result a strict ind-scheme.
Let's say that $X$ is an ind-affine ind-scheme if it is a filtered colimit of affine schemes. (Side question: Is this the right terminology? It seems there are several distinct notions with this name in the literature.)
Context:
It is well known that a presheaf $X \in \text{Psh}(\textbf{Aff})$ is represented by a scheme if and only if it satisfies the following two conditions: 1) $X$ is a Zariski sheaf and, 2) There is a covering of $X$ by open affine subfunctors.
Every ind-scheme is a Zariski sheaf (in fact satisfies the sheaf condition for the fpqc topology).
The affine Grassmannian (which is an ind-scheme) admits an open cover by ind-affine ind-schemes. See, e.g. the extremely helpful paper by Timo Richarz "Basics on Affine Grassmannians".
Question:
Is there a similar characterization for (strict or otherwise) ind-schemes? I.e. is something like the following true: a presheaf $X$ is an ind-scheme if and only if 1) $X$ is a Zariski sheaf and, 2) $X$ admits a covering by open subfunctors which are ind-affine ind-schemes? Is it at least sufficient? If not, is there a name for the presheaves that satisfy this property?
If you are a filtered colimit of schemes, aren't you also a filtered colimit of affine schemes?
@Asvin I think you are conflating two different kinds of colimits there. It is true that a scheme is a filtered colimit of affine schemes – in the category of schemes or in the category of Zariski sheaves. But I don't think it is true in the category of presheaves. Otherwise we could just be studying pro-rings instead of schemes.
Ah, I see. That's an interesting difference!
@ZhenLin Do you have a reference for the "It is true" claim? I am having difficulty writing $\mathbb{P}^1$ as a filtered colimit of affine schemes.
Ah, I confused cofiltered and filtered there, oops.
|
2025-03-21T14:48:31.661239
| 2020-07-31T20:56:31 |
368060
|
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|
Stack Exchange
|
A simple proof of non-convergence of a sequence
I'm looking for a proof of the following fact:
Suppose $a_n$ is a sequence of complex numbers of the form
$$a_n = \sum_{j=1}^m c_j \alpha_j^n$$
where $m$ is a positive integer and the $c_j, \alpha_j$ are complex numbers with $|\alpha_j|=1$ for each $j=1, \ldots, m$. Assume also that the $\alpha_j$ are distinct. Then, $a_n$ can not converge to $0$.
I think I can cobble together a proof using dynamics on the $m$ torus (i.e. using irrational rotations) but was hoping to find an easier proof. Any suggestions would be greatly appreciated - thanks.
Just evaluate $\lim \frac 1n\sum_{k=1}^n|a_k|^2=\sum_j|c_j|^2$.
Such a simple solution - thanks!
In a different way, if $\alpha_j=e^{i2\pi r_j}$, and all $r_j$ are rational, then there is a period (there exist $N$ such that all $r_jN$ are integer). Which implies that if $\lim \limits_{n \to \infty} a_n=0$, then $a_m=0$ for all $m \in \mathbb N$. If $r_i$ are not all rational, then also we can choose $N_0$ so that all $r_jN$ are arbitrarily close to an integer (${r_j N_0} \to 0$ for all $j=1,2...m$). Hence, for $a_{N_0+t}≈a_t, 0<t<N_0$. If $a_n$ converges to $0$, $a_{N_0+t}≈a_t=0, t=1,2...$. Hence, $a_n$ doesn't converge to $0$.
Another argument is as follows. For $j=1,\dots,m,$ let $u_j\in\ell_\infty$ denote the bounded sequence $(\alpha_j^n){n\ge0}$. If there is a linear combination $\sum{j=1}^m c_ju_j$ of $u_1,\dots,u_m$ which is in $c_0$ then $|\sum_{j=1}^m c_j\alpha_j^ku_j|\infty=o(1)$ as $k\to\infty$.
By compactness, a subsequence $(\alpha_1^{k_h},\dots, \alpha_m^{k_h})$ converges for $h\to\infty$ to some $(\beta_1,\dots,\beta_m)$ such that $\sum{j=1}^m (c_j\beta_j),u_j=0$, but since the $u_j$'s are linearly independent (e.g. by Vandermonde determinant) this implies $c_j=0$ for all $j$.
|
2025-03-21T14:48:31.661393
| 2020-07-31T21:37:45 |
368061
|
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"Chris Godsil",
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|
Stack Exchange
|
Chromatic number of regular graphs using spectra
There exist inequalities relating the maximum and minimum eigenvalues of the adjacency matrix of a graph with its chromatic numbers, i.e. the Wilf's and Hoffmann's inequalities, which put together state that $1-\frac{\lambda_{max}}{\lambda_{min}}\le\chi(G)\le 1+\lambda_{max}$, where $\chi, \lambda_{max}, \lambda_{min}$ respectively stand for the chromatic number, maximum and minimum eigenvalues of the adjacency matrix of $G$. But, for regular graphs, the upper bound by Wilf's inequality is quite trivial, that is same as the greedy coloring bound.
Is there a better bound for the chromatic number of a regular graphs using the spectra of, say the adjacency or Laplacian matrices? Thanks beforehand.
Because I wasn't able to find it easily, can you include the Wilf-Hoffmann inequality?
@M.Winter edited the post. See now
For regular graphs, the adjacency and Laplacian matrices give exactly the same information.
@ChrisGodsil ok. But, any better bound than the wilf and hoffmann's inequalities? especially upper bounds
|
2025-03-21T14:48:31.661499
| 2020-08-01T04:09:29 |
368066
|
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|
Stack Exchange
|
Composition in function fields
Let $k=\mathbb F_q\left(\!\left(\frac1T\right)\!\right)$. One has the map: $\circ:k\times\{v\in k\mid\deg(v)>0\}\to k$ defined by $f\circ g=\sum_{n\ge-m}a_ng^{-n}$ where $f=\sum_{n\ge-m}a_n\frac1{T^n}$ ($m\in\mathbb N_0$, $a_n\in\mathbb F_q$).
Denote by $\Omega$ a completion of $\overline k$ for the valuation $-\deg$. Can one extend $\circ$ continuously to $\Omega\times \{v\in\Omega\mid\deg(v)>0\}$?
|
2025-03-21T14:48:31.661561
| 2020-08-01T05:34:32 |
368067
|
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"Andreas Blass",
"Jesse Elliott",
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|
Stack Exchange
|
smallest ordinal $\alpha$ such that $L \cap P(L_\alpha)$ is uncountable
Let $V$ denote the von Neumann universe and $L$ Gödel's constructible universe. For any set $X$, let $P(X)$ denote the power set of $X$.
Assume that $0^\sharp$ exists (and ZFC).
What is the smallest ordinal $\alpha$ such that $L \cap P(L_{\alpha})$ is uncountable? (If $V = L$, then $\alpha = \omega$, but if $0^\sharp$ exists, then $\alpha > \omega$.)
It’s $\omega_1$.
I suspected so. Do you have a reference for a theorem that implies this?
If $0^\sharp$ exists, then every cardinal is inaccessible in L.
Ah, so $L \cap V_\alpha$ is countable for all countable $\alpha$, too, since $L \cap V_\alpha$ is the $V_\alpha$ of $L$. Thank you.
In ZFC alone, the $\alpha$ in the title can be described as: (1) If genuine $\omega_1$ is a successor cardinal of $L$, then $\alpha$ is its immediate predecessor cardinal of $L$. (2) If genuine $\omega_1$ is a limit (and therefore inaccessible) cardinal of $L$, then it is equal to $\alpha$. The additional hypothesis that $0^#$ exists implies that case (2) occurs.
Really, this was answered in the comments; I'm putting this answer down to move this off the unanswered queue. I've made this CW and will delete it if one of the original commenters adds their own answer.
We have in $L$, for each (infinite) $\alpha$, the following bijections:
$f_\alpha:\alpha\rightarrow L_\alpha$.
$g_\alpha: \mathcal{P}(L_\alpha)^L=\mathcal{P}(L_\alpha)\cap L\rightarrow L_{(\vert\alpha\vert^+)^L}$.
Hence $\vert\mathcal{P}(L_\alpha)^L\vert=\vert(\vert\alpha\vert^+)^L\vert$. Now assuming $0^\sharp$ we have that $\omega_1^V$ is a limit cardinal in $L$, so for each $\alpha<\omega_1^V$ we have $\vert\mathcal{P}(L_\alpha)^L\vert=\aleph_0$.
So the answer to your question is $\omega_1^V$.
Note that all this requires is that $\omega_1^V$ be a limit cardinal in $L$. More generally, let $\kappa$ be the supremum of the $L$-cardinals whose $L$-successor is $<\omega_1^V$; then the $\kappa$th level of $L$ is the first whose $L$-powerset is truly uncountable.
OK, except I think you meant $P(L_\alpha)^L = L \cap P(L_\alpha)$, not $P(L_\alpha)^L = P(\alpha) \cap L$.
@JesseElliott Whoops, quite right - fixed!
|
2025-03-21T14:48:31.661733
| 2020-08-01T05:58:58 |
368069
|
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"Alexandre Eremenko",
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|
Stack Exchange
|
Transcendence of Euler series
Is the Euler series $\sum_{n\ge0}n!X^n\in\mathbb C[[X]]$ transcendental over $\mathbb C(X)$?
Does not any algebraic power series has a positive radius of convergence? It looks that the bound $C^n/n^D$ for the coefficients should be provable by induction, looking at the relation for the new coefficient $a_n$.
@FedorPetrov Indeed, this is, e.g., proposition 2 in these notes (Peter Roquette, “On convergent power series”, 1996-07-16).
All series with zero radius of convergence are transcendental over $C(X)$.
Wahoo. Thanks for these very accurate answers. Roquette's paper is very impressive.
|
2025-03-21T14:48:31.661931
| 2020-08-01T06:30:56 |
368070
|
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|
Stack Exchange
|
Weisfeiler-Lehman test for hypergraphs
The Weisfeiler-Lehman test for graph isomorphism is based on iterative graph recoloring and works for almost all graphs, in the probabilistic sense. If we extend the domain to general hypergraphs, does there exist an analogous test for hypergraph isomorphism?
You can represent a hypergraph by its vertex-edge incidence graph and apply W-L to that.
|
2025-03-21T14:48:31.661991
| 2020-08-01T08:44:48 |
368072
|
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"Zhi-Wei Sun",
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|
Stack Exchange
|
Some conjectural congruences involving Domb numbers
The Domb numbers are given by
$$D_n=\sum_{k=0}^n\binom{n}{k}^2\binom{2k}k\binom{2(n-k)}{n-k}\ \ \ (n=0,1,2,\ldots).$$
Such numbers have combinatorial interpretation, see, e.g., http://oeis.org/A002895.
I have the following three conjectures on congruences involving the Domb numbers.
Conjecture 1 (2019). For any odd prime $p$, we have
$$\sum_{k=1}^{p-1}\frac{D_k}k\equiv\left(\frac p3\right)\frac 25pB_{p-2}\left(\frac13\right)\pmod{p^2},$$
where $(-)$ is the Legendre symbol and $B_{p-2}(x)$ is the Bernoulli polynomial of degree $p-2$.
Conjecture 2 (2020). For any prime $p>5$, we have
$$\sum_{k=1}^{p-1}\frac1k\left(D_k-\frac{4D_{k-1}}{64^{k-1}}\right)\equiv-\frac{16}3p^2B_{p-3}\pmod{p^3},$$
where $B_0,B_1,\ldots$ are the Bernoulli numbers.
Conjecture 3 (2013). For any prime $p>3$, we have
$$\det[D_{i+j}]_{0\le i,j\le p-1}\equiv\begin{cases}(\frac{-1}p)(4x^2-2p)\pmod{p^2}&\mbox{if}\ p=x^2+3y^2\ (x,y\in\mathbb Z),\\0\pmod{p^2}&\mbox{if}\ p\equiv 2\pmod 3.\end{cases}$$
Conjectures 1 and 3 appeared in Conjecture 79 of my published paper Open Conjectures on Congruences. I have not made Conjecture 2 public before, I can prove the congruence in Conjecture 2 modulo $p$.
QUESTION. Any ideas towards solving these conjectures?
how could you prove the conjecture 2 modulo $p$?
Because $D_k\equiv 64^kD_{p-1-k}\pmod p.$
|
2025-03-21T14:48:31.662091
| 2020-08-01T13:49:35 |
368079
|
{
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"David Roberts",
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"k.j."
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368079"
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|
Stack Exchange
|
A presentation of an algebraic stack is epi. in etale topology
Let $S$ be a scheme and $\mathscr{X}$ an Artin stack over $S$.
Let $X$ be a scheme and $P : X \to \mathscr{X}$ a representable morphism, which is smooth and surjective.
Then this $P$ is an epimorphism.
(i.e., for any $S$-schemes $U$ and $x \in \mathscr{X}(U)$, there exists an etale surjective morphism $U' \to U$ (of schemes), such that $x|_{U'}$ comes from $X(U')$)
I want to know it in order to show that $\mathscr{X} = [X/ R]$, the quotient stack.
($R = X \times_\mathscr{X} X$)
In 4.3. of Laumon, Moret-Bailly's Champs algébriques, the authors say that this is obvious.
But I don't understand.
I've heard that every Artin stack is a stack over fppf topology.
If we use it, then trivially $\mathscr{X} = [X/ R]$.
But in the proof of this proposition (10.7. of LMB's Champ algebriques), they use the highlighted statement, so it is circular reasoning.
How can I show the highlighted statement without using the stack-ness over fppf topology?
Take the fiber product $X \times_{\mathcal X} U$, which is a smooth scheme over $U$, then use that every smooth surjective morphism of schemes has an etale section.
Has étale-locally a section, no?
@WillSawin I did not notice that. Thank you!
@DavidRoberts Yes, that's what I should have said.
As Will Sawin pointed out in comments, every smooth surjective morphism of schemes admits étale-local sections (Tag 055U in the Stacks Project). See Tag 055S more generally for details and discussion. As mentioned at Tag 021Y the étale topology and the smooth topology give rise to the same topos, but it is useful to deal with the extra flexibility that smooth morphisms give, as evidenced by examples of Artin stacks that are not Deligne–Mumford stacks (for instance, IIRC, the classifying stack of $\mathbb{G}_m$).
A "similar" result along with proof can be found as Lemma 2.14 of Differentiable Stacks and Gerbes.
I would like to give more details if you want.
|
2025-03-21T14:48:31.662255
| 2020-08-01T13:49:42 |
368080
|
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"Lucia",
"Stanley Yao Xiao",
"Terry Tao",
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|
Stack Exchange
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Sequences of very well distributed integers
The set of integers $\mathbb{Z}$ has a peculiar property: it is extremely well-distributed modulo any positive integers. For an integer $m > 1$ and integer $a$, put $\mathbb{Z}(a; m) = \{n \in \mathbb{Z}: n \equiv a \pmod{m} \}$. Then for any interval $I \subset \mathbb{R}$, integer $a$, and positive integer $m$, one has the estimate
$$\displaystyle \#(I \cap \mathbb{Z}(a;m)) = \frac{|I|_{\mathbb{R}}}{m} + O(1),$$
where $|\cdot|_{\mathbb{R}}$ denotes the length of the interval $I$. Moreover, the implied constant in the big-$O$ is independent of $m$ and $I$: indeed, one has the sharper estimate that
$$\displaystyle |\#(I \cap \mathbb{Z}(a;m)) - |I|_\mathbb{R} m^{-1} | \leq 1.$$
In general, this kind of well-distribution cannot be expected for subsequences of integers, even when adjusted for "impossible" congruence classes. For example, even under GRH the best one can hope for for the analogous statement for primes is the estimate
$$\displaystyle \sum_{\substack{y < p < x \\ p \equiv a \pmod{q}}} \log p = \frac{x - y}{\phi(q)} + O_\epsilon \left(q^{-1/2} x^{1/2 + \epsilon}\right),$$
and while this is comparable to the integer case when $q$ is close in size to $x$ this is very bad if $q$ is tiny compared to $x$. Moreover this is a wide open problem: it is in some sense even stronger than GRH.
Let $g$ be a multiplicative function satisfying $g(p^k) = p^k(1 - \kappa p^{-1})$ for all primes $p$ and positive integers $k$ for some absolute constant $\kappa$. Indeed, the indentity function corresponds to the case when $\kappa = 0$ and the Euler $\phi$-function corresponds to the case $\kappa = 1$. For a given positive integer $m$ and an integer $a$, put $S(a; m) = \{n \in S : n \equiv a \pmod{m}\}$. We say that a subsequence $S \subset \mathbb{Z}$ is well-distributed if there is such a function $g$ such that for all $m \in \mathbb{N}$ and $a \in \mathbb{Z}$, we have for any interval $I \subset \mathbb{R}$ either $\# (I \cap S(a; m)) = O(1)$ or we have
$$\displaystyle \# (I \cap S(a;m)) = \frac{|I|_\mathbb{R}}{g(m)} + O_{g,\epsilon }\left(|I|_\mathbb{R}^\epsilon \right),$$
the implied constant depending at most on $g$ and $\epsilon$ but otherwise independent of $I,m$.
Does this property more or less characterize the integers? Are there any other natural sequences which can be shown to have this property?
@ChristianRemling indeed! Thanks for catching the (honestly glaring) mistake. I have fixed the problem in the question.
Presumably $#(I \cap {\mathbb Z})$ in your final display should be something else. Also, discrepancy theory results such as Roth's discrepancy theorem https://mathscinet.ams.org/mathscinet-getitem?mr=168545 will provide obstructions to any set that isn't close to full density or zero density from being as equidistributed as you wish here.
@TerryTao yes the last display suffered the same issue as the early part of the question, which has now been fixed. Thank you for the reference
You may also wish to look at Granville and Soundararajan "An uncertainty principle for arithmetic sequences" which shows that general sifted sets must either be poorly distributed in short intervals, or poorly distributed in arithmetic progressions. This is a generalization of the Maier type results on primes.
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2025-03-21T14:48:31.662479
| 2020-08-01T14:41:03 |
368082
|
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"folouer of kaklas",
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|
Stack Exchange
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Literature request: Schatten class difference of semigroups
Let $\mathcal{H}$ be a Hilbert space and $A,B$ two operators on it (not necessarily self-adjoint) such that $A, A+B$ are generators of strongly continuous one parameter semigroups $e^{-tA},e^{-t(A+B)}$, $t>0$. I would like to ask for some literature on the consequences of the condition
\begin{equation}
e^{-t(A+B)}-e^{-tA} \in S_{p}(\mathcal{H})
\end{equation}
where $p\geq 1$ and $S_{p}(\mathcal{H})$ is the p-Schatten class of operators on $\mathcal{H}$. In particular, I would like results crucially using that this difference is Hilbert-Schmidt, for example, and not just that it is compact as a result.
The problem is discussed in a more general setting (operator ideals in Banach spaces) for the so-called analytic semigroups (parabolic problems) in
Blunck, S.; Weis, L., Operator theoretic properties of differences of semigroups in terms of their generators, Arch. Math. 79, No. 2, 109-118 (2002). ZBL1006.47036.
The paper seems to be freely accessible. The idea is, if the differences (of some fractional powers) of the resolvents belong to the ideal and have a nice asymptotics, then the differences of the semigroups belong to the same ideal and have a nice decay.
The application they give is $e^{t(\Delta - V)} - e^{t\Delta}$ on the whole space, where the semigroups are not even compact, but the difference belongs to the Schatten classe you wish for.
Many thanks for your answer, it seems this fits and there are a few references in there I will study too.
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2025-03-21T14:48:31.662607
| 2020-08-01T14:57:59 |
368083
|
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"Mellon",
"Minseon Shin",
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|
Stack Exchange
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Distinguishing ample divisors by minimally intersecting curves on a smooth projective toric variety
My question has an easily formulated generalization, which I will state first. Let $\sigma \subseteq \mathbf{R}^n$ be a full-dimensional strongly convex polyhedral cone. For each lattice point $m \in \sigma^o \cap \mathbf{Z}^n$, minimally generating inside the interior cone $\sigma^o$, let $S(m) \subseteq \sigma^{\vee} \cap \mathbf{Z}^n$ denote the set of lattice points $u$ with $\langle u,m \rangle = 1$. The generalized question is:
Does $S(m) = S(m') \not = \varnothing$ imply that $m = m'$?
UPDATE: As Minseon Shin pointed out, there was a $2$-dimensional counter-example to the previous formulation of the above.
For my main question, as a special case of the above, assume that $\sigma$ is the nef cone of a smooth projective toric variety $X_{\Sigma}$. Then my question amounts to the following:
Let $D_1$ and $D_2$ be two ample divisors, minimally generating inside the ample cone. Suppose that there exists two effective curves $C_1,C_2$ such that $D_1 \cdot C_1 = D_2 \cdot C_2 = 1$. Then does $D_1 \cdot C = 1 \Leftrightarrow D_2 \cdot C = 1$ for all effective curves $C$ imply that $D_1 = D_2$?
Is the following a counterexample to the generalization? Let $n = 2$, let $s \ge 4$ be an integer, let $\sigma \subseteq \mathbf{R}^{2}$ be the cone generated by $(s,1)$ and $(0,1)$, let $m = (1,1)$ and $m' = (2,1)$. Then $S(m) = S(m') = {(0,1)}$.
Yes, thanks - I'll update my question.
|
2025-03-21T14:48:31.662722
| 2020-08-01T17:15:12 |
368086
|
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"authors": [
"Mark Wildon",
"Mikhail Borovoi",
"Will Sawin",
"curious math guy",
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"https://mathoverflow.net/users/4149",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368086"
}
|
Stack Exchange
|
Non-abelian representations and their induced abelian representations
$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Out{Out}\newcommand\ab{^{\text{ab}}}$Let $G$ and $H$ be a groups and assume that $H$ is non-abelian. Then we have a morphism
$$\Out(H)\rightarrow \Aut(H\ab).$$
For example if $H=F_2$, the free group with two generators, this map is an isomorphism, for $F_n$ with $n\geq 3$ this is only a surjection and for more general groups this is not a surjection, as was pointed out in the comments by Will Sawin and Mark Wildon.
My question thus is if we are given a morphism $$\rho:G\rightarrow \Aut(H\ab)$$
does there exist a way to tell if this comes from a morphism
$$G\rightarrow \Out(H)?$$
I am pretty sure this is not a surjection in general. Why should it be?
The map $\mathrm{Out}(H) \rightarrow \mathrm{Aut}(H^{\mathrm{ab}})$ is not in general surjective. I had an incorrect example and I see Will Sawin has now posted a correct one.
@MarkWildon I don't think your example is the right one, because $H^{\mathrm{ab}} =C_2$. However, $C_5 \rtimes C_4$ works, because its abelianization $C_4$ has an automorphism, which does not lift to $C_5 \rtimes C_4$.
You are both absolutely right! We do have a surjection if $H$ is a free group however, which is what I had in mind. I'll change the question accordingly.
First, you should require that $\rho$ lands in the image of ${\rm Out}(H)$.
If yes, then you get a cohomology class in $$H^2(G,{\rm ker}[{\rm Out}(H)\to {\rm Aut}(H^{\rm ab})]).$$
You can lift $\rho$ if and only if the cohomology class is neutral.
There may be more than one neutral class in nonabelian $H^2$.
You can find a down-to-earth treatment of nonabelian $H^2$ in my paper :M. Borovoi, Abelianization of the second nonabelian Galois cohomology. Duke Math. J. 72 (1993), 217-239, and also in the references therein (especially in Springer's paper) and in papers referring to my paper and to Springer.
|
2025-03-21T14:48:31.662874
| 2020-08-01T18:00:58 |
368088
|
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"url": "https://mathoverflow.net/questions/368088"
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|
Stack Exchange
|
How to show a contraction of singular moduli space is projective?
Let $\mathcal{H}$ be a certain kind of Hilbert scheme of curves on some smooth projective variety $X$ and $\mathcal{H}$ is projective and irreducible of dimension $3$. There is a divisor $\mathcal{D}\subset\mathcal{H}$ and $\mathcal{D}$ is a ruled surface over a curve $C$ (not necessarily smooth). Let $P\in C$ be a singular point and let $F$ be the correspondent fiber and $\mathcal{H}$ is only singular at $F$. Let $\pi:\mathcal{H}\rightarrow\mathcal{H}'$ such that $\pi$ contract $\mathcal{D}$ to $C\subset\mathcal{H}'$ and $\pi$ is isomorphism outside $\mathcal{D}$.
What information or property I should know to show that $\mathcal{H}'$ is also a projective. In the smooth setting, I can try to show that $\pi$ contracts negative $K_{\mathcal{H}}$ extremal ray and apply Cone theorem to show $\mathcal{H}'$ is projective. But now $\mathcal{H}$ has singularity and I do not know whether the singualr locus being a curve (actually $\mathbb{P}^1$ in my case) would belong to some class of singularities(say klt, lc,etc) to allow me apply Cone theorem in general.
|
2025-03-21T14:48:31.662980
| 2020-08-01T18:16:32 |
368090
|
{
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"authors": [
"Branimir Ćaćić",
"JP McCarthy",
"Konstantinos Kanakoglou",
"LSpice",
"Sam Hopkins",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/368090"
}
|
Stack Exchange
|
Axiomatic definition of quantum groups
This is a question I've discussed with a lot of mathematicians, and have read some mathematical texts about, and watched some conference talks about: what is, axiomatically, a quantum group?
There are many classes of noncommutative algebras that everybody agrees is a quantum group (or quantum algebra): quantizations of certain coordinate rings, quantizations of enveloping algebras, quantizations of semisimple algebraic groups, multiparameter quantizations of the Weyl algebra, etc; but what is the state of the art of attempts to give an axiomatic definition for this class of algebras?
A related MO question is What is quantum algebra?. A nice and leisure discussion, albeit not axiomatic, is Shahn Majid's 'What Is... a Quantum Group' (here).
What you'll find is variations on the theme of Hopf algebra, where the comultiplication is the “group law”, the antipode is the “group inverse”, and the counit is “evaluation at the group identity.” Different variations correspond to working with quantum “group objects” in different “categories”, as it were, though I can’t help but point out the $C^\ast$-algebraic theory of locally compact topological quantum groups as particularly well-developed from an axiomatic standpoint.
A quasi triangular Hopf algebra (https://en.m.wikipedia.org/wiki/Quasitriangular_Hopf_algebra) is a precisely defined mathematical object that comes close to giving an axiomatic framework for a quantum group. Unfortunately, it doesn’t quite capture the main examples, e.g., for q not a root of unity the R-matrix of the quantized universal enveloping algrebra doesn’t live in H tensor H but rather some completion of it.
I think it's much like "what is the field with one element?"—'the' definition is whatever someone who is investigating the subject needs it to be to prove interesting theorems, while still being related to the common (informal) theme, as articulated by @BranimirĆaćić.
Stupid correction: if you think of a Hopf algebra $H$ as the group algebra of a quantum group $\Gamma$, then the counit is “the (character of the) trivial irreducible representation” or equivalently, by Pontrjagin duality, “evaluation at the unit of the dual group $\widehat{\Gamma}$”.
@Branimir to be consistent with what you said above I would say that the counit is the inclusion of the group identity. In the finite group case, applying the free functor and then dual functor to the group law and inverse maps gives the comultiplication and antipode. The same functor composition applied to the map from the one element set to G, mapping to the identity element, gives the counit.
@JPMcCarthy you’re absolutely right—it’s then telling that the counit is the convolution identity for appropriate convolution algebras of functions on the Hopf algebra.
I am not sure about axiomatic but a "working" -though too broad- definition could be: "Non-(co)commutative, algebraic structures, whose presentation in terms of generators and relations depends on one ore more discrete, complex, parameters, in such a way that their limits for particular values of these parameters recover algebras, coalgebras, groups or lie algebras".
@Konstantinos would you not include quantum groups with commutative and cocommutative algebras of functions in any working definition? Furthermore quantum groups a la Woronowicz do not naturally have parameters and I don't see how they easily fit into your working definition?
@JP McCarthy, i was referring to the quantum groups not necessarily to their algebras of functions. Regarding the compact quant groups, i think that almost all examples in the literature do have parameters in their presentations with generators and relations.
@Konstantinos in my head we onpy talk of quantum groups through their algebras of functions? I do not think we (naturally) have such parameters for Wang's quantum permutation groups, nor the free/liberated versions of $O_N$ nor $U_N$.
I would say that if you are looking for a concrete definition then it's better to adopt the Tannakian point of view and to focus on the category of representations of the quantum group rather than on the algebra itself. So take as your fundamental object a tensor category (a special type of rigid abelian monoidal category - see here for details). A "quantum group" is then some way of realising the category as a category of representations or corepresentations. There can be different algebras which do this job, and they can come in different flavours, such as Hopf algebras or compact quantum groups in the sense of Woronowicz. This allows one to view the various quantum groups floating around as tools to study the category itself, removing the need for any axiomatic definition.
If the object is really deserving of the name quantum group, then the tensor category should be braided, as is the case for quasi-triangular Hopf algebras and their category of modules. (see the comment of Sam Hopkins above.)
I would have liked to write this as a comment, but with my points tally I can not. So writing this as an answer.
In quantum groups, we are probably at a stage group theory was, say in the first half of the 19th century (see here and here), where we have several important classes of objects that we more or less agree should qualify to be called quantum groups, but it is not clear yet if we are anywhere close to having a single set of axioms that will cover all these classes. It is in fact far from clear whether the union of all these `subclasses' will be part of one single meaningful class.
I think the voting (in which I participated) indicates that this should be an answer after all.
A cheap, soft, quick but objectionable meta-definition:
An algebra of functions on a quantum group is an algebra that satisfies some specific axioms such that whenever an algebra satisfying those same axioms is commutative, it is an algebra of functions on a group (and e.g. the group multiplication given as the transpose of the comultiplication); and whenever two commutative algebras satisfy those axioms are isomorphic as objects satisfying those axioms, their underlying groups are isomorphic.
Not satisfactory but a start.
I don't think that this is a great answer, but I think it's a shame silently to downvote it. Since a downvote means "This answer is not useful", why not explain why it's not useful?
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