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2025-03-21T14:48:31.322190
| 2020-06-22T14:42:45 |
363832
|
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|
Stack Exchange
|
How to show that $d_G(t)$ is decreasing in $t$ for a geometry mixing time?
Let $P$ be the transition matrix of a Markov chain with state-space $\mathcal{X}$, $\pi$ is the stationary distribution with $\pi=\pi P$, and $Z_t$ be a geometric random variable of parameter $1/t$ taking values in $\{1,2,\dots, \}$ and independent of $x$. Define
$$d_G(t):=\max_{x\in\mathcal{X}}\|P_x(X_{Z_t}=\cdot)-\pi\|_{TV}$$
where "TV" means total variation distance between two probability distributions $\mu$ and $\nu$ on $\mathcal{X}$ is defined by $$\|\mu-\nu\|_{TV}:=\max_{A\subset \mathcal{X}}|\mu(A)-\nu(A)|$$
How to show that $d_G(t)$ is decreasing in $t$?
This is true assuming that $Z_t$ is independent of the Markov chain. Indeed, then
$$d(t):=d_G(t)=\max_x E\|P_x(Z_t)-\pi\|_{TV},$$
where
$$P_x(n):=\delta_x P^n$$
and $\delta_x$ is the row matrix $([a_y]_{y\in\mathcal X})^T$ with $a_y:=1_{y=x}$. It is easy to see that for any probability measures $\mu$ and $\nu$
$$\|\mu-\nu\|_{TV}=\sup_{0\le f\le1} \int f\,d(\mu-\nu), \tag{0}$$
where $\sup_{0\le f\le1}$ is taken over all measurable functions $f$ such that $0\le f\le1$.
Take now any real $s$ and $t$ such that $1\le s\le t$. We have to show that then $d(t)\le d(s)$. So, it suffices to show that for each $x$
$$E\|P_x(Z_t)-\pi\|_{TV}\overset{\text{(?)}}\le E\|P_x(Z_s)-\pi\|_{TV}. \tag{1}$$
The random variable $Z_t$ is stochastically greater than $Z_s$. So, without loss of generality, $Z_t\ge Z_s$. Take now any column matrix $f=[f_x]_{x\in\mathcal X}$ with $f_x\in[0,1]$ for all $x$. Then the entries of the (random) column matrix $P^{Z_t-Z_s}f$ are in the interval $[0,1]$ as well and hence
$$(P_x(Z_t)-\pi)f=(P_x(Z_s)-\pi)P^{Z_t-Z_s}f\le\|P_x(Z_s)-\pi\|_{TV}$$
by (0).
So, again by (0),
$$\|P_x(Z_t)-\pi\|_{TV}\le\|P_x(Z_s)-\pi\|_{TV}.$$
Taking now the expectations, we get (1), as desired.
Thanks! How about $d(t)$ for general time $t$ rather than geometric time? It seems that we need to prove $|\mu P-\nu P|\leq |\mu-\nu|$.
@BobO. : The proof holds for any stochastically increasing family $(Z_t)$ of random variables. For all probability measures $\mu$ and $\nu$, the inequality $|\mu P-\nu P|{TV}\le|\mu-\nu|{TV}$ holds by (0).
Why $Z_t \geq Z_s$? Do not we need to find a coupling $(Z_t, Z_s)$?
@BobO. : The existence of such a coupling is well known. See e.g. https://projecteuclid.org/euclid.aop/1176995659
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2025-03-21T14:48:31.322359
| 2020-06-22T14:44:01 |
363834
|
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|
Stack Exchange
|
A conjecture on circular permutations of n elements in an abelian group of odd order
In 2013 I formulated the following conjecture in additive combinatorics.
Conjecture. Let $G$ be an additive abelian group of odd order, and let $A$ be a subset of $G$ with $|A|=n>2$. Then, there is a circular permutation $(a_1,\ldots,a_n)$ of all the elements of $A$ such that all the adjacent sums $ a_1+a_2,\ldots,a_{n-1}+a_n,a_n+a_1$
are pairwise distinct.
Recently, Mr. Yu-Xuan Ji, a student at Nanjing Univ., verified this conjecture for $|G|<30$.
I'm even unable to show the conjecture for $G=\mathbb Z/p\mathbb Z$ with $p$ an odd prime.
Any ideas towards the solution of this conjecture? Your comments are welcome!
Perhaps what was meant is "there is an enumeration" rather than "there is a circular permutation"?
I use the word "circular" since we consider $a_n$ and $a_1$ adjacent.
Given a choice of ordering of the elements of a set $A$ as $a_1,\ldots,a_n$, the standard understanding of "circular permutation" of that ordering is one of the form $a_i,\ldots,a_n,a_1\ldots,a_{i-1}$. If one understands the conjecture in this way, it does not make sense.
The conjecture was contained in my paper available from http://maths.nju.edu.cn/~zwsun/196a.pdf
Perhaps a better word is a cyclic permutation (in cycle notation) "$(a_1\cdots a_n)$".
For cyclic groups, replacing addition with subtraction in the question gives a relation to graceful labellings of graphs. For cyclic groups, these exist when $n \equiv 3 \mod 4$ but not when $n \equiv 1 \mod 4$ by a theorem of Rosa. Unfortunately this doesn't help answer the question...
|
2025-03-21T14:48:31.322509
| 2020-06-22T15:20:56 |
363839
|
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|
Stack Exchange
|
Are positivity for forms and that for currents consistent when talking about smooth forms?
Let $X$ be a complex manifold and $\theta$ a smooth $(1,1)$-form on $X$.
(1) If $\theta>0$ in the sense of currents, then can we deduce that $\theta>0$ in the sense of forms?
(2) If $\theta>0$ in the sense of forms, then can we deduce that $\theta>0$ in the sense of currents?
You should be more specific about what you mean by "> 0" for currents and forms. Do you mean > 0 as in strictly positive? In that case I would interpret "> 0 in the sense of currents" as \theta >= \epsilon \omega for some smooth strictly positive (1,1)-form \omega and \epsilon > 0, and if \theta is smooth, then then it is trivial that these notions of positivity are equal. If you mean "> 0" as in non-negative, then there are positive and strongly positive (k,k) forms and currents, but for a smooth (1,1)-form, all these notions of positivity coincide.
|
2025-03-21T14:48:31.322602
| 2020-06-22T15:36:20 |
363841
|
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|
Stack Exchange
|
Why is Haven's discovery important?
Today my attention was caught by one of those little stories that appear when you open a certain browser: an inmate achieved a number theoretic breakthrough
It is about continued fractions and I would like to understand the implications of the discovery.
So my question is whether someone could provide details about the mathematical aspects of the discovery, especially what further progress number theorists expect from the result and/or the author.
Have you read the paper? It is available for free on Springer: https://link.springer.com/article/10.1007%2Fs40993-020-0187-5 . It says "... similar relations are more hard to give and we think that other ways should be used for approaching the problem for general values of determinants. However, the results of this paper could help in finding a solution to the general case." and "The results obtained in this paper can also be used to determine the linear fractional transformation of certain Hurwitzian and Tasoevian continued fractions, as well as studying their leaping convergents."
@Favst yes, I have read it, but I would also like to know whether any other areas of number theory, that "only" utilize those continued fractions as a tool, will benefit from the result.
It isn't that important per se. It is a nice result which may be used for later work. But it wouldn't be getting this level of attention if some regular math PhD had written it. The attention is to a large extent due to the author in question.
Before we apply our prejudices, note that (as the story goes) the author has more mathematical desire and initiative than training, in my view. This is close to an amateur success story. Gerhard "With Moral: Keep On Trying" Paseman, 2020.06.22.
@GerhardPaseman that's the reason why I put the focus of my question on the mathematical result and not on the circumstances under which they were obtained or on moral aspects. Of course the story is remarkable and the therapeutic aspects of mathematics are yet a different universe to explore...
|
2025-03-21T14:48:31.322784
| 2020-06-22T16:06:14 |
363844
|
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"Gabriel Angelini-Knoll",
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|
Stack Exchange
|
THH of the tensor product of E_1 rings
Is there a formula or a spectral sequence relating the $THH$ of two $E_1$-rings and the $THH$ of their tensor product over $\mathbb{S}$? Krause and Nikolaus state without context on page 3 of their paper (Bokstedet's periodicity and quotients of DVRs) that $$THH(A) \otimes_{\mathbb{S}} THH(B) = THH(A \otimes_{\mathbb{S}} B).$$ It might be that they suppressed the requirement that $A$ and $B$ are $E_\infty$-rings, but I was wondering if there's something like that for $E_1$-rings in general?
Thank you in advance!
The category of $E_1$-algebra spectra admits a symmetric monoidal structure given by the smash product. Viewed as a functor $Alg(Sp) \rightarrow CycSp$, THH is symmetric monoidal. This claim essentially boils down to the fact that sifted colimits commute with the smash product. This is noted at the beginning of IV.2 of Nikolaus-Scholze. If one wishes to fully unpack the claim in the HA language, looking at HA 3.2.4 might be helpful.
One way to think about it is that THH(A)=dim(LMod(A)) where the symmetric monoidal dimension is taken in Pr_st (presentable stable infitnity-categories). Since symmetric monoidal dimension is multiplicative in general and since LMod(-) is symmetric monoidal w.r.t. the tensor product of associative algebras and tensor product of presentable categories, THH is symmetric monoidal.
Absolutely! I love the POV of THH as the trace of dualizable objects in Pr^L_St; it makes claims like this automatic.
@LiamKeenan Thank you! I will try to work out the details.
This was also known for S-algebras (in the sense of EKMM) since the early days of the subject and the proof is easy (see Lemma 3.1 in "Topological Hochschild homology" by Schwänzl-Vogt-Waldhausen).
|
2025-03-21T14:48:31.322982
| 2020-06-19T20:37:30 |
363572
|
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|
Stack Exchange
|
Important (but not too well known) inequalities
After seeing the question Important formulas in combinatorics, I thought it might be of interest to have a similar list of inequalities, although not restricted to combinatorics. As with that list, there should be some rules.
The inequality should not be too well known. This is to rule out things like Cauchy-Schwarz or the Sobolev inequalities. The inequality should be unfamiliar to a majority of mathematicians.
The inequality should represent research level mathematics. This is taken straight from the other list, and feels like a good rule.
The inequality should be important. Since it is easier to come up with inequalities versus exact formulas, this should be more restrictive than in the other list. The idea is to have inequalities which played an important role in the development of some field.
An answer can be a class of inequalities. As noted in the comments, often what is important is a family of inequalities which all convey the same idea but where no single result is the fundamental example. This is perfectly acceptable, and perhaps even encouraged since any such examples will likely have lots of applications.
To give an idea of what I mean, let me give an example which I think satisfies the first three criteria; the Li-Yau estimate.
The Li-Yau inequality is the estimate $$ \Delta \ln u \geq - \frac{
n}{ 2t}.$$
Here $u: M \times \mathbb{R} \to \mathbb{R}^+ $ is a non-negative solution to the heat equation
$ \frac{\partial u}{\partial t} = \Delta u, $ $(M^n,g)$ is a compact Riemannian manifold with non-negative Ricci curvature and $\Delta$ is the Laplace-Beltrami operator.
This inequality plays a very important role in geometric analysis. It provides a differential Harnack inequality to solutions to the heat equation, which integrates out to the standard Harnack estimate.
There are many results strengthening the original inequality or adapting it to a different setting. There are also results which are not generalizations of the original inequality but which bear its influence. For instance, Hamilton proved a tensor version of the Li-Yau inequality for a manifold which has non-negative sectional curvature and evolves by Ricci flow. Furthermore, one of Perelman's important breakthroughs was to prove a version of the Hamilton-Li-Yau inequality for a solution to time-reversed heat flow when the metric evolves by Ricci flow. These results are not at all corollaries of the original Li-Yau estimate, but they are similar in spirit.
Often in analysis it is a class of inequalities that is important, rather than a specific inequality from that class. For instance, the class of concentration of measure inequalities (Chernoff, Hoeffding, Bernstein, Azuma, McDiarmid, Levy, Talagrand, etc.) is extremely important in modern probability, combinatorics, random matrix theory, high dimensional geometry, and theoretical computer science, but I wouldn't single out a single inequality in this class as being particularly pivotal.
Thanks for the remark. Concentration of measure is definitely the sort of thing I had in mind when I asked this question, so I'll edit the question to allow for classes of inequalities.
@GabeK the Li-Yau inequality would be an example of all four criteria, not just the first three, wouldn't it? From scalar equations to Ricci flow and mean curvature flow, and also including the original Yau and Cheng-Yau gradient estimates in the elliptic setting
@QuartoBendir You're probably right. The fourth condition was added later and I was thinking of the original parabolic version as being the prototypical example.
This question might encourage answers that are well-known to everybody in a given subfield, but aren't needed by or relevant to people outside that area. Is that okay/desirable?
@usul I think that is basically the goal. The inequalities should be important enough to be widely used in some area, but not well known to people in an unrelated field.
The class of concentration of measure inequalities is a fundamental tool in modern probability (and any field that uses probability, e.g., random matrix theory, theoretical computer science, statistics, high-dimensional geometry, combinatorics, etc.). As explained in this blog post of Scott Aaronson, these are basic ways in which one "upper bounds the probability of something bad", and often the bounds are exponential or even gaussian in nature when one is far away from the mean (or median) and there are many independent (or somewhat independent) variables involved. Examples of such inequalities include
The Chernoff inequality and its relatives (Hoeffding, Bernstein, Bennett, etc.)
Azuma's inequality
McDiarmid's inequality
Levy's inequality
Talagrand's concentration inequality
Log-Sobolev inequalites are, strictly speaking, not concentration of measure inequalities, but are often closely related to them, thanks to techniques such as the Herbst argument.
A standard reference in the subject for these topics is
Ledoux, Michel, The concentration of measure phenomenon, Mathematical Surveys and Monographs. 89. Providence, RI: American Mathematical Society (AMS). x, 181 p. (2001). ZBL0995.60002.
I also have a blog post on this topic here.
These are certainly very important inequalities! Would you describe them all as "unfamiliar to the majority of mathematicians"? Anyone who has done a masters level probability or theoretical statistics course is going to know the Chernoff and Azuma inequalities -- even an undergraduate course I attended did Chernoff and McDiarmid (aka bounded differences). Maybe as a young probabilist myself I am not in a good place to judge how well-known these are, though!
If I may add a suggestion to McDiarmid's inequality, which can be found in his survey On the Method of Bounded Differences, various authors have considered 'typical' bounded differences (as opposed to worst-case). See, for example, On the Method of Typical Bounded Differences [cont...]
[...cont] Various people who desire these bounded difference inequalities may not be aware of such 'typical' bounded difference inequalities where the bounds do not hold for all points in the space, but only for 'typical' ones, in some precise manner
Some individual inequalities in this class (e.g., the Chernoff inequality) may be fairly well known, but my sense is that the broader concentration of measure phenomenon (in particular, its applicability to nonlinear (but still Lipschitz or convex) functions of independent variables) is not widely known outside of the fields of mathematics that rely heavily on probability.
The DKW inequality on deviation of an empirical CDF from the ground truth is a very nice one.
@TerryTao I see, thank you for your insight :)
@usul That is one of which I was not aware! It complements nicely the Berry--Esseen inequality
Gaussian Jensen's inequality:
Let $\boldsymbol{X}=(X_{1}, \ldots, X_{n})\sim \mathcal{N}(\boldsymbol{\mu}, \boldsymbol\Sigma)$ be a gaussian vector. The inequality
$$
\mathbb{E} B(f_{1}(X_{1}), \ldots, f_{n}(X_{n})) \leq B(\mathbb{E}f_{1}(X_{1}), \ldots, \mathbb{E}f_{n}(X_{n}))
$$ holds for all real valued (test functions) $f_{1}, \ldots, f_{n}$ if and only if $\mathrm{Hess}\, B\, \bullet \boldsymbol{\Sigma} \leq 0$.
Remarks: here $\bullet$ denotes Hadamard product; $B : \Omega \subset \mathbb{R}^{n} \to \mathbb{R}$ is a smooth function given on a rectangular domain $\Omega = J_{1}\times\ldots \times J_{n}$ for some intervals (rays, real line) $J_{k}$, and test functions map $f_{k} :\mathbb{R} \to J_{k}$. Inequality $\mathrm{Hess}\, B(s)\, \bullet \boldsymbol{\Sigma} \leq 0$ is required to hold for all $s \in \Omega$ and it means that the matrix is negative semidefinite.
Applications: (the list is far from complete!)
Prekopa--Leindler.
Ehrhard inequality - this might be not well-known. It is sharp analog of Brunn--Minkowski for Gaussian measure which implies Gaussian isoperimetric inequality
Hypercontractivity for the Ornstein--Uhlenbeck semigroup.
Brascamp--Lieb inequality (including Young's conviolution inequality etc). There is a very nasty limit passage from Gaussian to Lebesgue case.
Gaussian noise stability (it is better to google it).
How to think about this inequality:
If $X_{1}, ..., X_{n}$ are independent then $\mathrm{Hess}\, B\, \bullet \boldsymbol{\Sigma} \leq 0$ simply means that $B$ is separately concave. If $X_{1}=X_{2}=...=X_{n}$ then $\mathrm{Hess}\, B\, \bullet \boldsymbol{\Sigma} \leq 0$ is just concavity of $B$. The inequality improves on classical Jensen's inequality because $\mathrm{Hess} B \leq 0 \Rightarrow \mathrm{Hess}\, B\, \bullet \boldsymbol{\Sigma} \leq 0$ for any covariance matrix $\boldsymbol{\Sigma}$. If $\boldsymbol{X}$ is a random vector (with smooth density and different than Gaussian) then the "infinitesimal condition" $\mathrm{Hess}\, B\, \bullet \boldsymbol{\Sigma} \leq 0$ is always necessary for the "Jensen's inequality" but not always sufficient. So Gaussian vector is somehow universal.
what are good references for this?
Since there is no one good reference which would summarize the Gaussian Jensen's inequality in the way I posted in my answer, I decided to write a blog post with proofs and applications https://extremal010101.wordpress.com/2021/01/01/jensens-inequality/
I just skimmed through your post. This looks beautiful, both the math and the writing. Many thanks!
Esseen's anti concentration inequality is the basis of a lot of relatively recent (past 10-15 years) work on non asymptotic random matrix theory, particularly results on the smallest singular values of many random matrix models. It states that if $Y$ is a real valued random variable, then
$$\sup_{t \in \mathbb{R}} \mathbb{P}(|Y-t| \le 1)\le \int_{-2}^2 |\phi_Y(\theta)| \ d \theta $$
where $\phi_Y$ is the characteristic function of $Y$. It is mainly used to derive 'small ball' probability estimates where you want to control the dot product of a vector with another random vector. For a reference, see the excellent notes by Mark Rudelson here.
Strichartz estimates, which originated from
Strichartz, Robert S., Restrictions of Fourier transforms to quadratic surfaces and decay of solutions of wave equations, Duke Math. J. 44, 705-714 (1977). ZBL0372.35001,
are a family of inequalities that provide $L^p$ (or Sobolev) type control of solutions of linear dispersive or wave equations (such as the Schrodinger equation) in terms of the size of the initial data (usually measured in some sort of $L^2$-based Sobolev norm) as well as an inhomogeneous forcing term (also usually measured in some sort of $L^p$ or Sobolev norm). Through perturbative techniques (e.g., the contraction mapping theorem), Strichartz estimates can often be extended to nonlinear dispersive or wave equations, at least if the data and forcing term is small and/or one is working locally in time instead of globally. As such, Strichartz estimates form the backbone of modern local well-posedness theory for such equations, and often also play an important role in the global theory (e.g., scattering theory, or blowup analysis) of these equations. As a very crude measure of their impact, MathSciNet reports over a thousand papers devoted to the topic of Strichartz estimates. Very roughly speaking, Strichartz estimates are to dispersive and wave equations as Sobolev estimates are to elliptic equations.
Morawetz inequalities, which originated from the work of Cathleen Morawetz, and in particular
Morawetz, C. S., Time decay for the nonlinear Klein-Gordon equation, Proc. R. Soc. Lond., Ser. A 306, 291-296 (1968). ZBL0157.41502.
give global control of $L^p$ type on nonlinear dispersive or wave equations, and are usually proved using integration by parts arguments. In contrast to Strichartz estimates, they are often global in time and work in non-perturbative situations; on the other hand, they tend to be more restricted in the range of $L^p$ type quantities that can be controlled, and are also senstive to the focusing or defocusing nature of the nonlinearity. (The closest elliptic analogue to Morawetz inequalities would be Pohozaev type identities. There are also a useful variant of Morawetz inequalities known as viriel identities.)
Much of the modern global theory of nonlinear dispersive and wave equations (particularly for "critical" choices of exponents) relies heavily on a intricate combination of both Strichartz estimates and Morawetz inequalities (as well as other tools, such as conservation laws, Littlewood-Paley theory, and concentration compactness methods). See for instance my book on the subject.
Thanks for this excellent answer. It's interesting how the fundamental tools for hyperbolic equations versus elliptic/parabolic equations are so different. It really illustrates how different they are.
This is from Garding.
Let $P\in{\mathbb R}[X_1,\ldots,X_d]$ be a homogeneous polynomial. Assume that it is hyperbolic in some direction $e\in{\mathbb R}^d$ (with say the normalisation $P(e)=1$) and let $\Gamma$ be its cone of future, that is the connected component of $e$ in the complement of $\{P=0\}$. It is known that $\Gamma$ is convex. Then we have the inverse Hölder inequality: for every $v_1,\ldots,v_n\in\Gamma$,
$$M(v_1,\ldots,v_n)\ge(P(v_1)\cdots P(v_n))^{\frac1n},$$
where $M$ is the symmetric multiplinear form such taht $M(x,\ldots,x)=P(x)$.
Consequences occur in convex geometry, combinatorics, PDEs, ...
As a matter of fact, $P^{\frac1n}$ is concave over $\Gamma$. A simple example is that of quadratic forms of signature $(1,d-1)$. Another nice example is $P=\det$, where ${\mathbb R}^d={\bf M}_n({\mathbb R})$.
Very interesting. Is there a reference for further reading?
@VadimOgranovich. Well, there is the original paper by Garding in Acta Math., circa 1960. Otherwise, there are plenty of articles about hyperbolic polynomials. Still an active subject. I wrote a paper in Chinese Annals of Maths B (2009) DOI 10.1007/s11401-009-0169-3 .
A really simple but powerful inequality is the so-called improved Kato inequality. I first learned about it when I was studying Uhlenbeck's removable singularity theorem for self-dual Yang-Mills connections. However, when I explained the inequality to Duong Phong and Eli Stein in Phong's office, Stein reacted with "It's in my book! It's in my book!"
The expander Chernoff bound is a particularly nice generalization of the Chernoff inequality that is not so well known. It states the following: Let $G = (V,E)$ be a regular graph and consider a function $f : V \rightarrow [0,1]$. Perform a random walk $v_1, \cdots, v_t$ on $G$ by first picking $v_1$ uniformly at random. Then
$$ \mathbb{P}\left(\frac{1}t \sum_{i=1}^t f(v_i) \ge \mathbb{E}f + \epsilon + \lambda \right) \le e^{-\Omega(\epsilon^2 t)}$$
where $\lambda$ is the spectral gap of $G$.
Heuristically, this inequality is roughly stating that the random variables $f(v_i)$ satisfy a Chernoff like tail bound even though they are not independent! The major application of this inequality is in theoretical computer science where it can be used to replace multiple trials of a randomized algorithm with a walk on an expander graph which reduces the number of random bits needed.
I came across this inequality on the Gaussian space recently. I was not aware of its existence since it is not really a classical one in comparison to the Poincaré inequality or the Logarithmic Sobolev inequality but it seems to be useful in order to prove the analyticity of the Ornstein-Uhlenbeck semigroup in $L^p(\gamma)$. Let $\gamma$ be the standard Gaussian measure on $\mathbb{R}^d$. Let $p \in (1,+\infty)$, let $f\in \mathcal{S}(\mathbb{R}^d)$ and let $k \in \{1, \dots, d\}$. Then, \begin{align*} \|x_kf\|_{L^p(\gamma)} \leq C_{p,d} \left(\|f\|_{L^p(\gamma)}+\|\partial_k(f)\|_{L^p(\gamma)}\right), \end{align*} where $C_{p,d}>0$ only depending on $d$ and on $p$.
Another inequality which has not been mentioned but which is very useful is the Carbery-Wright inequality which gives anti-concentration bound for polynomials of Gaussian random variables.
Do you happen to have a reference with applications to analyticity of the OU semigroup?
It is well hidden. Lemma 2.3 of this paper: https://www.sciencedirect.com/science/article/pii/S0022123602939789. Regarding the analyticity, it follows from Bismut formula together with this inequality.
I am not a professional mathematician, thus feel free (you or your colleagues/professors of the site) tell me if my answer doesn't fit with your requirements that I can to delete it.
An important inequality in complex analysis and functional analysis is the statement of Hadamard three-lines theorem, see the section Statement from this link of the Wikipedia with title Hadamard three-lines theorem, also I add as a comment other inequality that I like, unrelated to my answer.
I tried do specializations for the nice Young inequality of the article Takayuki Furuta, The Hölder-McCarthy and the Young Inequalities Are Equivalent for Hilbert Space Operators, The American Mathematical Monthly, Vol. 108, No. 1 (Jan., 2001), pp. 68-69. One can to do the specialization $\lambda=1/n$ and after multipliying the inequality by a positive sequence, let's say $|G_n|$ with $G_n$ the Gregory coefficients or for example by $\psi(n)/n^3$ with $\psi(n)$ the Deedkind psi function, one gets a resulting inequality taking the sum $\sum_{n=1}^\infty$ from both sides of the inequality.
The Hadamard three-lines theorem counts as "well known", in my opinion, and therefore does not fit what the question is asking for
Many thanks @YemonChoi then if there are no more feedback in next few hours I think/agree that I should delete my answer. Also you're an excellent functional analyst, thus it makes sense (from my view) that you know it (there are few posts in MathOverflow dedicated to it).
I'm sorry @YemonChoi I've decided undelete the post. Many thanks for the feedback. In my opinion there are literature and lecture notes about Hadamard three-lines theorem, but I feel that this theorem isn't well-known (if I'm right there are five or six posts in this site MathOverflow for professional mathematicians about it).
Let $r(z)$ denotes the number of solutions in positive integers to $x+y\leq z$ with the unknown $x,y$ belonging to a set $S$ satisfying the following: the number of elements in $S$ less or equal to $x$ is asymptotically
$$N_S(x) \sim \frac{a x^b}{(\log x)^c}, \mbox{ with } 0<a, 0<b<1, \mbox{ and } c>0.$$
Then we have:
$$r(z) \sim \frac{a^2b z^{2b}}{(\log z)^{2c}}\cdot \int_0^1 (1-v)^b v^{b-1}dv =
\frac{a^2b z^{2b}}{(\log z)^{2c}}\cdot \frac{\Gamma(b)\Gamma(b+1)}{\Gamma(2b+1)}$$
This covers sums of two squares and sums of two primes. It has implications about the possibility to solve Goldbach's conjecture, see the third answer in my previous MathOverflow question, here.
This very post reminds me of the infamous Waring's problem (which has a solution due to Hilbert), and two infamous inequalities namely Weyl's inequality and Hua's inequality about which I have mentioned below.
Weyl's inequality, and Hua's inequality.
They are quite important from the viewpoint of analytic number theory.
I was interested in hearing about some number theoretic inequalities. Could you give more background in what these inequalities are and how they are used?
Hi, I shall ask you to read the book titled : "Analytic methods in Diophantine equations and inequalities" authored by Harold Davenport. In chapter 3 (it may differ across different editions) you can see Weyl's inequality followed by Hua's inequality.
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2025-03-21T14:48:31.324585
| 2020-06-19T21:16:33 |
363573
|
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Is there a name for and/or reasonably nice characterisation of "mixingly physical" measures?
Let $M$ be a Riemannian manifold with volume measure $\lambda$, let $f \colon M \to M$ be a diffeomorphism, and let $\mu$ be a probability measure on $M$ with compact support.
As stated in the question Is there a name for a "stable" physical measure?, there are several different definitions of what it means for $\mu$ to be a "physical" measure. One possible defintion is as follows:
Definition. The measure $\mu$ is a physical measure of $f$ is there exists a neighbourhood $U$ of $\mathrm{supp}\,\mu$ such that for every probability measure $\nu$ with $\nu(U)=1$ that is absolutely continuous with respect to $\lambda$, we have
$$ \frac{1}{n} \sum_{i=0}^{n-1} f^i_\ast\nu \ \overset{\text{weakly}}{\to} \ \mu \quad \text{as } n \to \infty.$$
A simple example is when $f$ has a stable fixed point or stable periodic orbit $A$, and $\mu$ is the equal-weight distribution on $A$.
However, I now want the convergence to hold without the Cesàro averaging; so, for example, this would exclude the case of a non-trivial periodic orbit.
Definition. The measure $\mu$ is a mixingly physical measure of $f$ is there exists a neighbourhood $U$ of $\mathrm{supp}\,\mu$ such that for every probability measure $\nu$ with $\nu(U)=1$ that is absolutely continuous with respect to $\lambda$, we have
$$ f^n_\ast\nu \ \overset{\text{weakly}}{\to} \ \mu \quad \text{as } n \to \infty.$$
For example, in the notes of Marcelo Viana at http://w3.impa.br/~viana/out/sdds.pdf in Section 4 (starting at p79), the SRB measure on a uniformly hyperbolic attractor is a "mixingly physical measure". (Proposition 4.9 on p97 gives this fact with a description of the rate of the mixing behaviour.)
My question:
Is there a name for what I have called a "mixingly physical measure"? And/or is there any simple characterisation of mixingly physical measures? (E.g. "$\mu$ is mixingly physical if and only if $\mu$ is both mixing and physical", or "$\mu$ is mixingly physical if and only if $\mu$ is physical and $f$ is a mixing nonsingular transformation of $(M,\mathcal{B}(M),\lambda(\,\cdot\,\cap U))$"?)
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2025-03-21T14:48:31.324754
| 2020-06-19T21:55:40 |
363576
|
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Stack Exchange
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Is there a theory in a finite language that is computably axiomatizable but not by a finite number of axiom schemas?
I was told to ask this question on mathoverflow. I asked on math stack exchange whether there is a computably axiomatizable theory that can't be axiomatized by a finite number of axiom schemas. I got an answer, but it was a theory in an infinite language. Now, I am asking whether there is an example in a finite language.
EDIT by non-OP: this is the above-mentioned MSE question, and this answer gives the definition of "scheme" being used.
If it’s computable, then can’t you write it down as one “schema”?
Hmm. I missed schema. Maybe the poster has a limited meaning of schema which (for my example below hopefully) excludes hyperidentities. There may be a theory which does not have a finite hyperbase. Gerhard "Look Up Padmanabhan And Penner" Paseman, 2020.06.19.
I believe there is a theorem of Kleene about this, saying that, if the language (or perhaps the theory) is rich enough then a computable set of axioms can be replaced with a schema. But I failed to find the paper now. There is a paper by Vaught, "Axiomatizability by a schema"; maybe this is what I remembered, and I was wrong about Kleene. I'd expect that the result fails if the language is very poor. Suppose you have only a constant 0, unary function S, and unary predicate P, with axioms $P(S^n0)$ for prime $n$ and $\neg P(S^n0)$ for composite $n$. That doesn't look schematic to me.
I've edited to link the reader to the relevant definition - remember that "scheme" is not actually a technical term, as the answers to your original question stated.
Following up on the comment by Andreas Blass: Vaught proved that if a theory $T$ is computable and has "a modicum of coding", then $T$ is axiomatizable by a scheme. Vaught's result was improved by Visser, in the paper below, who reduced "the modicum of coding" used by Vaught to "has a definable pairing function" A. Visser, Vaught's theorem on axiomatizability by a scheme, The Bulletin of Symbolic Logic, vol. 18 (2012), pp. 382-402.
@Ali I think you should post this as an answer.
@EmilJeřábek OK, I did that.
Let me give an example of a theory that is computably axiomatizable but isn't axiomatizable by finitely many schemas.
Fix any finite signature $\Omega$ with equality. Further by finite $\Omega$-models I'll mean models encoded by binary strings in a natural way. Observe that for any $\Omega$-theory $T$ axiomatized by finitely many schemas the set of all finite models of $T$ is $\mathtt{co}\text{-}\mathtt{NP}$. And observe that given a finite $\Omega$-model $\mathfrak{M}$ we could effectively construct an $\Omega$-sentence $\chi_{\mathfrak{M}}$ such that for any $\Omega$-model $\mathfrak{N}$ we have $$\mathfrak{N}\models \chi_{\mathfrak{M}}\iff \mathfrak{N}\simeq\mathfrak{M}.$$ Consider arbitrary computable set of finite $\Omega$-models $A$ that is closed under isomorphisms and isn't $\mathtt{NP}$. Let $U_A$ be the theory axiomatized by sentences $\lnot\chi_{\mathfrak{M}}$ for $\mathfrak{M}\in A$. Clearly $U_A$ is computably axiomatizable. However, the set of finite models of $U_A$ is the complement of $A$ and thus isn't $\mathtt{co}\text{-}\mathtt{NP}$. Hence $U_A$ couldn't be axiomatized by finitely many schemas.
Any set of $n$-tuples from a finite model $\mathfrak{M}$ is definable by some formula with parameters. Hence a schemata $S(P_1,\ldots,P_n)$ holds in a finite model $\mathfrak{M}$ iff it holds in $\mathfrak{M}$ under all possible interpretations of $P_i$'s as predicates of appropriate arities. This clearly leads to a $\mathtt{co}\text{-}\mathtt{NP}$ check of whether a finite models satisfy a given scheme.
This answers complements Fedor Pakhamov's, who provided an example of a computable theory that is not axiomatizable by finitely many schemas.
Following up on the comment by Andreas Blass to the question: Vaught proved that if a theory $T$ is computable and has "a modicum of coding", then $T$ is axiomatizable by a scheme. Vaught's result was improved by Albert Visser, in the paper below, where "the modicum of coding" used by Vaught is reduced to the modest demand that $T$ interprets a non-surjective unordered pairing, where pairing need not be functional.
A. Visser, Vaught's theorem on axiomatizability by a scheme, The Bulletin of Symbolic Logic, vol. 18 (2012), pp. 382-402.
A preprint of Visser's paper can be found here.
There are a number of algebraic theories (in equational logic specifically, no predicate symbols besides equality) which are of finite type (so the language has only finitely many function symbols) but not finitely axiomatizable. Often one demonstrates this with an infinite sequence of structures for the language, but in some cases one can establish this linguistically.
A simple example involves work with hyperidentities (check out my arxiv preprint for details, 1408.something something). We posit the theory given by the hyperidentity F(F(x)) ideq F(F(F(x))), which is short hand for an equational theory that says every unary function term t in a language has its square equal its cube, or forall x t(t(x)) = t(t(t(x)).
If we choose a language with one binary function symbol, we get a finitely axiomatizable theory. I forget what happens with two unary function symbols. With three, you get a recursively axiomatizable theory which is not finitely axiomatizable. You show this by looking at the Thue Morse sequence and use square free fragments to build long terms and show long instances are not derivable from short instances of the axioms.
It gets real fun with more complicated hyperidentities and larger sets of function symbols. Check out the preprint for other families of examples.
Gerhard "Is The Question Computably Decidable?" Paseman, 2020.06.19.
While this is intereting, as far as I can tell it doesn't address the question which is about axiomatizability by finitely many schemes, not finite axiomatizability itself.
That's right. (I missed schema.) When I have an example of a variety which has no finite hyperbasis, I will post it. That won't conclusively answer the question, but it will get closer. There may even be a semigroup variety which is not finitely hyperbased, but I don't see it yet. Gerhard "Has Not Finished Scheming Yet" Paseman, 2020.06.19.
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2025-03-21T14:48:31.325194
| 2020-06-20T00:21:46 |
363583
|
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Number of permutations with longest increasing subsequences of length at most $n$
Is there a known expression for, or a nontrivial upper bound on, the number of permutations in $S_k$ with longest increasing subsequence of length at most $n$?
Let $l(\sigma)$ denote the length of the longest increasing subsequence of a permutation $\sigma\in S_k$. It seems like a lot is known about $l(\sigma)$ for a random permutation (and its asymptotic scaling), but are there upper bounds on the number of permutations in $\sigma\in S_k$ with $l(\sigma)\leq n$.
Motivation/context for this question: the moments of traces of random unitaries. It is known that $\int dU |{\rm tr}(U)|^{2k} = k!$ for $k\leq n$, where we integrate over the unitary group $U(n)$ with respect to the Haar measure. More generally, for any $k$ and $n$ one may write the expression as [1]
$$
\int dU |{\rm tr}(U)|^{2k} = \sum_{\lambda \vdash k,~\ell(\lambda)\leq n} \chi_\lambda(\mathbb{I})^2\,,
$$
summing over integer partitions $\lambda$ of $k$ with length at most $n$, and where $\chi_\lambda(\mathbb{I})$ is the identity character with respect to $\lambda$. The RHS is then counting the number of pairs of Young tableaux with width $\leq n$, which is equivalent to counting the number of permutations in $S_k$ with no increasing subsequences longer than $n$.
I'm essentially interested in upper bounds on this quantity which are tighter than the trivial bound of $k!$.
[1] E. Rains, "Increasing Subsequences and the Classical Groups," Electron. J. Comb. 5 (1998) R12. http://eudml.org/doc/119270.
Somewhat related question: https://mathoverflow.net/questions/149263/unimodality-of-length-of-longest-increasing-subsequence
There is an explicit determinental formula for these numbers due to Gessel in Symmetric functions and P-recursiveness (JCTA, 1990). Asymptotics were known much earlier and appear in a paper by Amitai Regev Asymptotic values for degrees associated with strips of young diagrams (Adv. Math. 1981). The gross asymptotics are that the $k$th root of the number of such permutations approaches $n^2$. Note that in most of the literature, $k$ and $n$ will play the opposite roles, i.e., the question will be about enumeration of permutations in $S_n$ with no increasing subsequence of size greater than $k$.
Thanks! These references are helpful. I didn't know about the determinantal expression. I'll check the paper more carefully, but I guess you're saying the trivial upper bound of $\leq$min${k!,n^{2k}}$ (with $\leq n^{2k}$ from the unitary integral expression) is asymptotically tight in $k$ for fixed $n$.
Sorry the notation is unfortunate, talking about $k$-th moments of unitaries in $U(n)$ gives the reverse of the convention in the combinatorics literature.
An explicit formula is the hook-product formula, due to Schensted I believe.
This formula is used in the classical work of Logan and Shepp, as well as in Vershik-Kerov.
See for example equation (1.1) in the Logan-Shepp paper
Asymptotics will depend greatly on whether $n>2\sqrt{k}$ or not. I assume you meant $n<2\sqrt{k}$. In that case, the asymptotics (under the name of Large deviations principle)
are known, and involve the Logan-Shepp functional. See Increasing subsequences of iid samples and Large deviations for increasing sequences in the plane. There is also work in the moderate deviations regime due to Lowe.
The hook-product theorem is for fixed shapes under RSK. One has to sum over all relevant shapes, which is how Regev did the asymptotics.
Sure. At the scale of large deviations, this is also what Logan-Shepp did.This gives a variational formula for the rate function.
This explicit formula is the sum over integer partitions given above, i.e. counting Young tableaux and summing over all relevant shapes, as was mentioned. And thanks for the references. I'm interested in upper bounds for a fixed $n$ at all $k$, so I care about both regimes. But the large $k$ asymptotics are helpful.
If $n>2\sqrt{k}+\Omega(k^{1/6})$, I believe the number is asymptotic to $k!$, so what improvement would you be after in that regime?
This relates to the Stanley-Wilf Conjecture (now a theorem). More generally you can consider $S_k(\sigma)$ the number of permutations of $k$ elements which do not contain the pattern given by the permutation $\sigma$. Here you are looking at the particular case $S_k(12\cdots(n+1))=:u_n(k)$. Exhaustive references on the subject are the books "Combinatorics of Permutations" by Bóna and "Patterns in Permutations and Words" by Kitaev. Theorem 4.10 in Bóna's book gives a very elementary combinatorial proof for the bound
$$
u_n(k)\le n^{2k}\ .
$$
A similar bound was conjectured by Arratia for any pattern $\sigma$ of length $n+1$ but this is known to fail for $\sigma=1324$.
Note that
the bound is trivial from the Haar integral formula because $U$ has eigenvalues of modulus one and so $|{\rm tr} (U)|\le n$.
Also, the numbers form a supermultiplicative sequence by a result of Arratia (same article as above).
The supermultiplicative property also follows from the Haar integral: the $S_k(12\cdots(n+1))$ sequence in $k$ being a Stieltjes moment sequence is log-convex.
I first thought that this fact (Feteke's Subadditive Lemma) combined with Regev's asymptotic formula might give a better exponential upper bound (rather than asymptotic). However one ends up with the same upper bound.
That's because Regev's formula gives, after computing a Selberg integral,
$$
u_n(k)\sim
1!2!\cdots(n-1)!\ (2\pi)^{-\frac{n-1}{2}}
\ 2^{-\frac{n^2-1}{2}}\ n^{\frac{n^2}{2}}\ \frac{n^{2k}}{k^{\frac{n^2-1}{2}}}
$$
when $k\rightarrow\infty$ (I took the formula from Stanley's ICM survey).
So the correct exponential growth of $n^{2k}$ is already in the trivial bound.
It should be said that $12...(n+1)$ is very different from an arbitrary pattern: I believe this is the only infinite family of patterns with a something like a "formula" (+ precise asymptotics) known (the aforementioned Gessel's formula)
The funny thing is Gessel's formula for the Haar integral, using a determinant of Bessel functions was known to mathematical physicists, in particular Bars and collaborators.
Thanks for the response and suggestions! And yes, I should have mentioned that the trivial upper bound is min${k!,n^{2k}}$. Interesting that your suggestion essentially gives a similar behavior.
The paper by Bars and Green which already contains the Haar integral analogue of Gessel's formula is https://journals.aps.org/prd/abstract/10.1103/PhysRevD.20.3311 see the appendix.
@4xion: It seems $n^{2k}$ is indeed the best one can do. Is that bound good enough for your goals?
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2025-03-21T14:48:31.325666
| 2020-06-20T00:28:42 |
363585
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What were Ramanujan's standard tricks/approaches to solving problems?
While trying to formulate an answer to this question, I realized I really have no idea how Ramanujan came up with his formulas. Bruce Berndt has a number of great expository articles, e.g., this one, but I couldn't discern how Ramanujan approached problems. There are famous stories about how solutions seemed to just pop into his head, e.g., when he quickly solved a tricky problem and was asked how, he responded:
It was clear that the solution should obviously be a continued fraction; I then thought, which continued fraction? And the answer came to my mind.
According to Wikipedia, Hardy said Ramanujan's results were "arrived at by a process of mingled argument, intuition, and induction, of which he was entirely unable to give any coherent account." In the same article, Ramanujan is quoted as saying "An equation for me has no meaning unless it expresses a thought of God" and crediting his mathematical abilities to his family goddess Namagiri Thayar.
Previous MathOverflow questions have asked how he came up with specific results, and in this mathoverflow question, Tim Chow said "Ramanujan is legendary for having an extraordinary, uncanny intuition, and it is natural to try to understand this intuition better."
Question: Now that so many of Ramanujan's formulas have been verified, that his notebooks have been carefully studied, and that his results have been understood as a part of a larger theory, has anyone discerned a pattern or a set of standard tricks/approaches that might have been underlying how he came up with his results?
What I've read from Berndt suggests that Ramanujan's work focused heavily on continued fractions, partition functions, asymptotic formulas, modular forms, zeta functions, $q$-series, Eisenstein series, and mock theta functions. I'd be happy for an answer in any of these individual areas. Berndt suggests Ramanujan worked on slate, and erased his work when finished, recording only the final formulas he discovered, so we can perhaps deduce that Ramanujan had some fairly compact way to do his work.
Side note: while it's fun to have stories of mathematicians so brilliant that no one can understand them, I don't think this is the right point of view if we want to make the field welcoming to newcomers. It's also not very satisfying from the point of view of really understanding what's going on in a field. I hope that one day the mathematical community will understand everything Ramanujan did, and now, 100 years after his death, I'm hoping there has been some progress on this goal.
I debated making this CW, but the other two MO questions I linked to weren't CW, so I decided to hold off. I'm not really sure what's standard with ho.history-overview questions
No one knows. He never even used spectral techniques in formulating his conjectures (this means he did not go through usual pigeonhole way of thinking and he knew something else). Remember these are standard tools now.
Even if we understood everything he did, we will never know how he did them. Not even close.He obviously had a lot of intuition and computational/algebraic experience. Even if we found a few patterns and tricks, there is no way to know if he used them or else used alternate methods.
@Somos I am not willing to give up so easily. I am sure that when Cardano was solving cubic equations left and right it seemed like magic, but there was a pattern and a trick to it. When Galois characterized which polynomials can be solved by radicals, it seemed like magic, but now we understand it in the context of Galois theory and teach it to undergraduates. The same with Gauss's work on the fast Fourier transform. We should aspire to truly understand Ramanujan's way of thinking, rather than assuming it's beyond our collective power.
Ramanujan had a great skill in algebraic manipulation (much better than current symbolic software). Almost all his independent (of Hardy) work is based on algebraic manipulation. And note that the processes of calculus were also a part of algebraic manipulation for him.
Also I think there is another factor in case of Ramanujan. He was using powers of algebraic manipulation almost all the time 24x7. Ramanujan's approach therefore also involves a great deal of labour involving highly efficient manipulation. A person who used similar approach was Jacobi. It's rather unfortunate that the techniques of Jacobi and Ramanujan have been ditched in modern times.
This is an exposition of my comments via actual examples. I will present a few of the tricks which Ramanujan heavily used (all of these are algebraic manipulation and do not involve anything high brow).
Partial Fractions
Often Ramanujan used to derive partial fractions for many functions (usually made of circular/hyperbolic functions). He never mentioned explicitly the technique used but it appears it was based on analysis of the poles of the function. However this did not involve complex analysis and instead it was an extension of the method used for typical rational functions in such a way to avoid common pitfalls. Partial fractions were then heavily used to obtain many series by comparing coefficients. In particular his formula related to $\zeta(2n+1)$ is derived in this manner (see this thread). Another application of the technique is described here.
Multisection of Series
This involves splitting a power series into multiple series by grouping terms with powers modulo a given number $n$.
Ramanujan used the technique in a different manner by trying to analyze power series for $f(x^{1/n})$ and collecting terms which contain fractional powers of $x$. Using this approach he proved many properties of Rogers Ramanujan continued fraction and also obtained generating functions of $p(5n+4),p(7n+5)$. A nice application of this technique is presented here.
Simplification of algebraic expressions
If $x, y$ are two numbers connected by algebraic equation (in theory) of the form $P(x, y) =0$ where $P$ is a polynomial in $x, y$ with integer coefficients then Ramanujan would often try to guess simple functions like $u=f(x), v=g(y) $ so that the relation between $x, y$ could be transformed into a visually simple form as $F(u, v) =0$ where $F$ need not be polynomial but rather general algebraic function.
Here it appears that he worked by trial and error and put a lot of effort to simplify the form of the algebraic relation. This is clearly seen when one compares Ramanujan's class invariants with the corresponding ones given by Weber. His modular equations are also in much simpler form compared to those given by others.
In this connection it should also be noted that Ramanujan had discovered many algebraic identities which helped him to denest radicals. I don't think there was a technique involved here. The identities were developed in pursuit of specific goals like expressing a number as sum of two cubes in two different ways or in another case for finding simple expressions for singular moduli. Also he wasn't aware of any Galois theory and probably he did not need it. I guess he used his time and skill to figure out these by trial and error (discarding quickly anything which did not seem to meet the desired goal).
Note: Some of the examples presented above are available on Math.SE and I will add links to them after some time.
|
2025-03-21T14:48:31.326310
| 2020-06-20T00:37:28 |
363586
|
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"Carlo Beenakker",
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|
Stack Exchange
|
Path integrals on statistical mechanics
In (rigorous) statistical mechanics and statistical field theory one is usually concerned in giving meaning to integrals of the form:
\begin{eqnarray}
\langle \mathcal{O}\rangle = \frac{1}{Z}\int D\phi e^{-S(\phi)} \mathcal{O}(\phi) \tag{1}\label{1}
\end{eqnarray}
where $D\phi$ is some measure on the space of fields $\phi$. One way to approach the problem is to study the discretized version of the theory and look for the existence of some limits (thermodynamic and continuous). Well, I'm not interested in the technical aspects of the theory here. The point is: one of the most important actions in the literature is given by:
\begin{eqnarray}
S(\phi) = \int_{\mathbb{R}^{d}}d^{d}x\bigg{(}\frac{1}{2}\nabla\phi(x)^{2}+\frac{1}{2}m^{2}\phi(x)^{2}+\lambda\phi(x)^{4}\bigg{)}. \tag{2}\label{2}
\end{eqnarray}
This is called the $\phi^{4}$-model. If the field $\phi$ and its derivatives have enough decay, the integrand in (\ref{2}) can be rewritten in terms of the massive Laplacian $-\Delta+m^{2}$.
I'm starting to write some personal notes about my studies on this topic and I plan to write an introductory section where I give the motivations to study the $\phi^{4}$ model. But I'm having a hard time trying to find a nice way to do it. Let me elaborate a little more. I know that many interesting models can be realized as $\phi^{4}$-models. For instance, I know that the Ising model is some sort of limit of the above scenario and also that $\phi^{4}$ models are fundamental to study random walks and white noise. But I'd prefer not to attain myself to explicit models but rather to give a more general motivation. I've been thinking about it and I think the most natural motivation to this model is to consider the Landau free energy $F(\phi)$, which is given by:
\begin{eqnarray}
F(\phi) = \int_{\mathbb{R}^{d}}\bigg{(}\frac{1}{2}\alpha(T)\nabla \phi(x)^{2}+\frac{1}{2}\beta(T)\phi(x)^{2}+\frac{1}{4}\gamma(T)\phi(x)^{4}+\cdots\bigg{)}
\end{eqnarray}
where $T$ stands for the absolute temperature of the system. The resemblance of this expression with expression (\ref{2}) is obvious. Also, the partition function of the system in the Landau approach is supposed to be:
\begin{eqnarray}
Z = \int D\phi e^{-\beta F(\phi)},
\end{eqnarray}
which is a genuine path integral, in the same spirit as (\ref{1}).
The problem here is that I do not find anything rigorous about Landau's theory. I mean, I know that the general picture of Landau's theory is supposed to be purely phenomenological, but I'd expect to find, say, a rigorous way to turn spins into fields or something like this, but I didn't find anything so far. What I did find is how to discretize the theory once you have your $\phi^{4}$ action, but not the other way around. The process of turning spins into fields in the physics literature is done by using a process of coarse-grain, which is usually explained in a purely qualitative way.
So I'd like to know if there is something rigorous about how spins become fields or even about Landau's theory itself. Also, is this really the best approach to motivate the $\phi^{4}$ theory or is that better ways to do that (rigorously and in a general setup)?
NOTE: The more general version of Landau's theory (as I wrote above) is more commonly called Ginzburg-Landau theory. I think these ideas were published by L. Landau and V. Ginzburg to explain, among other things, superconditivity.
The functional $F(\phi)$ is an expansion in powers of $\phi$ and $\nabla \phi$, assuming the field $\phi$ is small and varies slowly; there is nothing fundamental about which terms to keep and which to discard, and in fact higher order functionals are also of interest.
If you don't want to discuss any kind of specific model as motivation, you could always argue that the $\phi^4$ theory is the only renormalizable theory that shares the $Z_2$ ($\phi\mapsto-\phi$) symmetry of the free theory.
A rigorous way to relate a general model of interacting spins (or interacting particles) to a field theory is by means of a Hubbard–Stratonovich transformation and Renormalisation Group (RG) theory. This is discussed in detail for example in:
Amit, D. J. (1984), Field Theory, The Renormalization Group And Critical Phenomena (2nd Edition), Singapore: World Scientific Publishing Company.
Ch. 2-5 REPRESENTATION OF THE ISING MODEL IN TERMS OF FUNCTIONAL INTEGRALS
First, the Hubbard–Stratonovich transformation is used to trade spins for continuous variables. Second, RG theory is applied to the transformed model to derive an effective field theory, where the microscopic lattice model is macroscopically described by its continuous limit with the field self-interaction potential expanded in powers of the field about its minimum and higher derivatives and higher powers being irrelevant in the RG sense.
|
2025-03-21T14:48:31.326695
| 2020-06-20T01:51:49 |
363591
|
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"Ira Gessel",
"Mark Wildon",
"T. Amdeberhan",
"https://mathoverflow.net/users/10744",
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"https://mathoverflow.net/users/66131",
"https://mathoverflow.net/users/7709",
"user44191"
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|
Stack Exchange
|
Seeking a combinatorial proof for a binomial identity
Let $n\geq m\geq0$ be two integers. The below binomial identity is provable by other means:
$$\sum_{j=0}^m(-1)^j\binom{n+1}j2^{m-j}
=\sum_{j=0}^m(-1)^j\binom{n-m+j}j.$$
QUESTION. Can you provide a combinatorial proof for the above identity? I would be thrilled to see as many as possible.
POSTSCRIPT. I enjoyed the two solutions by Ira & Fedor. Still, more alternating proofs are welcome.
Is there an obvious combinatorial interpretation of the "completed" version of the equation (with $m = n + 1$)? The left-hand side should be $1$ in that case.
These numbers are, up to sign, sequences A035317, A108561, A059259, A080242, A112555, A220074, and A279006 in the OEIS.
As a response to user44191's comment: When $m = n$ both sides are [$n$ is even] (Iverson notation.) Multiply by two and add the 'missing term' $(-1)^{m+1}$ to both sides. The left-hand side is then the inclusion-exclusion count of subsets of ${1,\ldots,m,m+1}$ not containing any of $1, \ldots, m+1$, and the right-hand side is $2\sum_{j=0}^{m} (-1)^j + (-1)^{m+1}$, so both sides are $1$. I admit even this special case proof is not entirely combinatorial.
I think I can, if you permit me to multiply it by $2^{n+1-m}$. Then we want to prove $$P:=\sum_{j=0}^m(-1)^j\binom{n+1}j2^{n+1-j}
=2^{n+1-m}\sum_{j=0}^m(-1)^j\binom{n-m+j}j=:Q.$$
Denote $X=\{1,2,\ldots,n+1\}$, then
$$
P=\sum_{B\subset A\subset X,|B|\leqslant m} (-1)^{|B|}.
$$
Fix $A$, denote $a=\max(A)$, and partition possible $B$'s onto pairs of the form $\{C,C\sqcup a\}$, where $C\subset A\setminus \{a\}$. All $B$'s are partitioned onto pairs except those for which $|B|=m$ and $a\notin B$. The sum in each pair 0, therefore
$$
P=1+(-1)^m|B\subset A\subset X,|B|=m,\max(A)\notin B|.
$$
Extra 1 comes from the case $B=A=\emptyset$, for which $\max(A)$ does not exist.
Now about $Q$. Consider $B\subset X$, $|B|=m$, and denote by $m-j+1$ the minimal element of $\overline{B}:=X\setminus B$. For fixed $j$, there exist exactly ${n-m+j\choose j}$ such sets $B$. Each of them has $2^{n+1-m}$ oversets $A$. Therefore
$$
Q=\sum_{B\subset A\subset X,|B|=m} (-1)^{\min(\overline{B})+m+1}.
$$
Consider the "dominos" $\{1,2\}$, $\{3,4\}$, $\ldots$, and take the first domino which is not contained in $B$. If it contains exactly 1 element from $B$, we may switch this element to the other element of the same domino, and $\min(\overline{B})$ changes its parity. This cancellation in the sum for $Q$ lefts only those $B$'s for which FNFDE (the first not-full domino is empty). Therefore
$$
Q=(-1)^m|B\subset A\subset X,|B|=m,FNFDE|.
$$
So $P=Q$ reduces to
$$
(-1)^m+|B\subset A\subset X,|B|=m,\max(A)\notin B|=|B\subset A\subset X,|B|=m,FNFDE|.
$$
Subtracting the common part, we should prove that
$$
(-1)^m+|B\subset A\subset X,|B|=m,\max(A)\notin B,\,\text{not}\, FNFDE|=\\
|B\subset A\subset X,|B|=m,\max(A)\in B,FNFDE|.
$$
Fix the first not full domino $\{s,s+1\}$ and $a=\max(A)$. If $a\leqslant s+1$, there is unique possibility which gives $(-1)^m$. Otherwise, if we fix also $B_0:=B\setminus \{s,s+1,a\}$ (it is some set of size $m-1$), and $A_0:=A\setminus \{s,s+1\}$ such that $B_0\subset A_0$, there exist exactly 4 ways to complete the choice of the pair $(B_0,A_0)$ to $(B,A)$ both for the condition $\{\max(A)\notin B,\,\text{not}\, FNFDE\}$ (choose which of $\{s,s+1\}$ belongs to $B$ where another guy from the domino belongs to $A$); and for the condition $\{\max(A)\in B,FNFDE\}$ (choose which of $s,s+1$ belongs to $a$). This proves the result.
Here are some observations, though not quite a combinatorial proof of the identity in question.
Let $A(m,n)$ be the value of the sums. Let $B(m,n)=(-1)^m A(m, m+n)$. Then $B(m,n)$ is nonnegative for all $m$ and $n$ (and is zero only if $m$ is odd and $n=0$).
It's not too hard to give a combinatorial interpretation to $B(m,n)$. It's easy to show that $B(m,n)$ has the simple generating function
$$
\beta(x,y) = \sum_{m,n=0}^\infty B(m,n) x^m y^n = \frac{1}{(1+x)(1-x-y)}.
$$
It follows that $B(m,n)$ satisfies the Pascal-like recurrence
$$B(m,n)=B(m-1, n) + B(m,n-1)$$
for $m\ge0$ and $n>0$ with initial values $B(-1,n)=0$, $B(m,0)=1$ for $m$ even and $B(m,0)=0$ for $m$ odd.
We can see that $B(m,n)$ is nonnegative by writing the generating function as
$$
\beta(x,y)=\frac{1-x}{(1-x^2)(1-x-y)}=\frac{1}{1-x^2}\left(1+\frac{y}{1-x-y}\right),
$$
or more simply,
$$\sum_{m=0}^\infty \sum_{n=1}^\infty B(m,n) x^m y^n = \frac{y}{(1-x^2)(1-x-y)},$$
which gives the simpler formula
$$B(m,n) = \sum_{0\le i\le m/2} \binom{m+n-2i-1}{n-1}$$
for $n>0$. From these generating functions we see that that $B(m,n)$ is the number of lattice paths from $(0,0)$ to $(m,n)$, with unit east and north steps, that start with an even number of east steps.
The OP's second sum gives
$$
B(m,n) = \sum_{j=0}^m (-1)^{m-j}\binom{n+j}{j}=\sum_{j=0}^m (-1)^j \binom{n+m-j}{m-j}.
$$
This comes from expanding $\beta(x,y)$ as
$$\frac{1-x+x^2-x^3+\cdots}{1-x-y}$$
and is easy to interpret combinatorially: $\binom{n+m-j}{m-j}$ is the number of paths from $(j,0)$ to $(m,n)$, or equivalently the number of paths from $(0,0)$ to $(m,n)$ that start with $j$ east steps (possibly followed by more east steps), or in other words, the number of paths from $(0,0)$ to $(m,n)$ that pass through $(j,0)$. Then for $j$ even, $\binom{n+m-j}{m-j}-\binom{n+m-j-1}{m-j-1}$ counts paths from $(0,0)$ to $(m,n)$ that start with $j$ east steps followed by a north step. Add this over all even $j\le m$ gives all the paths counted by $B(m,n)$.
The identity in question (with $n$ replaced $m+n$ and the order of the summations reversed) may be written as
$$
\sum_{i=0}^m (-2)^i\binom{m+n+1}{m-i}
=\sum_{j=0}^m (-1)^j \binom{m+n-j}{m-j}.
$$
This is the case $t=-2$ of the identity
$$
\sum_{i=0}^m t^i\binom{m+n+1}{m-i}=
\sum_{j=0}^m (1+t)^j\binom{m+n-j}{m-j}.
\tag{1}
$$
We can give a combinatorial interpretation of $(1)$, but I don't see that setting $t=-2$ has a simple combinatorial interpretation (though what I described above is a combinatorial interpretation of setting $t=-2$ in the right side).
The combinatorial interpretation of $(1)$ is made clearer by looking at the generating function for $(1)$, which is
$$\frac{1}{(1-(1+t)x)(1-x-y)}.$$
The right side of $(1)$ is obtained by expanding this in the most straightforward way; the left side is obtained by expanding it as
$$
\frac{1}{(1-x)^2} \frac{1}{1-tx/(1-x)}\frac{1}{1-y/(1-x)}=
\sum_{i,n}\frac{(tx)^i y^n}{(1-x)^{i+n+2}}.
$$
To interpret the right side of $(1)$, we consider paths from $(0,0)$ to $(m,n)$, which are “cut” at some point $(j,0)$ on the $x$-axis (so they must start with at least $j$ east steps) and some subset of the first $j$ (east) steps are “marked” and weighted by $t$. It is clear that the contributions from the paths cut at $(j,0)$ is $(1+t)^j\binom{m+n-j}{m-j}$: each of the first $j$ (east) steps contributes 1 or $t$, and $\binom{m+n-j}{m-j}$ counts paths from $(j,0)$ to $(m,n)$. For the left side, given such a cut and marked path, with $i$ marked east steps, we change each marked east step to a north step and insert an additional north step after the $j$th step, obtaining a path with $m-i$ east steps and $n+i+1$ north steps, and these are counted by $\binom{m+n+1}{m-i}$. It is easy to see that this transformation is bijective—to go back we change the first $i$ north steps to marked east steps, set $j$ to the number of steps before the $(i+1)$st north step, and delete the $(i+1)$st north step.
It may be noted that $(1)$ is a special case of a $_2F_1$ linear transformation; a generalization can be obtained easily be expanding
$$\frac{1}{(1-(1+t)x)^a (1-x-y)^b}$$
in the same two ways.
Thank you for your generous analysis.
|
2025-03-21T14:48:31.327180
| 2020-06-20T02:29:31 |
363595
|
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|
Stack Exchange
|
Is it ever useful to consider a long exact sequence as a chain complex?
Here are two common ways of obtaining chain complexes with vanishing homology:
Chain complexes that compute the reduced homology of a contractible space
Chain complexes that arise as a "long exact sequence in homology" induced by a short exact sequence of chain complexes
These two examples seem to be on different footing. It is easy to think of useful maps within a type:
for 1), a continuous map of contractible spaces induces a chain map
for 2), a map of short exact sequences induces a map on long exact sequences
but I cannot really imagine a map between these two types, mostly because the three-fold periodicity in type 2) seems utterly incompatible with the ethos of type 1). The two types seem like unrelated notions that happen to have the same mathematical definition.
So here is my precise question, trying to see if there is any crossover:
Suppose (X,A) is a pair of spaces, $A \subseteq X$ so that there is a short exact sequence of singular chains $0 \to C_\bullet A \to C_\bullet X \to C_\bullet(X,A) \to 0$. Is it possible to functorially construct a filtered contractible space $\Lambda = \Lambda(X,A)$ with $\Lambda_0 \subseteq \Lambda_1 \subseteq \ldots$ so that
$H_\bullet(\Lambda_{3k},\Lambda_{3k-1}) \cong H_k A$,
$H_\bullet(\Lambda_{3k+1},\Lambda_{3k}) \cong H_k X$, and
$H_\bullet(\Lambda_{3k+2},\Lambda_{3k-1}) \cong H_k(X, A)$,
so that the spectral sequence of the filtered space $\Lambda$ recovers the long exact sequence of (X,A) as the $E^1$-page?
Motivation: I am writing a Macaulay2 package that makes it easy for the user to generate long exact sequences. I originally considered returning such a sequence as a "ChainComplex" object, since this data type already exists in M2 and can be used by other functions. Then I decided that this is not so convenient, actually, because it obscures the threefold periodicity. But now I wonder if it could ever be useful to consider a long exact sequence as a chain complex.
|
2025-03-21T14:48:31.327371
| 2020-06-20T04:41:34 |
363598
|
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"Ali Enayat",
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"David Roberts",
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|
Stack Exchange
|
Does bounded Zermelo construct any cumulative hierarchy?
ZF is sufficient to construct the von Neumann hierarchy, and prove that every set appears at some stage $V_\alpha$. This is the basis for Scott's trick, for instance. But how much of ZF is needed? Is bounded Zermelo/Mac Lane set theory enough, no Choice assumed? I know Foundation is necessary, and I'm not getting rid of that. I've seen something called the "rank axiom" in discussion of second-order version of original Zermelo (these notes), but I'm sure people have finely calibrated what precisely is needed.
To be honest, all I really want is an ordinal-valued rank function such that sets of rank at most $\alpha$ form a set, for all $\alpha$, and all sets have a rank. So if the von Neumann hierarchy doesn't work, I'm happy to work with something else (and for 'ordinals', I don't need von Neumann ordinals).
You need some Replacement, at least in the von Neumann case. If you are willing to work with "Fregean ordinals", there might be a way to resolve this without too much Replacement.
@AsafKaragila In fact it requires considerable amount of Replacement. In particular even $(\mathsf{ZFC}-\mathsf{Replacement})+\Delta_0\text{-}\mathsf{Replacement}$ couldn't prove that for every ordinal $\alpha$ there is $V_\alpha$, i.e. the set of all elements of the rank $\alpha$. This is because $H\aleph_\omega$ (the set of all sets $x$ s.t. the transitive closure of $x$ is of the cardinality $<\aleph_\omega$) is a model of this theory.
@Fedor: If you read closely, I was referring to the remark at the end of the question, foregoing the von Neumann hierarchy and the von Neumann ordinals.
David, you can't get rid of the von Neumann part too easily: let $\alpha$ be the supremum of the von Neumann ordinals of the model, you might want to "pace yourself" with adding those to the levels of your hierarchy. But that means that the cofinality of your hierarchy must be the same as the cofinality of $\alpha$. Taking Mathias' $M_\lambda$ example of a model of Zermelo shows that this is impossible (as the rank function must be unbounded in all possible order types, i.e. the real Ord, but $\alpha=\lambda$ is a set).
@Asaf I was commenting on the first sentence of your reply that were about the case of von Neumann ordinals.
Hmm, so what do you two think is then the answer? I worry about some circularity if one tries to define the ordinals using a Scott-type trick, but then a hierarchy index by these ordinals.
@David I think that it is essential to have $\Pi_1$-Replacement, PowerSet, and Regularity as axioms. I don't think that the usage of some alternative presentation of ordinals would really help. Note that in the model $H\aleph_\omega$ for ranks $\alpha\ge\omega2$ the collections of all sets of the rank $\alpha$ form proper classes. And even if you would represent ordinals differently $V_\alpha$'s still would be proper classes in this model.
@Fedor ok, that seems fairly clear, but we can wait and see what Asaf comes up with.
@Asaf I don't even care about the image of the rank function being every well-order type, just that the domain is the universe, and only set-many sets have a rank at most a given value.
David, it's not about the image. It's about the possible ranks of your von Neumann ordinals. They cannot be exhausted before the you're done with the whole universe.
@Fedor: $\Pi_1$-Replacement is not a lot. A lot would be $\Sigma_3$.
@AsafKaragila Personally I consider any collection/replacement principle stronger than $\Sigma_1$-Collection to be strong. We could develop most of basic set-theoretic constructions in $\mathsf{KP}$ and it could only prove $\Sigma_1$-Collection.
@FedorPakhomov: Why do you consider KP(P) to be strong? It does not have any replacement but simply has inbuilt power-set, which is arguably the correct way to axiomatize bounded ZF.
@user21820 I guess it is fundamentally matter of perspective. I am a proof-theorist and for me the main metrics is the proof-theoretic strength. And from this perspective unbounded applications of powerset are the core to the strength of $\mathsf{ZFC}$. So when you allow $\mathcal{P}$ as an atomic operation for me the resulting theory $\mathsf{KP}(\mathcal{P})$ appears essentially as strong as $\mathsf{ZFC}$ itself. And from the perspective of formalization of individual theorems, the only substantial limitation left is that $\mathsf{KP}(\mathcal{P})$ doesn't prove Mostowski collapsing.
@FedorPakhomov: I do not understand your perspective. Unbounded applications of Powerset contribute strength only if you have other strong assumptions such as impredicative Specification and Replacement. How can we 'blame' Powerset for that increase in strength if the real issue is with impredicativity? I know Powerset is not predicative, strictly speaking, but it's not as bad as unbounded quantification over the universe. Also, for you to use proof-theoretic strength as a gauge, you would need to state the p.t.o. of KP(P) and show that it is intrinsically comparable to that of ZFC...
@user21820 Consider as the base system the theory $\mathsf{BST}_0^{\mathsf{set}}$ axiomatized by Extensionality, totality of all Rudimentary recursive functions, and Infinity (see Definition VII.3.3 from Simpson's book on 2-nd order arithmetic). If you add to this theory Powerset, the resulting theory will be proof-theoretically equivalent to order $\omega$ arithmetic and would be able to prove the consistency of $\mathsf{ZFC}-\mathsf{Powerset}$ which is proof-theoretically equivalent to just second-order arithmetic and to $\mathsf{BST}_0^{\mathsf{set}}+\exists \mathcal{P}(\omega)$.
@FedorPakhomov: So? SOA (2nd-order arithmetic) is not that strong. Way way below ZFC. And SOA doesn't even have Powerset, so by your own criterion it shouldn't be considered strong. HOA (higher-order arithmetic) doesn't have unbounded Powerset either. So I really don't get your point.
@user21820 I was refuting your statement that Powerset doesn't contribute strength in isolation by pointing out that when compared with other axioms of $\mathsf{ZFC}$ it provides more proof-theoretic strength (over weak base theory) than all of them combined. And by this substantiating my statement that Powerset is the core of proof-theoretic power of $\mathsf{ZFC}$.
@FedorPakhomov: You didn't refute a statement I didn't intend to make. I didn't mean it didn't contribute anything; I intended to say that impredicativity is critical to the strength of ZFC, not Powerset. Consider that ZFC without the underlying (classical) FOL rules cannot do anything, whereas ZFC−Powerset is as strong as SOA. You cannot by this infer that FOL contributes immense strength to ZFC compared to Powerset! To say in another way: Every system that can reach the strength of ZFC must have some impredicative reflection (e.g. unbounded Replacement).
To give you a concrete example, look at this theory, which has no Powerset but obviously has impredicative reflection.
@user21820 At this point I'll stop this conversation: Admittedly I have no taste in what seems to be going into a prolonged and fruitless argument on this very informal (and ultimately fairly obscure in my opinion) topic.
Of course you can stop and leave at any point, especially if you cannot refute my points, even though they are simple and pertinent. It is completely well-known to experts that the proper strength scale for fragments of ZFC is the one given by Σ[n]-Rep[lacement], very like the proper strength scale for fragments of PA is the one given by Σ[n]-Ind[uction]. Your misconception about the strength of Powerset is because you're not looking at anything near the full strength of ZFC. Σ[n]-Rep is the n-th rung of the real ladder, while Powerset is on the floor below the 1st rung.
@useer21820 just chill, please. None of this is relevant to my question. I may delete all this debate as it shedding more heat than light.
KP (Kripke-Platek set theory) is the most well-known fragment of $\sf{ZF}$ which suffices for the development of the rank function, thus $\sf{KPR}$ = $\sf{KP}$ + "for all ordinals $\alpha$, $V(\alpha)$ exists" is the usual minimal theory in which one can be assured of the stratification of the universe into $V(\alpha)$s.
On the other hand, as observed by Mathias, $\sf{KPR}$ proves that Zermelo set theory $\sf{Z}$ has a transitive model, so in particular, $\sf{KPR}$ proves that $\sf{Z}$ is consistent; see Lemma 6.31 of this preprint, which was later published in APAL (2001).
Therefore, by the second incompleteness theorem, even Zermelo set theory (let alone bounded Zermelo set theory) cannot interpret $\sf{KPR}$.
Finally, I will add that $\sf{KPR}$ is provable in the well-known extension $\sf{KP}^{\cal{P}}$ of $\sf{KP}$, which is also studied in Mathias' paper.
OK, cool. So it's probably $\sf{KP}^{\cal{P}}$ or Rathjen's $\sf{KP}(\cal{P})$ that I'm after, since I don't really want to give up power set right now. Or maybe $\sf{BZ}+\sf{KP}$...
But the theory $\sf Z + $$\forall \alpha : V_\alpha$ exists, (where $\alpha$ is a von Neumann) will stratifiy the universe into stages, and this is weaker than $\sf KPR$?
@ZuhairAl-Johar The two theories you mentioned are incomparable, the first has full separation but no collection, the second has some collection but not much separation.
@AliEnayat, Ah! Yes. But the point is why we need collection for? I mean take $\sf KP - collection + \forall x \exists \alpha: x \in V_\alpha$ , why that won't ensure stratifying the universe into $V(\alpha)'s$ ?
One can directly assume in Zermelo that every set belongs to a rank. This does not add any strength at all. But notice that Zermelo may not prove the existence of very many von Neumann ordinals: there may be well-orderings which do not have a von Neumann ordinal as order type. A natural model of this theory is the union
of the V^{omega + n}'s for n a natural number. Notice that in this structure the axioms of Zermelo hold, every object belongs to a V_alpha, but omega+omega does not exist. There are well-orderings with order types far higher than omega+omega (and in this context the Scott representation of the ordinals is available).
There are a couple of additional remarks. It is interesting to note
that the assertion that every set has a rank adds no strength at all to
Zermelo set theory (or to Zermelo set theory with bounded separation) but that adding this assertion to KP, a theory much weaker than Zermelo, gives a theory much stronger than Zermelo. The reason is that KP has a lot of replacement.
The natural way to describe the rank function in "Zermelo with ranks" is probably to use the Scott ordinals as values of the rank function but note that the rank function is not necessarily onto the ordinals. In the absence of replacement, the von Neumann notion of ordinal simply isn't the right notion of ordinal number.
Could you clarify what “adds no strength at all” means?
OK, this is nice. I'm not wedded to von Neumann ordinals, since ultimately what I want is something like Scott's trick, and I'm happy to get this from some class function from $V$ to a any old well-ordered class, where the preimage of every member is a set. But you are relying on Scott ordinals, and I haven't seen that before (at least, not outside ZF).
What do you mean by the Scott representation of the ordinals? Is it something related to Scott's trick, e.g. represent a well-order-type by the set of all well-orderings of least rank having that order type?
@MikeShulman I assume so, but I don't know how to get this without having some kind of hierarchy as in question, which was the whole point! My only lead is section 3.6.1 in https://randall-holmes.github.io/proofsetslogic.pdf, where there are three extra axioms listed: Axiom of Ordinals, Axiom of Levels and Axiom of Rank. I presume that Randall in his answer is implying that one only needs the last of these, but in the notes the notion of rank uses subhierarchies, and it's not clearly signposted that these don't use the Ordinals and Levels axioms.
Aha, I see there is a second definition of rank, on page 275, in "Comments on the alternative definition". I think I see how Scott ordinals are a workable definition under the Axiom of Rank, though as Emil pointed out, why this "adds no strength at all" is unclear. (A claim of equiconsistency is made in https://randall-holmes.github.io/Drafts/hierarchy.pdf)
I've seen a citation to Matthias' "The strength of Mac Lane set theory" for the claim that adding an axiom of rank "adds no essential strength to Zermelo or Mac Lane set theory", but all I see are results about KP.
|
2025-03-21T14:48:31.328392
| 2020-06-20T05:09:59 |
363599
|
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|
Stack Exchange
|
Direct image of the structure sheaf by an endomorphism of $\mathbb{P}^2$
Let us take 3 quadratic forms on $\mathbb{P}^2$ with no common zero; they define a map $\pi : \mathbb{P}^2\rightarrow \mathbb{P}^2$ of degree 4. It is not difficult to see that $\pi _*\mathscr{O}_{\mathbb{P}^2}\cong \mathscr{O}_{\mathbb{P}^2}\oplus \mathscr{O}_{\mathbb{P}^2}(-1)^3$. Does anyone know how to write down the algebra stucture of $\pi _*\mathscr{O}_{\mathbb{P}^2}$ in terms of this decomposition?
Here's a partial answer for a particular case. I suspect you probably already knew this. (So it is really just a comment, but it is too long for the comment box.)
If the polynomials are $x_0^2, x_1^2, x_2^2$, and the characteristic is not 2, then in affine coordinates the cover $\pi$ is $y_1=x_1^2, y_2=x_2^2$. So it is Galois, with Galois group $G=(\mathbb{Z}/2)^2$. You can then break up $\pi_*\mathcal{O}_{\mathbb{P}^2}$ using 4 characters of $G$, which I'll label $1, \sigma_1,\sigma_2,\sigma_1\sigma_2$. The last 3 corresponds to the involutions $(x_1\mapsto -x_1)$, $( x_2\mapsto -x_2)$ and, I'm guessing $(x_0\mapsto -x_0)$. The invariant part of $\pi_*\mathcal{O}$ is just $\mathcal{O}$, and the remaining isotypic factors are isomorphic to $\mathcal{O}(-1)$. Under the identification
$$\Gamma((\pi_*\mathcal{O})(1))\cong \Gamma(\mathcal{O}(2))$$
coming from the projection formula, the sections coming from the $\mathcal{O}(-1)$ factors correspond to $x_1^2,x_2^2, x_0^2$ (modulo the guess above). Let me label the factors accordingly by $1,2,0$. You presumably want the multiplication table. I'm guessing, for instance,
$$\mathcal{O}(-1)_{i}\otimes \mathcal{O}(-1)_{i}\to \mathcal{O}$$
is
$$\mathcal{O}(-1)_{i}\otimes \mathcal{O}(-1)_{i}\cong \mathcal{O}(-2)\stackrel{x_i^2}{\to} \mathcal{O}$$
Etc.
Thanks, Donu. Yes, I was more or less aware of this particular Galois case. I am intrigued by the general case...
I imagine you can do something similar if only one of the quadratic forms is a square.
In general, if $f \colon X \to Y$ is a quadruple cover of algebraic varieties, the structure of the algebra $f_* \mathcal{O}_X$ is studied in the paper
D. Hahn, R. Miranda: Quadruple Covers in Algebraic Geometry, Journal of Algebraic Geometry 8 (1999).
You can download the paper from the webpage of the second author.
Thank you @Francesco Polizzi, but I am not sure this helps for my problem. According to this paper, my $\pi$ corresponds to a certain section of $\bigwedge^2 \mathsf{S}^2 \mathscr{0}{\mathbb{P}^2}(1)^3\otimes \mathscr{0}{\mathbb{P}^2}(-3)$, that is, $\mathscr{0}_{\mathbb{P}^2}(1)^{15}$. How do I find such a section starting from 3 quadratic forms $P,Q,R$ in 3 variables?
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2025-03-21T14:48:31.328609
| 2020-06-20T06:08:54 |
363601
|
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|
Stack Exchange
|
From restricted root space to root space
I am reading Knapp's book "Lie Groups beyond Introduction" https://link.springer.com/book/10.1007/978-1-4757-2453-0. I do not understand the following argument in Page 377 (2nd edition). I write the details. Let $G$ be a semisimple Lie group with Lie algebra $\mathfrak g.$ Let $\theta$ be a Cartan involution of $\mathfrak g.$ Let $\mathfrak g=\mathfrak k\oplus \mathfrak p$ be the Cartan decomposition. One can check that because of the inner product induced by $\theta$ we must have that $\{\text{ad}H:H\in\mathfrak a\}$ where $\mathfrak a\subseteq \mathfrak g$ a maximal abelian subspace, is a commuting family of self-adjoint operators. Therefore, we have an eigenspace decomposition of $\mathfrak g$ as $\mathfrak g=\mathfrak g_0\oplus{\oplus}_{\lambda\in\Sigma}\mathfrak g_\lambda,$ which is called a restricted root space decomposition. Moreover, one can proe that $\mathfrak g_0=\mathfrak a\oplus Z_{\mathfrak k}(\mathfrak a).$ One can prove that if $\mathfrak t\subseteq Z_{\mathfrak k}(\mathfrak a)$ is maximal abelian, then $\mathfrak h:\mathfrak a\oplus \mathfrak t$ is a Cartan subalgebra of $\mathfrak g.$ Therefore, we have a rot space decomposition as $(\mathfrak g)^\mathbb{C}=(\mathfrak h)^{\mathbb C}\oplus\oplus_{\alpha\in\Delta}\mathfrak g_\alpha.$ Now one can get back the components of restricted root space decomposition as the following. We have $\mathfrak g_\lambda=\mathfrak g\cap \oplus_{\alpha|_{\mathfrak a}=\lambda}\mathfrak g_\alpha$. I understand clearly upto this. But how to show that $(Z_{\mathfrak k}(\mathfrak a))^\mathbb C=\mathfrak t^\mathbb C\oplus\oplus_{\alpha|_{\mathfrak a}=0}\mathfrak g_\alpha$. However, I can see that $\mathfrak t^\mathbb C\oplus\oplus_{\alpha|_{\mathfrak a}=0}\mathfrak g_\alpha$ commutes with $\mathfrak a^\mathbb C.$
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2025-03-21T14:48:31.328737
| 2020-06-20T06:12:43 |
363602
|
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|
Stack Exchange
|
Noncommutative symmetric spaces
I am recently studying ergodic actions of Lie groups acting on Riemannian symmetric spaces. Since I am also interested in operator algebras, it makes me wonder if there are some very natural noncommutative analogues of Riemannian symmetric spaces and measure preserving ergodic group actions by classical Lie groups?
You might be interested in the work of Lezter and others on quantum symmetric spaces. See for example this paper
https://arxiv.org/pdf/math/0406193.pdf
These objects seem to be closely connected to the representation theory of quantum groups. People have also considered $C^*$-completions, but I'm not sure about ergodic actions.
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2025-03-21T14:48:31.328811
| 2020-06-20T07:50:24 |
363605
|
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|
Stack Exchange
|
On the proof of "Mapping space is a Chen space"
According to the page 5 in the paper Convenient Categories of Smooth Spaces https://arxiv.org/pdf/0807.1704.pdf by Baez and Hoffnung, Chen space is defined as follows:
(Note:I used different notations from the paper Convenient Categories of Smooth Spaces https://arxiv.org/pdf/0807.1704.pdf)
A Chen space $X$ is defined as a set $X$ equipped with , for each convex set $U$ there exists a collection $\lbrace \phi_{i}: U \rightarrow X \rbrace_{i \in I}$ of set maps called plots in $ X$ satisfying the following properties:
(A convex set $U$ is defined as a convex set (with non-empty interior) in a Euclidean space $\mathbf{R^n}$ where $n$ can be any arbitrary non-negative integer. We call $ n$ the dimension of $U$.A map $f: U' \rightarrow U$ from convex set $U'$ to convex set $U$ is called smooth function if $f$ has continuous derivatives of all order.)
If $f: U' \rightarrow U$ is a smooth function from convex set $U'$ to convex set $U$ and if $\phi: U \rightarrow X$ is a plot in $X$ then $\phi \circ f$ is also a plot in $ X$.
Let $U$ be a convex set of dimension $n$. Suppose a collection of convex sets $ \lbrace U_j \subset U \rbrace_{j \in J}$ forms an open cover of $U$ with respect to the subspace topology of $\mathbf{R^n}$. Let $\lbrace I_j: U_j \rightarrow U \rbrace_{j \in J}$ be the collection of inclusion maps. Let $\phi: U \rightarrow X$ be a set map. Now if each $ \lbrace \phi \circ I_j \rbrace_{j \in J}$ are plots in $X$ then $\phi$ is also a plot in $X$.
Every function from the one point of $\mathbf{R^0}$ to X is a plot in $X$.
In page 6 they defined a set map $f: X \rightarrow Y$ to be smooth if for any plot $\phi: U \rightarrow X$ in $X$ the set map $f \circ \phi: U \rightarrow Y$ is a plot in $Y$.
In page 15 they mentioned that the mapping space $C^{\infty}(X, Y)= \lbrace f:X \rightarrow Y: f$ is smooth$\rbrace$ is a Chen space (where $X, Y$ are Chen spaces) whose plots are declared as those functions $\phi:U \rightarrow C^{\infty}(X, Y)$ such that the corresponding function $\tilde{\phi}:U \times X \rightarrow Y$ is smooth defined as $(\zeta, x) \mapsto \phi(\zeta)(x)$ (Note that there is a natural Chen space structure on both convex sets and Products ).
I was verifying $C^{\infty}(X, Y)$ is indeed a Chen space.
Property 1 and Property 3 are verified easily.
But I am not able to verify the property 2.
According to the definition of plot to verify Property 2, I need to show that if $\tilde{\phi \circ I_i}:U_i \times X \rightarrow Y$ is smooth for each $i$ then $\tilde{\phi}: U \times X \rightarrow Y$ is smooth. (where $U_i \subset U$ forms an open convex cover of $U$ and $I_i$ are inclusion maps). For that I need to show that if $\psi:V \rightarrow U \times X$ is any plot in $U \times X$ then $\phi \circ \psi$ is a plot in $Y$ .
I am not able to progess much after that!
I also note that there exist a collection of smooth maps $I_i \times 1_X : U_i \times X \rightarrow U \times X$ (where $I_i$ , $1_X$ are inclusion and identity map respectively).I feel somehow I need to use this fact also but not able to guess how.
I feel that I have to somehow express the plot $\psi$ in $U \times X$ in terms of plots of $U_i \times X$ so that I can use the smoothness of $\tilde{\phi \circ I_i}$. But I am not able to guess how!!
I feel it is some sort of local property of smoothness (as we have the analogue in case of finite dimensional smooth manifolds.)
I apologise priorly if this question is not upto the standard of MathOverflow. I am guessing I am mistaking or overlooking something.. But not able to guess what is that!!
Thank you.
I need to show that if $\tilde{\phi \circ I_i}:U_i \times X \rightarrow Y$ is smooth for each $i$ then $\tilde{\phi}: U \times X \rightarrow Y$ is smooth. (where $U_i \subset U$ forms an open convex cover of $U$ and $I_i$ are inclusion maps).
We have to show that a map $U⨯X→Y$ is a morphism of diffeological spaces if and only if its restrictions $U_i⨯X→Y$ are morphisms of diffeological spaces.
To show that $U_i⨯X→Y$ uniquely glue to a morphism $U⨯X→Y$,
consider some cartesian space $S$ and map it to both sides.
We have to define a map on $S$-points natural in $S$:
$$U(S)⨯X(S)→Y(S).$$
Fix an element in the left side, i.e., a smooth map $f\colon S→U$
and an element of $X(S)$.
Fix a cover $\{S_i=f^*U_i\}_{i∈I}$ of $S$.
The map $f|_{S_i}$ factors through $U_i$.
Thus, we have an $S_i$-plot of $U_i⨯X$, which maps to an $S$-plot of $Y$ via the map $U_i⨯X→Y$.
The constructed $S_i$-plots of $Y$ are compatible and glue to a unique $S$-plot of $Y$.
This means that $\tilde\phi$ is a plot, as desired.
Sir, According to the definition of Chen Space I mentioned above there is no assumption of topology on it. So how can you consider open sets of $U_i \times X$ and $U \times X$? Also a map $F:P \rightarrow Q$ between Chen spaces is defined to be smooth if for any plot $\phi:U \rightarrow P$ in $P$ , $F \circ \phi$ is a plot in $Q$. So how does this follow from your Answer?
@AdittyaChaudhuri: I added a few more details.
Thank you Sir for the edit. I am trying to understand the proof in detail.
|
2025-03-21T14:48:31.329397
| 2020-06-20T07:50:35 |
363606
|
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|
Stack Exchange
|
Group algebras and group automorphisms
Say, we have a countable ICC group $G$, a Hilbert space $H$ with a basis indexed by the group elements, the group algebra generated by the left regular representation of $G$ on this Hilbert space, and its norm and weak closures, the reduced C-* and the von Neumann algebras.
On the other hand, an automorphism of $G$ also permutes the basis of $H$ (when applied to the indices) and thus defines a unitary on $H$.
What's the relationship of this unitary to the group algebras? Is it ever in, say, the von Neumann group factor? How do the properties of this unitary reflect whether it comes from an inner group automorphism or not?
These unitaries, representing automorphisms of $G$, also give rise to an operator algebra. How is this algebra related to the group's, say, von Neumann algebra?
This is a collection of interesting questions, IMHO. But the questioner could have done some basic work before asking. For example, the group von Neumann algebra is the commutant of the right regular representation, and this gives a criteria for these "automorphism unitaries" to be members of the group von neumann algebra. As far as I can see, this is only the case for the identity automorphism.
If I interpreted your question correctly, what you are interested in is the following result:
Suppose $G$ is an i.c.c., discrete group and $\delta: G \rightarrow G$ is a group homomorphism. Define the automorphism of the group von Neumann algebra $\theta(\sum_g c_g \lambda_g)= \sum_g c_g \lambda_{\delta(g)}$. Then, $\theta$ is outer if and only if $\delta$ is an outer automorphism of $G$.
A reference for the aforementioned result is Remark 2.3 in the paper " A generalization of free action" by Robert R. Kallman.
I think the OP was rather asking to define a unitary $U$ by $U(e_g) = e_{\delta(g)}$ where $(e_g)$ is the standard o.n. basis of $\ell^2(G)$. Then the questions is about properties of $U$ and $\delta$.
Thanks, Darth, yes I am. However, in my question the setup is different. In the result you mentioned, the group automorphism acts on the group, which then maps to its representation, which then completes to a von Neumann algebra. So one direction is obvious -- if the group automorphism is a conjugation by a group element, then conjugation by that element's corresponding unitary in the left representation will be our automorphism on the algebra level. In my question, the automorphism permutes the Hilbert space basis and becomes another unitary, likely outside of the algebra.
@MatthewDaws: wouldn't ad(U) just implement the automorphism $\theta$ that I defined in my answer? $U$ is in $L(G)$ if and only if $\theta$ is inner, in that case, right? Am I missing something?
@Chilperic: Yes one direction iss obvious. The other direction is that $\delta$ is outer implies $\theta$ is outer. This is what's proved in Remarks 2.3 in the quoted paper. The proof is basically what Matthew pointed out in the comment to your question.
Sorry, Darth, posted before edits finished. Basically I m curious to what extent $Aut(G)$ and $Out(G)$ properties extend to the von Neumann factor. Can we even tell from an automorphism of the factor whether it comes from a group automorphism or not?
@Chilperic: That's a different and a harder question. Basically, the answer is no. As mentioned in my answer, we always have $Out(G) \subseteq Out(L(G))$. However, $Out(L(G))$ is typically larger: for instance, there are outer automorphisms arising from characters of the group. There are examples, however, where $Out(L(G))$ is completely determined by $Out(G)$ and $Char(G)$. These are the so called $W^*$-superrigid groups. The first such examples were obtained by Ioana, Popa and Vaes in 2010.
In general, $Out(L(G))$ is lot larger than $Out(G)$. For instance, if $G=S_{\infty}$, then any countable group is contained in $Out(L(G))$.
Yea, Popa and rigidity... I was hoping some of the industry over $Out$ of free groups would have at least something for free group factors....
@Chilperic: That would perhaps be too ambitious in my opinion :). Especially because $L(Z_2 \ast Z_3)$ is isomorphic to an interpolated free group factor- but I don't think $Out(Z_2 \ast Z_3)$ is related to $Out(F_n)$ for any $n$.
Reminds me, Shlyakhtenko (with someone, I think) worked out which groups correspond to the interpolated factors, right? Is this result from there?
@Chilperic: The result that I refer is due to Dykema. See for instance, https://mathoverflow.net/questions/361709/is-l-mathbbz-mathbbz-2-a-free-group-factor/361740#361740
The result that you are referring to is due to Popa and Shlyakhtenko: https://arxiv.org/pdf/1805.10707.pdf
Oh yea, thanks, I thought Dykema after posting. I'll go post a question on $Aut(L(G)) / Aut(G)$ and see what happens.
|
2025-03-21T14:48:31.329807
| 2020-06-20T09:24:11 |
363608
|
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|
Stack Exchange
|
Lattices on classical combinatorial families
I am asking for examples of lattices defined on classical combinatorial families, such as Permutations, Catalan objects, set partitions or integer partitions, graphs.
I am mosty interested in lattices defined on the objects of a fixed size.
To make later usage easier, I would like to combine all lattices defined on one set of objects into one answer, and identify sets of objects which are in bijection.
My motivation for creating this list is that I realized how limited my supply of examples of lattices is.
Here is a catalog of posets and lattices implemented in SageMath.
(I've been noting when the lattices are distributive, because if your goal is to have a collection of interesting lattices, distributive lattices are in a sense very boring.)
@SamHopkins, you are unstoppable :-) Noting distributivity is a great idea, but I must say that also distributive lattices may be very useful!
Reverse plane partitions or semistandard tableaux of a given shape with entries from a given set also form distributive lattices with respect to entrywise comparison. More generally, this holds for P-partitions of any given labelled poset.
The lattices I find most interesting are distributive! Specific distributive lattices are interesting when they can be naturally constructed on nice combinatorial objects (other than order ideals of a given poset).
In addition to identifying the set ob objects in bijection, it may be interesting to identify which are subsets and/or isomorphic to sub-orders or sub-lattices of others. As I try to illustrate in my answer focused on integer partitions below, there are many such relations!
I posted a related question here: https://mathoverflow.net/questions/389846/objects-in-bijection-with-integer-partitions-and-lattices
Lattices are prevalent when one deals with integer partitions.
Let me give a few examples with pictures, that I hope you will enjoy despite the poor quality due to bitmap conversion.
The dominance order, that you already cited, is a lattice; it is not distributive and not graded, but it has nice self-similar properties. Here is a picture of its Hasse diagrams for $n=7$ and $n=20$:
The reachable configurations of the linear Sand Pile Model (SPM), where one starts with a pile of $n$ grains, and they may fall under the condition that the sequence of piles remains decreasing, is also a lattice. It is a sub-order of the lattice above. It is not distributive either, but it is graded, and has nice self-similar features too. Here is the Hasse diagram for $n=40$.
Ice Pile Models (IPM) are generalizations of the two above cases, where the grains may slip from one column to another under some conditions. The induced orders are non-graded lattices, and one may obtain not only partitions but also compositions of a given integer.
The examples above are partitions of a given integer; one may also consider integer partitions of given maximal part and/or number of parts.
This leads to finite distributive sub-lattices of the Young Lattice, strongly related to Dyck paths, already cited in previous answers. For instance, if we consider partitions in at most $3$ parts of value at most $3$, we obtain the following representations (the rightmost one is the lattice of Dyck paths with three steps in each direction):
These can be generalized to plane partitions (or piles of cubes), solid partitions and actually generalized integer partitions on graphs introduced, I think, by Stanley in his seminal book "Ordered Structures and Partitions". Again, we obtain distributive lattices, and here is the case of planar partitions contained in a 3x3x2 volume:
As you may see above, these partitions are equivalent to tilings of an hexagon with lozenges, with the order induced by flips of three tiles. In higher dimensions, we obtain tilings of 2D-gons, with the same flip ordering. They are not always lattices, but they are disjoint unions of distributive lattices, because of their relation to generalized integer partitions. For instance, here is the set of all tilings of a unit decagon:
Other kinds of tilings with flips induce distributive lattices, like tilings by dominoes, and generalized tilings with height functions.
Let me add a last example, because I love its drawing. If one considers, for a given $b$, the different ways to write a given number $n$ as a sum of powers of $b$, then one obtains a self-similar distributive lattice. For instance, for $n=80$ and $b=2$:
Catalan objects, http://oeis.org/A000108
noncrossing partition lattice (also known as Kreweras lattice)
Tamari lattice - this lattice is not distributive but it is semidistributive
Dyck paths ordered by containment (also known as Stanley lattice) - this lattice is distributive
Bernardi, Olivier; Bonichon, Nicolas, Intervals in Catalan lattices and realizers of triangulations, J. Comb. Theory, Ser. A 116, No. 1, 55-75 (2009). ZBL1161.06001.
There's another weirder one which is the lattice of nonnesting partitions (where a partition is nonnesting if whenever $a < b < c < d$ and $a,d$ are consecutive elements of a block, then $b$ and $c$ do not belong to the same block).
Is there an existing way to obtain the Stanley lattice in Sage?
P = lambda n: Poset([DyckWords(n), lambda D1, D2: all(h1 <= h2 for h1, h2 in zip(D1.heights(), D2.heights()))])
@Mare: Or "P = Partition(reversed([1..n])).cell_poset().order_ideals_lattice()"
Permutations, http://oeis.org/A000142
weak (Bruhat) order - this lattice is not distributive, but it is semidistributive
shard intersection order
the ``bubble-sort'' order or more generally any of the sorting orders of Armstrong - this lattice (which sits between weak and strong order) is distributive
Reading, Nathan, Noncrossing partitions and the shard intersection order, Krattenthaler, Christian (ed.) et al., Proceedings of the 21st annual international conference on formal power series and algebraic combinatorics, FPSAC 2009, Hagenberg, Austria, July 20–24, 2009. Nancy: The Association. Discrete Mathematics & Theoretical Computer Science (DMTCS). Discrete Mathematics and Theoretical Computer Science. Proceedings, 745-756 (2009). ZBL1391.05283.
Of course, there is also the classical (strong) Bruhat order. More confidential is the parabolic support order, introduced by Bergeron, Hohlweg and Zabrocki.
@F.C.: (strong) Bruhat is not a lattice, however, and the question asked specifically about lattices.
oh, right. Missed that, sorry.
Set partitions, https://oeis.org/A000110
refinement order
And I recall being told that every lattice is isomorphic to a partition lattice? Or was that just for finite lattices?
@bof: No, there are many examples of finite lattices that are not partition lattices. Already a 3-element chain (totally ordered set) isn't.
@NathanReading Huh? There is a $3$-element chain in the lattice of partitions of a $3$-element set.
So do you mean every lattice is isomorphic to a sublattice or subposet of a partition lattice? Your original comment would seem to say that the 3 element chain should be a partition lattice, and that's certainly not true. (I hope none of this sounds disagreeable... just looking for the math. If there is some true sentence that says "Every finite lattice ... partition lattice", I'm interested.)
Integer partitions, http://oeis.org/A000041
dominance order
Alternating Sign Matrices, https://oeis.org/A005130
distributive lattice of monotone triangles ordered componentwise, which is the MacNeille completion of (strong) Bruhat order on the symmetric group.
$n^{n-2}$ objects (labeled trees or parking functions)
the Cartesian product of $n-2$ chains of length $n$ can be interpreted as a (distributive) lattice on the Prüfer codes of labeled trees on $n$ vertices.
@RichardStanley: if we consider the poset of regions of the Shi arrangement where $\hat{0}$ is the region containing $(1,2,3,...,n)$, does this form a lattice? The poset on parking functions you mention is related to the Shi arrangement via the Pak-Stanley labeling; the one I am talking about is more related to the Athanasiadis-Linusson bijection (https://arxiv.org/abs/math/9702224).
@RichardStanley: If I remember correctly, whether the "weak order" on a hyperplane arrangement is a lattice has something to do with whether the regions are simplicial? I'm not sure if this property holds for the Shi arrangement with this choice of base region; but I checked it in the case $n=3$.
I checked (using SageMath) the case $A_4$, in which the poset of regions (with an element adjoined so it has a top and a bottom element) is not a lattice, apparently.
@MartinRubey: the poset of regions I am talking about should already have a bottom and top element (regions containing $(1,2,3,...n)$ and $(2n,2n-2,2n-4,...,0)$).
@MartinRubey: Ah, but I agree with you the case $n=4$ of my construction does not give a lattice. I'm impressed Sage has this functionality built in so easily!
My description of the poset of regions of the Shi arrangement is incorrect, and I am deleting this comment.
@RichardStanley: I checked that it is a meet semilattice for $n=3,4,5$; but I also don't understand why your comment was wrong? Evidently parking functions ordered coordinnatewise should form a meet-semilattice (you can always do min of coordinates), and it seems to me you were right in saying that this poset is the same as the Shi arrangement poset with base region the one containing $(n\varepsilon,(n-1)\varepsilon,\ldots,\varepsilon)$: shouldn't the Pak-Stanley labeling show this?
@user61318: Thanks very much for the reference, which indeed has a lot of relevant stuff. Unfortunately the (cone over the) Shi arrangement is not supersolvable, IIRC. So this result does not seem to apply here, unless I'm mistaken?
Fibonacci numbers: Two examples of distributive lattice structures
$\bullet$ Dominance Order on $\Bbb{YF}_n$:
A fibonacci word is a string of the form $u = a_1 \cdots a_k$ with
digits $a_i \in \{1, 2\}$. Define its length by $|u| := a_1 + \cdots + a_k$ and the let $\Bbb{YF}_n$ denote the set of all fibonacci words of length $|u|=n$. Clearly the cardinality of $\Bbb{YF}_n$ is $F_n$ where $F_n$ is the $n$-th fibonacci number. Given two fibonacci words $u= a_1 \cdots a_k$ and $v = b_1 \dots b_\ell$ of equal length declare $u \unlhd v$ if $a_1 + \cdots + a_i \leq b_1 + \cdots + b_i$ for $1 \leq i \leq \mathrm{min}(k,\ell)$. This is a distributive lattice.
$\bullet$ Ideals of the Zig-Zag Poset $\mathcal{Z}_n$:
The Zig-Zag poset $\mathcal{Z}_n$ is the index set
$[1, \dots, n]$ where all even numbers are maximal and incomparable,
all odd numbers are minimal and incomparable, and $2i$ covers
its odd neighbors $2i-1$ and $2i+1$ (whenever the latter is present). Clearly the number of (lower) ideals is
$F_{n+1}$ and the set $\mathrm{J}\big( \mathcal{Z}_n \big)$ of all
(lower) ideals ordered by inclusion will be a distributive lattice.
Maybe I'll add more later, ines.
Are these two distributive lattices on Fibonacci number objects not the same?
They shouldn't be: When $n=3$ the dominance order is the total order $1111 \lhd 112 \lhd 121 \lhd 211 \lhd 22$ while the ideals ${1 }$ and ${3 }$ are incomparable with the respect to the inclusion order of $\mathrm{J}(\mathcal{Z}_3)$.
Interesting. So what is the $P$ for which dominance order on fibonacci words is $J(P)$?
Good question; I'm not sure. One interpretation of the dominance order is to associate a Moztkin path to each fibonacci word where $1$'s are horizontal steps and $2$'s are diagonal steps, with the proviso that you add enough downward diagonals at the end to make sure the path meets the x-axis. The dominance order is equivalent to ordering these Motzkin paths by inclusion (of the region bounded by the path and the x-axis).
I think I finally figured out the answer to my question in the comment above, thanks to a talk by Richard Stanley. It should be the "comb poset"; see page 57 of https://math.mit.edu/~rstan/transparencies/posets.pdf
@SamHopkins Hi, the distributive lattice of order ideals of the comb poset (of size $2n$) has a unique maximal element which cover exactly $n$ elements. On the other hand, the dominance order on fibonacci words of size $2n$ has a unique maximal element which covers exactly one element. So I don't see how you can recover the $\Bbb{YF}_{2n}$ dominance order from the comb poset in this way.
@SamHopkins Also the cardinality of the $\Bbb{YF}{2n}$ dominance order is of course the fibonacci number $F{2n}$ while the cardinality of the lattice of order ideals of the comb poset of size $2n$ is the sum of fibonacci numbers up to $F_{2n}$. Could you explain in a bit more detail what you're considering?
I have a nice .pdf file depicting the dominance order (or order six for example). It would be nice to somehow attach this to this discussion. Is there a way to do this?
I see I was just confused about what was being discussed in Richard’s talk then.
$\binom{n}{k}$ objects, i.e., $k$-element subsets of $[n]$
the interval $[\varnothing,(n-k)^k]$ of Young's lattice of partitions - this is a distributive lattice
I believe this is the same as dominance order on compositions of a fixed number into a fixed number of parts.
All $2^n$ subsets of $[n]$
Boolean lattice - (distributive)
the interval $[\varnothing, (n,n-1,...,1)]$ of the shifted version of Young's lattice, this is the same as the dominance order on compositions of $n+1$ - (distributive)
Excerpt from "Distributive lattices, polyhedra, and generalized flows" by
Stefan Felsner and Kolja Knauer:
Many researchers have constructed distributive lattices on sets of combinatorial objects, e.g.,
• domino and lozenge tilings of plane regions ([23] and others based on [27]).
• planar spanning trees ([11])
• planar bipartite perfect matchings ([16])
• planar bipartite d-factors ([8,22])
• Schnyder woods of planar 3-connected graphs ([4])
• Eulerian orientations of planar graphs ([8])
• α-orientations of planar graphs ([8,7])
• circular integer flows in planar graphs ([15])
• higher dimensional rhombic tilings ([18])
• c-orientations of graphs ([22]).
Perhaps you would find this answer by Vijay D to a MO-Q on dualities informative.
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2025-03-21T14:48:31.330864
| 2020-06-20T09:48:57 |
363613
|
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|
Stack Exchange
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Infinite-dimensional torsion-free $F_\infty$-group not containing $F$
Is there an example of a group $G$ that has the properties
the cohomological dimension of $G$ is infinite: $\operatorname{cd}(G) = \infty$,
$G$ is torsion-free,
$G$ is of type $F_\infty$,
$G$ does not contain (an isomorphic copy of) Thompson's group $F$?
Those who don't know about Thompson's group can replace 4 by the stronger requirement
4'. $G$ does not contain an infinite-rank free abelian group.
Motivation: there are various known groups that satisfy 1-3 but all of them contain $F$. If one is bold one might wonder whether this because of what we know or because of $F$.
On a less speculative level, this is a variation of this question: two good reasons for 1 are for $G$ to have torsion or to contain infinite-rank free abelian groups. So how to assure it if the two are ruled out? The cited question did not require 3 although in comments IgorBelegradek and IanAgol speculate in that direction.
There are some recent papers on variants on Thompson's group using the group $\mathbf{Z}+\alpha\mathbf{Z}$, $\alpha$ irrational, in lieu of $\mathbf{Z}[1/2]$ (https://arxiv.org/abs/1806.00108, https://arxiv.org/abs/2006.02401). Might they be candidates? (Btw for a group of homeomorphisms, "contain" $F$ has its literal meaning distinct from the one you have in mind, so I'd rephrase 4. as "...not contain any isomorphic copy of Thompson's group $F$".)
@YCor: I doubt it, it is hard to keep $F$ out of a Thompson-like group, see for example Theorem 1.1 of https://arxiv.org/abs/1610.04099. But more to the point, 4 is an attempt at formalizing "is not related to Thompson groups", so if there was a Thompson-like construction that avoided $F$ I would alter my question accordingly ;-). I don't see a big risk of misunderstanding "contain $F$" since my question is about abstracts groups, but I'll modify it nonetheless.
Very interesting reference! I ignored that $F$ was so unavoidable.
@PauPlummer: I wouldn't ask it if it was on an open problem list if that is what you mean. But I have talked to people who might know and didn't. I am not sure how I would know for sure that a question is "open" (if it's not, like, the Riemann Hypothesis or existence of non-residually finite hyperbolic groups).
Did you try central extensions of $F$?
@Tox: No, good idea. I know that $H^2(F) = \mathbb{Z}^2$ but I will need to think about what the extensions look like. The critical question is of course about property (4).
Here is an update on central extensions: the cohomology of $F$ is nicely described by Brown. We may restrict to $F'$ which has cyclic second cohomology, since it contains copies of $F$. The central extensions of $F$ that give rise to the ones of $F'$ can be worked out of this to be $\langle A, B, Z \mid [A,Z], [B,Z], [AB^{-1},A^{-1}BA] = Z^r, [AB^{-1},A^{-2}BA^2] \rangle$ for $r \in \mathbb{Z}$.
@Tox: The answer to you suggestion is: the central extensions of $F$ contain $F$, so no candidates there. The infinite version of the above presentation has the relations $X_i^{-1} X_{i+1} X_i = X_{i+2}Z^r$ but $X_i^{-1}X_jX_i = X_{j+1}$ for $j > i+1$. As a consequence the subgroup generated by the $X_{2i}^2$ is isomorphic to $F$. And then it contains a copy of $F$ that lies in the commutator subgroup. [Both of these facts were pointed out by Matt Zaremsky.]
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2025-03-21T14:48:31.331142
| 2020-06-20T09:55:25 |
363614
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Stack Exchange
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Summing infinitely many infinitesimally small variables makes sense in algebra
There is an identity $e^x=\lim_{n\to \infty} (1+x/n)^n$, and I always thought it is a purely analytic statement. But then I discovered its curious interpretation in pure algebra:
Consider the ring of formal infinite sums of monomials in infinitely many variables $\varepsilon_1, \varepsilon_2,\ldots$ satisfying $\varepsilon_i^2=0$.
$$
R=\mathbb{Q}[\![\varepsilon_1, \varepsilon_2, \ldots]\!]/(\varepsilon_i^2: i=1,2,\ldots).
$$
Then the sum $x=\sum_{i=1}^\infty \varepsilon_i$ makes sense and is not infinitesimally small, in fact we have
$$
x^n = n! \sum_{1\leq i_1<i_2<\ldots<i_n} \varepsilon_{i_1} \cdots \varepsilon_{i_n}.
$$
So the ring of polynomials $\mathbb{Q}[x]$ embeds into $R$. Moreover, in $R$ we have the identity
$$
\prod_{i=1}^\infty(1+\varepsilon_i) = \sum_{n=0}^\infty \frac{x^n}{n!}.
$$
So somehow we multiplied infinitely many elements infinitely close to $1$ and managed to get away from $1$ and obtain the right answer.
I was wondering if this is well-known and if there are applications of this idea. For instance, one can probably use it to recover the formal neighborhood of $1$ in an algebraic group from the Lie algebra.
In positive characteristic the right hand side doesn't make sense, but the left hand side still does. In fact, symmetric functions in $\{\varepsilon_i\}$ form a ring with divided power structure. Can one build $p$-adic cohomology theories based on this idea instead of divided power structures?
Related to the Trotter product formula (or Lie-Trotter-Kato) of use to physicists, I believe.
You start with an identity and leave it hanging. You could delete it without changing the title or rest of post. But since you put it first, it sounds like you want it addressed. You could make sense of that identity in $\mathbb Q[![x]!]$. It is false there, because the coefficient of $x^2$ is $(n-1)/2n$, which does not converge in $\mathbb Q$. As @EBz says, $\mathbb Q[![x]!]$ embeds in your ring, so it does not change the notion of convergence, so it is still false.
@BenWieland I don't understand your comment. Did you expect $e^x=\lim_{n\to\infty} (1+x/n)^n$ to hold in $\mathbb{Q}[[x]]$? Of course not, as you explain the limit doesn't even exist. This limit exists in the usual topology on $\mathbb{R}[[x]]$ or $\mathbb{R}$ for fixed $x$. The point of the first identity, because it is well known, is a motivation for the rest. Are you asking if there is a direct connection between this formula and the formula with epsilons?
@TomCopeland The analogue of the Trotter formula is $\prod_i (1+(A+B)\varepsilon_i) = \prod_i (1+ A \varepsilon_i)(1+ B \varepsilon_i)$ for some elements $A$,$B$ of a Lie algebra, which in this setting holds trivially because $\varepsilon_i^2=0$. Note that the products are not commutative. It is natural to ask for a proof of Baker-Campbell-Hausdorff, but in this setting another formula, Zassenhaus formula (https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula#Zassenhaus_formula) follows more easily, by reordering the product.
One has an identity of formal power series $e^{a+b} = e^a e^b$ so $e^{ \sum_{i} \epsilon_i} = \prod_i e^{ \epsilon_i} $ in the ring of formal power series in infinitely many variables, but $e^{ \epsilon_i}=1+\epsilon_i$ if $\epsilon_i^2=0$.
This identity is not a perfect analogue of the calculus identity because it doesn't reflect that there are infinitely many variables, just that the variables are infinitely small. The same identity holds for any finite number of variables that square to zero.
@WillSawin My question wasn't about proving the identity, it was about its uses. In any case, if you define the exponential using $\prod_i (1+\varepsilon_i)$, then the statement $e^{a+b}=e^a e^b$ is evident. About your second remark: the point to have infinitely many variables is to embed the polynomial ring: $\mathrm{Spec}(R)$ is dense in $\mathrm{Spec}(\mathbb{Q}[x])$ and $\mathrm{Spec}(\mathbb{Q}[[x]])$. For instance, it is a useful principle in comm. algebra: if you need to prove an identity between rational functions it is enough to prove identity of formal power series.
@WillSawin for instance, here is a proof of the Taylor expansion for rational functions. I only use $f(z+\varepsilon)=f(z)+f'(z)\varepsilon$ (for $\varepsilon^2=0$), which is actually the definition. Applying iteratively to $f(z+x)=f(z+\varepsilon_1+\varepsilon_2+\cdots)$ we obtain the following expansion: $f(z+x)=\sum_{n=0}^\infty f^{(n)}(z) \sum_{i_1<\cdots<i_n} \varepsilon_{i_1}\cdots \varepsilon_{i_1}$. In view of $x^n=n! \sum_{i_1<\cdots<i_n} \varepsilon_{i_1}\cdots \varepsilon_{i_1}$ this is Taylor expansion. I didn't have to divide in the proof, which may be useful in char=p.
@AntonMellit I was trying to argue that your identity is not much different from the usual identity $e^{a+b} =e^{a}e^b$ of formal power series. The characteristic $p$ idea is good, but note that the polynomial algebra doesn't embed into your algebra in characteristic $p$, precisely because $x^n =n!\sum_{i_1< \dots < i_n} \epsilon_{i_1}\dots \epsilon_{i_n}$.
@WillSawin From a naive point of view, I like this approach maybe just for aesthetic reasons. When I studied physics, everyone was using things like $f(x+\Delta x)$, and the integral was a sum of deltas. But in analysis it was emphasized a lot that these sort of things "are not allowed", they are not rigorous. So I find it surprising that you can go quite far by using the naive approach of physicists (maybe, just being more explicit about $(\Delta x)^2=0$), if you restrict yourself to pure algebra. For a non-commutative application see my comment about Zassenhaus formula above.
@WillSawin About characteristic $p$ things are a little bit more subtle here. There is this notion of the crystalline site, which uses divided power structures, which when I tried to understand I found unmotivated. I am hoping to find a "physical motivation" through this approach. Maybe something similar is going on here https://arxiv.org/abs/1803.00812
See "Dual Numbers and Operational Umbral Methods" by
Nicolas Behr, Giuseppe Dattoli, Ambra Lattanzi, Silvia Licciardi https://arxiv.org/abs/1905.09931
$\DeclareMathOperator{\ex}{ex}$
One way to look at this is through symmetric functions. To be consistent with standard notation I'll take infinitely many variables $x_1, x_2,\dots$ (instead of $\varepsilon_1, \varepsilon_2,\ldots$). I will follow the presentation in Richard Stanley's Enumerative Combinatorics, Vol. 2, page 304.
There is a homomorphism ex from the ring of (unbounded degree) symmetric functions to power series in $t$ that can be defined by $\ex(p_1) = t$, $\ex(p_n)=0$ for $n>1$, where $p_n$ is the power sum symmetric function $x_1^n+x_2^n+\cdots$. Then $\ex$ is the restriction to symmetric functions of the homomorphism on all formal power series in $x_1,x_2,\dots$ that takes each $x_i^2$ to 0 (where $t$ is the image of $p_1$). It has the property that for any symmetric function $f$,
$$\ex(f) = \sum_{n=0}^\infty [x_1x_2\cdots x_n] f \frac{t^n}{n!},$$
where $[x_1x_2\cdots x_n] f$ denotes the coefficient of $x_1x_2\cdots x_n$ in $f$. In particular, $\ex(h_n) = \ex(e_n) = t^n/n!$ where $h_n$ and $e_n$ are the complete and elementary symmetric functions.
This idea is well known and is very useful in enumerative combinatorics. It allows one to derive exponential generating functions for objects with distinct labels (e.g., permutations or standard Young tableaux) from symmetric function generating functions for objects with repeated labels (e.g., words or semistandard tableaux). There are related homomorphisms that preserve more information; see, for example, section 7.8 of Stanley.
So when you say "ring of (unbounded degree) symmetric functions", this is bigger than what most people use "ring of symmetric functions" to mean, right? E.g., $\sum_{i=0}^{\infty}p_i$ is in the ring you are considering?
@SamHopkins. Yes, this is the ring usually denoted $\hat\Lambda$.
Yeah, in combinatorics this is quite natural because if you interpret the ring of symmetric functions as the ring of representations of symmetric groups, then $\operatorname{ex}$ sends a representation of $S_n$ to its dimension times $t^n/n!$. So this is precisely the map that allows you to "forget" the $S_n$ action when you don't care about it.
Nice answer! Going the other way, thinking of the exponential specialization as evaluation in the quotient ring where $x_i^2 = 0$ makes me feel I understand it much better.
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2025-03-21T14:48:31.331703
| 2020-06-20T10:02:35 |
363615
|
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|
Stack Exchange
|
Lowerbounding expectation value of binomial tail
I'm trying to find a lower bound for the following expression for $q\ge p$:
$$f(q,p,n) := \sum_{v=0}^n \sum_{k=v}^n \binom{n}{v} \binom{n}{k}q^v(1-q)^{n-v}p^k(1-p)^{n-k}.$$
It can be thought of as the expectation value of the tail of the binomial distribution with parameter $p$ taken with respect to the binomial distribution with parameter $q$, an expectation value of a p-value.
It is easy to see that $f(q,0,n) = (1-q)^n$ and $f(1,p,n) = p^n$, and with a bit of work one can show that $\lim_{n \to \infty} f(\frac12,\frac12,n) = 1/2$, so I conjecture that the following bound holds:
$$f(q,p,n) \ge \frac12(1-(q-p))^n.$$
I have no idea how to prove it, though. Applying the standard lower bounds to the binomial tail leads to a terribly loose bound, and I couldn't use generating functions to get a bound on the overall expression. On the other hand, one can easily prove the upper bound $(1-(q-p)^2)^n$ using generating functions, so perhaps there is also some easy trick that applies to the lower bound.
I don't know if it is helpful, but one can also show that $f(q,p,n) = f(1-p,1-q,n)$.
Isn't it just the probability that a Binomial(p,n) is greater than or equal to an independent binomial(q,n)? This gives immediately the cases you wrote (no need for limits or work...), and a bit of work should give you a decent bound (with actually the correct exponential rate).
Yes, it is. I don't see how that helps, though (I'm a physicist, not a mathematician). For example, the exact expression for $f(\frac12,\frac12,n)$ is $\frac12 + 4^{-n}\binom{2n}{n}/2$. How can one get that without doing work?
By symmetry. You have two identically distributed variables $A,B$ and you ask for $P(A\geq B)$, which equals $1/2+P(A=B)$. The local CLT tells you that the probability that $A=B$ is of order $1/\sqrt{n}$
As to the actual question you asked - asymptotics then become very easy, it is a large deviations question, since the means are different. Exact computations I will leave to others, although I note that you are asking for $P(B\geq 0)$ where $B=\sum_{i=1}^N W_i$, $W_i$ iid, and $EW_i=p-q$ and variance of $W_i$ equal $2q(1-p)$. Asymptotics now follow...
I still don't see how that helps. For example, I can directly apply the Chernoff bound to $P(B\ge 0)$, which gives me the upper bound $f(q,p,n) \le (\sqrt{pq}+\sqrt{(1-p)(1-q)})^{2n}$, but this is exactly the same bound I got via generating functions. And the difficult is anyway finding a lower bound. Do you have any specific anti-concentration inequality in mind that would work? Also, the variance of $W_i$ is $p(1-p)+q(1-q)$.
This answer gives asymptotics as $q-p=\alpha>0$ fixed and $n\to\infty$. Other variants are of course also possible.
As mentioned in the comments, the question can be rephrased as follows. Let $W_i$ be iid random variables so that $P(W_1=1)=p(1-q)$, $P(W_1=-1)=q(1-p)$ and $P(W_1=0)=1-p-q+2pq$.
With $B=\sum_{i=1}^n W_i$, the question is to evaluate the probability $P(B\geq 0)$. Now, the log moment generating function of $W_1$ is
$$\Lambda(\lambda)=\log E(e^{\lambda W_1})=\log(1-p-q+2pq+p(1-q)e^\lambda+q(1-p)e^{-\lambda}).$$
The Legendre transform $\Lambda^*(x)=\sup_{\lambda}(\lambda x-\Lambda(\lambda))$ satisfies, if I am not mistaken,
$$\Lambda^*(0)=-\log\left(\left(\sqrt{pq} + \sqrt{(1-p)(1-q)}\right)^2\right)=:\beta.$$
Now, since $\alpha>0$, $EW_1<0$, and so we are in the large deviations regime.
By the Bahadur-Rao theorem,
$$P(B\geq 0)\sim \frac{c}{\sqrt{n}}e^{-n\beta},$$
where
$$c = \frac{\sqrt{pq} + \sqrt{(1-p)(1-q)}}{\sqrt{4\pi}\left(1-\sqrt{\frac{p(1-q)}{(1-p)q}}\right)\big(pq(1-p)(1-q)\big)^\frac14}. $$
The case $q=p$ is just a local CLT. If $q,p$ depend on $n$, asymptotics can be read from the proof of the BR theorem.
For more details see the original paper of Bahadur-Rao (1960), and the exposition in Dembo-Zeitouni LD textbook (Theorem 3.7.4).
It's nice to know the asymptotics, but I still don't see how that helps finding a lower bound. I checked your book, and the theorem doesn't provide a bound, it only talks about the limit. Also, is there any reason you wrote $P(B>0)$ instead of $P(B\ge 0)$? Since you're taking $\Lambda^*(0)$ you must mean the latter, which is indeed what I asked about.
Thanks for pointing out the $>$ vs $\geq$ - the asymptotics is actually the same. The lower bound is quantitative - that is, you can actually write $P(B\geq 0) =ce^{-n\beta}/\sqrt{n}(1+C_1/\sqrt{n})$, and with enough diligence get an upper bound on $C_1$. Note that the lower bound involves computing, under the tilted measure, an upper bound on the complementary event,
Here is a sketch of how to do it quantitative: make a tilt of the law of $W_1$ to have the mean at +ϵ. This will give you a factor$e^{-n\Lambda^(\epsilon)}$ and then a lower bound is $e^{-n \Lambda^(\epsilon) - n\lambda \delta} P_\lambda^n ([0,(\epsilon+\delta)n])$ which is greater or equal to $e^{-n(\Lambda^*(\epsilon)+\delta\lambda)} (1-P_\lambda^n([0, (\epsilon+\delta)n]^c)$
Now, by Chebycheff, $P_\lambda^n([0,(\epsilon+\delta)n]^c)\leq e^{-c'(\epsilon \wedge \delta)n}$ for some $c'$ that is explicit and one can compute. For $n$ such that the RHS is smaller than $1/2$, this completes the lower bound.
I calculated the value of $c$ and added it as an edit to your answer, but some random prick deleted it saying that it does not matter. For the record, it is $c = \frac{\sqrt{pq} + \sqrt{(1-p)(1-q)}}{\sqrt{4\pi}\left(1-\sqrt{\frac{p(1-q)}{(1-p)q}}\right)\big(pq(1-p)(1-q)\big)^\frac14}.$
I just noticed a few minutes ago that you did, and was going to check it when I get to it. Did you use the formula for the lattice case? I suspect that there is a logarithm missing, isn't there?
Yes, I used the formula for the lattice case, with $d=1$. I don't think there's a logarithm missing, $\eta$ is a logarithm, but it only appears through an exponential.
There is an $\eta$ in the numerator, no? and the $\eta$ is a logarithm, being the tilt?
There is an $\eta$ both in the numerator of the lattice formula, but it cancels out with the $\eta$ in the expression of $J_n$. It is given by $\eta=\log\sqrt{\frac{q(1-p)}{(1-q)p}}$.
OK, I approved the edit. Thanks. And please in the future do not use the language you used earlier in comments. Editors are volunteers.
|
2025-03-21T14:48:31.332135
| 2020-06-20T10:45:52 |
363617
|
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"url": "https://mathoverflow.net/questions/363617"
}
|
Stack Exchange
|
Ito's Lemma (CVF) on product of Poisson processes
I have the following stochastic differential equation:
$da(t)=\{r(t)a(t)+w(t)−pc(t)\}dt+βa(t)dq(t)$,
with $q(t)$ a Poisson process with arrival rate $λ$ and its increment $dq(t)$ is denoted by:
$dq(t)=0$ with probability $1−λdt$
$dq(t)=1$ with probability $λdt$
I know that using the Change of Variable formula (CVF) (=Ito's Lemma on Poisson processes), we can write the differential $dV(a(t))$ as:
$dV(a)=V'(a)\{ra+w−pc\}dt+\{V(a+βa)−V(a)\}dq$
with $V'(a)$ the differential of $V$ with respect to $a$.
Now, if I modify the initial SDE and write it:
$da(t)=\{r(t)a(t)+w(t)−pc(t)\}dt+βa(t)dq(t)+γa(t)dq_tdq^i_t$
with $dq^i(t)$ the increment of a Poisson process $q^i_t$ similar to the previous one:
$dq^i(t)=0$ with probability $1−λ^idt$
$dq^i(t)=1$ with probability $λ^idt$
What would be the result of applying the the CVF on that new SDE?
Replacing $dq_tdq^i_t$ by a process $dQ$,
$dQ(t)=0$ with probability $1−λλ^idt^2$
$dQ(t)=1$ with probability $λλ^idt^2$
I would say:
$dV(a)=V′(a)\{ra+w−pc\}dt+\{V(a+βa)−V(a)\}dq+\{V(a+γa)−V(a)\}dQ$.
But is this reasoning correct?
|
2025-03-21T14:48:31.332233
| 2020-06-20T10:45:53 |
363618
|
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|
Stack Exchange
|
Positive vector in the kernel of an skew-symmetric incidence matrix
Let $G=(V,A)$ be an oriented graph, stronlgy connected with $n\in\mathbb{N}^*$ vertices. Let $M\in\mathcal{M}_n(\mathbb{R})=(m_{i,j})$ be an skew-symmetric matrix of size $n$ and rank $r$, such that for all $i,j\in V$: $i\longrightarrow j\Longleftrightarrow m_{i,j}>0$ (hence $M$ is a skew-symmetric incidence matrix of $G$). Is there any necessary and sufficient condition on $G$ and $r$ ensuring that there exists a positive vector in $\text{Ker}(M)$?
I've tried several matrices for $n\in\{3,4,5\}$ and it seems that, given $G$ and $r$, the fact that there exists a positive vector in $\text{Ker}(M)$ does not depend on the exact values of the coefficients $m_{i,j}$, but I do not know how to predict that such a vector exists. I have just noticed that given $G$, the answer of the question can vary with $r$.
Sounds like a manifestation of (a weighted) Kastelyn's Theorem. $Det(M)$ is the square of the Pfaffian. However in some cases the latter is a sum of terms corresponding to matchings, involving the weights $m_{ij}$ and all contributing with the same sign. See, e.g., https://arxiv.org/abs/1409.4631
|
2025-03-21T14:48:31.332346
| 2020-06-20T11:30:46 |
363619
|
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"Ali",
"Emil Jeřábek",
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|
Stack Exchange
|
Lob theorem for Robinson arithmetic
If i'm not wrong, the theory which Lob theorem applies to should be sufficiently strong, satisfying 3 "derivability" conditions, like PA.
$Q$ is the Robinson arithmetic.
I'm afraid $Q$, is not sufficiently strong, so the Lob theorem doesn't help for the following question :
If $ Q \vdash (\sigma \leftrightarrow Prb{Q}\sigma) $, does $Q \vdash \sigma$?
if the answer is no, the counterexample should be given, I assume.
Remark: my question is directly about excercise 3.7.1 of Enderton Introduction to logic.
I now had I look at Enderton’s book, and I am quite puzzled: as far as I can tell, the contents of the book gets nowhere near the tools needed to (affirmatively) solve the exercise.
@Matt F: I checked the 1972 version of the book and, no, it wasn't. you made a good point. But still i don't know why Enderton has put this problem in the exercises. As Emil said, the book only covers 3 lob conditions which doesn't help.
The answer is yes, and indeed, $Q$ is enough for Löb’s theorem:
Theorem. Let $T\supseteq Q$, and let $\tau\in\Sigma_1$ define an axiom set for $T$ in $\mathbb N$. Then
$$T\vdash(\Box_\tau\phi\to\phi)\implies T\vdash\phi$$
for all sentences $\phi$, where $\Box_\tau$ denotes the formalized provability predicate for $\tau$.
This was proved by Pudlák [1], even in a stronger form using restricted provability predicates (he states it for the second incompleteness theorem, but the argument for Löb’s theorem is essentially the same).
I will sketch the proof below. We rely on a few properties of Buss’s theory $S^1_2$ (see e.g. Chapter V in Hájek and Pudlák [2]): $S^1_2$ is a finitely axiomatizable fragment of arithmetic, it is interpretable in $Q$ on a definable cut, and it proves suitable versions of the usual Hilbert–Bernays–Löb derivability conditions.
Assume that $T\vdash\Box_\tau\phi\to\phi$, and let $I$ be an interpretation of $S^1_2$ in $Q$ on a definable cut. We have
$$T+\neg\phi\vdash\bigl(\neg\Box_\tau\phi\bigr)^I$$
as $\Pi_1$ statements are preserved downwards to cuts. Since $S^1_2$ is finitely axiomatizable, there exists a finite theory
$$\tag{$*$}U\subseteq T+\neg\phi$$
such that
$$\tag{$**$}U\vdash(S^1_2+\neg\Box_\tau\phi)^I.$$
We have the following derivability conditions for all sentences $\psi$ and $\chi$:
$$\begin{gather}
\tag{1}U\vdash\psi\implies S^1_2\vdash\Box_U\psi,\\
\tag{2}S^1_2\vdash\Box_U(\psi\to\chi)\to(\Box_U\psi\to\Box_U\chi),\\
\tag{3}S^1_2\vdash\Box_U\psi\to\Box_U(\Box_U\psi)^I.
\end{gather}$$
Using Gödel’s diagonal lemma, let $\nu$ be a sentence such that
$$\tag{$*{*}*$}S^1_2\vdash\nu\leftrightarrow\neg\Box_U\nu^I.$$
Then
$$\begin{align}
S^1_2\vdash\neg\nu
&\to\Box_U\nu^I&\text{by }&(*{*}*)\\
&\to\Box_U\bigl(\Box_U\nu^I\bigr)^I&\text{by }&(3)\\
&\to\Box_U\bigl(\Box_U\nu^I\to\neg\nu\bigr)^I&\text{by }&(*{*}*),(**),(1)\\
&\to\Box_U\neg\nu^I&\text{by }&(2)\\
&\to\Box_U\bot&\text{by }&(2)\\
&\to\Box_\tau\phi,
\end{align}$$
where the last step follows using the formalized deduction theorem from the fact that the axioms of $U$ consist of $\neg\phi$ and a finite list of axioms that satisfy $\tau$ (provably in $S^1_2$ as $\tau$ is $\Sigma_1$).
Thus,
$$\begin{align}
S^1_2+\neg\Box_\tau\phi&\vdash\nu,\\
U&\vdash\nu^I,&\text{by }&(**)\\
S^1_2&\vdash\Box_U\nu^I,&\text{by }&(1)\\
S^1_2&\vdash\neg\nu,&\text{by }&(*{*}*)\\
U&\vdash\neg\nu^I,&\text{by }&(**)\\
U&\vdash\bot,\\
T&\vdash\phi.
\end{align}$$
References
[1] Pavel Pudlák: Cuts, consistency statements and interpretations, Journal of Symbolic Logic 50 (1985), no. 2, pp. 423–441, DOI: 10.2307/2274231.
[2] Petr Hájek, Pavel Pudlák: Metamathematics of first-order arithmetic, Springer, 1994, 2nd ed. 1998, 3rd ed. Cambridge Univ. Press 2017.
|
2025-03-21T14:48:31.332702
| 2020-06-20T12:42:49 |
363624
|
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|
Stack Exchange
|
When minimal prime ideals are maximal with respect to not containing an element
Let $\{ P_i \}$ be the set of all minimal prime ideals of a commutative ring $R $. Is there any conditions on $R $ under which there exists an element $x\in R $ such that $P_i $ is an ideal of $R $ maximal with respect to not containing $x $ for each $i $?
If $A$ is a domain and the intersection of nonzero ideals of $A$ is nonzero, then $A$ is a field. Indeed, if $y$ is nonzero in this intersection, $y\in y^2A$, say $y=y^2a$, so $1=ya$ using that $A$ is a domain, and hence $y$ is invertible, and hence this intersection is $A$, so $A$ is a field. Apply this to $R/P_i$ to deduce that each $P_i$ is maximal. Conversely if each $P_i$ is maximal, then $x=1$ works. So this holds iff $A$ has Krull dimension $0$.
|
2025-03-21T14:48:31.332781
| 2020-06-20T12:58:06 |
363626
|
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|
Stack Exchange
|
Parabolic inductions for p-adic reductive groups
So I wish to ask for articles/comments surveying conjectures and theorems about parabolic induction for p-adic (non-archimedean case) reductive groups, and how local Langlands behaves under such. That is:
For a quasi-split group G, what we know about the sub-representations of principle series.
Same setting, but what we know about the (enhanced) Langlands parameters they correspond to, both conjectures and theorems.
Same questions in 1 & 2, but for parabolic induction of supercuspidal representations from Levi assumed local Langlands for that Levi is understood.
The only things I know is the articles of Bernstein-Zelevinsky that addresses part of 1 and somewhat 2 for GL_n, as well as recent work of Aubert-Moussaoui-Solleveld (and extensions of some of them with others) about a nice conjectural framework of 3.
Any comment is greatly appreciated! I am also under the impression that the above questions are largely known for classical groups given the local Langlands established by Arthur and many others, but love to know a reference good for quick reading. If more is known in the local function field case, it will be fantastic to learn too. Thank you very much!!!
|
2025-03-21T14:48:31.332893
| 2020-06-20T14:53:20 |
363634
|
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"Arno Fehm",
"Dror Speiser",
"Fedor Petrov",
"LSpice",
"Manikanta Abburi",
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|
Stack Exchange
|
Is the congruence $x ^ x \equiv a\pmod p$ always solvable?
Let $p$ be a prime and let $a$ belong to $\mathbb Z$. How do I prove that the congruence $x ^ x \equiv a\pmod p$ is solvable?
This is not correct; $x^x \equiv 2 \pmod3$ has no solution. (Also, please put the question in the body, not just the title. I have edited accordingly.)
@Lspice $x=5$ works, does not it?
@FedorPetrov, good point. I was just checking $0 \le x < p$, but I forgot you can't reduce the exponent modulo $p$.
x^x ≡ 2(mod3) has a solution for x = -1.
Chinese remainder theorem combined with Fermat's little theorem should give the answer (choose $x$ to be $a$ mod $p$ and $1$ mod $p-1$).
Сhoose $x$ congruent to $a$ modulo $p$ and to $1$ modulo $p-1$, say $x=1+(p-1)(p-a+1)$.
The formula also works for $a=0$ :)
We'd probably better be careful saying $x = 0$ for $a = 0$, lest we run into concerns about the meaning of $0^0$. Maybe better go with @DrorSpeiser's suggestion and take $x = p^2$, or some other non-$0$ multiple of $p$?
@DrorSpeiser indeed, shorten the post
|
2025-03-21T14:48:31.333012
| 2020-06-20T16:33:37 |
363643
|
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"Adittya Chaudhuri",
"Arun Debray",
"Dmitri Pavlov",
"Praphulla Koushik",
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Stack Exchange
|
Results in “generalised smooth spaces” that did not hold in the case of smooth manifolds
Consider the category of smooth manifolds $\text{Man}$. I quote from n-lab page:
Manifolds are fantastic spaces. It’s a pity that there aren’t more of them.
I understand that this category $\text{Man}$ is not well behaved in more than one sense or do not have enough objects, for it to be
closed under pullback,
to have mapping space, an appropriate smooth structure on $\text{Map}(X,Y)$ for manifolds $X$ and $Y$.
Then, people added more spaces to the category of manifolds, in an attempt to make sure the resulting category has (some) of the nice properties which the category $\text{Man}$ did not had. Some examples are
Chen spaces (On the proof of "Mapping space is a Chen space"),
Differentiable spaces (I saw the first in the paper, section $2.7$) which are sheaves over the category $\text{Man}$ that are Differentiable stacks over the category $\text{Man}$ (recall that, any manifold is a sheaf over the category $\text{Man}$ that are Differentiable stacks over the category $\text{Man}$).
Frölicher spaces. These are introduced to have a Cartesian closed category (please correct me if I have misunderstood something).
Question : Are there any (What are the) results that hold in these generalised spaces whose counterparts does not hold true in the set up of smooth manifolds?
There is one result (Lemma $2.35$ in above paper) I am aware of that holds true for Differentiable spaces but there is no appropriate counterpart for smooth manifolds.
Sub questions :
It looks like diffeological spaces are introduced not to “enrich” (not sure if it is correct word) the category of manifolds, but actually to study sheaves on the category of manifolds. Is that correct? I am not sure to what extent this question is making sense, so feel free to ask for more clarification or ignore it.
I also observe similarity with the notion of “Algebriac spaces”. Those were also (roughly) defined (similar to Differentiable spaces) as sheaves of particular kind (over some appropriate site). I think there are more than a handful of results that holds true in Algebriac spaces but not in the category $\text{Sch}/S$. You can also add them, but I am not sure if I can appreciate them enough.
I could not find Lemma 2.7. In the link you provided it is Example 2.7.
Fixed @AdittyaChaudhuri it is Lemma 2.35
Maybe not exactly an answer to your question, but Freed-Hopkins prove that Chern-Weil forms are the only natural differential forms associated to a connection on a principal bundle. To both precisely formulate and to prove the theorem, they require the language of simplicial sheaves on the site of smooth manifolds. I don't know how that notion of "generalized manifold" compares to the ones you're asking about, though.
@PraphullaKoushik According to the question above you mentioned (3) "Frölicher spaces. These are introduced to have a Cartesian closed category (please correct me if I have misunderstood something)."... But category of Chen spaces(1) is also Cartesian Closed..... Also I guess(though don't have the reference or proof right now..so may be wrong) the same is true for Differentiable space(2).... So it is for sure that "forming a Cartesian Closed Category" cannot be the reason for introducing Frölicher spaces .
@AdittyaChaudhuri Yes, there are other categories which are Cartesian closed (and which have embedding from smooth manifolds)... I did not say it is the only reason. I only know that reason, so I mentioned it :)
@ArunDebray Hey, thanks.. I think your reference to Freed-Hopkins paper is very relevant (could be one of the motivations for Dimitri Pavlov's answer).. I have heard about that before but I have not read... It looks like it is now time to read that :)
There are many such results.
Consider some smooth manifolds M and N.
The internal hom Hom(M,N) is a sheaf on smooth manifolds.
We can compute its tangent bundle,
and it turns out that the tangent space at some point f in Hom(M,N),
i.e., f:M→N is a smooth map, equals the vector space
of smooth sections of the vector bundle f*TN.
This is the expected result, but the setting of sheaves allows us to make it completely rigorous and precise with minimal technicalities.
Now take M=N and consider the open subobject of Hom(M,M) consisting of diffeomorphisms.
This is a group object (i.e., an infinite-dimensional Lie group)
and its Lie algebra is precisely the Lie algebra of vector fields on M.
Differential k-forms form a sheaf Ω^k on smooth manifolds.
In particular, morphisms Hom(M,N)→Ω^k are differential k-forms on the infinite-dimensional space of smooth maps M→N.
We also immediately obtain the de Rham complex on Hom(M,N)
in the same manner, and it satisfies the expected properties.
Liekwise, we have a sheaf of groupoids B_∇(G) of principal G-bundles
with connection.
Maps Hom(M,N)→B_∇(G) are principal G-bundles with connection
over the infinite-dimensional space of smooth maps M→N.
Hopkins and Freed compute the de Rham complex of B_∇(G),
and it turns out to be the vector space of invariant polynomials
on the Lie algebra of G.
This means, for instance, that you can immediately start
computing Chern–Weil forms of principal G-bundles with connection
on Hom(M,N), for example.
Now, we can also take G to be any group object in sheaves,
such as, for example, the group Diff(M) of diffeomorphisms
of M considered above.
This immediately allows us to consider principal G-bundles
with connection for such groups.
Other objects that can be encoded in this setting
include the (higher) sheaves of bundle (n-1)-gerbes with connection
and structure abelian Lie group A, denoted by B_∇^n(A).
Morphisms M→B_∇^n(A) are precisely bundle (n-1)-gerbes with connection over M.
Now you can talk about bundle (n-1)-gerbes with connection over Hom(M,N).
The Cheeger–Simons differential refinement of the Chern character
in this language is a morphism B_∇(G)→B_∇^n(A), etc.
So in particular, not only de Rham cohomology, but also differential cohomology make sense in this framework.
This is interesting.. For manifolds $M.N$, as the mapping space $\text{Map}(M,N)$ is not expected to have a nice smooth structure, you would like to see $M$ and $N$ as objects of the category $Sh(\text{Man})$ and taking then take the internal-hom object of $M,N$ (I do not know why it exists, I will check details; I guess here "the tensor" is tensor product of sheaves)..
I have some other questions, but I should at least read the Freed-Hopkins paper first :) Thanks for this answer (really).. :) :)
@PraphullaKoushik: Just to be clear, Hom(M,N) denotes the sheaf whose S-points are smooth maps S⨯M→N. In particular, the points of Hom(M,N) are precisely the smooth maps M→N. Hom is right adjoint to the (categorical cartesian) product of sheaves of sets, defined objectwise: (F⨯G)(S)=F(S)⨯G(S).
Yes, I saw that explanation only recently.. I am still trying to look at the motivation behind such description of the inner-hom/internal-hom $\text{Hom}(M,N)$.. I am reading MacLane and Moedijk's Sheaves in Geometry and Logic (3.6 first properties of categories of sheaves).. Do you have any suggestion on any other book that I should be looking at?
@PraphullaKoushik: Patrick Iglesias-Zemmour, Diffeology. It's all about diffeological spaces (and sheaves of sets on manifolds).
That book looks interesting (too lengthy though:)) but I do not see where does they talk about the internal hom description :O
@PraphullaKoushik: Sections 1.57–1.65.
Sorry. I was looking at table of contents and was lost in 8+ table of contents. Thanks for the reference. I feel like I might enjoy reading parts of this book.
For anyone else looking for more references, they can see “17.7 Internal Hom” of Masaki Kashiwara and Pierre Schapira’s book titled “Categories and Sheaves”.
|
2025-03-21T14:48:31.333521
| 2020-06-20T17:31:47 |
363647
|
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|
Stack Exchange
|
A question about Dedekind schemes and proper morphisms
The following is Exercise 15.3 of Görtz-Wedhorn Algebraic Geometry I:
Let $S$ be a Dedekind scheme with function field $K$ and let $f: X\to S$ be a proper morphism of schemes. Then the canonical map $X(S)\to X(K)$ is a bijection.
For example, the map $X(S)\to X(K)$ being injective means ${\rm Spec}\ K\to S$ is an epimorphism of schemes. I'm no idea to prove it, even for $S$ affine. For subjectivity, I can only see (using valuation criterion for being proper) that a morphism ${\rm Spec}\ K\to X$ extends to a morphism ${\rm Spec}\ \mathscr{O}_{S, y}\to X$ ($y$ being the image of ${\rm Spec}\ K\to X\to S$ but don't know how to extends to the whole of $S$.
I'm also wondering if the maps need to respect morphisms to $S$, namely if we need to consider $X_S(S)\to X_S(K)$ instead of $X(S)\to X(K)$.
Just to tell that I asked the authors and they really mean $X_S(S)\to X_S(K)$.
Injectivity is because $X$ is separated. The locus where two morphisms $S \to X$ agree is a closed subscheme and if it contains the generic point, it's everything.
For surjectivity, we can "spread out" the morphism $Spec K \to X$ to a morphism $U \to X$ from an open subset $U \subset S$ and then use the valuative criterion to fill in the finitely many missing point.
Edit: Here are some more details. So spreading it out is the general procedure where if we have an integral ring $R$ with fraction field $K$, a finite type scheme $X$ and a $K$ point of $X$, we can extend it to a $R_{f}$ point because there will only be finitely many denominators. For an easy example, suppose $X = \mathbb A^1$ and $R = \mathbb Z$ and the rational point is $t \to 1/2$. Then, there is an obvious way to think of this as a $\mathbb Z[1/2]$ point.
So given a rational point of $X$, we can extend this to a map from an open subset $U \subset S$ to $X$. Since $S$ is a Dedekind scheme, there are finitely many height one primes in the complement of $U$. Consider the local ring $R_{\mathfrak p}$ at one of these primes which is a DVR.
By the valuative criterion, we can get a $R_{\mathfrak p}$ point of $X$. Again, by spreading out, we can extend this to a map from $V \to X$ for $V$ another open subset of $S$ that contains $\mathfrak p$. On the other hand, the maps from $V \to X$ and $U\to X$ match on their intersection (by the same argument as for injectivity) and so we can extend this to a map $U\cup V \to X$.
Repeating this process for any points still left over, we get a map $S \to X$ that extends our rational map.
Thanks! Injectivity agree! For surjectivity, I don't know how to "spread out" and "fill in the finitely many missing point", could you give more details? And do you use $X(S)\to X(K)$ or $X_S(S)\to X_S(K)$?
OK, I think by "spread out" you mean extending morphisms as in Proposition 10.52 of Görtz-Wedhorn.
I think it's useful to try and work it out yourself. Anyway, I will try and write more in a hour or two if you are still stuck (busy right now).
OK, thanks, I will try to work it out myself and tell you if I still stuck.
I used some results on extending rational morphisms and schematically density to show $X_S(S)\to X_S(K)$ is a bijection, but I still don't know if you are meant $X(S)\to X(K)$. Even so, I'm still wondering how to use the valuative criterion to fill in the finitely many missing point; I can't make it work. It would be grateful if you can also add that to your answer.
Hi, I added details. Let me know if you have questions.
I think again about your answer, I think you found some element in () maps to (), but may not under the canonical map ()→() induced by the inclusion of the generic point of $S$, e.g. that may happen if the image point $y$ of ${\rm Spec}\ K\to X\to S$ is not the generic point of $S$. Am I wrong somewhere?
No that's right. This is why everything should really be relative to the structure maps to S which answers the other question you had.
Sorry, I think I misread your comment. I think the element we found in X(S) maps to the given point in X(K) under the canonical map. I also don't think we need to use X_S(S) to X_S(K) but currently our proof works only in this case. For instance, if you allow the affine line with double origin as a dedekind scheme and take X to be the affine line over it, injectivity is no longer true in the absolute sense. (I might be making mistakes since I am tired, please check everything I'm saying)
Let us continue this discussion in chat.
|
2025-03-21T14:48:31.333822
| 2020-06-20T18:45:26 |
363652
|
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|
Stack Exchange
|
chromatic minimal cell structures
If $X$ is a finite $p$-local spectrum, then the minimal number of cells needed to construct $X$ is exactly $\dim_{\mathbb F_p} H_\ast(X,\mathbb F_p)$. Is there an analogous result in the $K(n)$-local setting?
One should be a bit careful in formulating this question. After all, the $K(n)$-local category $Sp_{K(n)}$ has a nontrivial Picard group; these are spectra $X$ with $\dim_{K(n)_\ast} K(n)_\ast(X) = 1$ which are not suspensions of the $K(n)$-local sphere spectrum $\mathbb S_{K(n)}$. There are also many possible notions of "finite". So there are several versions of this question.
First, some notation. Given a full subcategory $\mathcal C \subseteq \mathcal D$, define $\mathcal C_0 = \{0\}$, and inductively let $\mathcal C_{n+1} \subseteq \mathcal D$ be the full subcategory of objects $D \in \mathcal D$ such that there exists a cofiber sequence $C \to C' \to D$ with $C \in \mathcal C$ and $C' \in \mathcal C_n$. For $C \in \cup_{n \in \mathbb N} \mathcal C_n$, define
$$numcells_{\mathcal C}(C)$$
to be the minimal $n \in \mathbb N$ such that $C \in \mathcal C_n$; define $numcells_{\mathcal C}(X) = \infty$ if $X \not \in \cup_{n \in \mathbb N} \mathcal C_n$.
Clearly, if $X \in Sp_{K(n)}$, we have
$$numcells_{Pic(Sp_{K(n)})} (X) \geq \dim_{K(n)_\ast} K(n)_\ast(X)$$
and, more weakly,
$$numcells_{\{\Sigma^k \mathbb S_{K(n)}\}_{k \in \mathbb Z}}(X) \geq \dim_{K(n)_\ast} K(n)_\ast(X)$$
My questions are about whether equality is "always" attained.
Question:
Let $Y$ be a finite $p$-local spectrum. Then do we have $numcells_{\{\Sigma^k \mathbb S_{K(n)}\}_{k \in \mathbb Z}}(L_{K(n)} Y) = \dim_{K(n)_\ast} K(n)_\ast(Y)$?
More generally, let $X \in Sp_{K(n)}$ be such that $numcells_{\{\Sigma^k \mathbb S_{K(n)}\}_{k \in \mathbb Z}}(X)< \infty$. Then do we have $numcells_{\{\Sigma^k \mathbb S_{K(n)}\}_{k \in \mathbb Z}}(X) = \dim_{K(n)_\ast} K(n)_\ast(X)$?
In a slightly different direction, let $X \in Sp_{K(n)}$. Then do we have $numcells_{Pic(Sp_{K(n)})} (X) = \dim_{K(n)_\ast} K(n)_\ast(X)$?
I'd also be interested in these questions $T(n)$-locally (in fact, what I'm actually most interested in is the $T(n)$-local version of Question (1)) but probably such versions of the questions are much harder.
It might also be interesting to ask about $numcells_{\{\Sigma^k L_{K(n)} F(n)\}_{k \in \mathbb Z}}(X)$ where $F(n)$ is a finite type-$n$ spectrum.
This addresses none of your questions, but you (or onlookers interested in some of the subtleties you’ve incorporated into the question) might also be interested in the spectrum Y analyzed on page 75 of the Hovey-Strickland memoir.
Also, I don’t understand the difference between (3) and the start of the “Clearly…” paragraph.
@EricPeterson Ah, thanks. That was a typo.
Ah: then Y falsifies (3). :) Their eigenvalue argument is worth knowing.
@EricPeterson I see -- Hovey and Strickland show that $Y = L_{K(1)} Y$ and that $numcells_{Pic(Sp_{K(1)})}(Y) = \infty$ while $\dim_{K(1)\ast}K(1)\ast(Y) = 2$. Thanks!
|
2025-03-21T14:48:31.334020
| 2020-06-20T19:22:48 |
363655
|
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|
Stack Exchange
|
Relation between two homomorphisms from $SO(3)$ to the Möbius group $PGL(2,\mathbb{C})$
Let $f, g$ be two homomorphisms from $\mathrm{SO}(3)$ to $PGL(2,\mathbb{C})$. Does there exist $S \in \mathrm{PGL}(2,\mathbb{C})$ such that for all $X\in \mathrm{SO}(3)$ we have $g(X)=Sf(X)S^{-1}$?
Background: I have images which are stereographic projections of subparts of a sphere. According to Wikipedia the images transform into each other using Möbius transforms. I want to parametrize these transforms in terms of the rotations given by the actuator. If the above is true the parametrization would be quite simple: i can pick any homomorphism and then just need to parametrize $S$.
If you don't specify continuous homomorphism, then this is false, since the automorphism group of $PGL_2(\mathbb{C})$ contains $Gal(\mathbb{C}/\mathbb{Q})$.
There are two continuous group morphisms $SO_3\to PSL_2\mathbb{C}$ up to conjugacy: the obvious one as rotations of the Riemann sphere, and the trivial one with image the identity element. The proof: We see from the Lie algebra of $SO_3$ (cross product of vectors in $\mathbb{R}^3$) that $SO_3$ has simple Lie algebra. Hence the image $G$ of any morphism is a 0-dimensional or a 3-dimensional compact and connected group. If 0-dimensional, it is the identity element, because it is connected. So suppose that $G$ is 3-dimensional.
By $QR$ decomposition, $PSL_2\mathbb{C}$ retracts to $SO_3$, so $SO_3$ is a maximal compact subgroup, and is connected. All maximal compact subgroups are conjugate. So any maximal compact subgroup containing $G$ is a conjugate of $SO_3$. By dimension count, and connectivity, $G$ is a maximal compact subgroup, so a conjugate of $SO_3$, so unique up to conjugacy. We thus reduce to the problem of proving that $SO_3$ automorphisms are inner, which is clear from its Dynkin diagram, as it is rank one, so has a Dynkin diagram with only one node, hence no noninner automorphisms.
A much simpler linear algebra proof is probably not hard to cook up.
I have no idea what's meant by "hence no nontrivial automorphisms", since the Dynkin argument directly says the fact about the outer automorphism group (while of course there are nontrivial automorphisms, namely inner).
Why does the image have to be maximal compact? The image of the trivial homomorphism is not maximally compact
@SebastianSchulz, thanks for asking this! Certainly the argument meant to include the caveat 'non-trivial' on the morphism, but even then I can't see it. A non-trivial morphism from $\operatorname{SO}_3(\mathbb R)$ will have image isomorphic either to $\operatorname{SO}_3(\mathbb R)$ or its adjoint quotient, but is it obvious that the only way to embed either of those in $\operatorname{PSL}_2(\mathbb C)$ is as a maximal compact subgroup?
@LSpice for reasons of dimension. If $SO_3$ or its adjoint quotient is embedded, the image's dimension is the dimension of a maximal compact. The image is compact, so it must be maximal.
@JoshuaMundinger, but, still, connected components...? Some details...
@LSpice what you mean by "its adjoint quotient"? $\mathrm{SO}_3$ is its own adjoint quotient; it's abstractly a simple group. The whole thing is clear. If the (continuous) homomorphism is nontrivial, its image is 3-dimensional, compact, and since the maximal compact subgroups in $\mathrm{PSL}_2(\mathbf{C})$ are 3-dimensional and connected [the maximal compact subgroups of every connected Lie group are connected!], the image is maximal compact.
@YCor, I always mix up which SO's have centres. As to "the whole thing is clear", well, it wasn't to me, which is why I asked; but now it is, because you explained it clearly. (I didn't know that maximal compacts in the real setting are connected; it's not always true in the $p$-adic setting.)
@LSpice actually if $G$ is a connected Lie group with finitely many components and $K$ a maximal compact subgroup, then the inclusion $K\to G$ is a homotopy equivalence and $G$ is homeomorphic to $K\times G/K$, with $G/K$ homeomorphic to a Euclidean space; in particular $G$ is connected iff $K$ is connected. In the $p$-adic case only ${1}$ is connected...
@YCor, in the $p$-adic case we have the Bruhat–Tits-type scheme-theoretic models of certain compact subgroups, and it makes sense to talk about whether those underlying schemes are connected. (More prosaically, we are dealing with pro-Lie groups over finite fields, and one may ask whether the "top" quotient is the group of rational points of a connected algebraic group.) This is in the same slightly abusive spirit as that in which one refers to a connected $p$-adic Lie group. But, of course, that is afield of this question.
Any continuous homomorphism $SO_3 \to SL_2(\mathbb C)$ is trivial because $SO_3$ is compact and has only one irreducible representation of dimension at most two: the trivial representation.
EDIT: the revised question may still be answered by representation theory. A continuous homomorphism $f: SO_3 \to PGL_2(\mathbb C)$ has a unique lift to a homomorphism $\tilde f: SU_2 \to SL_2(\mathbb C)$. By the representation theory of $SU_2$, there are two representations of dimension two: the trivial representation and the defining representation on $\mathbb C^2$. Thus if $f,g: SO_3 \to PGL_2(\mathbb C)$ are both nontrivial and continuous, their lifts $\tilde f, \tilde g$ are conjugate and thus $f$ and $g$ are conjugate.
I had to change $SL$ to the Möbius group, sry. The special unitary subgroup is isomorphic to SO(3). Hence there are nontrivial homomorphisms. Furthermore by taking any $S$ and doing the transformation as above we can make new homomorphisms. The question is then if we can create all of them that way.
(Note: this addressed an early wrong version of the question). Here's an immediate proof that the only homomorphism $f$ [not assumed continuous] $\mathrm{SO}(3)\to\mathrm{SL}_2(\mathbf{C})$ is trivial. Since the only elements of order dividing $2$ in $\mathrm{SL}_2(\mathbf{C})$ are $\pm I$, all elements of order $2$ in $\mathrm{SO}(3)$ map into $\pm 1$. Since $\mathrm{SO}(3)$ is generated by its elements of order $2$, $f$ maps into ${\pm 1}$. Since moreover these elements of order 2 are commutators, everything maps to $1$. (I'm not even using abstract simplicity of $\mathrm{SO}(3)$.)
|
2025-03-21T14:48:31.334447
| 2020-06-20T20:11:18 |
363663
|
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|
Stack Exchange
|
Examples of why conditions for Novikov compact leaf theorem are necessary
Let $M^3$ be a smooth, closed 3-manifold. Given a smooth codimension-one foliation $\mathcal{F}$ of $M$, the Novikov compact leaf theorem asserts that, when the universal cover $\widetilde{M}^3$ of $M^3$ is not contractible, the foliation $\mathcal{F}$ contains a compact leaf of genus $g \leq 1$. What is the most basic example of a smooth codimension-one foliation with all noncompact leaves or leaves of higher genus on a smooth three manifold that has a contractible universal cover, such as the $3$-torus?
What about $\Sigma_g\times S^1$?
Thank you! That answers the question of higher genus, but I am still confused about examples with all noncompact leaves.
Just take the codimension one foliation of $T^2$ by non-compact leaves (of irrational slope), and cross with $S^1$. This gives a codimension one foliation of $T^3$ by $S^1\times\mathbb{R}$.
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2025-03-21T14:48:31.334653
| 2020-06-20T20:37:33 |
363665
|
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|
Stack Exchange
|
Example of a projective module with non-superfluous radical
Let $R$ be a ring with unit. A submodule $N$ of an $R$-module $M$ is called superfluous if the only sumbodule $T$ of $M$ for which $N+T = M$ is $M$ itself.
It is shown, for example, in
[1] F. W._Anderson, K. R. Fuller "Rings and Categories of Modules" (1974)
that if every submodule of $M$ is contained in a maximal submodule, then the radical of $M$ is superfluous (Proposition 9.18). This, in particular, implies that for every finitely generated module $M$ its radical is superfluous. In exercise 9.2. it is explained that divisible abelian groups coincide with their radicals, and therefore their radicals are not superfluous. Divisible abelian groups are not projective objects.
I was curious if it is possible to construct a projective module with non-superfluous radical.
Question: is there an example of a ring $R$ and a projective $R$-module
$P$ such that the radical $JP$ of $P$ is not superfluous?
The existence of such module (or, at least, that its non-existence is non-obvious) is somehow hinted by the formulation of Corollary 17.12 in [1]:
Let $J = J(R)$. If $P$ is a projective left $R$-module such that $JP$
is superfluous in $P$ (e.g., if ${}_RP$ is finitely generated), then
$J(End({}_RP)) = Hom_R(P,JP)$ and $End({}_RP)/J(End_RP) \cong
End({}_RP/JP)$.
According to Proposition 17.10 in the Anderson-Fuller book (I am using the 1992 second edition; don't know if the first 1974 edition is any different), for any projective module $P$ over any (unital associative) ring $R$, the radical of $P$ is computable as $Rad\,P=JP$, where $J$ is the Jacobson radical of the ring $R$ (just as you say).
Let $p$ be a prime number. Consider the commutative ring $R=\mathbb Z_{(p)}$, that is, the localization of the ring of integers $\mathbb Z$ at the prime ideal $(p)\subset\mathbb Z$. Alternatively, one can consider the ring of $p$-adic integers $R=\mathbb Z_p$, that is the completion of the local ring $\mathbb Z_{(p)}$ at its maximal ideal. In both cases, $J(R)=pR$ is the unique maximal ideal of $R$.
Consider the free $R$-module $F$ with a countable set of generators, $F=R^{(\omega)}$. Let us show that $JF$ is not superfluous in $F$. For this purpose, we will construct a proper submodule $T\subset F$ such that $JF+T=F$.
Consider the $R$-module $Q=R[p^{-1}]$. In other words, $Q$ is just the ring of fractions of the local domain $R$. The $R$-module $Q$ is generated by the sequence of elements $1$, $p^{-1}$, $p^{-2}$, $\dots$; so $Q$ is a countably generated $R$-module. Hence $Q$ is a quotient $R$-module of the $R$-module $F$.
Denote by $T\subset F$ a submodule such that $F/T\cong Q$. So we have a short exact sequence of $R$-modules $0\to T\to F\to Q\to 0$. We want to check that $T+JF=F$.
Indeed, we have $JF=pF$, since $J=pR$. The desired equation $T+pF=F$ is equivalent to $p(F/T)=F/T$. Now $F/T\cong Q$ and we have $pQ=Q$ by construction.
In fact, as it is clear now, any discrete valuation ring can be used in the role of $R$ in this construction (with a prime number $p$ replaced by any uniformizing element).
As far as I can see, it is used only that $p$ lies in $JR$ and that it is not a zero-divisor. It is easier for me to think in terms of basis and generators. If we write $e_0$, $e_1$, ... for the basis elements of $F$, then $T$ is generated by the elements $e_i - p e_{i+1}$. As a result $T + JF$ contains all the elements $e_i$ and thus coincides with $F$. On the other hand every element in $T$ is a linear combination $a_{i_1} (e_{i_1} - pe_{i_1 + 1}) + ... + a_{i_k} (e_{i_k} - pe_{i_k + 1})$ with $i_1 < ... < i_k$ and all $a_{i_s}$ non-zero.
Now since $p$ is not a zero divisor the coefficients of $e_{i_1}$ and of $e_{i_k + 1} $ are not zero in the above linear combinations. This shows that every element of $T$ has support of size at least $2$ and thus $T$ does not contain the basis elements $e_i$.
After thinking a little bit more, I realized that the construction can be generalized further, if there is an infinite sequence of elements $p_1$, $p_2$, .... in $JR$ such that each initial product $p_1 p_2 .... p_k$ is non-zero in $JR$. Namely, if such sequence exists, we can take as $T$ the submodule of $F$ generated by the elements $e_i - p_{i+1}e_{i+1}$.
Now, existence of such sequence means that $JR$ is not left T-nilpotent. Knowing this I immediately landed with Lemma 28.3 of Anderon-Fuller, which says, among other things, that $JR$ is left T-nilpotent if and only if $JF$ is superfluous in $F$.
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2025-03-21T14:48:31.335315
| 2020-06-20T21:12:02 |
363666
|
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|
Stack Exchange
|
pointwise purity for character sheaves on a wonderful compactification
Consider the minimal (Goresky-MacPherson) extension of a character sheaf on a semi-simple (say) adjoint group to its wonderful compactification. Is that extension pointwise pure?
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2025-03-21T14:48:31.335365
| 2020-06-20T21:31:37 |
363667
|
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|
Stack Exchange
|
Lower bounds on analytic functions connected to Fox H
The question is related to the one I asked before and never got an answer to. Fourier transform of $f_a(x)= a^{-2}\exp(-|x|^a)$, $a \in (0,2)$, is decreasing in $a$ . I need to demonstrate that the following function is nonnegative for $\alpha\in(1,2)$ and $\delta >1$.
$$
\sum_{m=0}^{\infty} { (-t^2/4)^{m}\over (m) !}
{
\Gamma^{\prime} \left( { \alpha +1\over \alpha}+ m { 1\over \alpha} \right)
+\delta \Gamma \left( { 1\over \alpha} +m { 2\over \alpha}\right) \over \Gamma \left( { 1\over 2} +m\right) },
$$
where $\Gamma(\cdot)$ is a Gamma functions. One can recognize Fox-Wright function in the second term.
Here ia the background. I would like to show that the Fourier transform of
${1 \over \alpha^{\delta} } \exp(-|x|^\alpha)$ is nonincreasing in $\alpha$. Taking derivative,
$$
{\partial \over \partial \alpha } \left\{ {1 \over \alpha^{\delta} }
\int_0^{\infty} \exp(-x^\alpha)\cos(xt) dx \right\}=
$$
$$-{1 \over \alpha^{\delta} }\int_0^{\infty} \exp(-x^\alpha)\{x^\alpha \log(x) + \delta/\alpha\}\cos(xt) dx.
$$
Using Taylor series expansion for $\cos(\cdot)$ leads to
$$
\int_0^{\infty} \exp(-x^\alpha)\{x^\alpha \log(x) + \delta/\alpha\}\cos(xt) dx =
$$
$$
{1\over \alpha^2} \sum_{m=0}^{\infty} { (-t^2)^{m}\over (2m) !}
\Gamma \left( { 2m+1\over \alpha}\right)
\left\{\gamma\left( { 2m+1\over \alpha}\right)
\left( { 2m+1\over \alpha}\right) +
1+\delta \right\}=
$$
$${1\over \alpha^2} \sum_{m=0}^{\infty} { (-t^2)^{m}\over (2m) !}
\left[\Gamma^{\prime} \left( { 2m+1\over \alpha}+1\right)
+\delta \Gamma \left( { 2m+1\over \alpha}\right) \right],
$$
where $\Gamma(\cdot)$ and $\gamma(\cdot)$ are gamma and digamma functions respectively.
There is nothing special about $\delta$ I found it numerically. The connection to Fox Wright function is
$$
\int_0^{\infty} \exp(-x^\alpha)\cos(xt) dx =
{1\over \alpha} \sum_{m=0}^{\infty}
\Gamma\{(1+2m)/\alpha\}
{ (-t^2)^{m}\over (2m) !}=$$
$$
{\sqrt{\pi}\over \alpha} \sum_{m=0}^{\infty}
{ \Gamma\{(1+2m)/\alpha\}\over \Gamma(1/2+m)}
{ (-t^2/4)^{m}\over m !},
$$
|
2025-03-21T14:48:31.335491
| 2020-06-20T21:49:56 |
363669
|
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|
Stack Exchange
|
Reference request: associative subalgebras of Cayley algebras are at most 4-dimensional
By a Cayley algebra I mean an 8-dimensional algebra (over an arbitrary field) formed in the Cayley-Dickson process. (They are also called octonion algebras, but I prefer to reserve the term octonion for the real case.) An equivalent characterization is as a simple alternative algebra that is not associative.
Recently (see https://arxiv.org/pdf/2006.01268.pdf, Thm 6.6) I proved the statement in the title, which is intuitively clear in the division algebra case, but not so much in the split case where 3- and 6-dimensional subalgebras exist. (The split Cayley algebra even has a commutative three-dimensional subalgebra, whose six-dimensional Dickson double is associative, but luckily does not appear as a subalgebra of the split Cayley algebra.)
Anyway, the statement 'feels' like it has been proven already before in the 1950s by e.g. Zorn and I would be more than happy to replace my somewhat lengthy proof by a short reference. In my paper the theorem is just a small step in a bigger argument of which the proof only distracts.
Hence the question: does the claim that associative subalgebras of Cayley algebras are at most 4-dimensional appear in the literature and if so: where?
I thought that normed algebras follow basically the same classification scheme as divison algebras. But it seems that you are saying that there is six-dimensional normed algebra.
Only in situations where the 'norm' is zero for certain non-zero elements, so i guess in these situation you won't call it a norm. We still have a quadratic form though, that in the division algebra case over $\mathbb{R}$ is the square of the norm.
The split quaternion algebra is isomorphic to the algebra of two by two matrices where the analogue of the square of the norm is the determinant. Inside the two-by-two matrices we have the three dimensional (non-commutative) subalgebra of upper triangular matrices. If we realize the split octonions as the Dickson double of the split quaternion algebra then the Dickson double of the 3D subalg of upper triangular matrices sits inside it as a (nonassociative) 6-dimensional subalgebra
Over $\mathbb{R}$ the 'ordinary' octonions are waaaaaaay more famous than their split counterpart, due to being a division algebra. However when people started constructing these algebras over general fields the split ones more or less got their revenge as they exist over every field while division algebras do not.
I think I got that terminology from Springer and Veldkamp but my memory is not particularly good. Anyway, what I meant was an algebra with (nondegenerate?) quadratic form $Q$ such that $Q(xy) = Q(x)Q(y).$ And I thought that these are classified over (reasonable? all? most?) fields.
Anyway in the questin you mention 6 dimensional associative algebra that arises as Dickson double of a subalgebra of a normed algebra so I thought that this would be a normed algebra of strange dimension different from $1, 2, 4, 8.$ But then again, I never really looked into the classification of normed algebras so perhaps these do appear. I suggest you look into some thick books on Jordan algebras (Springer, Veldkamp, McCrimmon) as these classifications are often treated there.
Yes with that definition these are indeed normed algebras of stange dimension.
But not all strange normed algebras we can write down do appear as subalgebras of all bigger ones
|
2025-03-21T14:48:31.335741
| 2020-06-20T23:10:02 |
363674
|
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|
Stack Exchange
|
Second summand to make projective module free
Suppose there's a projective $R$-module $P$ (non-free). We know that there is another $R$-module $M$ such that $P\oplus M$ is free over $R$. Is there a way to write down such an $M$ in terms of $P$?
If this is not always tractable, is it possible in certain specialized circumstances? The setting that comes to mind is where $P$ is a non-principal ideal of the ring of integers of a number field (as Wikipedia says this is an instance of $P$ being projective but not free).
Thanks!
(This was originally posted on math.stackexchange but I realized that here might be the better place to ask, apologies if not.)
I can imagine it's one of those "I'll know it when I see it", but do you have a rigorous definition of "in terms of $P$"? For example, one can take $M$ to be the kernel of the projection to the module $P$ from the free $R$-module on the set $P$. Does that count?
Yes, I'd say that this solution falls into "in terms of $P$", thanks! To really get a sense of what direct-summing $M$ does, I was also hoping to be able to think about what the module $P\oplus M$ "looks like", separately from the exact sequence context (since that seems to be the most natural way to actually prove the freeness of $P\oplus M$) --- is there a nice way to think about the isomorphism between $P\oplus M$ and the free module $FP$ on $P$? I'm having trouble picturing exactly what that might be, i.e. how to carry $p\oplus\sum r_ip_i$ to a $FP$-element.
The problem with a question like "how to carry $p \oplus m$ to an element of $F P$" is that, in the generality of a totally arbitrary (ring $R$ and) projective $P$, we just have no way of peeking into the structure of $P$ other than by using the universal property; and, in terms of the universal property, the answer is that we use a splitting $P \to F P$ of $M \hookrightarrow F P \twoheadrightarrow P$ to map $P \oplus M \to F P$, and I think that's all that can be said! Obviously @StevenLandsburg's answer is more interesting.
If you want something completely general, LSpice's comment is the answer.
For the special case of an ideal $P$ in the ring $R$ of integers of a number field (or more generally if $R$ is a Dedekind domain) you can take $M=\{x\in K|xI\subset R\}$, where $K$ is the fraction field of $R$. The keyword to Google is fractional ideal.
Thank you! Do you have advice on how to show that that $M$ suffices? e.g. in $K=\mathbb{Q}[\sqrt{-5}]$ where $\frak{a}=\langle 2,1+\sqrt{-5}\rangle$ gives a nontrivial element of the class group $Cl(\mathcal{O}_K)\cong\mathbb{Z}/2\mathbb{Z}$, I'm having some trouble seeing the way to view $\frak{a}\oplus\frak{a}$ as free (either as a $\mathcal{O}_K$-module or $\mathbb{Z}$-module), and more generally how to see direct-summing of ideals as being equivalent in some way to their product.
In a Dedekind ring $R$ the direct sum of two ideals is related to their product by the formula $P\oplus Q\approx PQ\oplus R$. This is quite standard material, and any commutative algebra textbook should give you all the details.
|
2025-03-21T14:48:31.335971
| 2020-06-21T01:40:09 |
363678
|
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|
Stack Exchange
|
On relating $l(A), l(B)$ and $l(A+B)$ for Weil divisors on a smooth projective curve where one of the divisors is effective
Let $X$ be a smooth projective curve over an Algebraically closed field $k$. Let $k(X)$ denote its function field.
If $A, B$ are Weil divisors on $X$ such that $A$ is effective (i.e. $A\ge 0$) , then is there any (in)equality between $l(A+B)$ and $ l(B), l(A)$ ?
Here, for a Weil divisor $D$ on $X$, by $l(D)$ we denote the $k$-vector space dimension of the Riemann-Roch space $L(D):=\{f\in k(X)^*: D+ div(f)\ge 0\}\cup \{0\}$.
For a divisor $D$ on $X$, the complete linear system $|D|$ be the collection of all effective divisors which are linearly equivalent with $D$. $|D|$ can be given the structure of a projective space by identifying it with $( L(D)\setminus \{0\})/k^*$ and by that structure, $\dim |D|=l(D)-1$. Now it is known (Hartshorne, Chapter IV, Lemma 5.5) that if $D,E$ are both effective divisors, then $\dim |D|+\dim |E|\le \dim |D+E|$ i.e. $l(D)+l(E)\le l(D+E)+1$ . What I'm basically asking is that if something similar holds if we assume only one of the divisors is effective...
If $D$ and $E$ are linearly equivalent to effective divisors, this is OK from what's in Hartshorne, as both sides are invariant under linear equivalence.
If $E$, say, is not linearly equivalent to an effective divisor, then $l(E)=0$, so your desired inequality is $l(D) \leq l(D+E) +1$. It is easy to produce counterexamples to this. We can take $E$ to be a very negative divisor, or, alternately, we can take $D$ to be $k$ times the hyperplane class on a hyperelliptic curve of genus $g>2k$ and $E$ to be a generic divisor of degree $0$, which will give $l(D) = k+1$ and $l(D+E)=0$. One can even make examples with $E$ of positive degree.
Agh yes of-course I should have added that both my divisors have positive degree (otherwise, as you say, it's very easy to get counterexamples) ... what would be a counterexample with a divisor of positive degree ?
@Louis The simplest example would be $D$ the hyperelliptic divisor on a hyperelliptic curve of genus $>3$, so that $l(D)=2$, and $E$ a generic divisor of degree $1$, so that $D+E$ is a generic divisor of degree $3$, which means $l(D+E)=0$ since $3<g$.
|
2025-03-21T14:48:31.336125
| 2020-06-21T03:25:10 |
363684
|
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|
Stack Exchange
|
A complex analytic interpretation of multiplicity on the special fiber of a flat family
Let $X$ be a variety over $\mathbb C$ and $\pi: X\to \Delta$ be a flat morphism over the unit disk $\Delta=\{z:|z|<1\}$. Let $Z$ be a component of $X_0=\pi^{-1}(0)$. The multiplicity of $Z$ is defined as the order of vanishing of $\pi^*(t)$ on $Z$, where $t$ is the local coordinate of $0\in\Delta$.
I'm trying to understand the multiplicity through a complex analytic point of view, via the following way.
Let $p\in Z$ be a general point (in particular, $p$ is a smooth point of $Z$ and does not meet other components of $X_0$). Let $\Delta_p$ is a holomorphic disk in $X$ which is transversal to $Z$ at $p$. Then the restriction
$$\pi|_{\Delta_p}:\Delta_p\to \Delta$$
is identified with $z\mapsto z^m$ for some $m\ge 1$.
Question: Is it true that $m$ coincides with multiplicity of $Z$ in general?
My motivating example is the family of plane curves $X=\{y^2-tx=0\}\subset \mathbb C^2_{[x,y]}\times \Delta$, take $p=(1,0,0)$ and $\Delta_p=\{(1,t,t^2):t\in \Delta\}$ which is 2-to-1 to $\Delta$, so the degree coincides with the multiplicity of $\{y^2=0\}$. Also the degree is not dependent on the choice of $p$ (except at $p=[0,0,0]$ where the total space is singular) and choice of normal disk.
This leads to the construction in my question as above, and I think the answer should be true. However, I haven't seen such a complex analytic description in any reference before. Is that true? I would appreciate it if anyone can show me proof or reference.
p.s., My original question was posted on MSE here but I received no comment/answer so far.
|
2025-03-21T14:48:31.336265
| 2020-06-21T05:05:57 |
363686
|
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|
Stack Exchange
|
An inequality related to the numbers of faces of polytopes with d+2 facets
I would like to prove an inequality related to the number of $k$-faces of two $d$-polytopes with $d+2$ facets; see (1) below.
Let $r>0$, $s>0$, $t\ge 0$, and $d\ge 2$ be such that $d=r+s+t$. We denote by $\Delta({r,s})$ the sum of an $r$-dimensional simplex and an $s$-dimensional simplex with $r,s>0$, lying in complementary subspaces. The polytope $\Delta(r,s)$ is a simple polytope of dimension $r+s$. A $d$-dimensional $P$ polytope with $d+2$ facets is a $t$-fold pyramid over $\Delta({r,s})$. The number $f_k$ of its $k$-dimensional faces is as follows.
$$f_k(P)= \binom{r+s+t+2}{k+2} -\binom{s+t+1}{k+2}-\binom{r+t+1}{k+2} +\binom{t+1}{k+2}.$$
I am interested in finding the $d$-polytopes with $d+2$ facets, $2d+1$ vertices, and a smallest number of faces. Having $2d+1$ vertices amounts to saying that $d=rs$.
Let $r_{2}>r_{1}>0$ and $s_{1}>s_{2}>0$ be such that $r_{i}\le s_{i}$ for $i=1,2$. And let $t_{1},t_{2}\ge 0$ and $d\ge 2$ be such that $d=r_{i}+s_{i}+t_{i}$ and $d=r_{i}s_{i}$ for $i=1,2$. Suppose that $P_{i}$ be a $t_{i}$-fold pyramid over $\Delta({r_{i},s_{i}})$. Experimentation has shown that
\begin{equation}
(1)\quad f_{k}(P_{1})< f_{k}(P_{2})\quad \text{for each $k\in [2,d-2].$}
\end{equation}
Is (1) known? I have tried unsuccessfully to prove (1) for every possible triple $(r,s,t)$.
Any help would be greatly appreciated.
Thank you, and regards,
Guillermo
|
2025-03-21T14:48:31.336367
| 2020-06-21T08:25:51 |
363690
|
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|
Stack Exchange
|
Maximality principle in symmetric extensions
Let $M$ be a ctm and $P\in M$ a forcing order.
In regular forcing extensions, we have the following well-known Principle:
$$p\Vdash_{M,P}\exists x\phi[x]\;\Longrightarrow\;\exists\sigma\in M^P\;p\Vdash_{M,P}\phi[\sigma]$$
(where $M^P$ is the class of $P$-names in $M$).
Given an automorphism $f$ of $P$, we can turn $f$ into a function mapping $P$-names to $P$-names, given by
$$\overline{f}(\tau)=\{(\overline{f}(\sigma),f(p))\;|\;(\sigma,p)\in\tau\}$$
Given a subgroup $H$ of $Aut(P)$, we say that $\tau$ is $H$-invariant iff $\overline{f}(\tau)=\tau$ for all $f\in H$.
We can also fix a filter $\mathcal{F}$ on the set of subgroups of $Aut(P)$ and consider the so-called $\mathcal{F}$-symmetric extension of $M$ given by the evaluation only of those names $\tau$ that are hereditarily $\mathcal{F}$-symmetric, i.e. for some $H\in \mathcal{F}$, $\tau$ is $H$-invariant and for all $(\sigma,p)\in\tau$, $\sigma$ is hereditarily $\mathcal{F}$-symmetric. It can be shown that the resulting model always satisfies ZF, but not always AC.
My question now is whether or not the following principle can be shown to hold (in $M$): Given $p\in P$ such that $p\Vdash_{M,P}^{\mathcal{F}}\exists x\phi[x]$ (i.e. for any $P$-generic filter $G$ containing $p$, there is a hereditarily $\mathcal{F}$-symmetric name $\tau$ such that $M[G]\models\phi[\tau]$), does there necessarily exist a hereditarily $\mathcal{F}$-symmetric name $\sigma$ such that $p\Vdash\phi[\sigma]$?
As far as i can tell, the proof for the usual case can not be modified to prove the symmetric case, since we "stitch together" witnesses along a maximal antichain, which might not be $H$-invariant for any $H\in\mathcal{F}$.
No. Not even remotely.
Consider the Cohen model, i.e. add $\omega$ Cohen reals, permute them amongst themselves, and take finite supports. Let $\dot a_n$ be the canonical name of a Cohen real, and let $\dot A$ be the name of the set of Cohen reals.
$$1\Vdash\exists x(x\in\dot A\land\check 0\in x)$$
But there is no symmetric name that we can instantiate. If $\dot x$ would be such name, then it is easy to see that its value has to be one of finitely many reals, but then just take a condition which decides that $\check 0\notin\dot a_n$ for any of those finitely many reals.
My work on symmetric iterations deals with this problem extensively. See "Iterations of symmetric extensions", J. Symb. Log. 84 (2019) no. 1, pp. 123–159. Or, https://arxiv.org/abs/1606.06718
(I also mention this issue in several other papers dealing with iterations.)
|
2025-03-21T14:48:31.336573
| 2020-06-21T08:34:38 |
363691
|
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Stack Exchange
|
On permanents and determinants of finite groups
$\DeclareMathOperator\perm{perm}$Let $G$ be a finite group. Define the determinant $\det(G)$ of $G$ as the determinant of the character table of $G$ over $\mathbb{C}$ and define the permanent $\perm(G)$ of $G$ as the permanent of the character table of $G$ over $\mathbb{C}$. Note that due to the properties of the determinant and the permanent, this definition just depends on $G$ and not on the ordering of the conjugacy classes etc.
I'm not experienced with character theory but did some experiments with GAP on this and found nothing related in the literature, which motivates the following questions (sorry, in case they are trivial). Of course finite groups are dangerous and it is tested just for all finite groups of order at most $n \leq 30$ and some other cases, which might not be too good evidence for a question on finite groups.
Question 1: Are $\perm(G)$ and $\det(G)^2$ always integers?
I was able to prove this for cyclic groups. Since the character table of the direct product of groups is given by their Kroenecker product, one can conclude that $\det(G)^2$ is also an integer for all abelian groups $G$. Maybe there is a formula for the permanent of the Kroenecker product of matrices to conclduge that $\perm(G)$ is also an integer for all abelian groups or even better a more direct proof that question 1 is true at least for abelian groups. Note that $\det(G)$ is in general not an integer, even for cyclic groups.
Now call a finite group permfect in case $\perm(G)=0$.
Question 2: Is it true that all finite groups of order $n$ are permfect if and only if $n=4r+2$ for some $r \geq 2$?
Being permfect could be seen as having a high symmetry. It seems symmetric groups are permfect and for the alternating groups I only found $A_6$ to be permfect yet.
Pretty sure this can be cracked with the usual Galois-theoretical pincers: Invariance under field automorphisms should show that the numbers $\operatorname{per} G$ and $\left(\det G\right)^2$ are rationals, while the algebraic integrality of the character values shows that these numbers are algebraic integers. Hence, they are integers.
The "if" direction of your Question 2 might have to do with the fact that if $n = 4r+2$, then any group $G$ of order $n$ has a normal subgroup of order $2$ (by https://math.stackexchange.com/questions/225987/show-group-of-order-4n-2-has-a-subgroup-of-index-2 ). This gives the character table some sort of irregular chessboard structure. Do we know something about how many conjugacy classes of $G$ belong to said subgroup?
@darijgrinberg Thanks, you probably mean index 2 instead of order 2. There are groups having normal subgroups of index 2 without permanent 0, for example the Klein four group. I do some more experiments with GAP on that.
Oops, yes, index $2$ is what I meant. I suspect something more is needed.
You might be interested in Lemma 3.1 of a paper availabel from http://maths.nju.edu.cn/~zwsun/124e.pdf
Permfect is a great term.
@darijgrinberg Since your comment got so many upvotes, but there is no answer for the question on the integrality of the permanent yet, I wanted to ask you: Does your idea work to give a proof that the permanent is indeed always an integer?
@SteinChen: Yes, it does. I've now expanded it to a full answer.
For the $\det^2$, see this MO question https://mathoverflow.net/q/78050.
@darijgrinberg Thank you!
I won't discuss the permanent part of the question, but I think the other part can be done easily, even without Galois theory. Let $X = X(G)$ denote the character table of $G$ (rows indexed by complex irreducible characters of $G$, columns indexed by conjugacy classes of $G$). Then, by the orthogonality relations, $X\overline{X}^{T}$ is an integer diagonal matrix with $i$-th main diagonal entry $|C_{G}(x_{i})|$, where $x_{i}$ is a representative of the $i$-th conjugacy class.
Hence we have $|\det(X)|^{2} = \prod_{i}|C_{G}(x_{i})|.$ On the other hand, note that replacing $X$ by $\overline{X}$ gives a matrix with the same rows as $X$, but permuted (under some permutation of order one or two), since the complex conjugate of an irreducible character is an irreducible character. Hence $\overline{X} = PX$, where $P$ is a permutation matrix associated to a permutation $\sigma$ of order at most two. Thus $\overline{\det(X)} = \operatorname{sign}(\sigma) \det(X)$.
If $\sigma$ is an odd permutation, we then see that $\det(X)^{2}$ is a negative integer, while if $\sigma$ is an even permutation, we see that $\det(X)^{2}$ is a positive integer.
Hence we have the curious fact that $\det(X)^{2}$ is a positive integer if the number of complex irreducible characters of $G$ with Frobenius-Schur indicator zero is divisible by $4$, and is a negative integer if the number of complex irreducible characters of $G$ with Frobenius-Schur indicator zero is congruent to $2$ (mod $4$).
The number of complex irreducible characters of $G$ with Frobenius-Schur indicator zero is just the number of complex irreducible characters which are not real-valued, but I write it as above to illustrate the link between the F-S indicator and the congruence (mod 4).
See also the MO question https://mathoverflow.net/q/78050.
@DenisSerre : Thanks for pointing that out. I see that your first proof is essentially the same as that above.
I'll address the remaining question 2.
Theorem: If $|G|=4r+2$ then the permanent of its character table vanishes.
Proof: We know $G$ has a normal subgroup $H$ of index $2$. Let's denote the cosets of $G/H$ by $\{H, aH\}$. Start by noticing that the conjugacy classes of $G$ either completely lie inside $H$ or inside $aH$. Let $k$ be the number of conjugacy classes that lie inside $aH$.
Next, we let $U$ be the 1 dimensional representation where elements of $H$ act as $1$ and elements of $aH$ act as $-1$. Tensoring with $U$ gives an involution on irreducible representations of $G$ and thus also an involution $\sigma$ of the columns of the character table.
Denoting by $S$ the set of all irreducible representations of $G$, and by $f$ the possible bijections between $S$ and the conjugacy classes we see that our permanent can be written as
$$\sum_f \prod_{V\in S}\chi_V(f(V))=\sum_f\prod_{V\in S} \chi_{\sigma(V)} (f(V))=(-1)^k\sum_f \prod_{V\in S}\chi_V(f(V))$$
so the theorem will follow once we establish that $k$ is odd.
There is probably a better way of doing this but I reasoned as follows: $k$ is equal to the number of conjugacy classes that lie inside $H$ that don't split when we restrict to the conjugation action by $H$ (see lemma 2 here for a more general statement). Now, if $x\in H$ and $C_G(x)$ denotes its centralizer, we have $[C_G(x):C_H(x)]\in\{1,2\}$.
If this index is $1$ then the conjugacy class of $x$ splits into two conjugacy classes of equal size when we restrict to the conjugation by $H$. In particular the size of the conjugacy class of $x$ is even. If the index is $2$ then the conjugacy class doesn't split and its size is $(4r+2)/|C_G(x)|$ which is odd.
Since the total number of elements in $H$ is $2r+1$ there must be an odd number of nonsplit conjugacy classes, so $k$ is odd and we are done.
One also has to check that for other orders it is possible to find groups with nonvanishing permanent. For odd order one can take the cyclic group $\mathbb Z/n\mathbb Z$, and for order divisible by 4, I believe you can take the product of an elementary abelian 2-group with a cyclic group of odd order.
I also wanted to mention that the permanent of the symmetric group is not always zero and it's first few values are recorded in OEIS, in particular this permanent for $S_8$ is $-20834715303936$. There you will find a reference by Schmidt and Simion that uses an argument similar to the one above for the case of the symmetric group/alternating subgroup to prove that the permanent of the symmetric group vanishes whenever half the number of non-self-conjugate partitions of $n$ is odd. In a follow-up note they show that this occurs infinitely often.
I think I found https://arxiv.org/abs/1802.08001 showing that the permanent of an elementary abelian 2-group of order $2^n \geq 4$ is indeed nonzero (and divisible by $2^n-1$). Together with your answer, this should completely answer question 2.
A more group theoretic argument that $aH$ is the union of an odd number of classis is as follows. Notice that $H$ has odd order by assumption, and that we might as well assume that $a$ has order $2$. Then since $\langle a \rangle$ is a Sylow $2$-subgroup of $G$, we see that every element of $G \backslash H$ is $G$-conjugate to an element of the form $ab$ where $b$ has odd order and $ab = ba$. Furthermore, if $b,c \in C_{H}(a)$ have odd order, then $ab$ and $ac$ are $G$-conjugate if and only $b$ and $c$ are already conjugate in $C_{H}(a)$....
So the number of $G$ classes in $aH$ is the number of conjugacy classes of $C_{H}(a)$. This is odd because $C_{H}(a)$ has odd order and only the identity is conjugate to its inverse in a group of odd order.
@GeoffRobinson Very nice argument!
For the sake of completeness, here is the answer to Question 1, part of which
is missing from the other answers:
Proposition 1. Let $G$ be a finite group. Consider the representations of
$G$ over $\mathbb{C}$. Let $\det G$ denote the determinant of the character
table of $G$. (Note that this is only defined up to sign, since the order of
the rows and of the columns of the character table can be chosen arbitrarily.)
Let $\operatorname*{perm}G$ denote the permanent of the character table of
$G$. Then, $\left(\det G\right)^2$ and $\operatorname*{perm}G$ are integers.
To prove this, we need the following lemmas:
Lemma 2. Let $G$ be a finite group. Then, there is a finite Galois field
extension $\mathbb{F}$ of $\mathbb{Q}$ such that all irreducible
representations of $G$ are defined over $\mathbb{F}$.
Proof of Lemma 2. There is a finite Galois field extension $\mathbb{K}$ of
$\mathbb{Q}$ such that all irreducible representations of $G$ are defined over
$\mathbb{K}$. Indeed, this is known as a splitting field of $G$; its
existence is part of Theorem 9.2.6 in Peter Webb, A Course in Finite Group
Representation Theory, 2016.
Consider this field extension $\mathbb{K}$. Let $\mathbb{F}$ be the Galois
closure of $\mathbb{K}$ over $\mathbb{Q}$ (or any other finite field extension
of $\mathbb{Q}$ that is Galois over $\mathbb{Q}$ and contains $\mathbb{K}$ as
a subfield). Then, all irreducible representations of $G$ are defined over
$\mathbb{F}$ (since they are defined over $\mathbb{K}$, but $\mathbb{F}$
contains $\mathbb{K}$ as a subfield). This proves Lemma 2. $\blacksquare$
Lemma 3. Let $G$ be a finite group. Let $\mathbb{F}$ be a field extension
of $\mathbb{Q}$ such that all irreducible representations of $G$
are defined over $\mathbb{F}$. Let $\chi:G\rightarrow\mathbb{F}$ be an
irreducible character of $G$. Let $\gamma:\mathbb{F}\rightarrow\mathbb{F}$ be
a $\mathbb{Q}$-algebra automorphism of $\mathbb{F}$. Then, $\gamma\circ
\chi:G\rightarrow\mathbb{F}$ is an irreducible character of $G$.
Proof of Lemma 3. This is a totally straightforward "isomorphisms preserve
all relative properties of objects they are applied to" argument, but for the
sake of completeness, let me spell it out (at least to some level of detail):
The map $\chi$ is an irreducible character of $G$, and thus is the character
of an irreducible representation $\rho$ of $G$. Consider this $\rho$, and WLOG
assume that $\rho$ is a representation over $\mathbb{F}$. (This can be assumed
since all irreducible representations of $G$ are defined over $\mathbb{F}$.)
Thus, $\rho$ is a group homomorphism from $G$ to $\operatorname*{GL}
\nolimits_{n}\left( \mathbb{F}\right) $ for some $n\geq1$. Consider this $n$.
The $\mathbb{Q}$-algebra automorphism $\gamma:\mathbb{F}\rightarrow\mathbb{F}$
induces a group automorphism $\widetilde{\gamma}:\operatorname*{GL}
\nolimits_{n}\left( \mathbb{F}\right) \rightarrow\operatorname*{GL}
\nolimits_{n}\left( \mathbb{F}\right) $ that transforms each matrix in
$\operatorname*{GL}\nolimits_{n}\left( \mathbb{F}\right) $ by applying
$\gamma$ to each entry of the matrix. The composition $\widetilde{\gamma}
\circ\rho:G\rightarrow\operatorname*{GL}\nolimits_{n}\left( \mathbb{F}
\right) $ is a group homomorphism (since $\widetilde{\gamma}$ and $\rho$ are
group homomorphisms), and thus is a representation of $G$. Moreover, the
character of this representation $\widetilde{\gamma}\circ\rho$ is $\gamma
\circ\chi$ (since $\operatorname*{Tr}\left( \widetilde{\gamma}\left(
A\right) \right) =\gamma\left( \operatorname*{Tr}A\right) $ for any matrix
$A\in\operatorname*{GL}\nolimits_{n}\left( \mathbb{F}\right) $). We shall
now show that this representation $\widetilde{\gamma}\circ\rho$ is irreducible.
Indeed, let $U$ be a subrepresentation of $\widetilde{\gamma}\circ\rho$ --
that is, an $\mathbb{F}$-vector subspace of $\mathbb{F}^{n}$ that is invariant
under the action of $\widetilde{\gamma}\circ\rho$. Consider the $\mathbb{Q}
$-module isomorphism $\overline{\gamma}:\mathbb{F}^{n}\rightarrow
\mathbb{F}^{n}$ that applies $\gamma$ to each coordinate of the vector. Since
$\gamma$ is a $\mathbb{Q}$-algebra homomorphism, we can easily see that
$\left( \widetilde{\gamma}\left( A\right) \right) \left( \overline
{\gamma}\left( v\right) \right) =\overline{\gamma}\left( Av\right) $ for
each $A\in\operatorname*{GL}\nolimits_{n}\left( \mathbb{F}\right) $ and each
$v\in\mathbb{F}^{n}$. Thus, we can easily see that $\overline{\gamma}
^{-1}\left( U\right) $ is an $\mathbb{F}$-vector subspace of $\mathbb{F}
^{n}$ that is invariant under the action of $\rho$ (since $U$ is an
$\mathbb{F}$-vector subspace of $\mathbb{F}^{n}$ that is invariant under the
action of $\widetilde{\gamma}\circ\rho$). In other words, $\overline{\gamma
}^{-1}\left( U\right) $ is a subrepresentation of $\rho$. Since $\rho$ is
irreducible, this entails that either $\overline{\gamma}^{-1}\left( U\right)
=0$ or $\overline{\gamma}^{-1}\left( U\right) =\mathbb{F}^{n}$. Since
$\overline{\gamma}$ is an isomorphism, we thus conclude that either $U=0$ or
$U=\mathbb{F}^{n}$.
Forget that we fixed $U$. We thus have shown that if $U$ is a
subrepresentation of $\widetilde{\gamma}\circ\rho$, then either $U=0$ or
$U=\mathbb{F}^{n}$. In other words, the representation $\widetilde{\gamma
}\circ\rho$ is irreducible (since its dimension is $n\geq1$). Thus, its
character is an irreducible character of $G$. In other words, $\gamma\circ
\chi$ is an irreducible character of $G$ (since $\gamma\circ\chi$ is the
character of $\widetilde{\gamma}\circ\rho$). This proves Lemma 3.
$\blacksquare$
Proof of Proposition 1. Lemma 2 shows that there is a finite Galois field
extension $\mathbb{F}$ of $\mathbb{Q}$ such that all irreducible
representations of $G$ are defined over $\mathbb{F}$. Consider this
$\mathbb{F}$. Let $\Gamma$ be the Galois group $\operatorname*{Gal}\left(
\mathbb{F}/\mathbb{Q}\right) $ (which consists of all $\mathbb{Q}$-algebra
automorphisms of $\mathbb{F}$). The Fundamental Theorem of Galois Theory shows
that the invariant ring $\mathbb{F}^{\Gamma}$ is $\mathbb{Q}$.
Let $\chi_{1},\chi_{2},\ldots,\chi_{r}$ be all irreducible characters of $G$
(listed without repetition). Note that these characters are maps from $G$ to
$\mathbb{F}$ (since all irreducible representations of $G$ are defined over
$\mathbb{F}$).
Let $c_{1},c_{2},\ldots,c_{r}$ be the conjugacy classes of $G$ (listed without repetition).
Let $\operatorname*{per}A$ denote the permanent of any square matrix $A$.
Let $C$ be the matrix $\left( \chi_{i}\left( c_{j}\right) \right) _{1\leq
i\leq r,\ 1\leq j\leq r}\in\mathbb{F}^{r\times r}$. This matrix $C$ is the
character table of $G$ (for the ordering of characters given by $\chi_{1}
,\chi_{2},\ldots,\chi_{r}$ and the ordering of conjugacy classes given by
$c_{1},c_{2},\ldots,c_{r}$). Thus, the definition of $\operatorname*{perm}G$
shows that $\operatorname*{perm}G$ is the permanent of $C$. In other words,
$\operatorname*{perm}G=\operatorname*{per}C$.
Let $\gamma\in\Gamma$. Thus, $\gamma$ is a $\mathbb{Q}$-algebra automorphism
of $\mathbb{F}$ (since $\gamma\in\Gamma=\operatorname*{Gal}\left(
\mathbb{F}/\mathbb{Q}\right) $). We shall show that $\gamma\left(
\operatorname*{perm}G\right) =\operatorname*{perm}G$.
The $\mathbb{Q}$-algebra automorphism $\gamma:\mathbb{F}\rightarrow\mathbb{F}$
induces a $\mathbb{Q}$-algebra automorphism $\widetilde{\gamma}:\mathbb{F}^{r\times
r}\rightarrow\mathbb{F}^{r\times r}$ that transforms each matrix in
$\mathbb{F}^{r\times r}$ by applying $\gamma$ to each entry of the matrix.
We define a map $f:\left\{ 1,2,\ldots,r\right\} \rightarrow\left\{
1,2,\ldots,r\right\} $ as follows:
Let $i\in\left\{ 1,2,\ldots,r\right\} $. Then, $\chi_{i}:G\rightarrow
\mathbb{F}$ is an irreducible character of $G$. Thus, Lemma 3 (applied to
$\chi=\chi_{i}$) shows that $\gamma\circ\chi_{i}:G\rightarrow\mathbb{F}$ is an
irreducible character of $G$. Hence, $\gamma\circ\chi_{i}=\chi_{j}$ for some
$j\in\left\{ 1,2,\ldots,r\right\} $ (since $\chi_{1},\chi_{2},\ldots
,\chi_{r}$ are all irreducible characters of $G$). This $j$ is uniquely
defined. We define $f\left( i\right) $ to be $j$.
Thus, we have defined a map $f:\left\{ 1,2,\ldots,r\right\} \rightarrow
\left\{ 1,2,\ldots,r\right\} $ with the property that
\begin{equation}
\gamma\circ\chi_{i}=\chi_{f\left( i\right) }\qquad\text{for each }
i\in\left\{ 1,2,\ldots,r\right\} .
\label{eq.darij1.1}
\tag{1}
\end{equation}
If two distinct elements $i$ and $j$ of $\left\{ 1,2,\ldots,r\right\} $
would satisfy $f\left( i\right) =f\left( j\right) $, then they would
satisfy $\gamma\circ\chi_{i}=\gamma\circ\chi_{j}$ (by \eqref{eq.darij1.1}
) and therefore $\chi_{i}=\chi_{j}$ (since $\gamma$ is invertible), which
would contradict the fact that $\chi_{1},\chi_{2},\ldots,\chi_{r}$ are
distinct. Thus, two distinct elements $i$ and $j$ of $\left\{ 1,2,\ldots
,r\right\} $ always satisfy $f\left( i\right) \neq f\left( j\right) $. In
other words, the map $f$ is injective. Hence, $f$ is a permutation (since $f$
is an injective map from $\left\{ 1,2,\ldots,r\right\} $ to $\left\{
1,2,\ldots,r\right\} $).
Now, from $C=\left( \chi_{i}\left( c_{j}\right) \right) _{1\leq i\leq
r,\ 1\leq j\leq r}$, we obtain
\begin{align*}
\widetilde{\gamma}\left( C\right) =\left( \gamma\left( \chi_{i}\left(
c_{j}\right) \right) \right) _{1\leq i\leq r,\ 1\leq j\leq r}=\left(
\chi_{f\left( i\right) }\left( c_{j}\right) \right) _{1\leq i\leq
r,\ 1\leq j\leq r},
\end{align*}
since each $i,j\in\left\{ 1,2,\ldots,r\right\} $ satisfy
\begin{align*}
\gamma\left( \chi_{i}\left( c_{j}\right) \right) =\left( \gamma\circ
\chi_{i}\right) \left( c_{j}\right) =\chi_{f\left( i\right) }\left(
c_{j}\right) \qquad\left( \text{by \eqref{eq.darij1.1}}\right) .
\end{align*}
Thus, the matrix $\widetilde{\gamma}\left( C\right) $ is obtained from the
matrix $\left( \chi_{i}\left( c_{j}\right) \right) _{1\leq i\leq r,\ 1\leq
j\leq r}$ by permuting the rows (since $f:\left\{ 1,2,\ldots,r\right\}
\rightarrow\left\{ 1,2,\ldots,r\right\} $ is a permutation). In other words,
the matrix $\widetilde{\gamma}\left( C\right) $ is obtained from the matrix
$C$ by permuting the rows (since $C=\left( \chi_{i}\left( c_{j}\right)
\right) _{1\leq i\leq r,\ 1\leq j\leq r}$). Hence, $\operatorname*{per}
\left( \widetilde{\gamma}\left( C\right) \right) =\operatorname*{per}C$
(since the permanent of a matrix does not change when its rows are permuted).
But the definition of $\widetilde{\gamma}$ yields that $\operatorname*{per}
\left( \widetilde{\gamma}\left( C\right) \right) =\gamma\left(
\operatorname*{per}C\right) $ (since $\gamma$ is a $\mathbb{Q}$-algebra
homomorphism). Hence, $\gamma\left( \operatorname*{per}C\right)
=\operatorname*{per}\left( \widetilde{\gamma}\left( C\right) \right)
=\operatorname*{per}C$. In view of $\operatorname*{perm}G=\operatorname*{per}
C$, this rewrites as $\gamma\left( \operatorname*{perm}G\right)
=\operatorname*{perm}G$.
Forget that we fixed $\gamma$. We thus have shown that $\gamma\left(
\operatorname*{perm}G\right) =\operatorname*{perm}G$ for each $\gamma
\in\Gamma$. In other words, $\operatorname*{perm}G$ belongs to the invariant
ring $\mathbb{F}^{\Gamma}$. In other words, $\operatorname*{perm}G$ belongs to
$\mathbb{Q}$ (since the invariant ring $\mathbb{F}^{\Gamma}$ is $\mathbb{Q}$).
But all values of the characters $\chi_{1},\chi_{2},\ldots,\chi_{r}$ are sums
of roots of unity (because they are traces of matrices $A\in\operatorname*{GL}
\nolimits_{n}\left( \mathbb{F}\right) $ that satisfy $A^{\left\vert
G\right\vert }=I_{n}$, and the eigenvalues of such a matrix are roots of
unity), and thus are algebraic integers. Hence, all entries of the matrix $C$
are algebraic integers (since all these entries are values of the characters
$\chi_{1},\chi_{2},\ldots,\chi_{r}$). Thus, the permanent $\operatorname*{per}
C$ of this matrix $C$ is an algebraic integer (since the algebraic integers
form a ring). In other words, $\operatorname*{perm}G$ is an algebraic integer
(since $\operatorname*{perm}G=\operatorname*{per}C$). Hence,
$\operatorname*{perm}G$ is an algebraic integer in $\mathbb{Q}$ (since
$\operatorname*{perm}G$ belongs to $\mathbb{Q}$). Since the only algebraic
integers in $\mathbb{Q}$ are integers (because the ring $\mathbb{Q}$ is
integrally closed), this entails that $\operatorname*{perm}G$ is an integer.
A similar argument shows that $\left( \det G\right) ^{2}$ is an integer.
(Here we need to use the fact that the square of the determinant of a matrix
does not change when its rows are permuted. This is because the determinant
gets multiplied by a power of $-1$.) Thus, Proposition 1 is proved.
$\blacksquare$
Note that use of the transpose inverse (dual) representation shows that the complex conjugate of an irreducible character is an irreducible character. Also, as a matter of interest, it is not necessary to have the realizability theorem of the representations over a suitable cyclotomic field to know that algebraic conjugates of irreducible characters are irreducible characters. The irreducible characters of $G$ are (by Schur's Lemma) in bijection with the one-dimensional representations of $Z(\mathbb{C}G)$, the center of the complex group algebra.
Since the conjugacy class sums form a basis for $Z(\mathbb{C}G)$ and products of class sums are integer combinations of class sums, it follows that the linear characters of $Z(\mathbb{C}G)$ take algebraic integer values on class sums, and that algebraic conjugates of linear characters of $Z(\mathbb{C}G)$ are still linear character of $Z(\mathbb{C}G)$.
@GeoffRobinson: "The irreducible characters of $G$ are (by Schur's Lemma) in bijection with the one-dimensional representations of $Z\left(\mathbb{C}G\right)$, the center of the complex group algebra." Nice observation! (Although I don't see it as a consequence of Schur's lemma, but rather as a consequence of the Artin-Wedderburn theorem.) I'm not sure to what extent it could shorten my proof, though -- I still need a splitting field, don't I? (Composing a group representation with a Galois automorphism of the field clearly yields a group representation, but composing ...
... an algebra representation with a Galois automorphism might not -- or at least I don't see a quick reason why it should.)
In general, composing a linear representation of a commutative with a Galois automorphism does not lead to another representation, but, as I indicated, the special basis of class sums for the center of the group algebra (which multiply to integral combinations of class sums) means that an algebraic conjugate of a linear character leads naturally to another linear character, in the following. The (non-zero) linear function $\lambda: Z(\mathbb{C}G) \to \mathbb{C}$ yields an algebra homomorphism if and only if we have $ \lambda(C_{i}) \lambda(C_{j}) = \sum_{m=1}^{k} a_{ijm} \lambda(C_{m})$
.whenever we have $C_{i}C_{j} = \sum_{m=1}^{k}a_{ijm}C_{m}$, where $C_{i},C_{j},C_{m}$ are class sums, and the $a_{ijm}$ are the (rational integer) class algebra constants. The $\lambda(C_{i})$ are always algebraic integers (even cyclotomic integers), and it is clear that if $\lambda $ satisfies this property, so does $\lambda^{\sigma}$ (defined on class sums), when $\sigma$ is a Galois automorphism. Burnside and Frobenius knew this fact I think.
|
2025-03-21T14:48:31.337947
| 2020-06-21T10:33:16 |
363696
|
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"Adrien",
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|
Stack Exchange
|
Idempotent completion of an additive category
Let $\mathcal{A}$ be an additive category, and a morphism $e: X \rightarrow X$ in it an idempotent, i.e. $e \circ e = e$. We say that $e$ splits, if we can decompose $X = \operatorname{ker}(e) \oplus \operatorname{im}(e)$, and if every idempotent in $\mathcal{A}$ splits, we call it idempotent complete.
Note that there is a similar definition of a splitting idempotent (and idempotent completeness) for general categories, here $e$ is split when it arises as a retraction on a subobject, followed by its inclusion.
My question if the following: How are these two definitions related? To make this more precise:
Is there an additive category and an idempotent $e$ in it that splits in the second, but not in the first sense? (The other direction should not be possible, I think)
Is an idempotent complete additive category the same thing as a an additive category that is Cauchy-complete (this is a notion of idempotent complete for enriched categories) as an $\operatorname{Ab}$-enriched category? Or are the two notions distinct?
I think all of this is well explained on the nlab, eg https://ncatlab.org/nlab/show/Karoubian+category#general for your first question. For your second, a Cauchy complete Ab-enriched category is also closed under finite direct sums.
Yes you are right, that solves it - as it seems, both definitions are indeed eqivalent. I also found the ncatlab page where it is stated that Cauchy completion for Ab-categories is the same thing as idempotent completion and completion under finite direct sums. However I am still searching through the references there for a proof of this - I'll let you know if I find one.
|
2025-03-21T14:48:31.338096
| 2020-06-21T12:06:54 |
363703
|
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"Yaakov Baruch",
"https://mathoverflow.net/users/2383",
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"url": "https://mathoverflow.net/questions/363703"
}
|
Stack Exchange
|
Strange and non-strange prime numbers, are there infinitely many of them?
Definition. A prime number $p$ is called strange if there exists $k>1$ such that each prime divisior of $p^k-1$ divides $p-1$.
Example 3. The prime number $p=3$ is strange as $3^2-1=8$ has the same prime divisiors as $3-1=2$.
Example 5. The prime number $p=5$ is not strange, since for every $k>1$ the number $5^k-1$ is not a power of $2$ (by the Mihailescu Theorem). By the same reason the prime number $p=17$ is not strange.
Example 7. The prime number $p=7$ is strange since $7^2-1=48$ has the same prime divisors as $7-1=6$.
Example 31. The prime number $p=31$ is strange because $31^2-1=2^6\times 3\times 5$ has the prime divisors as $31-1=2\times 3\times 5$.
Using the small Fermat Theorem, it is possible to prove the following characterization
Theorem. A prime number $p$ is not strange if and only if for every prime divisor $q$ of $p-1$ the number $p^q-1$ has a prime divisor that does not divide $p-1$.
This theorem implies that the prime numbers $11,13,19,23,29,37,41,43,47,53,61,67,71,73,79,83,89,97$ are not strange.
Therefore, among prime numbers $<100$ only 3,7, 31 are strange. All these numbers are Mersenne numbers. In his comment Yaakov Baruch observed that each Mersenne number is strange. So, we can ask
Question 1. Is each strange prime number Mersenne prime?
Question 2. Is the set of non-strange numbers infinite?
Question 3. Is it true that for any number $x$ and prime numbers $p_1,\dots,p_n$ that not divide $x$, the arithmetic progression $x+p_1\dots p_n\mathbb Z$ contains a non-strange prime number?
$\DeclareMathOperator\rad{rad}$Is it obvious what changes if you weaken the requirement that $\rad(p^k - 1) = \rad(p - 1)$ for some $k > 1$ to just $\rad(p^k - 1) = \rad(p^\ell - 1)$ for some $k > \ell$? (More broadly, is yours just a definition that one could make, or is there some reason that it is a natural or interesting condition?)
Clearly every prime of the form $2^n-1$ is strange, therefore proving that there are only finitely many such primes seems currently hopeless.
@YaakovBaruch You are right, each Mersenne prime number is strange (and it is not known if the number of Mersenne primes is infinite). So, I changed the first question asking if every strange number is Mersenne prime.
@LSpice This question (especially the last one) was mottivated by some topological problems concerning the Kirch topology on ntural numbers. So, I indeed need that $rad(p^k-1)\ne rad(p-1)$ for "majority" of prime numbers.
Zsigmondy's Theorem shows that the only strange primes are the Mersenne primes. Indeed, it shows that for any $n\geq 2$ the number $p^n-1$ has a prime factor not dividing $p^k-1$ for any $k<n$, including $k=1$, with the only exceptions being $n=2$ when $p+1$ is a power of two.
It follows there are infinitely many non-strange primes, and existence of infinitely many strange ones is a famous open problem.
Thank you very much (dziekuje!). This is exactly what I needed to proceed my struggle with the Kirch topology on positive integers.
Do you know any modern reference to Zsigmondy Theorem (with proof)? I mean somewhere in a textbook. Thank you.
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2025-03-21T14:48:31.338331
| 2020-06-21T14:02:33 |
363709
|
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|
Stack Exchange
|
A structure sheaf for real analytification of semialgebraic sets in the context of signed tropicalization
Let $X=Spec(A)$ be an affine scheme, where $A$ be a commutative algebra over a non-archimedean valued field $K$. Assume that $K$ is a real closed field with the unique ordering $<$, which should be compatible with the absolute value $|\cdot|_K$ (i.e. it holds: $0<f\leq g$ iff $|f|_K\leq |g|_K$).
Now in the paper Real tropicalization and analytification of semialgebraic sets by Philipp Jell, Claus Scheiderer and Josephine Yu, the authors introduced the concept of a real analytification $X^{an}_r:= $ of $X$, which is basically an amalgation of the two following two familiar spaces:
The Berkovich spectrum $X^{an}=\mathcal{M}(A)$, consisting of all bounded real semivaluations $|\cdot|_x$ on $A$ extending $|\cdot|_K$, and
the real spectrum $X_r=Sper(A)$, consisting of all prime orderings $<_x$ on $A$.
The authors have shown that in the affine case $X^{an}_r$ can be described by the signed seminorms $|\cdot|^{sgn}_x$ (satisfying $|f|^{sgn}_x=sgn_x(f)|f|_K$) of $A$, and study the relation two the two classical spaces above. In particular, they showed that our new space inherits the nice topological properties of the Berkokovich spectrum, which they used to define a notion of tropicalization of semialgebraic subsets of $X(K)$. There exists a canonical map $\psi: X^{an}_r\to X_r$. If $M\subseteq X(K)$ is a semialgebraic subset, we call $M^{an}_r:=\psi^{-1}(\tilde{M})$ the real analytification of $M$, where $\tilde{M}$ is the abstraction of $M$.
I'm now wondering if it might be possible to endow these spaces $M^{an}_r$ (or more generally real analytifications of constructible subsets) with a nice structure sheaf. One could try to combine the structure sheaves of its parents:
Let $U\subseteq \mathcal{M}(A)$ be an open subset. We call a subset $Z\subseteq U$ special, if it can be written as $Z=\cup^n_{i=1}D_i$ where $D_i$ is an $k$-affinoid subdomain (Definition 13.10 in Mattias Jonsson's course over Berkovich spaces). We associate to $Z$ the affinoid $K$-algebra $A_Z:=ker(\Pi_i A_{D_i}\to \Pi_{i,j}A_{D_i\cap D_j})$. Then by setting
$$\mathcal{O}(U):=lim_{Z\subseteq U special}A_Z$$
we obtain a sheaf on $\mathcal{M}(A)$ (with respect to the special G-topology).
On a constructible subset $S\subseteq sper(A)$ there exists a sheaf of semialgebraic function developed by Niels Schwart in his Habilitationsschrift The Basic Theory of Real Closed Spaces. I will give a sketch of his construction. Let $\alpha,\beta\in S$ where $\beta$ is a specialization of $\alpha$. Then $supp(\beta)/supp(\alpha)\triangleleft A/supp(\alpha)$ is a convex prime ideal, such that there exists the largest convex subring $C_{\alpha\beta}\subset k(\alpha)$ with $supp(\beta)/ supp(\alpha)=\mathcal{m}_{\alpha\beta}\cap A/supp(\alpha)$. Denote $k(\alpha,\beta):=C_{\alpha\beta}/\mathcal{m}_{\alpha\beta}$, which is a real closed field with the canonical mapping $\lambda_{\alpha\beta}:C_{\alpha\beta}\to k(\alpha,\beta)$. Note that the monotone embedding $A/supp(\beta)\to k(\alpha,\beta)$ extends uniquely to $k(\beta)\subset k(\alpha,\beta)$.
Now we define the ring of compatible families $U_{A,S}(S)$ as the set of all families $f\in\Pi_{\alpha\in S}k(\alpha)$ such that for all $\alpha,\beta\in S$, where $\beta$ is a specialization of $\alpha$, we have: $f(\alpha)\in C_{\alpha\beta}$ and $\lambda_{\alpha\beta}(f(\alpha))=f(\beta)$. Now consider the subring $$R(S):=\lbrace f\in U_{A,S}(S)\mid \forall g\in A(S)\left[ f\right]: P(g)~is~constructible\rbrace$$,
where $P(g):=\lbrace \alpha\in S\mid g(\alpha)\geq 0\rbrace$ and $A(S)$ denotes the image of the canonical homomorphism $$\rho_S: A\to \Pi_{\alpha\in S}k(\alpha), f\mapsto (f(\alpha))_{\alpha\in S}.$$
We can consider the affine scheme $Spec(R(S))$ associated to $R(S)$. By Theorem 3.10 auf Schwart' paper the canonical map $supp: Sper(R(S))\to Spec(R(S)), \alpha\mapsto supp(\alpha)$ is a homeomorphism, so that we can consider the structure sheaf $\mathcal{O}_{Spec(R(S))}$ as sheaf on $Sper(R(S))$. Now we call $R_{A,S}:=\mathcal{O}_{Spec(R(S))\mid S}$ the sheaf of semialgebraic function on $S$, which makes $S$ into a locally ringed space.
My question now is whether we can possibly combine these two sheaves to become a sheaf for a real analytified constructible subset $S^{an}_r$, or if there might be another way to achieve this goal.
|
2025-03-21T14:48:31.338729
| 2020-06-21T14:29:07 |
363711
|
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|
Stack Exchange
|
Itō formula for the solution of a SPDE in the distributional sense
Let
$d\in\mathbb N$
$\Lambda\subseteq\mathbb R^d$ be open
$(\Omega,\mathcal A,\operatorname P)$ be a probability space
$(Y_t)_{t\ge0}$ be an $L^2(\Lambda)$-valued process on $(\Omega,\mathcal A,\operatorname P)$
$k\in\mathbb N$
$(W_t)_{t\ge0}$ be an $\mathbb R^k$-valued Wiener process on $(\Omega,\mathcal A,\operatorname P)$ with covariance operator $\operatorname{id}_{\mathbb R^k}$
$Q$ be a linear operator from $\mathbb R^k$ to $L^2(\Lambda)$
Let $(X_t)_{t\ge0}$ be an $L^2(\Lambda)$-valued process on $(\Omega,\mathcal A,\operatorname P)$ with $$\langle X_t,\varphi\rangle_{L^2(\Lambda)}=\langle X_0,\varphi\rangle_{L^2(\Lambda)}+\int_0^t\langle Y_s,\varphi\rangle_{L^2(\Lambda)}\:{\rm d}s+\langle QW_t,\varphi\rangle_{L^2(\Lambda)}\tag1$$ for all $t\ge0$ almost surely for all $\varphi\in C_c^\infty(\Lambda)$.
Now let $f\in C^2(\Lambda)$. Are we able to deduce an Itō formula for the process $f(X)$? At least for the choice $f(x):=\left\|x\right\|_{L^2(\Lambda)}^2$ for $x\in L^2(\Lambda)$?
Note that the ordinary Itō formula can be stated in the following way: If $U,H$ are separable $\mathbb R$-Hilbert spaces, $(\mathcal F_t)_{t\ge0}$ is a complete right-continuous filtration on $(\Omega,\mathcal A,\operatorname P)$, $W$ is a $U$-valued $\mathcal F$-Wiener process on $(\Omega,\mathcal A,\operatorname P)$ with covariance operator $Q\in\mathfrak L_1(U)$, $X_0$ is a $\mathcal F_0$-measurable $H$-valued random variable $X_0$ and $\varphi,\Phi$ are $H$-valued $\mathcal F$-predictable processes which are suitably integrable, then $$X_t:=\int_0^t\varphi_s\:{\rm d}s+\int_0^t\Phi_s\:{\rm d}W_s\;\;\;\text{for }t\ge0$$ is well-defined and if $f:[0,\infty)\times H\to\mathbb R$ is Fréchet differentiable in the first and twice Fréchet differentiable in the second variable, then \begin{equation}\begin{split}f(t,X_t)&=f(0,X_0)+\int_0^t\frac{\partial f}{\partial t}(s,X_s)\:{\rm d}s\\&+\frac12\int_0^t{\rm D}_x^2f(s,X_s)\:{\rm d}[\![X]\!]_s+\int_0^t{\rm D}_xf(s,X_s)\:{\rm d}X_s,\end{split}\tag2\end{equation} where $[\![M]\!]$ denotes the tensor-quadratic variation of $M$.
|
2025-03-21T14:48:31.338871
| 2020-06-22T17:28:57 |
363850
|
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"Piotr Achinger",
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"url": "https://mathoverflow.net/questions/363850"
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|
Stack Exchange
|
Compactification of Picard variety over normal, projective varieties
Let $X$ be a normal, projective, integral variety (over $\mathbb{C}$) and $P$ be the Picard variety parametrizing invertible sheaves on $X$. Does there exist a compactification $\overline{P}$ of $P$ along with an universal family $\mathcal{U}$ defined over $X \times \overline{P}$, flat over $\overline{P}$, parametrizing the space of rank $1$, torsion-free sheaves on $X$ (probably with a fixed Hilbert polynomial)? Any hint/reference will be appreciated.
For $X$ normal the connected component of ${\rm Pic}_X$ is projective (and over $\mathbf{C}$ it is an abelian variety). This was proved by Grothendieck in FGA, see Kleiman's survey article "The Picard scheme".
|
2025-03-21T14:48:31.338958
| 2020-06-22T18:14:58 |
363858
|
{
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"authors": [
"Alessandro Codenotti",
"Logan Fox",
"Taras Banakh",
"YCor",
"erz",
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"url": "https://mathoverflow.net/questions/363858"
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|
Stack Exchange
|
Topological properties inherited by the Hausdorff metric space
Let $(X,d)$ be a metric space and $(K_X , h_d)$ be the associated metric space of nonempty compact subsets of $X$ with the Hausdorff metric. It is well known that $K_X$ inherits certain topological (and analytic) properties from $X$. For example, if $X$ is compact, then so is $K_X$; and if $X$ is complete, then so is $K_X$.
Is there a reference which further explores the properties that $K_X$ inherits from $X$? In particular, if $X$ is locally compact then is $K_X$ also?
"The associated metric space of compact subsets of with the Hausdorff metric": you mean of nonempty compact subsets. Otherwise I don't see how you measure the distance to the empty set. For $X$ is locally compact, you can indeed make the set of compact subsets (including the empty set) topological in a natural way, and $\emptyset$ is isolated in this space iff $X$ is compact; in all cases the space of nonempty compact subsets is locally compact.
Yes, nonempty. I'll update the prompt
this is the reference, especially section $4$, showing that many properties of $X$ are preserved in $K(X)$ including being locally compact. Another one not mentioned in the paper is being zero dimensional. Regarding the empty set issue you can just set the distance between an empty set and a nonempty one to be $1$ and use the Hausdorff metric otherwise to get a metric compatible with the Vietoris topology on $K(X)$
(I was assuming the metric of $X$ to be bounded by $1$ in the above comment concerning the empty set)
Check out books Illanes, Nadler - Hyperspaces and Beer - Topologies on closed and closed convex sets. Also, some basic useful facts can be found in Engelking in exercises
Thanks @AlessandroCodenotti! The Ernest Michael paper is exactly what I was looking for. I will check out Illanes-Nadler and Beer also.
Side remark: if $X$ is metrizable, locally compact but not $\sigma$-compact (e.g., discrete uncountable), the space $K_X$ of nonempty compact subsets of $X$ is locally compact and metrizable, while the space $K_X\cup{\emptyset}$ of all compact subsets is compact but not metrizable.
It is known (I hope this is in Kechris' book) that for a metrizable separable space $X$ the space $K(X)$ is analytic if and only if $X$ is Polish if and only if $K(X)$ is Polish.
Michael, Ernest. “Topologies on spaces of subsets.” Transactions of the American Mathematical Society 71 (1951): 152-182.
The above paper shows in section $4$ that many properties of $X$, including being locally compact, are inherited by $K(X)$ (the latter space is called $\mathcal C(X)$ in the paper, while the author uses $2^X$ to refer to the hyperspace of closed sets of $X$, for which a more modern notation is $F(X)$).
In my comment I said that being zero-dimensional is also preserved but not shown in the paper, I now noticed that it is in fact shown there, together with more connectedness properties. The relationship between $\dim X$ and $\dim K(X)$ is much harder in general and has been widely studied, there is a whole chapter concerning this topic in the book Hyperspaces by Illanes and Nadler.
|
2025-03-21T14:48:31.339195
| 2020-06-22T18:46:47 |
363860
|
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|
Stack Exchange
|
Efficient algorithm for matrix equation $AXB + BXA = F$
For $n\in\mathbb{N}$, let $A\in\mathbb{R}^{n\times n}$ and $B\in\mathbb{R}^{n\times n}$ be two symmetric positive definite matrices and let $F\in\mathbb{R}^{n\times n}$ be arbitrary.
Is there any efficient algorithm for solving the matrix system $AXB + BXA = F$ (with $X\in\mathbb{R}^{n\times n}$ being the unknown) provided it has a unique solution?
I have found similar questions on the site, but they involve more assumptions on $F$, such as symmetry.
Thank you very much.
If you multiply by $A^{-1}$ on both sides, you get the Sylvester-type equation $X(BA^{-1}) + (A^{-1} B) X = A^{-1} F A^{-1}$. Such equations can be solved by LAPACK's dlasy2 routine. Presumably it's been efficiently implemented.
For dense problems, the standard algorithm is a generalization of the Bartels--Stewart algorithm: see for instance https://doi.org/10.1016/S0024-3795(87)90314-4 and https://people.cs.umu.se/isak/recsy/ for an implementation.
The basic idea of the algorithm is: there are (QZ decomposition) two orthogonal matrices $Q,Z$ such that $QAZ$ and $QBZ$ are both upper triangular (over $\mathbb{C}$, or block upper triangular with $2\times 2$ blocks on $\mathbb{R}$). Hence with a suitable change of basis one can reduce to the case in which $A$ and $B$ are upper triangular. At this point, the entries of $X$ can be computed with back-substitution. Checking complexities, one sees that all the various parts of this algorithm require only $O(n^3)$ floating-point operations. This compares very favorably with $O(n^6)$ for the "big linear system" approach.
Remarks:
the typical case is the slightly more general problem $AXB+CXD=E$; in your case, since some of the coefficients coincide, you can use only one QZ decomposition instead of two, but you need to change the order in which you compute the entries of $X$ by back-substitution.
You will need to save some intermediate values to compute the RHS of (13) in that paper in linear time; I did not see it mentioned there at a first glance but it is worth noting.
This algorithm has good stability properties because it relies on orthogonal transformations and (backward stable) back-substitution, and, unlike the algorithm mentioned in the comments, it does not break down if $A$ is singular or ill-conditioned.
For $F$ symmetric (which I note you have already looked at) is a mild generalization of the Lyapunov equation, see, e.g. (5.2) of http://www.dm.unibo.it/~simoncin/matrixeq.pdf (published version is at https://doi.org/10.1137/130912839).
For generic $F$, and indeed more general forms of this equation altogether, see section 7 of the paper above.
You can treat this as a system of $n^2$ linear equations for the entries of
$X$. The coefficient of $x_{ij}$ in the equation whose right side is $f_{kl}$ is $a_{ki} b_{jl} + b_{ki} a_{jl}$. The only way I can think of to make this "efficient" is to introduce some sparseness by taking a basis in which, say, $A$ is diagonal: then
the entry is $0$ unless $k=i$ or $j=l$.
Thank you. Tbh, I have already reduced a large system to this matrix equation, so my goal would be to solve the matrix equation without switching back to the large system.
|
2025-03-21T14:48:31.339422
| 2020-06-22T18:52:19 |
363861
|
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|
Stack Exchange
|
Diophantine equations that involve Gregory coefficients: a computational exercise
In this post that I've asked three weeks ago with same title in Mathematics Stack Exchange and identificator 3692235, for integers $k\geq 1$, we denote the Gregory coefficients as $G_k$. Wikipedia has an article for Gregory coefficients, are known as reciprocal logarithmic numbers (I add this as additional reference). I was inspired in problems that I know from the literature (in particular [1], that is from the section of problems of a journal) to solve the following diophantine equation that involves the first few Gregory coefficients in the brackets from RHS
$$y^2=1+\left(\frac{1}{2}n-\frac{1}{12}n^2+\frac{1}{24}n^3\right)$$
where we consider that $y\geq 1$ is integer and $n\geq 1$ also is integer.
Question 1. Prove or refute that the previous diophantine equation $$y^2=1+\sum_{k=1}^3G_k \cdot n^k\tag{1}$$
have no solutions $(n,y)$ when $y\geq 1$ and $n\geq 1$ run over positive integers. Can you find a counterexample? Many thanks.
My claim here was the following, that summarizes the things that I can see here (I don't know if previous question is easy to get). Also I know that $(1)$ is an elliptic curve (but in this post I'm interested in integral solutions).
Claim. Our equation $(1)$ can be rewritten as $n((n-2)n+12)=24(y-1)(y+1)$ (with help of Wolfram Alpha online calculator). From here we get easily (by contradiction) than $n$ is an even integer. And $n\equiv 0\text{ mod }3$ or $n\equiv 2\text{ mod }3$.
I've tested the conjecture stated in previous question for humble sets of integers. On the other hand I'm curious if there is some diophantine equation of the form $y^2=1+\sum_{k=1}^ NG_k n^k$ for some integer $N>3$ for which we can compute at least an integral solution $(n,y)$.
Question 2 (A computational exercise). Can you show an example of diophantine equation $$y^2=1+\sum_{k=1}^N G_k \cdot n^k\tag{2}$$ with at least a solution $(n,y)$, for integers $n,y\geq 1$ as before, where $N>3$? Many thanks.
I tried with my computer the first few values of $N$, the lowest of these integers $N>3$, and for $1\leq n,y\leq 5000$ both integers. If you can to answer Question 2 with a family of integral solutions, or you can find several examples of $N$ for diophantine equations $(2)$ having solutions feel free to expand your answer of this question.
I don't know if my questions are in the literature. If you know some of these from the literature refer it ansewring the questions as a reference request.
References:
[1] Fuxiang Yu, An Old Fermatian Problem: 11203, Problems, The American Mathematical Monthly, Vol. 114, No. 9 (Nov., 2007), p. 840.
Other reference in which I am inspired is On the Diophantine Equation $1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}=g(y)$ by Manisha Kulkarni and B. Sury from the book Diophantine Equations edited by N. Saradha, Tata Institute of Fundamental Research, Narosa Publishing House (2008).
I've asked my questions, this post, previously in Mathematics Stack Exchange without an answer, but if isn't suitable/interesting for this MathOverflow, please add a commment in next few hours, many thanks.
On this MathOverflow I've edited the post with title Partial sums involving Gregory coefficients that cannot be an integer. I add this as invitation, that in case that these posts have a good/interesting mathematical content for you, you can visit it and provide a contribution as a comment or adding an answer.
I've deleted the MSE post.
Integral points on elliptic can often be computed routinely. In Question 1, the curve can be rewritten as
$$Y^2 = 5184 + 432 X -12X^2 + X^3,$$
where $X:=6n$ and $Y:=72y$. SageMath computes:
sage: EllipticCurve([0,-12,0,432,5184]).integral_points()
[(0 : 72 : 1), (21 : 135 : 1)]
So, the only integer solution is $(n,y) = (0,1)$.
For Question 2 with $N=4$, we get a hyperelliptic curve:
$$(60y)^2 = 3600 + 1800n -300n^2 + 150n^3 -95n^4,$$
where the integral points can be found by Magma:
> IntegralQuarticPoints([-95, 150, -300, 1800, 3600]);
[
[ 0, 60 ]
]
So, again $(n,y) = (0,1)$ is the only integer solution.
This does not rule out the possibility of non-trivial solutions for $N>4$, but at very least we know that for any fixed $N>4$ there are only finitely many solutions.
Many thanks for your excellent answer. I think that it is reasonable that if in next week (the next five or six days) there are no more feedback I should accept your answer. Tomorrow I'm going to study the details of your answer and code.
|
2025-03-21T14:48:31.339723
| 2020-06-22T19:10:21 |
363862
|
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|
Stack Exchange
|
Connection as a jet section
Let $\pi:E\longrightarrow M$ a smooth fibre bundle. A connection is a linear bundle homomorphism $\Phi:TE\longrightarrow TE$ such that $\Phi$ is a projection to the vertical bundle $VE\subset TE$.
I read that a connection in $E$ is equivalent to a section $\Gamma:E \longrightarrow J^1E$, and the space of connection is an affine space.
What is this relationship, explicitly? and why it's an affine space?
Thanks
It is clear affine linear combinations $(1-t)\Phi_0+t\Phi_1$ of connections form a connection, following your initial definition, because the projection condition holds when you make such a combination. Hence an affine space.
"A connection is a linear bundle homomorphism $\Phi:TE\longrightarrow TE$" - This can't be, cause otherwise it would be $\mathcal{C}^\infty$-linear.
Dear Qfwfq, of course $\Phi$ is not the covariant derivative of a section, but you can easily obtain the covariant derivative by defining $\nabla s=\psi\circ \Phi\circ Ds$, where $Ds$ is the differential and $\psi\colon s^*VE\to E$ is the natural identification of the vertical bundle along a section. In fact, the two notions of a (linear connection) and a projection operator $\Phi$ satisfying certain linearity conditions are equivalent.
Dear alexpglez98, a projection to the vertical bundle is equivalent to the projection to a complementary bundle, usually called the horizontal bundle. If you specify a section of the 1-jet bundle you get a horizontal bundle (as the image). This horizontal bundle is complementary to the vertical bundle and hence yields projections to the two summands.
Thanks Ben and Sebastian. I understand now more or less what you say about the image. If $\Gamma:E \longrightarrow J^1E$ is a section, then $\Gamma(\xi)=j^1_p\phi$ for some section $\phi$. The image is $d\phi_p(T_pM)\subset T_\xi E$, it's well-defined because depends of the germ of $\phi$ at $p$ not on the section. But I don't see why this collection forms a horizontal subspace.
|
2025-03-21T14:48:31.339888
| 2020-06-22T20:23:40 |
363869
|
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|
Stack Exchange
|
Compressed sensing for partitioning instead of recovery
Let $x_0 \in \mathbb{R}^{m}$ be a signal whose support $T_0 = \{ t \mid x_{0}(t) \neq 0\}$ is assumed to be of small cardinality. The recovery of $x_0$ from a small number of $n \ll m$ linear measurements is the compressed sensing problem
$$y = Ax_{0},$$
where $A \in \mathbb{R}^{n \times m}$ is a sensing matrix. Obviously this is a very famous problem, with equally famous and celebrated results concerning the conditions required for the recovery of $x_0$.
But what if we do not care to recover $x_0$ completely? Instead we wish to partition $\{1, \dotsc, m \}$ into subsets $U_{i},\; i=1,\dotsc, |T_{0}|$ where $U_{i} \subset \{1, \dotsc, m\}$, $U_{i} \cap U_j =\emptyset\;\forall i,j$ and $|U_{i}\cap T_{0}|=1\;\forall i$.
In other words, we wish to find disjoint subsets $U_i \subset\{1, \dotsc, m\}$ such that for all $i$ there is only one value of $t \in U_i$ where $x_0(t)$ is nonzero.
Clearly, the conditions for recovery are sufficient to generate such a partition. However, it is not clear to me whether we can generate such a partition with significantly weaker conditions on $A$. For example, is it possible to construct these subsets with fewer measurements than would be required to recover $x_0$?
Does anyone have any insights on this? Has such a problem been studied in literature before?
It's been about a week and seem like no one has any ideas.
The best insight I have is as follows. Let $x_0$ be known to be a $k$-sparse vector. My thought is that the "information content" of the support of $x_0$, if treated as a random variable is something like
$$H(x_0) = \log \left( \binom{m}{k}\right) = O(k \log(m)).$$
But the information content of this partition is much lower, as shown in this paper by Fredman and Komlos,
$$H(U_1,\dotsc, U_k) = O(k + \log\log(m))$$
This shows nothing, but perhaps lends some credence to the idea that the partition can be constructed with fewer measurements.
how is the support set input ?
The setup is meant to exactly parallel the standard compressed sensing setup, in, for example https://arxiv.org/pdf/math/0503066.pdf by Candes and Tao (Section 1.1). It would even be of interest to deal with the case of $|T_0| = k \ll m$, where $k$ is known. Not sure if that answers your question, let me know! :)
Use \mid instead of | in $T_0 = { t \mid x_{0}(t) \neq 0}$.
Do you actually know a way of doing this with $O(k \log m)$ measurements? Which is the traditional entropy estimate of the subset (as opposed to the partition from Fredman/Komlos). Note that there are many famous algorithmic problems where the information theoretic lower bound has not been attained.
Here is one of the fundamental results of CS. In section 4.1, it shows that UUP is satisfied with high probability for a gaussian i.i.d matrix when $n = O(k\log(m))$. This UUP implies that linear programing can be used to reconstruct $x_0$.
|
2025-03-21T14:48:31.340111
| 2020-06-22T20:34:23 |
363870
|
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|
Stack Exchange
|
hypersurface of degree d Hilbert polynomial
I am in trouble to solve part (2) Exercise 1.13 from "Moduli of Curves"
by Harris and Morrison on page 9:
Exercise (1.13)
2) Fix a subscheme $X \subset \mathbb{P}^r$. Show, by taking cohomology
of the exact sequence of $X \subset \mathbb{P}^r$, that $X$ is a hypersurface of degree $d$ if and only it's Hilbert polynomial is.
$P_X(m)= \binom{r+ m}{m}- \binom{r+m-d}{m-d}$.
One direction is trivial: If $X$ is hypersurface of degree
$d$ defined by polynomial $f$ of degree $d$ then the sequence
$0 \to O_{\mathbb{P}^r}(-d) \to O_{\mathbb{P}^r} \to O_X \to 0$ does the job.
Another direction is more complicated: let $X \subset \mathbb{P}^r$ with
Hilbert $P_X(m)$. It's degree is $r-1$ thus $\dim \ X=r-1$.
Let us decompose $X= M_1 \cup ... M_s \cup D_1 \cup ... \cup D_t$
with $\dim \ M_i=r-1$ and $\dim D_j <r-1$. Every $M_i$ is a hyperplane
of degree $d_i$ and the main problem is to show that $s=1$, ie
$X$ has only one component of dimension $r-1$.
Because: assume we know $s=1$. Then $P_{M_1} + P_{\bigcup_j D_j}
= P_X +P_{M_1 \cap (\bigcup_j D_j)}$, so $deg(M_1)= deg(X)$ and
therefore $M_1$ is a hypersurface of degree $d$ and thus $P_{M_1}=P_X$.
The decomposition $X= M_1 \cup \bigcup_j D_j$ induces an exact sequence
$$0 \to I_{\bigcup_j D_j}/I_X \to O_X \to O_{M_1} \to 0$$
The Hilbert polynomial is additive wrt exact sequences and we obtain
$P_{I_{\bigcup_j D_j}/I_X}=0$, so $X=M_1$.
But I don't know how to show that $s=1$, ie that $X$ has only one
irreducible component of dimension $r-1$.
Addendum: Essentially my question reduces to following problem: Let $X= M_1 \cup ... \cup M_s \subset \mathbb{P}^r$ a purely $r-1$-dimensional subscheme of $\mathbb{P}^r$ (ie every irreducible component $M_i$ of $X$ has dimension $r-1$) with Hilbert polynomial $P_X(m)= \binom{r+ m}{m}- \binom{r+m-d}{m-d}= \binom{r+ m}{r}-\binom{r+m-d}{r}$. And we have to show that $X$ is irreducible, ie $s=1$. We observe that every $r-1$-dimensional subscheme $M_i$ is a hyperplane of degree $d_i$ and by additivity of the Hilbert polynomial we obtain $d= \sum_i d_i ^s$. But I don't know how to continue from here to prove the claim...
If $M_i$'s are hypersurfaces of degree $d_i$, then their union is a hypersurface of degree $\sum d_i$. So, can't you always take $s=1$?
@Mohan: I don't think that it's obvious here. Doing similar argument (namely split the Hilbert polynomial into union of dimension $r-1$ components $M_i$ and lower dimension components) as before then $M_i$ are hypersurfaces of degrees $d_i$ then again a degree comparison gives just $d= \sum_i^s d_i$. So why can it eg happen that $s=5$ with $\sum_i ^5 d_i=d$? or in other words that $X$ has $5$ irreducible components of of dimension $r-1$? That's exactly the crucial part
or did I maybe misread your argument?
@Mohan: Essentially my question reduced to following problem: Let $X= M_1 \cup ... \cup M_s \subset \mathbb{P}^r$ a purely $r-1$-dimensional subscheme of $\mathbb{P}^r$ with Hilbert polynomial $P_X(m)= \binom{r+ m}{m}- \binom{r+m-d}{m-d}= \binom{r+ m}{r}-\binom{r+m-d}{r}$. Show that then $X$ is irreducible, ie $s=1$. We observe that every $r-1$-dimensional subscheme $M_i$ is a hyperplane of degree $d_i$ and by additivity of Hilbert polynomials we obtain $d= \sum_i d_i ^s$. And what to do next?
Why do you think $X$ would have to be irreducible? I don’t see where that idea is coming from. In the “easy” direction you use the degree of the hypersurface, but you don’t use irreducibility.
@ZachTeitler: That's exactly the claim in the exercise I have quoted above. I have to show that if I have a subscheme $X \subset \mathbb{P}^r$ with the Hilbert polynomial given above, then $X$ is a degree $d$ hypersurface (ie irreducible). And in the “easy” direction you are right: I only use the assumption that $X=V(f)$ for homogeneous polynomial of degree $d$.
A "hypersurface" is not necessarily irreducible; $V(xy)$ is a hypersurface, it is reducible. Plus: you are trying to prove something that is not true. Having the given Hilbert polynomial does not imply that the hypersurface is irreducible. I think you are supposed to prove that with the given Hilbert polynomial, the variety is a degree $d$ hypersurface (ie., defined by a single equation); but it can be reducible (the defining equation might factor as a product). You can't prove it's irreducible because that's not necessarily true (with only information of Hilbert polynomial).
@ZachTeitler: Yes I think I see my thinking error. I indeed wrongly assumed that hypersurface has to irreducible. So a hypersurface of $\mathbb{P}^r$ is by definition a closed subscheme of the structure $V(f)$ with homogeneous polynomial $f \in k[x_0,...,x_r]$. And of course as you said $f$ can be reducible and thus $X$ as well. On the other hand is it true (and why) that if $X \subset \mathbb{P}^r$ is irreducible subscheme of dimension $r-1$ then it has structure $X=V(f)$ for $f$ homogeneous polynomial?
(On phone, sorry for terseness) https://math.stackexchange.com/a/2638397
|
2025-03-21T14:48:31.340410
| 2020-06-22T21:48:38 |
363873
|
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|
Stack Exchange
|
Is there a function that is not absolutely integrable in [−π,π] so that its Fourier Series Exists?
For existence of Fourier coefficients of a function f is sufficient that f is absolutely integrable in [−π,π] but, is this condition necessary? that is, is there a function that is not absolutely integrable in [−π,π] so that its Fourier series Exists?
Consider the usual trigonometric system.
How do you define $\int_{-\pi}^\pi f(x)dx$, if $\int |f|dx = \infty$? More concrete: What is then its Fourier Series?
This is the problem, because if $\int_{-\pi}^{\pi} |f(x)|dx$ converges then Fourier Series of $f$ exists always , but is possible that is there a functión that $\int_{-\pi}^{\pi} |f(x)|dx$ diverges (no necessarily $\int_{-\pi}^{\pi} f(x)dx$ diverges) and its Fourier series exists?
if it is not possible why?
If $\int |f|d\mu = \infty$, then always $\int fd\mu = \pm \infty$ (if either $\int f^+d\mu < \infty$ or $\int f^-d\mu < \infty$) or this integral does not exist.
I consider that this is false, because $\int_{a}^{\infty} \dfrac{sin(x)}{x}dx$ converges $\forall a>0$ but $\int_{a}^{\infty} \dfrac{|sin(x)|}{x} dx$ diverges
@PieroD'Ancona yes, I know. In this theory all is very nice because $\mathcal{L}_2$ is a Hilbert space. However, Serie Fourier also can be defined for another type of function (Function $f$ such that $f \notin \mathcal{L_2}$), and in these cases is this problem.
It is difficult to answer this question unless you supply a precise definition of what you mean by saying "the Fourier series exists". Given a function on $[-\pi,\pi]$, how do you propose to define its Fourier coefficients? And assuming you can define each individual Fourier coefficient, what do you mean by a series "existing"? Do you require the series to converge in some way? If not, then aren't you just asking if each Fourier coefficient exists?
The question has also been posted on MathStackExchange https://math.stackexchange.com/questions/3730663/is-there-a-function-that-is-not-absolutely-integrable-in-pi-pi-so-that-it which I think is a more appropriate place for it
All $2π$ periodic distributions are temperate and since all temperate distributions have a Fourier transform, you have plenty of examples. Note that the previous statements prove the power of abstract nonsense in Mathematics: you define for $u\in \mathscr S(\mathbb R)$ the Fourier transform
$$
\hat u(\xi)=\int_{\mathbb R} e^{-2iπ x \xi } u(x) dx,
$$
and you can prove that it is an isomorphism of $\mathscr S(\mathbb R)$, with the inverse given by
$
u(x)=\int_{\mathbb R} e^{2iπ x \xi } \hat u(\xi) d\xi.
$
Well, not trivial but very standard with direct proofs. Then you dramatically increase the generality by defining the Fourier transform of a temperate distribution $T$, as
$$
\langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R),\mathscr S(\mathbb R)}
=\langle T,\hat \phi\rangle_{\mathscr S'(\mathbb R),\mathscr S(\mathbb R)},
$$
and with $\mathcal F$ standing for the Fourier transform and $\mathcal C$ for the mapping $T(x)\mapsto T(-x)$ (clearly defined for a function and also by duality for distributions). Then (and this is the power referred to above), you get trivially that for any tempered distribution $T$
$$
\mathcal C\mathcal F \mathcal F T=T.
$$
For instance you get with $H=\mathbf 1_{\mathbb R_+}$,
$
\hat H=\frac12\delta_0+\frac{1}{2π i}\textrm{pv}\frac1x.
$
Yes, of course you can extend the domain of definition and get the above mentioned results. But I think a beginner (?) is misguided by not answering the original question.
The OP is working on the torus $[-\pi,\pi]$, not on the whole real line.
@PhoemueX Periodic functions are defined everywhere and a classical way to obtain the classical expansion in Fourier series for a periodic distribution is to use the Fourier inversion formula above on the real line.
@Dieter Kadelka I tried to answer the question which was "Can we define the Fourier transform on a larger set than $L^1$?". The distributional framework with temperate distribution is providing a very large space on which the Fourier transform makes sense.
How about $1/x$, for distributional Fourier transform
|
2025-03-21T14:48:31.341107
| 2020-06-22T22:59:10 |
363880
|
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|
Stack Exchange
|
Limited rook moves
I have an algebra problem, that could be solved if I could answer the following combinatorial problem.
Let $S$ and $T$ be two nonempty sets. We think of $S\times T$ as the index set for the squares of a rectangular chessboard.
Let $\emptyset\neq P\subseteq S\times T$. We think of $P$ as indices of certain special squares of the chessboard. We can call them the on-squares, while $S\times T-P$ are the indices of the off-squares.
We think of the on-squares as places where rooks can land. A rook can pass over off-squares, but can only move from one on-square to another on-square. (In other words, a rook can move from $(s,t_1)$ to $(s,t_2)$; or from $(s_1,t)$ to $(s_2,t)$, assuming those pairs are in $P$.)
We make the following assumptions about $P$:
(1) Any two squares indexed by $P$ are connected by a finite sequence of rook moves.
(2) For every $s\in S$, there is some $t\in T$ such that $(s,t)\in P$; and similarly for every $t\in T$ there is some $s\in S$ such that $(s,t)\in P$. (In other words, no column or row of our chessboard consists only of off-squares.)
For example, consider the $3\times 3$ board below, where we have placed $\bullet$'s on the on-squares and $\circ$'s on the off-squares.
$$
\begin{array}{|c|c|c|}
\hline
\bullet & \circ & \circ \\ \hline
\bullet & \circ & \bullet \\\hline
\circ & \bullet & \bullet\\\hline
\end{array}
$$
We can get from any on-square to any other on-square by a sequence of no more than 4 rook moves.
We make one further, invariance assumption.
(3) If $(i,j),(k,l)\in P$, then the map $i\mapsto k$ extends to a permutation $\sigma$ of $S$, and the map $j\mapsto l$ extends to a permutation $\tau$ of $T$, such that $\sigma\times \tau$ induces an permutation of $P$. In other words, up to renaming the rows and columns, any on-square looks exactly like any other on-square.
That's the set-up. To motivate my question, consider the case when on-squares are connected by a sequence of at most two rook moves. This holds, for instance, when all squares are on-squares. However, consider the case when there is an off-square. Then there is a partial diagram of the form
$$
\begin{array}{|c|c|}
\hline
\bullet & \\ \hline
\circ & \bullet \\ \hline
\end{array}
$$
The upper right corner must then be an on-square, else we cannot connect the two on-squares in two moves. So, the diagram must fill in as
$$
\begin{array}{|c|c|}
\hline
\bullet & \bullet \\ \hline
\circ & \bullet \\ \hline
\end{array}
$$
The lower right square is connected by a column then row move to another on-square, but it isn't connected to that on-square by a row move followed by a column move. Thus, by condition (3), the upper left square also has to have this property. So the diagram expands as
$$
\begin{array}{|c|c|c|}
\hline
\bullet & \bullet & \\ \hline
\circ & \bullet & \bullet \\ \hline
&\circ & \bullet \\ \hline
\end{array}
$$
The upper right square is then forced to be on, and the lower left square is forced to be off.
Continuing this process, one obtains an infinite chessboard whose on-squares are all in the upper right part, and the off squares are in the lower left half. It still isn't invariant unless one makes some additional modifications, but it is possible to form such a board.
Here is my question:
If we instead assume that all on-squares are connected by 3 moves or less, is it possible for some of the on-squares to only be connected by row, then column, then row moves, but not column, then row, then column moves (including trivial moves)?
Maybe I'm slow today, but I am not groking your motivating example. Taking an 8x8 board as an example, let all the off squares be in a 4x4 corner. Granted this does not fill condition 3, but all the on squares are two-distant or one-distant from each other. What prevents an infinite example following a pattern of four by four blocks? Further, with three-distant configurations, perhaps you should look at vertical and horizontal "off" rectangles. Gerhard "Perhaps Forming Stair Step Pattern" Paseman, 2020.06.22.
@GerhardPaseman It seems to me that your 8x8 example does fulfill 3, but not 2, and your "4 by 4 blocks" example doesn't fulfill 1. However, it doesn't fulfill the condition in the question at the end.
I'm curious - what are the "additional modifications" you mention near the end? I'm pretty sure I can show that if all on-squares are connected by length-2 paths, then a square's on-status is determined by only one of its x-position and its y-position.
Say the board is $\mathbb{Z}^2$, and the on-squares are $(i,j)$ with $i>j$. Then the on-square $(2,0)$ does not look like $(2,1)$ (up to renaming rows and columns). Instead you have to pass to something like $\mathbb{Q}^2$.
And maybe make $(i,j)$ an on-square when $i>j\sqrt{2}$.
Ah - of course, I forgot that infinity allows some things. I think it shouldn't be hard to prove in the 2-length paths case that the on-squares give total orders (with equivalence allowed) on rows and columns separately, but that doesn't lead to what I said.
Yes, this is possible. The sets
$$\mathbb A := \mathbb Q \setminus \{i 2^j \mid i,j \in \mathbb Z\} $$
and
$$\mathbb B :=\{B_{i,j}:=(i2^j,(i+1)2^j) \mid i,j \in \mathbb Z\}$$
are countable - the important bit here is that for fixed $j$, the $B_{i,j}$ are a disjoint cover of $\mathbb A$, and that any $B_{i,j+1}$ is a union of two of the $B_{i,j}$.
Instead of a $(\mathbb Z \times \mathbb Z)$-board consider an $(\mathbb A \times \mathbb B)$-board with on-squares $(a,b)$ for $a \in b$.
Any two on-squares are connected by a sequence of three moves since any two intervals in $\mathbb B$ have a common upper bound (w.r.t. union) in $\mathbb B$, in particular (1) holds. Property (2) holds since any element of $\mathbb A$ is contained in infinitely many intervals in $\mathbb B$. To see that (3) holds is a bit trickier. Note that the following maps $\mathbb Q \to \mathbb Q$ setwise preserve $\mathbb A$ and $\mathbb B$ for $i,j\in \mathbb Z$, $i$ even:
$$
\phi_j \colon q \mapsto 2^j q \qquad \text{and} \qquad
\psi_{i,j} \colon q \mapsto
\begin{cases}
q+2^j & i2^j < q \leq (i+1)2^j \\
q-2^j & (i+1)2^j < q \leq (i+2)2^j\\
q&\text{otherwise}
\end{cases}
$$
A map between elements $(a_1,b_1)$ and $(a_2,b_2)$ can be constructed as a limit of the above maps:
First note that there is $j_0\in \mathbb Z$ such that $\phi_{j_0} (b_1)$ and $b_2$ have the same length. Next note that there is some $J \in \mathbb Z$ such that $\phi_{j_0} (a_1)$ and $a_2$ lie in the same $B_{i,J}$. If they don't lie in the same $B_{i,J-1}$ then $\psi_{i,J-1} (\phi_{j_0} (a_1))$ and $a_2$ do. Proceed inductively for j=J-2,J-3,etc. always applying an appropriate $\psi_{i,j}$ when necessary to ensure that the image of $a_1$ lies in the same $B_{i,j}$ as $a_2$.
Note that every $q \neq a_1$ is only affected by finitely many of the $\psi_{i,j}$ (depending on the distance between $a_1$ and $q$) and that the images of $a_1$ converge to $a_2$. In particular the limit map $f \colon \mathbb Q \to \mathbb Q$ is well defined with $f(a_1)= a_2$. Moreover, $f$ preserves $\mathbb B$ setwise since every $B_{i,j}$ is setwise fixed by all but finitely many $\psi_{i,j}$, and clearly $f(b_1) = b_2$.
Finally, the only paths of length 3 from $(a,B_{0,0})$ to $(a+1,B_{1,0})$ are of the form $$(a,B_{0,0}) , (a,B_{0,j}) , (a+1,B_{0,j}) (a+1,B_{1,0})$$
for some $j > 0$ (note that at some point we have to change the $a$ to an $a+1$, and we can only do this if we change the $\mathbb B$-coordinate before and after it).
I'm having trouble showing that $(-1/3, (-1/2,0/2))$ is connected to $(1/3, (0/2,1/2))$.
You are right, and the maps $\phi_j$ and $\psi_{i,j}$ also never swap positive and negative numbers which wrecks property (3) - but restricting to $\mathbb Q^+$ should fix it, i.e. $\mathbb A := \mathbb Q^+\setminus {i 2^j \mid i,j \in \mathbb Z, i \geq 0}$ and $\mathbb B := {(i2^j,i+12^j)\mid i,j \in \mathbb Z, i \geq 0 }$
|
2025-03-21T14:48:31.341643
| 2020-06-23T00:10:54 |
363882
|
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|
Stack Exchange
|
Presentation of the Monster as a Hurwitz group
The Monster group is the largest of the sporadic simple groups, and has been proven by Wilson to also be a Hurwitz group. It has a presentation in terms of Coxeter groups, specifically Y443 along with the "spider" relator, and quotienting out by the center. However, I am interested in a presentation as a Hurwitz group.
Specifically, what is a presentation of the monster in terms of two elements a and b such that a has order 2, b has order 3, and ab has order 7? Also, what is the smallest possible order of the commutator of a and b in such a presentation?
It doesn't look like this is known, at least none of the papers citing Wilson's paper seem to do it. I'm not an expert though so I suppose I could have missed something.
I would guess that it would be possible to compute such a presentation by applying a standard change of generator algorithm to the existing presentation, although the resulting presentation would be unlikely to be particularly illuminating. To do that you would need to be able to do basic computations with elements of the Minster, but that is possible - Wilson has software for that. Wilson would also be the best person to ask about the minimal order of $[a,b]$.
Actually, the last question does not depend on a presentation, but just on a generating subset $(a,b)$ satisfying the $(2,3,7)$ relations (presentation refers to an explicit set of relations). Whether there's a "short" presentation is intriguing, anyway.
I suppose such a presentation gives a closed surface with the Monster group acting on it, but it will be pretty big.
A related question: is there a presentation by three involutions whose products have order 2, 3, and 7 respectively?
I have computed two pairs of generators $(a,b)$ of the Monster satisfying
the relations $a^2 = b^3 = (ab)^7 = 1$ using [1]. In both cases $a$ is of
class 2B, $b$ is of class 3B in the Monster, and the commutator $[a, b]$
has order 39. This gives an upper bound for the minimal order of that commutator.
More details of the computation are given in subdirectory
applications/Hurwitz of [1].
[1] Martin Seysen, the python mmgroup package,
https://github.com/Martin-Seysen/mmgroup
This is amazing, thank you! I've had a look through the comments on the github package, that does seem to take quite a while to find a Hurwitz generating set. Are calculations for that sill ongoing? I'd be keen to see if there are other possible generator pairs, with a different order of the commutator.
Computations are still ongoing. I expect to find about one pair per week, and a friend of mine may find about one pair per day with a faster computer. I have not yet looked for pairs $(a,b)$ of classes (2B, 3C), which may exist according to the paper The Monster is a Hurwitz group by R.A. Wilson,
Have any new pairs been found? @Martin Seysen
Altogether I found 5 pairs pairs $(a,b)$ of classes (2B, 3B). The commutator has order 39 in all cases.
|
2025-03-21T14:48:31.341890
| 2020-06-23T00:55:53 |
363883
|
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|
Stack Exchange
|
Can a quotient space of a locally convex space have finer topology that its domain?
The following is related to this post.
Let $X=X'$, as sets, and let $T:X \rightarrow X'$ be a surjective map from a countably infinite-dimensional LCS $X$ to itself and equip $X'$ with the final topology. Thus, $T$ is a quotient map.
When/(it possible) for the topology of $X'$ to be finer than that of $X$ or contain open sets not in the topology of $X$?
Yes. Take a metrizable lcs $X_1$ and a non-metrizable lcs $X_2$. Then the projection operator $X_1\times X_2\to X_1$ is not a homeomorphism.
@TarasBanakh Following your comment, I realized my initial formulation was trivial. I rephrased the question to better express what I had in mind.
Ok, then take a metrizable countably-dimensional lcs $X_1$ and the countably dimensional lcs $X_2$ with the strongest lcs topology (it is usually called the space $\varphi)$. Then $X_2$ is the image of $X_1\times X_2$ and algebraically, $X_2$ is isomorphic to $X_1\times X_2$. The space $X_2$ has the finer topology than $X_1\times X_2$ since the topology of $X_2$ is the largest lcs topology on $X_2$.
|
2025-03-21T14:48:31.342022
| 2020-06-23T01:03:11 |
363884
|
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|
Stack Exchange
|
What is the significance of the fact that $L^\infty$ is a dual space?
Is there some significance to the fact that $L^\infty$ is a dual space? How might this be important or significant in some way?
More precisely, it's dual to $L^1$, but that's not a point I necessarily wanted to concentrate on in my question.
What do you view as significant or important? Put it this way, in your research have you ever found it useful to know that a particular Banach space is a dual space?
Also: what is wrong with the answer you received (and accepted) when you asked this question on MSE? https://math.stackexchange.com/questions/3727120/is-there-some-reason-that-the-fact-that-l-infty-is-the-dual-space-of-l1-is
Likely relevant context: Sullivan's statement that Connes is excited about this fact, at 2:45 in https://www.youtube.com/watch?v=ixc0TNfT0ks
@user_1789 "He has given us a shoe! The shoe is the sign!" https://www.youtube.com/watch?v=Ka9mfZbTFbk
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2025-03-21T14:48:31.342118
| 2020-06-23T02:18:41 |
363888
|
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|
Stack Exchange
|
Constructing exotic $\mathbb{R}^4$'s using vector fields on $\mathbb{R}^5$
I was reading a paper of Arnol'd ("Topological Properties of Eigenoscillations
in Mathematical Physics") where he gives the following claim (hopefully I am stating it correctly).
One way to produce smooth 4-dimensional manifolds is to take some smooth, non-vanishing vector field $v$ on $\mathbb{R}^5$. The flow of this vector field defines a smooth $\mathbb{R}$-action on $\mathbb{R}^5$, and then we can just take the quotient of $\mathbb{R}^5$ by this action to produce some smooth 4-manifold $M$.
Arnol'd makes the interesting claim that, given any exotic $\mathbb{R}^4$ we know about, it can be produced by a careful choice of this vector field).
Could anyone shed some light on the details of this construction?
$Exotic(\mathbb R^4) \times \mathbb R$ is diffeomorphic to $\mathbb R^5$ as we know that there exits unique smooth structure on $\mathbb R^5$ (proved by Stalling A reference for smooth structures on R^n). Now there exists a nice $\mathbb R$ action on $Exotic(\mathbb R^4)\times \mathbb R$, i.e, translation along the $\mathbb R$ axis. And push-forward of this action will generate a smooth action on $\mathbb R^5$ which is the one you are looking for.
Great, thanks! This is probably what Arnol'd meant, but I was under the impression that he had some explicit construction method in mind.
It would be nice to see if there exists any explicit construction. Most of the proofs that I know about exotic structures on $\mathbb R^4$ is in some-sense existential.
It would be nice if you could add references to your statements.
@PiotrHajlasz I added a link of MO which contains nice explanations.
|
2025-03-21T14:48:31.342271
| 2020-06-23T03:09:24 |
363890
|
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|
Stack Exchange
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When is a distribution having a finite support actually zero?
Let $D$ be a differential operator with smooth coefficients in $\mathbb{R}^n$. Suppose $E$ is a bounded open set in $\mathbb{R}^n$. Suppose $u$ is a function that is smooth up to the boundary of $E$. If $D(u\cdot\chi_{E})$, as a distribution in $\mathbb{R}^n$, has a finite support, is it true that $D(u\cdot\chi_{E})$ is actually zero?
The answer is no in the case $n=1$. An example is $D=d/dx$, $E=(0,1)$ and $u(x)=1$ for all $x$. In this case, the support of $D(u\cdot\chi_{(0,1)})$ is $\{0,1\}$. I hope that the answer is yes if $n\geq 2$. If the answer is yes, what weaker conditions in u are actually needed to guarantee that the answer is yes? Any suggestion for appropriate references would be greatly appreciated.
I think the answer is no even in higher dimension, since the gradient of any characteristic function $\chi_\Omega$, $\Omega\Subset\Bbb R^n$ as exactly the same properties of $D=d/dx$, i.e. $\nabla \chi_\Omega\subseteq\partial\Omega$ as it is shown in this Q&A. And if you need a single PDE satisfying this property, you can take the divergence of the gradient and form the laplacian of the characteristic function.
Let $E$ be the square $(0,1)^2$ in $R^2$, $D=\partial_x\partial_y$ and $u=1$. The support of $D(\chi_E u)$ is the set of corners of $E$.
Up to a factor $1/\sqrt\pi $.
$E$ seems unbounded in you example.
Perhaps you should edit your answer and only keep the $\partial_x\partial_y$ example? the first part is wrong not only because the domain is unbounded, but more importantly because the fundamental solution is NOT smooth up to the boundary
Right. I just liked much more the fundamental solution example :)
While I will accept your answer for the original question, if $E$ is the unit ball (or generally, $E$ has a smooth boundary), would there be an example to show that the answer is also no? The unit ball is what I need but since I (falsely) hoped that the answer were yes, my original question asked for arbitrary $E$.
If $E$ has a smooth boundary, we can assume it is a half space $x_n>0$ near a point (say the origin) of the support of $D(\chi_Eu)$. Then $\chi_E=H(x_n)$ and we can write $D(\chi_Eu)=\Sigma_{j=0}^k a_j(x)H^{(j)}(x_n)$. Then $a_k$ must vanish in a punctured nbh of the origin in $x_n=0$ and so also at the origin by continuity (assuming n>1). Using that $x_nH^{(k)}(x_n)$ is a multiple of $H^{(k-1)}(x_n)$ we can eliminate the highest order term in the sum and inductively continue to eliminate all (I guess).
In the smooth case I believe you are right, and it might be sufficient to consider the case $E$ is a half space (EDIT: see the previous comment :)
What weaker conditions on u can we have for the conclusion to hold still? Of course we only need derivatives of u up to the order of D but can we do less than that?
I think that kind of assumption is very close to necessary. If you consider again the heat kernel example, integrating once w.r.to time you get a bounded (Holder 1/2) counterexample for the operator $D=\partial_t(\partial_t-\partial_{xx})$
|
2025-03-21T14:48:31.342505
| 2020-06-23T04:29:55 |
363892
|
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"Harry Gindi",
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|
Stack Exchange
|
Is every space a classifying space?
Despite a pretty thorough look (I think) I can’t find the answer to the following question: Is every (reasonable?) path connected space weakly equivalent to the classifying space of some topological group?
I’ve found some related results, such as the Kan-Thurston theorem, but nothing that answers this question.
This is true. If X is a pointed, path-connected space, then its loop space $\Omega X$ is weakly equivalent to a topological group and $X$ itself is weakly equivalent to its classifying space. This is perhaps most easily proven by working with simplicial sets, see Goerss-Jardine V.5 and V.6.
To expand on what Piotr said, there is an equivalence of categories between so-called 'grouplike E_1-spaces' and pointed connected spaces. It is then a classical theorem that every grouplike E_1-space admits a strict model (due to Stasheff I think? Although he would have called an E_1 space an A_∞-space).
@PiotrPstrągowski Might as well post your excellent (and highly-upvoted) comment as an answer.
|
2025-03-21T14:48:31.342617
| 2020-06-23T05:35:36 |
363894
|
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|
Stack Exchange
|
If $A^TA \ge B^TB$ does this imply $AA^T \ge BB^T$?
If $A^TA \ge B^TB$ does this imply $AA^T \ge BB^T$?
If not, is there a counter example?
what is the ordering $\ge$ used here? and are the matrices all square?
A counterexample has already been given, but one can already guess the answer by noting that the two inequalities have different symmetries. The former inequality is invariant with respect to multiplying A,B on the left by arbitrary orthogonal matrices, whilst the latter is invariant with respect to multiplying A,B on the right by arbitrary orthogonal matrices.
Take
$$A = \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}, \quad B = \begin{pmatrix}1 & 0 \\ 0& 0 \end{pmatrix}.$$
|
2025-03-21T14:48:31.342766
| 2020-06-23T06:51:58 |
363897
|
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|
Stack Exchange
|
References for quivers and derived categories of coherent sheaves for a string theory student
I'm a student mostly from physics knowledge hoping to learn about the math involved the string theory paper Topological Quiver Matrix Models and Quantum Foam.
Context: The topological string theory partition function can be understood as computing the Donaldson-Thomas invariants of the variety over which the topological string is defined via the Donaldson-Thomas/Gromov-Witten correspondence. That's ok to me, the problem is that I've started to find fascinating, and apparently isolated examples of connections between quiver representations and the derived category of coherent sheaves of some varieties (especially toric cases) that I dont' fully understand.
The prototypical example of the aforementioned connections, is the well known realization of the moduli space of stable, rank $r$ (and $c_{2}=n$) and torsion-free sheaves on $\mathbb{P}^{2}$ as the $\mathcal{M}(n,r)$ quiver variety of the Jordan quiver under the Geiseker stability condition; the first data is precisely what the Donaldson-thomas theory of $\mathbb{P}^{2}$ actually computes. Another example is the Nakamura computation of the $G$-equivariant Hilbert scheme of points over $\mathbb{C}^{3}/G$ where $G$ is a finite $SL(3,\mathbb{C})$ subgroup as described in the page 14 of the paper "The McKay correspondence" using McKay quivers. This latter fact was used in Crystals and Black holes to enummerate tautological sheaves over a crepant resolution of $\mathbb{C}^{3}/G$ to compute the topological string partition function.
My problem: I supspect that the connections are not accidental but I'm uncapable to see what's the precise relation between moduli problems of quiver representations and those ones of sheaves, or where to start to investigate.
My background: I've sudied algebraic geometry from the first four chapters (Varieties,Schemes,Cohomology and Curves) in Hartshorne's textbook, I'm also familiar with the identification between the derived bounded category of coherent sheaves and the D-branes of the topological string B-model.
My weakness: I know very little about representation theory of quivers.
Questions: In Topological Quiver Matrix Models and Quantum Foam
is apparently assumed that we can associate to a given toric variety a quiver whose derived category of representations is isomorphic to the derived bounded category of coherent sheaves of the given toric scheme.
1.-Does anyone know a gentile reference to learn about the mathematical details of how this can be explicity achieved?
2.- What could be a good reference to start learning about quivers focused to understand the papers Topological Quiver Matrix Models and Quantum Foam and Crystal Melting and Black holes given my prior knowledge and physics orientation.
Any comment or reading suggestion is very welcome.
Without knowing too much about the paper (I do not know enough physics to dig into your primary reference) if the toric variety possesses a strong full exceptional collection then its derived category is equivalent to the derived category of quiver representations. For a more complete discussion, see this post.
As for quivers, I have found Kirillov's "Quiver Representations and Quiver Varieties" to be helpful and fairly explicit.
One further comment, full exceptional collections of sheaves on toric varieties have been constructed by Kawamata (he has several papers on this). However there is no mention of when they are strong. Hopefully someone more knowledgeable can chime in.
Thanks for your time, kindness and interesting comments DK. I was unaware about Kawamata's work, and indeed, searching for his name in ArXiv shows a lot of wonderful and truly useful references. That's precisely what I was searching for. Thanks again.
My pleasure, good luck with your work.
First, that review is somewhat depressing in that it's been over ten years since people figured out how to write down explicit boundary conditions in the B-model for objects in the derived category, but it's still talking about 'tachyon condensation' and locally-free resolutions, which do not always exist. I'm partial to the discussion in my old paper, but see also Kapustin et al and Herbst et al.
For what it's worth, the main bolded statement in the review is wrong -- the D-branes in the B-model do not need to be stable. Stability depends on the Kahler information of the target and has to do with the physical D-branes, not the topological ones.
Anyways, to answer your actual question, when you have an equivalence of categories between the derived category of coherent sheaves on noncompact CY and the derived category of representations of a quiver algebra, you often get that a component of the moduli space of representations with a specific dimension vector is the original noncompact CY. Physically, you can think of this as the moduli space of D0-branes being the cone itself. These D0-branes naturally correspond to representations of the quiver with a fixed dimension vector, and you can find the cone in the moduli stack pretty easily. With a little more work, you can get the GIT quotient too. You can see this in my two papers with Nick Proudfoot 1 and 2. There have been generalizations of this work, but I don't know if it has been proven for all the toric stuff (I've been away from this for a while). I'd start by looking at the work of Alastair Craw.
With respect to quivers, I was going to recommend Harm Derksen's lecture notes for a good introduction, but it looks like he took them down at some point. Sorry I don't have any good recommendation there.
I'm just a beginner trying to create a background in topological string theory, that's why I really appreciate the enormous value of your comments and clarifications regarding the category of B-branes. What a wonderful bonus surprise was to me to learn that any object in the derived category can be coupled to the B-model.
Concerning your answer to my question. Wonderful, that was precisely what I was searching for. Indeed, some of your papers seem to be very relevant to solve some gaps in the papers I want to understand, an example is the conjecture in page 19 of Crystal Melting and Black Holes and
Stability Conditions and Branes at Singularities. Thanks for your time and such a wonderful and illuminating answer.
|
2025-03-21T14:48:31.343310
| 2020-06-23T09:37:22 |
363906
|
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|
Stack Exchange
|
versal deformation ring of a p-divisible group with some tensors
I'm trying to read Kisin's paper about the Integral model of Shimura varieties. In section five he discusses versal deformation ring of a p-divisible group. Assume that $K$ is a number field with residue field $k$ and consider p_divisible group $G_0$ defined on $k$ and consider the cocharacter $\mu$ of $GL(D(G_0)(\mathbb{W}(k)))$ defined by the filtration on the $D(G_0)(\mathbb{W}(k))$ and. $R$ versal deformation ring of $G_0$ is defined as the completion of the local ring of the identity section of the nilpotent group $U^0(\mu)$
consider a reductive group $H\subset GL(D(G_0)(\mathbb{W}(k)))$ defined by tensors $\{s_\alpha\}$ such that cocharacter $\mu$ of the natural filtration on $D$ factors through $H$. consider $U_H^0(\mu)$ the unipotent subgroup defined by $\mu$ and define the $R_H$ as the completion of the local ring of the natural section of $U_H^0(\mu)$
define $S=\widehat{W[u,\frac{E[u]^n}{n!}}]$.
we want to know when $O_K$ points of $Spf(R)$ come from $Spf(R_H)$. theorem says that a point comes from $R_H$ if and only if for the associated p-divisible group $G$ you can lift $\{s_\alpha\}$ to $D(G)(S)$ satisfying certain condition.
I have some problems understanding this theorem. I want to know why we need the ring $S$ in this theorem instead of saying an $O_K$ point of $R$ is a point of $R_H$ if and only if you can lift $S_\alpha$ to an element of $D(G)(O_K)$ satisfying those condition.
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2025-03-21T14:48:31.343442
| 2020-06-23T10:43:07 |
363909
|
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|
Stack Exchange
|
On a possible attempt to prove the invariant subspace problem
This question involves a possible method to prove the invariant subspace problem for (separable) infinite dimensional Hilbert spaces. The idea comes from various results on this topic; more precisely, first let me recall one version of Lomonosov Theorem:
Let $T \in B(H)$ be a nonscalar bounded linear operator which commutes with some nonzero compact $K$. Then, $T$ has a nontrivial hyperinvariant (and thus also invariant) subspace.
It is well know that there exist operators for which there is no nonzero compact operator commuting with them (see, for instance, D. W. Hadwin, E. A. Nordgren, H. Radjavi, and P. Rosenthal, An operator not
satisfying Lomonosov's hypothesis, J. Funct. Anal. 38 (1980), no. 3, 410-415). The next step is then to consider operators for which there exists at least one nonzero compact $K$ such that
$$ \operatorname{rank} (TK-KT) = 1 $$
(the case with rank $0$ has been established with Lomonosov result). This result has actually been shown, as for Lomonosov Theorem, in the general setting of Banach spaces. Some references for this are the following ones:
Daughtry, J. An invariant subspace theorem. Proc. Amer. Math. Soc. 49 (1975), 267–268
Kim, H. W.; Pearcy, C; Shields, A. L. Rank-one commutators and hyperinvariant subspaces. Michigan Math. J. 22 (1975), 193-194
Kubrusly, C. S. Hilbert space operators. Birkhauser, Boston, 2003 (Problem and Solution 12.4)
In this case, I wonder if any progress has been done in establishing the existence/nonexistence of operators for which no nonzero compact $K$ such that the rank of the commutator is $1$ exist. I'm aware of the paper Kim, H. W.; Pearcy, C.; Shields, A. L. Sufficient conditions for rank-one commutators and hyperinvariant subspaces. Michigan Math. J. 23 (1976), no. 3, 235-243, in which some results related to this are obtained, but I would be interested in something more recent, if possible.
Furthermore, it has been noted even in the above papers that the case with
$$ \operatorname{rank} (TK-KT) = 2 $$
if established, would prove the invariant subspace problem. Indeed, for rank $1$ (compact) operators $K$, we always have the rank of the commutator less than or equal to $2$. I have tried to follow this way, and this is what I've come up with. First of all, the result does not hold for general Banach spaces, so I have tried to extend a proof in the setting of Hilbert spaces (more precisely, the one in Kubrusly's book Hilbert space operators). Everything can be reproduced verbatim, except the claim:
$\mathcal{R}(R) \subseteq \mathcal{R}(S)$
where $\mathcal{R}$ denotes the range of the operator, $S$ is any operator such that $0 < \dim \ker S < \infty$, and $R:=[T,S]=TS- ST$. Indeed, the claim is not necessarily true when the rank is equal to $2$. To simplify a bit, I will restrict the attention to the case we are actually interested in: in the proof of this statement, following Kubrusly, we will let $S:= Id - LK$, where $Id$ is the identity operator, $K$ is a nonzero compact operator, and $L$ is an operator commuting with $T$ such that $\exists g \neq 0 :$
$$ LK g = g $$
(the existence of $g$ follows from the assumption that there is no nontrivial invariant subspace, which implies that there is no nontrivial hyperinvariant subspace, and from Lomonosov Lemma). Since we want to deal with rank $1$ operators, we can write:
$$ Kx:= \langle x, z \rangle u $$
for some fixed nonzero $z,u \in H$. Now let $A$ denote the linear subspace of the points $x$ in $H$ such that $TS x \in \mathcal{R}(S)$. We will start proving that $R(A)$, that is, the linear subspace
$$ R(A):= \lbrace Rx, x \in A \rbrace $$
is equal to $\operatorname{span} \lbrace STy \rbrace$
where $y \in \ker S$ is such that $Ty \not \in \ker S$. The existence of such a point is prove as in Kubrusly's book: it follows from the finite dimensionality of the $\ker S$. Since $\operatorname{rank} R = 2$ by assumption, this implies that there are points in $\mathcal{R}(R)$ not belonging to $\mathcal{R}(S)$, thus implying that the above claim is not anymore true when the rank of the commutator is $2$. We know that there is a nonzero $g \in \ker S$: we can consider any such $g$, so let $g=y$. Then:
$$ LKg = \langle y, z \rangle L u = y \Rightarrow Lu = \frac{y}{\langle y, z \rangle} $$
(the denominator is nonzero, otherwise we would have $L(0) = y \neq 0$, absurd since $L$ is linear). Let $v \in A$. Then, this means that:
$$ TSv = Tv - TLKv = Tv - \langle v, z \rangle TLu = Tv - \frac{\langle v, z \rangle}{\langle y, z \rangle} Ty = St = t - \frac{\langle t, z \rangle}{\langle y, z \rangle} y $$
for some $t \in H$. This implies that:
$$ \langle Tv,z \rangle = \frac{\langle v,z \rangle \langle Ty, z \rangle}{\langle y, z \rangle} $$
We will now show that, for all $v \in A$, there is always $\rho \in \mathbb{C}$ such that $Rv=\rho Sty$, that is, such that:
$$ TSv= ST(v+ \rho y) $$
Writing again the equation using the expression for $S$ given above, we arrive at:
$$ (\rho + \frac{\langle v, z \rangle}{\langle y, z \rangle}) T( y ) = \frac{\langle T(v + \rho y), z \rangle}{\langle y, z \rangle} y $$
By the previous equation, it is easy to see that:
$$ \rho= - \frac{\langle v,z \rangle}{\langle y, z \rangle} $$
makes both sides equal to $0$ (recall that $v \in A$). Thus, $R(A) = \operatorname{span} \lbrace STy \rbrace$, and the claim does not anymore hold when the rank is $2$. Now, two possible options seem to be available to continue: we either look for another subspace which results to be invariant and closed (here, the closure of the range of $S$ would have done this, but the claim does not hold, so we cannot proceed with this space), or we find a contradiction with the space $A$ formed this way. I have attempted to follow this second idea. I have shown that $A$ is infinite dimensional, and there are infinitely many lineraly independent vectors not belonging to $A$. Moreover, $\forall v, \omega \not \in A$, $\exists ! \lambda \in \mathbb{C} :$
$$ (v - \lambda \omega) \in A $$
I do not know if this leads to any contradiction. My question is: has been done some work on the direction that I have shown above? And if the answer is yes, could you please give me some references?
Interesting post, but I'm skeptical that this approach will work. The existence of a compact operator with ${\rm rank}(TK - KT) \leq 2$ is trivial, so it's unlikely to be of much help. For instance, fix a unit vector and let $K$ be the orthogonal projection onto its span. Now you give me an arbitrary bounded operator $T$ and I notice that the commutator with my $K$ has rank at most 2. Would that really help me to show that $T$ has a nontrivial invariant subspace?
That said, even if you don't solve the ISP you might well discover something interesting in this direction ...
@NikWeaver Nice point! Indeed, my idea was to try to construct a particular rank one operator $K$ which could help to show the existence of a nontrivial invariant subspace, because as you noticed the use of any $K$ would probably be not so useful
|
2025-03-21T14:48:31.343861
| 2020-06-23T11:42:30 |
363912
|
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|
Stack Exchange
|
Subdivision of closed homology manifold reference request
I am interested in the barycentric subdivision of closed homology manifolds.
Definition A (finite) simplicial complex $K$ is a closed homology manifold of dimension $n$ if for every $k$-simplex, its link has the homology of $\mathbb{S}^{n-k-1}$.
I wonder whether the barycentric subdivision of a closed homology manifold remains a closed homology manifold and if case it does not, if there are some extra (minimal) conditions I can add to guarantee that.
Do you know of a reference where this or related questions are addressed?
(I have not found an answer in Rourke and Sanderson's "Introduction to Piecewise linear topology", which was my first try.)
Thanks in advance for your time
Using excision, and the decomposition of the star of a simplex $\sigma$ as $\sigma * \mathrm{Lk}(\sigma)$, you can easily show that your definition is equivalent to asking that
$$H_i(|K|, |K| \setminus \{x\} ;\mathbb{Z}) = \begin{cases}\mathbb{Z} & \text{ if $i=n$}\\
0 & \text{ else}.\end{cases}, \text{ for every $x \in |K|$}.$$
But this (re)definition of homology manifold only depends on the topological space $|K|$, which is unchanged under barycentric subdivision.
|
2025-03-21T14:48:31.343960
| 2020-06-23T12:34:31 |
363915
|
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"url": "https://mathoverflow.net/questions/363915"
}
|
Stack Exchange
|
Polynomial isometries of $\mathbb{A}^2_\mathbb{C}$
I have the following question, which I'm sure must be explored somewhere.
Consider a group of polynomial automorphisms of $\mathbb{A}^2_\mathbb{C}$ preserving a standard hermitian metric. Is there any description of this group?
I know that analogous question for symplectomorphisms has been studied, for example, here the authors prove infinite transitivity of the group of polynomial symplectomorphisms.
Still, I was unable to find anything similar for the group of polynomial isometries.
Yes, this is just $\mathrm{U}(2)\ltimes \mathbf{C}^2$, that is, all such automorphisms are affine.
Indeed, let $f$ belong to your group. After composing by a translation, we can suppose that $f$ fixes zero. The tangent map of $f$ at zero preserves the Hermitian scalar product, and hence, after composing by some element of $\mathrm{U}(2)$, we can suppose that the differential of $f$ at zero is identity. Now we use that for an arbitrary connected Riemannian manifold, if the differential of some isometry at some fixed point is identity, then the isometry is the identity.
(This is not surprising: for an arbitrary connected Riemannian manifold, the isometry group is a [finite-dimensional] Lie group.)
Two good references for finite dimensionality results of symmetry groups of various structures on manifolds: Kobayshi, Transformation Groups, and d'Ambra, Gromov, Lectures on transformation groups: geometry and dynamics.
|
2025-03-21T14:48:31.344081
| 2020-06-23T13:58:45 |
363923
|
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|
Stack Exchange
|
Fiedler vector, what else?
In the spectral analysis of a graph with 1 connected component, the first non-trivial eigenvector (corresponding to the non-zero smallest eigenvalue) is also called the Fiedler vector. This vector is useful in graph partitioning because it minimizes the distance between the connected vertices in the original graph. In other ways, it can creates partitions in which nodes within the same partitions have minimal distance/high similarity and nodes between partitions have minimum edges connecting them.
Now, my question is, as soon as we go to higher eigenvalues, what do eigenvectors mean? Do they represent other partitions of the same graph? Should they be taken into consideration?
Try to look up "nodal domains" in the context of graphs. The rough idea (but wrong in this exact form) is that the subgraph induced on the non-negative (or non-positive) entries of an eigenvector to the $k$-th largest eigenvalue has $k-1$ connected components. There exist some estimates on how wrong this number can be.
Yes. See e.g. the paper Multi-way spectral partitioning and higher-order Cheeger inequalities by Lee, Oveis-Gharan and Trevisan. They show how the first $k$ eigenvectors can be used to find a useful $k$-way partitioning.
Thank you for your quick and helpful answer. Do you think 3rd,4th,5th... eigenvectors by themselves can be used or the k partitioning is supposed to happen in k-dimensions(so with eigenvectors up to k)?
The Fiedler vector refers to the second smallest eigenvalue, here is a study of
The third smallest eigenvalue of the Laplacian matrix (2001).
The relationship between the third smallest eigenvalue of the
Laplacian matrix and the graph structure is explored. For a tree the
complete description of the eigenvector corresponding to this
eigenvalue is given and some results about the multiplicity of this
eigenvalue are given.
|
2025-03-21T14:48:31.344234
| 2020-06-23T13:59:01 |
363924
|
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|
Stack Exchange
|
Harnack inequality for $F(D^{2}u, Du,x)=f$
I would like to know if there is an harnack inequality for the viscosity solutions of the following pde in the literature: $F(D^{2}u,Du,x)=f \in L^{\infty}(B_{1})$, for a function $u \in C(B_{1})$, and $F$ is uniformly elliptic in matricial entrie.
If so, where could you find it?
|
2025-03-21T14:48:31.344288
| 2020-06-23T14:42:08 |
363928
|
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|
Stack Exchange
|
Modelling fluid flows with mean curvature flow
A while ago I was wondering if the displacement of fluid described in this blog post
could be modelled with mean curvature flow or some other flow, but when I asked someone in Engineering they replied that industry is generally not interested in simple mathematical models.
However, I noticed in this article that they do actually model bouncing of droplets and the associated coalescence phenomenon using mean curvature flow and get quite good agreement with experiment and numerical simulations. Are there are any other possibilities for modelling fluid flows with mean curvature flow? I am thinking along the lines of contracting bubbles and things like that.
I don't think the "correct" description is a simple mean curvature flow. Things like contracting bubbles you have really a two-phase fluid with a free boundary interface. The standard model for surface tension is given by the mean curvature. The motion of the free boundary then is by pressure difference between the two phases minus the mean curvature (or something like that). Mean curvature flow by itself is probably too naive.
|
2025-03-21T14:48:31.344392
| 2020-06-23T14:53:55 |
363930
|
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|
Stack Exchange
|
Gaussian bounds with exponential decay for discrete (graph) Dirichlet heat kernel
Let $\Omega$ be a finite, connected subset of $\mathbb{Z}^n$, $W_t$ a standard random walk on $\mathbb{Z}^n$ started at $x$, and $T_\Omega$ the first time at which $W_t$ leaves $\Omega$; consider
$$
P^D_\Omega(x,y;t) := \mathbb{P}[W_t=y \text{ and } T_\Omega>t],
$$
the discrete or graph heat kernel on $\Omega$ with Dirichlet boundary conditions.
By analogy with some known results on infinite graphs and continuous regions with boundaries one would expect a bound of the form
$$
P_\Omega^D(x,y;t)
\le
C_\Omega
\frac{\phi_\Omega(x,t) \phi_\Omega(y,t) e^{-\lambda_\Omega t}}{t^{n/2}}
e^{- c |x-y|^2/t},
$$
with $\phi:\mathbb{Z}^n \times \mathbb{N} \to [0,1]$ vanishing outside $\Omega$ (also with some bounds near the boundary of $\Omega$).
Is this known?
I assume the question pertains to continuous time random walk; the counterexamples are even simpler in discrete time. There is no reason to expect the power law factor $t^{-n/2}$ in this setting. For the simplest example, consider the case where $\Omega$ consists of two adjacent points $x,y$ in $\mathbb{Z}$. Then
$$ P_\Omega^D(x,y;t)=\sum_{k \ge 0} 2^{-2k-1} P({\rm Poisson }(t)=2k+1)=
\sum_{k \ge 0}\frac{(t/2)^{2k+1}}{(2k+1)! \, e^t}=\frac{\sinh(t/2)}{e^t}$$
which is asymptotically $\exp(-t/2)\cdot (1/2-o(1))$.
A good discussion of this topic when $\Omega$ is an interval and the walk is discrete can be found in page 243 of [1]. This is easily converted to continuous time, see e.g.
Exit time estimate for a simple continuous-time random walk
[1] Spitzer, Frank. Principles of random walk. GTM Vol. 34. Second edition, Springer.
Is that not compatible with the bound I wrote, with a suboptimal exponent? I will clarify in a couple of days when I am back to my keyboard.
This is not compatible with the bound you wrote because $\lambda_{\Omega}$ equals $1/2$ in this case.
Ah yes; looking more closely the bound I wrote is in poor correspondence with the second paper I cited for a closely related reason. I will edit the question to something more reasonable. Thank you!
Once your question is answered, it is best practice to acknowledge that and then ask another question if needed. Editing a question after it is answered will not motivate people to think about your questions.
|
2025-03-21T14:48:31.344576
| 2020-06-23T15:24:56 |
363932
|
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|
Stack Exchange
|
How does the scalar TV invariant of a 3-manifold with boundary fit into the TQFT picture?
Chen and Yang have a more general version of the volume conjecture that they state for all hyperbolic $3$-manifolds (Conjecture 1.1 of [2]) including those with boundary. To do this, they have to define (citing Benedetti and Petronio [1]) a version of the Turaev-Viro invariant that assigns a manifold with boundary a number instead of a vector space.
My understanding is that they do this by triangulating the manifold and its boundary, and then dropping all the singular vertices that correspond to lower-dimensional strata and taking the usual state-sum over the remaining triangulation.
I think this means defining
$$
\mathrm{TV}_r(M, q) = \mathrm{TV}_r(M \setminus \partial M, q).
$$
If that's the case, is there a way to fit these numerical invariants into a TQFT-like structure?
One idea I had was as follows: If we have a link complement $M \setminus L$, the obvious thing to do would be to think of it as an embedded link $L \to M$, and I know how to think of an embedded link in a closed manifold as giving a number. In particular, $\mathrm{TV}_r(L \to S^3)$ would agree with the norm-square of the Jones polynomial of $L$ evaluated at an appropriate root of unity. However, that's not what's going on in the construction, because the invariant Chen-Yang are considering takes different values on link complements.
[1] Benedetti, Riccardo; Petronio, Carlo, On Roberts’ proof of the Turaev-Walker theorem, J. Knot Theory Ramifications 5, No. 4, 427-439 (1996). ZBL0890.57029.
[2] Chen, Qingtao; Yang, Tian, Volume conjectures for the Reshetikhin-Turaev and the Turaev-Viro invariants, Quantum Topol. 9, No. 3, 419-460 (2018). ZBL1405.57020, arXiv:1503.02547.
Given a surface, one may consider the formal linear span of all the manifolds bounding it. https://arxiv.org/abs/math/0503054 One may pair two such manifolds together by gluing along the surface, then evaluate $TV_r$ on the closed manifold. Then $TV_r$ gives a pairing on this space, by bilinear extension. One can quotient by the null-space of this pairing, maybe this gives a finite-dimensional vector space associated to the surface? However, I think the pairing is indefinite in the Chen-Yang case.
Applying your comment, a manifold with boundary would be a linear functional on the vector space associated to its boundary. I suppose this is also what happens in the usual formalism, but here the vector space has a much more topological interpretation than the usual combinatorial state space of TV theory.
I'm still not sure how to get a number instead of a covector in a natural, topological way. Maybe you close up the manifold in some canonical fashion?
Actually, what I suggested probably doesn’t work, given that TV is the square of RT. https://www.sciencedirect.com/science/article/pii/0040938394000530
Maybe for a manifold with boundary, it’s the inner product of the appropriate RT invariant with itself (in the RT vector space associated to the boundary, which may have an indefinite inner Hermitian inner product)?
That might be the right direction: it suggests how you obtain the formulas in Theorem 1.1 of https://arxiv.org/pdf/1701.07818.pdf. Maybe I should read that paper more carefully: I think Section 3.2 answers at least part of my question.
I think https://arxiv.org/pdf/1807.03327.pdf Proposition 5.3. may answer your question.
@TianYang Thanks! That also looks relevant.
Based on the discussion in the comments with Ian Agol, here's a draft answer. I would welcome corrections/confirmation from anyone who knows more.
Let $M$ be an orientable manifold with possibly nonempty boundary, viewed as a cobordism $\emptyset \to \partial M $.
Then its $r$th Reshetikhin-Turaev invariant $\mathrm{RT}_r(M)$ is a vector in $\mathrm{RT}_r(\partial M)$, a vector space.
The TQFT axioms say that we can regard $\mathrm{RT}_r(\overline M)$ as an element of the dual space $\mathrm{RT}_r(\partial M)^*$, where $\overline M$ is $M$ with opposite orientation.
We can pair the vector and covector to obtain an invariant
$$
\mathrm{TV}_r(M) := \left \langle \mathrm{RT}_r(\overline M), \mathrm{RT}_r(M) \right \rangle \in \mathbb C
$$
which is in $\mathbb C$ even when $\partial M \ne \emptyset$. (Actually, I think it's always in $[0, \infty)$, and should be nonzero for any nontrivial $M$.) I believe that this is what Chen-Yang call the Turaev-Viro invariant of a manifold with boundary. This is closely related to the results in arXiv:1701.07818, which discusses this construction for knot complements.
The idea is that, while $\mathrm{RT}_r(\partial M)$ isn't quite a Hilbert space, there's at least an inner product on vectors coming from cobordisms, and we can exploit this to define $\mathrm{TV}_r(M)$ as the norm of the vector $\mathrm{RT}_r(M)$.
However, there now seems to be little relation between $\mathrm{TV}_r(S^3 \setminus L)$ and the norm-square of the $r$th colored Jones polynomial of $L$ evaluated at a $r$th root of unity; as per arXiv:1701.07818, the former involves a sum over the lower-order colored Jones polynomials at a different root of unity. Understanding this relationship better would be helpful in comparing the volume conjectures of Chen-Yang and of Kashaev-Murakami-Murakami.
|
2025-03-21T14:48:31.344953
| 2020-06-23T16:13:47 |
363934
|
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|
Stack Exchange
|
Sheaf cohomology commutes with colimits of sheaves
Let $X$ be a Noetherian scheme over a Noetherian ring $R$ and $(F_{\alpha})_{\alpha \in I}$ a direct system of $O_X$-module sheaves on $X$. I'm looking for source literature where I can find a proof of the fact that colimits of sheaves commute with sheaf cohomology, ie that for all $i\ge0$ the canonical morphism of $R$-modules
$$\varinjlim_i H^{i}(X, F_{\alpha}) \to H^i(X, \varinjlim_{\alpha} F_{\alpha}) $$
induced by $F_{\alpha} \to \varinjlim_{\alpha} F_{\alpha}$, applying the naturality of cohomology functor and the universal property of colimits. That's of course a well known fact used by people involved in research on algebraic geometry literally by reflex but I nowhere found a well explained proof of this isomorphism.
See the last paragraph of section II.4.12 in Godement, Topologie algébrique et théorie des faisceaux.
Or Tag 01FF (which implies the result for quasi-compact quasi-separated schemes).
I think the canonical reference (true for any coherent topos, so for every qcqs scheme, as Remy mentions) is SGA 4 II, Expose VI, Corollaire 5.2.
Let $E'$ be a coherent topos. For every integer $q$, the functor $H^q(E', -)$ commutes with filtered inductive limits of abelian sheaves.
Hartshorne Chapter 3 Proposition 2.9
can be apply to this case
|
2025-03-21T14:48:31.345098
| 2020-06-23T16:15:37 |
363935
|
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|
Stack Exchange
|
What is the role of topology on infinite dimensional exterior algebras?
Wedge products and exterior powers are discussed in W. Greub's book Multilinear algebra as follows.
Definition: Let $E$ be an arbitrary vector space and $p \ge 2$. Then a vector space $\bigwedge^{p}E$ together with a skew-symmetric $p$-linear map $\bigwedge^{p}: E\times \cdots \times E \to \bigwedge^{p}E$ is called a $p$-th exterior power of $E$ if the following conditions are satisfied:
(1) The vectors $\bigwedge^{p}(x_{1},\dotsc,x_{p})\mathrel{:=} x_{1}\wedge \dotsb \wedge x_{p}$ generate $\bigwedge^{p}E$.
(2) If $\psi$ is any skew-symmetric $p$ linear mapping of $\overbrace{E\times \dotsb \times E}^{\text{$p$ times}}$ into an arbitrary vector space $F$, then there exists a linear map $f\colon \bigwedge^{p}E \to F$ such that $\psi = f\circ \bigwedge^{p}$.
Now, we set:
\begin{equation}
\bigwedge E \mathrel{:=} \bigoplus_{n=0}^{\infty} \bigwedge^{p}E, \tag{1}\label{1}
\end{equation}
where $\bigwedge^{0}E \mathrel{:=} \mathbb{C}$ and $\bigwedge^{1}E \mathrel{:=} E$.
Identifying each $\bigwedge^{p}E$ with its image under the canonical injection $i_{p}\colon\bigwedge^{p}E \to \bigwedge E$, we can write $\bigwedge E = \sum_{p=0}^{\infty}\bigwedge^{p}E$. In other words, elements of $\bigwedge E$ can be thought as sequences $(v_{0},v_{1},\dotsc)$ where $v_{p} \in \bigwedge^{p}E$ for each $p\in \mathbb{N}$. Furthermore, there is a uniquely determined multiplication on $\bigwedge E$ such that the following rules hold:
\begin{gather*}
(x_{1}\wedge \cdots \wedge x_{p})(x_{p+1}\wedge \dotsb \wedge x_{p+q}) = x_{1}\wedge \cdots \wedge x_{p+q} \\
1(x_{1}\wedge \cdots \wedge x_{p}) = (x_{1}\wedge \dotsb \wedge x_{p})1 = x_{1}\wedge \cdots \wedge x_{p}.
\end{gather*}
This turns $\bigwedge E$ into an algebra, which is called exterior (or Grassmann) algebra.
Note that Greub's construction considers arbitrary vector spaces, so that, in particular, we can take $E$ to be infinite dimensional.
Grassmann algebras are used by physicists to study fermionic systems. While searching for some material on Grassmann algebras of infinite dimensional vector spaces, I found lecture notes Fermionic functional integrals and the renormalization group by Feldman, Knörrer and Trubowitz, which has an appendix (page 75) on this topic. Their construction seems interesting, but I'm having trouble trying to relate it with Greub's construction.
The first part of their notes discusses Grassmann algebras of finite dimensional vector spaces. Then, the cited appendix starts with the statement that in order to further generalize it to infinite dimensional vector spaces we need to add a topology on these spaces. This seem not to be necessary in the general case, since Greub's construction does not consider topological vector spaces. However, I think they might have physical motivations in which the addition of a topology might be important. Their construction is as follows.
Let $I$ be a countable set. The Grassmann algebra will be generated by vector from the vector space:
$$E\mathrel{:=} \ell^{1}(I)\mathrel{:=}\{\alpha\colon I \to \mathbb{C}\mathrel: \sum_{i\in I}\lvert a_{i}\rvert < +\infty\}.$$
$E$ is a Banach space with norm $\|\alpha\| \mathrel{:=}\sum_{i\in I}\lvert a_{i}\rvert$. Let $\mathcal{J}$ be the set of all finite subsets of $I$, including the empty set. Take
$$\mathcal{U}(I) = \ell^{1}(\mathcal{J}) \mathrel{:=}\{\alpha\colon \mathcal{J} \to \mathbb{C}\mathrel: \sum_{I\in \mathcal{J}}\lvert a_{I}\rvert<+\infty\}$$
where $a_{I} \mathrel{:=} a_{i_{1}}\dotsb a_{i_{p}}$, $I=\{i_{1},...,i_{p}\}$. Then $\mathcal{U}(I)$ is a Banach space with norm $\|\alpha\| = \sum_{I\in \mathcal{J}}\lvert a_{I}\rvert$ and, when equipped with the product:
$$(\alpha \beta)_{I} \mathrel{:=}\sum_{J\subset I} \operatorname{sign}(J, I\setminus J)\alpha_{J}\beta_{I\setminus J},$$
it becomes an algebra which is called the Grassmann algebra.
With all this being said, let me get to the questions.
Feldman, Knörrer and Trubowitz's construction might not be the most general construction there is (I don't know actually, but I think it's not as I justified before). However, I'd expect their construction to be at least a particular case of Greub's general construction. However, I don't seem to be able to relate these two since the definition of $\mathcal{U}(I)$ strongly depends on its topology. So is the second construction a particular case of the first one? If not, why not? Does it have to do with the hypothesis of $E$ to be a topological vector space? Does the topology on $E$ change the definitions of objects used on Greub's construction?
NOTE: When I ask "does the topology on $E$ changes the definitions of objects on Greub's constructions?", I mean the following. If $E$ is a vector space, $\bigoplus_{n=0}^{\infty}E$ is the space of all sequences $x=(x_{0},x_{1},\dotsc)$, $x_{i} \in E$, with all but finitely many nonzero entries. If $E= \mathcal{H}$ is a Hibert space, on the other hand, $\bigoplus_{n=0}^{\infty}\mathcal{H}$ is the space of sequences with $\|x\|^{2}:=\sum_{n=0}^{\infty}\| x_{i}\|^{2}_{\mathcal{H}}<+\infty$. Thus, although $\mathcal{H}$ is itself a vector space, the norm on $\mathcal{H}$ allows us to define the direct sum in alternative way. In other words, the topology on $\mathcal{H}$ makes the difference when we define direct sums. Maybe the use of Banach spaces by Feldman, Knörrer and Trubowitz implies some modifications like this, say, to define the direct sum (\ref{1}) in an alternative way, so these two constructions might be isomorphic or something like this.
ADDED: Does anyone know this particular construction from Feldman, Trubowitz and Knörrer? Any references on this approach would be really appreciated!
You might be misinterpreting the notation. The $\bigoplus_{n=0}^\infty E$ that you mention as the algebraic direct sum is not the same object as the "infinite sum of Hilbert spaces". The point is that the usual direct sum construction, when applied to a sequence of complete normed spaces, usually does not yield a complete normed space. So if you want your algebraic constructions to start with complete things and produce complete things, they need to be modified
@YemonChoi I think I get it. So, in summary, in the second construction I need to consider the space of sequences such that $||x||=\sum_{n=0}^{\infty}|x_{n}| <+\infty$, right? But if (\ref{1}) is modified to account to topological spaces $E$ in this way, does Greub's construction remain the same and equivalent to the second one?
TeX notes: $\wedge$ \wedge is the binary operation; you want $\bigwedge$ \bigwedge for the unary prefix operation, like $\bigwedge E$. Also, \mbox doesn't size well in superscripts, whereas \text does: compare $\overbrace{E \times \dotsb \times E}^{\mbox{$p$ times}}$ (with \mbox) to $\overbrace{E \times \dotsb \times E}^{\text{$p$ times}}$ (with \text). I have edited accordingly.
@LSpice thanks! These are very good tips!!
Focusing on exterior powers here is a distraction. The main problem already appears when considering the tensor algebra $T(E)=\oplus_{n\ge 0}E^{\otimes n}$. Once the issue is understood for the tensor algebra, figuring out what to do for the exterior or symmetric algebras (e.g., Fermion or Boson Fock spaces) is trivial, because we are in characteristic zero. In positive characteristic, this becomes subtle as can be seen for example in the recent work "Koszul modules and Green’s conjecture" by Aprodu et al. where a positive characteristic Hermite Reciprocity map is constructed.
Given a vector space $E$, the first step is to consider tensor products like $E^{\otimes n}$. This can be done algebraically as in the mentioned book by Greub. However, when $E$ is an infinite dimensional TVS (topological vector space) the resulting algebraic tensor product $E\otimes\cdots\otimes E$ is a rather unsuitable object for the purposes of analysis. One typically needs to enlarge this space using a completion procedure (topology is essential for that), and one then obtains a topological tensor product $E\widehat{\otimes}\cdots\widehat{\otimes}E$. The caveat is: even when working with Banach spaces, there are lots of ways of doing that. This was Alexander Grothendieck's Ph.D. thesis work. He considered a dozen or so inequivalent reasonable definitions for these completions/versions of the tensor product which depend on the topological structure. In other words, in the course of his explorations Grothendieck found Hell. Luckily, he kept exploring and he eventually also found Paradise: the class of nuclear spaces for which all these different constructions become the same and therefore acquire a cananical feel to them.
Likewise, for the sum $\oplus_{n\ge 0}$ one typically starts with the algebraic direct sum (only finite sums allowed, i.e., we look at almost finite sequences where after a while all the terms are zero) and one then enlarges the space by taking a completion.
The construction by Feldman, Knörrer and Trubowitz is an explicit way (just a choice that works for their purposes) of doing a succession of algebraic constructions followed by topological completions, as explained above.
Now one might think that the algebraic construction as in Greub's book is more general/powerful/etc. than the topological procedure. This is a misconception. For infinite dimensional spaces that are not too big, one could in fact argue the opposite is true.
Take for example the simplest infinite dimensional space: $E=\oplus_{n\ge 0}\mathbb{R}$ which can be viewed as the space of almost finite sequences of real numbers, or the space of polynomials in one variable with real coefficients. Then $T(E)$ constructed algebraically à la Greub is a particular case of the topological completion construction. Indeed, equip $E$ with the locally convex topology defined by the set of all seminorms on $E$. This is also called the finest locally convex topology. With this topology, the space is nuclear in the sense of Grothendieck's general definition (but not nuclear in the sense of the more restrictive definition used by the Russian school around Gel'fand et al., namely, the notion of countably Hilbert nuclear spaces).
So that is a good sign: pretty much any reasonable completion will give you the same $E\widehat{\otimes}\cdots\widehat{\otimes}E$ which will also coincide with the algebraic tensor product (without hats). Finally for the sum one has several possible choices, but one of them will give the algebraic construction. Let us say that a seminorm on the algebraic direct sum $T(E)$ is admissible if and only if it restricts to a continuous seminorm on each summand. Take the locally convex topology on $T(E)$ defined by the set of all admissible seminorms. Take the completion. This will give nothing new. Note that all seminorms are admissible for the case $E=\oplus_{n\ge 0}\mathbb{R}$ but I wanted to introduce a more general construction which can be applied for example to $E=\mathscr{S}(\mathbb{R})$, the Schwartz space of rapidly decaying smooth functions.
Then the $T(E)$ will be isomorphic as a TVS to $\mathscr{D}(\mathbb{R})$, the space of compactly supported smooth functions.
Moral(s) of the story:
For infinite dimensional spaces ordinary bases (Hamel bases) are no good. You need Schauder bases which allow infinite linear combinations.
You will need to base your construction on topology. Even when topology seems to be absent, and one uses purely algebraic direct sums and tensor products, topology is still there hiding behind the scenes as in the $E=\oplus_{n\ge 0}\mathbb{R}$ example.
Recommended reading:
The excellent vignette "Schwartz kernel theorems, tensor products, nuclearity" by Paul Garrett.
July 2020 Edit:
Let me give more details on the relation between the above general methodology and the particular FKT construction.
First some notation: I will write $\mathbb{N}=\{0,1,2,\ldots\}$, and I will denote the set functions from the set $X$ to the set $Y$ by $\mathscr{F}(X,Y)$.
We start from the $\ell^1$ space $E$ defined as the set of functions $f\in\mathscr{F}(\mathbb{N},\mathbb{C})$ such that
$$
||f||_E:=\sum_{i\in\mathbb{N}}|f(i)|
$$
is finite.
The first step is to understand the algebraic tensor product $E\otimes E$. The general construction proceeds via the free vector space with basis indexed by symbols $f\otimes g$ with $f,g\in E$ and quotienting by relations $(f_1+f_2)\otimes g-f_1\otimes g-f_2\otimes g$ etc. Another equally uninspiring construction is to take an uncountable Hamel basis $(e_i)_{i\in I}$, for $E$, produced by the Axiom of Choice, and realize $E\otimes E$ as the subset of $\mathscr{F}(I\times I,\mathbb{C})$ made of functions of finite support (equal to zero except for finitely many elements of $I\times I$). The proper definition is as a solution to a universal problem: $E\otimes E$ together with a bilinear map $\otimes:E\times E\rightarrow E\otimes E$ must be such that for every vector space $V$ and bilinear map $B:E\times E\rightarrow V$, there should exist a unique linear map $\varphi:E\otimes E\rightarrow V$ such that $B=\varphi\circ\otimes$. One can construct such a space more concretely as follows.
Let $E_2$ be the subset of $\mathscr{F}(\mathbb{N}^2,\mathbb{C})$ made of functions $h:(i,j)\mapsto h(i,j)$ which are finite sums of functions of the form $f\otimes g$ with $f,g\in E$. Here $f\otimes g$ is the function $\mathbb{N}^2\rightarrow\mathbb{C}$ defined by
$$
(f\otimes g)(i,j)=f(i)g(j)
$$
for all $i,j\in \mathbb{N}$. Note that the definition I just gave also provides us with a bilinear map $\otimes:E\times E\rightarrow E_2$.
Proposition 1: The algebraic tensor product of $E$ with itself can be identified with $E_2$.
The proof relies on the following lemmas.
Lemma 1: For $p,q\ge 1$, suppose $e_1,\ldots,e_p$ are linearly independent elements in $E$ and suppose $f_1,\ldots,f_q$ are also linearly independent elements in $E$. Then the $pq$ elements $e_a\otimes f_b$ are linearly independent in $E_2$.
Proof: Suppose $\sum_{a,b}\lambda_{a,b}e_a\otimes f_b=0$ in $E_2$. Then $\forall i,j\in\mathbb{N}$,
$$
\sum_{a,b}\lambda_{a,b}e_a(i) f_b(j)=0\ .
$$
If one fixes $j$, then one has an equality about functions of $i$ holding identically.
The linear independence of the $e$'s implies that for all $a$,
$$
\sum_{b}\lambda_{a,b}f_b(j)=0\ .
$$
Since this holds for all $j$, and since the $f$'s are linearly independent, we get $\lambda_{a,b}=0$ for all $b$. But $a$ was arbitrary too, so $\forall a,b$, $\lambda_{a,b}=0$ and we are done.
Lemma 2: Let $B$ be a bilinear map from $E\times E$ into some vector space $V$.
Suppose $g_k,h_k$, $1\le k\le n$ are elements of $E$ satisfying
$$
\sum_{k}g_k\otimes h_k=0
$$
in $E_2$, i.e., as functions on $\mathbb{N}^2$. Then
$$
\sum_k B(g_k,h_k)=0
$$
in $V$.
Proof: This is trivial if all the $g$'s are zero or if all the $h$'s are zero. So pick a basis $e_1,\ldots,e_p$ of the linear span of the $g$'s and pick a basis $f_1,\ldots,f_q$ of the linear span of the $h$'s (no Axiom of Choice needed). We then have decompositions of the form
$$
g_k=\sum_a \alpha_{k,a}e_a
$$
and
$$
h_k=\sum_b \beta_{k,b} f_b
$$
for suitable scalars $\alpha$, $\beta$.
By hypothesis
$$
\sum_{k,a,b}\alpha_{k,a}\beta_{k,b}\ e_a\otimes f_b=0
$$
and so $\sum_k \alpha_{k,a}\beta_{k,b}=0$ for all $a,b$, by Lemma 1.
Hence
$$
\sum_k B(g_k,h_k)=\sum_{a,b}\left(\sum_k \alpha_{k,a}\beta_{k,b}\right) B(e_a,f_b)=0\ .
$$
Now the proof of Proposition 1 is easy. The construction of the linear map $\varphi$ proceeds as follows.
For $v=\sum_{k}g_k\otimes h_k$ in $E_2$, we let $\varphi(v)=\sum_k B(g_k,h_k)$.
This is a consistent definition because if $v$ admits another representation $v=\sum_{\ell}r_{\ell}\otimes s_{\ell}$, then
$$
\sum_k g_k\otimes h_k\ +\ \sum_{\ell}(-r_{\ell})\otimes s_{\ell}=0
$$
and Lemma 2 implies
$$
\sum_k B(g_k,h_k)=\sum_{\ell} B(r_{\ell},s_{\ell})\ .
$$
The other verifications that $E_2$ with $\otimes$ solve the universal problem for the algebraic tensor product pose no problem.
The second step is to construct a topological completion $\widehat{E}_2$ for $E_2$. I will use the projective tensor product construction $E\ \widehat{\otimes}_{\pi}E$.
For $h\in E_2$, I will use the $l^1$ norm
$$
||h||_2=\sum_{(i,j)\in\mathbb{N}^2}|h(i,j)|\ .
$$
I will also use the seminorm
$$
||h||_{\pi}=\inf\ \sum_k ||g_k||_E\times||h_k||_E
$$
where the infimum is over all finite decompositions $h=\sum_k g_k\otimes h_k$.
The projective tensor product is the completion with respect to $||\cdot||_{\pi}$.
The $||\cdot||_1$ is an example of cross norm, i.e., it satisfies
$||f\otimes g||_2=||f||_E\times||g||_E$. Moreover, one has the following easy result.
Proposition 2: For all $h\in E_2$, we have $||h||_2=||h||_{\pi}$.
For the proof use the cross norm property and triangle inequality for $\le$, and for the reverse inequality, approximate $h$ by the truncation where $h(i,j)$ is replaced by zero unless $i,j\le N$.
Now it is clear that the abstract topological tensor product $\widehat{E}_2$ is nothing but the familiar $\ell^1$ space of functions on $\mathbb{N}^2$.
Likewise (but with heavier notations) one can construct $\widehat{E}_n=E\ \widehat{\otimes}_{\pi}\cdots\widehat{\otimes}_{\pi}E$, $n$ times,
with the corresponding $\ell^1$ norm
$$
||h||_n=\sum_{(i_1,\ldots,i_n)\in\mathbb{N}^n}|h(i_1,\ldots,i_n)|\ .
$$
The topological exterior power $\widehat{E}_{n,{\rm Fermi}}$ can be identified with the closed subspace of antisymmetric functions inside $\widehat{E}_n$, namely functions $h:\mathbb{N}^n\rightarrow\mathbb{C}$ which satisfy
$$
h(i_{\sigma(1)},\ldots,i_{\sigma(n)})=\varepsilon(\sigma)\ h(i_1,\ldots,i_n)
$$
for all $(i_1,\ldots,i_n)\in\mathbb{N}^n$ and all permutations $\sigma$.
We will equip the space with restriction of the norm $||\cdot||_n$.
Now consider the algebraic direct sum $W=\oplus_{n\ge 0}\widehat{E}_{n,{\rm Fermi}}$.
Given (for the moment unspecified) positive weights $w_n$, let us define the norm
$$
||H||_{\rm Big}=\sum_{n\ge 0}w_n||h_n||_n
$$
where $H$ is an element of $W$ seen as an almost finite sequence $(h_0,h_1,\ldots)$
of functions in $\widehat{E}_{0,{\rm Fermi}},\widehat{E}_{1,{\rm Fermi}},\ldots$
Clearly the completion $\widehat{W}$ is obtained by removing the almost finite restriction but still requiring convergence of the sum defining $||\cdot||_{\rm Big}$.
Finally, to make contact with FKT, to $H=(h_0,h_1,\ldots)\in\widehat{W}$
we associate the set function $\alpha:\mathcal{J}\rightarrow\mathbb{C}$ where $\mathcal{J}$ is the set of finite subsets of $\mathbb{N}$ (including the empty set), as follows.
For $I=\{i_1,\ldots,i_n\}\in\mathcal{J}$ with $i_1<\cdots<i_n$ we let by definition
$$
\alpha(I)=h_n(i_1,\ldots,i_n)\ .
$$
If we pick the weights $w_n=\frac{1}{n!}$, then this correspondence is a bijective isometry with the giant $\ell^1$ space of FKT.
Remark: One can do the same long construction with $\ell^2$ norms instead of $\ell^1$ norms,
and this will produce the Fermionic Fock space of the Hilbert space $\ell^2(\mathbb{N})$, as in the mathematical literature on second quantization, e.g., in the book by Reed and Simon.
Note that the corresponding topological tensor products of Hilbert spaces were introduced by Murray and von Neumann in "On rings of operators", Ann. of Math. 1936,
and further developed by Cook in "The mathematics of second quantization", PNAS 1951, for the needs of Quantum Field Theory.
And just to close the loop back to the question: To get from $T(E)$ to $\bigwedge E$ one uses the structure of a topological algebra and quotients out a closed ideal: $\bigwedge E$ is the quotient of $T(E)$ by the closure of the ideal generated by $v\otimes v$ for all $v\in E$.
Instead of quotienting, I would prefer using the symmetrizer or antisymmetrizer projections. Again we're in good shape because of the zero characteristic.
@AbdelmalekAbdesselam this was very useful! I think I'm starting to clear my thoughts now. If I may, I'd like to make some questions/observations. I'd like to focus on the explicit construction of Feldman, Trubowitz and Knörrer. If I understood it correctly, I should first think about the definitions of tensor products $\ell^{1}(I)\otimes\cdots\otimes\ell^{1}(I)$ and then figure out the exterior powers. As you pointed out, there is no recipe for defining these tensor products and since these are not nuclear (at least I think) spaces, I should basically chose one.
I still got a little confused about: (1) what is the proper way to define these tensor products, so my construction is compatible with Feldman,Trubowitz and Knörrer? (2) How to further define the exterior power? I have some guesses if $\mathcal{I}$ were, say, $\mathbb{N}$ but for arbitrary countable $\mathcal{I}$, I have no idea on how to define permutations and so on. Can you give me some hints, here? I'd like to first address this explicit construction.
In summary: I want to reproduce the construction of Feldman, Trubowitz and Knörrer (it's better to me to work with a explicit construction). Where should I begin? Seems there are a lot of ways to do it (by your answer) and I don't what path to take here.
I see. You want to see explicitly the FKT construction as a particular case of the general method I outlined. That's a good question, but the dirty secret here is: it is probably not worth your time. You would be better off taking FKT's definition as is and use it. Working out the completion approach is rather long. For example you will have to learn https://en.wikipedia.org/wiki/Topological_tensor_product At least FKT work with Banach spaces so the easier notion of completion for metric spaces will suffice and you will not need...
...https://en.wikipedia.org/wiki/Uniform_space#Hausdorff_completion_of_a_uniform_space I don't have time now, but if I do later I will try to explain how to relate FKT to my more general answer.
@AbdelmalekAbdesselam I already accepted your answer because it was undoubtely helpful. But I'd appreciate if you add more details on the explicit FKT construction if you have the chance. Just to clarify a little more: the exposition of FKT is supposed to be a generalization of the finite dimensional case. The first part of the text discusses finite dimensional Grassmann algebras in detail and this appendix is supposed to be a way to define the same objects but in a more general way by taking infinite dimensional vector spaces. However their two approaches seem really disconnected (cont)
Their first construction (for finite dimension) seem very alike to Greub's one, so I thought it would be natural to connect those two if their approach to the infinite dimensional case is, in fact, a generalization of the finite dimensional one. I know that the algebraic approach and FKT approach for infinite dimensional already differ by the fact that one is a topological approach and the other one is not. That's why these connections are so confusing to me and I think an explicit construction would clarify these points.
Thanks for the bounty. As promised, I added more details about the connection to the FKT construction.
@AbdelmalekAbdesselam sorry I didn't answer you before. Idk why but sometimes MO doesn't notify some comments on my posts. I've just seen it by accident. Your answer was incredible, thank you so much for it and also for the attention to my question!
@AbdelmalekAbdesselam just a little comment. When the connection to FTK is made, it seems possible to take $\omega_{n}=1$ and $\alpha(I) = h_{n}(I)$ for each $I$ with $|I|=n$. This seems to be an isometry too since $||\alpha_{I}|| :=\sum_{I\in \mathcal{J}}|\alpha(I)| = \sum_{n\in \mathbb{N}}\sum_{|I|=n}|\alpha(I)| = \sum_{n\in \mathbb{N}}\sum_{(i_{1},..,i_{n})\in \mathbb{N}^{n}}|h_{n}(i_{1},...,i_{n})|$. What do you think?
No. Your last equality is false because you are confusing the set $I={i_1,\ldots,i_n}$ with the sequence $(i_1,\ldots,i_n)$. There are $n!$ sequences that appear in the $h_n$ side that correspond to the same set $I$. In fact the best way to think of a sequence is as a function from $[n]$ to here $\mathbb{C}$. The set $I$ is the image of this function.
ah, true! Indeed, my mistake!
|
2025-03-21T14:48:31.346895
| 2020-06-23T16:28:37 |
363937
|
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Stack Exchange
|
Should I expect functions to have analytic continuations?
I spend lots of time working with Dirichlet series with bounded coefficients, and I often need to find whether or not they have analytic continuations to the full complex plane. When proving that some mathematical object has some property, I like to know whether I'm working to prove that the object I'm looking has some strange property or if I'm working to prove that it's normal and the numbers aren't just conspiring against me.
For example, when trying to prove whether a number is irrational or not, I know that $100\%$ of numbers are irrational and so I'm trying to show that I didn't happen to pick one of those $0\%$ of numbers.
Sadly, I have no such intuition for analytic continuation. I think that my guess would be that either $100\%$ or $0\%$ of Dirichlet series have analytic continuations, but I could be wrong. To make my question more concrete,
If $\{a_n\}$ is a sequence of complex numbers chosen uniformly randomly in the unit disk, and $F(s)=\sum_{n=1}^{\infty}\frac{a_n}{n^s}$ is its Dirichlet series, what is the probability that $F(s)$ has an analytic continuation past the the line $\Re(s)=1$ (not necessarily to the entire plane).
Answers to variants of this question are also greatly appreciated, like if $a_n$ is chosen uniformly randomly of $[0,1]$ or if we are looking for continuations to the entire complex plane.
EDIT: If it's too complicated to analyse analytic continuation, what about meromorphic ones? Should I expect functions to have meromorphic continuations to $\Re(s)=1$?
I don't know any formal statements which say this, but the intuition definitely should be that analytic continuation beyond the obvious domain of convergence is rare, definitely not a norm.
@Wojowu why does your intuition tell you this?
I expect the situation is the same as for power series. With poser series, analytic continuation beyond the radius of convergence "almost always" fails. ("Almost always" in the sense of category, and in the sense of measure.)
@GeraldEdgar interesting. What is the main idea of the proof that analytic continuation almost always fails?
This title sounds promising... MR3457245
Bhowmik, Gautami(F-ULIL-LM); Matsumoto, Kohji(J-NAGO-GM)
Analytic continuation of random Dirichlet series.
Proc. Steklov Inst. Math. 282 (2013), suppl. 1, S67–S72.
I would actually expect continuation to $\Re(s)>1/2$ but no further (by looking at the variance)
@WillSawin from looking at Gerald Edgar's link I believe that you are correct, but I am not entirely sure. The paper is very dense.
Maybe this old result of Hideo Sakashita is pertinent to your investigation (don't let the description page fools you: the paper is in English and is freely downloadable). Considering the set of power series on the unit disk, the author proves an extended version of a result of Fatou and Polya: The set of the continuable power series is the boundary of the set of the non continuable ones. This implies that if you naively "choose at random" a power series on the unit disc, the probability that this is non continuable is 1.
I don't think it is reasonable to use "random" Dirichlet series as a guide if you are working with examples that are expected to have some actual structure to them (like most Dirichlet series that arise in practice in number theory). If you are working with Dirichlet series for reasons unrelated to number theory, then perhaps your question is reasonable. What are some reasons you are looking at Dirichlet series with bounded coefficients?
Let's describe a probabilistic model for Dirichlet $L$-functions and see what probability theory predicts about them.
The coefficients of a Dirichlet $L$-function are roots of unity (or $0$), which are on the unit circle, so should we consider a random
Dirichlet $L$-function to be $\sum z_n/n^s$ where $\{z_n\}$ is a sequence chosen independently and uniformly on
the unit circle? That doesn't reflect the multiplicativity of the coefficients of a Dirichlet $L$-function, so we will use a random Euler product, as follows: define a "random" Dirichlet $L$-function to be $L(s) = \prod_p 1/(1 - z_p/p^s)$ where $z_p$ for each prime $p$ is
chosen from a uniform distribution on the unit circle.
For a random number $z = \cos \theta + i\sin \theta$ on the unit circle,
its real and imaginary parts have average value 0 ($\int_0^{2\pi} \cos \theta \,d\theta/2\pi = 0$ and
$\int_0^{2\pi} \sin \theta \,d\theta/2\pi = 0$) and variance 1/2 ($\int_0^{2\pi} \cos^2 \theta \,d\theta/2\pi = 1/2$ and
$\int_0^{2\pi} \sin^2 \theta \,d\theta/2\pi = 1/2$). Note we are not computing
the variance of $z^2$ on the unit circle, which would be 0: $z^2$ is not $\cos^2\theta + i\sin^2\theta$!
The product $L(s)$ converges absolutely and uniformly on compact
subsets of ${\rm Re}(s) > 1$. What happens if $0 < {\rm Re}(s) \leq 1$?
When ${\rm Re}(s) > 1$, a logarithm of $L(s)$ is
$$
\sum_{p^k} \frac{z_p^k}{kp^{ks}} = \sum_{p} \frac{z_p}{p^s} + \sum_{\substack{p^k \\ k \geq 2}} \frac{z_p^k}{kp^{ks}},
$$
where the sum involving $k \geq 2$ is absolutely convergent if ${\rm Re}(s) > 1/2$ since $|z_p^k/kp^{ks}| = 1/kp^{k\sigma}$. The series over primes is an integral:
$$
\sum_p \frac{z_p}{p^s} = s\int_1^\infty \frac{Z(x)}{x^{s+1}}\,dx,
$$
where $Z(x) = \sum_{p \leq x} z_p = \sum_{n \leq \pi(x)} z_{p_n}$. Applying the law of the iterated logarithm to the real and imaginary parts of $z_p$ (or $2z_p$ to make the variance 1), $|Z(x)| = O(\sqrt{\pi(x)\log\log \pi(x)})$ for almost all sequences $\{z_p\}$, which makes the integral above absolutely convergent for ${\rm Re}(s) > 1/2$ since $\pi(x)\log\log \pi(x) \sim (x/\log x)\log \log x$.
Therefore for almost all sequences $\{z_p\}$,
$\sum z_p/p^s$ converges for ${\rm Re}(s) > 1/2$, so a random Dirichlet $L$-function as defined here is almost certain to have an analytic continuation from ${\rm Re}(s) > 1$ to ${\rm Re}(s) > 1/2$ (as the exponential of
the analytic continuation of its logarithm) and no zeros
with ${\rm Re}(s) > 1/2$.
There is something inconsistent between random Dirichlet $L$-functions as defined above and actual Dirichlet $L$-functions that go by this name in the "real world" of number theory: for the sequences with $z_p \in \{\pm 1\}$ (a random quadratic Dirichlet $L$-function), with probability 1 the function $L(s)$ does not have an analytic continuation to ${\rm Re}(s) > 1/2 - \delta$ for $\delta > 0$ by Theorem 2 p. 550 of Queffélec's paper on random Euler products; see
"Propriétés presque sûres et quasi-sûres des séries de Dirichlet et des produits d'Euler" Canad. J. Math 32 (1980), 531-558. (I am not aware of a treatment of this issue for random Dirichlet $L$-functions with non-real $z_p$.
The treatment of random Euler products $\prod_p 1/(1 - z_p/p^s)$ for $z_p \in S^1$ in Kowalski's course notes https://people.math.ethz.ch/~kowalski/probabilistic-number-theory.pdf focuses on ${\rm Re}(s) > 1/2$.) Actual
Dirichlet $L$-functions in number theory are not random objects, but highly structured ones, and being able to extend them beyond ${\rm Re}(s) > 1/2$ is a significant feature that you can't predict by only thinking of a "random" $L$-function in a probabilistic sense.
In addition to @KConrad's good answer: there were results already from the early 20th century (maybe partly due to Harald Bohr, but I can't find the ref) about natural boundaries of Dirichlet series with random coefficients from among $\{-1,0,1\}$ or similar.
More interesting to me is the Estermann phenomenon, from T. Estermann, "On certain functions represented by Dirichlet series", Proc. London Math. Soc. 27 (1928), 435-448. For example, he shows that $\sum d(n)^3/n^s$ has a natural boundary (where $d(n)$ is the number of divisors of $n$), while $\sum d(n)/n^s$ and $\sum d(n)^2/n^s$ have meromorphic continuations.
N. Kurokawa published considerable generalizations of this in 1986, in two papers in Proc. London Math. Soc. ("On the meromorphy of Euler products" (I) and (II)), showing, for example, that for three modular cuspforms with coefficients $a_n$, $b_n$, $c_n$, the naïve "triple product" $\sum a_nb_nc_n/n^s$ has a natural boundary. (In contrast to the naïve Rankin–Selberg $\sum a_nb_n/n^s$.)
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2025-03-21T14:48:31.347533
| 2020-06-23T16:37:35 |
363938
|
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|
Stack Exchange
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Inequality for trace of a symmetric product?
Let $A$ be a real, positive-definite, symmetric operator on an $n$-dimensional space $V$. Write $\odot^k A$ for the action of $A$ on the symmetric power $\odot^k V$. Let $v_1,\dotsc,v_n$ be a basis for $V$ (an orthonormal basis, if you wish). Write $\alpha_i$ for $\langle v_i, A v_i\rangle$.
Is it the case that $\mathrm{Tr} \odot^k A\geq \sum_{i_1\leq i_2\leq \dotsc \leq i_k} \alpha_{i_1} \dotsb \alpha_{i_k}$ for $k\geq 1$ arbitrary? Or for $k$ even?
Note the answer is yes (a) for $k=2$, (b) when $v_1,\dotsb,v_n$ are eigenvectors of $A$.
If the $v_i$ are orthonormal, then yes. If the $v_i$ aren't assumed to be orthogonal, they can cluster around the dominant eigenvector and cause a counterexample.
Let the eigenvalues of $A$ be $\{\lambda_i\}_{i=1}^n$. The trace of $\odot^k A$ is the complete homogeneous symmetric polynomial $h_k(\lambda_1,\dots,\lambda_n)$. In the (orthonormal) $v_i$ basis, the diagonal entries of $A$ are the $\alpha_i$. By the Schur-Horn theorem, the $\lambda$'s majorize the $\alpha$'s. Since $h_k$ is Schur-convex (see here or here), $h_k(\lambda_1,\dots,\lambda_n)\ge h_k(\alpha_1,\dots,\alpha_n)$. Both references prove Schur-convexity for even and odd $k$.
Can you send me your actual name, so that I can add you to the acknowledgements?
After applying an orthogonal transformation, we may and will assume that $v_1, \dots, v_n$ is the canonical basis, so $ \langle Av_i, v_i \rangle= a_{ii}$. So, the question boils down to showing that
$$ \sum a_{11}^{m_1} \cdots a_{nn}^{m_n} \le \textrm{tr } (\textrm{Sym} ^k A)= \sum \lambda_{1}^{m_1} \lambda_2^{ m_1} \cdots \lambda_n^{m_n} . $$
Both sums run on the set of vectors $(m_1, \dots, m_n)$ of non-negative integers summing up to $k$. Write $F( x_1, \dots, x_n)= \sum x_1^{m_1} \cdots x_n^{m_n}$ with the same condition. Note that $F$ takes is constant set of vertices of the polytope $P$ whose extreme points are all permutations of $(\lambda_1, \dots, \lambda_n)$. By Schur-Horn inequality, the point $(a_1, \dots, a_n)$ belongs to $P$. So, to prove the inequality, it suffices to show that $F$ is convex on $P$. Now, since $A$ is positive, we have $x_i \ge 0$ on $P$, so to show that $F$ is convex, it suffices to show that it is convex in $x_i \in [0, \infty)$ when the other variables are fixed and non-negative. But this is clear, since it is a polynomial with non-negative coefficients, so its second derivative is non-negative.
UPDATE: MTyson pointed out below that my proof of convexity is not correct. At the moment I don't see how to fix it.
How do you show $F$ is jointly convex in that region? For instance, $x^2+3xy+y^2$ has nonnegative coefficients but is not convex in the region $x,y\ge 0$.
You are right. I thought for a second that it should follow trivially from convexity with respect to each variable, but it does not. I will add an update.
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2025-03-21T14:48:31.347776
| 2020-06-23T17:09:50 |
363940
|
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|
Stack Exchange
|
Supercuspidals of GL(n) and the induced representation $Ind_{T}^{G}1_{T}$
Let $G$ be the group $GL(n,F)$, where $F$ is a non-archimedean local field. Is it true that for any smooth irreducible supercuspidal representation (with trivial central character) $(\pi,V)$ of $G$ we have
$$\dim_{\mathbb{C}}Hom_{T}(\pi,1_T)=1,$$
where $T$ is the diagonal torus of $G$ ? By Frobenuis reciprocity the result is equivalent to say that
$$\dim_{\mathbb{C}}Hom_{G}(\pi,Ind_{T}^{G}1_{T})=1$$
Is there any reference on the subject ?
Certainly not if the central character of $\pi$ is nontrivial - it will be zero instead.
In fact, I want the result in exactly this case, that is when $\pi$ have trivial central character.
https://doi.org/10.1007/BF01445212
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2025-03-21T14:48:31.347862
| 2020-06-23T17:23:06 |
363942
|
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|
Stack Exchange
|
Natural appeareances of (commutative) algebras in $\mathfrak g$-modules
$\newcommand{\g}{\mathfrak g}$
Let $\g$ be a Lie algebra, and observe that since $U(\g)$ is a cocommutative Hopf algebra, it makes sense to look for (naturally arising and perhaps commutative?) algebras in the (symmetric monoidal) category of $\g$-modules (call these $\g$-algebras).
My question is quite simple: let $G$ be a Lie group with Lie algebra $\g$. Is there a natural way to construct commutative $\g$-algebras? For example, suppose that $M$ is a manifold with a left $G$-action. Can one produce on $C^\infty(M)$ a $\g$-action compatible with the product (in the sense that $g(xy) = (gx)y + x(gy)$) that encodes such $G$-action, in the way Lie-Rinehart pairs encode Lie algebroids?
Can you think of any one natural/geometric context where $\g$-algebras arise?
Hi Pedro. Any context in mind? If $\mathfrak g=0$ then Tour question is "what is a natural comm algebra?". If the answer is "a space", then with $\mathfrak g$ then answer is "a space with some vector fields".. I know you know that answer, but what are you looking for? The symmetric algebra in some natural representation?
Deriving the action of G in C infty you always get an action of the Lie algebra by derivations on functions
Hola @MarcoFarinati ! Of course. I am looking for something similar to: a Lie algebroid on $M$ is the same as a Lie—Rinehart pair $(C^\infty(M),\mathfrak g)$. So I guess the equivalent I want here is just some vector fields that satisfy the defining relations on $\mathfrak g$, yeah. Perhaps there are more interesting examples...
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2025-03-21T14:48:31.348015
| 2020-06-23T17:40:12 |
363943
|
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|
Stack Exchange
|
Bounds on number of "non-metric" entries in matrices
Question:
what upper bounds are known on the number of non-metric entries of finite dimensional square matrices $\boldsymbol{A}\in\mathbb{R}^{n\times n}$ with strictly positive off-diagonal elements $a_{ij}$?
In this context $a_{ij}$ is defined to metric iff $\quad a_{ij}\leqq a_{ik}+a_{kj}\,\forall k\notin\lbrace i, j\rbrace\quad $ and non-metric otherwise.
In the other direction (i.e., forming a metric matrix), you may find https://doi.org/10.1137/060653391 interesting
@SteveHuntsman's reference: Brickell, Dhillon, Sra, and Tropp - The metric nearness problem.
I think the requirement $k\not\in{i,j}$ is slightly confusing: It competes with the quantifier $\forall$ in the attention of the reader. I think the question can be clarified by adding : "The requirements $k\not\in{i,j}$ is equivalent to the requirement that all diagonal entries are maximal (or something similar)".
I had posted the problem because I couldn't see how to solve it for some time but for some strange reason I found a simple answer not long after I put it on MO, so it is rather to be seen as a comment:
The basic idea for constructing an extremal example is to take a densest triangle-free graph and set its edgeweights to $1$ and augment it to a complete graph by adding edges of weight $3$.
Densest triangle-free graphs are $K_{n,n}$ with $n^2$ edges, implying that the number of non-metric edges that augment it to $k_{2n}$ is $\ n\cdot(2n-1)-n^2\ =\ n^2-n$ if the number of vertices is $2n$
Nice. If you write up a proof sketch of optimality and include it here, that will turn this into a bona-fide answer. Gerhard "Always Willing To Upgrade Comments" Paseman, 2020.06.24.
@GerhardPaseman I currently don't see how I can provide a proof. The reason being that things seem to be contradictory: adding edges to the densest triangle-free graph appears to allow for adding the least number of edges; on the contrary, not every edge that is added to a sparser graph will generate a triangle and thus need not increase the count of non-metric edges - I'm puzzled.
If I understand your definition right, then the inequality $a_{ij}>a_{i1}+a_{1j}$ already makes $a_{ij}$ with $i,j>1$ non-metric, so in this case you can easily have only $2(n-1)$ metric elements in the whole $n\times n$ matrix: just those in the first row and the first column off the diagonal. However, judging from the discussion in this thread so far it looks like I misunderstand something, so, please, correct me if I'm wrong.
@fedja: I think you got it right. Thanks for pointing out my error. (My answer, now hidden, gives only a bad lower bound!)
@fedja I can only confirm that you are right; things become clearer to me now...
could you please formulate your solution as an answer so I can indicate the question as settled?
It is almost settled. It is clear now that the minimal number of metric elements is $\ge n$ (every row/column contains at least one) and $\le 2(n-1)$ (my example), but there is still a 2-fold discrepancy between these bounds...
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2025-03-21T14:48:31.348258
| 2020-06-23T18:04:18 |
363946
|
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|
Stack Exchange
|
Sheaf of smooth functions and restriction to a divisor
My question is targeted towards a very particular detail in my research that I am trying to understand. I will therefore break it down into some more general questions.
Let $X$ be a smooth variety, $i:D\hookrightarrow X$ a smooth divisor. Let $E$ be a holomorphic vector bundle, $E|_D =i_*i^* E$ and $\mathcal{A}_X$ the sheaf of smooth functions on $X$.
The tensor product $E^{\infty}=E\otimes_{\mathcal{O}_X}\mathcal{A}_X$ is the sheaf of smooth sections of $E$ as a vector bundle. How do I understand the sheaf $E|_D\otimes_{\mathcal{O}_X}\mathcal{A}_X$?
Let $\rho: E\to E|_D$ be the restriction of holomorphic sections. Consider $\rho\otimes\text{id}:E^{\infty}\to E|_D\otimes_{\mathcal{O}_X}\mathcal{A}_X$. Is there some way to understand how this map acts on the smooth sections of $E$?
Now the actual question that I would like to understand.
Let $T$ be the tubular neighborhood of $D$, $\phi:E|_D\to E|_D$ an automorphism of the vector bundle on $D$. Let $\tilde{\phi}: E^\infty|_T\xrightarrow{\sim} E^\infty|_T$ be a smooth extension of $\phi$ to $T$ (as $T$ is smoothly retractible to $D$). Does the following diagram commute?
Clearly, if $T$ is a holomorphic tubular neighborhood, then question 3 would have a positive answer. But existence of a such neighborhood is very restrictive. Therefore, I am hoping that understanding the first two questions, will help me better understand the last one. I appreciate any comments, corrections and suggestions.
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2025-03-21T14:48:31.348383
| 2020-06-23T19:35:22 |
363950
|
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|
Stack Exchange
|
Curves on potatoes
On twitter recently, Robin Houston brought up this problem from a mathematical puzzle book of Peter Winkler:
The puzzle is attributed to the book "The mathemagician and pied puzzler", and appears on p. 56 in a chapter on "Puzzles from around the world" by Richard I. Hess (so presumably this problem did not originate in this book).
The point of this question is how does one make this puzzle precisely formulated mathematically? Specifically, does it hold if one interprets "potato" to mean a compact surface embedded in $\mathbb{E}^3$ (3-dimensional Euclidean space), and "curve" to mean an embedded loop, and “identical” to mean differing by an orientation preserving isometric of $\mathbb{E}^3$?
I don't want to give any hints as to the "solution" to the puzzle, so I will leave further discussion to the comments (so don't look there if you don't want a hint).
Robin Houston's comment that "the solution is incredibly simple" is a great hint!
The most obvious mathematical interpretation of the puzzle is that the surface of the potatoes should be smoothly embedded spheres. Then one can show by transversality that there will be a translate of one potato that intersects the other in a smooth closed curve. However, this argument doesn't work in the topological case. One could imagine that the surface is an Alexander horned sphere, or even worse an Alford sphere, and that the intersections between translates might be indecomposable continua, or maybe with non-trivial $\pi_1$ but no closed embedded curve, or a topologist's sine curve.
@IanAgol Unless I am misunderstanding the constraints of the puzzle you don't necessarily need to take intersecting translates of the two surfaces. If you rotate one first and then intersect can't you avoid the bad intersections to get a closed curve? This seems possible in the horned sphere case at least.
"identical" means that there's a (positive) isometry of the ambient Euclidean space mapping the first to the second?
@YCor: yes, do you have a possible other interpretation?
@IvanMeir: Yes, it might work for a specific example, but it's not clear how to do that e.g. for an Alford sphere. https://books.google.com/books?id=Qo2DAwAAQBAJ&lpg=PA91&ots=7pG0sAhVNy&dq=alford%20sphere&pg=PA89#v=onepage&q=alford%20sphere&f=false
@IanAgol Yes I do, and you certainly can imagine so. I mentioned the one I found most likely.
@ycor: what other interpretation can you think of? That would help me clarify the question if it is ambiguous.
@IanAgol for instance equal modulo possibly negative isometry, or equal modulo similarity (I've already seen "equal" to mean "the same mod positive similarity" for planar triangles).
@YCor Okay, I see what you mean. Given that the original problem as stated was ambiguous, I should try to make it more precise.
I guess one could also interpret “identical” to mean an isometry of the induced metric. For connected arcs I think this should be equivalent to an ambient isometry (but possibly orientation reversing).
Well, "smooth compact surfaces" and "embedded loop" should be enough. Just move one surface toward the other and let them intersect transversally. Probably Sard theorem is necessary at a certain point in order to formalize things, but the idea seems to work
@FrancescoPolizzi: Yes, it follows from Sard. Given two smooth oriented surfaces $S_1,S_2 \subset \mathbb{E}^3$, take a map $\phi: T^1S_1\times T^1S_2\to Isom^+(\mathbb{E}^3)$ which is the unique isometry $g$ taking $v_1\in T^1S_1$ to $v_2 \in T^1S_2$, and the unit normal to unit normal. $S^1$ acts on the fibers of this map, since rotating the vectors does not affect the isometry. Hence the image of $\phi$ is $5$-dimensional in a $6$-dimensional Lie group. Any isometry $g \in Isom^+(\mathbb{E}^3)-im(\phi)$ with $g(S_1) \cap S_2\neq \emptyset$ will be transverse and hence contain a loop.
@IanAgol A very weird potatoe :-)
Real potatoes aren't what I would call "smooth," so it would be be nice to have a topoogical version of the result. Though I've never seen an Alexander horned potato! Does it make things easier to assume that the potato is a sphere whose exterior is simply connected?
I always skip potatoes with infinite surface area at the store - they take too long to peel.
@AntonPetrunin I think the problem is well-defined as given. Sorry, I didn’t know (or forgot) about your older question.
@AntonPetrunin the smooth case is easy to solve, so not an open question.
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2025-03-21T14:48:31.348704
| 2020-06-23T19:46:34 |
363952
|
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|
Stack Exchange
|
Chromatic t-structures?
Questions: Fix a prime $p$ and $n \in \mathbb N_{\geq 1}$.
Does the category $Sp_{K(n)}$ of $K(n)$-local spectra admit a nontrivial $t$-structure?
By "nontrivial", I simply mean that $\{0\} \subsetneq Sp_{K(n),\geq 0} \subsetneq Sp_{K(n)}$.
Does $Sp_{K(n)}$ admit a nontrivial monoidal $t$-structure?
"Monoidal" means that (1) $\mathbb S_{K(n)} \in Sp_{K(n),\geq 0}$ (where $\mathbb S_{K(n)}$ is the $K(n)$-local sphere) and (2) $Sp_{K(n),\geq 0}$ is closed under the $K(n)$-local smash product. (Evidently I am using homological, rather than cohomological, indexing.)
As usual, the corresponding $T(n)$-local questions are also interesting, though presumably harder. For context, I'd also be interested in hearing about the $E(n)$-local or $T(0) \vee \dots \vee T(n)$-local versions of these questions.
In the above, feel free to interpret "the category $Sp_{K(n)}$" as either "the triangulated category $Sp_{K(n)}$" or as "the stable $\infty$-category $Sp_{K(n)}$" -- whichever is most comfortable.
There's an easier question which has a negative answer: for $n \in \mathbb N_{\geq 1}$, the category $Mod_{K(n)}$ of $K(n)$-module spectra does not admit a nontrivial $t$-structure. For every object of $Mod_{K(n)}$ is a coproduct of shifts of $K(n)$. So if $0 \neq X \in Mod_{K(n),\geq 0}$, then there is a retract $\Sigma^k K(n)$ of $X$ which is a shift of $K(n)$, so that $\Sigma^k K(n) \in Mod_{K(n),\geq 0}$. Then because $K(n)$ is periodic, every object $Y \in Mod_{K(n)}$ is a coproduct of nonnegative shifts of $\Sigma^k K(n) \in Mod_{K(n),\geq 0}$, and so $Y \in Mod_{K(n),\geq 0}$.
But of course, the category $Sp_{K(n)}$ is much more complicated than the category $Mod_{K(n)}$.
When $n = 0$ (so that $K(n) = H \mathbb Q$) or $n = \infty$ (so that $K(n) = H \mathbb F_p$), $Mod_{K(n)}$ does admit a monoidal $t$-structure given by usual connectivity, and $Sp_{K(n)}$ inherits a monoidal $t$-structure by pullback along the free functor $Sp_{K(n)} \to Mod_{K(n)}$ (which is an equivalence for $n = 0$, of course). I don't think these cases shed much light on the case $n \in \mathbb N_{\geq 1}$, though.
To expand on Tim's answer, the arguments generalize to show that $Sp_{K(n)}$ admits no non-trivial t-structures in general.
The crucial ingredient is that $Sp_{K(n)}$ has no non-trivial localising or colocalising subcategories, see 7.5 in Hovey-Strickland. Thus, to finish the argument it is enough to show that any subcategory $C \subseteq Sp_{K(n)}$ which is closed under limits is in fact colocalising, ie. it's also closed under suspension.
If $X$ is any $K(n)$-local spectrum, then it is well-known (see 7.10 in the aforementioned book) that it can be written as a limit $X \simeq lim \ X \wedge F_{i}$ of its smash products with type n generalized Moore spectra. If $X \in C$, then some desuspension of $X \wedge F_{i}$ is contained in $C$ as well, but as Tim observed these spectra are periodic and so $\Sigma^{n} X \wedge F_{i} \in C$ for all $n \in \mathbb{Z}$. It follows that $\Sigma^{n} X \in C$, ending the argument.
Thanks! Do you know whether the limit expression $X = \varprojlim X \wedge F_i$ holds in $Sp_{T(n)}$ as well? That would imply at least that any $t$-structure on $Sp_{T(n)}$ is colocalizing.
Not sure, but this sounds plausible. To be more precise, there is a dual description of the $K(n)$-local sphere as a colimit of type n Moore spectra (obtained by dualizing the previous one) and I think this should hold $T(n)$-locally as well. For example, at $n=1$ this is saying that $K(1)$-locally we have $colim \ \Sigma^{-1} S^{0}/p^{k} \simeq S^{0}$. Indeed, the cofibre of this map is the rational sphere - but this also vanishes $T(1)$-locally. Then at $n = 2$ you would write each of $\Sigma^{-1} S^{0}/p^{k}$ itself as a colimit of $\Sigma^{-2} S^{0}/(p^{k}, v_{1}^{l})$, and so on.
The second question turns out to have a surprisingly easy negative answer. This is depressing on two counts: both that the answer is negative and that it's so easy.
Suppose that $Sp_{K(n)}$ has a $t$-structure such that $\mathbb S_{K(n)} \in Sp_{K(n),\geq 0}$. Let $F(n)$ be any finite type-$n$ $p$-local spectrum. Then for some $k \geq 0$ we have that $\Sigma^k F(n)$ is in the closure of $\mathbb S_{(p)}$ under finite colimits in the category $Sp_{(p)}$ of $p$-local spectra. Therefore, $\Sigma^k F(n)_{K(n)} \in Sp_{K(n),\geq 0}$. But we also have $\Sigma^k F(n)_{K(n)} \simeq T(n)_{K(n)}$, which is a periodic spectrum. Thus $\Sigma^l F(n)_{K(n)} \in Sp_{K(n),\geq 0}$ for all $l \in \mathbb Z$. Since $F(n)$ was an arbitrary finite type-$n$ spectrum, we see that all $K(n)$-localizations of finite type-$n$ spectra are in $Sp_{K(n),\geq 0}$.
Now I'm pretty sure that every object of $Sp_{K(n)}$ is a colimit of $K(n)$-localizations of finite type-$n$ spectra. It follows that every object is in $Sp_{K(n),\geq 0}$ and the $t$-structure is trivial. But I can't find a reference for this fact at the moment, so here's an alternate argument. It's at least the case that $\mathbb \Sigma^l \mathbb S_{K(n)}$ is a (sequential) colimit of $K(n)$-localizations of finite type-$n$ spectra for all $l \in \mathbb Z$, and so $\Sigma^l \mathbb S_{K(n)} \in Sp_{K(n),\geq 0}$ for all $l \in \mathbb Z$. If the $t$-structure is monoidal, it follows that $Sp_{K(n),\geq 0}$ is closed under desuspension, i.e. $Sp_{K(n),\geq 0} \subseteq Sp_{K(n)}$ is a a stable subcategory. This kind of $t$-structure is not very interesting, and anyway I believe that since $Sp_{K(n)}$ doesn't admit any nontrivial Bousfield localizations, it doesn't admit any such $t$-structures which are nontrivial either.
|
2025-03-21T14:48:31.349041
| 2020-06-23T20:00:17 |
363953
|
{
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"Christian Remling",
"MightyPower",
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"https://mathoverflow.net/users/152731",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/48839"
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}
|
Stack Exchange
|
The limit of the operator norm in a Hilbert space
I am not familiar with functional analysis. Could you tell me please, how to prove the following statement (if it is true)?
$$
\lim_\limits{M \to \infty} \|T_A - T_b \| = 0,
$$
here operator norm defined as $\|A\| = \sup_\limits{\| x \| = 1}\| Ax \|$, $x \in \ell^2$. $T_A$ and $T_B$ are operators defined as
$$
T_A = \exp\{\alpha B\},
$$
$$
T_B = \left(E - \frac{\alpha}{M}B \right)^{-(M - p)},
$$
here $\alpha \in \mathbb{R}$, $M \in \mathbb{N}$, $p \in \mathbb{N}$ , and $B$ is the following linear operator
$$
B = \sum_{k} |k\rangle \langle k+1|.
$$
Is $E$ the identity operator?
@Nate Eldredge, Yes E - identity operator.
@Nate Eldredge, Why? I compare this with a similar limit for functions and $\lim_\limits{n \to \infty} \left(1 - \frac{a}{n} \right)^{- (n-p)} = e^a$, not $e^{pa}$.
Oh you're right, never mind.
This should have the same proof as the scalar calculus identity that you mention since all the operators involved here ($1,B$) commute, so the fact that we're dealing with operators rather than numbers can't really make itself felt.
|
2025-03-21T14:48:31.349150
| 2020-06-23T20:31:59 |
363956
|
{
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"url": "https://mathoverflow.net/questions/363956"
}
|
Stack Exchange
|
Conjugate operator
Assume that $$T(f) = \int_D K(z,w) f(w) du dv+\int_D H(z,w) \overline{f(w)} du dv$$ is an operator. It seems to me that its conjugate operator is $$T^*(f) = \int_D \overline{K(w,z)}f(w) du dv+\int_D \overline{H(w,z)} \cdot \overline{f(w)} du dv$$ however something is wrong. Here $D$ is for example the unit disk in the complex plane. What is wrong?
Is it an exercise?
Let me define by $C$ the operator of complex conjugation, then $T= K+HC$; I presume by the conjugate operator $T^\ast$ you mean the Hermitian adjoint of $T$; as discussed here, $C^\ast=C$, so $T^\ast=K^\ast+CH^\ast=K^\ast+H^{\rm t}C$, with $H^t$ the transpose. Hence
$$T^*(f)[z] = \int_D \overline{K(w,z)}f(w) \,dw+\int_D H(w,z) \overline{f(w)}\, dw.$$
|
2025-03-21T14:48:31.349224
| 2020-06-23T22:03:41 |
363963
|
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"url": "https://mathoverflow.net/questions/363963"
}
|
Stack Exchange
|
Convolution Integral Equation on a compact subset of the real line
I am dealing with the following equation: $$ f(x) = g(x) + \intop_{X} dt K(x-t)f(t) \;,\qquad \left\lbrace \begin{array}{c}f(x)>0\;,\;x\in X \\ f(x)<0\;,\;x\notin X \end{array}\right.$$ where $X$ is a compact, not necessarily connected, subset of $\mathbb R$ and $$ g(x) = -\cosh x \;,$$ and $K(x)$ is a function in $L^2(\mathbb R)$. The subset $X$ is to be treated as an unknown, to be determined self-consistently by the requirements on the positivity of the function $f(x)$. Hence, by solution to the above equation I mean a function $f(x)$ together with a subset $X$.
A specific case is $X = [-A,A]$ and $$ K(x) = \frac{1}{2\pi}\sum_{\sigma,\sigma'=\pm1}\frac{1}{\cosh\left(x+\sigma\theta+i\sigma'\gamma\right)}\;,$$ but I would like to keep things more general.
My question consists of two parts:
there exist known results on existence and uniqueness of solutions?
is there a known procedure to solve this equation for $f(x)$ and $X$? I tried to extend the method of Wiener and Hopf, but I do not seem to be getting anywhere useful.
Thanks,
Stefano
|
2025-03-21T14:48:31.349332
| 2020-06-23T22:27:38 |
363964
|
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"Eric Towers",
"Igor Belegradek",
"Willie Wong",
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|
Stack Exchange
|
Gauss-Bonnet Theorem on noncompact surface without boundary
Let $(M,g)$ be a simply connected 2-dimensional Riemannian manifold without boundary, and let $K$ be the Gausssian curvature defined on $M$. If $M$ is compact, then by Gauss-Bonnet Theorem, we have
$$\int_M K dA = 4\pi,$$where $dA$ is the area element of $M$ under the metric $g$.
If $M$ is not compact, then the above equality is no longer true. For example, let $M=\mathbb{R}^2$ and we define the conformal metric $g=e^{2u}\delta$, where $u=\ln(sech x)$ and $\delta$ is the Euclidean metric, then one can verify that $K \equiv 1$ on $M$, but the total area of $M$ is $\infty$, not $4\pi$.
Motivated by this example, my question is that, if we assume $(M,g)$ is a noncompact simply connected surface without boundary, and the total area of $M$ is finite, then is it true that
$$\int_M K dA =4\pi?$$
Or is it true at least for the conformal case?
Consider the metric on $\mathbb R^2$ that is rotationally symmetric metric outside a compact set, namely, it is $dr^2+m(r)^2 d\phi^2$ for $r>R>0$. Here $m$ is a positive function on $[R,\infty)$.
The area form at points with $r>R$ is $dA=m(r)drd\phi$, so the surface has finite area if and only if $m$ is integrable on $[R,\infty)$. The total curvature of the rotationally symmetric end is $$\int_R^\infty -\frac{m^{\prime\prime}}{\!\!\! m} dA=-2\pi\int_R^\infty m^{\prime\prime} dr=2\pi\left(m'(R)-\lim_{r\to\infty} m'(r)\right).$$
It is easy to find examples where the limit on the right hand side does not exist but $m$ is integrable. If the limit exists and $m$ is integrable, the limit is zero, and the total curvature of the end is $2\pi m^\prime(R)$.
The metric is arbitrary if $r<R$, and the total curvature of this region can be computed with the usual Gauss-Bonnet for surfaces with boundary. The geodesic curvature of the boundary is easy to compute from the $r\ge R$ side (I don't remember the answer).
EDIT: As Willie Wong points out by Gauss-Bonnet every smooth metric on the region $\{r<R\}$ will have the same total curvature. So just extend $m$ to a smooth function on $[0,R]$ so that $m(r)=r$ near $0$ and consider the metric $dr^2+m(r)^2d\phi^2$ for all $r>0$. Its metric completion is smooth at the origin. (More generally, the metric is smooth at the origin if and only if $m^\prime(0)=1$ and $m$ extends to an odd smooth function on $\mathbb R$). Now the above computation gives total curvature as $2\pi(m^\prime(0)-m^\prime(\infty))=2\pi(1-m^\prime(\infty))$ and if $m^\prime(\infty)$ exists and the area is finite, the total curvature is $2\pi$.
To be lazy, you can set $m'(R) = 0$ and you have that the circle $r = R$ is geodesic. Then the total curvature is exactly $2\pi$.
Right, but it would be nice to compute the general case. Here is a sketch (to be checked !). The vector field $X=\frac{1}{m(R)}\frac{d}{d\phi}$ is a unit vector field tangent to the curve $r=R$. The geodesic curvature is the length of $\nabla_X X=-\frac{m^\prime(R)}{m(R)}$. So it looks like the geodesic curvature of the boundary is $-2\pi \frac{|m^\prime(R)|}{m(R)}$.
Thank you very much for this inspirational post. I'm just curious that, does there exist a positive function $m(r)$ such that $m(r)$ is integrable, while $\lim_{r \rightarrow \infty} m'(r) \ne 0$? Also, I think the metric cannot be arbitary if $r< R$, since we require the metric is smooth, right?
Taking my laziness further: by Gauss Bonnet it doesn't matter what the portion $< R$ is, so might as well be a spherical cap and assume $R < \pi \rho$ where $\rho$ is the radius of the spherical cap. The curvature integral of the cap is $2 \pi (1- \cos(R/\rho))$. Then since $m(r) = \rho \sin(r/ \rho)$ we have $m'(R)$ is simply $\cos(R)$. So in fact even in the general case you will have that that any surface that you constructed will have total curvature integral exactly $2\pi$.
@student: if the limit $\lim m'(r)$ exists, it must be zero (otherwise $m(r)$ is unbounded). The limit may fail to exist (imagine of $m(r)$ containing a train of skinnier and skinnier bumps). // Sufficient regularity of the metric (for Gauss Bonnet to apply) is implicitly assumed. You just need the metric to be piecewise $C^2$ (IIRC) for Gauss-Bonnet to work.
@student : Such $m(r)$ do exist. For instance, let $m(r)$ be a smooth bump version of a (successively less) truncated rationals identification function. Say on each $[z,z+1)$ for each $z \in \Bbb{Z}_{\geq 0}$, we place a bump of height $z^{-2}$ and width $z^{-4}$ (narrow enough that no two bumps overlap) only on rationals whose reduced form has a denominator less than $z$. The number of bumps goes as $z^2$, their areas go as $z^{-6}$, and $\sum_z z^{-4}$ converges. This $m$ is integrable, but the limit of $m'$ does not exist.
If $M$ has finite total curvature, then $$\int_M KdA\leq 2\pi$$ by the Cohn-Vossen inequality. The $2\pi$ arises since the Euler characteristic of a noncompact simply-connected surface without boundary is $1$.
I don't think this is right. A simple example is to consider $\mathbb{R}^2, \frac{4}{1+x^2+y^2}\delta$. This is simply-connected without boundary, and it is not compact (we don't compatify it at $\infty$), then the total curvature is $4\pi$. I checked th eCohn-VOssen inequality, and it requires that the manifold is complete.
For Cohn-Vossen you need to assume that the surface is complete.
Since you asked about the conformal case: consider the metric $g = e^{2\phi} \delta$. The area element is $e^{2\phi} ~dx$. The Gauss curvature is
$$ K = - e^{-2\phi} \Delta \phi $$
and so the curvature integral is equal to
$$ - \int_{\mathbb{R}^2} \Delta \phi ~dx $$
Suppose now that $\phi = \phi(|x|)$ is radial. Then the total integral can be evaluated using Gauss-Green as
$$ - \lim_{R\to\infty} 2\pi R\phi'(R). $$
Consider the case that $\phi$ is a smooth function such that for all $|x|$ sufficiently large we have $\phi(|x|) = - \kappa \ln(|x|)$. Notice that when $\kappa > 1$ we have that $M$ has finite total area. Observe also that by a direct computation that the total curvature integral can be evaluated to equal exactly $2\pi \kappa$.
And hence we have that in the conformal case the valid range of values of the total curvature integral contains at least the full range $(2\pi, \infty]$. (The $\infty$ endpoint is attained for, e.g., $\phi = - |x|^2$.)
Thank you very much!
|
2025-03-21T14:48:31.349859
| 2020-06-23T22:37:29 |
363965
|
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"Federico Poloni",
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|
Stack Exchange
|
Updating the null space of a matrix
I am facing a problem where I have to find any (nontrivial) vector x such that Ax=0, where A is a rectangular nxm matrix with m>n, so the problem is underdetermined. I must find this x for A, but also for a new matrix A' = (A with the column j removed), and so on...
It would be very helpful to find a way to obtain the new solution x' for the matrix A' knowing the solution x for A without recomputing a whole null space by SVD or QR at each iteration. I managed to find x' with a Newton Raphson method (as x' is close to x with the element j removed), but I have that problem of inverting the Jacobian matrix at each iteration once again.
There is literature on updating various matrix factorizations under rank-1 modifications (which includes row/column insertions and removals). See for instance Secton 6.5 on Golub--Van Loan 4th edition. In particular, QR updating is already implemented in Matlab and Scipy. I am not familiar with updates of the SVD, but a Google search for "svd update" returns various articles that treat this exact problem.
You will probably want to make sure that the factorization you update is a rank-revealing one; note that QR without column pivoting does not always work: there are counterexamples where all the diagonal entries of $R$ are large, but the matrix is numerically singular. For a specific example, see for instance Golub--Van Loan, 4th ed, sec. 5.4.3: there is an example of a 300x300 upper triangular matrix where the smallest diagonal entry is $\approx 0.05$, and yet the matrix has a singular value $\approx 10^{-19}$.
Hi Federico and also this is a question for the OP too. I was wondering, if the task is to find any non-trivial vector in the nullspace of A why would one choose the SVD or QR? Gaussian Elimination is faster, stable in practice and easy to update given a change in a column vector.
@IvanMeir To make Gaussian elimination stable, you need pivoting, but I don't see immediately how you can combine pivoting and column updates. Do you have a specific procedure in mind? Also, standard Gaussian elimination with partial pivoting is known not to be "rank-revealing"; i.e., one can find examples where all pivots are large but the matrix has a very small singular value.
For a specific example, see for instance Golub--Van Loan, 4th ed, sec. 5.4.3: there is an example of a 300x300 upper triangular matrix where the smallest diagonal entry is $\approx 0.05$, and yet the matrix has a singular value $\approx 10^{-19}$.
Thank you Federico for your reply, I agree these are considerations and I guess the best option it will depend a lot on the data. If the column removed contains a pivot you need to perform more steps to repair the echelon form and recompute the solution but only to the right of the pivot. How expensive this will be overall depends on precisely how many of these columns are eventually removed. In terms of stability I agree there can sometimes be problems though my experience is that you generally do much better than the worst case scenario! Again it generally depends on the application :-)
True but if your system is over determined and you don't have a weird case where most columns lie in a low dimensional space just picking a random set of columns should give you a reasonable starting point for elimination. Again it depends on the data distribution and the matrix dimensions.
Nice example though, thank you for sharing :-)
One way to compute a particular vector that solves such an undetermined system is to perform gaussian elimination and compute the row echelon form. If you then eliminate one column from the leading columns - the one's with 0's you can simply update the form with a few more operations and quickly update your solution.
Note that removing a column could require an update in most elements of your solution. For example if you have the following matrix with solution $(1,1,1,-1,0)$ then if you remove the first column the solution becomes $(0,0,0,1,-1)$ which is unique up to multiplication by a non-zero scalar.
$$\left[ \begin{array}{ccccc}
1 & 0 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 & 1 \\
0 & 0 & 1 & 1 & 1
\end{array} \right]$$
|
2025-03-21T14:48:31.350141
| 2020-06-24T00:05:20 |
363969
|
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|
Stack Exchange
|
Ihara zeta function and closed paths and trails
Let $\Gamma$ be a finite graph. There seem to be two definitions of closed path in the literature. In one, a closed path is just a walk whose starting vertex is the same as the ending vertex. (Let us call that a closed walk.) In the other one, which we will follow, a closed path is a closed walk visiting vertices $v_0, v_1,\dotsc,v_k=v_0$ with all $v_i$ distinct (except perhaps $v_k=v_0$.) Let us follow the second definition.
A closed trail is a closed walk that does not repeat edges.
A closed geodesic is a closed walk without backtracking or tails, i.e., $v_i\ne v_{i+2 \mod k}$ for all $i$, where $k$ is the length of the path, just as above. A closed geodesic is said to be prime if it is not equal to another closed geodesic repeated $d\geq 2$ times.
The Ihara zeta function is defined to be
$$\zeta_\Gamma(z) = \prod_{[P]} (1- z^{|[P]|})^{-1},$$
where $[P]$ runs over all prime closed geodesics. Here $|[P]|$ is the length of $P$.
Is it possible to write $\zeta_\Gamma(z)$ as a product over (some) closed paths of given lengths (following the second definition)? Is it possible
to write it as a product over closed trails?
I think I had a similar thought/problem here...
What's the difference between closed trails and a closed geodesics? If you don't allow backtracking and tails, you won't repeat vertices or edges...
|
2025-03-21T14:48:31.350267
| 2020-06-24T00:09:40 |
363970
|
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|
Stack Exchange
|
Not triangulirazable over $\Bbb R$ implies diagonalizable over $\Bbb C$
Let $A$ be a real matrix. Suppose $A$ is not triangulirazable over $\Bbb R$ then $A$ is diagonalizable over $\Bbb C$.
My proof: Since $A$ is not triangularizable over $\Bbb R$ it has a complex eigenvalue. But complex eigenvalues occur in pairs for real matrices. Hence, not all the eigenvalues of $A$ are equal. is this sufficient to conclude that $A$ is diagonalizable over $\Bbb C$?
Thank you.
Just because there is more than one eigenvalue of $A$, that doesn't necessarily imply that all the eigenvalues of $A$ are distinct
The question belongs on MSE, but since I don't have an MSE account I have supplied an answer here before the question gets migrated
Wait! did I ask this in MO? very sorry. Just noted it. I thought I was in MSE.
I was going to sketch a construction for a counterexample in the comments, but on reflection it may be more sensible to write it out in full as an answer.
Let $B=\begin{pmatrix}
i & 0 \\ 0 &-i
\end{pmatrix}$ and let $T=\begin{pmatrix} 1 & 1 \\0 & 1 \end{pmatrix}$. Then the $4\times 4$ matrix
$$
B\otimes T = \begin{pmatrix} iT & 0 \\ 0 & -iT \end{pmatrix} = \begin{pmatrix} i & i & 0 & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & -i & -i \\ 0 & 0 & 0 & -i \end{pmatrix}
$$
has no real eigenvalue and is not diagonalizable over ${\mathbb C}$. On the other hand, note that for a suitable complex unitary matrix $U\in M_2({\mathbb C})$, $A_0:=UBU^*\in M_2({\mathbb R})$. We could take
$A_0 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.
Therefore if we take
$$
A:= A_0 \otimes T = \begin{pmatrix} 0 & - T \\ T & 0 \end{pmatrix} =
\begin{pmatrix}
0 & 0 & - 1 & -1 \\
0 & 0 & 0 & -1 \\
1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0
\end{pmatrix}
$$
then $A$ is a real matrix, which is similar in $M_4({\bf C})$ to $B\otimes T$, and hence has no real eigenvalue nor has any diagonalizable over ${\bf C}$.
is it possible to find such a matrix in $3 \times 3$ order? Thank you.
Consider
$$A = \begin{bmatrix}1 & 1 & & \\ 0& 1 & & \\ & & 0& 1 \\ & & -1 & 0\end{bmatrix}$$
$A$ has complex eigenvalues, so it is not triangulizable over $\mathbb R$. $A$ is not diagonalizable over $\mathbb C$ since $\mathrm{rank} (A-I)^2 < \mathrm{rank} (A - I)$.
is it possible to find such a matrix in $3 \times 3$ order? Thank you.
|
2025-03-21T14:48:31.350428
| 2020-06-24T00:42:10 |
363972
|
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|
Stack Exchange
|
Probabilistic inequality for sum of squares of zero mean Gaussian random variables
Let $X_1,...,X_n$ be i.i.d. standard normal random variables. How to show that there is constant $c>0$ such that for every $a_k>0$ and for every $n>0$: $P(\sum_{k=1}^{n}a_kX_k^2>\sum_{k=1}^{n}a_k+c\cdot\sqrt{\sum_{k=1}^{n}a_k^2})>c$
Such problems are known as "small ball probabilities". Checkout Theorem 8.1 of this manuscript by Rudelson et al. http://www-personal.umich.edu/~rudelson/papers/rv-smallball.pdf
Also related are chi-squared variables[1] and their tail inequalities[2]. [1] https://en.wikipedia.org/wiki/Chi-square_distribution [2] https://stats.stackexchange.com/questions/4816/what-are-the-sharpest-known-tail-bounds-for-chi-k2-distributed-variables
|
2025-03-21T14:48:31.350512
| 2020-06-24T01:27:46 |
363977
|
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|
Stack Exchange
|
Any finite flat commutative group scheme of $p$-power order is etale if $p$ is invertible on the base
This question is immediately related to Discriminant ideal in a member of Barsotti-Tate Group
dealing with Barsotti–Tate groups and here I
would like to clarify a proof presented by
Anonymous in the comments from linked thread which I do not completely understand. Although meanwhile I found another proofs of the claim below I have a big interest on understanding this proof below.
Assume $G$ is a finite flat commutative group scheme
over a field $k$ of
order $p^N$. Assume $p$ prime and $p \in k \setminus \{0\}$,
equivalently invertible on the base.
Claim: Any finite flat commutative group scheme of $p$-power order
is etale if $p$
is invertible on the base.
Anonymous' proof works as follows:
Firstly we reduce to case over a field (because a
finitely presented flat map is etale if it is
fibrewise etale). Since we assumed $G$ commutative
the multiplication by $p^N$ map $f_{p^N}: G \to G$
is well defined and by Deligne's theorem
$p^N$ kills $G$ since it's the order of $G$.
That means that $f_{p^N}$ is the zero map: equivalently
it factorizes over $\operatorname{Spec}(k)$.
What now comes I do not understand:
It is claimed that $f_{p^N}$ is unramified
"as the map on the tangent spaces is given
by $p^N$, which is invertible".
Question I: why is the induced by $f_{p^N}$
maps on tangent spaces given
by $p^N$?
Question II: assume we understand Question I.
Why does this imply $G$ is unramified?
when we can answer these two question we are done because unramified finite type schemes over a field are etale.
You don't need Deligne's theorem to prove this. For the standard proof, see, for example, 11.31 of Milne, Algebraic Groups, CUP, 2017.
Nor do you need to assume it's commutative. Another reference is Waterhouse Introduction to affine group schemes, §11.4, §14.4.
But if you insist on this argument, I follows since the derivative of $\mu \colon G \times G \to G$ is addition $T_e G \times T_e G \to T_e G$ (use $\mu \circ (\operatorname{id} \times e) = \operatorname{id} = \mu \circ (e \times \operatorname{id})$ where $e \colon \operatorname{Spec} k \to G$ is the unit). II is a near immediate consequence of the definitions (see e.g. Tags 00UT and 00RS).
I finished the title (which ended mid-sentence), and added a link to the comment I think that you meant. If I got the wrong comment, please fix it or let me know.
@LSpice: you linked the correct comment
@R.vanDobbendeBruyn: This is the important point: the map between tangent spaces $T_e G$ over $e$ induced by $f_{p^N}= \mu^{p^N}$ is as you correctly remarked the multiplication by $p^N$. Since $p^n$ is inverible in $k$ this map $T_e G \to T_e G$ is an isomorphism and dualizing the (1) in
https://stacks.math.columbia.edu/tag/00UT implies only that $f_{p^N}$ is unramified at the stalk $\Omega_e$; that is at the unit point $e$ (!) . But Anonymous clamed that this should imply that $f_{p^N}$ is unramified; that means is unramified at every point of $G$.
Ok, you're right, but differentials on a group scheme are determined by what happens at the identity. See for example Prop 3.3 of Bhatt's notes on abelian varieties. To be precise, $\Omega_{G/S} = \pi^e^\Omega_{G/S}$ for $\pi \colon G \to S$ a group scheme with unit $e \colon S \to G$. Similarly for a morphism of group schemes $\phi \colon G \to H$, the map $\phi^\Omega_{H/S} \to \Omega_{G/S}$ is determined by $e^\Omega_{H/S} \to e^*\Omega_{G/S}$ under pullback to $G$. This is standard and used all the time.
I understand. And how now we deduce from this observation that $G$ is unramified. As you said $f_{p^N}^\Omega_{G,e} \to \Omega_{G,e}$ determine $f_{p^N}^\Omega_{G} \to \Omega_{G}$. The first one was an isomorphism ($p^N$ invertible) therefore $f_{p^N}^*\Omega_{G} \to \Omega_{G}$ as well. Why this imply $G$ unramified?
I think you can apply it to the morphism of group schemes $G \to 1$, which is both the map $f_{p^N}$ (when viewing $1 \subseteq G$ as a closed subscheme) and the structure morphisms $G \to \operatorname{Spec} k$.
Yes, I think I got it: $f_{p^N}^\Omega_{G,e} \to \Omega_{G,e}$ is simultaneously in isomorphism ($P^N$ invertible) and zero map (because induced by zero map $f_{p^N}$). With the argument you remarked from Bhatt's notes then same must be true for $f_{p^N}^\Omega_{G} \to \Omega_{G}$ as well. Consequently $\Omega_{G}=0$. Fini. Thank you a lot!
|
2025-03-21T14:48:31.350803
| 2020-06-24T04:10:10 |
363981
|
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"KConrad",
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"url": "https://mathoverflow.net/questions/363981"
}
|
Stack Exchange
|
Dirichlet's unit theorem in finite characteristic
I'm looking for a source of the following analog of Dirichlet's unit theorem for finite characteristic fields:
Let $\mathbb{F}_p$ be a finite field and denote $K=\mathbb{F}_p(x)$ to be the field of rational functions over $\mathbb{F}_p$, i.e. functions of the form $\frac{\sum_{n=0}^{N}a_nx^n}{\sum_{n=0}^Mb_n x^n}$ where $a_n,b_n\in \mathbb{F}_p$. This is the field of fractions of $\mathbb{F}_p[x]$, the ring of polynomials over $\mathbb{F}_p$. Here, we have a norm defined by $\Vert \sum_{k=-\infty}^{N}a_kx^k\Vert=p^N$ if $a_N\neq 0$. Let $L$ be a field extension of $K$ of finite rank and let $\sigma_1,\dots \sigma_n:L\rightarrow \bar{K}$ be the embeddings. Denote by $\mathcal{O}_L$ the algebraic integers in $L$, i.e. the elements $t\in L$ which are roots of polynomials with coefficients in $K$. We can look at the $\mathcal{O}_L^*$, the group of units in $\mathcal{O}_L$ and we would like to say something about its rank.
In Dirichlet's unit theorem for number fields over $\mathbb{Q}$, we know that $\mathcal{O}_L^*$ is of rank $r_1+r_2-1$, where $r_1$ is the number of real embeddings and $r_2$ is the number of complex embeddings without conjugates. There are many sources for the case of number fields over $\mathbb{Q}$ but I couldn't find a source for the unit theorem over fields of finite characteristic.
The "embeddings" $L \to K$ should be $L \to \overline{K}$. Anyway, there is a unit theorem for all global fields (number fields and function fields over finite fields) called the $S$-unit theorem. It is about the structure of the group of $S$-units: it is finitely generated of rank $|S| - 1$. If $S$ is the infinite places of a number field, $|S| = r_1 + r_2$. See Theorem 5.3.10 in Weiss' "Algebraic Number Theory" or the chapter on $S$-units and the $S$-class group in Rosen's "Number Theory in Function Fields", or the theorem in Section 18 of Chapter II in Cassels & Frohlich (using ideles).
Your MO page mentions that you're in Tel Aviv. Among the faculty at Tel-Aviv Univ. is Bary-Soroker, who works in function field arithmetic. If you have questions about number theory in function fields then you could reach out to him.
|
2025-03-21T14:48:31.350960
| 2020-06-24T04:27:12 |
363983
|
{
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"authors": [
"Denis Serre",
"Rajesh D",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/363983"
}
|
Stack Exchange
|
Bounding the condition number of a matrix associated with an even symmetric positive definite function
Define a set $A = \{x_i/x_i\in\mathbb{R}^m, i = 1,2,3..n\}$. Let $f:\mathbb{R}^m\to(0,\infty)$ be an even symmetric positive definite function.
Let $D = [d_{i,j}]$ be an $n\times n$ matrix such that $d_{i,j} = f(x_i-x_j)$
Let $\epsilon = \min\limits_{i,j} \|x_i-x_j\|_2$ and assume $\epsilon > 0$.
Consider the matrix $D+\alpha I$, where $I$ is an identity matrix and $\alpha>0$. Naturally $D$ is a positive semi definte matrix as the function $f$ is a positive definite function. So $D+\alpha I$ is positive definite.
I am looking for an upper bound on the condition number of the matrix $D+\alpha I$ in terms of $\alpha$, $\epsilon$ and the function $f$. Does such a bound exist?
Condition number defined as the ratio of magnitude of largest eigenvalue to the magnitude of the least eigenvalue.
"$D$ is positive semi-definite" is a false assertion. Its diagonal is actually made of zeros !
@DenisSerre : Sorry its a mistake. Editing the question.
@DenisSerre : Edited. Thanks for pointing it out. Hope its good now.
|
2025-03-21T14:48:31.351064
| 2020-06-24T04:46:35 |
363984
|
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"Arun Debray",
"Branimir Ćaćić",
"Quarto Bendir",
"S. carmeli",
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"site": "mathoverflow.net",
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"url": "https://mathoverflow.net/questions/363984"
}
|
Stack Exchange
|
Chern-Weil theory in the cohomological Atiyah-Singer theorem
I am interested in the following piece of data appearing in the cohomological Atiyah-Singer theorem. My reference is "The index of elliptic operators. III" by Atiyah and Singer.
Let $D:\Gamma(E)\to\Gamma(F)$ be an elliptic operator where $E\to X$ and $F\to X$ are complex vector bundles over a closed manifold $X$. The principal symbol of $D$ is a bundle isomorphism $\pi^\ast E\to\pi^\ast F$, where $\pi$ is the restriction of the projection $T^\ast X\to X$ to the complement of the zero section.
Given a Riemannian metric on $X$, the Thom space $\operatorname{Th}_X$ of $T^\ast X\to X$ is the topological space arising as the one-point compactification of the unit ball subbundle of $T^\ast X$. Using the above bundle isomorphism (and the obvious extension across the zero section), one extends $\pi^\ast E-\pi^\ast F$ to an element of $K(\operatorname{Th}_X)$. The Chern character gives an element of cohomology on the Thom space, and the cohomological Thom isomorphism gives an element of cohomology on $X$.
Now, the de Rham theorem says that this final cohomology element can be represented by a smooth differential form.
My question is:
can such a smooth form be prescribed in differential-geometric terms from the principal symbol, the (choice of) Riemannian metric on $X$, and perhaps a choice of hermitian bundle metrics and metric-compatible connections on $E\to X$ and $F\to X$?
This seems non-obvious since the Thom space does not seem to have a natural smooth structure [edit: see the comment below - it is not even a topological manifold], and so the use of the Chern character in the above presentation doesn't seem immediately amenable to the Chern-Weil approach. But, given the context and the conclusion, it seems unnatural to be forced to reach into the theory of topological characteristic classes.
If I understand correctly, the answer to my question is "essentially yes" according to Quillen's article "Superconnections and the Chern character," which identifies such a differential form via a choice of "superconnection" on $\pi^\ast E\oplus\pi^\ast F.$ However, Quillen's article seems to be answering a more general question, and has nothing to do with, for instance, the Thom space. Can one make use of the more special situation above to give a simpler answer than Quillen's?
I asked the same question on math.stackexchange with no response https://math.stackexchange.com/questions/3721536/chern-weil-theory-in-the-cohomological-atiyah-singer-theorem
I can't comment on most of the post, but re: "the Thom space does not seem to have a natural smooth structure" -- in general it is not a smooth manifold at all. The Thom space of a trivial bundle $\underline{\mathbb C}^r\to X$ is homeomorphic to $\Sigma^{2r}X$. Suspension shifts the cohomology groups of $X$ in a way that even mod 2 Poincaré duality usually does not hold, so the Thom space cannot even be a topological manifold.
When $D$ is a Dirac-type operator (i.e., an odd symmetric first-order differential operator $D$ on a $\mathbb{Z}_2$-graded Hermitian vector bundle $E$ such that $D^2 = -g^{ij}\partial_i\partial_j + \text{lower order terms}$ for some Riemannian metric $g$), isn’t this exactly what the heat kernel proof for Atiyah–Singer gives you? Berline–Getzler–Vergne is basically the canonical reference for this, but there are many others besides.
I see... I suppose that does answer my question as I phrased it (in the case of Dirac operators), but I was hoping for a direct prescription, via curvature polynomials or such. If I understand correctly, the heat kernel proof determines the form by more transcendental methods, something like as a term in the asymptotic expansion of a solution of a differential equation?
I'm not sure that I understand you, because the heat kernel proof gives you the index of a Dirac-type operator $D$ on a $\mathbb{Z}_2$-graded Clifford module bundle $E \to (M,g)$ as $\int_M \hat{A}(M,g) \wedge \mathop{ch}(E/\mathbb{S})$, where $\hat{A}(M,g)$ is the $\hat{A}$-genus form of $(M,g)$ and $\mathop{ch}(E/S)$ is the relative Chern character of the Clifford module bundle $E$, which is explicitly computable, Chern-Weil-style, from the curvature of the Quillen superconnection corresponding to $D$ and the Riemannian curvature of $(M,g)$
Indeed, when $(M,g)$ is spin and $E = S \otimes W$ is a twisted spinor bundle, $\operatorname{ch}(E/S)$ is just the Chern character of $W$.
Maybe I had just incorrectly understood the nature of the heat kernel proof. So does the heat kernel proof offer a simplification beyond the use of Quillen superconnections, as in the last paragraph of my question?
The heat kernel proof is independent of the cohomological proof. You're really using the McKean-Singer formula $\operatorname{index}(D) = \operatorname{supertrace}(e^{-tD^2})$ (for $t > 0$) together with a very delicate analysis of the small $t$ asymptotics for the integral kernel of the heat semigroup $e^{-tD^2}$, and it's that asymptotic analysis that picks out the appropriate differential forms on $M$. If you like, it's only after the fact that you notice that the differential forms you get this way happen to represent the cohomology classes you get from the cohomological proof.
McKean–Singer is true even for pseudo-differential operators, but the asymptotic analysis of the heat semigroup $e^{-tD^2}$ makes essential use of the fact that $D^2$ is a Laplace-type operator (i.e., $D^2 = -g^{ij}\partial_i \partial_j$ up to lower order terms for $g$ the Riemannian metric on $M$). See, for instance, this MO answer.
just one more comment, you can interpret the cohomology and K-theory of the Thom space as cohomology of the cotangent bundle with proper support along the projection to the base. So you can try to represent it by differential forms that decay fast at the vertical direction hopefuly.
BranimirĆaćić At the moment I'm not so interested in the proof, I just want to properly understand the meaning of the topological terms in the formula. So what I understand now is that there's three interpretations- one via k-theory on the thom space, one via Quillen superconnections, and another as an asymptotic term in a solution of the heat equation. I like both of the latter two, but the first is still somewhat mysterious to me. So I suppose what I'm most interested in is understanding the K-theoretic viewpoint in 'smooth' terms, something like what @S.carmeli suggests
|
2025-03-21T14:48:31.351627
| 2020-06-24T05:10:33 |
363985
|
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|
Stack Exchange
|
Books/Lecture notes which contrast Risch algorithm with basic standard procedure of finding an antiderivative
I vaguely remember a book/some lecture notes which introduce integration algorithms such as Risch algorithm by first giving a list of quasi-algorithmic way of evaluating symbolic integrals. (For example, when integrating a rational trignometric function, use one of the substitutions $u=\sin \theta$, $u=\cos \theta, u=\tan \frac12\theta$ and so on; when dealing with a rational function, express it in partial fractions.) Then it goes on to develop the theory of differential algebra.
I could not find the text now. After searching for quite a few sources, I have not found any books which contrast those easy procedures with Risch algorithm.
Where could I find a book which is similar to the one I describe above?
I'm not sure if this is exactly what you're looking for, but my go-to volume for these kinds of question is Symbolic Integration I by Manuel Bronstein. Risch's original treatment is sketchy in many places, and Bronstein did a lot of work to flesh out the details and actually implement Risch's methods. From the Foreword by B. F. Caviness:
With the advent of general computer algebra systems, some kind of symbolic integration facility was implemented in most. These integration capabilities opened the eyes of many early users of symbolic mathematical computation to the amazing potential of this form of computation. But yet none of the systems had a complete implementation of the full algorithm that Risch had announced in barest outline in 1970. There were a number of reasons for this. First and foremost, no one had worked out the many aspects of the problem that Risch's announcement left incomplete.
Starting with his Ph.D. dissertation and continuing in a series of beautiful and important papers, Bronstein set out to fill in the missing components of Risch's 1970 announcement. Meanwhile working at the IBM T. J. Watson Research Center, he carried out an almost complete implementation of the integration algorithm for elementary functions. It is the most complete implementation of symbolic integration algorithms to date.
The book is a very nice blend of practical algorithms and general theory.
As the title indicates, Bronstein planned to write at least one more volume,
about the integration of algebraic functions. Sadly, he passed away before he could complete this task. But I think you'll find a lot of relevant information in this first volume.
J. H. Davenport, On the integration of algebraic functions. Lecture Notes in Computer Science, 102. Springer-Verlag, Berlin-New York, 1981.
J. H. Davenport, Integration in closed form. Computers in mathematical research (Cardiff, 1986), 119–134, Inst. Math. Appl. Conf. Ser. New Ser., 14, Oxford Sci. Publ., Oxford Univ. Press, New York, 1988.
|
2025-03-21T14:48:31.351842
| 2020-06-24T05:36:24 |
363986
|
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"Sunny",
"abx",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/363986"
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|
Stack Exchange
|
Commutative square of module of differential is cartesian?
Is it true that the following square is Cartesian? $\require{AMScd}$
\begin{CD}
R @>{d}>> \Omega^{1}_{R} \\
@VVV @VVV\\
\widehat{R} @>{ \widehat{d}}>> \Omega^{1}_{\widehat{R}}
\end{CD}
where R is any regular local algebra over rational number. Any help would be great.
The answer is no, already for the second square. Take for $R$ the ring $\mathbb{Q}[T]$ localized at $(T)$. The differential form $(1+T)^{-1}dT$ becomes exact in $\Omega ^1_{\widehat{R}}$, but is not exact in $\Omega ^1_{R}$.
Doesn't the above comment answer your question???
Yes. I just edited and removed the unnecessary part of question.
My comment answers your question (negatively) as it stands now.
I think your answer still stands correct. Because anyway you gave counter example to second square and it had nothing to do with first square.
|
2025-03-21T14:48:31.351932
| 2020-06-24T06:33:43 |
363989
|
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"Ben McKay",
"Piero D'Ancona",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/7294"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363989"
}
|
Stack Exchange
|
Rellich-Kondrachov variant for compact manifold with piecewise $C^1$ boundary?
In this Wikipedia article, the Rellich-Kondrachov theorem says that whenever $M\subset\mathbb{R}^n$ is a compact manifold with $C^1$ boundary then
$W^{k,p}(M)$ embeds compactly in $W^{\ell,q}(M)$ if $k>\ell$ and $k-n/p>\ell-n/q$. Does a similar result follow for manifolds with piecewise $C^1$ boundary such as a rectangle?
For a rectangle, or a domain in Euclidean space with piecewise smooth boundary, clearly you can quote theorems from Adams book, which is probably still the standard reference. But that doesn't quite answer your question.
Bounded Lipschitz domains should be fine since you have a bounded universal extension operator in that case
I quote the following remark from "Aubin, Nonlinear Analysis on Manifolds. Monge-Ampère Equations":
2.35 Remark. We have given only the main results concerning the theorems of Sobolev and Kondrakov. These theorems are proved for the compact manifolds with Lipschitzian boundary in Aubin [17]..
[17] Aubin T.-Espaces de Sobolev sur les varietes Riemanniennes. Bull. Sc. Math. 100,
(1976) 149-173.
|
2025-03-21T14:48:31.352034
| 2020-06-24T06:59:12 |
363993
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363993"
}
|
Stack Exchange
|
Minimal and maximal degrees in critical graphs
Let $G=(V,E)$ be a finite simple undirected graph. We say that $G$ is critical if for all $v\in V$ we have $$\chi(G\setminus\{v\}) < \chi(G).$$ By $\Delta(G)$ and $\delta(G)$ we denote the maximum and minimum degrees of $G$, respectively.
Let ${\cal C}$ be the set of all finite critical graphs $G=(V,E)$ with $V \subseteq \mathbb{N}$ and $|V|>1$. Is the set $$\big\{\frac{\Delta(G)}{\delta(G)}:G \in{\cal C}\big\}\subseteq \mathbb{Q}_{\geq 0}$$
bounded?
(Note that $G$ critical and $|V(G)|>1$ implies $\delta(G) > 0$.)
No, it is unbounded: for the wheel graph $W_n$ of even order $n$, we have $\Delta(G)=n-1$, $\delta(G)=3$, $\chi(G)=4$, and $\chi(G\setminus\{v\})=3$ for any vertex $v$.
|
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