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2025-03-21T14:48:30.381617
| 2020-04-20T17:56:32 |
358057
|
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|
Stack Exchange
|
For what sets $X$ do there exist a pair of functions from $X$ to $X$ with the identity being the only function that commutes with both?
It is not too difficult to show that if $X$ is an infinite set, then there exists a two-element subset of the group $\operatorname{Sym}(X)$ with trivial centralizer iff $\lvert X\rvert \leq \lvert\mathbb{R}\rvert$.
My question is if this is true if we replace $\operatorname{Sym}(X)$ with $\operatorname{End}(X)$.
I.e., for what infinite sets $X$ do there exist functions $f,g: X \rightarrow X$, such that if $h:X \rightarrow X $ satisfies $fh = hf$ and $gh = hg$, then $h = I$? The same argument from the $\operatorname{Sym}(X)$ case shows that it is true when $|X| \leq \mathbb{R}$ (and was given as a problem in the 6th Romanian Masters of Mathematics competition). But is it false for $|X| > |\mathbb{R}|$?
I think $h = I$ is too much to ask for in general, as you can always set $g = f^{\circ k}$ for some $k\in\mathbb{N}$. Then by re-associating you have that any $h = f^{\circ n}$ works, but may not be the identity.
@Mark OP is right that there exists a pair in $\mathrm{Sym}(c)$ whose centralizer is trivial. If the idea you have in mind contradicts this, it means it does not work.
Don't use Sym$(X)$ Sym$(X)$ for upright text in math mode. $\operatorname{Sym}(X)$ $\operatorname{Sym}(X)$ is best. I have edited accordingly.
It's very interesting; I'm curious what your motivation is. One is that the sentence, in a group "there exists a pair with trivial centralizer" can be written as a first-order formula, and for the symmetric groups $\mathrm{Sym}(X)$, it indeed characterizes $|X|\le c$ (while for $|X|>c$ every countable subset has centralizer of cardinal $|X|$).
The answer is no: for every set $X$ there exists a pair in the monoid $X^X$ of self-maps of $X$, with centralizer reduced to $\{\mathrm{id}\}$.
(I first left my original "groupwise" answer because it's easier and because it has other follow-up questions. It's now deleted and copied as an answer to another question).
For $X$ empty take $(\mathrm{id},\mathrm{id})$. For $X$ finite nonempty, take a constant, and a cycle. So henceforth I assume that $X$ is infinite.
(a) First I use Sierpiński-Banach theorem [cf. here and here] that every countable subset (here just finite is fine) of $X^X$ is contained in the subsemigroup generated by a 2-element subset. This reduces to proving that there is a finite (actually 6-element) subset $\Sigma\subset X^X$ with trivial centralizer.
(b) Next I split $X$ as union of two subsets $Y,Z$ of the same cardinal. Let $f,g\in X^X$ have image equal to $Y$ and $Z$ respectively. If $u$ commutes to $f$, then $u$ stabilizes $\mathrm{Im}(Y)$, and similarly with $g$, $Z$. I'll therefore assume $f,g\in\Sigma$, and hence every $u$ in the centralizer of $\Sigma$ stabilizes both $Y$ and $Z$.
(c) It was proved in [VPH] that there exists a "strongly rigid" binary relation on $Y$: a subset $R\subset Y^2$ (actually, $R$ being subset of a well-ordering) such that the only endomorphism $u$ of $(Y,R)$ is the identity. (Here endomorphism means that $u\times u:Y^2\to Y^2$ maps $R$ into itself.) Clearly the cardinal of $R$ is that of $|Y|=|X|$.
Choose a partition $Z=Z'\sqcup Z''$ of $Z$ in subsets of the same cardinal.
Choose a bijection $i$ from $R$ to $Z'$. Define self-maps $p,q$ of $X$ as follows. On $Y$, $p$ and $q$ are chosen as injective maps into $Z''$. Also $p$ and $q$ are defined on $Z'$ by: for $(y,y')\in Y^2$ and $z=i(y,y')$, $q(z)=p(y)$ and $p(z)=q(y')$. Finally, extend $p,q$ arbitrarily choosing maps $Z''\to Y$.
Then, for $(y,y')\in Y^2$, we have $(y,y')\in R$ if and only if there exists $z_1,z,z_2\in Z$ such that $p(y)=z_1$, $q(z)=z_1$, $p(z)=z_2$, $q(y')=z_2$. [Intuition: this is a "$\stackrel{p}\to\stackrel{q}\leftarrow\stackrel{p}\to\stackrel{q}\leftarrow$ path" from $y$ to $y'$]
Indeed $\Rightarrow$ works by construction with $z_1=p(y)$, $z=i(y,y')$, $z_2=q(y')$. Conversely, suppose that such elements exist; write $(Y,Y')=i^{-1}(z)$, so $(Y,Y')\in R$. By definition $p(z)=q(Y')$ and $q(z)=p(Y)$. So $q(Y')=q(y')$ and $p(Y)=p(y)$. By injectivity of $p$ and $q$ on $Y$, we have $(y,y')=(Y,Y')\in R$.
As a consequence, if $u$ stabilizes $Z$ and $Y$ and commutes with $p$ and $q$, then $u$ preserves $R$ on $Y$.
Next we define similarly $p',q'$ from a strongly rigid binary relation on $Z$.
Then the above proves that the centralizer of $\{f,g,p,q,p',q'\}$ in $X^X$ is reduced to $\{\mathrm{id}\}$.
[VPH] Vopěnka, P.; Pultr, A.; Hedrlín, Z.
A rigid relation exists on any set.
Comment. Math. Univ. Carolinae 6 (1965), 149–155.
Informal outline: the hard step is the above reference (existence of a strongly rigid binary relation). Then, the 0th step is Sierpinski-Banach (which is not hard) to pass from 6 to 2. The second is quite trivial: there exists a pair such that centralizing this pair implies preserving each component of a partition into two moieties. The third step is to encode a binary relation into a pair of self-maps using such a $\stackrel{p}\to\stackrel{q}\leftarrow\stackrel{p}\to\stackrel{q}\leftarrow$ path and the "coloring" by the 2-component partition.
One related question I couldn't answer so far is whether there exists $f\in X^X$ whose centralizer is reduced to ${f^n:n\ge 0}$. Or if at the opposite, is it true that for $|X|>c$ every $f\in X^X$ has a centralizer cardinal $2^{|X|}$.
Eventually I made it a separate question: https://mathoverflow.net/questions/359660/
I used above the existence of a strongly rigid binary relation on every set (in ZFC). Let me mention that Hamkins–Palumbo proved that it is consistent with ZF that there exists a set with no rigid binary relation, that is, in which every binary relation has a nontrivial automorphism group, see this post.
I took the liberty to add references to the result by Sierpiński and Banach (which I did not know).
|
2025-03-21T14:48:30.382002
| 2020-04-20T18:59:03 |
358064
|
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|
Stack Exchange
|
Intrinsic numerical methods on Riemannian manifolds
I am interested in numerical methods for ordinary differential equations on a Riemannian manifold $M$. The general form of such an equation is $\dot x(t)=V(x(t)), x(0)=x_0 \in M$, where $V$ is a vector field in $M$. I have spent many hours checking the literature and I cannot find any intrinsic general method. I have found many good methods for submanifolds of $\mathbb{R}^n$, e.g. https://www.unige.ch/~hairer/poly-sde-mani.pdf (projected methods, coordinate methods) and many intrinsic methods for specific manifold structures, for instance Lie groups and homogeneous spaces, like https://hal.archives-ouvertes.fr/hal-01328729/document. Recently I found out a reasonable extension of explicit Euler method here https://onlinelibrary.wiley.com/doi/epdf/10.1002/cnm.516, i.e. $x_{k+1}=exp_{x_k}(hV(x_k))$. However there is no analysis for the order of the method, i.e. of its local error.
My question is whether there are already established general intrinsic Taylor methods on Riemannian manifolds and more specifically an intrinsic Euler scheme like above, with guaranteed rate for the local error. This is equivalent with the existence of Taylor expansions for curves on manifolds.
The local error estimate for Euler's method can have a few forms depending on how "local" you choose to make it. On a small enough scale your Riemann manifold is (for all intents and purposes), just Euclidean space. So there is a reasonable formulation of a local error estimate that is functionally identical to the Euclidean one. If you want something more like a Gronwall inequality though you would need to do a bit more work.
Seems related: https://people.ee.ethz.ch/~bsaverio/papers/Allerton2016.pdf
One small comment : the problem $\dot x=V(x)$ is independant of the Riemannian structure. An auxiliary Riemannian metric can be used to devise a geometric numerical scheme as in the example you gave, but the Riemannian metric does not play any role in the original problem.
Ryan: I want to measure the local error intrinsically, e.g. if $X$ is the solution of the ODE and $x_{k+1}=\Phi(x_k)$ a discretization of $X$ with step-size $h$, I care to estimate $dist(X((k+1)h),\Phi(X(kh))$. user35593: This looks like a continuous-time dynamical system without guarantees for the discretized version, it should be named projected gradient flow. Thomas: I assume a fixed Riemannian metrix from the beginning, so practically I would like to have an estimation for any Riemannian metric.
|
2025-03-21T14:48:30.382188
| 2020-04-20T19:29:47 |
358069
|
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"Pavel Čoupek",
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|
Stack Exchange
|
Locally isomorphic algebras over a Dedekind domain
Let $R$ be a Dedekind domain. Let $A$ and $B$ be two finitely generated domains over $R$. Assume that for every maximal ideal $\mathfrak{p}\subset R$ the $R_{\mathfrak{p}}$-algebras $A_{\mathfrak{p}}$ and $B_{\mathfrak{p}}$ are isomorphic. Are $A$ and $B$ isomorphic?
Does subscript $\mathfrak{p}$ stand for localization or completion on the adic topology?
@kneidell localization.
Counterexample. Let $R$ be a Dedekind domain with $\operatorname{Cl}(R) \neq 0$. Let $I \subseteq R$ be an ideal that is not principal (in algebraic geometry language, let $\mathscr L$ be a nontrivial line bundle), and let $J = R$ be the trivial ideal (let $\mathcal O$ be the trivial line bundle). Then $I_{\mathfrak p} \cong R_\mathfrak p \cong J_\mathfrak p$ for every $\mathfrak p$.
Then $A = \operatorname{Sym}^*(I^\vee)$ and $B = R[t] = \operatorname{Sym}^*(J^\vee)$ are not isomorphic. Geometrically, this is saying that the geometric vector bundles $\mathbf V(\mathscr L) = \operatorname{Spec}(\operatorname{Sym}^* \mathscr L^\vee)$ and $\mathbf V(\mathcal O) = \mathbf A^1_R$ are not isomorphic as $R$-schemes. For example, $\mathbf V(\mathcal O)$ has a pair of disjoint sections $0, 1 \colon \operatorname{Spec} R \rightrightarrows \mathbf V(\mathcal O)$, but a nontrivial line bundle $\mathscr L$ does not have two disjoint sections (the difference gives a nowhere vanishing section, which is an isomorphism $\mathcal O \stackrel\sim\to \mathscr L$). But they are locally isomorphic at every prime since every vector bundle over a local ring is trivial.
(In principle you could unwind this argument algebraically if you want: $B$ has a surjection to $R \times R$ by $f(t) \mapsto (f(0),f(1))$, but $A$ does not admit such a surjection. Basic geometric operations like 'take the difference of two sections' and 'a nowhere vanishing section is an isomorphism $\mathcal O \stackrel\sim\to \mathscr L$' become Hopf algebra stuff, so you have to do some work.)
Remark. If you want an example over a PID, just take any PID $R$ that has a finite extension $R \subseteq R'$ of Dedekind domains such that $\operatorname{Cl}(R') \neq 0$ (such an extension exists in many cases, e.g. if $R = \mathbf Z$ or $R = k[t]$ for any field $k$). Take $I, J \subseteq R'$ and $R' \to A$ and $R' \to B$ as above. They are still isomorphic around any $\mathfrak p \subseteq R$, because a line bundle on $\operatorname{Spec} R'$ is trivialised on the finite set of primes above $\mathfrak p$.
Given an isomorphism $\phi \colon A \stackrel\sim\to B$ of $R$-algebras, the integral closure $R'$ of $R$ in $A$ and $B$ is preserved by $\phi$, hence up to composing with an $R$-automorphism of $R'$ we may assume $\phi$ is an $R'$-algebra isomorphism, which is impossible by the argument above.
are there counterexamples if the class group is trivial?
One thing you can do in many cases is take an extension $R \subseteq R'$ of Dedekind domains with $\operatorname{Cl}(R') \neq 0$ and proceed as in my answer. Then you have to convince yourself that an isomorphism of $R$-algebras can be upgraded to an isomorphism of $R'$-algebras, which should be fine for example if $R = k[t]$ and $R'$ is an affine part of an elliptic curve (use that maps $\mathbf A^1 \to E$ are constant).
There might be some subtlety in the "for example because..." part: The left hand sides "should be" just sets, not $R$-modules, so to show that they are not "equal", i.e. bijective, one should not invoke the module structure. On the geometric side, I believe that the same can be formulated as: we would like to show that $\mathbf{V}(\mathscr{L})$ and $\mathbf{V}(\mathcal{O})$ are not isomorphic as actual $R$-schemes, while the argument shows that they are not isomorphic as (geometric) line bundles.
@PavelČoupek: ok, you're right. The trivial line bundle has two disjoint sections $B \rightrightarrows R$, but no nontrivial line bundle does.
|
2025-03-21T14:48:30.382439
| 2020-04-20T20:46:38 |
358073
|
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|
Stack Exchange
|
Can a torsion-free group be quasi-isometric to a torsion group?
I have looked around in the literature on group theory and geometric group theory and this looks to be an open question as far as I can tell (by torsion group, I mean as usual a group in which every element has finite order).
I was wondering if anyone has recently made any progress on this question or if there is some review article which looks at the possibilities?
The trivial group is both torsion-free and torsion. But apart from this exception I think it's an open question. Already, one can ask if being torsion is a QI-invariant.
If torsion is in fact an invariant under quasi-isometries, wouldn't that give a negative answer to my question?
Yes of course. But on the other hand, proving that my question has a negative answer might be easier. Of course which question is harder depends on what's true!
One can also ask if every finitely generated group is quasi-isometric to a torsion-free group.
@MoisheKohan The answer is no. Indeed it follows from Eskin-Fisher-Whyte that every group QI to a lamplighter group (finite)$\wr\mathbf{Z}$ has an infinite locally finite normal subgroup. I also guess that $\mathrm{SL}_d(\mathbf{F}_p[t])$ is a counterexample for $d\ge 3$ but I doubt it's known.
@YCor: Good point!
This is one of many open questions in geometric group theory related to quasi-isometries. Proving things about invariance under quasi-isometries is generically quite tricky, as quasi-isometries do not even need to be continuous. Some other open questions:
Is the Haagerup property invariant under quasi-isometries? (see comment by YCor for recent work on this one)
Is the rapid decay property invariant under quasi-isometries?
Is the property of having uniform exponential growth invariant under quasi-isometries?
Are random finitely presented groups quasi-isometry rigid?
How can fundamental groups of compact $3$-manifolds be classified up to quasi-isometry?
The first one (on Haagerup) has been settled by Carette by combining work of Whyte [on QI between generalized Baumslag-Solitar groups) and myself–Valette (on the Haagerup property for generalized Baumslag-Solitar groups). But I'm not sure Whyte result on which it relies is confirmed. It roughly says "if 2 generalized BS groups (+ assumptions) have their associated acting groups at finite Hausdorff distance, then they are QI", so in principle it's not the hardest part of Whyte's preprint (the hardest part, not needed here, goes the reverse direction).
|
2025-03-21T14:48:30.382619
| 2020-04-20T21:08:05 |
358075
|
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"Filippo Bianchi",
"Sebastian Goette",
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|
Stack Exchange
|
Is it possible to tessellate a torus minus a disk using hyperbolic right-angled pentagons?
I am trying to construct a compact hyperbolic surface tessellated with hyperbolic right-angled polygons with $n \ge 5 $ edges. I found quite easily a way to do it for $n$ even, but the odd case seems more tricky. For example, if n=5, my idea was to take two right-angled pentagons, with edges say $a$, $b$, $c$, $d$, $e$, and glue the couples of edges $(a,a)$, $(c,c)$ and $(d,e)$ preserving the orientation, and $(e,d)$ reversing it. My guess is that I should obtain a torus without a disk, that is, with one boundary component. However, I cannot see the tessellation on the torus, which makes me wonder if my construction actually works.
Idle thought: can you construct a 'symmetric' tiling of a figure-eight (genus-2 surface) and then cut it down the middle to get the one you're after?
The problem is that I cannot see the tiling I would like to have on the torus, so I cannot see it on the figure-eight either
I am not sure if I understand your construction. Are there two corner points of each pentagon on the boundary after gluing? Then there are six right angles left to construct the interior vertices. That would be too much for one interior vertex, but not enough for two.
If you glue four regular right-angled pentagons (side-lengths $\ell$) at one corner, you get a right-angled hyperbolic octogon with alternating side-lengths ($\ell$ and $2\ell$). You can glue either along pairs of opposite long edges,
or along pairs of opposite short edges to get the figure you want.
|
2025-03-21T14:48:30.382757
| 2020-04-20T23:14:37 |
358080
|
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"Asaf Karagila",
"Monroe Eskew",
"Todd Eisworth",
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|
Stack Exchange
|
On the utility of Silver machines
This question arises out of having Devlin's Constructibility [1] in my collection of books at home during the lockdown. Chapter IX of the book deals with Silver machines, which are presented as Silver's attempt at capturing the fine structural content of $L$ using a coherent hierarchy of functions.
Searching for "Silver machine" on MathSciNet yields four results, one of which is Devlin's book and the other three which are currently inaccessible to me on the internet.
Devlin uses Silver machines to prove a global $\square$ principle in $L$, and has an exercise that asks the reader to prove the Covering Lemma(!) using them as well. It appears that this may be done in one of the four references from MathSciNet, and the title of another advertises the construction of a gap-1 morass using Silver's ideas. This shows that such "machines" should have a lot of power, but they also seem to have not attracted the attention of set-theorists.
At an abstract level, I am wondering "Why is this?" and even "What are Silver machines good for?"
For a more concrete question:
Are there significant consequences of V=L for which there is no known
"Silver machine proof"? In particular, can one establish the existence
of gap-n morasses using Silver machines?
Any references would be appreciated. Devlin's discussion mentions unpublished work of Silver, and also unpublished notes of Litman.
[1] Devlin, Keith J., Constructibility, Perspectives in Mathematical Logic. Berlin etc.: Springer-Verlag. XI, 425 p. DM 158.00; {$} 57.60 (1984). ZBL0542.03029.
This is not my bread and butter, so excuse this comment if it is off. Looking at the wikipedia page, it seems like these are just abstracting some properties of "most" $L_\alpha$'s (with $\Sigma_1$-Skolem functions in place of the the $h_i$'s?). So I wonder, is it just a reformulation of basics, like $L_\alpha$ vs. $J_\alpha$? Is there is some kind of translation between "standard" proofs and Silver machine proofs? I don't see the combinatorial mojo they bring to the table.
I know that silver has a better conductivity than copper, but copper has better antibacterial properties. In this day and age, did you consider using copper machines?
That may be the answer to the question @MonroeEskew, and that they only reformulate the salient properties of the hierarchy.
The existence of morasses using Silver machine in L is proved in the PhD thesis of Thomas Lloyd Richardson, ``Silver Machine Approach to the Constructible Universe''.
See also Silver machines and Singular cardinals and analytic games. In this last paper the author relativises the Silver machine concept to an arbitrary $L[a], a⊆ω$, and using this relativised notion he establishes a relativised version of the Jensen covering lemma.
To begin to answer your 'abstract question', "What are Silver Machines good for', one need only look at the title of Prof. Silver's unpublished manuscript
"How to eliminate the fine structure from the work of jensen"
to find the beginning of the answer.
Why does one wish to eliminate the fine structure of Jensen? For the reason given by Boris Piwinger in section 0 of his Diploma Thesis found by clicking on the link "Silver Machines" found in Prof. Golshani's answer (I am quoting from an English translation I found on the Web--on pp. 2-3 of the translation):
The idea [of fine structure theory--my comment] is to have a closer look at the passage from $L_{\alpha}$ to $L_{\alpha + 1}$ and to describe the process with some "bookkeeping device". Even today--after 25 years of development--, this method is extremely complicated and wearisome [is it really?--my question].
This is the correct context in which to frame the OP's question (at least in my opinion).
Addendum: I also find the following passage found on pg.3 of Piwinger's Thesis of interest and relevance here:
Also in the early seventies, Jack Silver found a different approach--the Silver machines. These machines reduce the considerations to calculations on sets of ordinals.
This leads me to wonder whether Silver machines can be considered as a type of Ordinal Turing Machine (OTM) and by the following theorem
A set $x$ of ordinals is ordinal computable form a finite set of ordinal parameters if and only if it is an element of the constructible universe $L$.
one can reduce the primitive recursive set functions of Jensen and the Silver machines to this seemingly irreducible form (OTM's). This would serve to eliminate (hopefully) the concern regarding the "reformulating the salient properties of the (Jensen) hierarchy" Prof. Eisworth has regarding the Silver machines (though reformulating to simplify is certainly an important gain) and perhaps provide a means of classifying the various Fine Structure Hierarches as particular classes of OTM's
|
2025-03-21T14:48:30.383116
| 2020-04-21T01:12:13 |
358086
|
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|
Stack Exchange
|
Non-parametric regression and curvature
Given a finite set of points $(x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)$ in the plane, Linear Regression tells us how to find the straight line "$y=a+bx$" best approximating the given points, in the sense that the quantity
$$
E(a, b)= \sum_{i=1}^n\big (ax_i+b-y_i\big )^2
$$
is as small as possible. However, when the given points are believed to be generated by a nonlinear phenomenon, perhaps the time series of an exponential process, one might prefer to replace "$y=a+bx$" with some other class of functions, often one that is parametrized by a small number of parameters, in which case one is often interested in finding the values of such parameters that minimize some sensible error estimate replacing our $E(a, b)$ above.
On the other hand, according to the Wikipedia, Nonparametric Statistics is the branch of Statistics that is not based solely on parametrized models, although the term non-parametric is not meant to imply that such models completely lack parameters but that the number and nature of the parameters are flexible and not fixed in advance.
I believe that one of the reasons for the above disclaimer is that, should one adopt a completely non-parametric approach for fitting a function to a given set of points $(x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)$, as above, there will be too much freedom and hence the problem will become too easy (a piecewise linear function joining the points being a trivial solution), while I think it is safe to assume that such a solution will likely shed no light on the phenomenon under study.
In order to avoid such trivialities one must therefore either choose a model (parametrized family of functions) beforehand or else impose extra conditions on the fitted function. One possible approach is to require that the fitted function should not wiggle too much and, since the curvature of the graph of a function is related to its second derivative, a possible measure of wiggleness could be taken as
$$
W(f) = \int_a^bf''(x)^2\, d(x).
$$ Note that, if $W(f)=0$, then $f$ is necessarily a straight line, which certainly does not wiggle at all.
Question: Given a finite set of points
$(x_1, y_1), (x_2, y_2),..., (x_n, y_n)$ in the plane, such that the $x_i$ all lie in the
interval $[a, b]$, does there exist a twice differentiable function
$f$ defined on $[a, b]$, such that the quantity $$L(f) = \sum_{i=1}^n\big (f(x_i)-y_i\big )^2 + \int_a^bf''(x)^2\, d(x)$$
is minimum among all such functions? In other words, does the
functional $L$ defined above attain a minimum on $C^2([a,b])$?
Yes, that is the cubic smoothing spline with $\lambda$ (multiplier on the integral, which controls the amount of smoothing) ) = 1. See https://en.wikipedia.org/wiki/Smoothing_spline#Cubic_spline_definition .
Thanks a lot Mark! I came up with an argument based on seeing L as a Lagrangean and immitating the deduction of the Euler Lagrange equation, and which implied that the solution is a cubic polynomial throughout. However this was based on the mistake of inadvertently assuming the solution to be of class $C^4$. After reading your answer everything is now clear!
I may start answering by pointing out that the term "nonparametrics statistics" is essentially "parametric". The existing methods (e.g. Smoothing splines) in nonparametrics, are somehow all parametrized by some (finite dimensional) set of parameters.
The term "flexible" is true. However, from a applied perspective, you need to conduct a model selection to choose a fixed parameter space to do statistical inference. Alternatively, if you go for a Bayesian nonparametric modeling, instead of model selection, people usually will do model averaging (e.g. RJ-MCMC used for Bayesian modeling)
...there will be too much freedom and hence the problem will become too easy (a piecewise linear function joining the points being a trivial solution), while I think it is safe to assume that such a solution will likely shed no light on the phenomenon under study...
Again, let's consider the smoothing splines. The main restriction we attempt to impose is "knot conditions" that lead to certain order of smoothness at certain sub-domains. Once you put these restraints and choose the splines as basis, the linearity arise from the space of these basis functions. Therefore, "parametric linear" statistical inference like ANOVA can again be conducted. The answer to your question can also be found in the same book (or following wikipdeia's answer here).
In a more general sense, you can do some probabilistic inference without assuming linearity, for example, in a Banach space (e.g. Probability in Banach Space). But in that case, most object you derive will not have specific forms like splines.
Dear Henry and Mark, thanks very for your answers. It is great to know that what I was trying to do is already in a well developped state! As a humble contribution, here is a video I just made showing smoothing splines under a decreasing wiggleness penalization https://vimeo.com/410670012
@Ruy Glad it helps, there’re some more of my answers in mathematical statistics, please see if any of them helps.
|
2025-03-21T14:48:30.383572
| 2020-04-21T01:48:34 |
358090
|
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|
Stack Exchange
|
Splitting principle in algebraic geometry and ample line bundles
Splitting theorem in algebraic geometry claims that if we have a vector bundle $V$ on $X$ (we consider a smooth projective variety for this question), if we pull-back $V$ to $\mathbb{P}(V)$, we get a sub-line bundle of the pulled back vector bundle. Repeating this pull-back procedure we will have a filtration of sub-bundles $V_0\subseteq V_1 \subseteq \cdots \subseteq V_n=V $ such that $L_i=V_{i+1}/V_{i}$, is a line bundle. Is there a way that we do a splitting similar to this but with some more control on $L_i$'s? I'm particularly interested whether it is possible to find $L_i$'s such that for any two $L_i$ and $L_j$, if $L_i\ncong L_j$, then either one of $L_i\otimes L_j^{-1}$ or $L_j\otimes L_i^{-1}$ is an ample line bundle. (We can replace ampleness with weaker conditions like big and nef and ask the similar question.)
Edit: In this paper, end of lemma 7.6, it is claimed that the regular splitting procedure for a point with a trivial vector bundle on it, satisfy the property above. More precisely, if we pull-back it to $\text{Flag}(V)$, the $L_i$'s that we get satisfy the property, $L_i\otimes L_j^{-1}$ is ample for $i>j$. Although I do not understand the given reference at this point, I wonder whether just the regular splitting in general does satisfy this property or not?
Are you asking if there is a variety $Y$ that can replace $\mathbb{P}(V)$ in the statement of the splitting principle (which has the additional property that there is a way to filter pullbacks so that your condition on the quotients holds) or are you asking if there is a different short exact sequence on $\mathbb{P}(V)$ (i.e.\ not the Euler exact sequence) that gives you such a filtration with the given property?
The variety $Y$, doesn't need to be the same $\mathbb{P}(V)$. I want some pullback that is injective on rational algebraic $K$-theory, which for $\mathbb{P}(V)$ it is. Some of these types of pullbacks include finite etale covers and Frobenius (in the positive char case).
Another example of good morphism could be a blow-up morphism. But everything in blow-up needs to be smooth. The variety $X$, the closed sub-variety $Z$, the blow-up along $Z$, and the exceptional divisor.
|
2025-03-21T14:48:30.383756
| 2020-04-21T03:15:32 |
358095
|
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|
Stack Exchange
|
Infimum of weakly dependent Gaussian process?
Consider some collection of weakly dependent Gaussians $\{w_i\}$ with a uniform bound of $r$ on the magnitude of their covariances. Are there any bounds or techniques towards:
$$E[\inf_i|w_i|] \le f(r)$$
for some function $f$?
Alternatively--this would be even better--are there any comparison theorems such as:
if for two Gaussian processes $\{w_i\}$ and $\{z_i\}$, if $E[|w_i - w_j|^2] \le E[|z_i-z_j|^2]$, then we can say something about $E[\inf_i|w_i|]$ relative to $E[\inf_i |z_i|]$?
(Of course, if the inf is replaced with a sup, we have Sudakov-Fernique. But, replacing the sup with an inf makes this object seem quite different; am I being silly?)
I haven't come across this type of problem even though I have been very interested lately in Gaussian processes and read a few books and references. Still, I am not an expert.
I propose a bound in an overly simplified setting that is nearly sharp at its endpoints. I also started studying the possibility to make a comparison lemma. I'll keep you posted if I find something in this direction.
To simplify things a bit, first assume that $w$ is a centred Gaussian vector with $\operatorname{Var}(w_i)=1$ for all $i$ and $\mathbb{E}(w_iw_j)=r>0$ for all $i$ different from $j$.Say that this vector has cardinality $n$. $Z$ denotes a one-dimensional standard Gaussian independent of all other r.v.'s appearing. I will also assume that all the elements I use (probability spaces, etc.) were well-defined.
First, using Berman's lemma, there exists an isotropic Gaussian vector $\tilde w$ such that
$$
\mathbb{E}(\min_{i\le n} \vert w_i\vert )= \mathbb{E}\left(\min_{i\le n} \vert (1-r)^{1/2} \tilde w_i + r^{1/2}Z\vert \right).
$$
Using triangular inequality,
\begin{align*}
\mathbb{E}(\min_{i\le n} \vert w_i\vert )&\le \mathbb{E}\left( (1-r)^{1/2} \min_{i\le n}\vert\tilde w_i \vert + \vert r^{1/2}Z\vert \right)\\
&\le (1-r)^{1/2} \int_0^\infty \mathbb{P} \left(\min_{i\le n}\vert\tilde w_i \vert>u\right)du + \left(\frac{2r}{\pi}\right)^{1/2} \\
&\le (1-r)^{1/2} \int_0^\infty \left[\mathbb{P} (\vert\tilde w_i \vert>u)\right]^ndu + \left(\frac{2r}{\pi}\right)^{1/2} \\
&\le (1-r)^{1/2} \int_0^\infty \frac{1}{u^n} \exp\left(-\frac{1}{2}nu^2\right) du + \left(\frac{2r}{\pi}\right)^{1/2}\\
&\le (1-r)^{1/2} \left(\frac{2}{\pi}\right)^{n/2} 2^{-n/2-1/2} n^{(n-1)/2} \ \Gamma(1/2 -n/2) + \left(\frac{2r}{\pi}\right)^{1/2},
\end{align*}
where Mill's inequality was used in the fourth line.
I believe that the general case will be extremely more complicated.
May I ask in what context this question came up ?
Edit: I tried to develop a Sudakov-Fernique type of theorem.
The idea is to rely on the smooth min function, construct an interpolation between the two Gaussians you want to compare and study the derivative.
Formally, I will set the function $$F_\beta(x):= -\beta^{-1} \log\left(\sum_{i=1}^n \exp(-\beta \vert X_i \vert)\right).$$
Then, given two centred multivariate normal random vectors X and Y with constant variance equal to 1, you can construct
$$
Z_t=\sqrt{1-t}X +\sqrt{t} Y
$$
and set $\phi(t)= \mathbb{E}(F_\beta(Z_t))$.
To be able to compare the two variables, one should find the sign derivative of $\phi(t)$. We further have
$$
\phi'(t)= \mathbb{E}\left(\sum_{i=1}^n \frac{\partial F_\beta(Z_t) }{\partial x_i} \left(\frac{Y_i}{2\sqrt{t}}-\frac{X_i}{2\sqrt{1-t}}\right)\right).
$$
And we can use the property that for $f$ sufficiently smooth
$$
\mathbb{E}\left(X_i f(X)\right) = \sum_{j=1}^n \mathbb{E}\left(X_i X_j\right) \mathbb{E}\left(\frac{\partial F_\beta(X) }{\partial x_i} \right).
$$
Then, if am not mistaken after some quite tedious calculations, it holds
$$
\phi'(t)= \frac{1}{2} \sum_{1\le i,j\le n} \mathbb{E}\left( \frac{\exp(-\beta(\vert Z_{t,i}\vert+\vert Z_{t,j}\vert)}{(\sum_l\exp(-\beta(\vert Z_{t,l}\vert ))²} sign(Z_{t,i})\times sign(Z_{t,j}) (\sigma_{i,j}^Y -\sigma_{i,j}^X) \right),
$$
where $\sigma_{i,j}^Y$ is the covariance between elements $i$ and $j$ in the r.v. $Y$.
Should this last quantity always have the same sign, you would be able to prove an inequality. I am still stuck at this stage. Maybe some reader will figure a way out. The fact that we use the smooth min is not so important as the limit in $\beta$ will finally be taken.
Thank you for the answer! I find this Berman lemma very interesting but am having trouble finding it online. Do you have a reference for this?
(I was thinking about this question in the context of matrix discrepancy)
It appears on p. 134 in Extremes and related properties of random sequences and processes by Leadbetter et al. I remember seeing it in Kevin Tanguy's thesis. You can also derive it directly by noticing that $\tilde w$ is a multivariate Gaussian with identity covariance matrix while $Z$ could be rewritten as an independent multivariate Gaussian with all-ones covariance matrix. The claim follows using the properties of the Gaussian distribution.
I didn't see your edit with the comparison principle until just now. This is a good start; I will try this in the next few days for my application and will comment again if I find anything nice. Thanks!
There could even be more to it. Through adapting the proof, I noticed that the original proof proposed by Chatterjee could in fact be rewritten as a function of a random Laplacian (I wrote it up as a short note on my homepage accessible from my MO profile) and this fact is convenient to skip calculations. It may be that the structure appearing here is also tractable for what you need...
|
2025-03-21T14:48:30.384090
| 2020-04-21T03:22:11 |
358096
|
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|
Stack Exchange
|
Books to develop a unified view of statistics and information theory?
I hope to understand the connection between statistics and information theory in a deep philosophical sense.
I suppose the best place to start would be David MacKay's Information Theory, Inference, and Learning Algorithms, but I was curious what else there may be. Are there any other good books out there like this? What are your recommendations?
Having read a few of Jaynes' papers but not his book, I would still suggest looking at the latter.
What precisely, mathematically speaking, does "in a deep philosophical sense" mean here?
The booklength tutorial by Shannon award winner Imre Csiszár and Paul Shields is freely available online here:
Information Theory and Statistics: A Tutorial
I. Csiszár, Rényi Institute of Mathematics, Hungarian Academy of Sciences
There is also an article in the International Encyclopeadia of Statistical Science by E. Haroutunian which seems to be behind a paywall.
|
2025-03-21T14:48:30.384187
| 2020-04-21T04:18:57 |
358098
|
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|
Stack Exchange
|
Quantum analogue of certain property of compact groups
Let $\mathcal{A}$ be the category of $C^*$ algebras. For a group $G$ let $\tilde{G}$ be the space of all conjugacy classes of $G$.
What is a precise description of a maximal ,or in some sense widest as possible, category of compact topological groups $\mathcal{TG}$ for which the following functor is well defined:
$$\mathcal{F}:\mathcal{FG}\to \mathcal{A} \\ \mathcal{F}(G)=C(\tilde{G}) $$ where $C(\tilde{G})$ is the algebra of all continuous functions on Haussdorf space $\tilde{G}$ where the latter is equiped with the quotient topology imposed by the natural quotient map $G\to \tilde{G}$
Now how can this functor be naturally and maximally extended to an appropriate category of quantum groups?
|
2025-03-21T14:48:30.384265
| 2020-04-21T07:11:12 |
358102
|
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Stack Exchange
|
Counting adjacency matrices
Here is a question that has come up in the context of a problem that involves counting partially ordered sets.
For an adjacency matrix $A$, let $p$ be the sum of elements in the strict upper triangle (upper triangle minus the diagonal) of $A$, and $q$ be the sum of elements in the strict upper triangle of $A^2$. For fixed values of $p$ and $q$, is it possible to compute the cardinality of the set of all such matrices $A$?
If yes, how does one go about it? My guess is that the problem may be easier to tackle if we demand that $A$ has some extra symmetry, but I have not been able to arrive at any definite conclusion. Though I am interested in the generic case (without any added symmetries), solutions for any special cases will also be helpful. So will be any references that deal with similar problems.
EDIT (4/21/20): new link re: function inserted at end
Source: my tweets, with minor errors removed (https://twitter.com/krzhang/status/1252529588049072128)
Let's assume $A$ is symmetric and 0 on the diagonal. (disclaimer: I'm guessing this is not what you meant by "symmetry" because of poset context, but it may still be helpful) This means we are really working with unlooped undirected graphs, where
- p is how many edges the graph has, and
- q is now many 2-paths (disregarding order) that are not loops.
Now, cool observation: 2-paths that are not loops can be identified with their middle point and 2 neighbors. So for each vertex $i$ of deg. $d_i$, it contributes $d_i(d_i-1)/2$ 2-paths that are not loops.
So our problem becomes: "How many ways are there to split $p$ into nonnegative integers $d_1 + ... + d_n,$ such that $\sum d_i(d_i - 1)/2 = q$?"
Some manipulation gives $\sum d_i^2 = 2q + p$, so this problem really reduces to
"Given the first and 2nd power sums of $d_1 ... d_n$, how many sets of nonnegative $d_n$ are there?" or the quite-beautiful probabilistic form:
"How many nonnegative integral distributions are there of a fixed mean and variance?"
There're number theory constraints here, so I guess this is hard (which means original problem is even harder). However, computationally this isn't bad. Here's a solution:
construct a 3-d infinite array so that $P[x][y][z]$ = "the number of ways to solve this with $x$ numbers such that their 1-power sum (sum) is $y$ and their 2-power sum is $z$
use dynamic programming to build this layer by layer by $x$. So compute everything with $x = 1$ first, then reduce each problem with $x+1$ to those with $x$ by summing over different values for the first element.
This gives an $O(n^2p(2q+p))$ algorithm.
Link (h/t Boris Alexeev): This last function is explored at https://mathworld.wolfram.com/SumofSquaresFunction.html as "sum of squares function." As I predicted, it seems number-theoretical and thus needs to be written as sums of modular functions for larger $n$ and $k$. Therefore, finding a closed form seems super hard.
Do you know of any references where the nature of the cardinality in the asymptotic limit of $n$ has been studied?
|
2025-03-21T14:48:30.384491
| 2020-04-21T07:19:26 |
358103
|
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|
Stack Exchange
|
What kind of object are the solutions of the Knizhnik-Zamolodchikov Equations
I am reading about the KZ equations in Kassel's Quantum groups. In definition XIX.3.1 (page 455) he defines the differential system $(KZ_n)$ as
$$
dw = \frac{h}{2\pi\sqrt{-1}} \sum_{1 \leq i <j \leq n} \frac{t_{ij}}{z_i-z_j}(dz_i- dz_j)w $$
where the $t_{ij}$ are some element of the universal enveloping algebra $U(\mathfrak{g})$ for some Lie algebras $\mathfrak{g}$. In this definition "$w = w(z_1, \dots, z_n)$ is a function on $Y_n$ with values in $W^{\otimes n}$" where $Y_n = \{(z_1, \dots, z_n) \vert i \neq j \Rightarrow z_i \neq z_j\}$ and $V$ is a finite dimensional $\mathfrak{g}$-module.
At this point the solutions are then simply foncitons on some complex parameters taking value in the $V^{\otimes n}$. However, when doing the proof of the Drinfled-Kohno theorem , Kassel makes some change of coordinates for the case $n=3$ and we have the following equation (equation 7.3, page 469 in Kassel's book):
$$ w(z_1,z_2,z_3) = (z_3 - z_1)^{\hbar(t_{12} + t_{23} + t_{13})}G(z)$$
where $z = \frac{z_2-z_1}{z_3-z_1}$ and $G(z)$ is a formal serie in $t_{12}$ and $t_{23}$ with coefficients being analytic functions in $z$. This seems to indicate that the solutions are actually taking value in in $U(\mathfrak{g})^{\otimes 3} [[h]]$ and this is where my confusion comes from. Especially since it does not seem to be coming from the case where $V = U(\mathfrak{g})$ for this is not finite dimensional.
Maybe the explanation for this is in the paragraph at the end of page 457 but I don't really understand what he is doing here neither.
There are really 3 levels at which you can define this equation:
for functions
$$G:Y_n \longrightarrow V^{\otimes n}$$ where $V$ is a f.d. module and $h$ is a complex number. You can then take a formal expansion around $h=0$ and regard $G$ as being valued in $V^{\otimes n}[[h]]$.
for functions
$$G:Y_n \longrightarrow U(\mathfrak g)^{\otimes n}[[h]].$$
This only makes sense formally (by which I mean that you need formal power series in $h$).
for $n=3$ and up to some change of variable as you say, you get a version where $G$ takes values in the algebra of formal power series in two non commuting variables $A,B$. This is the one appearing in section 6 of Kassel's book. More generally, for arbitrary $n$ you have a version of the KZ equation taking values in the algebra of so-called horizontal chords diagrams.
The general theory of this kind of equation really has been done only in case 1, so in the other 2 cases, in the literature, existence and uniqueness of solutions are proved in a somewhat ad hoc way but this isn't hard.
The relation between 3 and 2 just boils down to the substitution $A\mapsto h t_{1,2}, B\mapsto h t_{2,3}$. So in equation 7.3 that you mention in your message, $G$ does indeed takes value in $U(\mathfrak g)^{\otimes 3}[[h]]$ (so it's a formal power series in $h$, not in $t_{1,2}$ and $t_{2,3}$).
The relation between 2 and 1 at the level of equations is that the former specializes to the latter through the algebra map
$$U(\mathfrak g)^{\otimes n}[[h]] \rightarrow End(V^{\otimes n}[[h]])$$
coming from the action of $\mathfrak g$ on $V$.
In terms of solutions, if $z_0 \in Y_n$ then there is a unique solution $G$ of $2$ defined in a neighborhood of $z_0$ such that $G(z_0)=1$, the unit of $U(\mathfrak g)$. Now if $v \in V^{\otimes n}$, then the function $$\tilde G:z\longmapsto G(z)\cdot v$$
is by construction the unique solution of equation 1 satisfying $\tilde G(z_0)=v$.
Thank you for your answer, it certainly helps a lot. I am just wondering what you mean exactly by "You can then take a formal expansion around h=0 ". This seems to be closely related to the paragraph at the of page 457 in section 3 of Kassel's book.
If I understand correctly, you say that what Kassel is doing in his book is that you can understand equation 7.3 with $G$ taking values in $U(\mathfrak{g})^{\otimes n}[[h]]$ from which one can build a solution of this same equation this time taking value in $V^{\otimes n} [[h]]$, which leads to a solution of $(KZ)$.
That's right, you get a function taking values in $U(g)^{\otimes n}[[h]]$, which can further be specialized to a solution in $V^{\otimes n}[[h]]$ for any choice of $v \in V$. Now it just so happen that if $V$ is finite dimensional this is actually analytic in $h$, ie it converges to an actual (as opposed to formal) solution in $V^{\otimes n}$, but this doesn't really play a role in this story (it does play a role if you want the stronger version of the KD theorem where $h$ is specialized to a generic complex number).
In fact you don't need all the $V$'s to be the same. At the end of the day, the bottom line is you get the KZ associator as an element in $U(g)^{\otimes 3}[[h]]$ the action of which describes the holonomy from 0 to 1 of solutions valued in $(V \otimes W \otimes Z)[[h]]$ for any f.d. modules $V,W,Z$, and this is natural w.r.t $g$-morphisms.
Can we then understand $w(z_1,z_2,z_3) = (z_3 - z_1)^{\hbar(t_{12} + t_{23} + t_{13})}\tilde{G}(z)$ to be a solution of $(KZ_3)$? I ask this because we want to link this to the monodromy of the $KZ$ system.
Well, yes, it is a solution of the (formal) KZ equation valued in whatever $G$ is valued in. Nothe monodromy can also be described at either of those level, that was the point of my previous comment somehow.
Just to be clear: all of those are rightfully called "a solution of the KZ equation". What does depends on the context is what "the KZ equation" means, sorry if this led to some confusion. People don't always clearly distinguishes between 2 and 3, since there isn't much lost in working formally and imply that whatever result you care about also holds for the specialized version whenever it makes sense.
I think my confusion comes more from a question which is implicit in my post. At some point we want to show that the representation of $B_n$ given by the monodromy of $dw = \frac{h}{2\pi\sqrt{-1}} \sum_{1 \leq i <j \leq n} \frac{t_{ij}}{z_i-z_j}(dz_i- dz_j)w$ is equal to the one coming from the top. quasi. bialgebra $U(\mathfrak{g})[[h]]$ with associator $\phi_{KZ}$ and R-matrix $e^{ht/2}$. This is why I feel like I need some kind of concrete solution of $dw = \frac{h}{2\pi\sqrt{-1}} \sum_{1 \leq i <j \leq n} \frac{t_{ij}}{z_i-z_j}(dz_i- dz_j)w$ to check that those two are somewhat related.
I'm not sure what you mean by "concrete" here, but you do have particular solutions whose monodromy can be expressed using $\Phi$, that's of course the whole point, namely those taking the value 1 in some appropriate "asymptotic zone" in Drinfeld's language. Btw this discussion might be starting go be too long for comments, feel free to send me an email (the address of my webpage is in my profile).
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2025-03-21T14:48:30.384958
| 2020-04-21T08:20:27 |
358107
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Stack Exchange
|
Subgroup of $\mathrm{GL}_n$ stabilizing linear subspace skew-symmetric matrices
I am currently reading "Schiffer variations and the generic Torelli theorem for hypersurfaces" by Voisin, where it is claimed that the subgroup of $\mathrm{SL}_{2m}$ ($m \geq 3$) which preserves a generic subspace of $\bigwedge^2 \mathbb{C}^{2m}$ of dimension bigger than $3$ must be finite. There are no references given in the paper, where it is claimed that this fact is easily checked. Unfortunately (for me), I am not able to prove it.
Here "preserves" means that $g\cdot w\cdot{\vphantom g^tg} \in W$, for any $w \in W$. Is this fact indeed well-known? I am looking for a reference of this fact, or a quick proof.
Presumably easier to check at the Lie algebra level, in any case.
@paulgarrett : probably, but that means finding a "generic" $A \in W \otimes \bigwedge^2 \mathbb{C}^n$ and computing the isotropy Lie sub-algebra of $\mathfrak{sl}{W} \oplus \mathfrak{gl}{2m}$. I am not sure how this can be easily checked.
What does "generic" mean? The condition is very surprising to me, because it seems easier to preserve a bigger subspace. For example, the preserver of $\bigwedge^2\mathbb C^{2m}$ is all of $\operatorname{GL}_{2m}$ …. Could you point to the precise statement?
@LSpice It is assumed in the paper that the codimension of this subspace is more than $2(2m-2).$ (By duality really the only relevant condition is that the codimension is $\geq 3$ (speaking of, I believe in this question it should be dimension $\geq 3$, not $>3$, no?)). Generic here means generic (in the Zariski sense) among subspaces of that fixed dimension.
@LSpice : end of page 11, proof of lemma 2.2
@dhy, it seems to me that the entire subspace is generic in the Zariski sense among subspaces of its dimension …. Are you saying that the condition of dimension bigger than 3 should be codimension bigger than 3?
@Libli, thanks. I see conditions there that look like yours, but I don't see the claim that the group is finite. I do see a lot of notation I don't recognise, so maybe you've just translated into concrete language ….
Well, I might be mistaken, but I think that the sentence is badly written and "this space is zero" refers to the stabilizer being finite. Furthermore, it is claimed at the beginning of the proof that the space of infinitesimal automorphisms of a linear section of the Pfaffian will be proven to vanish. So following her proof, I think that she deduces that from the fact that the tangent space to the stabilizer is zero.
It just occurred to me: linearizing to Lie algebras, is a highest-weight computation feasible?
@paulgarrett : As I told you in my first comment, we know the action and the linearization is obtained easily. But computing the isotropy subalgebra at a generic point is not trivial. Otherwise, why would Kimura and Sato have had witten a 156-pages long paper on prehomogeneous vector spaces?
I'm sure that I'm not the first person to point this out, but I didn't see it in the previous comments: The statement as you have made it is false. Clearly the stabilizer subgroup contains all the multiples of the identity in $\mathrm{GL}{2m}(\mathbb{C})$, so it's never of dimension $0$. Maybe you meant to write $\mathrm{SL}{2m}(\mathbb{C})$ or $\mathrm{PGL}_{2m}(\mathbb{C})$?
@RobertBryant : you're obviously right. I will edit that.
Here is the worst possible proof of all but one case, namely $m=r=3$ (but this is not included in your question as stated, though I think it is claimed in the paper.)
Let $r$ denote the dimension of a generic subspace $W$ of $\bigwedge^2\mathbb{C}^{2m}$. We assume $3\leq r\leq\frac{1}{2}\operatorname{dim}(\bigwedge^2\mathbb{C}^{2m})$ (Voisin assumes a possibly weaker upper bound, but this is OK by duality.) Our goal is to show that no nontrivial element of $\operatorname{PGL}_{2m}$ stabilizies $W$.
The first observation we make is that if $g$ stabilizes $W$, then the semisimple and unipotent parts $g_{ss},g_{un}$ also stabilize $W$ (because they are polynomials in $g$). If $g_{ss}$ is nontrivial, we can take some polynomial in it to get a semisimple element with exactly two eigenvalues. If $g_{un}$ is nontrivial, we can take some shifted power $(g_{un}-1)^k+1$ which is nontrivial and only has Jordan blocks of length $1$ and $2$. In conclusion: it suffices to show that $W$ is not stabilized by any semisimple element with two eigenvalues nor any unipotent element with Jordan blocks of length $\leq 2$. We treat these two cases separately.
In both cases, we will do an incidence correspondence argument. Namely, we look at the variety of pairs $(g,W)$, with $g$ nontrivial and semisimple w/ two eigenvalues (or unipotent w/ $(g-1)^2=0$) and show this variety has dimension less than the dimension of $\operatorname{Gr}(r,m(2m-1)).$ We do this by splitting this variety into many many (but finitely many) strata, and showing the dimension bound for each strata.
The semisimple case. Our semisimple element $g$ with two eigenvalues gives a partition $d+e=2m$, given by the dimensions of each eigenspace. The action of $g$ on $\bigwedge^2\mathbb{C}^{2m}$ decomposes it as a sum of eigenspaces $\mathbb{C}^{\binom{d}{2}}\oplus\mathbb{C}^{de}\oplus\mathbb{C}^{\binom{e}{2}}.$
A subspace $W$ is stabilized by $g$ iff it is a direct sum of subspaces in each component. Letting the dimension of these subspaces be given by $f_1,f_2,f_3$, we see that we have $f_1(\binom{d}{2}-f_1)+f_2(de-f_2)+f_3(\binom{e}{2}-f_3)$ degrees of freedom to choose $W$.
Consider the strata given by fixing d,e, and the $f_i$. We have a $2de+1$-dimensional space of choices of $g$; $de$ for a subspace of dimension $d$, another $de$ for a subspace of dimension $e$, and a final degree of freedom for the ratio between the two eigenvalues. So our desired inequality becomes
$$2de+1+f_1(\binom{d}{2}-f_1)+f_2(de-f_2)+f_3(\binom{e}{2}-f_3)<r(m(2m-1)-r).$$
Assume this inequality is false. Consider the choice of $f_i$ that maximizes the left hand side. As $f_1+f_2+f_3=r\leq\frac{1}{2}m(2m-1),$ we must have $f_1\leq\frac{1}{2}\binom{d}{2}$, $f_2\leq\frac{1}{2}de$, $f_3\leq\frac{1}{2}\binom{e}{2}$ (by the maximizing assumption). Now rewrite (the negation of) the inequality as:
$$2de+1\geq (f_2+f_3)(\binom{d}{2}-f_1)+(f_1+f_3)(de-f_2)+(f_1+f_2)(\binom{e}{2}-f_3).$$
Assume WLOG that $d\geq e$. As $f_2\leq\frac{1}{2}de$, we must have $f_1+f_3\leq 4.$ The right hand side contains $f_2(\binom{d}{2}+\binom{e}{2}-f_1-f_3)$ as a summand, and because $m\geq 3$, we have $\binom{d}{2}+\binom{e}{2}\geq\frac{2}{3}de,$ so $\binom{d}{2}+\binom{e}{2}-f_1-f_3\geq\frac{1}{3}de,$ so $f_2\leq 6$.
We see in particular that $2de+1\geq f_2(\binom{d}{2}+\binom{e}{2}-4)+(f_1+f_3)(de-6),$ and since $(f_2+f_3)+(f_1+f_3)\geq r\geq 3$, either $\binom{d}{2}+\binom{e}{2}-4$ or $de-6$ is $\leq\frac{2de+1}{3}$. If $de-6\leq\frac{2de+1}{3}$, then $de\leq 19$ and $m \leq 10$. while if $\frac{2de+1}{3}\geq\binom{d}{2}+\binom{e}{2}-4,$ then we have $\frac{2m^2+1}{3}\geq\frac{2de+1}{3}\geq\binom{d}{2}+\binom{e}{2}-4=\frac{d^2+e^2}{2}-m-4\geq m^2-m-4,$ so $m\leq 5.$
In summary, we have $m\leq 10.$ Using a computer to check all such cases (as I said, this is the worst possible proof), we see that the only possible case is $f_1=0,f_2=3,f_3=0,d=e=m=3$, but this is exactly the case I excluded at the start.
Now let do the same argument for unipotent $g$ with $(g-1)^2=0$. Let $n$ denote the number of nontrivial Jordan blocks, or equivalently, the dimension of $\operatorname{im}(g-1).$ Then the action of $g$ on $\bigwedge^2\mathbb{C}^{2m}$ has $n(2m-n-1)$ nontrivial blocks (each of length $2$).
The data of a preserved subspace $W\subseteq\bigwedge^2\mathbb{C}^{2m}$ is equivalent to the data of a subspace $W'\cong W\cap\operatorname{im}(g-1)\subseteq\operatorname{im}(g-1)$ and a subspace $W/W'\subseteq (g-1)^{-1}W'/W'.$ If the dimension of $W'$ is given by $d$, then there are $d(n(2m-n-1)-d)+(r-d)(m(2m-1)+d-n(2m-n-1)-r)$ degrees of freedom. Meanwhile, the data of a choice of $g$ is equivalent to the choice of the subspace $\operatorname{im}(g-1)$ and of a map $\mathbb{C}^{2m}/\operatorname{im}(g-1)\rightarrow \operatorname{im}(g-1)$, which gives $n(2m-n)+n(2m-n)=2n(2m-n)$ degrees of freedom. So the inequality we want is
$$d(n(2m-n-1)-d)+(r-d)(m(2m-1)+d-n(2m-n-1)-r)+2n(2m-n)< r(m(2m-1)-r).$$
We can rearrange this to
$$2d^2+(m(2m-1)-2r-2n(2m-n-1))d-2n(2m-n)+rn(2m-n-1) > 0.$$
Assume otherwise. Note that $\frac{2m-n-1}{2m-n}\geq\frac{2}{3}$, so $-2n(2m-n)+rn(2m-n-1)\geq 0,$ with equality only when $m=n=r=3$ (and so we land in the excluded case again.) So we must have $m(2m-1)-2r-2n(2m-n-1)<0$ and, by the discriminant test,
$$8(rm(2m-n-1)-2n(2m-n))\leq(m(2m-1)-2r-2n(2m-n-1))^2.$$
Note that we have $m(2m-1)-2r-2n(2m-n-1)\geq m-2r$ and $m(2m-1)-2r-2n(2m-n-1)\geq -2n(2m-n-1)$. Applying these inequalities, we see that
$$8(rm(2m-n-1)-2n(2m-n))\leq(2r-m)2n(2m-n-1)$$.
Again applying $2(2m-n)\leq 3(2m-n-1),$ we get
$$8(rm-3n)(2m-n-1)\leq(2r-m)2n(2m-n-1)$$
or
$$8(rm-3n)\leq(2r-m)2n.$$
This implies
$$8(rn-3n)\leq(2r-m)2n$$
and
$$8r-24\leq 4r-2m$$
$$4r\leq 24-2m$$
so $m\leq 6$. Again, we check all these cases by computer. All of them either have $m=r=3$ or have equality in the dimension bound. But in this case, equality suffices, because all subgroups of $G$ containing a unipotent element have positive dimension.
Are "worst possible proofs" collected in a book kept by the devil?
@dhy : thanks for the answer. I think I can follow the argument in the semi-simple case, the unipotent case is still a bit difficult for me. But at least, I understood that the general idea of decomposing into semi-simple and unipotent will provide bounds on the possible exceptions. That's a nice idea. I am just waiting a day or two to see if someone comes up with a simpler proof or a reference to a simpler proof.
I don't know whether the case $m=r=3$ is claimed in the paper, but if it is, this is an error. The $\mathrm{SL}_6(\mathbb{C})$-stabilizer of a generic 3-plane in $\Lambda^2(\mathbb{C}^6)$ has dimension $1$.
@RobertBryant : Very interesting! Do you have a proof or a reference for this?
@Libli: I have a proof, but since you omitted this case in your question, I am guessing that this may already be known. On the other hand, in dhy's answer, it was stated that the case m=r=3 was claimed in the paper. Can you tell me whether this is so? (I don't have a copy of the paper to look at myself.)
@RobertBryant : Well, the proof of the lemma which uses this result is done in a rather offhand way, to say the least. It's not clear at all what cases she needs precisely. At least, she seems to claim that it must be true for any $r \geq 3$. You can have a look at the paper here : https://arxiv.org/pdf/2004.09310.pdf page 11, proof of lemma 2.2.
@RobertBryant : still interested by your proof. And it is included in my question (I wrote bigger meaning bigger or equal).
@RobertBryant: Upon closer inspection, it appears that the paper may be assuming $m\geq 4$. In any case, the $m=3$ case is not needed for Voisin's argument.
@Libli: Now that Friedrich's Knop's answer has appeared, feel free to unaccept mine and accept his instead. Interestingly, the Andreev-Vinberg-Èlašvili argument also does this split into semi-simple and unipotent, but then they cleverly (almost magically) apply the classification of invariant inner forms on $\mathfrak{g}$... it seems plausible that my long calculations are somehow encoding their more conceptual Lie-theoretic argument.
Here is an outline of the argument that shows that the $\mathrm{SL}_6(\mathbb{C})$-stabilizer of the generic $3$-plane $W\subset\Lambda^2(\mathbb{C}^6)$ has dimension $1$, not $0$, as (apparently) claimed.
First, note that the cone $C_2\subset \Lambda^2(\mathbb{C}^6)$ consisting of the elements $b\in \Lambda^2(\mathbb{C}^6)$ that satisfy $b^3 = 0$ is a hypersurface of degree $3$ in $\Lambda^2(\mathbb{C}^6)\simeq\mathbb{C}^{15}$. It is irreducible but not smooth, as it is singular along the locus $C_1\subset C_2$ consisting of the elements that satisfy $b^2=0$, and $C_1$ itself is a smooth cone of dimension $9$ (i.e., its only singularity is the origin $b=0$).
Thus, a generic $3$-plane $W$ in $\Lambda^2(\mathbb{C}^6)$ will only meet $C_1$ at the origin and will not be tangent to $C_2$ anywhere. Thus, the intersection $W\cap C_2$ will be a smooth $2$-dimensional cone that projectivizes to a smooth cubic curve in $\mathbb{P}W\simeq \mathbb{P}^2$.
The group $G\subset \mathrm{SL}_6(\mathbb{C})$ that stabilizes $W$ must act on $\mathbb{P}W$ as symmetries of the nonsingular cubic curve and hence must act as a finite group on $\mathbb{P}W$. By passing to a subgroup $G'\subset G$ of finite index in $W$, we can assume that $G'$ acts trivially on $\mathbb{P}W$ and hence as a scalar multiple of the identity on $W$. However, if we let $b\in W$ be an element that satisfies $b^3\not=0$, then $G'$ must preserve $b^3$ and hence, it can at most multiply $b$ by a nontrivial cube root of unity. Again passing to a subgroup $G''\subset G'$ of index at most $3$, we can arrange that $G''$ acts trivially on $W$.
Let $P\subset W$ be a plane that intersects $W\cap C_2$ in three distinct lines. This means that $P$ has a basis consisting of two elements $b_1\not\in C_2$ and $b_2$ such that the polynomial $p(\lambda)$ that satisfies
$$
(b_2-\lambda b_1)^3 = p(\lambda) b_1^3
$$
has $3$ distinct roots. After a change of basis in $P$, I can assume that those roots are $1$, $2$, and $3$.
Then, by an elementary argument, there is a basis $e_1,\ldots,e_6$ of $\mathbb{C}^6$ such that
$$
b_1 = e_1\wedge e_2 + e_3\wedge e_4 + e_5\wedge e_6
\quad\text{and}\quad
b_2 = e_1\wedge e_2 + 2e_3\wedge e_4 + 3e_5\wedge e_6
$$
Moreover, the fact that $G''$ fixes $b_1$ and $b_2$ implies that it also fixes the individual monomial terms $e_1\wedge e_2$, $e_3\wedge e_4$, and $e_5\wedge e_6$.
Let $Q_i\subset\mathbb{C}^6$ be the $2$-plane spanned by $e_{2i-1},e_{2i}$. Then $G''\subset \mathrm{SL}(Q_1) \times\mathrm{SL}(Q_2)\times \mathrm{SL}(Q_3)$, and so, of course, $Q_i\simeq Q_i^*$ as $G$-modules. Now let $b_3\in W$ be linearly independent from $b_1$ and $b_2$. Using the $G''$-module decomposition
$$
\begin{aligned}
\Lambda^2(\mathbb{C}^6) &= \Lambda^2(Q_1\oplus Q_2\oplus Q_3) \\
&= \Lambda^2(Q_1)\oplus\Lambda^2(Q_2)\oplus\Lambda^2(Q_3)\oplus Q_1{\otimes}Q_2\oplus Q_2{\otimes}Q_3\oplus Q_3{\otimes}Q_1\\
&\simeq\mathbb{C}\oplus\mathbb{C}\oplus\mathbb{C}\oplus Q_1{\otimes}Q_2^*\oplus Q_2{\otimes}Q_3^*\oplus Q_3{\otimes}Q_1^*,
\end{aligned}
$$
one can decompose $b_3$ into a sum of the three basic monomials that occur in $b_1$ and $b_2$ plus a triple of linear maps $L_i:Q_{i+1}\to Q_i$. For generic $W$ with basis $b_1$, $b_2$, and $b_3$ as above, these $L_i$ will be isomorphisms, and they will have to be $G''$-module isomorphisms in order for $b_3$ to be fixed by $G''$, whose elements can be thought of as triples of elements $(g_1,g_2,g_3)$ where $g_i\in\mathrm{SL}(Q_i)$. However, the relations $L_ig_{i+1}=g_iL_i$ (necessary for $G''$ to fix $b_3$) will then determine $g_2$ and $g_3$ in terms of $g_1$ and the $L_i$. Moreover, $L = L_1L_2L_3:Q_1\to Q_1$ must commute with $g_1$. Conversely, if $g_1$ commutes with $L$ then it determines an element of $G''$. However, $L$ will always have a positive dimensional commutator in $\mathrm{SL}(Q_1)$ (generically of dimension $1$), so $G''$ always has dimension at least $1$.
Very nice indeed. By the way, should the relation $L_ig_{i+1}=g_i$ instead be $L_ig_{i+1}=g_iL_i$ (in your last paragraph)? Also, does this argument also prove triviality for $r=3$ and general $m$? It seems like it should to me, but maybe I am missing some obstruction.
For results in positive characteristic, see for example this preprint: link (this example is Proposition 3.2.17 on page 123).
@dhy: Oh, yes. Thanks for the correction. I've fixed it. Also, yes, as soon as $m\ge 4$, the stabilizer is generically finite when $r = 3$.
@MikkoKorhonen: Thanks for the reference! I would never have guessed that this example had anything to do with $\mathrm{E}_6$! That's pretty amazing.
In the following, a linear group is a (closed) subgroup $G$ of some $GL(V)$. Now the stabilizer of a generic $d$-space in $V$ is the same as the stabilizer in $G\times GL(d)$ of a generic vector of $V\otimes\mathbb C^d$. Thus, in our case we have to investigate the generic stabilizer of $GL(n)\times GL(d)$ acting on $\wedge^2\mathbb C^n\otimes\mathbb C^ d$ with $1\le d\le n(n-1)/4$ and determine when it is infinite.
Determining the generic isotropy group $H$ for a linear action of a reductive group $G$ is a classical problem of invariant theory. It was pretty much settled by the Vinberg school approx. 50 years ago. More specifically: In
Andreev, E. M.; Vinberg, È. B.; Èlašvili, A. G.: Orbits of highest dimension of semisimple linear Lie groups. (Russian) Funkcional. Anal. i Priložen. 1 1967 no. 4, 3–7
the authors derive a numerical criterion for $H$ to be infinite (the first theorem of that paper). The argument is really ingeneous and I recommend reading the paper. From their criterion they derive the classification of irreducible simple linear groups with infinite $H$. The table became later notorious because the very same table showed up in various different classification projects (see, e.g., the appendix to Mumford-Fogarty(-Kirwan)).
Probably, the numerical criterion alone is already enough to settle the finiteness of $H$ in the case at hand. But that is not necessary. In the follow-up paper
Èlašvili, A. G. Stationary subalgebras of points of general position for irreducible linear Lie groups. (Russian) Funkcional. Anal. i Priložen. 6 (1972), no. 2, 65–78
Èlašvili (almost, see below) settles the case of arbitrary irreducible linear groups. The answer for non-simple groups is given in Tables 5 and 6. The cases $\wedge^2\mathbb C^n\otimes\mathbb C^ d$ with infinite $H$ are $(n,d)=(n,1), (n,2), (4,3), (5,3)$, and $(6,3)$. So $d\ge4$ does not occur, as claimed.
There is more to say. Sifting through the tables one will notece that Robert Bryant's case $(n,d)=(6,3)$ is missing in Table 5. This case along with three more missing cases was pointed out in
Popov, A. M. Irreducible semisimple linear Lie groups with finite stationary subgroups of general position. (Russian) Funktsional. Anal. i Prilozhen. 12 (1978), no. 2, 91–92.
See the second paragraph. As for reliability of these results, the whole classification was recovered in
Knop, Friedrich; Littelmann, Peter Der Grad erzeugender Funktionen von Invariantenringen. (German) [The degree of generating functions of rings of invariants] Math. Z. 196 (1987), no. 2, 211–229
There we dealt with a slightly broader classification problem where we nevertheless had to do all calculations from scratch. So we were not using Èlašvili's tables directly. That's why I am pretty positive that the tables are complete.
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| 2020-04-21T08:51:52 |
358108
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Stack Exchange
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What aspects of math olympiads do you find still useful in your math research?
I was rereading the book Littlewood's Miscellany and this passage struck me:
It used to be said that the discipline in 'manipulative skill' bore
later fruit in original work. I should deny this almost absolutely - such
skill is very short-winded. My actual experience has been that after a
few years nothing remained to show for it all except the knack, which has
lasted, of throwing off a set of (modern) Tripos questions both suitable
and with the silly little touch of distinction we still feel is called for;
this never bothers me as it does my juniors. (I said 'almost' absolutely;
there could be rare exceptions. If Herman had been put on to some
of the more elusive elementary inequalities at the right moment I can
imagine his anticipating some of the latest and slickest proofs, perhaps
even making new discoveries.)
I would like to ask a question to former math olympiad students who now are actively involved in math research. Do you find the training for olympiads useful in later research career as a mathematician?
Olympiad training in Germany 15 years ago (and probably still to these days) included active classes in several subjects that are commonly done in university. We learned quadratic residues, max-flow-min-cut with applications (including Hall), some basic convexity and majorization, elementary linear algebra including the $\mathbb{F}_p$ case, generating functions and discrete Fourier transforms IIRC. And this all was supported with better exercises than in most university classes. This alone makes it useful, nevermind the olympiad-specific methods some of which are more and some less helpful.
When I was an undergraduate I knew several of the best contest takers in the country. At least half of them dropped out of grad school before getting a PhD, although those that stayed succeeded brilliantly (Jeremy Kahn, Bjorn Poonen). It seemed to me that having put a great deal of effort into developing a skill and then discovering that it wasn't really that helpful in doing research must have been very disappointing.
One of my professors said that according to him, problem solving and doing research required very different aptitudes and skill sets. He said he had known an IMO gold medallist who had failed quite spectacularly as a researcher (well, maybe what made it so spectacular was the fact that he had won a gold medal). Anyway, when I was a mathematician, on occasion I found it useful to switch over to "problem-solving mode", and hack my way to results which I could not obtain by (let's say) "pure thought". But I was only ever a postdoc...
John Rickard represented Britain in the IMO three times. He went up to Cambridge (Trinity) at the age of 15 and skipped Part 1A. But he never submitted his thesis and became a programmer. He died from liver cancer at the age of 41. I write this not to answer the question but as a memorial to a friend who should have had better pastoral care. I hope that his brother @JeremyRickard will excuse me.
Not getting through the preliminary selection was humbling. Handling disappointments is crucial in research.
My experience: Research requires infinitely more resilience (and this cannot be overstated), but there is basically no mathematical difference between how I approach research and how I approached olympiads. But I learned deeper math and did olympiads at the same time and each influenced (positively) how I approached the other. I imagine if I had focused on one exclusively the patterns of thought would not translate as well. It seems many others found it helpful to alter their mathematical personalities for research.
In the early days of MathOverflow, I learned in an offline conversation about some of the contributors that "in Russian there exists a special word to describe a person obsessed with IMO-style mathematical problems: olimpyadnik (usually male, hence the masculine gender). And it is not a complement." (maybe олимпядник)
@PaulTaylor, your comment is misleading. "Олимпиадник" is "olympiad" plus a standard suffix denoting a person related to something; it does have a natural feminine form. It's a completely neutral term to refer to a high school student who frequents olympiads and attends a training circle, in any subject not necessarily maths. It has nothing to do with "obsession with IMO style problems". Of course, when applied to a senior mathematician, it will sound ironic and some people may use it to denigrate colleagues they don't like.
It's my belief that a large part of mathematical research, perhaps more than we would like to admit, comes down to finding clever elementary arguments. This is particularly true in my own area (combinatorics and related fields) but it is true of many other fields as well. Of course there's usually some machinery to be mastered, but at the end of the day, you're trying to come up with something new, and it's not so often that you're building some gigantic new machine out of whole cloth. Typically you're taking various known ideas and trying to figure out how to adapt them and put them together in a new way to prove something new. When the pieces of the puzzle finally fall into place, I find the experience to be not unlike the process of solving an Olympiad problem. The Olympiad training is useful for building a sense of confidence that something nontrivial can emerge with a bit of persistence and cleverness. I find that some of my colleagues without this kind of problem-solving background will sometimes give up too quickly, because they are at a loss as to how to proceed when their usual box of tools doesn't apply.
A second way in which I find Olympiad training useful is that when I am confronted with a new and difficult problem that seems too hard to tackle directly, I can often find a way to invent a toy version of the problem whose solution may give some insight. Experience with Olympiad problems has given me a sense of what a "bite-sized" problem looks like—subtle enough to be nontrivial, but simple enough to be tractable. Interestingly, I often find that some of my colleagues who are better than I am at solving Olympiad problems can often solve my bite-sized problems when I can't; at the same time, I often seem to be better than those same colleagues at coming up with the bite-sized problems in the first place. This may partly explain why some Olympiad stars don't become good mathematical researchers. Research requires several skills, and those who only know how to solve tractable problems and don't know how to formulate them in the first place may not do so well at research. But I think that experience with Olympiad problems can help with the formulation process as well.
A quote from http://timothychow.net/cv.pdf:
"Princeton University Class of 1939 Prize for the senior with the highest academic standing"
-- Interesting!
Excellent answer !
I'd point out a trivial thing: often, you just need to sit down and do a calculation, or a case-by-case analysis, etc. I mean something that can't be directly fed to Mathematica, requiring higher-level reasoning, still technical enough so that you need to crunch it with pen and paper.
Olympiad kids are trained to concentrate on such tasks and do them quickly and cleanly. Or course, any research mathematician, knowing the problem boils down to a computation, will be eventually able to do it. But they might spend more time, get distracted, make a mistake, spend even more time because of that, etc. Even more importantly, you often don't know in advance whether the outcome of your computation will solve the problem at hand. The quicker and more confident you are at such things, the more you can try.
My personal experience and view is that there are certain olympiad type problems that, if you can solve them, demonstrate genuinely useful skills applicable in research.
Often in research you encounter similar questions perhaps as sub-problems to your main problem or in the search for special cases or counterexamples. Unless there is an obvious approach by existing theory progress usually then involves simply making an educated guess and trying to prove it. The ability to "feel" ones way to a good guess in reasonable time say by raw instinct, clever use of heuristics or analogy is vital in this case. Similar skills are needed I believe in solving many Olympiad Questions, especially under time pressure.
For example the "Pentagon Game" discussed on Matt Baker's Math Blog involves a pentagon with integers at the vertices and a rule to evolve those. You have to prove that the game ends in a finite time - the solution involves finding an positive integer invariant that always decreases. Finding this invariant quickly is non-trivial and requires the ability to guess some good options and/or exclude many which won't work. There is no standard theory to fall back on you which is often the case for genuine research problems.
(See this question for a discussion of Olympiad Questions with connections to real mathematics, many of which fit the above criteria and this paper for more mathematical developments from the "Pentagon Game".)
More generally concerning the question of whether problem solving skills are important to learn and practise for budding mathematicians you might be interested in John Hammersley's view which was certainly an outlier amongst mathematicians at the time - he believed that manipulative skills, problem solving skills were much more important than abstraction and theory which often did not help in solving real world problems.- see his article "On the enfeeblement of mathematical skills by "Modern Mathematics" and by similar soft intellectual trash in schools and universities"
About the pentagon game, you write: "There is no standard theory to fall back on you which is often the case for genuine research problems." But as Chazelle's solution (sketched in Matt Baker's blog) shows, there is: The problem can be more or less restated as "any element of the affine symmetric group has a reduced expression" (i.e., if you remove its descents one by one through multiplication by $s_i$'s, it eventually becomes the identity). It's not quite about the affine symmetric group, ...
... because the $\sum_{i=1}^n \omega\left(i\right) = \dbinom{n+1}{2}$ condition does not hold in general in the pentagon game, but at this point you can take the argument for the affine symmetric group and generalize it straightforwardly.
@darijgrinberg Thank you for your comment. I was aware of Chazelle's solution but the statement in my answer was comparing the skills required to solve problems solved at an olympiad level against those needed for genuine research questions. It is hard to imagine that any student attempting the "Pentagon Game" would be able to draw on the theory you describe and hence would be forced to guess an invariant and try to prove it. I wasn't saying that this was the only approach to this question and clearly it isn't as you rightly point out.
However I think it's also true that most research mathematicians if faced with the Pentagon game during their research would not probably follow a theoretical route like Chazelle but rather guess and prove the invariant. Also I think that in this case coming up with Chazelle's approach seems a lot harder though it does obviously lead to a more general and informative solution.
|
2025-03-21T14:48:30.387446
| 2020-04-22T23:21:29 |
358257
|
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|
Stack Exchange
|
When is a set defined by multivariate polynomial inequalities convex?
Consider the set of real numbers given by
$$S = \{(a,b,c,d,e,f,g,h) \in [0,1]^8 : 0 \le \frac{e(g-h)}{b(g-f)} \le 1 \text{ and } 0 \le \frac{e(h-f)}{(1-b)(g-f)} \le 1\}$$
Note that this set can also be defined as
$$S = \{(a,b,c,d,e,f,g,h) \in [0,1]^8 : g \ge h \ge f \text{ and }e(g-h) - b(g-f) < 0 \text{ and } eg-eh-bg +bf \le 0\}$$
I'd like to show that there is a unique projection from any element of the 8 dimensional hypercube $[0,1]^8$ onto $S$. My understanding is that one requirement for this is convexity. But how do we show $S$ is convex when it is defined by multivariate polynomial inequalities? Moreover, might there be a faster way to show that there is a unique projection onto $S$ (or that there is not)? Thank you.
Where does this set come from?
Your set $S$ is not convex. E.g., the octuples $x:=(0, 5/64, 0, 0, 1, 1/2, 1, 1)$ and $y:=(0, 1/2, 0, 0, 0, 0, 1, 0)$ are in $S$, but $(x+y)/2$ is not: both conditions $e(g-h)-b(g-f)<0$ and $eg-eh-bg+bf\le0$ fail to hold for the octuple $(x+y)/2$.
As for the unique projection, for its existence you will first of all need to replace $S$ by its closure, say $\bar S$ -- which latter will also be non-convex, by what was noted above. So, by the Bunt--Motzkin theorem Theorem E.<IP_ADDRESS> on page 607, the projection here will not be unique.
|
2025-03-21T14:48:30.387550
| 2020-04-22T23:28:27 |
358259
|
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|
Stack Exchange
|
Degree of a divisor along a subscheme
I'm curious about a computation of Prop2.3 in The gonality conjecture on syzygies of algebraic curves of large degree by Ein and Lazarsfeld. Let $C$ be a smooth projective curve carrying a pencil $\alpha$ with degree $p+1$. Let $pr:C\times C_p\to C$ be the projection and $\sigma:C\times C_p\to C_{p+1}$ be defined by $(x,\xi)\mapsto x+\xi$, where all the subscripts refer to the symmetric product. Let $L_d$ be a divisor on $C$ of degree $d$ in the form $dx$ with $x\in C$ and $N_d=\det\sigma_*pr^*L_d$ a divisor on $C_{p+1}$. My question is why the degree of $N_d$ along $|\alpha|=\mathbb{P}^1$ is $d+$constant? Because we cannot distinguish invertible sheaves via fibers, I want to calculate $H^0(|\alpha|,i^*N_d)$, where $i:|\alpha|\hookrightarrow C_{p+1}$ is the embedding. Any help would be appreciated.
|
2025-03-21T14:48:30.387637
| 2020-04-23T01:09:26 |
358266
|
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|
Stack Exchange
|
Minimize overlap penalty between paths in graph
Suppose we have a weighted undirected graph $G(V,E)$. We are given the information that $V_a \cap V_b = \emptyset$ and $V_a,V_b \subset V$.
We want to find paths from all vertices in $V_a$ to all vertices in $V_b$ given the following conditions:
Primary objective: paths minimize the number of shared edges
Secondary objective: sum of weighted path lengths are minimized
Sharing an edge gives a penalty equivalent to the weight of the shared edge
Consider the following example:
If path $s_1 - t_1$, $s_2 - t_1$, and $s_3 - t_1$ all use edge $e_2$ then the penalty is $3\cdot weight(e_2)$.
Paths from any particular $s \in V_a$ to all $t \in V_b$ do not acquire a penalty for overlapping with other paths from that particular $s$. That is to say paths from $s_i$ to $t_j$ for all $t_j \in V_b$ cannot acquire a penalty from each other.
Consider the following examples:
If path from $s_1$ to $t_1$ uses edges $e_1, e_2, e_3$ and path from $s_1$ to $t_2$ uses the edges $e_1, e_2, e_4$, then there is no penalty.
If path from $s_1$ to $t_1$ uses edges $e_1, e_2, e_3$ and path from $s_2$ to $t_2$ uses the edges $e_1, e_2, e_4$, then there is a penalty of $\mathit{weight}(e_1) + \mathit{weight}(e_2)$.
Pictured Example:
In the pictured example we can find the following solutions for each of the desired six paths:
$s_1 - t_1$: $s_1,a,t_1$
$s_1 - t_2$: $s_1,a,t_2$
note there is no penalty for sharing the edge $(s_1,a)$ because both paths so far both start from $s_1$
$s_2 - t_1$: $s_2,b,a,t_1$
Note this would have a penalty for sharing edge $(a,t_1)$ with the path from $s_1-t_1$
$s_2 - t_2$: $s_2,b,a,t_2$
Note there is no penalty from this path
$s_3 - t_1$: $s_3,c,t_1$
Note this has no penalty and we could have chosen path $s_3,c,d,t_1$ for no penalty as well, but it is a longer path so we don't choose it
$s_3 - t_2$: $s_3,c,d,t_2$
Note this has no penalty no matter what path was chosen for $s_2-t_1$
The practical application of this problem is loosely defined. These constraints are not rigid if they cause too many complications. For example, instead of having the penalty in the first example being $3 \cdot \mathit{weight}(e_3)$ it might be acceptable to have the penalty as $\mathit{weight}(e_3)$. I'm mostly looking for new ideas to handle this or a slight variation of this problem.
This is an extension to Algorithm for finding minimally overlapping paths in a graph
You can formulate this as a multicommodity flow problem and solve it via linear programming. The commodities are $K = V_a \times V_b$. Let $A$ be the arc set, with one arc in each direction for each edge in $E$. For $(i,j)\in A$ and $k\in K$, let variable $x_{i,j}^k \ge 0$ be the flow along arc $(i,j)$ of commodity $k$. For $\{i,j\}\in E$ and source node $s\in V_a$, let variable $z_{i,j}^s \ge 0$ be the maximum amount of flow (in either direction) across edge $\{i,j\}$. Let variable $y_{i,j}\ge 0$ be the excess number of sources that use edge $\{i,j\}\in E$. Let $b_{i,k}$ be the supply at node $i$ of commodity $k$; explicitly, $b_{i,k}$ is $1$ for the source node of commodity $k$, $-1$ for the sink node of commodity $k$, and $0$ otherwise. Let $c_{i,j}$ be the weight of edge $\{i,j\}$. The primary problem is to minimize $\sum_{\{i,j\}\in E} c_{i,j} y_{i,j}$ subject to:
\begin{align}
x_{i,j}^k + x_{j,i}^k &\le z_{i,j}^s &&\text{for all $\{i,j\}\in E$ and $k=(s,t)\in K$}\tag1\\
\sum_{s:(s,t)\in K} z_{i,j}^s &\le 1 + y_{i,j} &&\text{for all $\{i,j\}\in E$}\tag2\\
\sum_j (x_{i,j}^k - x_{j,i}^k) &= b_{i,k} &&\text{for all $i\in V$ and $k\in K$}\tag3
\end{align}
The secondary problem is to minimize $\sum_{(i,j)\in A}\sum_{k\in K} c_{i,j} x_{i,j}^k$ subject to $(1),(2),(3)$ and:
$$\sum_{\{i,j\}\in E} c_{i,j} y_{i,j} \le p^*,$$
where $p^*$ is the optimal primary objective value.
Any advice for the last desired condition: That is to say paths from a particular $s_i$ to $t_j$ for all $t_j \in V_b$ cannot acquire a penalty from each other.
This is already part of the model. The $z$ variable is shared by all $(s,t)$ pairs with the same $s$. By the way, the side constraints almost certainly imply that you will need to impose that $x$ is binary instead of just nonnegative. Otherwise, the resulting flows will not necessarily be paths.
|
2025-03-21T14:48:30.388022
| 2020-04-23T01:46:58 |
358269
|
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|
Stack Exchange
|
Understanding a step in proof of sheaf version Verdier duality
Warning: This question is likely low-level for MathOverflow. My apology that there is almost surely something basic I miss.
So all proofs I can find factors through a particular statement, which goes to Kashiwara-Shapira, which leaves it to readers as exercise ((2.6.25) on page 114).
The statement: Let $f:X\rightarrow Y$ be a map of ``nice'' spaces (finite CW-complex or even manifolds if you like, so every cohomological dimension ever appear is finite). Let $\mathcal{F},\mathcal{G}\in D^b(Sh(X))$ be two bounded complexes of sheaves (of abelian groups) on $X$. The statement is question asserts that we have a map
$$Rf_*R\underline{Hom}_X(\mathcal{F},\mathcal{G})\rightarrow R\underline{Hom}_Y(Rf_!\mathcal{F},Rf_!\mathcal{G})$$
that is functorial in $\mathcal{F}$ and $\mathcal{G}$. Here $\underline{Hom}$ is the sheaf Hom and $R\underline{Hom}$ its derived functor. This map (or rather natural transformation) is supposed to be derived from the natural map for sheaves
$$f_*\underline{Hom}_X(\mathcal{F},\mathcal{G})\rightarrow \underline{Hom}_Y(f_!\mathcal{F},f_!\mathcal{G})$$
which is fairly straightforward. But how do we derive it? In the $f_*$ version we can use that $\underline{Hom}_Y(f_*\mathcal{F},f_*\mathcal{G})$ is flabby for some injective $\mathcal{G}$. I can't see the same for $f_!$ ...
|
2025-03-21T14:48:30.388134
| 2020-04-23T02:23:28 |
358271
|
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|
Stack Exchange
|
General Tarski-Seidenberg Theorem
The Tarski-Seidenberg Theorem states that the polynomial image of a semi-algebraic set is semi-algebraic. A semi-algebraic subset of a Euclidean space $\Bbb{R}^n$ is by definition a finite union of subsets of the form
$$
\{P_1=\dots=P_k=0, Q_1>0,\dots,Q_l>0\}
$$
where $P_i$'s and $Q_j$'s belong to $\Bbb{R}[x_1,\dots,x_n]$. I wonder is there a general coordinate-free version of this theorem for morphisms of real varieties? (By a real variety I mean the set of $\Bbb{R}$-points of a variety over $\Bbb{R}$.)
What does 'coordinate-free' mean? The original theorem seems coordinate-free to me ….
@LSpice I mean a version for abstract varieties that are not defined as subsets of a Euclidean space $\Bbb{R}^n$.
The most abstract version of the Tarski-Seidenberg theorem I know of is the following
Let $f:A\to B$ be a morphism of finite presentation of commutative rings. Then the induced map
$$f^*:\operatorname{Sper}B\to \operatorname{Sper}A$$
sends constructible sets to constructible sets.
Here $\operatorname{Sper}$ is the real spectrum of the ring, i.e. the set of all pairs $(p,<)$ where $p$ is a prime ideal and $<$ is an order on the residue field at $p$.
It is well known (e.g. theorem 7.2.3 in Bochnack-Coste-Roy Real Algebraic Geometry) that if $A$ is an algebra of finite presentation over $\mathbb{R}$, the boolean algebra of constructible subsets of $\operatorname{Sper}A$ is in natural bijection with the semialgebraic subsets of the real points of the variety $\operatorname{Spec}A$.
Thanks! Out of curiosity, is there a notion of a "real scheme", a ringed space which is locally isomorphic with $Sper A$?
@KhashF Yes, they are called real closed spaces and they have been developed by Niels Schwartz (they are not really locally modeled on $\operatorname{Sper}A$, rather on proconstructible subsets of it, but the idea is the same)
|
2025-03-21T14:48:30.388312
| 2020-04-23T04:08:22 |
358276
|
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"Gerhard Paseman",
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"skd"
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|
Stack Exchange
|
Number of polytopes formed by connecting points on a hypercube
Fix an integer $d\geq 1$, and let $n\geq 1$. Drawing hyperplanes between all the $d$-sets of lattice points on the boundary of the hypercube $[0,n]^d\subseteq \mathbf{R}^d$ defines a partition of $[0,n]^d$ into several distinct polytopes; let $a(n,d)$ denote the number of such polytopes. (Note that $a(n,d)$ is divisible by $2^d$.) For instance, $a(1,2) = 4$ and $a(2,2) = 56$. What can be said about the sequence $a(n,d)$ as $n$ and $d$ vary? (I'd originally asked about the generating function, but this seems way too hard. I would be interested in asymptotics with $n$ or $d$ fixed.)
May I ask: How did you calculate $a(2,2)=56$?
@JosephO'Rourke I just drew it and counted! I don't know how to code this up, but it's probably possible to do so and compute more values. It'd be way more efficient than drawing and counting...
There is a combinatorial formula for this, it gets cumbersome to generalize for higher dimension. Its easy for a(2,2), it also validates 56 as the answer. Its a double sum, inner sum over outer lattice points - 2, and outer sum over increasingly fewer lattice points as the starting point.
Just realized, the question needs a little more precision, when you say partition into several distinct polytopes....are you counting polytopes that are solids, facets, ridges, planes, lines....is there any restriction to the kind of polytopes? I've been making assumptions. Also, for some reason the phrasing with lines instead of hyperplanes made a lot more sense.
Joseph O'Rourke calculated that a(3,2) = 340. Searching the sequence 4, 56, 340 on the OEIS led to this: http://oeis.org/A255011. There doesn't seem to be a formula recorded there, though.
For d=2, use Euler's formula to compute these values programmatically. When ever you add a new line segment, count the edges it gets broken into, and the number of new vertices created. The end result is a planar graph whose edges you have tallied (be sure to add extra edges when creating a new vertex) and vertices, and then compute faces using the formula for a planar graph.
I imagine a similar approach is used for higher dimensions. I would ask Joseph O'Rourke about it.
Gerhard "Is This Really Computational Topology?" Paseman, 2020.04.23.
Here is $a(3,2)$, which confirms Gerhard's count of $340$
regions:
Didn't your mother teach you to use symmetry? Really, I had higher hopes. Starting with half of the lower right square ( (3,3) in my indexing) I count 20 regions (more than half triangular) which gives 160 for all four corner squares. Picking half of an edge square ((1,2), say) I count 31 non central and four central regions, giving 264 for the edge squares. Take a triangular fourth of the central square and use symmetry to get 24 for the quarter and 96 for the square. 520 in all. Gerhard "Taught Himself Symmetry When Younger" Paseman, 2020.04.23.
Does anyone else experience the optical illusion of thin, white horizontal and vertical lines passing through the center of the square? For me they come and go depending on my visual focus...
yes, I'm seeing them as well
I think the pixellation makes a one pixel high contrast which suggests part of a horizontal line segment, and having several them at the same y ordinate reinforces the idea that there is a horizontal"ground line" occurring. I'm not seeing the vertical as much, possibly be sure of astigmatism. Gerhard "We Should Ask Scott Kim" Paseman, 2020.04.23.
I just noticed there are sixteen extra lines in the picture above, creating extra regions. (Dividing lines need to not stop at internal points.) Using symmetry, we see one line has ten segments and another has fourteen. Taking into account the (up to symmetry) two kinds of intersections involving only internal lines, we get 8*(10+14 - 1) - 4 .= 180 fewer regions, for a new total of 340. Gerhard "Surely That's Easier To Count?" Paseman, 2020.04.24.
OK, as long as you caption the picture as having to do with something other than a(3,2), it should be OK. Remember the problem speaks of boundary points. Gerhard "Likes Getting A Good Picture" Paseman, 2020.04.24.
@GerhardPaseman: Thanks for your several corrections. I confirm your count of $340$.
You should check out the pictures behind the OEIS link. Gerhard "Step Up Your Game Time" Paseman, 2020.04.24.
@GerhardPaseman: Beautiful! Same diagram, colored.
This is a counting problem, first we need to know a one thing:
How many lattice points exist on the boundary of our $L=[0,n]^d$ dimensional cube?
We only need count lattice points $x \in [o,n]^d$ such that a given coordinate $x_j \in \{0,n\}$, for some $1 \le j \le d$. This tells us $x$ lies on the boundary of $S$.
We count all lattice points and substract the internal points;
$$ Q:=|\{ x \in L | \;\; x\in \text{ surface}\}| = (n+1)^d - (n-1)^d $$
The way to reason about that count is as follows,
Every $x\in L$ has the vector form $(x_1,x_2,...,x_{d+1})$.
For every coordinate we have $n+1$ choices of possible values.
Choices are independent, so we multiply by $n+1$ for every coordinate.
There are $d$ coordinates so we get $(n+1)^d$
The reasoning behind how many inner points there are so we can subtract them is almost exactly the same, only in step 2 we have $(n+1) - 2$ choices; that is, no coordinate is allowed to be $n,0$ as that would place the point on the surface of the lattice.
Let's say, $L(n,d) := [0,n]^d$, and $$ \gamma( L(n,d) ) = |\{ A \in L(n,d) | A \text{ is a polytope as described in the question } \}| $$ then
$$\gamma(L(n,2))=\sum_{k=2}^{Q-1} \binom{Q-1}{k}= (2)^{(Q-1)}-(Q-1)-1 $$.
$$\gamma(L(n,3))=\sum_{k=3}^{Q-1} \binom{Q-1}{k}= (2)^{(Q-1)}-\frac{Q(Q-1)}{2}-(Q-1)-1 $$.
$$\cdots$$
$$\gamma(L(n,d))=\sum_{k=d}^{Q-1} \binom{Q-1}{k}= (2)^{(Q-1)}-\sum_{k=0}^{d-1} \binom{Q-1}{k} $$.
The origin is fixed, so we have $Q-1$ points to choose from, we may choose any number of these to form a polytope with some exceptions. For $d=2$, except 0 points, which leaves only the origin; or one point, which would make only lines. Similar for $d=3$, we may not choose just two points as this only makes a plane, and so on.
EDIT: The underlying assumption of this count is that the origin is a "corner" of every polytope.
Thanks for the response. I'm not sure I fully understand; I don't parse the string "L(n,d) := [0,n]^d" and then the equations for L(n,d). Perhaps I've not calculated/understood your answer correctly, but it doesn't seem like any of the equations listed recover the sequence 4, 56, 340 for an n-by-n square in two dimensions.
Fixed. Thanks! (Also, the 15 character comment length minimum is silly.)
Thanks. Again, apologies for possibly being silly, but it seems like the sequence γ(L(n,2)) goes 4, 120, 2036, ... It's not clear to me how this sequence relates to the question.
I think we can easily see the first case is indeed 4: 3 triangles all with one corner at the origin and then the entire square. For $n=3$, that is the right count if you realize that polytopes need not be a triangle, you could choose 4 points including the origin and get a polypote with 4 sides where an edge would be part of the boundary, there are many such cases that only counting triangles leaves out..
He is counting a different quantity, which may (or may not) bear some relation to your quantity. I suggested using Euler's formula for d=2. For higher d, one might start out with his formula to count related quantities. By the way, boundary is not clear yet in higher dimensions for me. I can see counting only lattice points on edges or on edges and faces as qualifying, but you might mean something more general. Gerhard "Not Used To High-Dimensional Combinatorics" Paseman, 2020.04.24.
@GerhardPaseman, looking at the vector representation and counting those with properties that place them on the boundaries is the only "reasonable" way I know of looking at it. A face of the hypercube has all point lying on some hyperplane at $x_i=0$ or $x_i=d$. Using higher level geometry, or fancier math, seems unnecessary. Just like in 3D $y=0$ means zx-plane, or $z=0$ means xy-plane, and so on.
|
2025-03-21T14:48:30.388884
| 2020-04-23T04:47:21 |
358280
|
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|
Stack Exchange
|
Examples of constant scalar curvature kähler metric that is not kahler einstiein
It is well known that if the first Chern class is proportional to the kähler class given, then every cscK in that class has to be kähler Einstein. So there are two directions to generate examples as far as I can think of:
1). It is a result if we have a cscK in $[w]$ then for any kähler class sufficiently close to $[w]$ the existence of cscK is guaranteed. So maybe we can try perturb our class on manifolds where $C_1(M) < 0$.
2). May be we can consider manifolds where $C_1(M)$ is semiample.
I have no idea. Are there any standard examples which are relatively simple to calculate?
The easiest examples are products of two KE manifolds with different Einstein constants. For example $M=\mathbb{P}^1\times\mathbb{P}^1$ with the product of two Fubini-Study metrics with different volumes. This is clearly cscK and not KE. Note that $M$ here is Fano.
Also, your "fact" number 1) above is wrong. It is true if $M$ doesn't have nontrivial global holomorphic vector fields, but false in general, see e.g. LeBrun-Simanca
Here is a non-compact example.
Consider the half-space $ \mathbb{C} \times \mathbb{H}$ as a subset of $ \mathbb{C}^2$. We use the Kahler potential
$$\Psi = \frac{x_1^2}{x_2} - \log(x_2).$$
Here, $z_1 =x_1+ \sqrt{-1} y_1$ and $z_2 =x_2+ \sqrt{-1} y_2$.
This metric has constant scalar curvature but the Ricci potential is $3 \log(x_2)$, so is not Kahler-Einstein.
This Kahler manifold is known as the Siegel-Jacobi space and has been studied in quite a bit of detail. To give two examples, Mathieu Molitor wrote a paper "Gaussian distributions, Jacobi group and Siegel-Jacobi space" describing the geometry in considerable detail. It was also studied by Jae-Hyun Yang in "Geometry and Arithmetic on the Siegel-Jacobi Space."
Are there any examples that are compact?
I don't know of any examples off the top of my head. Presumably they exist since cscK seems like a much weaker condition than Kahler-Einstein. However, any example cannot be Fano.
Why can't it be Fano?
For Fano manifolds, csck metrics are precisely Kahler-Einstein and their existence depends on an algebro-geometric condition known as k- poly stability. This is a famous result of Chen, Donaldson, and Sun.
This is wrong -- there are many Fano examples, see my comment above. What is true (and has nothing to do with Chen, Donaldson, Sun) is that on a Fano manifold a cscK metric which is cohomologous to $c_1(M)$ must be KE (proof extremely easy). But there are in general many other Kahler classes, e.g. on $\mathbb{P}^1\times\mathbb{P}^1$.
Ah okay. Thanks for the correction. I'll leave the mistake up so that it's clear what your comment is referring to.
$\newcommand{\Proj}{\mathbf{P}}$Examples of compact constant scalar curvature Kähler manifolds that are not Einstein are constructed by solving explicit ODEs in On existence of Kähler metrics with constant scalar curvature, Osaka J. Math., 31 (1994), 561–595. If $M$ is a compact almost-homogeneous space with real hypersurface orbits and two ends, and if the connected automorphism group is reductive, then the set of Kähler classes containing a Kähler metric of constant scalar curvature is a real-algebraic hypersurface that separates the Kähler cone.
The simpest non-product example is $\Proj^{3}$ blown up along two skew lines. We may view this Fano threefold as the completion of the line bundle $\mathcal{O}_{\Proj^{1}}(-1) \otimes \mathcal{O}_{\Proj^{1}}(1) \to \Proj^{1} \times \Proj^{1}$ (with the line bundles pulled back to $\Proj^{1} \times \Proj^{1}$ by projection to the respective factors). The Kähler cone has three parameters, which may be viewed loosely as the sizes $a_{1}$ and $a_{2}$ of the $\Proj^{1}$ factors and the size $2b$ of the $\Proj^{1}$ fibre. [See note below] If we scale to make $b = 1$, the resulting slice of the Kähler cone is the open quadrant $a_{1}$, $a_{2} > 1$.
The point $(a_{1}, a_{2}) = (2, 2)$ is proportional to the anticanonical class, which contains an Einstein-Kähler metric by work of Koiso and Sakane (Non-homogeneous Kähler-Einstein metrics on compact complex manifolds, in Springer Lecture Notes 1201, 1986, 165-179).
The set of classes containing a constant scalar curvature representative consists of the diagonal $a_{1} = a_{2}$ and a branch of the right hyperbola $(a_{1} - 1)(a_{2} - 1) = 1$.
Note: More literally, let $\omega$ denote the unit-area Fubini-Study form on $\Proj^{1}$. In the Kähler class $(a_{1}, a_{2}, b)$, the zero section has Kähler form cohomologous to $(a_{1} - b)\omega \times (a_{2} + b)\omega$ and the infinity section has Kähler form cohomologous to $(a_{1} + b)\omega \times (a_{2} - b)\omega$.
|
2025-03-21T14:48:30.389190
| 2020-04-23T05:22:13 |
358282
|
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|
Stack Exchange
|
Intersection of principal ideals
Let $x,y$ be nonzero elements in a commutative ring $R$. Is $(x)\cap (y)$ always finitely generated?
What if we further assume that $R$ is an integral domain? Can we construct non-Noetherian non-local semi-local integral domains whose every maximal ideal is principal (in the local case, there are certain valuation rings.)?
To your first question about $R$ being a domain and $(x)\cap (y)$ not being finitely generated, see https://math.stackexchange.com/q/296653
Excellent, thanks!
One way to force finite generation of $(x)\cap (y)$ is to make this ideal equal to $(xy)$ , and one nice case where this happens is that $x$ is not a zero divisor on $R/(y)$
|
2025-03-21T14:48:30.389274
| 2020-04-23T06:29:43 |
358285
|
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"Bruno Kongawi",
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|
Stack Exchange
|
A conjecture/algorithm about prime numbers
I am not a mathematician, but I used to love reading Euclid's elements. In 2013 I discovered a very inefficient way of generating prime numbers exactly using a statement I deduced while reading BOOK VII:
If a set of $n$ prime numbers, $p_1, p_2, \cdots, p_n$, contains a complete continuum of primes such that $p_g$ marks the (inclusive) upper bound of this continuum and all primes below $p_g$ are present in the set, then all numbers of the form
\begin{gather*}
\left|\prod_{i=1}^m \,\,\,p_i^{s_j} \pm \prod_{i=m+1}^n p_i^{s_k}\right|
\end{gather*}
that are less than ${p_{g}}^2$ and greater than the unit are also prime, where $m$ is less than $n$ ($n \geq 2$) and $s_j$ and $s_k$ denote any natural number.
I have written about it here, where you can also find few programs in Kotlin which demonstrate this up to a certain point (very inefficiently):
https://github.com/EtherBit/On-Immeasurable-Magnitudes
Is this known? Could somebody point me to more helpful resources.
Pardon the terminology and verbosity.
It's pretty easy to see that your number can't be divisible by any of your primes, so if it's less than the square of your largest prime (and isn't $1$) then it must be prime. You can probably learn more Number Theory from an intro Number Theory text, say, Niven, Zuckerman, and Montgomery, than you can from Euclid. Congrats on finding this by yourself, that's the main thing.
You might be interested in Guy, Lacampagne, and Selfridge, Primes at a Glance, Mathematics of Computation, Volume 48, Number 177, January 1987, pages 183-202, available at https://www.ams.org/journals/mcom/1987-48-177/S0025-5718-1987-0866108-3/S0025-5718-1987-0866108-3.pdf
@GerryMyerson thanks for the link! Briefly skimmed through it. Seems to be similar.
|
2025-03-21T14:48:30.389440
| 2020-04-23T07:56:10 |
358294
|
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|
Stack Exchange
|
Definability in countable nonstandard models of Peano arithmetic
I know that every element of $\mathbb{N}$ is definable the standard model of Peano Arithmetic. Does there exist a countable non-standard model of PA where the same is true?
Yes. PA has definable Skolem functions, hence the set of definable elements of any model is an elementary submodel, which is nonstandard as long as the original model is not elementarily equivalent to $\mathbb N$.
Why does having definable Skolem functions imply that the set of definable elements of any model is an elementary submodel?
@MarcusDubious To see that the definable elements form an elementray submodel (in the presence of definable Skolem functions), apply the so-called Tarski-Vaught test of elementarity, the test is explained on: https://math.stackexchange.com/questions/3226832/proof-of-the-tarski-vaught-test/3226853#3226853
To move this off the unanswered queue, the answer is yes. For any $\mathcal{M}\models\mathsf{PA}$, the set $\mathcal{M}_{\mathit{def}}$ of parameter-freely-definable elements of $\mathcal{M}$ is an elementary substructure of $\mathcal{M}$ by the Tarski-Vaught test. (More generally this occurs in any theory with definable Skolem functions, another important example being $\mathsf{ZFC+V=L}$.)
This gives us the following quite nice result:
$Th(-)$ provides a bijection between isomorphism types of pointwise-definable models of $\mathsf{PA}$ and completions of $\mathsf{PA}$.
That is, every completion $T$ of $\mathsf{PA}$ has exactly one pointwise-definable model up to isomorphism, and this model is nonstandard iff $T\not=\mathsf{TA}$. A key step to proving the above is the following:
If $\mathcal{A},\mathcal{B}$ are elementarily equivalent and each is pointwise-definable, then $\mathcal{A}\cong\mathcal{B}$.
|
2025-03-21T14:48:30.389583
| 2020-04-23T08:20:25 |
358297
|
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|
Stack Exchange
|
Turan numbers of r-partite hypergraphs
Let $H$ be a balanced $r$-partite $r$-uniform hypergraph with $nr$ vertices. (Each part of this hypergraph consists of $n$ vertices; every hyperedge has exactly one vertex in each part.) Denote a complete balanced $r$-partite $r$-uniform hypergraph with $nr$ vertices as $K_{n}^r$.
Question: What is the maximum number of hyperedges in a hypergraph $H$, if it doesn't contain a copy of $K_{l}^r$?
I know that there is a theorem by Erdős ("On extremal problems of graphs and generalized graphs", 1964), which states that if an $r$-uniform hypergraph doesn't contain a copy of $K_{l}^r$, then it can't have more than $n^{r-1/l^{r-1}}$ hyperedges. This theorem gives a good bound for the case $l^{r-1}=o(\log n)$. But I'm interested in a bound for $l=n^{\varepsilon}$. This bound should probably have a form $n^{r}-f(n, r, l)$, where $f(n, r, l)=o(n^r)$.
Here is a construction for $r = 2$ and $l = \Omega(n^{3/4})$ with $n^2 - O(n^{3/2})$ edges. You can further extend the construction for $r > 2$, which I omit here. However I do not know how to deal with smaller $l$, for example, $l \approx n^{1/2}$.
Taking the complement (with respect to a complete $r$-graph) of the graph in your question, the question itself is equivalent to the following one:
Equivalent question: What is the minimum number of edges in an $n$ by $n$ bipartite graph with parts $P$ and $L$ such that every $n^\varepsilon$-subset of $P$ has at least $n - O(n^\varepsilon)$ neighbors in $L$?
When $\varepsilon = 3/4$, the minimum number of edges in the above question is $O(n^{3/2})$. Here is how to construct such a graph.
Let $P$ and $L$ be the points and lines in the projective plane $PG(2,q)$ over $\mathbb{F}_q$, and let the bipartite graph be the point-line incidence graph. In this case the number of vertices $n = q^2 + q + 1$ and the number of edges is $n(q+1) \approx n^{3/2}$. We want to understand the smallest number of the neighbors of $l$ points in $P$. This is known as the isoperimetric problem in $PG(2,q)$.
In The isoperimetric problem in finite projective planes by Harper and Hergert, the problem is solved precisely when $l$ is of the form $1 + (m-1)(q+1)$, and there exists $l$ points (known as a maximal $(l,m)$-arc) such that no $m+1$ points of the the arc lie on the same line. When $q$ is a power of $2$ and $m \mid q$, a maximal $(l, m)$-arc exists (see https://en.wikipedia.org/wiki/Maximal_arc).
Thus take $q = 2^{2r}$ and $m = 2^r$. We know that the smallest number of neighbors of $l := 1+(m-1)(q+1) \approx n^{3/4}$ points in $P$ is at least $l(q+1)/m \approx n - n^{3/4}$ given by a maximal $(l, m)$-arc.
Acknowledgment: I benefited a lot from discussing the problem with Ryan Alweiss.
|
2025-03-21T14:48:30.389787
| 2020-04-23T08:29:56 |
358299
|
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|
Stack Exchange
|
Homomorphisms from higher rank lattices with infinite center to $\mathbb{Z}$
Suppose that $\Gamma$ is an irreducible lattice in a semi-simple real Lie group $G$ of higher rank (with infinite center!), is every homomorphism $\Gamma \to \mathbb{Z}$ trivial?
The case where $G$ has finite center follows easily from Margulis Normal subgroup Theorem. The simplest example I can think of where this question is relevant is the lift of $SL_2(\mathbb{Z}(\sqrt{2}))$ to the universal covering of $SL_2(\mathbb{R})\times SL_2(\mathbb{R})$.
Also, any reference where a discussion about lattices in semi-simple real Lie group of higher rank with infinite center would be appreciated. I only know of Ch.9 Sec.6 in Margulis Book, where I couldn't find an answer to this question.
Thank you!
It's true if $G$ has a noncompact simple factor with Property T. So the remaining case is that when $G$ is a product of $\ge 2$ rank-1 groups without Property T, as in your specific example.
Assume for simplicity that that the center is virtually cyclic (we can boil down to this case), so one has to prove that $\Gamma\times\mathbf{Z}$ and $\tilde{\Gamma}$ are not virtually isomorphic, where $\tilde{\Gamma}$ denotes the lattice and $\Gamma$ is its projection modulo the center. One approach would be to prove that $\Gamma\times\mathbf{Z}$ and $\tilde{\Gamma}$ are not IME (integrably measure equivalent). A result in this direction (for cocompact lattices in $\mathrm{SL}_2(\mathbf{R})$) is due to Das-Tessera.
In the case of $\Gamma=SL_2(Z[\sqrt{2}])$, the central extension of $SL_2(\mathbb{R})$ will be induced by the holomorphic 2-form on $\mathbb{H}^2$. I think this defines a holomorphic Hilbert modular form on $(\mathbb{H}\times\mathbb{H})/\Gamma$ of weight $(2,0)$, and it should give a non-trivial 2nd cohomology class on this Hilbert modular surface for each factor. So I think that the extension by $\mathbb{Z}\times\mathbb{Z}$ should be non-trivial, even for any finite-index subgroup. Hence it should lie in the kernel of a homomorphism of $\mathbb{Z}^2\rtimes \Gamma \to \mathbb{Z}$.
@IanAgol you're using a semidirect product notation for a nontrivial central extension...
Thanks for the insight @IanAgol and for the references Yves.
Yes, every homomorphism $\Gamma \to \mathbb{Z}$ is trivial.
We may assume that $G$ is simply connected, thus it decomposes as a product of simple factors. Let's consider two cases:
$G$ has exactly one non-compact simple factor.
$G$ has at least two non-compact simple factors.
In case 1 $G$ has property (T), so also does $\Gamma$ and the result follows.
In case 2 the result follows from theorem 0.8 in
Shalom, Yehuda
Rigidity of commensurators and irreducible lattices.
Invent. Math. 141 (2000), no. 1, 1–54.
Formally, the above theorem applies only for $\Gamma<G$ cocompact, but in fact the proof shows that you need 2-integrability of $\Gamma$ in $G$,
which holds by Proposition 7.1 here, see the preceding discussion for the definition.
The above is an edit of an earlier partial answer I gave, based on the answer of Mikael de la Salle.
See Mikael's answer and YCor's comments for further details.
Some references about square integrability of lattices are listed in Example 6.8 here.
Hi @YCor, you have a mistake in this reference. A non-compact lattice in $\text{SL}_2(\mathbb{R})$ cannot be square integrable, see Lemma 5.4. in https://arxiv.org/pdf/1006.5193.pdf.
Thanks, fortunately it's the only exception and harmless there. Indeed I relied on a Shalom's (Annals 2000) Theorem 3.6-3.7, proving square integrability of lattice in some Lie groups. Theorem 3.6 says "in $\mathrm{SO}(n,1)$ for $n\ge 4$" and Theorem 3.7 says "if $\mathrm{SO}(n,1)$ is replaced with any other rank one simple Lie group". The proof indeed seems to be done for $\mathrm{SO}(n\ge 4,1)$, then $\mathrm{SO}(3,1)$, and $\mathrm{SU}(n\ge 2,1)$, so only $\mathrm{SO}(2,1)$ should be excluded. It's hard to believe that such an ambiguous formulation survived in the published version.
@YCor, almost. I disagree regarding $\text{SO}(3,1)$. The computation by the end of section 3 in Shalom's paper that you mentioned shows that for $\text{SO}(n,1)$ there is a $p$-integrable domain for every $p<n-1$, in particular a non-uniform lattice in $\text{SO}(3,1)$ is $p$-integrable for every $p<2$. While I do not know a proof that such a lattice in not 2-integrable, I find the converse unlikely. As I mentioned above, for $n=2$ such a proof exists. It could be a nice research project to settle this for a general $n$.
Thanks again, I hoped you would correct me if necessary.
About your edit: do you mean that the quite intuitive statement "If $G$ is a LC-group, $N$ a discrete normal subgroup, $\Gamma$ a lattice in $G$ containing $N$. Then $\Gamma$ is square-integrable in $G$ iff $\Gamma$ is square-integrable in $G/N$." is not clear (esp. the $\Leftarrow$ direction, which is needed here)?
Yes, this is what I meant. Initially I assumed it is obvious without checking, then I thought again and it didn't seem obvious anymore. I haven't put much thought on it altogether, anyway.
OK, I guess it can be fixed. Roughly, let $\tilde{G}$ be the group, $G$ modulo its center $Z$, $T$ be the radical of a minimal parabolic (so $T$ is cocompact, and simply connected solvable). Choose a fundamental domain $D$ that makes $\Gamma$ square-integrable: write $D=\bigsqcup g_nD_n$ with $D_n$ inside a fixed compact subset and $g_n\in T$. Lift $D_n$ to $\tilde{D}_n$, uniformly bounded, and choose $\tilde{D}=\bigsqcup g_n\tilde{D}_n$ as fundamental domain for $\tilde{\Gamma}$, where $g_n$ was lifting according to the continuous lift $T\to\tilde{G}$. Then it seems to work...
Thanks Yves and @UriBader, I think this gives me enough info to figure it out by myself. When I have a clean proof of the square integrability, I'll add it to the answer.
Thanks Yves, this seems plausible. I am leaving this to Sebastian to check the details... @shurtados, please let me know if it doesn't go smoothly.
@UriBader Here's an amended version of my paper (I completed Example 6.8(2) adding missing restrictions, and a footnote. [I'll arXiv it later but just in case you have a look before— I'll post a clone of this message then] Thanks very much for pointing out the flawed statement!
To be precise the statement of Proposition 7.1 in [dL-dlS] doesn't apply. But what's really proved there is that if $\Gamma/Z$ is a lattice in a semisimple group and has center $Z$ and $\Gamma/Z$ is square-integrable in $G/Z$, then $\Gamma$ is square-integrable in $G$. The 2nd ingredient is: $Z$ has finite index in $Z(G)$ (i.e., $G/Z$ has finite center), which is automatic but needs a proper reference. The 3rd ingredient is: in $G$ semisimple with no compact factor but not simple, every irreducible lattice is square-integrable. The reference seems to be Chap VIII in Margulis' book, Prop. 1.2.
@Ycor if all you want is the integrability condition given in the formula before Prop 1.2 in Chap VIII in Margulis' book, there is a much easier argument (by Maucourant) giving a much stronger result (exponential integrability), which holds for every "weakly cocompact lattice", and in particular for any lattice in any Lie group. See Theorem 5.3 in my paper https://arxiv.org/abs/1711.01900.
Note however that this integrability is much weaker than what Uri means by square-integrability. Not only because Margulis has $L^1$ where you require $L^2$. But mostly because to deduce the integrability of the lattice, you need that the lattice is quasi-isometrically embedded [Lubotzky-Mozes-Raghunathan]. This is what Shalom realized, I think... and why the reference should be Shalom.
@MikaeldelaSalle what I needed (for me) is Shalom's theorem 4.1 in his 2000 Inv Math paper, which I understood to require square-integrability. Are you saying that QI-embedded + integrable implies square-integrable? and that this was first observed by Shalom? or is it an ingredient among other specific assumptions?
@Ycor Sorry, I think I missread part of this comment thread, and I am still a bit lost. I was reacting to the fact that the reference you give for square integrability of lattices is Margulis' book. But I do not see anything like this in Margulis' book.
... The inequality says something about the rate of decay of the measure of the complement of balls in $G/\Gamma$. What I was saying is that this (and much more) has a very easy proof. And the fact that, when a lattice is QI-embedded (and only then), this rate of decay can be translated to integrability properties of the lattice was, I think first observed by Shalom.
... and last : the inequality in Margulis' book is formally an $L^1$-integrability condition, whereas Shalom needs $L^2$-integrability. Of course, as Shalom writes, the sketch of proof that Margulis provides also proves at the same time the $L^p$-integrability for all $p$.
This is a follow-up of Uri's answer. My goal is just to confirm that (for any $p$) the $L^p$-integrability of lattices in a connected semisimple Lie group $G$ follows from the $L^p$-integrability of lattices in $G/Z(G)$. The non-trivial ingredient that is needed is that the central extension $G\to G/Z(G)$ is represented by a bounded $2$-cocycle. The argument (which I think I learned from Nicolas Monod) is at least in Proposition 7.1 of my paper with Tim de Laat https://arxiv.org/abs/1401.3611
The fact that this central extension is represented by a bounded $2$-cocycle follows, for simple Lie groups, from the well-known classical work of Guichardet-Wigner, see also the paper Shtern, A. I. Bounded continuous real 2-cocycles on simply connected simple Lie groups and their applications. Russ. J. Math. Phys. 8 (2001), no. 1, 122–133.
The case of semisimple Lie groups follows by decomposing into simple parts.
Actually that at the QI-level the extension $Z\to G\to G/Z$ is trivial (and hence represented by a bounded 2-cocycle) is essentially immediate, using that $Z$ is discrete and that $G/Z$ has a simply connected closed cocompact (solvable) subgroup (this is why I used $T$ in my sketch of argument as a comment to Uri's answer).
Oh, I had never realized that. (and I had not read the comments...) Thanks.
Thanks for the reference Mikael.
Thanks, Mikael, I should have remembered that you had this figured out. I incorporated this into my answer for the benefit of possible future readers. I hope you don't mind...
@UriBader Sure, it it better if your answer is complete.
|
2025-03-21T14:48:30.390642
| 2020-04-23T08:40:51 |
358302
|
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|
Stack Exchange
|
Convergence of heavy-ball method for non-convex optimization
The heavy-ball method (also called gradient descent with momentum) is commonly used in optimization. The update rule can be written as:
$$x_{t+1}=x_t-\eta\nabla f(x_t)-\beta (x_t-x_{t-1})$$
Suppose $\eta$ and $\beta$ are constants. It has been proven that for convex function $f$, the Heavy-ball method converges globally. However, for non-convex setting, I can not find any literature which proved the convergence of the algorithm. Here convergence for a non-convex function means that for any $\epsilon>0$, we can find a point $x$ such that $\|\nabla f(x)\|_2\le \epsilon$. Does the algorithm really have convergence guarantee in non-convex settings? If it does, what is the convergence rate of the algorithm? Thank you!
Have you looked at this paper "Heavy-ball method in nonconvex optimization problems"?
It seems that the I must spend a lot of money buying the pdf. Does this paper prove the convergence of heavy-ball method?
https://sci-hub.tw/https://link.springer.com/article/10.1007/BF01128757
Thank you very much. I have browsed the paper successfully. This paper proved the convergence of heavy-ball method, but it did not show a convergence rate. Can we show that under $L$ smooth property, after $T$ steps, we can find a point $x$ such that $|\nabla f(x)|^2\le O(L/T)$, for example? (This bound applies for traditional gradient descent.)
Please read this paper "iPiano: Inertial Proximal Algorithm for Nonconvex Optimization". In Section 4, heavy ball method reduces the to case where $g=0$. Moreover, in this paper, the asymptotic convergence rate that $\min\limits_{0\leq i\leq N}|x_{i+1}-x_i|^2=O(1/N)$ was given. And if $\alpha_n$ is constant, you can prove that $|\nabla f(x_n)|^2$ has the same asymptoticc convergence rate. For non asympototic convergence rate, I think you need the KL inequality.
Thank you very much!
May I consult you further about the stochastic case of heavy ball method? For SGD algorithm, we can prove convergence if one decays the step size gradually. However, I can not prove convergence of heavy ball method using the same technique as in the deterministic case.
|
2025-03-21T14:48:30.390824
| 2020-04-23T09:04:45 |
358303
|
{
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|
Stack Exchange
|
Enumerating antichains modulo permutation
I encountered the following combinatorics problem in my research, and I'd like to know if there is a reference or an easy solution for such a problem.
Given a partially ordered set $\mathscr P$, an antichain is a subset of $\mathscr P$ such that no two elements can be compared.
Fix positive integers $n$ and $k$. Define $\mathscr A_{n,k}$ to be the set of all size-$k$ antichains taken from the poset of subsets of $[n] = \{1, \cdots n \}$. Define $A_{n,k} = \mathscr A_{n,k} / S_n$ where the permutation group $S_n$ acts on $\mathscr A_{n,k}$ by extending the natural action of $S_n$ on $[n]$; for $\sigma \in S_n$ and $\mathscr U \in \mathscr A_{n,k}$, $\sigma( \mathscr U) := \{\sigma(U) | U \in \mathscr U \}$. Alternatively,
\begin{align*} & \mathscr A_{n,k} := \left\{ \{ U_1, \dots, U_k \} | \emptyset \neq U_i \subseteq [n], \forall i \neq j, U_i \not\subseteq U_j \right\} \\ & A_{n,k} := \mathscr A_{n,k} / S_n \end{align*}
Problem. Enumerate all elements of $A_n = \bigcup_{k=1}^\infty A_{n,k}$.
The problem is in data science context; I'd like to have an algorithm that computes, say, $A_{10}$ that doesn't take too long. But I'll be computing it only once, so it's fine if the computation takes a whole week (but not a whole year!). Naive enumeration and checking for all overlaps will take quite long; for example, the set of all size-5 antichains of $[10]$ has size $(2^{10})^5 \approx 10^{15}$.
Remark. One way to understand $\mathscr A_{n,k}$ is to see it as the collection of all hypergraphs on $[n]$ such that one hyperedge never includes another; it is of completely opposite nature to abstract simplicial complexes. Also, following Brendan McKay's observation, I edited the question and used the antichain terminology.
You do realize that all size $k$ sets automatically satisfy the non-inclusion condition.
@kodlu, $|\mathscr{U}|=k$, not $|U_i|=k$.
ok, now I see it...
For $k=1,2,\ldots$, $|A_{10,k}| = 10,95,1627,51194,2599518,170816889,\ldots$. That much took about 70 seconds but $|A_{10,7}|$ took 40 mins so far and I'm going to bed. I'll report tomorrow. I think you won't actually be able to store the full set.
$|A_{10,7}|=11988870351$ in 80 mins. Estimate 800 billion for $A_{10,8}$ in about 130 hours -- I won't do it. Incidentally these
are called antichains.
There are 252 sets of size 5 and trivially none of them is a subset of another. There are $2^{252}-1$ ways to choose a non-empty collection of 5-sets. Moreover, the group action can reduce the number by at most $10!$. This means that $A_{10}$ is greater than $10^{69}$.
I think $A_7$ would be fairly easy and $A_8$ would be very difficult.
See http://oeis.org/A306505 which includes the antichain with just the empty set.
Thanks a lot for the comments and computations, @BrendanMcKay. This seems quite hard after all.
@BrendanMcKay The lower bound to $A_{10}$ should be smaller than that because we have to eliminate inclusion comparability.
If you build it out of sets that all have the same size (I used size 5), then there is no inclusion possible.
oh, I thought you meant having five subsets, rather than subsets of size five.
The number for 8 points is about 1.4e18. Too many unless you have 1000 years of cpu time handy.
The following is basic information which should be part of an introductory combinatorics (or even computer science) course. It should help give you a good sense of scale.
It is clear that the power set of the power set of the base set contains (an inverse image of) your desired collection, and so an upper bound on the size of your collection is $2^n$ where $n=2^{10}$. As observed by Brendan McKay, quotienting out by all permutations puts a small dent in $n$, the exponent, but we still get $2^{1000}$ as a rough estimate after quotienting.
As Brendan also observed, one can take the middle sized sets and look at antichains formed from them. Here one has about $2^m$ choices, each of them mapping to part of your collection, with $m$ being the nth central binomial coefficient. Again being optimistic about the quotient map, $m$ is roughly 230 for a base set of size 10.
Finer estimates are available, but note that including a small or large set takes a good sized bite out of allowed middle sized sets for the antichain, and so the real exponent won't be much more than 230. Even if it is as high as 500, what are you going to do with it?
Gerhard "Sometimes Enough Is Too Much" Paseman, 2020.04.24.
|
2025-03-21T14:48:30.391177
| 2020-04-23T10:02:56 |
358305
|
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|
Stack Exchange
|
How can we know the well-foundedness of $\epsilon_0$?
I think the question can be quite philosophical, but I see that $WF(\epsilon_0)$ is widely accepted as one of the attributes of the natural numbers.
Gentzen proved $Con(PA)$ with $PRA+WF(\epsilon_0)$.
The proofs of some theorems of arithmetic, such as Goodstein's theorem or the termination of Hydra Game, essentially rely on $WF(\epsilon_0)$.
However, I'm curious if there ever is an justification about this. I'm aware of that ZFC provides such justification, but also I couldn't convict myself whether the set $\omega$ in ZFC (one of the interpretation of it) really gives us the natural number $\mathbb{N}$.
(Just to be clear: the statement of $WF(\epsilon_0)$ itself doesn't require any set theory - it can be coded into arithmetic statement.)
On the other hand, highly unlikely, but if ever $WF(\epsilon_0)$ turns out to be equivalent with $Con(PA)$ or $Con(PA+Con(PA))$, all of which have $\mathbb{N}$ as a model, we know that it is true.
If I understand formalism correctly, even the strictest formalists wouldn't deny these consistency statements because they can't make any deduction without having actual natural numbers or strings, which is equivalent to having PA.
I am relatively new on metamathematics field, and I learned logic from a formalist. ZFC seems just another random formal theory to me, except that I can't do second-order logic without some decent set theory.
So my question is this: is there any non set-theoretical justification for $WF(\epsilon_0)$, which involves the natural number $\mathbb{N}$?
Second-order arithmetic can prove $WF(\varepsilon_0)$ too.
Timothy Chow has written an expository article on the consistency of arithmetic and the induction up to $\epsilon_0$. It appeared in the Mathematical Intelligencer (http://timothychow.net/consistent.pdf). Maybe he can say something about this.
@Wojowu Actually I mentioned in OP about second-order scheme and I'm not satisfied with it. I believe that proving something with second-order arithmetic is technically equivalent to proving it with a set theory, which must be backed up with a first-order characterization such as ZFC.
@godelian Not exactly. I've read the article a month ago. Perhaps I skimmed a lot, but I couldn't get any information further than "$PRA+WF(\epsilon_0)$ is neither stronger nor weaker than $PA$, and it proves $Con(PA)$." No claim that $WF(\epsilon_0)$ actually holds for the natural numbers.
Let me use the notation $\omega_n$ for an exponential tower of $\omega$'s of height $n$, so $\omega_{n+1}=\omega^{\omega_n}$. Then $\epsilon_0$ is the supremum of $\{\omega_n:n\in\omega\}$. PA proves well-foundedness of $\omega_n$ for each individual $n$, but it needs a separate proof for each $n$. If you believe PA (which you apparently do) and you believe in the natural numbers (which, unlike your formalist teacher, you also apparently do), then you should accept that "for all $n$, $\omega_n$ is well-founded." The well-foundedness of $\epsilon_0$ then follows, because, if there were an infinite decreasing sequence in $\epsilon_0$, its first term and therefore all its terms would be below $\omega_n$ for some $n$.
The funny thing is that you can even prove in PA that PA separately proves well-foundedness of $\omega_n$ for each n. You just can't go from this to "$\omega_n$ is well-founded for all $n$" in PA (by Lob). But if PA is sound then this conclusion does follow and therefore $\epsilon_0$ is well-founded.
@NikWeaver Can you give a source for the claim in your first sentence or a brief sketch of how one would show that?
Might take a while to fully understand the scheme, but I think this is the answer I was looking for. Thanks!
@JoshuaZ I don't have a source ... it's "clear" because you can give a uniform description of the $n$th proof $P_n$, and seeing that each of them is a proof in PA is easy. So I guess it would be pretty easy to write down a proof of this in PA, just very tedious.
@NikWeaver yes, somehow it's very hard to accept PA without implicitly accepting it's soundness, but this is strictly stronger!
@cody I'm not sure what you mean. The consistency of PA + all true $\Pi_1$ sentences implies $WF(\epsilon_0)$. This is strictly weaker than "PA is sound".
Looks like I got the point. PA can prove $WF(\omega_n)$ for all $n$, and accepting the natural numbers gives us the $\omega$-conjunction rule(which lives outside PA), validating $\forall n. WF(\omega_n)$, that is $WF(\epsilon_0)$. I almost forgot about these $\omega$-stuffs.
@NikWeaver I just mean that psychologically it's hard to imagine accepting PA without accepting PA + "PA is sound" which also implies $WF(\epsilon_0)$.
@cody Oh, I see, you just meant that "PA is sound" is stronger than PA. That's right.
|
2025-03-21T14:48:30.391798
| 2020-04-23T10:41:06 |
358307
|
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|
Stack Exchange
|
What is the meaning of lifting property? and some question in $\infty$-category
When I learn model category, the important compute tool is the lifting property between $(\operatorname{Cof}, \operatorname{Fib} \cap W)$ and $(\operatorname{Cof} \cap W, \operatorname{Fib}) $, where $W$ is class of weak equivalence. This means for commutative diagram whose vertical map are $f$, $g$, $f \in \operatorname{Cof}$, $g \in \operatorname{Fib}\cap W$. So in localization of $C$ by $W$, we must have this filler of $f$, $g$.
so my first question is why we need filler of $g$, $f$ in localization of $C$ which must exist one in $C$, this property means what in the category special or in general.
we can regard filler of $f$, $g$ as every $\Delta^1 \times \Delta^1 \rightarrow N(C)$, which two vertical map are $f$, $g$, which is restriction of $\Delta^3 \rightarrow N(C)$. So we say commutative diagram has the structure of the 3-dimension. but in $\infty$-category. we proof many statements relative composition, homotopy using $\operatorname{sk}_1(\Delta^3) \rightarrow X$ in which $d_0(\sigma)$ , $d_3(\sigma)$ is commute, then $d_1(\sigma)$ iff $d_2(\sigma)$, $\sigma: \Delta^3 \rightarrow X $ by definition of $\infty$-category. So this is also question existing the structure of 3-dimension.I know model category and $\infty$-category using extension in different way. in model category means commutative diagram extension to $\Delta^3$. in $\infty$-category, i have two $x_1,x_2:\Delta^2 \rightarrow X$ which $d_0 (x_1) = d_2(x_2)$. when $d_0(x_2) \circ d_1(x_1) = d_1 \circ d_2(\sigma)$ this composition in meaning of $\infty$-category. we can extension $x_1 ,x_2$ in $\Delta^3 \rightarrow X$, then we have diagram commutative. this two way i feel some adjoint, but I don't know mean what adjoint. But I know there are some redicular. because $N(C)$ is infinity category. So just is a question of $\operatorname{sk_1}(\Delta^3) \rightarrow X$ where X is just simplicial set. in this morphism, this two extension have what relevant?
in my view $\Delta^3$ like association, maybe $\Delta^n$ like higher association. this intuition is true? if it is true what is special description.
|
2025-03-21T14:48:30.391959
| 2020-04-23T10:59:42 |
358308
|
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|
Stack Exchange
|
Characterisation of minimal projective resolutions via the Euler characteristic
Let $A$ be a finite dimensional $K$-algebra (where $K$ is a field) and $M$ a finitely generated $A$-module.
Let $\psi: 0 \rightarrow P_r \rightarrow ... \rightarrow P_0 \rightarrow M \rightarrow 0$
be a complex of $A$-modules such that $P_0 \rightarrow M$ is the projective cover of $M$ and $P_r \rightarrow P_{r-1}$ is injective and where the $P_l$ are projective for $l=0,1,...,r$ and the maps $d_i :P_i \rightarrow P_{i-1}$ are minimal, that is $d_i(X) \neq 0$ for any direct summand $X$ of $P_i$. (so $\psi$ is nearly a minimal projective resolution of $M$, the only thing missing is exactness)
Set $P_{-1}:=M$, then it is well known that the Euler characteristic $\chi(\psi):=\sum\limits_{i=-1}^{r}{dim(P_i)}$ is equal to zero in case $\psi$ is exact.
I remember that there is a converse to that, namely that $\psi$ is exact in case $\chi(\psi)=0$ under some extra conditions on $\psi$. I forgot where I saw that. Maybe someone knows a reference for what I have in mind.
Question: Are the conditions here enough to prove that $\psi$ is exact in case $\chi(\psi)=0$ or what other conditions are needed (is there a reference in case this is well known?)?
Without more conditions it's not true.
Take the Nakayama algebra with two simples and indecomposable projectives
$$P(1)=\matrix{1\\2\\1}\hspace{1cm}\text{and}\hspace{1cm}P(2)=\matrix{2\\1}$$
Then there is a complex
$$0\to P(2)\to P(1)\to P(1)\to P(1)\to P(1)\to\matrix{1\\2}\to 0$$
which satisfies your conditions, is not exact, but has zero Euler characteristic.
Thanks, I leave the thread open for some suggestions how one could make this true. Maybe one can hope this is true at least for acyclic quiver algebras, since there the possible summands of the projective terms "get smaller" in higher degrees. So one probably cant take repretitions.
@Mare For the quiver $1\to2\to3\to4\to5$ with $\text{rad}^3=0$ consider the complex $0\to P(5)\to P(3)\to P(1)\to S(1)\to0$.
|
2025-03-21T14:48:30.392121
| 2020-04-23T13:10:44 |
358313
|
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|
Stack Exchange
|
Kneser subgraph with high chromatic number
For positive integers $n\geq 2k$, it is known that the chromatic number of the Kneser graph $K_{n,k}$ is $n-2k+2$. Moreover, the Schrijver graph $S_{n,k}$ (definition in the same link), which is a subgraph of $K_{n,k}$, also has chromatic number $n-2k+2$. The number of vertices of $S_{n,k}$ is $\binom{n-k+1}{k}$.
Is it known whether there is a subgraph of $K_{n,k}$ with chromatic number $n-2k+2$ whose number of vertices is polynomial in $n$ and $k$?
This is related and may be of interest. https://arxiv.org/pdf/1502.00699.pdf
The number of vertices of $S_{n,k}$ seems to be wrong. For instance, when $k=2$ it should be ${n\choose 2}-n\ne\binom{n-2+1}{2}$.
The general answer is no, at least when $n$ is close to $2k$.
Theorem 3 in this paper shows that a graph with odd girth $\geq 2d+1$ and chromatic number $>m$ contains more than
$$
\frac{(m+d)(m+d+1)\dots(m+2d-1)}{2^{d-1}d^d}
$$
vertices. The parameters for the Kneser graph $K_{n,k}$ are $m=n-2k+1$ and $d=\bigl\lceil\frac k{n-2k}\bigr\rceil$, so, say, for $n\approx 2k+\sqrt k$ the bound is already exponential in $\sqrt k$.
|
2025-03-21T14:48:30.392223
| 2020-04-23T14:25:23 |
358318
|
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|
Stack Exchange
|
If $n$ is not a power of 2 then the dual Stiefel-Whitney class $\bar{w}_{n-1} = 0$
Stiefel-Whitney classes are invertible and for $w$, the Stiefel-Whitney class of the tangent bundle of $M$, we have its inverse $\bar{w}$. I want to prove that if $n$ is not a power of 2 then the dual Stiefel-Whitney classes $\bar{w}_{n-1} = 0$.
For the Steenrod square $\mathrm{Sq} : H^\Pi (M) \rightarrow H^\Pi(M)$ we can define its inverse $\bar{Sq}$. Then we can see easily that $\langle \bar{Sq}(x), \mu \rangle = \langle \bar{w} \cup x , \mu \rangle$. Set $x$ as suitable element in $H^0(M)$ we can prove that $\bar{w}_n = 0$. And I also think of proving $\bar{w}_{n-1} = 0$ by letting $x$ be Poincaré dual to $\bar{w}_{n-1}$. However I can't check that $\langle \bar{Sq}(x) , \mu \rangle = 0$.
The proof is in Massey's "On the Stiefel-Whitney classes of a manifold" American Journal of Mathematics 1960; see Corollary 1: https://www.jstor.org/stable/pdf/2372878.pdf.
|
2025-03-21T14:48:30.392321
| 2020-04-23T15:16:19 |
358325
|
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|
Stack Exchange
|
Strictly isotropic and strictly coisotropic submanifolds
Let $M$ be a $2n$-dimensional symplectic manifold. A question: are there special terms for isotropic submanifolds of $M$ of dimensions $<n$ (i.e., isotropic submanifolds that are not Lagrangian) and for coisotropic submanifolds of $M$ of dimensions $>n$ (i.e., coisotropic submanifolds that are not Lagrangian)? Of course, the terms “strictly isotropic” and “strictly coisotropic” come to mind, but they are already used to mean other things, see e.g. R.Baer, Linear Algebra and Projective Geometry, second ed., Dover Publ., Mineola, NY, 2005; M.S.Borman, F.Zapolsky, Quasimorphisms on contactomorphism groups and contact rigidity, Geom. Topol. 19 (2015) 365-411.
I think I've seen "non-maximally isotropic" somewhere.
I've seen "subcritical(ly)" used in the isotropic case, in the context of an h-principle for "subcritical isotropic immersions/embeddings" (see e.g. the book "Introduction to the h-principle" by Eliashberg and Mishachev). So if you said "subcritically isotropic" or "supercritically coisotropic," some subset of the symplectic population would know what you were talking about, and my vote would be that this become standard terminology.
I must be getting conservative in my old age: I'd advise against coining new terminology unless it makes your paper substantially more readable. Definitely "subcritical" (and possibly "supercritical") would be understood by today's symplectic community as KSackel says, but in 50 years' time the community may no longer exist, and it will make your paper much easier to understand if you say "coisotropic of dimension > n" rather than introducing a new adjective which may or may not survive.
Of course, I reserve the right to ignore my own advice (in the past, present and future).
As a counterpoint to this: "non-Lagrangian" would work (e.g. non-Lagrangian coisotropic).
Note that in the context of KAM theory, given a Hamiltonian system with $n$ degrees of freedom, isotropic invariant tori of dimensions $<n$ are often said to be lower dimensional, and coisotropic invariant tori of dimensions $>n$ are sometimes said to be higher dimensional.
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2025-03-21T14:48:30.392596
| 2020-04-23T15:16:37 |
358326
|
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|
Stack Exchange
|
An orientable surface that cannot be embedded into $\Bbb R^3$?
I previously asked this question on MSE, without success.
By Whitney's embedding theorem, every 2-dimensional manifold (aka. a surface) can be embedded into $\Bbb R^4$.
Now, Wikipedia states in this paragraph that we can even embedd into $\Bbb R^3$ if the surface is
compact and orientable, or
compact and with non-empty boundary.
In the second bullet point, it is clear that I cannot drop either of the conditions.
It is not clear to me why I can't drop "compact" in the first bullet point.
Question: Is there an orientable but non-compact surface that does not embedd into $\Bbb R^3$?
Total space of canonical bundle is orientable but can't be embedded in $\mathbb{R}^3$.
@XTChen Geat, Thanks! But I checked Wikipedia on this, and I admit, I wasn't able to understand much of it. Can you provide a more accessable explanation as an answer?
Sorry. Total space of canonical bundle on $\mathbb{P}^1$ is non orientable.
It is important to specify clearly which definition of "surface" one is using. There is one definition that I believe is at least in the plurality, was what Ian Richards used as his definition: a Hausdorff space that is locally homeomorphic to the plane and has a countable basis for its topology. If instead one considers, e.g., the Prüfer manifold (which has no countable basis but is Hausdorff and locally 2-dimensional), is known not to be metrizable. I doubt that it can be embedded in 3-space.
It seems to me that every orientable surface is indeed embeddable in $\mathbb{R}^3$. By Ian Richards' classification theorem (https://www.ams.org/journals/tran/1963-106-02/S0002-9947-1963-0143186-0/S0002-9947-1963-0143186-0.pdf), we know that a non-compact orientable surface is determine by the pair $Y\subset X$ where $X$ is its space of ends (homeomorphic to a compact subset of the Cantor space) and $Y$ is the closed subset of ends with genus.
When genus is finite, i.e. $Y=\varnothing$, the embedding is easily realised as a genus-$g$ surface with a copy of $X$ removed. When genus is infinite, one simply starts with a sphere with a copy of $X$ removed, and adds smaller and smaller handles accumulating to all points of $Y$ (but not to points of $X\setminus Y$).
To ensure this can be done, observe that $Y$ has a countable dense subset. By taking a countable basis of neighborhoods of each of them, we get a countable family $(U_n)_{n\in\mathbb{N}}$ of open subsets of $\mathbb{S}^2\setminus X$ and we want to put one handle in each of them, but no family of handles should approach any point outside $Y$. We put a handle in $U_1$ (i.e. we remove two discs from $U_1$ and glue a handle to them), then inductively add one in the smallest $U_n$ not containing both gluing circle of any previous handle.
I don't think you need to use the classification of noncompact surfaces --- only the existence of a proper real-valued function (and hence an exhaustion by compact manifolds), and the fact that if $\Sigma$ is a compact surface and we are given any embedding of $\partial \Sigma \hookrightarrow \partial\left([0,1] \times \Bbb R^2\right)$, we may extend it to an embedding $\Sigma \hookrightarrow [0,1] \times \Bbb R^2$. This follows from classification of compact surfaces and isotopy extension.
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2025-03-21T14:48:30.392835
| 2020-04-23T15:20:14 |
358327
|
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|
Stack Exchange
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Separation of balls in the torus
Let $X_1, \dots, X_N$ be $N$ balls of radius $R<<1$ in $[0,1]^d$ such that $N R^d \leqslant R^{\alpha}$ for some $\alpha > 0$ and $d(x_i,x_j)\geqslant 2R$ for any $i\not=j$.
The assumption means that there are very few balls and I would like to know if one can separate them : finding $\delta_1<\delta_2$ (depending only on $\alpha$ and $d$) such that for all $x_i, x_j$, $d(x_i,x_j)\leqslant R^{1-\delta_1}$ or $d(x_i,x_j)\geqslant R^{1-\delta_2}$.
Obiously, this is false in general (consider balls of radius $R$ around $\{(x,0) , \;x\in [0,1]\}$ in the two-torus) but is this is the only example ?
Is it true that one can always find a partition of the balls such that two partition are distant and, in a given partition, the balls are "in the same direction" ?
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2025-03-21T14:48:30.392921
| 2020-04-23T15:38:05 |
358328
|
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|
Stack Exchange
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Leaves of stable foliation of holomorphic Anosov diffeomorphism
I'm trying to understand the first half of the paper "Holomorphic Anosov systems" by E. Ghys (the journal reference is Inventiones mathematicae volume 119, pages 585–614(1995)). My question is about a particular claim that Ghys makes. I am a finishing undergraduate, so I suspect my main issue is missing background.
The setup is a compact complex manifold $M$ equipped with a holomorphic Anosov diffeomorphism $\phi$, and it is assumed that the unstable foliation has complex dimension 1. In the Proof of Proposition 2.2 (in the penultimate sentence of the second paragraph), Ghys says that the leaves of the stable foliation are simply connected, and even diffeomorphic to some euclidean space. He later also uses this fact in the proof of his Theorem B.
I don't understand why this is true and I wasn't able to find some hint or explanation in Ghys' references. Is there some simple explanation, or some good reference, for this fact? Is it special to all of the assumptions I stated in the setup, or is it something more general?
I am not an expert on several complex variables, but I suspect that it is a linearisation result. If you have a hyperbolic fixed point, then you have a local linearisation, meaning your stable manifold is locally $\mathbb{C}^k$. The functional relation allows you to extend to a global diffeomorphism, since your map is invertible. In the Anosov setting, you should have a similar uniformisation for all stable leaves.
The fact that a stable manifold is diffeomorphic to a (real) Euclidean space is a consequence of the Stable Manifold Theorem, see for instance [Katok&Hasselblatt, Introduction to the modern theory of dynamical systems. chap 6 sec 4]:
By the Stable Manifold Theorem for every $p\in M$, there is a local stable manifold $W^s_{loc}(p)$ which is diffeomorphic to an Euclidean ball, and the global stable manifold satisfies $W^s(p)=\bigcup_{n\geq 0} f^{-n}( W^s_{loc}(f^n(p)))$. Thus $W^s(p)$ is a monotone union of Euclidean balls, hence $W^s(p)$ itself is diffeomorphic to an Euclidean ball. Here is nothing to do with holomorphic map.
To determine the complex structure of a stable manifold when we have a holomorphic diffeomorphism, is more difficult. When $W^s(p)$ has dimension 1, it can be shown that $W^s(p)$ is biholomorphic to $\mathbb{C}$ but not $\mathbb{D}$, see for instance [Bedford&Smillie, Polynomial diffeomorphisms of $\mathbb{C}^2$: currents, equilibrium measure and hyperbolicity. Thm 5.4]. It is conjectured that if $f:M\to M$ is a holomorphic diffeomorphism with an invariant hyperbolic set $\Lambda$, then a $k-$dimensional stable manifold of $p\in\Lambda$ is biholomorphic to $\mathbb{C}^k$. As far as I known this is still open.
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2025-03-21T14:48:30.393108
| 2020-04-23T15:39:38 |
358329
|
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|
Stack Exchange
|
The Hopf conjecture for products and slight modification
Perhaps this post won't get too much attention, and I apologize if this is deeply charged with a self perspective more than general facts.
I would like to know why do people believe that the standard Hopf conjecture for $S^2\times S^2$ is true. I have a few knowledge about the recent works on the literature, in particular the works of Bettiol-Mendes, and which concerns the generalized Hopf conjecture
$M\times N$ does not admit admit a metric with positive sectional curvature,
the work of Lee Kennard (although I must admit I am not quite familiar with this).
Still, I haven't understood yet the plausibility of the Hopf conjecture for $S^2\times S^2$. It is probably more a deep flaw on my background than anything else. However, although I can't justify too much, apart from my own tries on these conjectures, that perhaps this conjecture may be not true. Instead, what if the following was true:
Let $(M,g)$, $(N,h)$ closed Riemannian manifolds with positive sectional curvature. Then, $M\times N$ admits a metric of positive sectional curvature if, and only if, $\chi(M\times N) > 0$.
Note that this conjecture, if true, asserts that $M^{2n}\times M^{2m+1}$ would not care a metric of positive sectional curvature, but $S^2\times S^2$ may be.
This conjecture is false provided $M$ or $N$ do not care a metric of positive sectional curvature (for example, there are exotic spheres that do not bound spin manifolds that don't even admit a metric of positive scalar curvature).
I haven't seen this last conjecture on literature, but it relates two of the known Hopf conjectures, and I would like to know: what do you think about the plausibility of it? Is there more reason to believe in it than the standard Hopf conjecture for spheres?
I wouldn’t say that it’s believed to be true. There have been serious efforts to construct a counterexample. What is perhaps surprising is how hard this turns out to be.
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2025-03-21T14:48:30.393254
| 2020-04-23T16:01:10 |
358333
|
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|
Stack Exchange
|
One-sided Jacobi SVD and Divide&Conquer SVD stability and cost
I'm studying SVD, in particularly the Jacobi SVD and Divide&Conquer SVD algorithms. I can't find anything on the stability and error analysis on these methods. Also can someone show show me what's the computational cost of both algorithms?
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2025-03-21T14:48:30.393305
| 2020-04-23T16:17:37 |
358336
|
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|
Stack Exchange
|
Does there exist a non-degenerate symmetric combinatorial 3-design?
Is there a non-degenerate 3-design where the number of blocks equals the number of points?
Non-degenerate in this context means that a point is incident with at least 2 and at most #blocks-2 blocks.
No. If $b=v$ and $t>2$ the block design equations give that either $k=v$ (in which case there may only be one block and the design is trivial) or $k=v-1$ so that the design is also trivial (see Lemma 2.6 in the paper of Hughes cited below). This is why only quasi-symmetric $t$-designs are discussed in texts when $t>2$.
Hughes, D. R. On $t$-designs and groups. Amer. J. Math. 87 (1965), 761-778.
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2025-03-21T14:48:30.393379
| 2020-04-23T16:25:11 |
358337
|
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|
Stack Exchange
|
When are all signatures represented by units?
What conditions on a number field imply that every signature is represented by a unit? Are there conditions that imply that it is not the case that every signature is represented by a unit?
For example, if K is a cubic cyclic number field with odd class number then all signatures are represented by units by Theorem V in [Armitage, J., & Fröhlich, A. (1967). Classnumbers and unit signatures. Mathematika, 14(1), 94-98. doi:10.1112/S0025579300008044].
What are some references for articles on this topic?
The structure of the group of units modulo totally positive units is studied here; see also the references in this article. Since then, relevant articles by Dummit and Dummit & Voight have appeared.
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2025-03-21T14:48:30.393457
| 2020-04-23T16:31:16 |
358338
|
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|
Stack Exchange
|
Low-complexity method for sub-matrix inversion
Assume that $\mathbf{A}$ be an $N\times N$ matrix. We know that the complexity of the computation of matrix inversion is $\mathcal{O}(N^3)$. Let define $\mathbf{D}=\mathbf{A}^{-1}$. Now, assume that $\mathbf{D}_n$ is any $n\times n$ sub-matrix of $\mathbf{D}$, where $n\ll N$. Is there any way to compute $\mathbf{D}_n$ exactly with complexity oreder related to $n$ (not $N$)? If not, is there any better approximation than $\mathbf{D}_n\simeq\mathbf{A}_n^{-1}$, where $\mathbf{A}_n$ is corresponding sub-matrix?
related: https://mathoverflow.net/q/297567/11260
Complexity completely unrelated to $N$ seems impossible, since $D_n$ depends nontrivially on all entries of $A$, so we have at least to read them all in $O(N^2)$.
The complexity of matrix inversion is not $O(N^3)$. Using Coppersmith-Winograd it is $O(N^{2.376})$.
@RobertIsrael $\text{True complexity of matrix inversion (unknown)} \subset O(N^{2376}) \subset O(N^3)$, so technically OP is correct (or at least "as wrong as you", depending on how you interpret $O()$).
Good point. I should have said it is better than $O(N^3)$.
Unfortunately, the following solutions are not independent of $N,$ (they can't be, see comment to your question) but may still help reduce the computations:
By using Schur complement, you can compute the upper $n\times n$ matrix $(\boldsymbol{D}_n)$ by using an inverse of an $(N-n)\times (N-n)$ matrix and an $n\times n$ matrix. (You can also try a few interesting approximations based on your matrix structure which may turn out to be better than direct inversion of the $n\times n$ submatrix.)
See: https://en.wikipedia.org/wiki/Schur_complement
You may use the method (in the link below) based on matrix inversion using Cholesky decomposition but stop during the second step (equation solving) at the required number of elements.
See: https://de.mathworks.com/matlabcentral/fileexchange/41957-matrix-inversion-using-cholesky-decomposition
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2025-03-21T14:48:30.393613
| 2020-04-23T17:26:16 |
358343
|
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|
Stack Exchange
|
What is the exceptional divisor of a divisorial contraction?
Let $\pi \colon X\to Y$ be a divisorial contraction between $\mathbb{Q}$-factorial terminal varieties. Let $E\subset X$ be the exceptional divisor and let $\pi(E)=D$ be its image. It is true that $\rho(E/D)=1$ ? Or at least over an open subset $U$ of $D$? More precisely: if $C$ and $D$ are two curves contracted by $\pi$ onto points of $U$, is it true that some multiples of $C$ and $D$ are numerically equivalent in $E$ ? (or course this holds if one asks for a numerical equivalence in $X$).
If $D$ has codimension $2$ in $Y$, it is true as $\pi$ is just the blow-up of $D$ on an open subset of $D$. If $\dim(X)=3$ and $D$ is a smooth point or a point with mild singularities, this is also true as $E$ is a weighted projective space (by a result of Kawakita). In "On extremal rays of the higher dimensional varieties" of Tetsuya Ando (Inventiones, 1985), it is said that if $X$ is smooth and $\dim(D)=\dim(X)-3$, the general fibres are either $\mathbb{P}^2$ or $\mathbb{P}^1\times \mathbb{P}^1$; in the second case, are the two fibrations of $\mathbb{P}^1\times \mathbb{P}^1$ numerically equivalent in $E$ ?
Isn't the blowup of an ordinary 3-fold nodal singularity a counterexample? The exceptional divisor is a quadric, i.e. isomorphic to $\mathbb{P}^1\times\mathbb{P}^1$.
@TomDucat: Your example is not $\mathbb Q$-factorial.
Of course I mean an algebraically irreducible singularity which is locally analytically isomorphic to the standard ordinary node $x^2+y^2+z^2+t^2=0$. To be precise, this is case (E3) of Theorem 1.32 on page 28 of Kollár & Mori.
@TomDucat thanks, that is a good example. I guess that one can also do a similar example with a curve $C$ in a fourfold, with a divisor birational to $\mathbb{P}^1\times\mathbb{P}^1\times C$. Do you have some specific example like this?
I guess the obvious 4-fold example would be to take a product of my example with a curve $C$, but I don't have anything more interesting than that off hand. And just a comment on why my example wasn't covered by Kawakita's result: he classifies 3-fold divisorial contractions to non-Gorenstein points.
Thanks. I would be very interested to see an example on rational fourfolds, but with $C$ being any curve.
@TomDucat: I still don't think your example is $\mathbb Q$-factorial: It is the cone over a smooth quadric surface $S$. The two Weil-divisors divisors corresponding to the two rulings do not have a Cartier multiple.
No, it is only locally analytically isomorphic to that. In other words, take something like $x^2+y^2+z^2+t^2+t^3=0$.
|
2025-03-21T14:48:30.393810
| 2020-04-23T18:19:07 |
358349
|
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|
Stack Exchange
|
Minimal number of "words" that contain all possible pairs of letters in all position pairs
Defining a word as sequence of ordered letters ($1$..$q$ letters) of length L,
what is the minimal number of words such that among the entire list of words,
at every pair of positions, I can find any two letters?
for example, for $q=3$ and $L=2$
here is the minimal list:
$$1 1,
2 2,
3 3,
1 2,
2 3,
3 1,
1 3,
2 1,
3 2,$$
total $q^2$ words are needed.
but for $L=3$ the minimal number is still $q^2$, obtained by:
$$1 1 1,
2 2 1,
3 3 1,
1 2 2,
2 3 2,
3 1 2,
1 3 3,
2 1 3,
3 2 3,$$
for $L=4$ the number is different...
what is the minimal number of words for $(q,L)$ and specifically, what is the asymptotic value for $L\gg 1$?
Thanks for you answers!
This is not research level and may be closed. Anyway what you want is an orthogonal array of strength $t=2$ and repetition $\lambda=1$ over a $q-$ary alphabet.
This is an array of symbols from the alphabet, where looking down any two columns each possible pair of symbols appear once. There is a large literature on this subject.
@kodlu if there is a large literature, why is not it research level?
The OP's response to @kodlu's question indicates that the object in question is related to, but distinct from, orthogonal arrays. I think this question should get a fair shot at being answered rather than getting closed.
@FedorPetrov If I play devil's advocate, there is a large literature on computing derivatives of elementary functions too.
@NajibIdrissi true, and even more literature on unhappy love. But I guess the literature we did not all study at school is mentioned. At least I did not study these arrays. The question looks non-trivial and I would like to see the answer.
For $q$ a power of a prime, the maximal word length for which $q^2$ words suffice is $L=q+1$. In particular, for $q=3$, $L=4$ there are still 9 words 0000, 0111, 0222, 1102, 1210, 1021, 2201, 2012, 2120. The construction comes from the projective plane.
A Possible Solution: Actually given your family for $q=L=3,$ we can duplicate it (side by side) with the goal of achieving the goal for $L=6.$ Here, all $(i,j)$ with $i\leq 3,j\geq 4,\quad i\neq j\pmod 3$ positions will also achieve all 2-tuples by design.
So you can insert extra rows at the bottom with x incidating don't care to cover those $i,j$ in the left and right halves with $i=j \pmod 3.$
$$
\begin{array}{c}
111~111\\
221~221\\
331~331\\
122~122\\
232~232\\
312~312\\
133~133\\
213~213\\
321~321\\
\\
111~222\\
222~111\\
111~333\\
333~111\\
222~333\\
333~222\\
\end{array}
$$
If I'm not missing something this gives a gain in efficiency.
Now recursively double the new full array, this will then require you covering some other pairs whose indices are the same mod 6.
Earlier Discussion:
What you may use is an orthogonal array of strength $=2$ and repetition $\lambda=1$ over a
$q−$ary alphabet. This is an array of symbols from the alphabet, where looking down any two columns each possible pair of symbols appear once ($\lambda=1)$.
The exactly once property is not necessarily needed for your purposes, but it yields constructions that are uniform with respect to the symbol pair positions. This may be natural feature of an optimal solution.
Hedayat and Sloane have a nice book on orthogonal arrays. The talk at the Isaac Newton institute, video and slides available here is a nice overview.
Rao’s and Bose-Bush’s bounds apply generally, but can be improved in special cases.
From page 10 of the slides, the general lower bound (translated into your variables with $N$ the number of rows)
$$
N\geq L(q-1)+1
$$
can be obtained. Strengthening this in general is hard since the standard results rely on the condition $\lambda-1$ being nonzero modulo some parameter but your $\lambda=1.$
Thank you for a thorough answer! My question however is slightly different than what the answer would suggest. In the scenario that I describe, you don't have to have the same repetition of every pair. It merely requires the existence of every pair...
Also, that lower bound seems pretty weak to me. Given for example L=2, the minimal N for such an array is q^2 words. Given q>>2 this would make this bound entirely irrelevant, am I missing anything? Thanks again!
I do not understand your naive bound. Could you please elaborate?
@FedorPetrov, it was wrong, deleted. Should teach me not to post too late at night.
So, your answer asymptotically provides about $q(q-1)\log_2L$, right?
First, let me elaborate the upper bound from @kodlu’s answer. If $q$ is a power of a prime, then $q^2$ words suffice for $L=q+1$ (for $L=q$ it is almost trivial, improving by $1$ needs just a bit more). Then, doubling $L$ increases the number of words by $q(q-1)$, so for those values of $L$ it suffices to take $q(q-1)\log_2\frac L{q+1}+q^2$ words.
Let me show a somewhat close lower bound. Let $w$ be the number of words; set $k=w-q(q-1)+1$. Take any $k$ words. Let $v_i$ be a vector composed from all $i$th entries of the $k$ words. If two of those vectors, say $v_i$ and $v_j$, coincide, this means that at most $w-k=q(q-1)-1$ words differ in positions $i$ and $j$, so not all pairs are covered.
Thus, we have $L$ distinct vectors in $[q]^k$, so $L\leq q^k$, or $w\geq q(q-1)+1+\log_qL$.
Therefore, the growth rate is indeed logarithmic (but the constant at the logarithm is yet unclear).
Addendum. Let me present an example for $L=q+1$, when $q$ is a power of a prime.
Consider an affine plane $\mathbb F_q^2$. All lines in it are partitioned into $q+1$ classes $C_1,\dots,C_{q+1}$ of mutually parallel lines (one class consists of all lines with equations of the form $ax+by=c$ with fixed $(a:b)$). Enumerate the lines in each class by numbers from 1 to $q$.
For each point $p\in \mathbb F_q^2$, take a word $w_1\dots w_{q+1}$ where $w_i$ is the number of the line in $C_i$ passing through $p$. Then, for any two classes $C_i$ and $C_j$ and for any two lines in them, the lines meet at a unique point, which means exactly what we need to get in the $i$th and $j$th positions.
The base of logarithm can be improved to $q-1$ (for $q>2$ by looking at $k=w-q(q-1)+1+2t$ vectors and saying they should differ in at most $2t+1$ positions. But the improvement is not that large...
Thank you! Could you clarify "for = it is almost trivial" ?
I’ve added an example even for $L=q+1$.
BTW, a ‘trivial’ example I meant consists of all arithmetic all progressions module $q$, if $q$ is prime.
|
2025-03-21T14:48:30.394249
| 2020-04-23T18:46:02 |
358350
|
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|
Stack Exchange
|
Implementation of Koebe–Andreev–Thurston circle packing?
The circle packing theorem (Koebe–Andreev–Thurston theorem) claims for a planar graph, we can pack disjoint circles, such that: the circles correspond to vertices and the disks are tangent if the vertices are adjacent.
I would like to implement this algorithm in computer code (input: graph, output: circle packing). Where can I find a readable (not too complex) version of the procedure? I do not need the theory behind it. I know it has been implemented (e.g. in KnotTheory package in Mathematica), but I'm interested in the algorithm itself, not a software that does it.
It might help to examine existing code, e.g., Robert Syzymanski's C++ code, or Ken Stephenson's MatLab GOPack code.
Actually, I found an excellent Python implementation by David Eppstein in his PADS library. Works perfectly!
Here is Ken Stephenson's book on the subject: "Introduction to Circle Packing: The Theory of Discrete Analytic Functions". The wikipedia page has many more references - https://en.wikipedia.org/wiki/Circle_packing_theorem
Ah, and see the references given at this very closely related post - Koebe–Andreev–Thurston theorem - where can I find a proof?
|
2025-03-21T14:48:30.394378
| 2020-04-23T19:17:44 |
358351
|
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|
Stack Exchange
|
Residue of the canonical sheaf along subvariety
Let $S$ be a smooth projective surface over an
algebraically closed field $k$ and $C \subset S$ a singular curve. Let us denote by $K_S$ the class of canonical divisor of $S$ and $\mathcal{O}(K_S)$ the canonical sheaf. If $\Omega_{S/k}$ is the sheaf of Kähler differentials, then smoothness of $S$ and $\mathcal{O}(K_S)=\Omega_{S/k}^2$ imply that $\mathcal{O}(K_S)$ is generated locally by an the element $\frac{1}{f} dx \wedge dy$ where $dx, dy$ are local basis elements of $\Omega_{S/k}$ and certain $f \in K(S)$.
I found recently a statement (I forgot the source, I think it was Kollar's Resolutions of Singularities) that locally the residue of $\frac{1}{f} dx \wedge dy$ gives a generator $r \vert _C$ of restriction $\mathcal{O}_S(C+K_S) \vert _C$.
The residue $r$ is characterized by equation
$$f^{-1}dx \wedge dy= \frac{df}{f} \wedge r $$
My questions are quite general:
I'm looking for sources dealing with this residue construction and it's properties in detail. How far can this constrution be generalized? ie is it possible to construct for instead of a smooth surface for an arbitrary smooth projective variety $V$ of algeb closed field from the local generator of it's canonical sheaf $O_V(K_V)$ a residue along a codimension one subscheme $D$? And why gives this residue a local generator of $O_V(D +K_V) \vert _D$?
It looks like a far generalization of the concept of residues from basic complex analysis but since I still have nowhere found any books treating this construction in detail I don't know which interesting properties the residue by this constructions obtains and why the construction can be considered as a generalization of the story from complex analysis.
Poincaré residue. You can also search for « Poincaré residue » in Griffiths-Harris.
@abx: I see now that by the construction the restriction of the residue $r \vert _C$ lands in $\mathcal{O}_S(C+K_S) \vert _C$. But it is moreover claimed that $r \vert _C$ is locally a generator. Do you know an argument?
@abx: Your notation $K_C$ confuses me. $C$ is a singular curve, thus we cannot expect that it has a canonical divisor $K_C$ (or canonical sheaf $\mathcal{O}_C(K_C)$). What do you mean by $K_C$?
Oops! Sorry, I overlooked that you take $C$ singular. I will delete my comment which is totally irrelevant.
I believe that what you wrote is not entirely correct and that might be the reason that it does not seem to work out. As a response to one of your comments, $K_C$ actually makes sense in this case, because it is a Cartier divisor in $S$ and hence Gorenstein. I'll elaborate on this below.
So, I don't think $\frac{1}{f} dx \wedge dy$ generates $\mathscr O(K_S)$ (locally). If you think about it, essentially by definition $dx \wedge dy$ generates that sheaf (locally), so the additional $\frac 1f$ gives it a twist. My guess is that what you saw said that $f$ should be a local equation for $C$ and if it is that then $\frac{1}{f} dx \wedge dy$ generates $\mathscr O(K_S+C)$ locally. If this is OK with you, then it is no surprise that its restriction to $C$ generates the restricted sheaf.
And, indeed, there is a more general framework for this. Using your notation, let $V$ be a Cohen-Macaulay (CM) scheme (or variety, or complex analytic space) and let $D\subseteq V$ be an effective Cartier divisor (this can be weakened to $D$ being a pure codimension $1$ subscheme/subvariety, but in that case you have to be a bit more careful about the sheaves that appear, so I'll stick to this case for simplicity).
I will use the notation $\omega_X=\mathscr O_X(K_X)$ for any $X$ for which this makes sense.
Then we have a short exact sequence:
$$
0\to \mathscr O_V(-D)\to \mathscr O_V\to \mathscr O_D \to 0
$$
The long exact $\mathscr Ext$ sequence associated to the functor $\mathscr Hom_V(\__, \omega_V)$ (and the fact that $V$ is CM) gives the
short exact sequence:
$$
0\to \omega_V\to \omega_V(D)\to \omega_D \to 0
$$
The map $\omega_V(D)\to \omega_D$ is essentially what you are looking for and can be interpreted so that $\omega_V(D)|_D\simeq \omega_D$.
If $V$ is Gorenstein and $D$ is a Cartier divisor (then $D$ is also Gorenstein), then all of these sheaves are locally free of rank $1$ on their support, so you can find local generators that look pretty much the ones you wrote down. If $V$ is smooth, then it is Gorenstein and then every codimension $1$ subvariety is a Cartier divisor, so this covers the case you were wondering about.
A few explanations, mainly sparked by the questions in the comments:
The canonical sheaf, Hom long exact sequence, and all that jazz: I wasn't entirely sure how much to include and what generality to go to. The formula involving Ext and projective space and the codimension of V in that projective space is a special case of Grothendieck duality and it holds any time a Gorenstein scheme is contained in another. (The proper way to write this would be writing out the "usual" (total) derived functor format and then take cohomology of that). In any case, this implies that $\omega_D\simeq \mathscr Ext^1_V(\mathscr O_D, \omega_V)$ if you start writing out the Hom-Ext sequence, the first Hom will be zero, then you have two terms (the first two terms of that ses) and then this Ext. All other terms are zero, because if the first term is locally free, then all higher Exts are zero. This is how you get that short exact sequence.
As far as using the Ext formula vis-a-vis the projective space, you can do that, you'll get the same thing. Grothendieck duality is nice in this regard, it is natural, so you can apply it consecutively for a composition. This is actually a nice exercise if you have never done it.
Being Gorenstein is equivalent to being CM and the canonical sheaf being locally free of rank 1. The definition of CM implies that (local) hypersurfaces inherit being CM (essentially the definition of depth does this). And the above short exact sequence implies the local freeness of the canonical sheaf.
The author not explicitely says that that $f$ is the equation for $C$ but it seems as you explaned to be reasonable to assume that this was what he has in mind. One remark: you used several times that if a nice enough scheme $V$ is Gorenstein and $D$ a Cartier divisor, then $D$ is also Gorenstein. Could you give a reference for a proof of this fact?
And could you explain why the application of $\mathscr Hom_V(__, \omega_V)$ to $0\to \mathscr O_V(-D)\to \mathscr O_V\to \mathscr O_D \to 0$ and using that $V$ is CM give us the sequence $0\to \omega_V\to \omega_V(D)\to \omega_D \to 0$?
Applying Hom fonctor we obtain l.e.s.
$$ ... \to \mathscr{Ext}^{i}{O{V}}(\mathcal{O}D, \omega{V}) \to \mathscr{Ext}^{i}{O{V}}(\mathcal{O}X, \omega{V}) \to \mathscr{Ext}^{i}{O{V}}(\mathcal{O}X(-D), \omega{V}) \to \mathscr{Ext}^{i+1}{O{V}}(\mathcal{O}D, \omega{V}) \to ...$$
How to continue? By the the way are you sure that you mean $\mathscr Hom_V(_, \omega_V)$ and not $\mathscr Hom{O_{\mathbb{P}^N}}(_, \omega{\mathbb{P}^N})$?
I'm asking this, because by definition $\omega_V:= \mathscr{Ext}^{N-n}{O{\mathbb{P}^N}}(\mathcal{O}{\overline{X}}, \omega{\mathbb{P}^N}) \vert _X$ where $V \subset \mathbb{P}^N$ and application if long Ext sequence suggest that some of them we intend to identify with dualizing sheaves.
On the other hand I don't know how to identify the terms in the long exact sequence from above.
@TimGrosskreutz: I think I responded to all of your comments. I hope this helps. :)
|
2025-03-21T14:48:30.394969
| 2020-04-23T19:39:55 |
358353
|
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|
Stack Exchange
|
Inequality on the Hellinger distance between Poisson and mixture of Poisson
Let $H$ denote the Hellinger distance; i.e., for two discrete distributions $p,q$ (identified with their pmf) over $\mathbb{N}$,
$$
H(p,q)^2 = \frac{1}{2}\sum_{n=0}^\infty \left(\sqrt{p(n)}-\sqrt{q(n)}\right)^2 = 1-\sum_{n=0}^\infty \sqrt{p(n)q(n)}
$$
and $\operatorname{Poi}(\lambda)$ for the Poisson distribution with parameter $\lambda$.
I would like to know whether there exists a simple proof of the following result:
Let $\lambda>0$, and $\alpha\in[0,1]$. Define $Q= \frac{1}{2}(\operatorname{Poi}((1+\alpha)\lambda) +\operatorname{Poi}((1-\alpha)\lambda) )$. Then $H(\operatorname{Poi}(\lambda),Q) \lesssim \alpha^2\lambda$.
I can easily prove that the square Hellinger distance satisfies $H(\operatorname{Poi}(\lambda),Q)^2 \lesssim \alpha^2\lambda$, but the only proof I know for the correct (and tight) statement above is Lemma 4.1 of [VV17]; and I feel there should (?) be a neater and more generalizable argument to establish this seemingly simple bound.
As far as I can tell, the crux of the difficulty (at least in my attempts) is to tightly (upper) bound the quantity
$$
\sum_{n=0}^\infty \frac{\lambda^n}{n!}\sqrt{\frac{e^{\lambda\alpha}(1-\alpha)^n+e^{-\lambda\alpha}(1+\alpha)^n}{2}}
$$
where the obvious attempts (e.g., AM-GM) only yield the loose bound mentioned above.
[VV17] Valiant, Gregory; Valiant, Paul. An automatic inequality prover and instance optimal identity testing. SIAM J. Comput. 46 (2017), no. 1, 429--455.
https://theory.stanford.edu/~valiant/papers/instanceOptFull.pdf
I have an indirect (but simple) proof, which relies on the chi-squared distance $\chi^2(p,q) = \sum_n \frac{(p(n)-q(n))^2}{q(n)}$ as a proxy, along with the fact that $H(p,q)^2 \leq 1\land \chi^2(p,q)$ for all discrete distributions $p,q$.
Lemma. Let $\lambda>0$, and $\alpha\in[0,1]$. Define $Q= \frac{1}{2}(\operatorname{Poi}((1+\alpha)\lambda) +\operatorname{Poi}((1-\alpha)\lambda) )$. Then $1\land \chi^2(\operatorname{Poi}(\lambda),Q) \leq \alpha^2\lambda$.
Proof. We can assume in the rest of the proof that $\alpha^2\lambda \leq 1$, as otherwise there is nothing to prove. For convenience, write $P:= \operatorname{Poi}(\lambda)$. We can express the pmf of $Q$ as
\begin{equation}
Q(n) = P(n)\cdot\frac{e^{-\lambda\alpha}(1+\alpha)^n + e^{\lambda\alpha}(1-\alpha)^n}{2} \tag{1}
\end{equation}
for $n\in \mathbb{N}$. It follows that
$$\begin{align*}
\chi^2( Q , P ) &= -1 + \sum_{n\in\mathbb{N}} \frac{Q(n)^2}{P(n)} \\
&= -1 + e^{-\lambda}\sum_{n\in\mathbb{N}} \frac{\lambda^n}{n!}\left( \frac{e^{-\lambda\alpha}(1+\alpha)^n + e^{\lambda\alpha}(1-\alpha)^n}{2}\right)^2 \tag{2}
\end{align*}$$
Focusing on the last sum, we expand the square and compute it explicitly:
\begin{align*}
\sum_{n\in\mathbb{N}} &\frac{\lambda^n}{n!}\left( \frac{e^{-\lambda\alpha}(1+\alpha)^n + e^{\lambda\alpha}(1-\alpha)^n}{2}\right)^2\\
&= \sum_{n\in\mathbb{N}} \frac{\lambda^n}{n!} \frac{e^{-2\lambda\alpha}(1+\alpha)^{2n} + e^{2\lambda\alpha}(1-\alpha)^{2n} + 2(1-\alpha^2)^n}{4}\\
&= e^{\lambda}\cdot\frac{e^{\lambda\alpha^2} + e^{-\lambda\alpha^2}}{2} \tag{3}\,.
\end{align*}
Plugging this in (2), we get
$$
\chi^2( Q , P ) = -1 + \frac{e^{\lambda\alpha^2} + e^{-\lambda\alpha^2}}{2} \leq \lambda^2\alpha^4 \tag{4}
$$
where for the last inequality we used our bound $\lambda\alpha^2\leq 1$, and the fact that $\cosh x \leq 1+x^2$ for $|x|\leq 1$ (I didn't try to optimize the constant).
|
2025-03-21T14:48:30.395176
| 2020-04-23T21:36:21 |
358362
|
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|
Stack Exchange
|
Is Axiom of Choice equivalent to its version for families of sets, indexed by ordinals?
Is Axiom of Choice equivalent to the following statement?
Axiom of Ordinal Choice: For any ordinal $\lambda$ and any indexed family of sets $(X_\alpha)_{\alpha\in\lambda}$ there exists a function $f:\lambda\to\bigcup_{\alpha\in\lambda}X_\alpha$ such that $f(\alpha)\in X_\alpha$ for all $\alpha\in\lambda$?
https://www.math.purdue.edu/~hrubin/JeanRubin/Papers/WO-choice.pdf
https://math.stackexchange.com/questions/3140508/how-strong-is-the-axiom-of-well-ordered-choice
@Wojowu Thank you for the link. Does the well-ordered connection of sets $C(WO,\infty)$ in their paper mean the same as in my question?
Yes, I believe it does.
It is strictly weaker than choice. This is explained in Asaf Karagila's answer at MSE: the $L(\mathbb{R})$ of $L$ + $\aleph_1$-many Cohen generics witnesses this.
(There the principle is phrased for well-ordered index sets, but you can always pass from a well-ordered index set to an ordinal index set.)
As Taras Banakh comments below, there is a superficially-similar fact: namely that over NBG without global choice, global choice is equivalent to $Ord$-indexed choice. The proof is simple: for $\alpha\in Ord$ let $C_\alpha$ be the set of well-orderings of $V_\alpha$. From a choice sequence $(\triangleleft_\alpha)_{\alpha\in Ord}$ we get a well-ordering of $V$ as follows: set $x\prec y$ iff $rk(x)<rk(y)$ or $rk(x)=rk(y)$ and $x\triangleleft_{rk(x)}y$.
Note the key role of set-sized choice above. More abstractly (and being a bit sloppy with details), what's being used here is that in NBG without global choice every class is "locally well-orderable" in the sense of being a well-ordered union of well-orderable sets (via Foundation plus Set-Choice). We deduce global choice from well-ordered-class choice by first applying WOCC to well-order the locally-well-ordered indexing class of our global choice instance, and then applying WOCC to that.
However, in ZF we don't have anything like this. For example, amorphous sets are not locally well-orderable in the relevant sense. Instead, we get the following:
(ZF) Suppose $\lambda$ is a limit ordinal and full choice holds in $V_\alpha$ for each $\alpha<\lambda$. Then the following are equivalent:
Every $s\in (V_\lambda\setminus\{\emptyset\})^\lambda$ has a choice function.
$V_\lambda$ is well-orderable.
Very interesting. My question was motivated by a (simple) observation that the Axiom of Global Choice in NBG is equivalent to the standard AC + Axiom of choice for families indexed by the class of all ordinals. (Yhis follows from von Neumann cumulative hierarchy).
@TarasBanakh I've added a bit about why that analogy breaks down.
|
2025-03-21T14:48:30.395380
| 2020-04-24T00:44:13 |
358373
|
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|
Stack Exchange
|
Strict inequality in decoupling inequality
I am working on the decoupling inequality developed by Bourgain and Demeter: https://arxiv.org/abs/1604.06032.
Is there an example where we have strict inequality in Theorem 1.1, say in the case $n=2$ with $\delta^{-\alpha}$ losses in the power? Here by stricty inequality I mean $\alpha>0$ is a fixed positive constant depending on $p$ only.
This question may seem stupid, but I have not found a reference for that.
Edited: precisely, I am asking the following question.
For $g\in L^1([0,1])$ and $I\subseteq [0,1]$, define $E_I g(x,y)=\int_I g(s)e^{2\pi i (s x+s^2 y)}ds$. Is it true that for every $2\leq p\leq 6$ and every $\epsilon>0$, there is a constant $c_\epsilon>0$, depending on $p$ and $\epsilon$ only, such that for any $g\in L^1([0,1])$, any$\delta\in 2^{-2\mathbb N}$ and any ball $B$ of radius $\delta^{-1}$, we have
$$
\left\| Eg\right\|_{L^p(w_B)}\geq c_\epsilon \delta^{\epsilon}\left(\sum_{j=1}^{\delta^{-1/2}}\left\|E_{[(j-1)\delta^{1/2},j\delta^{1/2}]}g\right\|^2_{L^p(w_B)}\right)^{1/2}?
$$
The answer to your edited question is "no", as follows by applying the Bourgain-Demeter inequality with $\epsilon$ replaced by (say) $\epsilon/2$.
(Assuming of course that you meant to write $\delta^{-\epsilon}$ instead of $\delta^\epsilon$.)
@TerryTao Thanks for the answer! I actually kind of figured out the answer in some other way. We may choose $g$ such that $Eg_j$ has very sparse physical support. Then the LHS becomes essentially $\left|\left|E_j\right|{L^p}\right|{l^p(j)}$, and this can be arbitrarily smaller than the RHS for $p>2$. Is that correct?
If you are quantifying over all $g$ in your question, then yes. If you are asking whether the inequality stated holds for at least one $g$ (which is the usual interpretation of what it means for the opposing inequality valid for all $g$ to be "sharp"), then no.
@TerryTao Sorry for my unclear presentation. I actually meant that $c_\epsilon$ is independent of $g$; I re-edited my question. Also, I do think that I meant $\delta^{\epsilon}$ instead of $\delta^{-\epsilon}$ since I would allow an $\epsilon$-loss in the reverse direction.
Certainly for $n=2$ the $\delta^{-\epsilon}$ factor can not be dispensed with entirely. It is known that
$$ || \sum_{n\leq N} e(nx + n^2 t) ||_{L^4(dxdt)} \approx N^{1/2} \log^{1/6} N $$
This is essentially the two dimensional case of Vinogradov's mean value theorem, and is discussed in Bourgain's original 1991 discrete restriction papers on the periodic Schrodinger equation. A precise asymptotic for this quantity can be found in: https://mathscinet.ams.org/mathscinet-getitem?mr=2661311.
It is unclear what one should expect in the higher dimensional cases. It is also unclear, for instance, if the $N^{\epsilon}$ factor can be removed at the endpoint in the multi-linear restriction theorem.
Even reducing the $\delta^{-\epsilon}$ factor to a logarithmic factor for $n=2$ (for decoupling or the discrete parabolic restriction theorem) is open, interesting, and seems to require new ideas.
Thanks for your answer. However, I guess you are not exactly answering my question. I am considering an example which gives strict inequality, not that $\epsilon$ can be removed or not.
I don't understand what you mean by "gives strict inequality", then.
I still don't understand what you are looking for (following your revision to the question). The claim in Theorem 1.1 is that $\epsilon$ can be taken to be any real number greater than $0$ provided, of course, the implied constant is taken sufficiently large depending on $\epsilon$.
Basically I am asking whether or not the reverse inequality is true for decoupling.
Thomas: I suggest you write down the precise inequality you would like to see disproved. If there was an extermal or almost extremal for some fixed $\epsilon$, then the result wouldn't hold for all $\epsilon>0$, contradicting the statement of the theorem.
Sure. I will do that.
|
2025-03-21T14:48:30.395661
| 2020-04-24T00:53:06 |
358374
|
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Equivalence between Ramanujan and Selberg conjectures
At first the Ramanujan conjecture for automorphic forms and the Selberg conjecture appear to be understood as totally independent. However, they are now known to be tighyly connected once viewed in the light of global adelic automorphic representations. I would like to understand how far this is true. I recall the different versions, say in $GL(2)$.
Ramanujan-Petersson conjecture. Let $f$ be a modular form (we could state it for a Maass form too) of weight $k$ and level $1$, if the $a_n$ denotes its Fourier coefficients then
$$a_n \ll n^{(k-1)/2}.$$
Selberg conjecture. Let $f$ be a Maass form and $\lambda_1$ its smallest nonzero eigenvalue for the Laplacian. Then
$$\lambda > \frac 14.$$
Now, there is a "more modern" version in terms of automorphic representations.
"Automorphic" Ramanujan conjecture. Let $\pi$ be an automorphic cuspidal representation of $GL(2, \mathbb{A})$. It decomposes by Flath's theorem as $\pi \simeq \bigotimes_v \pi_v$. Then $\pi_v$ is tempered for all place $v$.
Are these three formulation equivalent, or only in a mild sense? More precisely,
does the automorphic version imply the two others?
do the two "local" versions imply the automorphic one?
are the two "local" version equivalent? (or: does one of them imply the automorphic version, by a kind of rigidity of the automorphic representations)
(I haven't found proper answer to these questions and would be glad to know if I missed papers or notes about it.)
Let $f$ be an automorphic form corresponding to an automorphic representation $\pi =\otimes_v \pi_v$ of $GL_2(\mathbb A_{\mathbb Q})$.
For an unramified prime $p$, the following are equivalent (for $f$ holomorphic of weight $k$):
$|a_p| \leq 2p ^{(k-1)/2 }$.
For all $n$, $|a_{p^n} |\leq (n+1) p^{n (k-1)/2} $.
For all $n$, $|a_{p^n} |\ll p^{n ((k-1)/2 +\epsilon) } $.
$\pi_p$ is tempered.
There is an analogous local equivalence for Maass forms, where for the standard normalization you should take $k=1$. There is also, a more complicated, statement about ramified primes.
For the place $\infty$, the following are equivalent (for $f$ Mass)
$\lambda \geq \frac{1}{4}$
$\pi_{\infty}$ is tempered.
For holomorphic forms, $\pi_{\infty}$ is automatically tempered.
Putting this together, we see that $\pi$ is tempered at all places if and only if $a_n \ll n^{ (k-1)/2+ \epsilon}$ for all $n$, plus some additional conditions at the ramified primes, and (if $f$ is Maass) $\lambda_1 \geq 1/4$.
Because all cuspidal automorphic representations of $GL_2(\mathbb A_{\mathbb Q})$ correspond to cuspidal holomorphic or Maass forms, the Ramanujan conjecture for $GL_2(\mathbb A_{\mathbb Q})$ is equivalent to these statements for all holomorphic or Maass forms simultaneously.
In summary, the automorphic one implies the two local ones, the two local ones imply the automorphic (at least away from ramified primes), but the local ones are not equivalent in any meaningful sense.
An additional comment: No rigidity of the kind you suggest is expected to be provable over number fields as far as I know, but over function fields, one knows from work of V. Lafforgue and P. Deligne a very strong rigidity: A cuspidal automorphic form tempered at one unramified place is tempered at every unramified place.
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2025-03-21T14:48:30.395881
| 2020-04-24T01:11:45 |
358376
|
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|
Stack Exchange
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Is there an orbit map without path lifting property?
I am looking for an example of a topological group $G$ acting by homeomorphisms on a metrizable space $X$ such that the orbit map $X\to X/G$ doesn't have the path lifting property, that is, there is a path in $X/G$ that cannot be lifted to $X$.
EDIT: Let's also assume that every $G$-orbit is a closed subset.
Non-example: If $G$ is a compact Lie group, then the orbit map has the path lifting property (in fact, there is a a slice). Palais gave a generalization to proper Lie group actions.
How about $\mathbb{Z}$ acting on $S^1$ by an irrational rotation? The induced topology on $S^1/\mathbb{Z}$ is indiscrete, so we can make a path that just jumps from one point to another. This cannot be lifted to $S^1$ because every path in $S^1$ is either constant or its range is all of $S^1/\mathbb{Z}$ under the quotient map.
Thank you! Dense orbit is enough. I was focused on the case of closed orbits.
I think, there are even examples even when $G$ is compact and the action is free, but, of course, $G$ is not a Lie group. Check the example I gave in my answer here.
@MoisheKohan: I don't know why a path lifting property (for the orbit map of a free action by a compact group) has to imply the map is a principal bundle (or fibration). Ruling our principal bundles is much easier, see Remark 2 in https://mathoverflow.net/questions/57015/which-principlal-bundles-are-locally-trivial.
Remark: the question doesn't refer to any topology on $G$ (still, the "non-example" says that the lifting property holds assuming that $G$ carries some group topology such that the action is continuous, plus some extra-assumptions).
@YCor: not sure what you are saying. As stated, the question is about any topological group, and non-example says that the desired group cannot be a compact Lie group, or more generally Lie group that acts Palais-properly. Naturally, a simpler example would be better. All I want is to gain intuition on paths that don't lift.
This is not what I am saying: I am saying that there is a good chance that this example does not have a path-lifting property. In this example, a totally disconnected group acts on a 1-dimensional space such that the quotient is 2-dimensional.
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2025-03-21T14:48:30.396080
| 2020-04-24T04:06:03 |
358384
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Element of Weyl chamber contracting $\mathbb{A}^n_k$ to a point
Let $G$ be a connected reductive group over an algebraically closed field $k$ of characteristic 0. Fix a Borel subgroup $B$ and a maximal torus $T \subset B$. Let $P \subset G$ be a parabolic subgroup containing $B$, and let $L \subset P$ be a Levi subgroup containing $T$.
Suppose given an action of $L$ on $\mathbb{A}^n_k$ such that (1) the commutator subgroup $[L,L]$ fixes $\mathbb{A}^n_k$, and (2) $\mathbb{A}^n_k$ is a toric variety for some quotient $T'$ of the torus $L/[L,L]$. In other words, $T' \cong \mathbb{G}_m^n$, and under this isomorphism, $T'$ acts on $\mathbb{A}^n_k$ in the natural way. The origin $0 \in \mathbb{A}^n_k$ is fixed by $T'$, and there are "many" one-parameter subgroups $\lambda': \mathbb{G}_m \to T'$ such that
$\lim_{t \to 0} \lambda(t) \cdot z = 0$ for all $z \in \mathbb{A}^n_k$. On the other hand, the composition $T \to L \to T'$ of the inclusion map and the quotient map allows us to associate to any one-parameter subgroup $\lambda: \mathbb{G}_m \to T$ a one-parameter subgroup $\lambda': \mathbb{G}_m \to T'$ (given by the composition of $\lambda$ with the map $T \to T'$).
My question is this: does there exist some choice of $\lambda: \mathbb{G}_m \to T$ such that (1)
the corresponding one-parameter subgroup $\lambda': \mathbb{G}_m \to T'$ contracts all of $\mathbb{A}^n_k$ to $0$, and (2) $\lambda$ lies in the Weyl chamber of our fixed Borel subgroup $B$ (equivalently, $\langle \lambda, \alpha \rangle > 0$ for all positive roots $\alpha$ of $G$ with respect to $T$ and the base given by $B$).
In the case where $G = P = \mathrm{SL}_n$, $B = L$ is the subgroup of upper triangular matrices, and $T$ is the subgroup of diagonal matrices, everything works out very nicely. In this case, $T = T'$ and $\lambda(t) = \mathrm{diag}(t^{m_1},\dots,t^{m_n})$ for some $m_i \in \mathbb{Z}$, and the condition that $\lim_{t \to 0} \lambda(t) z = 0$ for all $z \in \mathbb{A}^n_k$ is the statement that $m_i > 0$ for all $i$. The positive roots of $G$ with respect to $T$ are the maps $T \to \mathbb{G}_m$ sending $\mathrm{diag}(t_1,\dots,t_n) \mapsto t_i/t_j$ for $i < j$, so we may pick for $\lambda$ any choice of $m_i > 0$ such that $m_i > m_j$ for $i < j$.
In general, I think if we embed $G$ in $\mathrm{SL}_n$ and pick a basis where $T$ and $T'$ are diagonalized, we might be able to do something similar. I'm not convinced that this works, though. Essentially, the condition that $\lambda$ contracts all of $\mathbb{A}^n_k$ to a point says that
$\langle \lambda, e_i \rangle > 0$ for all $1 \leq i \leq r$, where the characters $e_i: T \to \mathbb{G}_m$ factor through $T'$ and form a certain basis for the character group $\mathcal{X}(T')$. It's not clear to me that these hyperplanes will in general intersect all the hyperplanes
$\langle \lambda,\alpha \rangle > 0$ for $\alpha$ a positive root. I'm not particularly comfortable with arguments about roots of reductive groups and one-parameter subgroups, so I would really appreciate any suggestions you may be able to give me!
This (somewhat convoluted) setup comes from a situation involving the local structure theorem on spherical varieties. I didn't think it was essential to the question, but I'd be happy to provide more context about that if it would help!
In your example with ${\rm SL}_n$ you seem to abuse the notation for $n$. In this case the dimension of $T'=T$ is $n-1$, so it should act in ${\Bbb A}_k^{n-1}$, not in ${\Bbb A}_k^{n}$.
The answer to your question, as stated, is No. Take $G={\rm SL}_2$, $P=B$, $T'=T$.
Note that you do not specify the isomorphism $T'\to {\Bbb G}_m^n$. Let us take the following isomorphism:
$$T'\to {\Bbb G}_m\colon\,{\rm diag}(s,s^{-1})\mapsto s^{-1}\text{ for }s\in k^\times.$$
Then our torus $T=T'$ acts on $\Bbb A^1$ by
$${\rm diag}(s,s^{-1})\colon\, x\mapsto s^{-1}x.$$
Now if you take $s=\lambda(t)=t^m$, then your condition (1) that $\lambda$ contracts all of $\mathbb{A}^1_k$ to $0$ means that $m<0$, while your condition (2) that $\lambda$ lies in the Weyl chamber of our fixed Borel subgroup $B$ means that $m>0$. Contradiction....
Oh, of course! I was worried there was some basis compatibility problem, but I got lost in the details and didn't realize it was so simple. Thank you for your help!
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2025-03-21T14:48:30.396414
| 2020-04-24T05:21:14 |
358385
|
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List of long open, elementary problems which are computational in nature
I would like to ask a question of a similar vein to this question.
Question: I'm asking for a list of long open problems which are computational in nature which a beginning graduate student can understand. One problem per answer, please.
Meaning of "beginning graduate student": anyone who can solve all the problems on a pure mathematics qualifying exam at a top 30 institution in the U.S.
Meaning of "computational in nature": By this, I do not mean a computational task which can be executed by a computer, but rather a problem where one must compute some object (e.g. topological invariant, closed formula, etc.) associated to some mathematical object. Example: calculating the homotopy groups of a sphere.
Meaning of "not too famous": (Same as in this question.): Roughly, if there exists a whole monograph already dedicated to the problem (or narrow circle of problems), no need to mention it again here. I'm looking for problems that, with high probability, a mathematician working outside the particular area has never encountered.
Meaning of "long open": (Same as in this question): The problem should occur in the literature or have a solid history as folklore. So I do not mean to call here for the invention of new problems or to collect everybody's laundry list of private-research-impeding unproved elementary technical lemmas. There should already exist at least of small community of mathematicians who will care if one of these problems gets solved.
This doesn't quite fit the bill - so I am leaving it as a comment - but perhaps there is something of interest in the computational number theory text by Hutz: https://www.maa.org/press/maa-reviews/an-experimental-introduction-to-number-theory
Close to duplicate: https://mathoverflow.net/q/112097/30186
Question: Is this intended to include aspects of famous problems which have non-famous computational parts? For example, obviously the existence of odd perfect numbers clearly falls into your too famous category, but there's also a body of literatre of proving results of the form "the kth largest prime factor of an odd perect number must be at least n" which are highly computational and would be likely be unknown to most people. My guess is that you don't want problems like this that are closely connected to famous problems.
@JoshuaZ : I think user676464327 wants objects that are known to exist but that we don't know as "explicitly" as we would like. Moreover, there should be some "interesting" reason we don't know; it's not that nobody has bothered to leave a computer running long enough, or that no computer is big enough. So something like the Atlas of Lie Groups except that that's probably too famous, and a lot of the computations have been completed now.
Finding more Hadamard matrices might fit the bill, though I don't know if that would be considered too famous of a problem.
@TimothyChow That was my impression also but it wasn't completely clear to me from the phrasing of the question.
Question for the OP: when you say "computational", can this include problems for which there is not a known algorithm that would find the desired number, but one where we now the desired number (minimizing something like "distance" or energy) must exist by some kind of compactness principle?
@YemonChoi I was hoping for problems for which there is a way of computing the actual number rather than just knowing the existence of the number.
Problem: extend the table of known van der Waerden numbers from 7 to 8 entries.
Given $K\geq 2$ colors, the length $N=W(L,K)$ of the smallest set of colored integers $\{1,2,3,\ldots N\}$ with a monochromatic arithmetic progression of length $L\geq 3$ is only known in 7 cases.
The seventh entry on the list was computed in 2012: $W(3,4)=293$, meaning 293 is the smallest integer $N$ such that whenever the set of integers $\{1,2,3,\ldots N\}$ is 3-colored, there exists a monochromatic arithmetic progression of length 4.
Adding one more entry to this table seems to meet the four criteria in the OP: a problem which is "understandable", "computational", "not too famous" (unlike the Ramsey numbers), "long open" (van der Waerden's paper, which started the search for $W(2,L)$, is from 1927).
This is a cool problem, because you can approach it in a purely computational way and try to find N non-deterministically and then verify it as an upper bound (which shouldn't be too hard since for any given N there are a lot very trivial colorings), or you can try and build the number combinatorially, and both are interesting.
Radziszowski's dynamic survey on small Ramsey numbers contains further examples with a similar flavor, but graph-theoretic rather than number-theoretic.
Note sure whether it is too famous (it has a monograph).
Find the Moore graph of girth $5$ and degree $57$, if one exists.
That means, find a graph with diameter $=2$ (i.e., the distance between any two vertices is at most two), girth $=5$ (i.e., the shortest cycle has length five) and degree $=57$ (i.e., any vertex has exactly 57 neighbors).
All Moore graphs are known, except this one.
If it exists, it must have 3250 vertices, so still quite accessible.
Finding the set of forbidden minors for the class of toroidal graphs (finite graphs that can be embedded in the torus with no crossings). By the Robertson–Seymour theorem, this set is finite, but it is only partially known, and the finiteness proof is ineffective. A recent paper by Myrvold and Woodcock states, among other things, that the current list of known obstructions (over 17000 forbidden minors!) is unlikely to be complete.
I imagine that the proof can be mined to give an upper bound though right? Obviously, it's likely that that bound is hilariously large though.
@cody : I don't think so. Robertson and Seymour proved Wagner's conjecture that any minor-closed family has a finite forbidden minor characterization. If all you know is that a family is minor-closed, that does not imply any upper bound on the number of minors or the size of the minors. To get a bound, you would have to use some special fact about toroidal graphs, and I don't think anything like that is known.
Certainly, the proof that it is finite is a (non-constructive) proof that it is bounded by a natural number, based on the fact that it is closed by minors. My understanding is that such proofs can be proof mined to obtain constructive bounds, in this case, depending on the proof that it is minor closed, and the ambient logic of the proof. Again, this number is likely to be astronomical.
@cody : You can't always get a bound, since the proof is by contradiction. Assume the set S of forbidden minors is infinite. Then there must be 2 members of S such that one is a minor of the other; contradiction. This uses excluded middle and is ineffective. Now if you can prove that the set of forbidden minors must grow at some controlled rate r, then you can mine the proof to get a bound f(r) on how far you have to go to get the requisite 2 graphs. Here f is your astronomical function. But you still need control over r in the first place, and I don't think this is known for toroidal graphs.
Isn’t it more or less true that the logical strength needed for the proof imposes an upper bound on how the number of forbidden minors can grow, since the logical strength is basically equivalent to the fastest growing function which the logic can define? And there is of course a feature of toroidal graphs which is relevant: they all admit an almost-embedding on a surface of genus 1 (which is the induction parameter in Robertson-Seymour). So one might be able to get something relevant by really looking at the proof.
@user36212 : I don't think you can argue that way. At the end of the day, the set of forbidden minors is finite, so its growth rate is zero. It's only in the reductio ad absurdum part of the proof that an infinite set is postulated. It's the same story with all those ineffective theorems in number theory, like Faltings's theorem. Just because the proof can be formalized in some weak system doesn't mean that you can get bounds on the size of provably finite sets. As long as you're using classical logic rather than intuitionistic logic, there's no guarantee of effectivity.
@cody: The issue is not that it's non-constructive (proof mining often gets bounds from non-constructive proofs). The issue is that the statement is Sigma2 ("there exists a finite set of minors such that for all finite non-toroidal graphs..."), so proof mining gives a weaker statement than a bound on the finite number. We know this to optimal in this case, because there's a diagonalization argument showing that the bound can't be computed. One thing that is possible is an assignment of ordinals to finite families such that each time we add a new forbidden minor, the assigned ordinal decreases.
By the way, if Robertson-Seymour seems like an intimidating black box that gives you the nagging worry that it's hiding something, I recommend that you look at something simpler, such as Kruskal's tree theorem, where you can grasp the whole proof. Then you can see for yourself that in general you can mine the proof until the cows come home, but you can't get an effective upper bound on the size of a finite forbidden set.
@TimothyChow I was actually thinking about the $\mathrm{TREE}(n)$ numbers as an example! but Henry's answer has clarified things.
@HenryTowsner can you point me to the diagonal argument? I'm not sure I can follow it, but I do seem to recall that classical $\Sigma^0_2$ statements are conservative over intuitionistic ones, so I want to at least try to clarify my understanding.
Intuitionistically speaking, I believe that all that has been proved is that the set of forbidden minors is not infinite. I don't think that there is an intuitionistic proof that the set is finite.
@cody: Theorem 8 of "On search, decision, and the efficiency of polynomial-time algorithms". Classical $\Pi^0_2$ statements are conservative over intuitionistic ones; $\Sigma^0_2$ statements are generally not.
@HenryTowsner : There's a potentially misleading sentence in that paper: "It has been shown that Theorem 1 is independent of constructive axiomatic systems." By this is meant that FRS showed independence from (say) RCA$_0$, but that's a separate question from ineffectivity. RCA$_0$ uses classical logic and can still prove ineffective theorems, and conversely, unprovability in RCA$_0$ does not imply ineffectivity.
By the way, I should correct something I said about Faltings's theorem. In this case there are effective bounds on the number of solutions, just no effective bounds on their size. But the point remains that these results require extra insight into the problem in question; they don't follow from generalities about logic.
There exists a (99,14,1,2)-strongly regular graph? That is a graph with 99 vertices, each vertex connected with 14 other vertices, each edge entering in a unique triangle, and such that for each non-connected pair of vertices $a$, $b$, there exist other two $c$ and $d$, and only those two, connected simultaneously with $a$ and $b$?
All the restrictions studied do not rule out the existence, but nobody has been able to construct it. E. Berlekamp, J. H. van Lint y J. J. Seidel have constructed a (243,22,1,2)-strongly regular graph. (A strong Regular Graph Derived from the Perfect Ternary Golay Code, in the book A Survey of Cominatorial Theory, ed. by J. N. Srivastava, North Holland, 1973, p.~25–30.)
I think the determinant spectrum problem should be more well known.
I've written about this elsewhere on MathOverflow, e.g.
Determinants of binary matrices . Briefly:
Fix n, look at the set of n by n binary (I like 0-1 matrices, others prefer 1,-1, but they are morally equivalent) matrices, and compute their determinants over the integers. What is the set of values thus obtained? This is open for n=11, 13, and larger. (Unfortunately, Will Orrick's website at Indiana.edu is down at present, so you have to find an archived copy. The index n shifts by one sometimes, so it may be reported as 12, and 14 or larger.)
There are related questions, one of which might be computationally resolved: find a brief description, uniform in the parameter n, which gives better than exponentially many matrices whose spectrum subset is large and contiguous. I got exponentially many which hit all determinants in (-2F(n-1),2F(n-1)) using Fibonacci matrices; can someone do better?
Gerhard "Thanks Again To Roger House" Paseman, 2020.04.24.
I should add useful to brief and uniform in n. Ideally, given n and d in a given range, use the description to produce quickly an order n binary matrix with determinant d. I can do this for all n and exponentially (in n) many d. How much can this be improved? Gerhard "Not Just For Breakfast Anymore" Paseman, 2020.04.24.
I was able to improve this to $[-c\cdot 2^n/n, c\cdot 2^n/n]$ from $[-c\cdot 1.618^n,c\cdot 1.618^n]$ using a generalization of Fibonacci matrices. I just uploaded the preprint here https://arxiv.org/abs/2006.04701
@rikav I will check it out. Thanks for letting me know. I don't know if Robert Craigen is still interested, but I'm confident Will Orrick is interested in this. You might notify him by email. If you prefer, you can wait till I check it out. If it looks good to me (and my first skim suggests it will be), I'll ping Will myself. Gerhard "Looking Forward To Good Reading" Paseman, 2020.06.09.
Černý conjecture was stated in 1964 and it's not very famous (no monograph, but a special number of Journal of Automata, Languages and Combinatorics last year), but probably is not "computational in nature", strictly speaking. Anyway, there are many open problems related to such conjecture which are also less known or studied.
E.g., let $f_1, \ldots, f_m$ be functions from $\{1,\ldots, n+1\}$ to itself, and let $h$ a function obtaind by composing $f_1, \ldots, f_m$ as many times as you want, so $h$ is a word on the alphabet $\{f_1, \ldots, f_m\}$. If the image of $h$ has cardinality $1$, then the set $\{f_1, \ldots, f_m\}$ is $n$-compressible and $h$ is a $n$-collapsing word.
Sauer and Stone proved that there exist words like $h$ that are $n$-collapsing for every $n$-compressible set of $m$ functions from $\{1,\ldots, n+1\}$ to itself: such words are called $n$-synchronizing words.
Find $s(n,m)$, the lenght of the shortest $n$-synchronizing word over an alphabet with $m$ functions (letters).
This is obviously "computational in nature" as for fixed $n$ and $m$ there is only a finite number of $n$-compressible sets $\{f_1, \ldots, f_m\}$ and an upper bound for $s(n,m)$ it's known.
We know that $s(3,2)=33$ and $s(2,3)=22$, try to find other values. (Note that the problem can be stated in a more effective way using automata, see here for many other details and the upper bound for $s(n,m)$).
I think this one fits the profile, since it computational in nature, understandable by an undergrad student and still an open problem:
The envy-free cake-cutting: the problem of cutting a heterogeneous "cake" that satisfies the envy-free criterion, namely, that every partner feels that their allocated share is at least as good as any other share, according to their own subjective valuation.
How many queries are required for cutting this cake into $n$ slices?
Whether it is "not too famous" might be disputable. Please take it with a grain of salt (I have never heard of it until a while ago, but I am not a mathematician). According to wikipedia and this other question:
The continuous "moving knife" algorithms for envy free cake cutting into connected pieces is only mentioned for up to 4 players. The general case is still an open problem.
I believe (I'm not a professional mathematician) that the problem concerning aliquot sequences could fit your requirements. Wikipedia has the article Aliquot sequence and the online encyclopedia Wolfram MathWorld has the article (Catalan's Aliquot Sequence Conjecture and) Aliquot Sequence both provide remarkable references.
In my view two important articles that I've known in the past are [1] and [2]. If I refer well, I've known it in the past, the professor Juan Luis Varona (Universidad de La Rioja) has a page/website dedicated to this subject.
For example this conjecture was as an answer on this Mathoverflow from the post What are conjectures that are true for primes but then turned out to be false for some composite number?, question with identificator 117891 on MathOverflow (Jan 2 '13), where is added more information in a concise way.
References:
[1] Richard K. Guy and J. L. Selfridge, What Drives an Aliquot Sequence?, Mathematics of Computation, Vol. 29, No. 129, (January, 1975), pp. 101-107.
[2] P. Erdös, On Asymptotic Properties of Aliquot Sequences, Mathematics of Computation, Vol. 30, No. 135, (July, 1976), pp. 641-645.
Feel free in comments to tell me if my answer is useful. In other case I can to delete it. Many thanks.
Are there any algebraic integers of degree $d \geq 3$ with bounded partial quotients?
It is a theorem of Dirichlet that for every irrational number $\alpha$, there exists infinitely many rational numbers $p/q$ with $\gcd(p,q) = 1$ and $q > 0$ such that
$\displaystyle \left \lvert \alpha - \frac{p}{q} \right \rvert < \frac{1}{q^2}.$
This can be improved with the constant on the right hand side improved to $1/\sqrt{5}$, and then this is sharp (Hurwitz's theorem). The reason is that there are badly approximable numbers, whose partial quotients in their continued fraction expansions are bounded, for which it is possible to prove a lower bound of the form $|\alpha - p/q| \geq c(\alpha) q^{-2}$ for all rational numbers $p/q$ and the constant $c(\alpha)$ depending only on $\alpha$. Note that since quadratic irrationals have eventually periodic continued fraction expansion, all quadratic irrationals are badly approximable.
It is a theorem of Roth, for which he was awarded a Fields Medal in 1958, that for any algebraic integer $\alpha$ having degree $d \geq 2$ and for any $\varepsilon > 0$, the number of reduced fractions $p/q$ such that
$\displaystyle \left \lvert \alpha - \frac{p}{q} \right \rvert < \frac{1}{q^{2 + \varepsilon}}$
is finite. In other words, all algebraic integers are almost badly approximable.
The question is, are there any algebraic integers $\alpha$ having degree $d \geq 3$ which has bounded partial quotients, or equivalently, badly approximable? This question, shockingly, remains open even for degree 3.
What is the computational aspect of this problem?
The question allows for the prospect of computing the existence of an object, and constructing such an algebraic integer could be considered as such. Further, computing the continued fraction expansion of a given real number is, by nature, computational.
While computing a continued fraction is certainly a computational task, showing that a continued fraction has bounded coefficients certainly is not, and I don't see how you could in any reasonable sense "compute" such a number.
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2025-03-21T14:48:30.398094
| 2020-04-24T05:53:15 |
358386
|
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|
Stack Exchange
|
self dual character of local fields and global fields
There are two concepts of self dual character, one is for global and another is for local.
Let $K$ be an imaginary quadratic number field, and a Hecke character $\chi : \mathbb{A}_K^{\times}/K^{\times} \to \mathbb{C}^{\times} $ is called self-dual if for any $x \in \mathbb{A}_K^{\times}$
$$ \chi(x \cdot \bar{x}) = |x|_{\mathbb{A}_K}$$
where the bar means conjugate and $|x |_{\mathbb{A}_K} = \prod_v|x_v|_v$ .
Fix a prime $p$. For any $K_v,\ v|p$, we konw that $K_v/Q_p$ is a quadratic extension. Let $\omega : K_v^{\times} \to \mathbb{C}^{\times} $ be a character, then it's called self-dual if $\omega|_{Q_p^\times}$ is the quadratic character of $Q_p^{\times}$ associated to the quadratic extension $K_v/Q_p$.
My question is, if a global Hecke character is self-dual, can we say it's every local component is self-dual?
Form the name I think that's true, but I have problem in proving this satement.
I consider the case that $K_v/Q_p$ is ramified, and let denote the quadratic character of $Q_p^{\times}$ induced from quadratic extension $K_v/Q_p$ by $\omega$. Then because in such a case $p \in Nm_{K_v/Q_p}K_v$, we know that $\omega(p) = 1$. Then we ask will a self-dual character of $K$, say $\chi$, always give $\chi_v(p) = 1?$
And I think that can't be true at all ! Since we can embed $p$ into $\mathbb{A}_K$ by $x=(1,1,...,1,p,1,...)$, only the $v$-place is $p$ and other places be $1$. Then for a self-dual Hecke character $\chi$, we have
$$ \chi(x) = |x \cdot \bar{x}|_{\mathbb{A}_K} = |p^2|_v = p^{-4}$$
And we also have
$$ \chi(x) = \prod_w \chi_w(x_w) = \chi_v(p) $$
Thus we always have $\chi_v(p) = p^{-4}$, which means $\chi_v$ can't be self-dual !
Did I make some mistakes?
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2025-03-21T14:48:30.398242
| 2020-04-24T06:04:46 |
358387
|
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Stack Exchange
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Is the minimum of a constraint optimization problem differentiable in the constraint parameter?
Let $h:\mathbb R^{>0}\to \mathbb R^{\ge 0}$ be a smooth function, satisfying $h(1)=0$, and suppose that $h(x)$ is strictly increasing on $[1,\infty)$, and strictly decreasing on $(0,1]$.
Let $s>0$ be a parameter, and define $
F(s)=\min_{xy=s,x,y>0} h(x)+ h(y)$.
If I am not mistaken, the map $s \to F(s)$ is continuous.
Question: Is $F$ differentiable everywhere on $(0,\infty)$? We cannot expect more than $F \in C^1$ for sure, as the example below shows.
There are examples where the minimum points cannot be chosen in a differentiable manner in $s$, yet $F$ is still differentiable:
Take $h(x)=(x-1)^2$. Then
$$
F(s) =
\begin{cases}
2(\sqrt{s}-1)^2, & \text{ if }\, s \ge \frac{1}{4} \\
1-2s, & \text{ if }\, s \le \frac{1}{4}
\end{cases}
$$
is $C^1$, and in particular, differentiable at $s=\frac{1}{4}$, even though the points of minima $(a(s),b(s))$ are given by
$$
\begin{cases}
\sqrt{s}, & \text{ if }\, s \ge \frac{1}{4} \\
\frac{1}{2}(1 \pm \sqrt{1-4s}), & \text{ if }\, s \le \frac{1}{4}
\end{cases}
$$
which is not differentiable at $s=\frac{1}{4}$. These points of minima are unique up two permuting $a$ and $b$.
Note that $F \in C^1$, but is not twice differentiable at $s=\frac{1}{4}$, so we had some loss of regularity, as we started with smooth objective function, and a smooth constraint.
Is there any "standard theory" for when the minimum of a contraint optimization problem differentiable in the parameter? I tried to google in various ways, but couldn't find the relevant material I guess.
Perturbation Analysis of Optimization Problems by Bonnans and Shapiro would come to mind as a comprehensive resource (Chapter 4.3). They also have a SIAM Review article.
The answer to your question is: No, in general $F$ is not differentiable everywhere on $(0,\infty)$.
First, to simplify the notations a bit, consider the change of variables $x=e^u$, $y=e^v$, $s=e^t$, $g(u)=h(x)=h(e^u)$, and $G(t)=F(s)=F(e^t)$, induced by the smooth increasing correspondence $\ln\colon(0,\infty)\to\mathbb R$.
Then the problem can be rewritten as follows:
Let $g\colon\mathbb R\to\mathbb R$ be a smooth function with $g(0)=0$, and suppose that $g$ is strictly increasing on $[0,\infty)$ and strictly decreasing on $(-\infty,0]$.
For each real $t$, let
$$G(t):=\min_{u\in\mathbb R}[g(u)+g(t-u)].$$
Is then $G$ differentiable everywhere on $\mathbb R$?
Note that any minimizer $u$ of $g(u)+g(t-u)$ satisfies the equation $g'(u)=g'(t-u)$. Therefore, with the implicit function theorem in mind, the main idea -- in order to produce a promised counter-example -- is to get a function $g$ with the the equation $g'(u)=g'(t-u)$ having, for some real $t$, appropriate multiple roots $u$.
It turns out that
$$g(u):=\frac{u^6}{6}+\frac{2 u^5}{5}-\frac{3 u^4}{4}-\frac{4 u^3}{3}+2 u^2,$$
with $g'(u)=u(u-1)^2(u+2)^2$ will do. Indeed, first of all here, clearly this function $g$ satisfies all the conditions: $g$ is smooth, $g(0)=0$, $g$ is strictly increasing on $[0,\infty)$, and strictly decreasing on $(-\infty,0]$. Moreover, for this function $g$ we have
$$G(t)=\begin{cases}
G_1(t) & \text{ if }t\geq 2\text{ or } t_*\leq t\leq \frac{4}{5}\text{ or }t\leq -4, \\
G_2(t) &
\text{otherwise},
\end{cases}
$$
where
$$G_1(t):=\frac{1}{960} \left(5 t^6+24 t^5-90 t^4-320 t^3+960 t^2\right),$$
$$G_2(t):=\frac{1}{60} \left(55 t^6+264 t^5+390 t^4+60 t^3-345 t^2-5 \sqrt{(t+1)^6 \left(5 t^2+6 t-7\right)^3}-300 t+225\right),$$
and $t_*=-1.958\ldots$ is the only negative root of the polynomial $P(t):=55 t^4+176 t^3+156 t^2-32 t-148$.
Finally,
$${G^{\,}}'(t_*+)={G^{\,}}'_1(t_*)=-3.995\ldots\ne-0.0492\ldots={G^{\,}}'_2(t_*)={G^{\,}}'(t_*-).$$ So, $G$ is not differentiable at $t_*$, as claimed.
Here are the graphs $\{(t,g'(t))\colon-2.5<t<1.5\}$:
and $\{(t,{G^{\,}}'(t))\colon t\in(-3,3)\setminus\{t_*\}\}$:
A few more details: Recall the main idea: that (i) any minimizer $u$ of
$$H_t(u):=g(u)+g(t-u)$$
satisfies the equation $g'(u)=g'(t-u)$ and (ii) we want the equation $g'(u)=g'(t-u)$ to have, for some real $t$, appropriate multiple roots $u$.
Indeed, then we will have
\begin{equation*}
G(t)=H_t(u_j(t))\quad\text{for}\quad t\in T_j
\end{equation*}
for some natural $k$ and all $j=1,\dots,k$, where the $u_j$'s are different branches of the roots $u$ of the equation $g'(u)=g'(t-u)$ and the $T_j$'s form a subdivision of the real line; if $g$ is algebraic, then the $T_j$'s will be intervals, say $[t_{j-1},t_j]$.
Then for $t\in(t_{j-1},t_j)$
\begin{equation*}
G\,'(t)=g'(u_j(t))u'_j(t)+g'(t-u_j(t))(1-u'_j(t))=g'(t-u_j(t)).
\end{equation*}
So, there is no reason for $G\,'(t_j-)=G\,'(t_j+)$ if $j<k$. That is, in the presence of multiple roots $u$ of the equation $g'(u)=g'(t-u)$, it should be expected that $G\notin C^1$. What is then a bit surprising to me (and what I cannot explain) is that in most of the simple cases I have considered we have $G\in C^1$.
Note also that $t/2$ is always a ("trivial") root $u$ of the equation $g'(u)=g'(t-u)$. Further, if $u$ is a root of $g'(u)=g'(t-u)$, then $t-u$ is obviously a root, too. So, we should be interested in the pairs $(u,v)$ of roots of $g'(u)=g'(t-u)$ such that $u<v\le t/2$. All these pairs are as follows:
\begin{equation}
\begin{aligned}
(u_1(t),t/2)&\quad\text{if}\quad -4<t\leq -2,\\
(u_1(t),u_2(t))\text{ or }(u_1,t/2)\text{ or }(u_2,t/2)&\quad\text{if}\quad -2<t<-t_{**},\\
(u_1(t),t/2)&\quad\text{if}\quad t=t_{**},\\
(-2,-1/2)&\quad\text{if}\quad t=-1,\\
(u_1(t),t/2)&\quad\text{if}\quad 4/5<t<2,
\end{aligned} \tag{1}
\end{equation}
where
$$t_{**}:=-(3+2\sqrt{11})/5=-1.926\ldots,$$
$u_1(t)$ is the smallest real root of the polynomial
$$Q_t(u):=u^4-2 t u^3+\left(4 t^2+4 t-3\right) u^2+t \left(-3 t^2-4 t+3\right) u+\left(t^2+t-2\right)^2,$$
and $u_2(t)$ is the second smallest real root of the polynomial
$Q_t(u)$ (for $t$ in the corresponding intervals); we see that such pairs $(u,v)$ exist only for $t\in(-4,t_{**}]\cup\{-1\}\cup(4/5,2)$.
Below are the graphs (left panel) of the functions $u_1$ (red), $u_2$ (green), and $t\mapsto u_3(t):=t/2$ (blue), with the fragments (right panel) of these graphs over the most interesting interval, $(-2,t_{**})$.
It is plausible that the discontinuity of $G\,'$ occurs at a point $t$ where some of the distinct branches $H_t(u_i(t))$ ($i=1,2,3$) meet, that is, at a point $t$ such that $H_t(u_i(t))=H_t(u_j(t))$ for some distinct $i$ and $j$ in the set $\{1,2,3\}$.
In fact,
$$\{t\in\mathbb R\colon H_t(u_1(t))=H_t(u_3(t))\}=\{-4,4/5,2,t_*\}$$
(with $t_*=-1.958\ldots$ as before),
$$\{t\in\mathbb R\colon H_t(u_2(t))=H_t(u_3(t))\}=\{-4,-2,4/5,2\},$$
$$\{t\in\mathbb R\colon H_t(u_1(t))=H_t(u_2(t))\}=[-4,-2)\cup\{t_{**},-1\}\cup[4/5,2];$$
concerning the latter two results of the three, note that $u_2(t)$ actually appears in the description (1) of the pairs of roots of $g'(u)=g'(t-u)$ of interest only for $t\in(-2,-t_{**})$.
The actual point of discontinuity of $G\,'$ is $t_*$, as was noted before. Here, one may also note that $t_*=-1.958\ldots$ is in the most interesting interval, $(-2,t_{**})=(-2,-1.926\ldots)$.
Thank you very much. This is a very interesting solution. If I am not mistaken, I was able to verify that $G$ is differentiable at all the transition points except at $t=t_$, although twice differentiable at none of them. I wonder whether this can be deduced directly somehow, or is it surprising? In particular, I am trying to understand what property distinguishes the transition point $t=t_$ from the other transition points. Is it special because only there the equation $g'(u)=g'(t-u)$ has multiple solutions $u$, like you mentioned at the beginning? I guess you tried to find...
a function such that the minimizers will change non-smoothly from different directions of the transition point, but as my example in the question shows, this is certainly not enough. To summarize- I would like to have a better idea about how you came up with this counter-example, since the fact that the minimum points cannot be chosen in a differentiable manner does not in general imply non-differentiability of the minimal value function. Finally, I find it very interesting that there are multiple transition points, as this answers...
this question... as well. I must say the phenomena you discovered is very non-trivial for me. I wasn't expecting such complexity. Thank you again for all your many insights and help.
@AsafShachar : I am glad you liked this answer. As was mentioned there, the main idea is that the equation $g'(u)=g'(t-u)$ should have multiple roots $u$
for some real $t$. Next, for some reasons that I don't remember now, it occurred to me that for some $t$ the equation $g'(u)=g'(t-u)$ should have a root $u$ such that $u$ and $t-u$ be of the opposite signs. But the monotonicity pattern of $g$ implies that $g'\le0$ on $(-\infty,0]$ and $g'\ge0$ on $[0,\infty)$.
Previous comment continued: So, I thought, there must be real $u$ and $v$ such that $u<0<v$ and $g'(u)=g'(v)=0$. I then tried $g'(u)=u(u-1)^2(u+1)^2$, but it did not work. After that, I tried $g'(u)=u(u-1)^2(u+2)^2$, and it worked.
Thank you. I think that the last thing that surprised me is that the transition is only between two functions $G_1,G_2$, and not between $G_1,G_2,G_3,G_4$...Do you think that this could be explained somehow in advance?
@AsafShachar : For me to be sure, can you specify what you mean by $G_3,G_4,\dots$, and also by "the transition"? (Thank you for generously awarding me with the bounty.)
You are welcome. By $G_3,G_4,...$ I meant for some "unknown functions" (placeholders) that could in principle arise in the solution. To be more specific, the minimum $G(t)$ "switches" between the functions $G_1(t),G_2(t)$ at $t=-4,t_*,\frac{4}{5},2$. I wonder if there is some conceptual explanation for why we don't see here transition(s) between more functions, i.e. could it have been that $G(t)$ equals to a different function $G_i$ on each subinterval, where all the $G_i$ are "distinct" (that is given by different formulas).
@AsafShachar : I have added some details that may throw more light on the matter.
|
2025-03-21T14:48:30.398974
| 2020-04-24T07:50:39 |
358395
|
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|
Stack Exchange
|
real analytic function with given shape
I am looking for a 5 parameter family of analytic functions $f:[0,1]\to \mathbb{R}$ such that
(0) $f$ has zeros at $0,p,1$.
(1) $f$ is convex in $[0,p]$ and concave in $[p,1]$.
(2) The five parameters, $p$ with $0<p<1$ and arbitrary positive values of $-f'(0)$, $-f'(1)$, $f''(0)$, and $-f''(1)$, can be independently prescribed, and $f$ depends continuously on these parameters.
A closed form solution in terms of rational operations and elementary functions is preferred. (I know a solution using a cubic spline with 7 knots, but due to its piecewise construction it is nonanalytic.)
I would already be happy with a 4 parameter family where $p$ is determined implicitly or explicitly by the other 4 parameters, since in spite of many attempts, I could not even satisfy this weaker goal.
Sorry to sound naive, but can't you do that with a polynomial ?
@LoïcTeyssier: I don't know. In any case, it seems to be very difficult to give a 5-parameter family of such polynomials.
I think that for polynomials, the degree would have to depend on the data pescribed in (2) and could not be bounded.
I agree, with bounded degree you couldn't get all freedom degree without introducing a root in $[0,1]$ for some values of the parameters.
You should specify more precisely what is a 5-parametric family: a) parametrized by what ($R^5$?) and b) how should a function depend on parameters.
@AlexandreEremenko: I thought that was already clear enough, but I made it more precise in (2).
How about $a_0(x-a_1)(x-a_2)(x-a_3)x(x-p)(x-1)$ with real numbers $a_0\ne0, a_i\not\in(0,1)$ for $i\in{1,2,3}$?
@Hans: This does not work for all required choices of derivatives in (ii). Polynomials of fixed degree cannot bend very much on one side and very little on the other. I already tried many intuitively suggestive (polynomial and nonpolynomial) expressions with 5 parameters and none was flexible enough. That's why I asked.
Yes, I agree. Just saw what @LoïcTeyssier said earlier.
There must be some papers on the rigidity of an analytic function based on discrete data aside from its analytic continuation property.
@Hans: Such a rigidity theorem, which would forbid a solution to my problem, would be very interesting!
This is a very nice question - an innocent-looking problem, easily described - basically construct a horizontal s-shaped analytical function under some well-defined conditions.
But I must admit I had changed my mind several times, bouncing back and forth, in trying to decide (and prove) whether it can or cannot be done.
I now believe that the problem can be solved with polynomials as @MattF wrote in his answer,
and as you and @LoïcTeyssier suggested in the comments, the degree $n$ of the polynomial depends on the constraints.
The problem is very suited to be represented with polynomials in Bernstein/Bezier form (see also this comprehensive survey).
The Bernstein polynomial $P(x)$ is defined as $P(x) = \sum_{i=0}^{n} c_i B_{i}^{n}(x)$,
where $B_i^{n}$ are the Bernstein basis polynomials $B_i^{n}(x) = \binom{n}{i} (1-x)^{n-i} x^{i}$.
Below, I will sometimes use freely the term "curve" (or "polynomial curve" or "Bezier curve") for the graph of the polynomial.
Also, the points $p_i = (i/n, c_i)$, denoted the control points or control polygon of the curve, have a geometric meaning, which I will be using.
The first thing to notice is that the conditions on the first and second derivatives totally define the first and last three control points.
Since the Bezier curves interpolate their endpoints, we have from condition (0):
$$c_0 = c_n = 0$$
The first derivative of a Bernstein polynomial of degree $n$ is itself a Bernstein polynomial of degree $n-1$ of the form:
$P'(x) = n \sum_{i=0}^{n-1} (c_{i+1} - c_i) B_{i}^{n-1}(x)$
Thus $P'(0) = n (c_1 - c_0)$ and from condition (2) we get:
$n (c_1 - c_0) = f'(0)$,
and so:
$$c_1 = f'(0)/n$$
and in a similar manner:
$$c_{n-1} = -f'(1)/n$$.
The second derivative is a Bernstein polynomial of degree $n-2$ of the form:
$$P''(x) = n(n-1) \sum_{i=0}^{n-2} (c_{i+2} - 2 c_{i+1} + c_i) B_{i}^{n-2}(x)$$
Therefore,
$$P''(0) = n(n-1)(c_2 - 2 c_1 + c_0)$$
and from condition (2) we get:
$$n(n-1) (c_2 - 2 c_1) = f''(0)$$
and (with some manipulation):
$$c_2 = \frac{f''(0)}{n(n-1)} + 2 c_1 = \frac{f''(0)}{n(n-1)} + 2 \frac{f'(0)}{n}$$.
and in a similar manner:
$$c_{n-2} = \frac{f''(1)}{n(n-1)} - 2 c_{n-1} = \frac{f''(1)}{n(n-1)} - 2 \frac{f'(1)}{n}$$.
Furthermore, if the control polygon intersects the $x$-axis only once in $(0, 1)$ then, by the Bezier variation diminishing property, there is only one inner root.
This is useful since we want (from condition (0)) only a single inner root (at $x=p$),
and the roots at $0$ and $1$ are immediate since we set $c_0 = c_n = 0$.
Another important concept is what I call the "convex-concave property" of the control polygon.
I say the control polygon is "convex-concave" if every consecutive control point
tuple $(p_i, p_{i+1}, p_{i+2})$ make a left-turn up to some index $i=k$ and from then on all consecutive tuples make right-turns.
The important insight here is that if the control polygon is convex-concave, then so is its Bezier curve.
The reason is that, as we saw above, the second derivative of the polynomial is itself a Bezier curve (of degree $n-2$), where its coefficients are $d_i = c_{i+2} - 2c_{i+1} + c_i$.
A left-turn tuple gives a positive control point $d_i$, whereas a right-turn tuple gives a negative one.
Thus, up to $d_k$ all $d_i$ are positive, and afterwards all are negative.
Therefore, by the variation diminishing property,
there is only one zero-crossing of the second-derivative control polygon,
which means that there is just a single inflection point in $[0, 1]$.
So a convex-concave control polygon certifies that the curve is convex up to some point and concave afterwards.
One thing to notice is that if $c_2 > c_1$ then it may be impossible to satisfy the convex-concave condition.
To see this, take for example $c_2 = 0$. Then $c_3$ must be positive or the tuple $(c_1, c_2, c_3)$ will be concave and violate the convex-concave property.
If this happens on both sides (e.g., $c_2 = 0$ and $c_{n-2} = 0$) then the convex-concave property cannot be satisfied.
If, however, $c_2 < c_1$ and $c_{n-2} > c_{n-1}$, then you can always construct a convex-concave polygon.
For example, we can construct such a polygon by taking all $c_i$ to be on
the line between $c_1$ and $c_2$ for $i < k$,
and all $c_i$ on the line between $c_{n-1}$ and $c_{n-2}$ for $i \geq k$.
We can derive a lower bound on $n$ for which this will be the case.
Since $c_2 = \frac{f''(0)}{n(n-1)} + 2 c_1 = \frac{f''(0)}{n(n-1)} + 2 \frac{f'(0)}{n}$,
then if we set $n > |f''(0)/f'(0)| + 1$, we will have $c_2 < c_1$.
Similarly, for $n > |f''(1)/f'(1)| + 1$, we will have $c_{n-2} > c_{n-1}$
So setting the lower bound $n > \max(|f''(0)/f'(0)| + 1, |f''(1)/f'(1)| + 1)$
guarantees that we can construct convex-concave control polygons with the required end conditions.
This already takes us a long way.
Condition (2) is satisfied by the first and last three control points.
Condition (1) is satisfied in the weak sense (i.e., with an implicit $p$) by the convex-concave property.
And condition (0) is also partially satisified by the first and last control points and the fact that there is only one inner root.
What is missing to fully satisfy all the conditions (at least in the weak sense)
is to find a way to force the root and the inflection point
(where the curve turns from convex to concave) to be the same point $p$.
We can now think of an algorithm to construct the polynomial, which starts with the first and last three control points,
and continues by trying to build a convex-concave control polygon, which satisfies the additional constraints.
Apart from the constraints on the first and last three coefficients described above, there are the following constraints.
The polynomial zero-value at $p$:
$$P(p) = \sum_{i=0}^{n} c_i B_{i}^{n}(p) = 0$$
The zero-value of the second derivative at $p$:
$$P''(p) = \sum_{i=0}^{n-2} (c_{i+2} - 2 c_{i+1} + c_i) B_{i}^{n-2}(p) = 0$$
The inequality constraints to satisfy the convex-concave property:
$$(c_{i+2} - 2 c_{i+1} + c_i) > 0 \text{, for } 0 < i < k$$
$$(c_{i+2} - 2 c_{i+1} + c_i) < 0 \text{, for } k \leq i < n-2$$
These constraints are linear equalities and inequalities in the variables $c_i$ and can therefore be solved using a Linear Programming solver (with varying k-values).
Note that the above lower bound on $n$ may not be sufficient.
For example, if $p$ is very close to $0$ or $1$, e.g., smaller than $1/n$, then
we will need to increase $n$.
However, the LP-algorithm also reports if a feasible solution is not found and one can then increase $n$ and try again.
Even though I believe this algorithm is theoretically correct (after all, there are so many degrees of freedom..) I haven't succeeded in proving it.
I wrote an initial implementation of the algorithm, and while it works fairly well on some inputs, on others the LP solver fails due to numerical errors (I used scipy's LP solver).
This looks to me like an implementation issue (possibly a bug in my code), but there might be something fundamental hiding there, which I'm missing.
Below is a figure of polynomials and their second derivatives for $p=0.5$ and varying end conditions, computed by the algorithm.
For the weaker version of the problem (i.e., with an implicit $p$), I believe I can show that a solution exists.
Given the first and second derivative constraints on the endpoints, I'll show that there exists a convex-concave polynomial $P(x)$ (of a sufficiently large degree $n$) that satisfies the end constraints,
and for which there exists some value $p$ that satisfies $P(p)=0$ and $P''(p)=0$.
This means that $x=p$ is both the root and the inflection point and with the convex-concave property satisfies the
weaker conditions of the problem. Note, I prove the existence of $p$ for a sufficiently large $n$, but don't explicitly compute $p$ nor $n$.
I will use the Weierstrass approximation theorem, which states that any continuous function on a closed interval can be approximated with a polynomial.
More specifically, I will use Bernstein's constructive proof of the Wierstrass theorem, which gives the following theorem:
For a continuous function $f$ over $[0, 1]$, for any $\epsilon$, for a sufficiently large $n$,
$|f(x) - \sum_{i=0}^n f(i/n) B_{i}^{n}(x)| < \epsilon$.
From Bernstein's theorem, by setting the control polygon to be the function $f$, the following corollary follows:
For any $\epsilon$, for a sufficiently large $n$, the distance between the Bernstein polynomial and its control polygon is smaller than $\epsilon$.
From the corollary it follows that (for a suitable $n$)
the root $p$ of $P(x)$ is within a $\delta$-environment of the intersection point of the control polygon and the $x$-axis.
Similarly, the inflection point (the root of $P''(x)$) is within a $\delta$-environment of the intersection of the control polygon of $P''(x)$ and the $x$-axis.
For any $n$, I assume we can construct a convex-concave control polygon, which satisfies the end conditions,
and has an inflection point above the $x$-axis, i.e., to the right of the root.
For a sufficiently large $n$, by the corollary, the curve inflection point will also be to the right of the curve root.
We can similarly construct such a configuration that has an inflection point to the left of the root.
Below is are example figures of two such corresponding configurations, where the inflection point is the blue dot, and the root is black (the control points are the blue crosses).
Moving continuously from the first configuration to the second (while maintaining the convex-concave property),
we will encounter (at least one) value $x=p$ where the inflection
point and the root coincide, which satisfies our request.
So, to summarize, I assume we can draw for any $n$ a pair of convex-concave control polygons that satisfy the end conditions and for which one has an inflection point above the $x$-axis and the other has it below (I think this is a reasonable assumption).
Then, for a sufficiently large $n$, the polynomial curves will be close enough to their control polygon so that their inflection point is also to the right/left of the polynomial root.
Then there should be at least one configuration where the root and inflection point coincide.
Some open issues:
Try and convert the weak-constraint proof to an algorithm. Basically, if you have the two configurations with their roots on opposite sides of the inflection point, then a bisection algorithm might be used.
Prove that there is always a solution for the problem with the strong constraints (for a given $p$ value).
The consequence is that the LP algorithm will eventually give the correct solution.
Find a (hopefully analytical/concise/easily computable) upper bound on the polynomial degree $n$.
Thanks for your efforts. But this does not quite answer my question, as it is a many-parameter family containing some function for each case of my 5 parameters.
I'm not sure I understand what you mean by a many-parameter family. Given the five parameters (or four in the weak version) there are indeed many polynomials that can satisfy the constraints.
I'd like to have a closed form 5-parameter family of solutions, not a recipe for constructing one solution (or many) for each set of parameters that may jump discontinuously when the parameters are varied.
Too long to comment.
Here is a suggestion. Consider transforming the question to finding an analytic non-positive convex function $f$ over [0,1] with (i) zeros at 0 and 1, (ii) the required first and second order derivatives and (iii unimodal at $p$. In that case, one can then multiply by $g(x) = \frac{2}{1+e^{K(x-p)}}-1$, $K$ sufficient large and positive, and see that for $h(x)=f(x)g(x)$, $h(p)=0$ and
$$
h'' = f''g + 2f'g' + g''f
$$
is non-negative (and hence conves) in $[0,p]$ and non-positive in $[p,1]$. Reason for signs of curvature: (i) for $x \in [0,p]$, $f''g$ is non-negative since $f''$ and $g$ are positive, $f'g'$ is non-negative since $f$ reaches minimum at $p$ and $g'$ is non-positive, and $g''f$ is non-negative since $g''$ is non-positive and $f$ is non-positive. The same can be reasoned out for concavity in $[p,1]$.
Could you please explain how to find $f$? This is still a nontrivial problem.
I wish I had an answer for that. I should have put it as a comment, but was just too long.
$C^\infty$ is not analytic.
@Hans, thanks for catching that. My bad. Have made the change.
|
2025-03-21T14:48:30.399877
| 2020-04-24T09:51:19 |
358399
|
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|
Stack Exchange
|
Does a morphism of etale sheaves restricting to a closed subscheme $Z$ induce a morphism of their subsheaves of sections supported on $Z$?
Let $X$ be a locally Noetherian scheme and $i:Z\to X$ be an immersion of closed subschemes.
Let $\mathcal{F},\mathcal{G}$ be two etale abelian sheaves over $X_{et}$.
We can define the subsheaf $\mathcal{H}_Z(\mathcal{F})\subset \mathcal{F}$ of sections supported over $Z$, i.e. for any etale morphism $h:V\to X$
$$\mathcal{H}_Z(\mathcal{F})(V):=\{s\in\mathcal{F}(V)|\text{ Supp}(s)\subset h^{-1}(Z)\}$$
(check out this tag on stacks project).
Assume that we have a morphism of etale abelian sheaves over $Z_{et}$
$$\phi:i^* \mathcal{F}\to i^* \mathcal{G}$$
Can we induces a map $i^*\mathcal{H}_Z(\mathcal{F})\to i^*\mathcal{H}_Z(\mathcal{G})$ over $Z_{et}$?
If so, does it naturally follow that a map $i^*\mathcal{F}\to i^*\mathcal{G}$ can induce a map $\text{H}^n_Z(X_{et},\mathcal{F})\to \text{H}^n_Z(X_{et},\mathcal{G})$ which is my ultimate goal, and can the argument safely transfer to fppf sheaves?
If I understand the first question correctly, the answer is no. Assume $Z\neq\emptyset$ and $X\smallsetminus Z$ is dense in $X$. Let $A$ be any nonzero abelian group and take $\mathcal{G}=\underline{A}_X$ (the constant sheaf), and $\mathcal{F}=i_*(\underline{A}_Z)$.
We have $\mathcal{H}_Z(\mathcal{F})=\mathcal{F}$ and $\mathcal{H}_Z(\mathcal{G})=0$. The natural map $\mathcal{G}\to\mathcal{F}$ induces an isomorphism $\psi:i^*\mathcal{G}\to i^*\mathcal{F}$ (both are canonically isomorphic to $\underline{A}_Z$).
Now consider $\phi:=\psi^{-1}:i^*\mathcal{F}\to i^*\mathcal{G}$. This is an isomorphism which does not map the subsheaf $i^*\mathcal{H}_Z(\mathcal{F})=i^*\mathcal{F}$ into $i^*\mathcal{H}_Z(\mathcal{G})=0$.
|
2025-03-21T14:48:30.400005
| 2020-04-24T09:54:02 |
358400
|
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|
Stack Exchange
|
Recurrence relation between $a(p^{2n})$'s
Let $f$ be a Hecke eigenform of weight $k,$ and $a(p^n)$ be the $p^{n}$th Fourier co-efficient of $f$. I need to a determine a recurrence relation (Hecke type) between $s(n)=a(p^{2n})'s$. We can write $1-a(p)X+p^{k-1}X^2=(1-\alpha X)(1-\beta X).$ Doesn't it imply $a(p^n)=\alpha^n+\beta^n ?$ (well, at least for $n\geq 1$). In that case, I am getting a relation like:
$$s(n+1)=a(p^2)s(n)-p^{2(k-1)}s(n-1),$$ this I am getting by simply doing $\alpha^{n+2}+\beta^{n+2}=(\alpha^2+\beta^2)(\alpha^n+\beta^n)-(\alpha\beta)^2(\alpha^{n-2}+\beta^{n-2})$. But someone provided me a counter example with $n=1$ i.e. my derived recurrence relation doesn't correctly compute $a(p^4)$. I am frustrated because, I don't get what's going wrong here. Could anyone please help me out ?
$a(p^n)$ can be written as a linear combination of $\alpha^n,\beta^n$ but not necessarily their sum. For instance, note that your formula for $n=0$ gives $a(1)=2$, which is wrong assuming your eigenform is normalized. You can find the right coefficients by writing $a(1)=c\alpha^0+d\beta^0,a(p)=c\alpha^1+d\beta^1$ and solving for $c,d$.
So putting all together $c+d=1, c\alpha+d\beta=\alpha+\beta \implies \alpha(c-1)=\beta(1-d)=\beta c.$ So $c,d's$ are now expressed interms of $\alpha, \beta,$ you mean this ?
I think this is giving me $a(p^n)=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta},$ really I should get this ? It matches with right choice of $a(p)$ and $a(1)$, and everything is determined by them. So shouldn't be wrong,I guess.
Looks like the correct result to me. Regarding your first reply comment, I assume the flaw in your inductive argument comes from only checking one base case (e.g. $n=1$ which is obvious), which is not sufficient for a recursion like the one defining $a(p^n)$ which depends not on one, but on two previous terms.
|
2025-03-21T14:48:30.400170
| 2020-04-24T10:58:13 |
358403
|
{
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|
Stack Exchange
|
Continuity upgrade for nonlinear maps
Let $E,F,G$ be topological vector spaces such that $F\subset G$ with continuous embedding.
By continuity upgrade I mean the following phenomenon: In some circumstances a continuous linear map $f:E\rightarrow G$ with $f(E)\subset F$ is automatically continuous as map $f:E\rightarrow F$. This is true e.g. if all involved spaces are Fréchet (and then follows from the closed range theorem).
I am interested in nonlinear versions of this phenomenon. One example for this is if $f$ is a quadratic form, i.e. $f(x)=b(x,x)$ for a bilinear map $b:E\times E\rightarrow G$. In this case one can use the linear result for $b(x,\cdot)$ and $b(\cdot,y)$ to conclude that $b:E\times E \rightarrow F$ is separately continuous in each variable and (at least in the Fréchet setting) this implies joint continuity and thus continuity of $f:E\rightarrow F$.
Question. Let $f:E\rightarrow G$ be a continuous (possibly non-linear) map with $f(E)\subset F$. Which conditions on $f$ (e.g. growth conditions, algebraic conditions) and on the involved spaces ensure that $f$ is continuous as map $f:E\rightarrow F$?
There is a simple closed graph theorem for non-linear maps $\varphi:X\to Y$ from topological spaces $X$ to compact Hausdorff spaces $Y$. You can apply this to all all restrictions $\varphi=f|_X$ with $X\subseteq E$ for some class $\mathscr C$ of subsets which determines continuity, e.g., all convergent sequences in $E$ if $E$ is metrizable. It would be thus sufficient that $f(X)$ is relatively compact in $F$ for all $X\in \mathscr C$.
Edit: I have rewritten my answer after learning more details
I have been wondering for a while whether there is an interesting answer concerning analytic functions. I have looked at the papers suggested by Jochen Glueck in the comments and I am happy that I have found a nice answer in the following theorem.
Theorem A. Let $E,F,G$ be complex Banach spaces, $F\subset G$ with continuous (injective) embedding. Let $U\subset F$ an open subset, and $f:U\to G$ a holomorphic map such that $f(U)\subset F$. Then, $f$ is also holomorphic from $U$ to $F$ assuming that is locally bounded when seen as an $F$-valued map.
The proof relies on a result by Arendt and Nicolski contained in this paper:
Theorem 3.1. Let $f : \Omega \to X$ be a locally bounded function such that $\phi\circ f$
is holomorphic for all $\phi\in W$ where $W\subset X^*$ is a subspace that separates points in $X$. Then f is holomorphic.
Proof of Theorem A. W.l.o.g., we assume $0\in U$ and we prove holomorphicity around $0$. Since $f$ is locally bounded by assumption, to check holomorphicity, by known facts (see, e.g., L. Nachbin, “Holomorphic maps and invariant distances”, Theorem II.3.10.) it is enough to prove that for any $x\in E$, $a\in U$ close enough to $0$ and $\phi\in F^*$, the map $g:\mathbb C\to \mathbb C$,
$$ g(\lambda):= \phi\circ f(a+\lambda x) $$
is holomorphic in the usual sense (the domain of $g$ is actually some subset of $\mathbb C$).
We note now that since $F\subset G$ is a continuous injection, $W:=\{\phi|_{F}\,|\,\phi\in G^*\}\subset F^*$ separates points in $F$. By Theorem 3.1, it is thus enough to show that $g$ is holomorphic assuming that $\phi$ is in $G^*$. But this is true by assumption, so the theorem is proved. $\square$
As noted in the paper, we can actually choose $F$ an $G$ to be Fréchet spaces (maybe also $E$, I am not sure).
Remark: the theorem is not true for real Banach spaces. For instance, consider the following ODE in $\mathbb R^2$:
$$ \left\{\begin{aligned} &\dot x=-y(x^2+y^2),\\ &\dot y=x(x^2+y^2).\end{aligned}\right.$$
Solutions are bounded for any initial datum (the distance from the origin is conserved), they travel with constant velocity, and the data-to-solution map is analytic from $\mathbb R^2$ to the Banach space
$$ G=L^\infty([0,\infty),e^{-|t|^2}dt;\mathbb R^2). $$
It is also a bounded map from $\mathbb R^2$ to $$F=L^\infty([0,\infty),dt;\mathbb R^2),$$
but it is not Lipschitz, since there exist pair of points that are arbitrarily close to a third point $p\neq(0,0)$, but whose solutions get $2|p|$ apart in finite time. One can notice that it is possible to extend the ODE to complex data by extending the vector field analytically on $\mathbb C^2$, but solutions with complex data are not bounded anymore, so it is not possible to try to "complexify" the problem and use our Theorem for the complex case.
Remark. Unfortunately the local boundedness is necessary, as noted in this other post of mine, where I asked a related question. I know that this assumption destroys the spirit of your question, but I found it interesting.
It might be trivial to point out, but note that if $f$ is just a polynomial (so, an analytic function with a finite number of terms), one simply needs $E,F,G$ to be Fréchet spaces, since one can use some clever trick (involving Vandermonde's matrix) to prove that each monomial maps $E$ into $F$, and then apply your argument above. The problem of course is when the series has infinitely many terms.
An easy example to keep in mind. Note that there are explicit examples of continuous maps which are "formally" analytic but they fail to be smooth. One example is the Schrödinger flow $S:\mathbb R \to L^2(\mathbb R^d)$,
$$ S: t \mapsto e^{it\Delta} u_0, $$
where $u_0\in L^2(\mathbb R^d)$. This is a smooth, analytic map from $\mathbb R$ into $H^{-\infty}$ (...if a good definition of analytic map in that setting exists), as
$$ S(t)=\sum_{n\in\mathbb N} \frac{(it\Delta)^n }{n!} u_0, $$
but of course not necessarily smooth when seen as an $L^2$-valued map. By the above theorem, if $u_0\not\in H^\infty(\mathbb R)$, there is actually no Banach space containing $L^2$ such that the map $S$ is analytic from $\mathbb R$ into that space (but maybe it can be smooth...? I don't know).
(Fun fact: I am not claiming the two examples are related, but if you look closely, the ODE example I gave above is a Schördinger equation in disguise ;))
The assumption (1) can be can be completely dropped from the theorem. The functionals in $G^*$ separate the points in $F$ since the embedding of $F$ into $G$ is injective. This together with the local boundedness gives the holomorphy; that's a theorem by Arendt and Nikolskii.
@JochenGlueck Thank you so much for your comments. Would you mind if I asked you to point me to the precise article by Arendt and Nikolskii? I had a quick look at the ones I found online, and they all seem to deal with the case where $E$ is $\mathbb C$ or finite dimensional (or maybe there is something easy I am missing).
Ouch, sorry, you're of course right. I shouldn't comment so late at night. I'll delete my last two comments
Regarding Arendt and Nikolskii: Yes, they do it for one-dimensional $E$. But then you can do the same trick as in your answer to reduce the case of general $E$ to the case of one-dimensional $E$.
Ok, now I was wondering for a moment "But your argument combined with the Krein–Smulian theorem should directly give the result by Arendt and Nikolskii." Then I looked up their article again (Theorem 3.1) and that's just precisely what they do. Apparently I had never bothered to look at the proof and had somehow been under the impression that this result were more involved. So thanks a lot!
@JochenGlueck I have rewritten my answer, thank you!
|
2025-03-21T14:48:30.400687
| 2020-04-24T12:13:58 |
358408
|
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|
Stack Exchange
|
Understanding a deduction in research paper of Sprang, Fischler and Zudilin ("Many Odd zeta values are irrational")
I am a master's student interested in number theory. I am reading a research paper in Analytic Number theory which is "Many Odd zeta values are irrational" by Stephane Fischler, Johannes Sprang and Wadim Zudilin.
I have a question about article 5 of this research paper on page 14.
A screenshot of which is posted below:
In the highlighted part of screenshot screenshot I am unable to deduce
$ \hat{r}_{n, d} = ( d+o(1) ) r_{n, 1} $ as $n \rightarrow \infty$, and
$\tilde{r}_n = \left(\sum_{d\epsilon D } w_d d +o(1) \right) r_{n,1} $
despite trying a lot.
Any help would be greatly appreciated.
I think you need to give me the bounty separately. Thank you in advance!
@GH from MO yes, you answered 2 questions. By Stackexchange rules bounty can be awarded only after 24 hours of starting it. So, It can't be given to you before 24 hours. But don't worry I will not forget to award you. I am really thankful to you for helping.
Thanks for the bounty (twice)!
I am not sure where the difficulty lies here. By (3.2) and the display above the first highlighted item,
$$\hat r_{n,d}=\sum_{j=1}^d r_{n,jD/d}=\sum_{j=1}^d (1+o(1))r_{n,1}=(d+o(1))r_{n,1}.$$
Combining this result with the second display above the second highlighted item,
$$\tilde r_n=\sum_{d\in\mathcal{D}}w_d\hat r_{n,d}=\sum_{d\in\mathcal{D}}w_d(d+o(1))r_{n,1}=\left(\sum_{d\in\mathcal{D}}w_d d+o(1)\right)r_{n,1}.$$
@YannicMuller: Well, Lemma 3 is emphasized below the first highlighted item, and (3.2) is the statement of Lemma 3. I answered your other question as well. Two further remarks. First, please always use a high-level tag like "nt.number-theory". Second, if you are a master's student, then you have an advisor who should (and should be able to) help you out with reading this paper.
|
2025-03-21T14:48:30.400947
| 2020-04-24T12:16:09 |
358409
|
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|
Stack Exchange
|
On Glaeser's Theorem for non-smooth functions
Glaeser's Theorem says that a $C^\infty$ function $F$ on $\mathbb R^n$ which is invariant under permutation of the variables is a smooth function of the symmetric polynomials of $(x_1, \dots, x_n)$.
Question 1: What remains (if anything) of this statement if $F$ is $C^k$ ?
Question 2: In the statement above, is it clear that you can replace $\mathbb R^n$ by a symmetric open subset of $\mathbb R^n$?
Can you provide a reference for Gleaser's theorem?
For 1), take $F(x_1,\ldots ,x_n)=\sum |x_i|^k$ with $k$ odd.
https://www.jstor.org/stable/1970204?origin=crossref&seq=1
@abx Your example is $C^{k-1}$ and invariant under permutation. Is it not a $C^{k-1}$ function of the symmetric polynomials? In the case $n=2$, $k=1$, $|x_1| + |x_2| = \sqrt{(x_1+x_2)^2 - 2 x_1 x_2 + 2 |x_1 x_2|}$. Or are you just saying it's not a smooth function?
@Robert Israel: maybe I was too hasty. Do you think you could do that for any $k$?
@piotr MR188382 26.40
Glaeser, Georges Fonctions composées différentiables. (French) 1963 Séminaire d'Analyse, dirigé par P. Lelong, 1962/63, No. 2 4 pp. Secrétariat mathématique, Paris
A more direct reference (quoted in Rumberger's paper) is: G. Barbançon, Le théorème de Newton pour les fonctions de classe $C^r$. Ann. Sci. École Norm. Sup. 5 (1972), 435–458. He proves that a symmetric function of class $C^{nr}$ on $\Bbb{R}^n$ is a function of class $C^{r}$ of the symmetric polynomials.
My collegue Armin Rainer pointed me to the following paper, which has some answers, also in its references. There is a loss of differentiability involved.
Matthias Rumberger: Finitely differentiable invariants. Math. Z. 229, 675–694 (1998)
|
2025-03-21T14:48:30.401097
| 2020-04-24T12:45:41 |
358411
|
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|
Stack Exchange
|
On the moduli stack of abelian varieties without polarization
(I am especially interested in abelian surfaces and characteristic 0).
How bad is the moduli stack of abelian varieties (with no polarization or level structure)? Is it an Artin stack? DM (Deligne-Mumford) stack?
How bad is the is the stack of abelian varieties with full 2 level structure (so with a basis for $A[2]$)?
Consider the maps from either 2) above or stack of principally polarized abelian varieties to 1) above. Are these maps smooth, are the geometric fibers finite (ie, are there only finitely many principal polarizations on an abelian variety)?
Neither moduli space is a stack because every point has the automorphism $-1$ but the same is true for the moduli stack of elliptic curves and that is still a DM stack and not too bad.
Even in characteristic 0, the CM locus is higher dimensional so the "especially stacky" locus has high dimension but I don't know how serious the problem is.
For the second question, while $-1$ fixes the two level structure, I suppose a generic CM automorphism doesn't fix it so perhaps the second stack is very nice, or atleast almost as nice as that of elliptic curves?
I don't even know how to prove that the prestack of abelian varieties is a stack, i.e., satisfies descent. So before you ask whether this is a `reasonable algebraic stack' it might be a good idea to prove that it is a stack at all. For example the prestack of genus 1 curves (so elliptic curves without a fixed point) is not a stack.
I agree, I was primarily thinking about the level 2 structure version (which I do think is a stack?). Anyway, it looks like none of these ideas work anyway.
Why would adding a full level $2 structure help with descent?
Now that I think about it, probably not. The point with elliiptic curves is that picking a point gives an ample divisor. I guess one would have to stackify in both cases.
Thinking about it a little bit more, I wonder if an argument along the following lines works: Define a prestack that takes a test scheme T to the groupoid of isomorphism classes of `abelian algebraic spaces', that is smooth proper group algebraic spaces over $T$, with connected geometric fibers. This prestack is certainly a stack, by descent for algebraic spaces and because all the above properties can be checked (fpqc) locally. But it is a theorem of Raynaud (c.f. Theorem 1.9 of Chai-Faltings) that every abelian algebraic space is actually isomorphic to an abelian scheme.
Intrresting! This would work for both the stacks, right?
First, when defining the stack you will have the issue that there are formal deformations of abelian varieties which do not extend to families of abelian varieties over any reduced scheme. These are the deformations that do not respect any polarization. (In the complex analytic world these correspond to deformations of complex tori) So unless you have some very strange definition of the functor, the local structure of this stack will be at least as bad as the formal limit $\lim_{n\to \infty} \operatorname{Spec} k[x]/x^n$. I think this rules out ever having a smooth morphism from a scheme, and thus rules out being an Artin stack.
For $E$ a non-CM elliptic curve, the automorphism group of $E^n$ is $GL_n(\mathbb Z)$. This shows that for $n>1$, points of this moduli stack can have infinitely many automorphisms. In particular, the diagonal is not quasicompact.
Level $2$ structure doesn't help with this at all, you just get the group of $n\times n$ matrices congruent to $1$ mod $2$.
The map from the stack of principally polarized abelian varieties to this stack is not smooth because deformations can kill a principal polarization, and the fibers are not finite, again because of examples like $E^n$, whose principal polarizations are in bijection with $n \times n$ symmetric positive definite integer matrices with determinant $1$.
In summary: There's a reason you haven't heard of this stack before.
P.S. You shouldn't worry so much about automorphisms of the generic point, which almost never cause problems in practice. It's everything else that you should worry about!
Thanks, that's very helpful! Is the stack with full level structure also not an artin stack? This might be a different question and it's a little off the cuff but how about if I consider the stack of abelian surfaces in char 0 with 1 geometric non torsion point with finite stabilizer group?
@Asvin (1) The third paragraph of my answer discusses full level $2$ structure. (2) The stack of abelian surfaces with one geometric non-torsion point with finite stabilizer is not even a stack, because these conditions are not reasonable conditions in a family of points. Consider which points in $E \times E$ have this property - it's the complement of a union of infinitely many curves.
Yes, I see now. Thanks again!
|
2025-03-21T14:48:30.401443
| 2020-04-24T13:46:32 |
358415
|
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|
Stack Exchange
|
Is there any name/occurence to this sequence of numbers?
I am curious if there is any name for this sequence of numbers, or any occasion that this sequence is used.
The sequence is $(c_1,c_2,c_3,\cdots)$ with recursive formula
$$c_n=\frac{1}{2n+1}\sum_{i=1}^{n-1} c_i \cdot c_{n-i}$$
and in my case $c_1=\frac{1}{3}$, but my question is not limited to this initial condition.
My example: $c_1=\frac{1}{3}$, $c_2=\frac{1}{3^2 \cdot 5}$, $c_3=\frac{2}{3^3\cdot 5 \cdot 7}$, $c_4=\frac{27}{3^4\cdot 5^2 \cdot 7 \cdot 9}=\frac{1}{3^3 \cdot 5^2 \cdot 7}$, $c_5=\frac{90}{3^5\cdot 5^2 \cdot 7 \cdot 9 \cdot 11}=\frac{10}{3^5\cdot 5^2 \cdot 7 \cdot 11}$ and so on.
Thanks in advance for help.
It’s the Catalan numbers divided by something. You’d probably have success finding a generating function for them.
The generating function seems to be $1 - \sqrt{x} \cot(\sqrt{x})$. But your computations are a bit off: I get $c_4 = 1/4725$ and $c_5 =2/93555$.
@RobertIsrael Thanks! and yes, I miscomputed starting from $c_3$. I am correcting the computation.
What @RobertIsrael says is correct: If we set $f = \sum_{n\geq 1} c_n x^{2n}$, then the recursive equation translates into $f^2 = \left(xf\right)' - x^2$, and this ODE indeed has solution $f = 1 - x \cot x$. You can now enter "1 - x cot x" in Google Scholar and find some surprisingly relevant hits, eventually getting to the Langevin function.
Why did you tag this question "dg.differential-geometry"? Did I miss something?
@SebastianGoette I did not mention it in my question but those numbers appeared while I was doing computation with dg manifolds. I was hoping maybe I could find any differential geometric situation where those sequence appear.
@darijgrinberg Thanks a lot. Learnt a lot to deal with this kind of questions
As mentioned in comments, the sequence comes from the Taylor series of $\cot.$ They are implicit in OEIS sequence A002431 "Numerators in Taylor series for cot x." Note that $c_n/(3c_1)^n$ does not depend on $c_1$ so without loss of generality $c_1=1/3$ is generic.
|
2025-03-21T14:48:30.401616
| 2020-04-24T13:55:36 |
358416
|
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|
Stack Exchange
|
Sub optimal algorithm for linear programming
Consider the linear programming problem
\begin{align}
f^* = \max_{x}&~p^Tx~\\~&A^Tx\leq b~,~0\leq x_i\leq 1
\end{align}where $c$ is a $n\times 1 $ vector, $A$ is a $n\times c$ matrix and $b$ is a $c \times 1$vector. Here $x=[x_1,\dots,x_n]$ is the $n \times 1$ vector to be found. Thus, in addition to the box constraint $0\leq x_i\leq 1$, we have $c$ constraints. It is also known that $p,A,b$ are all element-wise positive. Assume that this problem is feasible and let $(x^*,\lambda^*)$ be the optimal solutions for the primal and dual respectively. Note that $\lambda^*$ is a $c\times 1$ vector.
Consider the function
\begin{align}
d(\lambda)=\max_{x}(p-A\lambda)^Tx+\lambda^Tb~,~0\leq x_i \leq 1
\end{align}
where $\lambda$ is a $c \times 1$ vector. It is not hard to see that $d(\lambda^*)=f^*$. Define the vector $r$ such that $$r_i=[p-A\lambda^*]_i$$ where $[]_i$ denotes the $i^{th}$ entry.
Thus
\begin{align}
d(\lambda^*)=\max_{x}r^Tx+\lambda^Tb~,~0\leq x_i \leq 1
\end{align}
Now, consider the following strategy constructing the $n\times 1$ vector $y$
\begin{align}
y_i = \begin{cases}1 ~,& r_i>0 \\0~,& r_i \leq 0\end{cases}
\end{align}
You can see the motivation of defining $y$ from $d(\lambda^*)$. Given this,
Can $y$ violate the constraints?
Can we comment on the gap $$f^*-p^Ty$$ in terms of $p,A,b,n,c$?
if anyone is interested, background:
This is from an engineering problem. Typically $c$ won't be more than 2, thus that many constraints. $n$ will run into millions of variables. $p,A,b,n,c$ comes from our data. In some sense, finding $\lambda^*$ is an easier problem for us. And also from an engineering perspective not related to linear programming, $y$ is much easier to implement in our system rather than solving a linear programming system. Often, empirically, for our data (for smaller sampled sets), we have seen that it does well also. Is there any justification?
Yes, $y$ can violate the constraints. The one about box constraints ($0 \leq y \leq 1$) is OK by how $y$ is defined, the problem is the other one ($A^T y \leq b$). First I tried getting a feasibility proof unsuccessfully, then I decided to build a reproducible counterexample using Python:
import cvxpy as cp
import numpy as np
# Generate a random instance according to the conditions stated by OP (small c, larger n).
n = 150
c = 2
# Build p, A and b, which are elementwise positive.
np.random.seed(1)
p = np.random.uniform(0, 1, n)
A = np.random.uniform(0,1, (n,c))
b = np.random.uniform(0,1, c)
# Define and solve the CVXPY problem.
# Note that we'll solve the primal problem only to get the dual variable lambda^*
x = cp.Variable(n)
prob = cp.Problem(cp.Maximize(p.T@x),
[A.T @ x <= b,
x <= 1,
x >= 0 ])
prob.solve()
print("A dual solution is")
print(prob.constraints[0].dual_value)
# Construct the suggested heuristic solution, starting from the dual solution:
lambda_opt = prob.constraints[0].dual_value
r = p - np.matmul(A, lambda_opt)
y = (r > 0).astype(int)
# Verify feasibility for y:
print("Check nonpositive:", (A.T @ y) -b )
This piece of code outputs the array [0.138349, 0.11109537] at the last line, therefore, $A^T y - b \leq 0$ doesn't hold and $y$ is unfeasible. As a conclusion, without further information or hypotheses about $A, b$ and $p$, the proposed approach won't work in general.
Regarding the second part of your question, it reminds me about Approximation Algorithms. I cite from the Wikipedia article:
In computer science and operations research, approximation algorithms are efficient algorithms that find approximate solutions to optimization problems (in particular NP-hard problems) with provable guarantees on the distance of the returned solution to the optimal one.
Assuming the problem of finding $\lambda^*$ easier as you mention, there are two scenarios:
1) If you can find $\lambda^*$ by other means and not solving the linear problem nor its dual, then you could use complementary slackness conditions to try find $x^*$. This somewhat relates to what you are building with the binding/nonbinding dual constraints (i.e. $r_i$ values). There are several examples available on how to procede, like here, here or here.
2) If the dual problem is "easier" to solve, your approach sounds similar to using a primal-dual method. You can read an explanation on how it works at Section 5.1 of this document, while the original source where it was proposed for solving linear programs is several decades old.1
The general idea will be as follows:
Find a feasible dual solution $\lambda$.
Given $\lambda$, find some $x$ that minimizes the violation of complementary slackness in the primal problem. (This is the step that reminds me of what you are trying to approach when constructing $y$).
If complementary slackness holds, $y$ is optimal, and the algorithm terminates.
Otherwise, change $\lambda$ so as to improve the dual objective, and go to 2.
1 Dantzig, George Bernard, Lester Randolph Ford Jr, and Delbert Ray Fulkerson. A PRIMAL--DUAL ALGORITHM. No. P-778. RAND CORP SANTA MONICA CA, 1956.
|
2025-03-21T14:48:30.401956
| 2020-04-24T14:02:35 |
358418
|
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|
Stack Exchange
|
Euler–Maclaurin formula in $\mathbb{Z}^d$
I was wondering whether there is a Euler–Maclaurin formula of sorts for expressions such as
$$
\sum_{x \in [a,b]^d\cap \mathbb{Z}^d} f(x) - \int_{[a,b]^d}f(x)
$$
where $d\ge 2$ is an integer, $a,b \in \mathbb{R}$ and $f:\mathbb{R}^d \longrightarrow \mathbb{R}$ is a smooth function in $[a,b]^d$. I am particularly interested in such expansion with a control of the error term.
I appreciate any reference or suggestions.
See e.g. Y. Karshon, S. Sternberg, and J.Weitsman. The Euler-Maclaurin formula for simple integral polytopes and Y. Karshon, S. Sternberg, and J. Weitsman. Euler-Maclaurin with remainder for a simple integral polytope.
For an alternative summation formula -- in terms of integrals only, without using derivatives -- see Approximating sums by integrals only: multiple sums and sums over lattice polytopes.
In these papers, you will also find explicit expressions for and/or simple bounds on the remainder.
sorry for necromancing this thread but I wanted to give an elementary (with physics style abuse of notation) answer to this:
Yes there are some simple generalizations that support this. We need to recall where the Euler-Maclaurin formula comes from:
Let $I$ denote the identity operator. Let $H$ denote the shift operator $H[f] = f(x+1)$. Then (with abuse of notation) we can compute the following:
$$ \frac{I}{I-H} = I + H + H^2 + H^3 ... = f(x)+f(x+1)+f(x+2)+f(x+3)...$$
So then it follows that
$$ - \frac{I}{I-H} |_{x=a}^{x=b} = f(a)+f(a+1)+ \ ... \ + f(b-1) $$
Now if you recall that $H = e^{\frac{d}{dx}} $
We then have
$$ \sum_{k=a}^{b-1} f(x) = - \frac{I}{I-e^{\frac{d}{dx}}} |_{x=a}^{x=b} $$
Now recall on the right hand side that
$$ - \frac{1}{1-e^x} = \frac{1}{x} - \frac{1}{2} + \frac{1}{12}x + ... $$
So we conclude then that:
$$ - \frac{I}{I-e^{\frac{d}{dx}}} |_{x=a}^{x=b} = \int_{a}^{b} f dx - \frac{1}{2}(f(b)-f(a)) + \frac{1}{12} (f'(b)-f(a)) + ... $$
Which is the Euler Maclaurin formula.
Now we go to the multivariable case. Suppose we have a function $f(x,y)$ we have shift operators $H_x = H(x+1,y)$ and $H_y = H(x,y+1)$
It's easy to see that $H_x, H_y$ commute so that with abuse of notation we can write
$$ \sum_{n=a}^{b-1} \left[ \sum_{k=c}^{d-1} f(n,k) \right] = \frac{I}{I-H_x } \frac{I}{I-H_y}|_{x=a}^{x=b} |_{y=c}^{y=d} = \frac{I}{I-e^{\frac{\partial}{\partial x}}} \frac{I}{I-e^{\frac{\partial}{\partial y}}}|_{x=a}^{x=b} |_{y=c}^{y=d} $$
Thereby giving you that:
$$ \sum_{n=a}^{b-1} \left[ \sum_{k=c}^{d-1} f(n,k) \right] = \int_{c}^{d} \left( \int_{a}^{b} f(x,y) \partial x - \frac{f(b,y)-f(a,y)}{2} + \frac{ \partial_x[f]_{x=b} - \partial_x[f]_{x=a} }{12} + ... \right) \partial y + \\ -\frac{1}{2} \left( \int_{a}^{b} f(x,y) \partial x - \frac{f(b,y)-f(a,y)}{2} + \frac{ \partial_x[f]_{x=b} - \partial_x[f]_{x=a} }{12} + ... \right)|_{y=c}^{y=d} + ... $$
Which is just the Euler Maclaurin Formula for variables $x$ and $y$ composed with each other.
In general moving to $\mathbb{Z}^d$ follows naturally from this by just consider each variable $x_1 ... x_d$ separately and composing them just like we did for $\mathbb{Z}^2$ above.
As a more general corollary. If we want to evaluate
$$ f(x+c_0) + f(x+c_1) + f(x+c_2) + ... $$
Then we can let $T(x) = \sum_{n=0}^{\infty} x^{c_n}$. Then the operator $T(e^{\frac{d}{x}})$ when expanded via the Laurent series for $T(e^x)$ will give you the corresponding euler maclaurin formula.
As an example if we want to evaluate
$$ f(x) + f(x+1) + f(x+\sqrt{2}) + f(x+\sqrt{3}) + ... $$
We look at $\sum_{n=0}^{\infty} x^{\sqrt{n}} = \frac{\Gamma(1+2)}{(-z)^2} + \sum_{k=0}^\infty \zeta\left(-\frac{k}{2}\right)\frac{z^k}{k!}$ as per here: and therefore we conclude
$$ f(x+\sqrt{a}) + f(x+\sqrt{a+1}) + f(x+\sqrt{a+2}) + ... f(x+\sqrt{b-1}) = \\ \underbrace{-2\int \int f(x) dx}_{\text{bounds selected carefully}} - \frac{f(b)-f(a)}{0!2} - \frac{\zeta(-\frac{1}{2})(f'(b)-f'(a))}{1!} + \frac{(f''(b)-f''(a))}{2!12} + ... $$
|
2025-03-21T14:48:30.402185
| 2020-04-24T14:10:12 |
358419
|
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|
Stack Exchange
|
Dimension of the socle of the first local cohomology module
Let $M$ be a graded $\mathbb{C}[z_0,\dots,z_n]$-module. Using local duality one can show that
$$
\dim_\mathbb{C} (\text{soc} H_\mathfrak{m}^1(M))_k = \beta_{n,k+n+1}(M).
$$
Here $H_\mathfrak{m}^1(M)$ denotes the first local cohomology module of $M$ with respect to the ideal $\mathfrak{m} = (z_0,\dots,z_n) \subset \mathbb{C}[z_0,\dots,z_n]$, and $\beta_{i,j}(M)$ denotes the graded Betti numbers of $M$.
Has this formula appeared in print somwhere? The closest thing I could find was in https://arxiv.org/pdf/0809.1458.pdf, where is says on page 17 that "...the socle of the first nonzero local cohomology precisely reflects the top Betti numbers."
Edit: I also found an ungraded version of the formula in Theorem 12.4 in Combinatorics and Commutative Algebra by Stanley.
I would check the Brodmann-Sharp book on Local Cohomology.
|
2025-03-21T14:48:30.402274
| 2020-04-24T14:20:31 |
358422
|
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|
Stack Exchange
|
Comparing Euclidean norm of two normal vectors
Let $X_i$ ($i = 1,2$) be two random vectors in $\mathbb R^n$, with normal distribution with scalar covariance matrix $\sigma_i^2$ and center $\mu_i$ (in my case, $n = 2$). Is there a way to estimate the probability that $\Vert X_1 \geq \Vert X_2\Vert$ assuming, for instance, that $\Vert\mu_1\Vert \gg \Vert \mu_2 \Vert$ ? (this is the Euclidean norm).
What I tried so far:
directly integrating the joint distribution of $(X_1, X_2)$ on the cone: this does not seem possible.
writing the probability distribution for $\Vert X_1 \Vert^2$: this leads me to integrals of the form $\int_{-\pi}^{\pi} e^{\alpha \sin(\theta) + \beta \sin^2(\theta)} \sin^n(\theta) d\theta$, which I cannot make anything of.
What is a circular variance?
This means that the covariance matrix is scalar. (Since $n=2$ in my case, I meant that the variables are circularly symmetric). I edited the question to clarify.
What does "$|\mu_1|\gg |\mu_2|$" mean? That $\sigma_1+\sigma_2+|\mu_1|=o(|\mu_2|)$? -- You need to relate $|\mu_2|$, not only to $|\mu_1|$, but also to $\sigma_1$ and $\sigma_2$ in order to get a specific answer.
Do you want to upper or lower bound the probability?
@IosifPinelis You are obviously right; I left this “open” on purpose because I would be interested even in a result that needs a small tweak on the exact domination hypothesis. In my case I certainly have at least $\Vert \mu_2\Vert + \sigma_1 = o(\Vert \mu_1 \Vert)$, while $\sigma_2$ might be somewhat larger (though still $O(\mu_1)$).
Your question is still too "open". It needs to be more specific.
|
2025-03-21T14:48:30.402421
| 2020-04-24T15:18:32 |
358428
|
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|
Stack Exchange
|
Examples of noetherian local rings $R$ such that $K'_0(R)$ is not isomorphic to $\mathbb Z$
Does there exist a simple example of a commutative noetherian local ring $R$ such that $K'_0(R) = K_0(\mbox{Mod-}R)$ (by $\mbox{Mod-}R$ I mean the abelian category of finitely generated $R$-modules) is not isomorphic to $\mathbb Z$?
Do you mean $R$ to be commutative?
$\mathbb Q[X,Y]/(XY)$ localized at the origin. Or maybe you wanted a domain.
$\Bbb{Q}[[t^2,t^3]]$ should work.
Does it? It's clear that the group is generated by the free module of rank one and the residue field. But the latter is trivial in the group. (There is a principal ideal of codimension $2$ and another of codimension $3$.)
@TomGoodwillie How one can compute K' of your example?
@YCor Yes, but if you have a good noncommutative example, it will be interesting too.
By $\operatorname{Mod}_R$, do you mean finitely generated $R$-modules?
In my example there are two independent maps to $\mathbb Z$ given by doing two things to a module $M$: look at the rank of $M/YM$ as a module for localized $\mathbb Q[X]$, and look at the rank of $M/XM$ as a module for localized $\mathbb Q[Y]$.
@TomGoodwillie Thank you! But how we can see that they are really independent? It sounds very natural and intuitive, but how one can provide the formal argument?
@R.vanDobbendeBruyn Yes, you are right, I would have to write more accurately.
@TomGoodwillie: Sorry, I don't understand your argument. Are you saying that the residue field is the quotient of a principal ideal by another one? That would mean that it has finite projective dimension, which is not the case.
@abx: $M=\mathbb Q[[t]]$ is a module over your ring, and the exact sequence $M \to M\to \mathbb Q$ shows that $\mathbb Q$ is zero in the Grothendieck group.
@Hailong Dao: Ah, nice! Thank you.
This is just to flesh out @Tom Goodwillie example.
For any reasonable scheme $X$ and an open set $U$, one has a natural exact sequence,
$$K_0(X-U)\to K_0(X)\to K_0(U)\to 0.$$
Taking $X=\operatorname{Spec} (\mathbb{Q}[x,y]/xy)_{(x,y)}$ (or a number of similar examples) and $U$ the punctured spectrum, we note that the punctured spectrum is two points and thus $K_0(U)=\mathbb{Z}^2$. The kernel is generated by the closed point, but going mod $x+y, x+y^2$, one can easily see that 2 and 3 times the closed point is zero in $K_0(X)$. So, we get $K_0(X)=\mathbb{Z}^2$.
|
2025-03-21T14:48:30.402618
| 2020-04-24T15:55:06 |
358431
|
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|
Stack Exchange
|
Explicit check of the invariance of the Weil-Petersson form
Using Fenchel-Nielsen coordinates, the Weil-Petersson metric can be written as
$\omega_{WP} = \sum_{i} d\ell_i \wedge d \tau_i,$
where $i$ is an index labelling the curves of a pants decomposition of the surface in question and the pairs $(\ell_i,\tau_i)$ form the associated Fenchel-Nielsen coordinates. Although it is not manifest from the above expression, $\omega_{WP}$ is actual independent on the choice of pants decomposition.
Now, in this question it was asked and answered how the Fenchel-Nielsen coordinates change upon performing the so-called A-move or S-move on the pants decomposition. The answer is provided in this paper (and references there) by explicit transformation laws.
Let us consider the case of a once punctured torus, then a pants decomposition consists of a single curve and therefore we have $\omega_{WP} = d\ell \wedge d\tau$. I tried to check that $\omega_{WP} = d\ell \wedge d\tau = d\ell' \wedge d\tau'$, using the expressions given by Proposition 3.1 of the above paper (and setting $\ell_0 = 1$), but the equality does not seems to hold. Indeed, by putting numerical values one easily finds that the determinant of the transformation is not 1.
Where is the problem?
(Let me also add that it seems to me that there is a typo in the expression for the twist parameter, the last factor should be $\{ \dots \}^{(-1/2)}$ in order to agree with the same expression given in terms of the quantities $A$ and $B$ at pag. 7).
Apparently both the original paper by Okai and Proposition 3.1 in the paper mentioned by the OP contain typos in the expressions. Corrected formulas can be found in equations (5.13) and (5.22) of this recent arXiv submission (see also footnote 9 on page 75):
Jørgen Ellegaard Andersen, Gaëtan Borot, Séverin Charbonnier, Alessandro Giacchetto, Danilo Lewański, Campbell Wheeler, On the Kontsevich geometry of the combinatorial Teichmüller space, arXiv:2010.11806
Starting with their expressions it is easily checked with Mathematica that the determinant of the transformation is equal to 1:
Det@D[{2 ArcCosh[ Cosh[t/2]/Sinh[l/2] Sqrt[(Cosh[l] + Cosh[l0/2])/2]],
-2 ArcCosh[ Cosh[l/2] Sqrt[
((Cosh[t/2]^2 (Cosh[l] + Cosh[l0/2]) - 2 Sinh[l/2]^2) /
(Cosh[t/2]^2 (Cosh[l] + Cosh[l0/2]) + Sinh[l/2]^2 (Cosh[l0/2] - 1)))]]},
{{l, t}}] // FullSimplify[#, t > 0 && l > 0 && l0 > 0] &
1
Another doubt: if you set the boundary lengths to 0 (or just equal) in the formula for the sphere, the formula becomes completely symmetric in i=1,2,3,4. But then it says that the length of the curve separating 12|34 and 13|24 are equal which doesn't make sense (for instance you could send them both to 0 simultaneously although they intersect). How is it possible?
Can you elaborate? What formulae are you referring to? $\ell'(\ell, \tau)$ should depend non-trivially on $\tau$.
I was thinking of formula 5.12, which gives $\ell'(\ell,\tau)$, where $\ell'$ is the length of the curve 12|34 and $\ell$ that of the curve 14|23. What I meant is that the formula seems to give the same result for the curves 12|34 and 13|24, because it is symmetric in the four boundary components and I would imagine that the formula giving the length $\ell''$ of the curve 13|24 is given by the same formula upon permuting the boundary labels?
You cannot simply permute the boundary components, because you would have to transform $\tau$ accordingly as it is tied to the boundary components.
But then, is there a formula that gives both $\ell'$ and $\ell''$ using a single $\tau$? After all, if $\ell$ and $\tau$ are fixed one should be able to write down the lengths of all geodesics in the sphere
If I'm not mistaken, you have $\ell''(\ell,\tau) = \ell'(\ell,\tau + (k+\tfrac12) \ell)$ when the boundary components are of equal length, where $k$ is an integer parametrizing the possible choices of the curve 13|24. This is because interchanging the boundaries 2 & 3 is like performing half a Dehn twist on the 14|23 curve.
Thanks, I had the same feeling too.
|
2025-03-21T14:48:30.403284
| 2020-04-24T16:20:41 |
358432
|
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|
Stack Exchange
|
Mostow Rigidity Theorem and reconstruction from fundamental group
The Mostow Rigidity Theorem is phrased in terms of a relationship between isometries and isomorphisms of fundamental groups, which raises an obvious question. Given the fundamental group of a complete finite-volume hyperbolic manifold of dimension $> 2$, is it possible to reconstruct the hyperbolic manifold?
Yes, but it's quite hard and is not algorithmic to the best of my knowledge. By the way, you should specify what you mean by "reconstruct."
What do you mean by reconstruct? Do you have an algorithmic question in mind? If so, how do you encode the group, and how do you encode the manifold? (a possible output is a triangulation but it doesn't say what the metric is)
I was thinking more generally about a mathematical construction rather than an effective algorithm, so any answer is fine. I don't have any particular application in mind; I was more thinking about why rigidity theorems are often stated in isomorphism form.
At a theoretical level the Mostow rigidity theorem precisely say that $\Gamma$ determines the manifold $X_\Gamma$ up to isometry. If I understand correctly, the question is to have a more "constructive" description of $\Gamma\mapsto X_\Gamma$? Maybe asking the manifold only should be more reasonable, because if you can output the metric easily, basically you reprove Thurston's hyperbolization conjecture (in this case in dim 3, from my very rough understanding the metric is output by the Hamilton-Perelman Ricci flow procedure).
... reading Toffee's answer, I understand that something is doable without reproving Thurston-Hamilton-Perelman... I understand (again, roughly) that if we know beforehand that hyperbolization exists, then it can be done in some kind of effective way, by "retrieving" the relevant representation $\Gamma\to\mathrm{PSL}_2(\mathbf{C})$.
The algorithm given by Toffee in their nice answer was described by Jason Manning in this 2002 G&T paper: https://arxiv.org/abs/math/0102154 .
How is your fundamental group $\Gamma$ given to you? As a presentation in terms of generators and relations? Here is an answer for hyperbolic $3$-space that can probably be generalized to $\mathbb{H}^n$, $n \ge 4$, with some effort.
In short, you compute the $\mathrm{SL}_2(\mathbb{C})$ character variety. The relevant ideas and facts I use below are in Culler and Shalen's famous paper [CS].
Here is an algorithm.
Compute the algebraic set $\mathrm{Hom}(\Gamma, \mathrm{SL}_2(\mathbb{C}))$ from your presentation. This is naturally an affine algebraic set in $\mathbb{C}^{4 d}$, where $\Gamma$ has $d$ generators.
Then, there is an explicit set of elements $\sigma_1, \dots, \sigma_n \in \Gamma$ for which the trace function
$$
t(\rho) = \left(\mathrm{tr}(\rho(\sigma_1)), \dots, \mathrm{tr}(\rho(\sigma_n))\right) \in \mathbb{C}^n
$$
has image the $\mathrm{SL}_2(\mathbb{C})$ character variety, $X(\Gamma)$. See Proposition 1.4.1 in [CS].
Compute this algebraic set in $\mathbb{C}^n$ and find its irreducible components (maybe in Macaulay?).
Mostow rigidity tells you that a discrete and faithful representation $\rho : \Gamma \to \mathrm{SL}_2(\mathbb{C})$ determines an isolated point of $X(\Gamma)$. You might get several isolated points in $X(\Gamma)$ from different lifts from $\mathrm{PSL}_2$ to $\mathrm{SL}_2$ and from different Galois conjugates of the discrete and faithful representation, not to mention possibly other random isolated points. (Small nitpick: In the finite volume, noncompact case you need to also cut out the equations that traces of peripheral elements are all $2$ in order to get isolated points. You need to cut away deformations related to Dehn filling.)
There is a standard way described in [CS] to algorithmically lift a point on $X(\Gamma)$ back to a representation $\rho$. Basically you choose a lift so that your first generator is upper-triangular with the right trace and $1$ in the upper right entry. You make the second generator lower-triangular of the right trace, where now the lower left entry is going to depend on the trace of the product of the two generators, etc. Compute such a $\rho$ for each isolated point of $X(\Gamma)$.
Now, compute a Dirichlet or Ford domain in $\mathbb{H}^3$ for each $\rho(\Gamma)$ from your list of isolated points. One of these will terminate to give you a fundamental domain for the complete structure.
Now you have the representation associated with the complete structure and a fundamental domain. So if that is your desired notion of "reconstructing" the manifold, there you go.
A last remark on the finite volume case. Your presentation oracle also needs to enumerate the conjugacy classes of $\mathbb{Z}\oplus \mathbb{Z}$ subgroups - these are precisely the peripheral subgroups. (The manifold is homeomorphic to the interior of a compact manifold with boundary a union of essential tori.) As described in [CS], Thurston proved that the dimension of the character variety at the complete structure equals the number of cusps, and the complete structure is cut out by making each peripheral subgroup unipotent (in fact, setting the traces of these elements all equal to $2$ cuts out the desired point). You can probably make finding these algorithmic in just the presentation. The rank of the abelianization of $\Gamma$ is an upper bound for the number of peripheral subgroups (by Poincaré duality), so one knows when enough subgroups have been found. This point is irrelevant in higher dimensions.
[CS] Culler, Marc; Shalen, Peter B. Varieties of group representations and splittings of 3-manifolds. Ann. of Math. (2) 117 (1983), no. 1, 109–146.
Thanks for this great answer. It's more than what I was expecting to get.
the rigidity result which gives discrete+faithful => isolated in character variety would be local (Calabi--Weil) rigidity rather than strong (Mostow) rigidity.
Have the isolated points that do not come from the monodromy representation or its Galois conjugates ever been looked at? I can't even think of an example.
@JeanRaimbault That's right about local rigidity, except for the fact that local rigidity fails for nonuniform lattices in $\mathrm{SL}_2(\mathbb{C})$. Local rigidity only holds once you deem the parabolic subgroups must stay parabolic (i.e., traces of peripheral elements are $\pm 2$).
@JeanRaimbault As for other isolated points, I don't know an example off the top of my head. You could probably build one with a dominant map between closed hyperbolic $3$-manifolds that isn't a covering. Such an example is probably sufficiently complicated that cranking out the character variety isn't practical though. There are known examples of dominant maps between knot complements, so you see the canonical component of the target knot complement in the character variety of the other one. I can't remember one off-hand, but papers by Boyer and Zhang are a likely source.
|
2025-03-21T14:48:30.403737
| 2020-04-24T17:00:25 |
358436
|
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|
Stack Exchange
|
Is the point giving the width in strictly convex surface a cut point?
Assume that $\Sigma$ is a stricly convex surface in $\mathbb{E}^3$ homeomorphic to a sphere. Further, assume that $p_0,\ p_1\in \Sigma$ are intersection points with planes $z=0,\ z=1$ and the surface $\Sigma$ is between the two planes. Then $p_1$ is a cut point of $p_0$ ? (I do not know whether or not this is true)
Definition : Consider a intrinsic metric on $\Sigma$, a length of simple path.
Then $p_1$ is not a cut point of $p_0$ if there is unique shortest
path from $p_0$ to any point sufficiently close to $p_1$.
Could you please define "cut point"? Thanks.
Thanks for the definition.
Counterexample : Consider a chart $T(u,v)=(0,u,0)+f(u)(\cos\ v,0,\sin\ v),\ f(u)=1-u^4,\ -1\leq u\leq 1,\ v\in [-\frac{\pi}{2}-\epsilon,\frac{\pi}{2}+\epsilon]$ for the surface. Here $T(0,v)$ is a geodesic and sectional curvature is zero along the geodesic so that Jacobi field is $t(0,1,0)$ along $T(0,t+\frac{\pi}{2})$
No.
It is clear that this is true for a sphere, but cutting of a spherical cap from the sphere leaves the intersection points with the planes the same, but now there is a unique shortest path through this new flat area.
$\qquad\qquad\qquad\qquad\quad$
Of course you can make this object strictly convex, differentiable etc. by smoothing out the cut.
Intuitively, it seems that there is no reason that $p_1$ in a tilted
ellipsoid is a cut point of $p_0$.
Because the orientation of the ellipsoid is a global issue
(i.e., the location of $p_0$ and $p_1$ change with the tilt), whereas
cut-point-ness is intrinsic.
In this symmetric example, perhaps there are a pair of equal-length geodesics
connecting $p_0$ to $p_1$, but if I arranged $\Sigma$ to be asymmetric
with respect to the coordinate axes, there would
be a unique shortest geodesic connecting those two $z=0,1$ tangency points.
|
2025-03-21T14:48:30.403896
| 2020-04-24T17:02:56 |
358437
|
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|
Stack Exchange
|
Almost one-to-one endomorphism of minimal subshift?
Let $(X,T)$ be a minimal subshift. Can it happen that an endomorphism $\varphi\colon (X,T) \to (X,T)$ is almost 1-to-1 but not 1-to-1? Can it happen that a factor $\pi\colon (X,T) \to (Y,T)$ between minimal subshifts is almost 1-to-1 but not 1-to-1? I know that, for example, Toeplitz subshifts are almost 1-to-1 extensions of a Cantor system (some odometer), but in the subshift case I couldn't find anything on internet.
PD. Here, a factor $\pi\colon(X,T)\to(Y,T)$ is almost 1-to-1 if $\exists y \in Y$ such that $\#\pi^{-1}(y) = 1$. From this, one can prove that $\#\pi^{-1}(y) = 1$ for all $y$ in a residual set.
@YCor I added the definition!
Isn't Downarowicz' royal couple an example of this?
IIRC only a special point contains a royal couple (and has multiple preimages), others have only normal couples and thus unique preimages.
The answer is yes. In this paper Downarowicz proves the following theorem
There exists a regular Toeplitz sequence $\omega$ such that the induced
Toeplitz flow $(\bar O(\omega), S)$ is noncoalescent, more precisely, it admits an endomorphism $\pi : \bar{O}(\omega) \to \bar{O}(\omega)$ of the first type.
Here, $\bar O(\omega)$ is the orbit closure (so a Toeplitz subshift since $\omega$ is Toeplitz), $S$ the shift map, noncoalescent means not injective, and first type means every Toeplitz point has a unique preimage, in particular some point does.
|
2025-03-21T14:48:30.404023
| 2020-04-24T17:24:12 |
358441
|
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|
Stack Exchange
|
Mac Lane's planarity criterion and 2-bases
Mac Lane's planarity criterion states that a graph is planar if and only if its cycle space has a basis such that each edge is contained in at most two cycles. We call a basis with this property a 2-basis. Finding a 2-basis for a planar graph is easy; just find an embedding of the graph in the plane and use the faces of the graph as the cycle basis.
I am interested in the more general algebraic version of this problem. Given a finite dimensional vector space $V$ over $\mathbb{Z}_2$ and a basis $B$, can I determine if $V$ has a 2-basis? That is, does there exist an algorithm that takes $B$ as input and produces a 2-basis if one exists?
I have not encountered any references to the idea of a 2-basis outside of planar graphs. Have these bases been studied in any other context? What is known about them?
Let's see if I can get this straight.
So a binary vector space has a 2-basis if and only if it is a cographic matroid (i.e., the dual of a graphic matroid).
An equivalent matroid-terminology-free statement is that a binary vector space has a 2-basis if and only if it is the cocycle space of a graph (i.e. the vector space generated by all edge-cuts of the graph).
MacLane's result arises because if a binary vector space is simultaneously the cycle space of a graph and the cocycle space of a graph then it is planar.
As far as I know, this is due to Dominic Welsh in the paper "On the hyperplanes of a matroid" (https://doi.org/10.1017/S0305004100044017) where he called it a "2-complete basis".
|
2025-03-21T14:48:30.404158
| 2020-04-24T17:35:58 |
358442
|
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Stack Exchange
|
Nonnegativity implies $\langle Lf,f\rangle\geq \int f^2-(\int fg)^2$ for $g\geq 0$
I have been a lot of time trying to understand a key step on a paper about spectral analysis but I have no clue how to prove it (and the authors only said "by standard analysis"). Let me state the question (this is how I interpret actually, I tried to clean it since I prefer to avoid all the details and definitions of the paper). Consider the following 1D-Schrödinger operator $$
L:=-\partial_x^2+c_1-c_2\Phi
$$
where $c_1,c_2>0$ and $\Phi$ is a positive Schwartz function.
Now, I can prove that due to the specific structure of the operator and $\Phi$, $L$ is a nonnegative operator (zero is its first eigenvalue, which is simple). Moreover, the eigenfunction associated to its first eigenvalue (zero), let's say $\zeta$, is a positive Schwartz function (it's some power of $\Phi$ actually). Here is where everything become a bit dark for me. Is it true that if I consider a nonnegative function $g\in L^2\setminus\{0\}$, then, due to the spectral information of $L$ above, there exist $\lambda>0$ (depending on my election of $g$) such that for all $f\in H^1(\mathbb{R})$ it holds: $$
\langle Lf,f\rangle\geq \lambda\int f^2-\dfrac{1}{\lambda}\left(\int fg\right)^2?
$$
I am quite surprised that it seems that I can "choose" this $g$ (as soon as restricting myself to nonnegative functions not identically zero). Does anyone has any idea on how to prove it? Or maybe some recommended references?
PS: Notice that since the eigenfunction associated to the zero eigenvalue is positive, and we are also
assuming $g$ positive, then $g$ cannot be orthogonal to $\zeta$. Thus, you cannot choose $g\perp \zeta$ and then try to choose $f=\zeta$ to make the left-hand side equals zero (while the right-hand side would remain strictly positive).
Since you are mentioning Schwartz functions and functions $f \in H^1(\mathbb{R})$, do I get this right that your problem is posed on the whole real line rather than on a bounded interval?
@JochenGlueck Yes, in the whole space, sorry for the misunderstanding
It would be nice if you linked the paper from which the doubt arose.
Under your assumptions the essential spectrum is $[c_1, \infty[$, hence the part of the spectrum in $[0,c_1[$ is discrete. Let $\mu>0$ be the second eigenvalue (the first is 0), if it exists, or $c_1$. Then $(Lh,h) \ge \mu (h,h)$ if $h$ is orthogonal to $\zeta$. Next assume by contradiction that $(Lf_n, f_n) +(f_n,g)^2 \le n^{-1}\|f_n\|^2$ and $\|f_n\|=1$ and split $f_n=c_n \zeta+h_n$ with $(\zeta, h_n)=0$. Then $(Lf_n,f_n)=(Lh_n,h_n) \to 0$, hence$\|h_n\| \to 0$ and $|c_n| \to 1$. Next $(f_n,g)=(c_n \zeta+h_n,g) \to 0$, which is impossible since $|c_n| \to 1$ and $(\zeta,g)>0$.
|
2025-03-21T14:48:30.404371
| 2020-04-24T17:56:32 |
358443
|
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Stack Exchange
|
Is there a known closed form expression for this integral?
I am interested in the following integral:
$$f(x,y) = \int_{\mathbb{S}^d} \max(0,x^Tw)\cdot\max(0,y^Tw) \, dw, \qquad x,y\in\mathbb{S}^d,$$
where $\mathbb{S}^d\subset\mathbb{R}^{d+1}$ is the $d$-dimensional unit sphere. Here, $x^Ty$ denotes the dot-product/inner-product between $x$ and $y$.
Is there a known closed-form expression for this integral? It is clear that $f(x,y)$ is isotropic, i.e., if $Q$ is an orthogonal transformation then $f(x,y) = f(Qx,Qy)$. Therefore, I believe that $f(x,y) = \phi(x^Ty)$ for some function $\phi : [-1,1]\rightarrow \mathbb{R}$. Ideally, I would know $\phi$.
As you've pointed out, the only parameter that matters here is the angle $\theta$ between $x$ and $y$. To see how, consider instead the Gaussian integral:
$$
I(x,y)=\frac{1}{(2\pi)^{(d+1)/2}}\int_{u\in\mathbb{R}^{d+1}}\max(0,x^Tu)\cdot\max(0,y^Tu)\exp(-\frac{1}{2}u\cdot u)du
$$
The integral you are interested in is obtained by changing to $(d+1)$ dimensional spherical coordinates, and easily integrating over the radial coordinate. As such, it will be enough to just focus on the Gaussian integral.
I like computing these types of integrals by forming an orthonormal basis of $\mathbb{R}^{d+1}$ like so:
$$
b_1=(x+y)/\|x+y\|\\
b_2=(x-y)/\|x-y\|
$$
...and where the remaining basis elements are chosen to be orthogonal to $b_1$ and $b_2$. With respect to this basis we have:
$$
x=(a,b,0,...,0)\\
y=(a,-b,0,...,0)
$$
where $a=|\cos(\theta/2)|$ and $b=|\sin(\theta/2)|$. Of course the components of $u$ change as does the Gaussian measure, but the Gaussian measure is invariant under rotations so I'll suppress relabeling the components of $u$ with respect to the new basis. After easily integrating over the last $(d-1)$ coordinates of $u$ we write the integral $I(x,y)$ as:
$$
\frac{1}{2\pi}\int_{(u_1,u_2)\in R}(a\cdot u_1+b\cdot u_2)\cdot (a\cdot u_1-b\cdot u_2)\exp(-\frac{1}{2}(u_1^2+u_2^2))du_1du_2,
$$
where $R=\{(u_1,u_2)|(a\cdot u_1+b\cdot u_2)>0\textrm{ and }(a\cdot u_1-b\cdot u_2)>0\}$ is the region where the two maximums are nonzero. After a change of coordinates $v_1=au_1$ and $v_2=bu_2$ we obtain:
$$
\int_{(v_1,v_2)\in R'}\frac{\left(v_1^2-v_2^2\right) e^{\frac{1}{2}
\left(-\frac{v_1^2}{a^2}-\frac{v_2^2}{b^2}\right)}}{2 \pi a b}dv_1dv_2
$$
where $R'=\{(v_1,v_2)|(v_1+v_2)>0\textrm{ and }(v_1-v_2)>0\}$. From here, change to polar coordinates $(r,t)$ and integrate over the radial coordinate to obtain:
$$
\int_{-\pi/4}^{\pi/4}\frac{a^3 b^3 \cos (2 t)}{\pi \left(a^2 \sin ^2(t)+b^2 \cos ^2(t)\right)^2}dt
$$
Using Mathematica for example, we obtain:
$$
\int_{-\pi/4}^{\pi/4}\frac{a^3 b^3 \cos (2 t)}{\pi \left(a^2 \sin ^2(t)+b^2 \cos ^2(t)\right)^2}dt = \frac{a b+(a-b) (a+b) \tan ^{-1}\left(\frac{a}{b}\right)}{\pi }
$$
From here, you can then express everything in terms of the inner product of $x$ and $y$ along with the norms of $x$ and $y$.
Of course, hopefully the result can be checked! May I ask where this integral came up? I bet there are others who can obtain something similar using some pretty slick symmetry arguments.
Cheers!
|
2025-03-21T14:48:30.404561
| 2020-04-24T18:22:22 |
358444
|
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Stack Exchange
|
Algebraic property of a transition matrix
Consider the simple random walk on $\mathbb{Z}^2$. Given a finite $\Sigma \subset \mathbb{Z}^2$, one can induce the random walk on $\Sigma$: set $\tau_0 = 0$, and define recursively $\tau_{n+1} := \inf \{k > \tau_n : \ S_k \in \Sigma\}$. Then $(X_n) = (S_{\tau_n})$ is a Markov chain on $\Sigma$. Let $P$ be its transition matrix.
The matrix $P$ can be computed using potential theory, in a way akin to the XKCD "nerd sniping" problem (see here).
I decided to do this for a simple example, $\Sigma = \{A, B, C\}$ with $A = (0,0)$, $B = (-1,0)$ and $C=(1,1)$. The computation is somewhat lenghty, but hopefully I made no mistake:
$$P = \frac{1}{-\pi^2+8\pi-4} \left( \begin{array}{ccc}
-\frac{1}{2}\pi^2+4\pi-4 & -\frac{1}{2}\pi^2+3\pi & \pi \\
-\frac{1}{2}\pi^2+3\pi & -\pi^2+6\pi-4 & \frac{1}{2}\pi^2-\pi \\
\pi & \frac{1}{2}\pi^2-\pi & -\frac{3}{2}\pi^2+8\pi-4 \\
\end{array}
\right)$$
So, a few remarks:
The result is a symmetric matrix, which comes from the fact that the transition kernel of the simple random walk is itself symmetric.
It can be written as $F(\pi)$ where $F$ is a rational function with rational coefficients. This is not surprising, because, using Fourier transform, it can be evaluated from trigonometric integrals such as
$$\frac{1}{\pi^2} \int_{[-\pi, \pi^2]} \frac{\sin^2 (\xi_1)}{\sin^2 (\xi_1)+\sin^2 (\xi_2)} \ \text{d} \xi_1 \text{d} \xi_2,$$
and using algebraic operations on matrices with coefficients in $\mathbb{Q} (\pi)$.
The rational function $F$ above satisfies $F(0) = I$.
And the last fact I don't understand. It seems too perfect to be random chance, but I don't see where it comes from. So, why should we have $F(0)=I$ ?
|
2025-03-21T14:48:30.404696
| 2020-04-24T18:46:09 |
358449
|
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|
Stack Exchange
|
If the convolution of two functions $f\star g$ is equal to $g$, $f$ is even with compact support and $g$ is bounded, implies that $g$ is constant?
Let $f$ be an even continuous function with compact support such that
$$
\int f(t)\,\mathrm{d}t=1,
$$
and let $g$ be a bounded continuous function such that the convolution $f\star g$ satisfies the following equality
$$
(f\star g)(x)=g(x).
$$
How to prove that if $g$ has global minimum then $g$ is constant? I thought about using the Convolution Theorem, but it seems it doesn't work in this case. Maybe, without the requirement that $g$ has global minimum, there is a way to prove that $g$ is linear?
If the global minimum for $g$ occurs on a set $E$, then write out $f\star g(x)$ for $x\in E$.
@Anthony Quas, could you add some details? What should I get if I write f*g(x)?
@AntonSorokovskiy, you'll get something greater than $g(x) = \min g$, unless $E = \mathbb R$ (where I assume all this lives). This is not research level.
@LSpice, I thought about it, but what if g can be negative on part of its support?
@LSpice, could you help me to get this inequality?
Oh, I was, and perhaps @AnthonyQuas was also, assuming that $f$ was non-negative.
The sought-after statement is wrong: $g$ can be non-constant. Fourier transforming your conditions we see that the Fourier transform $\hat{g}$ is supported at points where $\hat{f}$ is equal to $1$. We also see $\hat{f}(0)=1$ and $\hat{f}(-\xi)=\hat{f}(\xi)$. This motivates the following condition.
It should be trivial to construct an even function $f$ with compact support and such that $\int f(x)dx=1$ and $\int f(x)\cos(x)\,dx=1$. Then take $\hat{g}$ to be supported at $0$ and $\pm 2\pi$, for instance $g(x)=\cos(x)$. Then
$$
(f\star g)(y) = \int f(x) g(y-x) dx = \cos(y) \int f(x) \cos(x)\,dx + \sin(y) \int f(x)\sin(x)\,dx = \cos(y) = g(y)
$$
where I used $\int f(x)\cos(x)\,dx = 1$ by construction of $f$, and $\int f(x)\sin(x)\,dx = 0$ because $f$ is even.
Per Fourier inversion, we could take for $f$ the extension by $0$ of the function $x \mapsto \frac1{2\pi}(1 + 2\cos(x) + \cos(2x))$ on $[-\pi, \pi]$.
In order to use the machinery of the Fourier transform and deduce that $\hat f \hat g = \hat g$ from $f \star g = g$, you tacitly assume that $g$ is a function that admits a Fourier transform which is itself a function. But this excludes, among others, precisely the constant functions (the Fourier transforms of which are Dirac distributions).
|
2025-03-21T14:48:30.404887
| 2020-04-24T19:25:09 |
358450
|
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|
Stack Exchange
|
Bialynicki-Birula decompositions and fixed points
I was reading Luna's paper Toute variété magnifique est sphérique and stumbled on a few facts about Bialynicki-Birula decompositions and fixed points that I don't understand.
Here is the setup. Let $G$ be a connected reductive group over an algebraically closed field $k$ (of characteristic 0, though I'm not certain if that matters), and fix a Borel subgroup $B \subset G$ and a maximal torus $T \subset B$. Denote by $B^-$ the opposite Borel group to $B$ containing $T$. Let $X$ be an irreducible, normal, complete $G$-variety, and suppose that $X$ has finitely many $G$-orbits. Given any one-parameter subgroup $\lambda: \mathbb{G}_m \to T$ and any $y \in X^T$, we write
$$X(\lambda,y) = \{x \in X\ |\ \lim_{t \to 0} \lambda(t)x = y\}.$$
Here are the claims Luna makes that I don't understand:
(1) The fixed point set $X^T$ is finite.
(2) We are mainly interested in the case where $\lambda$ is in the Weyl chamber of $B$ (i.e.\ where $\langle \lambda, \alpha \rangle > 0$ for all positive roots $\alpha$), so that $X(\lambda,y)$ is $B$-stable. Luna states that for a "sufficiently general" such $\lambda$, we will have $X^{\mathbb{G}_m} = X^T$, where $\mathbb{G}_m$ acts on $X$ via its image under $\lambda$. He also states that in this case, the $X(\lambda,y)$ for various $y \in X^T$ form the Bialynicki-Birula decomposition of $X$.
(3) With $\lambda$ satisfying the conditions in (2), if $X(\lambda,y)$ is open, then $y$ is fixed by the opposite Borel subgroup $B^-$. (Luna doesn't say anything about this, but I'm also curious: is it true that if $y$ is fixed by $B^-$, then $X(\lambda,y)$ is open?)
All of these statements seem pretty reasonable to me, and I've worked them out in the case where $X = \mathbb{P}(V)$, $G = \mathrm{SL}(V)$, and $B$ (resp. $T$) is the subgroup of upper triangular (resp. diagonal) matrices. In this case, everything is clear using projective coordinates, but I don't know how to make these types of arguments without appealing to coordinates like that. Any proofs (or references to proofs) would be much appreciated!
I think all of these should be easy enough to resolve. First note that (1) is a triviality from your assumption that $G$ has finitely many orbits on $X$, because a maximal torus of $G$ can only have finitely many fixed points on $G/H$.
Now recall that by a beautiful result of Sumihiro (in the case $G$ is an connected linear algebraic group) given a normal $G$-variety $X$ we and an orbit $Y \subset X$ we can find a $G$-stable opens $Y \subset U \subset X$ such that $U$ is isomorphic to a $G$-stable locally closed subset of $\mathbb{P}(\rho)$ for $\rho$ a finite dimensional representation of $G$, a well-known corollary allows us to let $U$ be affine when $T$ is a split $k$-torus. Since all of your questions are about the local structure of orbits in $X$ unless I am misreading you, I will just assume $X \subset \mathbb{P}(\rho)$ is locally closed for $(V, \rho)$ some fixed representation from now on.
For (2), $T$ acts on $V$ as such $V \cong \oplus_i V_{\chi_i}$ for $\chi_i$ various distinct characters of $T$, then let $\lambda$ be sufficiently general in the sense that $\langle \lambda, \chi_i - \chi_j \rangle \neq 0$ for $i \neq j$. Then the eigenvalues $\langle \lambda, \chi_i \rangle$ are distinct so $\lambda$ and $T$ induce the same eigendecomposition of $V$ and thus a point $x \in \mathbb{P}(V)$ is fixed by one iff it is fixed by the other. To me the definition you gave is the definition of the B-B decomposition, so you will have to elaborate on what definition you are working with if you want me to show that this induces the B-B decomposition.
For (3) lets take $y \in X^T$ such that $X(\lambda, y) = U$ is the big cell, and take $\lambda$ as above to be a regular cocharacter of $T$ wrt $B$. We know that for any $b \in B^-$ we have that $\text{lim}_{t \to 0} \lambda(t)^{-1}b\lambda(t) \in T$ by explicit computation of the BB decomposition for $G$ (alternatively if you use the dynamic approach to parabolics this is by definition).
Further considering $A: B^- \to X$ the action map $A(b) = b\cdot y$ we know that there is an open subscheme $V$ of $B^-$ such that $V \cdot y \subset U$. But for $b \in B^-$ $\text{lim}_{t \to 0} \lambda(t)^{-1} b \cdot y = \text{lim}_{t \to 0} b^{\lambda(t)^{-1}} \lambda(t)^{-1} \cdot y = y$, since $y$ is torus-stable. But then for $b \in V$ we have that $b \cdot y$ has both limits $\text{lim}_{t \to 0} \lambda(t) \cdot b \cdot y$ and $\text{lim}_{t \to \infty} \lambda(t) \cdot b \cdot y$ defined, which defines a map $\mathbb{P}^1 \to X$, the image of this map lies inside of $B^- \cdot y$, which is affine because $B^-$ is solvable, therefore it is constant. Thus $b\cdot y$ is a fixed point for $\lambda$, thus by $(2)$ it is a fixed point for $T$, for any $b \in V$. Because the stabilizer of $y$ is closed and $V$ is dense in $B^-$ we are done, $B^- \cdot y = y$.
Hope this helps, let me know if anything is unclear.
Thanks so much for your reply! This is really helpful to me. Two minor questions: (1) When you say that $B^- \cdot y$ is affine, are you using that $y$ is $T$-stable and that orbits of unipotent groups are affine, or is there some other result like this? (2) I'm not familiar with the statement that $T$ has finitely many fixed points on $G/H$. I flipped through Borel's book on algebraic groups but couldn't find it. Can you explain more and/or provide a reference?
Hi Michael, it’s a general result that solvable affine linear algebraic groups have affine orbits, the result is just predicated on the fact that it’s true for unipotents and for tori, and one can do induction by dimension. But you are right that in this case it’s simpler.
So if T has a fixed point on G/H then there is some g in G such that T is conjugate by g to a subgroup of H, whence a maximal torus of H. As all maximal tori of H are conjugate we may as well assume g sends T to a specific maximal torus inside of H. But for T, T’ maximal tori of G, the set of g in G which send T to T’ is clearly a torsor for the normalizer of either torus, whence finiteness modulo H.
|
2025-03-21T14:48:30.405400
| 2020-04-24T19:31:12 |
358451
|
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|
Stack Exchange
|
Advice for graphic tablet for math
With the current Coronavirus disease (COVID-19), many of us had to switch all our activity to full online mode.
I am wondering whether some of you had the chance to use graphic tablets. I am looking in particular for advice regarding the best model of graphic tablet for math (in particular, for research, giving online seminars, teaching, etc).
I am not looking for expensive tablets with a screen (such as ipad or similar), but simple, possibly not too expensive, graphic tablet.
Wacom intuos is perhaps the standard one? I'm using one for my teaching this semester.
There are better forums for this question. Academia and Math Educators are two, and there may be hardware recommendation forums also on StackExchange. I won't vote it down, but some of the community may find it too far off the scope of this forum. Gerhard "Every Question Has Its Place" Paseman, 2020.04.24.
I think that the main point is to practise. Writing is very slow and untidy at first, but improves a bit with time.
you might want to consider an analogue solution: a white board which you place in front of your web cam when you give a seminar or lecture.
You can have a look at this older question (it is rather old and it is closed - but perhaps you can find there some useful information): Best tablet computer for mathematics. This is a slightly different question, but some users mentioned what they use: Tools for long-distance collaboration. (Again, this comes with the warning that the information there might be dated.)
|
2025-03-21T14:48:30.405545
| 2020-04-24T19:48:33 |
358454
|
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|
Stack Exchange
|
Generators and relations for the 2-dimensional unoriented cobordism category
It is very well known in the field of TQFT that the 2-dimensional oriented cobordism category is generated by the disk and the pair of pants (each going in both directions), subject to a finite set of relations. Those generators and relations are equivalent to the morphisms and axioms of Frobenius algebras.
What is the analogue statement in the unoriented case? It is easy to see that it suffices to add the Moebius strip to the generators, but is there a provably sufficient set of relations for them?
I feel like this must be worked out somewhere, but I'm having trouble to find anything. If someone could give a reference where generators and relations are listed, this would be helpful!
My initial answer was wrong, here's the correct version plus a reference: Turaev-Turner
New generating morphisms: The Mobius strip $\emptyset \rightarrow S^1$ and the "orientation reversing" diffemorphism of the circle $S^1 \rightarrow S^1$.
New relations: Orientation reversal is an involution. Orientation reversal plays well with all the other generators. Orientation reversal composed with a twice punctured unoriented surface is itself. The punctured Klein bottle is both the product of two Mobius strips, and a composition of copairing, orientation reversal, and pants.
Algebraically, you have a Frobenius algebra involution $\phi: A \rightarrow A$ and an element $\theta \in A$ such that:
$\phi(\theta v) = \theta v$
$(m \circ (\phi \otimes id) \circ \Delta)(1) = \theta^2$.
It actually follows from these definitions, see Lemma 2.8, that $\theta^3 = (m \circ \Delta)(\theta)$. That is the simplest relation you can state without making reference to the orientation reversing map, but you need the orientation reversing map in order for $\mathrm{Hom}(S^1,S^1)$ to be correct, which is why my initial guess was wrong.
|
2025-03-21T14:48:30.405695
| 2020-04-24T19:58:02 |
358455
|
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|
Stack Exchange
|
Has the exponentiation of ordinals a nice geometric model?
It is well known that the sum $\alpha+\beta$ of two ordinals $\alpha,\beta$ can be defined "geometrically" as the order type of the sum $(\{0\}\times \alpha)\cup(\{1\}\times\beta)$ endowed with the lexicographic order.
Also the product $\alpha\cdot\beta$ of ordinals $\alpha,\beta$ is the order type of the Cartesian product $\beta\times\alpha$ endowed with the lexicographic order.
What about the exponentiation of ordinals?
Does $\alpha^\beta$ have some nice "geometric'' or combinatorial model?
Maybe as some set of (partial) functions endowed with a suitable well-order?
"Maybe as some set of (partial) functions endowed with a suitable well-order?" Does "the set of partial functions $\beta\rightarrow\alpha$ which are nonzero at only finitely many values, ordered by comparing the point of least difference" constitute such a characterization? (See Rosenstein.) D
@NoahSchweber Yes, this is exactly what I am looking for. Thank you.
It is not exactly what you asked for, but I'll mention that some visualization of $\omega^\omega$ can be found here: Visualizations of ordinal numbers and Intuition for $\omega^\omega$.
Adding to the suggestions of @Martin, https://math.stackexchange.com/questions/278992/how-to-think-about-ordinal-exponentiation might be helpful as well.
Related: martini's answer to Ordinal Exponentiation and transfinite induction AND How do you define ordinal exponentiation without induction? (the comments might be of interest).
Ordinal exponentiation is a special case of linear order exponentiation. For any linear order $L$, element $a\in L$, and ordinal $\beta$ we can define the $\beta$th power of $L$ at $a$, which I'll call "$L_a^\beta$," as the set of functions $f:\beta\rightarrow L$ such that all but finitely many $\alpha\in\beta$ have $f(\alpha)=a$. The ordering on this set is given by looking at the last point of difference: $$f\trianglelefteq g\iff f=g\mbox{ or } f(\max\{x:f(x)\not=g(x)\})<g(\max\{x:f(x)\not=g(x)\}).$$
For ordinals $\alpha,\beta$ we have $\alpha^\beta=\alpha_0^\beta$. Rosenstein's book treats this in some detail (and is generally an awesome book all-around - it's a huge tragedy that it's so hard to find).
What I like about this definition is that it very nicely complements the definition of cardinal exponentiation: they start with the same basic idea of counting functions between sets, but ordinal exponentiation is the "finite support" version.
It's not well-ordered. $(1,0,0,...) > (0,1,0,...) > ... > (0,...,0,1,0,...) > ...$. Maybe replace min with max?
@MonroeEskew It would make sense to use $\max$ instead if you want to generalize ordinal exponentiation, but it won't be well-ordered if $L$ isn't anyways.
@nombre That’s ok... I’m just saying that the above definition doesn’t define $\alpha^\beta$ properly. One needs to use max, by my example. (Isn’t it right? The definition given is the same as the lex order, which corresponds to the rationals for $\alpha=2, \beta=\omega$.) But it is easy to show by induction that max works.
If one thinks of $\omega^\omega$ as the ordertype of the semiring of polynomials $\mathbb N[X]$, then one has to look to the largest degree where two polynomials have different coefficients to see which one dominates, not the smallest.
@MonroeEskew Derp, fixed - thanks!
Do you need $\beta$ to be well-ordered? It looks to me like this definition still makes sense when $\beta$ is any linear order, so you've really defined a way of exponentiating with two linear orders . . . right?
|
2025-03-21T14:48:30.405982
| 2020-04-24T20:37:51 |
358460
|
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"Phil Tosteson",
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|
Stack Exchange
|
How related are Fourier transforms on finite groups and Fourier transforms on graphs?
Here are two generalizations of the notion of a Fourier transform. I am also aware of the Pontryagin Duality generalization for locally compact abelian groups, though I am personally more concerned with discrete cases.
Given a finite group $G$, we decompose the natural representation of $G$ in the space of functions $G \to \mathbb{C}$ into irreducibles. The Fourier coefficient at an irreducible $\rho$ is then the weighted sum $\sum_{g \in G} f(g) \rho(g)$. We recover the classical discrete transform when $G = C_n$.
Given a graph $\Gamma$, we can write any function on its vertices in the basis of eigenfunctions $\{\phi_i\}$ of its Laplacian $\mathcal{L}(\Gamma)$, with the coefficients in this basis being the Fourier coefficients. As far as I can tell, this definition is called a Fourier transform because the standard Fourier series rewrites a function in the basis of sine waves, the eigenfunctions of the continuous Laplacian $\frac{\partial^2}{\partial x^2}$.
Is there any real sense in which these two generalizations describe the same phenomenon? Can the former definition be rephrased in terms of eigenfunctions of some Laplacian? For the latter definition, maybe the natural convolution operation $f \ast g = \sum_{i}\hat{f}(i)\hat{g}(i)\phi_i$ has some nice algebraic meaning? While a truly formal common generalization would be nice, I am also okay with a conceptual one. I have seen the metaphor of "Fourier transforms break signals into frequencies" but it is unclear to me why these two types of generalized "frequencies" are both equally deserving of that name.
Maybe this is too simple, but in the graph case you have a representation of the group $\mathbb Z$ on the functions of the graph, and you are decomposing this representation into irreducibles.
Also, if your graph has a symmetry group, then the eigenfunction decomposition will be compatible with the decomposition of functions on the graph into irreducibles. So you may be interested in the laplacian of Cayley graphs: https://en.wikipedia.org/wiki/Cayley_graph.
The connection that @PhilTosteson mentions is described in more detail in "Laplacian eigenvectors of graphs: Perron-Frobenius and Faber-Krahn type theorems" by Biyikoglu, Leydold, and Stadler, section 2.7.
|
2025-03-21T14:48:30.406164
| 2020-04-24T21:46:26 |
358465
|
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|
Stack Exchange
|
Isoclinism for Lie groups: existing accounts of basic properties?
Philip Hall introduced the relation of isoclinism between two groups. One statement of the definition (not Hall's original statement) is to introduce a category whose objects are the canonical maps
$$c_G: G/Z(G) \times G/Z(G) \to G'\quad; \quad c_G( xZ(G), yZ(G) ) = xyx^{-1}y^{-1}$$
and where a morphism from $c_G$ to $c_H$ is given by a pair of group homomorphisms $(G/Z(G) \to H/Z(H),\, G' \to H')$ that make the obvious square commute.
Note that every group homomorphism $\theta:G\to H$ gives rise to a morphism $c_G\to c_H$, but we don't require every morphism $c_G\to c_H$ to be of this form.
We say that $G$ and $H$ are isoclinic if $c_G$ and $c_H$ are isomorphic in this category. Every abelian group is isoclinic to the trivial one; more generally, if I understand the literature correctly, the derived series of two isoclinic groups (not necessarily finite) have the same length.
For reasons motivated by noncommutative harmonic analysis, I was wondering if Hall's notion had ever been taken up in the setting of Lie groups (with the natural choice of morphisms, and defining $G'$ as the kernel of $G \to G_{ab}$). To my surprise, a quick search online did not show me any sources discussing this. In particular, one would expect that requiring two connected Lie groups to be isoclinic says something about their adjoint representations and hence about their Lie algebras, but I have not yet sat down to work through the details.
My question is this: are there any sources/articles that describe the basic properties of isoclinism in the world of Lie groups, or which make use of the fact that two groups are isoclinic to say something about their Lie algebras?
It might be the case that "there are some obvious statements which are easy to check; there are facts from the setting of finite groups, which extend in the obvious way by adding the right adjectives; and beyond that there is little to say." But if anyone knows of references in the literature, which would save me needlessly repeating straightforward calculations, that would be helpful.
|
2025-03-21T14:48:30.406331
| 2020-04-24T22:02:51 |
358466
|
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|
Stack Exchange
|
Strichartz estimate for the Schrödinger equation
Estimates of the extension operator can be seen as estimates of the initial value problem for the evolution Schrödinger equation. If $u(x,t)=e^{it\Delta}u_0$ is the solution to the IVP:
$$i\partial_t u(x,t) + \Delta u(x,t)=0$$
$$u(x,0)=u_0$$
Then the following estimates hold:
$a)$ if $D^\gamma$ is the fractional derivative $D^\gamma _x =\int e^{ix\epsilon } |\epsilon|^{\gamma} \hat{f(\epsilon)}d\epsilon$ then $\|D^\gamma _x e^{it\Delta}u_0\|_{L^q_x(L^2_t)}\leq C\|u_0\|_{L^2}$
$b)$ $\|D^\gamma _x e^{it\Delta}u_0\|_{B^*_{1/2}(L^2)}\leq C\|u_0\|_{L^2}$
And this is know as Strichartz Estimate for the Schrödinger equation. I know this was proved by Strichartz and also by Thao for more general family of functions but the way to do this with polar coordinates is writing $D^{\gamma}_xe^{it\Delta}u_0(x)=\int e^{ix\epsilon} e^{it|\epsilon|^2}|\epsilon|^{\gamma} \hat{u_0}(\epsilon)d\epsilon $ and then writing this in polar coordinates. I dont know how to apply polar coordinates in this step $(r,\theta)$ would it be just:
$\int e^{ix\cdot (r\cos(\theta),r\sin(\theta))} e^{it|r|^2}r^{\gamma} \hat{u_0}(r,\theta)rdr d\theta $ but then changing the variable $s=r^2$ would difficult me the sinus and cosine terms... so I think I am doing something wrong.
Then I have to use Plancherel in $t$ and then Minkowski integral inequality but I don't know how to follow from here.
Thanks in advance.
|
2025-03-21T14:48:30.406444
| 2020-04-24T22:24:54 |
358467
|
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|
Stack Exchange
|
Existence of a maximal rank CR Lie subalgebra
Let $\mathfrak{g}$ be a real Lie algebra of dimension $2n+1$ and let $\mathfrak h \subset \mathfrak g \otimes \mathbb C$ be a subalgebra of complex dimension $n+1$ satisfying $\mathfrak h + \overline{\mathfrak h} = \mathfrak g \otimes \mathbb C.$ When does there exist a subalgebra $\mathfrak h' \subset \mathfrak h$ of dimension $n$ such that $\mathfrak h' \cap \overline{\mathfrak h'} = \{0\}$?
I also would not mind requiring that $\mathfrak{g}$ is the Lie algebra of a compact Lie group.
What sort of conditions on $\mathfrak h$ would you want in your answer? I mean to say, there's lots and lots of real Lie algebras (though not so many reductive ones), not parameterised uniformly in any way that I know (though the reductive ones are), and probably even more "near-Lagrangian subalgebras"; are you hoping that this is some familiar condition in disguise?
I do not want to impose any condition on $\mathfrak{h}$. It is just a complex Lie algebra in $\mathfrak{g} \otimes \mathbb{C}$ satisfying the conditions above.
@LSpice, I just noticed that I didn't answer your last question. Yes, I was hoping that the problem I wrote, that are related to the theory of involutive structures and CR geometry, could be equivalent to some other algebraic problems.
|
2025-03-21T14:48:30.406571
| 2020-04-24T22:41:44 |
358469
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Emil Jeřábek",
"Fedor Pakhomov",
"Gerhard Paseman",
"Noah Schweber",
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"https://mathoverflow.net/users/3402",
"https://mathoverflow.net/users/36385",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358469"
}
|
Stack Exchange
|
The lattice of analogues of Robinson's $Q$
This question was asked and bountied at MSE without response.
Call a sentence $\varphi$ in the language of arithmetic $Q$-like iff $\mathbb{N}\models\varphi$ and $\{\varphi\}$ is essentially incomplete. The standard example is of course the conjunction of the finitely many axioms of Robinson's $Q$, but this is of course not unique - and indeed the partial order $\mathfrak{Q}$ of equivalence classes of $Q$-like sentences under entailment as in the Lindenbaum algebra ($\varphi\le\psi\iff\vdash\varphi\rightarrow\psi$) is not linear. On the positive side, $\mathfrak{Q}$ is clearly a distributive lattice, and every countable partial order embeds into each element of $\mathfrak{Q}$'s lower cone (see my comment below).
My question is:
What exactly is $\mathfrak{Q}$, up to isomorphism?
There's an obvious candidate, based on the idea that everything that can happen does in this sort of situation: the (countable) random distributive lattice (that is, the Fraisse limit of the set of finite distributive lattices). However, I'm having trouble proving this. Even showing that $\mathfrak{Q}$ has no greatest element isn't trivial, as far as I can see. (EDIT: by "isn't trivial" I now mean "I can't.")
(As a quick remark, note that essentially undecidable theories need not come from elements of $\mathfrak{Q}$: Robinson's $R$ is essentially undecidable but each of its finitely axiomatizable subtheories has a computable completion.)
Can one embed the rationals inside this lattice? Gerhard "Pardon The Unstated Notational Pun" Paseman, 2020.04.24.
@GerhardPaseman Yes, embedding is easy enough. Specifically, take your favorite sentence $P\in\mathfrak{Q}$ and pick an appropriate collection of sentences $\Phi=(\varphi_q){q\in\mathbb{Q}}$ such that $P\vdash\varphi_q\rightarrow\varphi{q'}$ iff $q\le q'$. The existence of such a $\Phi$ isn't hard to establish. Now consider ${P\wedge \varphi_q: q\in\mathbb{Q}}$. More generally, every countable partial order embeds into every (nonempty) upper cone in $\mathfrak{Q}$ viewed as a partial order.
Is the order upside down? Because if I use the order $\phi\le\psi\iff{}\vdash\phi\to\psi$, it’s obvious that the funny-letter-lattice has no least element, and every nontrivial interval is the countable atomless Boolean algebra, but I don’t immediately see how to prove it has no largest element (though it is probably true). Is the random distributive lattice completely described by these three conditions?
(The answer to my second question is yes: moreover, the theory of nontrivial distributive lattices such that each nontrivial interval is an atomless Boolean algebra has exactly four countable models, depending on whether top or bottom elements exist. Thus, there are only two candidates for the lattice in question, depending on if it has a largest element or not.)
@EmilJeřábek "Is the order upside down?" I would consider the right way up to be $\varphi\le\psi\iff\vdash\psi\rightarrow\varphi$ ("stronger sentences are bigger"). But is the other way around standard?
The order I wrote is the standard order on the Lindenbaum–Tarski algebra (of which your lattice is a convex sublattice). This makes the algebraic operations ($\land,\lor,\neg,\top,\bot$) agree with logical connectives.
@EmilJeřábek Oh duh, of course - I've edited my question to fix this.
$\mathfrak{Q}$ is the countable random distributive lattice.
Emil Jeřábek has already pointed in his comments that there are only two possibilities for $\mathfrak{Q}$. Either there are no greatest element in $\mathfrak{Q}$ and it is the countable random distributive lattice. Or there is the greatest element in $\mathfrak{Q}$ and $\mathfrak{Q}$ is the countable random distributive lattice with appended greatest element. So I'll only need to show that there exist no sentence $\varphi_0$ such that $\mathbb{N}\models\varphi_0$ and for any $\varphi$, if $\mathbb{N}\models \varphi$, then
$$\varphi\text{ is essentially undecidable }\iff \vdash \varphi\to \varphi_0.$$
Indeed assume for a contradiction that $\varphi_0$ exist.
To simplify things as much as possible here I'll consider $\mathbb{N}$ to have the signature consisting of the constant $0$ and predicates $\mathsf{Succ}(x,y)$, $\mathsf{Add}(x,y,z)$, $\mathsf{Mul}(x,y,z)$, and $x\le y$; it is possible to modify the argument so that it will work with the standard signature $0,S,+,\times$, but it would add additional complications. Let us consider the class $\Pi_1^{-}$ of all formulas of the form $\forall x\;\theta(x)$, where all quatifiers in $\theta$ are $x$-bounded. Note that the set of all true $\Pi_1^{-}$ sentences is $\Pi_1$-complete.
For any $\Pi_1^{-}$ arithmetical sentence $\psi$ of the form $\forall x \;\theta(x)$ let us consider the sentence $\psi^\star$:
$$\mathsf{Q}^{-}\land \forall x\;(\theta(x)\to \exists y\;(\mathsf{Succ}(x,y)).$$
Here $\mathsf{Q}^{-}$ should be a version of $\mathsf{Q}-\text{"totality of $S,+,\times$"}$ in our signature. The key properties of $\psi^\star$ that we will need are the following:
if $\psi$ is false, then $\psi^\star$ has a finite model;
if $\psi$ is true, then any model of $\psi^\star$ contains $\mathbb{N}$ as an initial segment;
$\mathbb{N}\models \psi^\star$, regardless of whether $\psi$ were true or not.
Notice that any sentence $\varphi$ (in our finite signature) with a finite model isn't essentially undecidable. And that by the standard argument (that uses a pair of recursively inseparable sets) we see that if any model of a sentence $\varphi$ contain $\mathbb{N}$ as an initial segment, then $\varphi$ is essentially undecidable. To conclude, $\psi^{\star}$ is always true and is essentially undecidable iff $\psi$ is true.
Under the assumption that $\varphi_0$ exists we see that $$\{\psi\in \Pi_1^{-}\mid\mathbb{N}\models \psi\}=\{\psi\in \Pi_1^{-}\mid \psi^{\star}\text{ is essentially undecidable}\}=\{\psi\in \Pi_1^{-}\mid \vdash \psi^{\star}\to \varphi_0\}$$ is $\Sigma_1$. But on the other hand it should be $\Pi_1$-complete, contradiction.
For the sake of completeness let me sketch my reconstruction of Emil's argument. Observe that by Gödel's first incompleteness theorem $\mathfrak{Q}$ has no least element. By Rosser's theorem, for any pair $a<_{\mathfrak{Q}}b$ the interval $[a,b]$ is a countable atomless Boolean algebra. By a standard back and forth argument it is easy to show that for a countable distributive lattice $K$, if all non-trivial intervals in $K$ are countable atomless Boolean algebra, then there are only 4 possibilities for $K$:
$K$ is the random distributive lattice;
$K$ is the random distributive lattice with appended $0$;
$K$ is the random distributive lattice with appended $1$;
$K$ is the random distributive lattice with appended $0$ and $1$.
Lovely! My original attempt to show that $\mathfrak{Q}$ had no greatest element was along these lines but not focusing on as small a syntactic class as $\Pi^-_1$, which made coding the necessary complexity into the sentence no longer easy (or doable for me).
I think that something like $\Pi_1^{-}$ is essential for this kind of argument.
Yes, I think that's right. (Certainly we can't go much higher and get a naive index set argument: $Th_\Gamma(\mathbb{N})$ quickly becomes too complicated as $\Gamma$ gets bigger.)
A potentially-interesting followup question: what if we look at the ordering of essentially incomplete sentences (possibly not true of $\mathbb{N}$) ordered by interpretability? Given any $\varphi$ the set of e.i.-sentences interpreting $\varphi$ is d.c.e. (interpreting $\varphi$ is c.e. and consistency is co-c.e.) while a priori essential incompleteness looks $\Pi^0_3$, so there should be no minimal element in this order either. But now d.c.e. seems a bit too big to get this kind of argument to work.
(I can get $\Pi^0_2$-hardness of essential incompleteness assuming this question has an affirmative answer - basically attach finite pieces of $\mathbb{N}$ to "stages," where the pieces grow each time we see another piece of evidence for the $\Pi^0_2$ property - but I haven't convinced myself that the answer given there is correct yet.)
I think that my argument about non-existence of the greatest element could be adopted to this case with almost no changes. We just need to note that the interpretability relation $\psi^{\star}\triangleright \varphi_0$ is $\Sigma_1$. Although I don't know enough about the lattice of interpretability degrees of arbitrary sentences to say whether it would be enough to figure out the answer to your question.
|
2025-03-21T14:48:30.407156
| 2020-04-24T22:48:56 |
358470
|
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|
Stack Exchange
|
Is this lower bound for the size of minimal vertex cover new/interesting?
I have found this lower bound for the size of minimal vertex cover (and proved it).
If a simple connected graph G on n vertices has largest and smallest eigenvalues $\lambda_1,\lambda_n$, respectively, and $\theta_{n-1}$ is the second smallest Laplace eigenvalue, then
$$
\tau(G)\geq\frac{n\theta_{n-1}^{2}}{\theta_{n-1}^{2}-4\lambda_{1}\lambda_{n}}
$$
When $\tau(G)$ is the minimal size of minimal vertex cover.
I checked it for some graphs and it was very tight for most of them.
There are some graphs for which the estimation is not close, but I think it works pretty well most times.
I tried to look for it and didn't find anything. Anyone saw that before? Are there better bounds?
Thank!
I found a better bound for regular graphs- the Hoffmann bound:
Let G be a d-regular graph on $n$ vertices with minimal eigenvalue $\lambda_{min}$. Then $$
\alpha\left(G\right)\leq\frac{-n\lambda_{min}}{d-\lambda_{min}}
$$
It is known that for a graph $G$ on $n$ vertices, $\alpha\left(G\right)+\tau\left(G\right)=n$. Thus
$$
\tau\left(G\right)\geq\frac{nd}{d-\lambda_{min}}
$$
For regular graphs, this bound is always tighter than mine.
|
2025-03-21T14:48:30.407383
| 2020-04-24T23:02:27 |
358471
|
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"Harry Richman",
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"url": "https://mathoverflow.net/questions/358471"
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|
Stack Exchange
|
How long for Brownian motion to "fill-out" a torus in d-dimensions?
I've been taken by the concise result1
that (roughly!), on a $2$-dimensional torus $\mathbb{T}^2$, the time it takes
to visit nearly every point (within $\epsilon$, as $\epsilon \to 0$) is: $\frac{2}{\pi}$.
My question is:
Q. Is the same situation known for the $d$-dimensional
torus, $\mathbb{T}^d$? What is the time it takes for a Brownian-motion particle to visit within $\epsilon \to 0$ of
every point of $\mathbb{T}^d$?
This is probably known—or known to be unknown—so this is just
a reference request.
1
Dembo, Amir, Yuval Peres, Jay Rosen, and Ofer Zeitouni. "Cover times for Brownian motion and random walks in two dimensions." Annals of Mathematics (2004): 433-464.
Annals link.
Wouldn't it be more accurate to say that the time it takes is $2/ \pi \cdot |\log \epsilon|^2$? Not sure if I am misinterpreting the abstract of their paper.
@HarryRichman: What I see is $$\lim_{\epsilon\to0} \frac{C_\epsilon}{({\ln \epsilon})^2} = \frac{2}{\pi}$$ where $C_\epsilon$ is the time to come w/in $\epsilon$.
A very general answer, in dimension $d\geq 3$,
is in the following paper of Dembo, Peres and Rosen
https://projecteuclid.org/euclid.ejp/1464037588:
for compact $d$-dimensional manifolds,
$$C_\epsilon(M) /\epsilon^{2-d}\to V(M)$$
where $V(M)$ is the volume.
The $d$ dimensional case for $d\geq 3$ is much simpler than $d=2$ because in the latter, you have
long range (=logarithmic) correlations arising from the recurrence of BM.
In passing, let me mention that Belius's paper mentioned in Josiah Park's answer gives a more refined answer, namely a limit law for a centered version of
$\sqrt{C_\epsilon}$ (with no scaling),
for the random walk case. Similar results can be written
for the manifold case in $d\geq 3$. For the
critical case of $d=2$, higher precision than the law of large numbers is available but as of this writing, the limit law for $\sqrt{C_\epsilon}-E\sqrt{C_\epsilon}$ has not been derived.
It looks like the $d$-dimensional case is easier generally than the $d=2$ case according to the excerpt below from this paper (see page three).
...the two-dimensional model is also more difficult than its
higher-dimensional counterparts. This is because dimension two is critical for
the walk, resulting in strong correlations. To illustrate the dimension-based
comparison, observe that very fine results are available for d ≥ 3, see e.g. 2
and references therein... In contrast, in two dimensions the first-order asymptotics of the
cover time was completed only recently, after a series of intermediate steps
over a decade of efforts.
Looking at Aldous and Fill's book (Corollary 7.24) there is an expression for the cover time for $(\mathbb{Z}/n\mathbb{Z})^d$ for $d\geq 3$, $T\sim R_d n \log n$, where
$$R_d=\int_{0}^1\dots\int_{0}^1\frac{1}{\frac{1}{d}\sum_{u=1}^d(1-\cos{2\pi x_u})}dx_1\dots dx_d.$$
In the limit as $n\rightarrow \infty$ with appropriate normalization, it appears this constant $R_d$ is the cover time.
2: D. Belius (2013) Gumbel fluctuations for cover times in the discrete
torus. To appear in: Probab. Theory Relat. Fields.
Prelininary arXiv abs.
|
2025-03-21T14:48:30.407633
| 2020-04-24T23:31:57 |
358472
|
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"Noah Schweber",
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|
Stack Exchange
|
Comparing "axiomatized function spaces"
This was previously asked and bountied at math.stackexchange with no response. I've also tweaked the language for clarity; see the edit history for the broader context, and note that the existing answers were aimed at the earlier version of the question.
Let $C(\mathbb{R}^2,\mathbb{R})$ be the space of all continuous functions $\mathbb{R}^2\rightarrow\mathbb{R}$ with the compact-open topology, and consider the following two subspaces:
$\mathcal{Asso}$ = the subspace of all associative functions.
$\mathcal{Comm}$ = the subspace of all commutative functions.
Of course these are distinct as subsets; my question is whether they are topologically distinguishable:
Is $\mathcal{Asso}\cong\mathcal{Comm}$?
I strongly suspect that the answer is negative - in particular, $\mathcal{Comm}$ is connected and I suspect $\mathcal{Asso}$ is not - but I don't see how to prove this. (Another nice feature of $\mathcal{Comm}$ is that it is an $\mathbb{R}$-vector space in a natural way; however, it's not clear how to extract any power from this observation.)
Generally, $\mathcal{Asso}$ seems rather mysterious, and I'm interested in any information about it.
The following result from Section 6.2 of János Aczél's classic book Lectures on Functional Equations and their Applications seems to be closely related, and the long list of references on the solutions of the associativity equation (p.253/4) may provide further entry points into the literature. (Although that list is way out of date, with the book being from 1966.)
Theorem: Let $f \in C(\mathbb{R}^2,\mathbb{R})$ satisfy the associativity equation
$$f(x,f(y,z)) = f(f(x,y),z).$$
Suppose moreover that $f$ is cancellative, i.e. that
$$
f(x,y) = f(x,z) \quad \Rightarrow \quad y = z,
$$
and similarly in the other argument. Then $f$ is conjugate to addition $+ : \mathbb{R}^2 \to \mathbb{R}$ via a monotone homeomorphism. In particular, $f$ is commutative.
It seems plausible that this result can be used to prove that if your theory $T$ consists of associativity and the cancellativity laws, then $T_{\mathbb{R}}$ is homemorphic to the space of monotone homeomorphisms $\mathbb{R} \to \mathbb{R}$. But I haven't worked out the details.
+1 - although that doesn't answer the question, it's a very neat observation (and the reference is great, I'd never heard of that book).
|
2025-03-21T14:48:30.407831
| 2020-04-25T00:19:37 |
358477
|
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|
Stack Exchange
|
How to compute a more general version of Vandermonde / Cauchy double alternant determinant
Consider some variables $\{X_i\}_{1\le i \le n}$, $\{Y_i\}_{1\le i \le n}$, and $\{W_i\}_{1\le i \le n}$. Does anyone know how to compute the following determinant?
$$
\det ~ \left(\frac{W_j^{i-1}}{X_i+Y_j}\right)_{1\le i,j\le n}.
$$
Update: If you could also provide an answer for the case where $W_j=1$ for $j\ge2$, that would be sufficient for the problem I have encountered in my research.
It's easy to come up with families of matrices; you had another one at https://mathoverflow.net/questions/358175/how-to-prove-the-determinant-of-a-hilbert-like-matrix-with-parameter-is-non-zero . Before asking multiple questions of this sort, I think that it is appropriate to give some more information about how this arises, and why it is of particular significance (as opposed to just one of a sequence of close but not identical questions).
Thanks for your comment @LSpice. As I mentioned in my question, these matrices are coming up in some polynomial regression that I'm doing for my research. I have been going over Christian Krattenthaler's "Advanced Determinant Calculus", and I realized I can state my question in a more general format as you can see in the new edition of my question.
Maybe you can get something out of the technique of displacement equations. It works as follows.
Notice, first, that a matrix $A$ is a Cauchy-like matrix if and only if it satisfies the so-called displacement equation $LA-AR = vu^T$, where $L$ and $R$ are diagonal matrices (containing the nodes) and $vu^T$ is a generic rank-1 term.
Suppose you are given a matrix $A$.
Find (if you can!) two matrices $L$ and $R$ such that $LA-AR = vu^T$ has rank 1. If $A$ is a Cauchy matrix $A_{ij} = \frac{1}{X_i + Y_j}$, then $L = diag(X_i)$ and $R = diag(Y_j)$ work, while for a Vandermonde matrix $A = W_j^{i-1}$ then $L$ is a shift matrix and $R$ is $diag(W_j)$.
Diagonalize (if you can do it explicitly) $L = VD_LV^{-1}$ and $R=UD_RU^{-1}$, and then with some algebra you get $D_L V^{-1}AU - V^{-1}AUD_R = V^{-1}vu^TU$.
Then, $V^{-1}AU$ is a Cauchy matrix with nodes the entries of $D_L$ and $D_R$, because of that displacement equation. You can compute its determinant, and use it to get the determinant of $A$.
My conjecture is that the answer is close to this:
$$
\frac{\prod_{1\le i < j \le n} (X_j - X_i) (W_jY_j - W_iY_i)}
{\prod_{1\le i ,j \le n}(X_i + Y_j)}.
$$
For $n = 2$, the determinant is a fraction whose numerator is an irreducible degree-$3$ polynomial. So I don't think the answer is that easy.
@darijgrinberg For $n=2$, my conjectured answer has a numerator which is an irreducible degree-3 polynomial too. Am I missing something?
Yours is not irreducible.
@darijgrinberg I see, Thanks!
|
2025-03-21T14:48:30.408035
| 2020-04-25T01:20:15 |
358480
|
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|
Stack Exchange
|
A complex analytic version of the eigencurve
I am very much a beginner to the theory of eigencurves so there might be many mistakes in what follows, especially since it is all very speculative.
My understanding of the eigencurve $\mathcal C_{N,p}$ (of Coleman and Mazur) is that it is a rigid analytic family that parametrizes and interpolates p-adically the classical eigenforms of level $Np$. There is a map to weight space $\mathcal C_{N,p} \to \mathcal W_{N,p} = \text{Hom}_{\text{cont}}(\mathbb Z_p^\times,\mathbb C_p^\times)$.
Moreover, the points should give rise to Galois representations, as ordinary forms do.
This is supposed to be an extension of the naive family of Eisenstein series (by Serre/Swinnerton-Dyer?):
$$E^*_{k} = \frac{(1-p^{k-1})\zeta(1-k)}{2} + \sum_{n\geq 1}\sigma_{k-1}(n)q^n$$
where we use the p-adic analytic continuation of the zeta function to define the above for $k \in \mathbb Z_p$.
Now, I read in this paper of Buzzard the idea that one could also define a complex analytic family of Eisenstein series by just letting $k$ vary over complex numbers in the above formula instead:
$$E_{k}(q) = \frac{\zeta(1-k)}{2} + \sum_{n\geq 1}\big(\sum_{d|n}d^{k-1}\big)q^n.$$
Question: Is it also possible to find a complex analytic interpolation or parametrization of the eigenforms (fixing level), analogous to the eigencurve?
In the case of $GL_1$ over $\mathbb Q$, we want to interpolate Dirichlet characters of fixed level (=period). In the p-adic case, by class field theory, the space we should look are characters from $\mathbb Z_p^\times \to \mathbb C_p^\times$ for $p$ not dividing the level which is a one-dimensional family.
So the complex analogue should also probably be one dimensional and an obvious analogue would be the set of all functions from $(\mathbb Z/N)^\times \to \mathbb C^\times$.
Please use a high-level tag like nt.number-theory (added now).
Thanks! I wasn't aware of that convention but it makes sense.
This won't work. $\mathbf{Z}$ is discrete in $\mathbf{R}$ or $\mathbf{C}$ and that kills off the hope for any meaningful notion of family -- there's no meaningful way for two cusp forms of different weights can be "close" to one another. Just another one of the many ways that the nonarchimedean world is richer and nicer than the archimedean one.
I guess you are assuming that classical forms have to be dense in the the moduli space?
There are actually interesting examples of p-adic eigenvarieties in which classical points are not dense. However, the non-density "means something", in the following sense: there's a Zariski-dense set of points in weight space such that the fibre of the eigenvariety over weight space has a classical interpretation; the fact that almost all of those fibres are actually empty is telling you some deep non-obvious global property of the relevant automorphic forms space.
(cont'd) If you try to do this in the real-analytic world, then because cusp-forms only make sense for integer weights and the integers are discrete, then either your parameter space will be 0-dimensional and convey no geometric information, or classical points will be boringly non-dense in the parameter space -- boringly non-dense because it wouldn't be meaningful for it to be any other way.
|
2025-03-21T14:48:30.408296
| 2020-04-25T01:46:35 |
358483
|
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|
Stack Exchange
|
Why are we interested in operators that share a basis of eigenfunctions?
I hope this is an appropriate question for this forum. If not, I apologize. Before stating my question (which may be found at the end of this post), I will attempt to provide sufficient context.
I am studying eigenfunctions of the Laplacian (both Dirichlet and Neumann cases) on a specified bounded domain $D\subseteq \mathbb{R}^2$ which has piece-wise smooth boundary. Next, consider the billiard map
$$\phi : M \to M,$$
where $M$ is the subset of the unit cotangent bundle $S^\ast(\partial D)$ containing only the inward pointing directions. The billiard map models the behaviour of a free particle in the space $D$. Suppose we are base point $x\in \partial D$ and a unit direction vector $w$ pointing inwards. Then a free particle starting at x and travelling in direction $w$ will eventually hit the boundary $\partial D$. Let $x^\prime$ be the point of incidence and suppose that $w^\prime$ is the new direction of the particle upon hitting the boundary. Then $\phi(x, w) = (x^\prime, w^\prime)$. The Billiard map can also be described in terms of the billiard flow. That is, suppose that $\varphi_t : S^\ast D \to S^\ast D$ is the billiard flow. The billiard flow $\varphi_t$ solves the equation
$$
\partial_t\varphi_t(x, \omega) = \begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix}\varphi_t(x, w).
$$
Then
$$
\phi(x, w) = \varphi_{\tau(x,w)}(x, w)
$$
where
$$
\tau(x, w) = \inf\{t > 0: \varphi_t(x, w) \in M\}.
$$
Let $p$ be a given function that is invariant with respect to the billiard map. i.e, $$p:M \to \mathbb{C}$$ is such that
$$
p\circ \phi = p \quad \text{on } M.
$$
Supposed that the function $p$ is defined on all of $\mathbb{R}^2\times \mathbb{R}^2$. By quantizing $p$, we obtain a pseudo-differential operator $P_h = p(x, hD)$. That is,
$$
P_hu(x) = \frac{1}{\left(2\pi h\right)^2}\int_{\mathbb{R}^2}\int_{\mathbb{R}^2} e^{\frac{i}{h}\langle x-y, \xi\rangle} p(x, \xi) u(y)\,\mathrm{d}{y}\mathrm{d}{\xi}
$$
for every sufficiently nice function $u$.
I have shown that $P_h$ shares a complete (in $L^2(D)$) collection of eigenfuntions with the Laplacian $\Delta$. We also note that, $P_h$ commutes with the Laplacian $\Delta$.
Is there an intuitive reason why the operator I obtained shares an orthonormal basis of (Dirichlet/Neumann) eigenfunctions with the Laplacian? Can anyone provide information regarding the significance (or applications) of such an operator?
Hello! It might be helpful to people who might answer this if you provided a definition of both "billiard map" and "quantizing" (I'm probably not one of them. But for what it's worth, from a quantum-mechanics perspective your construction seems to produce symmetries of the free particle on $D$ (whose hamiltonian is, by definition, $\Delta$.))
@AlexArvanitakis I added some rather informal definitions. Please let me know if you think I should add additional context.
Those helped me at least, thanks.
Since people have not ventured an answer yet I will make another vague quantum-mechanics related comment. Insofar as the billiard map actually determines the PDE problem fixed by the laplacian $\Delta$ and domain $D$, it's not unexpected that symmetries of the billiard map (roughly specified by your $p$) actually descend onto symmetries $P_h$ of the free quantum-mechanical particle on $D$. (In quantum mechanics, an operator $\mathcal O$ is a symmetry if it commutes with the Hamiltonian $H$. Here $H=\Delta$ and ${\mathcal O}=P_h$.
@AlexArvanitakis I appreciate it! I have no physics background (only pure math) so anything helps!
I'm not sure what precisely you are looking for but let me try the guess an answer on the soft end (basic intuition stuff) of what you are asking. I'm not qualified for the hard end of this question, for this you want an expert on partial differential operators and microlocal analysis. I'm in the applied/physics corner.
To me the relationship of billiard constructions to PDEs is most evident when one considers a dynamical case, e.g. the d'Alembertian over the domain. This then is the physical case of the dynamics of a drum-head (for a compact domain). One way to attach this problem was to hope that separation of spacial (Laplacian) and temporal (second partial derivative with respect to the time signature) and then a further separation in over the domain with respect to some coordinate system yield something we can say a lot about. The general strategy here is "global" (in the sense of the size of the domain) Fourier decomposition of the correct eigenfunctions defined by the solutions of separated ODEs. For a set of simple domains (rectangle, circle, ellipse) this was possible but in general it's not.
The reason why the d'Alembertian is nice here for intuition is because it more strongly motivates a more local approach. Rather than study the behavior "globally" over the range of the domain, what if we model local behavior and observe "evolution" of the solution. Early attempts at this on hyperbolic PDEs is the theory of characteristics. If you inject a local non-trivial initial condition (say a impulsive distribution), the solution will propagate in a certain way (along characteristics). Think of the initial impulse position to be an infinite set of infinitessimal billiard balls, and think of the d'Alembertian imposing the dynamics all these balls have to follow, and the initial directions of each ball corresponds to the characteristic cone. Then you will get a billiard decomposition of this local disturbance. If you just study the trajectories of these billiard balls you get the billiard maps with their reflection laws. If however you also study either local value of displacement, you end up studying what is known the fundamental solution of the PDE (the impulse response of that PDE). The billiard maps in this setting captures the directions of the characteristics for one initial location. The full solution would be recovered by convolution with an arbitrary initial function over the domain. Billiards at least intuitively are dynamical objects which is why I like to think of them in the hyperbolic case like this.
Given that we are in a linear operator theory here we have some nice transformation and decomposability properties. So we can reduce to the Laplacian by separating out the time component. In the classical PDE theory this is separating temporal oscillations from oscillatory shapes (eigenvalues of complex exponentials over time and eigenfunctions over the domain shape). So the study of the Laplacian here is merely the reduction to only study the eigen"shapes" of say a drum-head. So when we use billiard-type decompositions we are in a sense only looking for the stationary solutions and not the temporal dynamics. But we can still do this kind of decomposition regardless. Though we consider billiard maps to be a flow the purpose is to represent the Laplacian alone.
The goal of course is for your the full collection of functions over billiard maps to contain all the information that is the solution of the laplacian over the domain. So the eigenfunctions (in the re-composition) have to be identical!
But the well-known cases show that one has to be careful here. The typical eigenfunctions of a single billiard map is the complex exponential. Yet the eigenfunction of the Laplacian over a circular domain is Bessel. So your precise treatment of eigenfunctions in the re-composition of all billiard maps with all the local behavior has to be equal to the Bessel function to solve the problem. But this is an easy case because we have the Bessel ODE and lots (though not all) useful properties so we can compare. In many non-trivial cases all we can do is move forward and make sure all our steps are correct. (So when you say that you have the eigenfunction of the Laplacian over your domain, I'm a bit confused how you got it, but there is lots of detail missing). It turns out that local singularities of the billiard map bundle relate to "phase" contributions. We count the number of these with Keller-Maslov (or more generalized) indices, and the nature of the phase change is subject to local integration around the singularity.
Basic topic areas and important contributors (apologies for any omissions!) for related literature:
Physics: Geometric optics, wave optics, semi-classical physics.
Keller's work on geometric optics and correction to quantization conditions. Maslov's work. Ballian Bloch. Gutzwiller. See also Berry on the relation of geometric and wave optics and Nye's beautiful book on the topic.
Mathematics:
Microlocal analysis. Index theories. Singularity theory, particularly in the context of Lagrangian and Legendrian manifolds. I also recommend some Arnold for a highly geometric view of what happens at singularities (his work on wavefronts and caustics). Integration theory of singularities in particular Picard-Lefschetz.
Video:
Zeldich recently gave a recently talk on his result that mild elliptic perturbations of the circle are spectrally determined (not isospectral) contains a lot of related background:
https://www.youtube.com/watch?v=sL73RYikETw
So when you say that you have the eigenfunction of the Laplacian over your domain.
I am working in one explicitly defined domain where I am able to find all the eigenfunctions. Of course, this is not possible in general. I didn't write which domain in order to make it clear that I am looking for a general theory, rather then something that would only apply to my case.
Understood. In short there are very few domain boundary shapes where we understand things fully, so rather than a coherent theory we have different lines of attack to the problem.
Thank you! I'll look into the sources you provided.
|
2025-03-21T14:48:30.409339
| 2020-04-25T02:18:47 |
358484
|
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|
Stack Exchange
|
Bounds on convergence of two orbits in the limit set of a Schottky group
Suppose we have two points in the limit set of a Schottky group, $x,y\in \Lambda(\Gamma)$. Consider the orbits of those points under a primitive subset $\Gamma'$ of $\Gamma,$ that is, one that does not contain the identity, and for a pair of $g,g^{-1}\in \Gamma$ contains only one of them.
If a point in $\Lambda(\Gamma)$ is an attracting point for some element $g$ of $\Gamma'$, then, quite obviously, the Euclidean distance $|g^n(x)-g^n(y)|$ decreases exponentially fast.
However, what if a point is not an attractive point of any element, but, nevertheless, there exists a sequence $g_n$ such that $|g_n(x)-g_n(y)|$ converges to 0; how fast does this converge?
I assume this corresponds to the following geometric picture: if $x,y$ are repelling and attracting points of some element of the Schottky group $T_{x,y}$. Then I presume that the images of these points under $g_n$ correspond to non-trivial closed geodesics in the handle body that converge to some kind of foliation, and it would be nice to know how fast the numbers of leaves grows, or else,how well this point is approximated by the corresponding sequences.
Since we are talking about Diophantine approximations now, can the bound be expressed purely in terms of the Hausdorff dimension of the limit set?
EDIT: Upon closer inspection, I presume that the bound must again be something like exponential, because if $|g_n(x)-g_n(y)|$ converges to $0$, then the points must lie in smaller and smaller circles eventually. Moreover, these circles must be contained in the circles, corresponding to powers of one of the generators that must occur in $g_n$ an increasing number of times (more or less pigeonhole principle,) so we can even have a lower bound on the speed of convergence.
|
2025-03-21T14:48:30.409515
| 2020-04-25T02:45:24 |
358485
|
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"Jeremy Brazas",
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|
Stack Exchange
|
A spectral characterization of path connected spaces
Let $X$ be a compact Haussdorf topological space with the following property:
For every continuous function $f:X\to \mathbb{C}$ the image $f(X)$ is a path connected subset of $\mathbb{C}$.
Is such a space $X$ necessarilly a path connected space?
This question is inspired by (and realated to) the following post:
Unital $C^{*}$ algebras whose all elements have path connected spectrum
I think the extended long line $\bar L$ (add one point at each end) is a counterexample. Every map $\bar L \to \mathbf R$ (or to $\mathbf C = \mathbf R^2$) is eventually constant so the image is path connected, but $\bar L$ itself is too long to be path connected.
The one-point compactification of $\omega_1\times [0,1)$ with the order topology is indeed a counterexample. This question might be more interesting with an extra hypothesis on $X$, e.g. sequential, first countable, or metrizable.
@JeremyBrazas What about metrizable counterexamples?
@TarasBanakh That is the question I would ask...As stated, the question is answered. I cannot tell if the OP is satisfied or wants other cases addressed. My comment is meant to encourage the OP to revise the question. I will consider thinking about it more when I know what the revised question really is.
|
2025-03-21T14:48:30.409646
| 2020-04-25T04:23:26 |
358489
|
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|
Stack Exchange
|
Torsion divisor on non-hyperelliptic curve
Let $P$ and $Q$ are two points on a curve $C$. We say $P-Q$ is a torsion divisor if $dP-dQ$ is a principle divisor for some positive integer $d$.
For hyperelliptic curves, Grant proved a Nagell-Lutz like theorem in this paper for torsion divisors. I wonder if there is an easy-to-use criteria for torsion divisors on non-hyperelliptic curves.
|
2025-03-21T14:48:30.409702
| 2020-04-25T08:37:28 |
358495
|
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"Carl-Fredrik Nyberg Brodda",
"Yi Wang",
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|
Stack Exchange
|
Reference request for finite simple exceptional group of lie type $E_7(q)$ and its Schur covering group $2.E_7(q)$?
Does anyone have the paper named 'Génerateurs, relations et revêtements de groupes algébriques' written by Robert Steinberg in 1962, or any other reference for simple groups of Lie type $E_7(q)$ and its Schur cover $2.E_7(q)$?
Thank you in advance!
The paper appears on page 113 of "Robert Steinberg Collected Papers" published by the AMS. A link to the Google books version is here, with the paper in question appearing there in full text.
Thank you very much!
|
2025-03-21T14:48:30.409776
| 2020-04-25T12:23:05 |
358502
|
{
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"authors": [
"Sylvain JULIEN",
"Will Sawin",
"Wolfgang",
"https://mathoverflow.net/users/12481",
"https://mathoverflow.net/users/13625",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/29783",
"joro"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358502"
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|
Stack Exchange
|
Some statistics related to the abc conjecture
We did some statistics about the 14 million good abc triples below 10^18
taken from Bart de Smith site.
This was examining just the top of the iceberg, since the
interesting triples grow very likely well above 10^18.
We were interested in $\#\{(a,b,c) :a \le c^\alpha\}$ for real $\alpha$,
as $a,b,c$ are taken from the good triples, assuming $a < b$.
Here are the numbers:
alpha , #a<=c^alpha in percents
0 (a=1) 0.31
1/4 2.38
1/2 11.75
3/4 38.59
9/10 70.34
The arithmetic mean of $\log{a}/\log{c}$ is 75.894%,
which is very close to $3/4$.
We did the same computation for the 234 (give or take few)
high quality triples (quality > 1.4) and the data is:
0 4.27
1/4 16.24
1/2 35.47
3/4 68.38
9/10 91.03
The arithmetic mean of $\log{a}/\log{c}$ is 58%.
Can we get some heuristic arguments about the distribution
of good abc triples?
58% is not far from $\gamma$.
@SylvainJULIEN I strongly doubt this holds in general, we are examining very small set.
"good" = $\operatorname{rad}(abc)<c$?
@WillSawin Yes. Or in other words quality >1.
If one assumes $\operatorname{rad}(a), \operatorname{rad}(b), \operatorname{rad}(a+b)$ behave as independent random variables, the key quantity is understanding the number of $n \approx N$ with $\operatorname{rad}(n) \approx n^\alpha$. Surely this must be known/studied. I think the highest-order terms in the asymptotic are $n^\alpha e^{ \sqrt{\log n}}$ but I don't know what the next ones are.
@SylvainJULIEN also note that the threshold of 1.4 used here is completely arbitrary...
@WillSawin May I have your opinion about abc and surjective polynomials:
https://mathoverflow.net/questions/332189/abc-conjecture-and-surjective-polynomials
|
2025-03-21T14:48:30.410081
| 2020-04-25T14:12:13 |
358506
|
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"Claudio Rea",
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"Josiah Park",
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|
Stack Exchange
|
Acting with all rational rotations on a subset of the circle having positive measure do you fill almost the whole circle?
Set $\Gamma$ for the group of the roots of the identity: $\Gamma=\{z\in \Bbb C | z^n=1$, for some $n\geq 0\}$ and for $E\subset S^1$ set
$\Gamma E=\{z\zeta, z\in \Gamma, \zeta\in E \}$
A trivial but very frequently used fact is that if $E$ is open then $\Gamma E=S^1$.
A measure-theoretic version of that is the following
QUESTION. If $E$ has positive Lebesgue measure, does $\Gamma E$ have measure $2\pi$?
If there is a counterexample, E can be taken to be a union of orbits of Gamma without being of full measure. I think one can show a la Vitali that such an E is non-measurable, but I am unsure. If not, consider Bernstein sets. Gerhard "Taking Measure Of This Question" Paseman, 2020.04.25.
By the aswer of Nik Weaver we have the same result for any infinite $\Gamma$.
An easy way to see this is by using the Lebesgue density theorem. Any set of positive measure has a density point $t$ (indeed, almost every element of the set is a density point). This means that for any $\epsilon > 0$ there is an interval $I$ containing $t$ such that $m(E \cap I) > (1-\epsilon)m(I)$. This pretty much immediately implies that $m(\Gamma E) > (1-\epsilon)2\pi$.
Thanks for answering. Could you please explain better the last step? In particular which lower bound you have for $m(I)$?
$I$ is an interval of the circle and $m(I)$ is its arc length ... not sure what you're asking.
In the first $4,5$ rows of your answer you tell us what are density points as we learnt in the kindergarten. The last $0,5+0,5$ rows are the conclusion which should be proved. Where is the proof? The only nontrivial step is that $m(E \cap I) > (1-\epsilon)m(I)$ implies $m(\Gamma E) > (1-\epsilon)2\pi$. But this is absent. Could you write, please, an extended proof of that?
Okay, if you learned about density points in kindergarten then you really shouldn't have trouble with this last very easy step. Hint: if $m(E\cap I) > (1-\epsilon)m(I)$ then $m(\Gamma E\cap I') > (1-\epsilon)m(I)$ for any arc $I'$ of the same length as $I$.
Let $S$ be the circle with Haar-Lebesgue measure and let $G$ be the group of rotations of $S$ through angles that are rational multiples of $\pi$. The action of $G$ is ergodic (according to pg. 69 in The Legacy of John von Neumann
edited by James Glimm, John Impagliazzo, Isadore Singer).
So for any measurable subset $E\subset S$ of positive measure, $m(\Gamma E)=2\pi$.
"The action of G is ergodic " Does this mean that for any $E\subset S$ either $m(\Gamma E)=2\pi$, either $m(\Gamma E)=0$ ?(sorry about my inculture)
@jcdornano It means that any positive non full-measure set must map to a larger measure set under the action of $G$, and this in turn implies $m(\Gamma E)=2\pi$.
sorry for my bad english, but what i mean by "does it mean ..." is if one can quickly deduce from ergodicity that we cannot have $0<m(\Gamma E)< 2\pi$ for some $E\subset S$. If not, is it true?
An action is ergodic if the only sets $E$ which are mapped to themselves (up to zero measure sets) are either of full or zero measure. Since rational rotations include the identity, $E\subset \Gamma E$, and so for $E$ of neither full nor zero measure, $m(\Gamma E)>m(E)$. But this shows in turn that $m(\Gamma E)=2\pi$ by repetition.
No claims here about the nonmeasurable case. Maybe ask your question separately?
sorry i deleted the bad comment. i will ask a new question and might continue the discution in the chat , if you don't mind....
if $m(E)=0$ $m(\Gamma E)=0$ by sigma additivity
Let us continue this discussion in chat.
I just found this proof. Take a sequence $\Bbb Q\ni x_n\to x\notin\Bbb Q$ and set $R_n$ and $R$ for the rotations of angles $x_n$ and $x$. Set $f$ for the characteristic function of $\Gamma E$. Use continuity of the $L_1$ norm $||\cdot||$ with respect to rotations and have $m(\Gamma E)=||f||=||f\circ R_n||\to ||f\circ R||=m[R(\Gamma E)]$. The sequence on the left is constant! Thus $\Gamma E$ has stable measure under the action of $R$ which is ergodic. Hence $m(\Gamma E)=2\pi$ because
$m(\Gamma E)>0$
So, according with Josiah Park a gave a problem which is already at page 69 of a book. I Really apologize! I am going to give an harder one
I just found this proof. Take a sequence $\Bbb Q\ni x_n\to x\notin\Bbb Q$ and set $R_n$ and $R$ for the rotations of angles $x_n$ and $x$. Set $f$ for the characteristic function of $\Gamma E$, hence $f\circ R_n=f$ . Use continuity of the $L_1$ norm $||\cdot||$ with respect to rotations, so for $n\to\infty$ we have $m(\Gamma E)=||f||=||f\circ R_n||\to ||f\circ R||=m[R(\Gamma E)]$. Thus $\Gamma E$ has stable measure under the action of $R$ which is ergodic. Hence $m(\Gamma E)=2\pi$ because
$m(\Gamma E)>0$
|
2025-03-21T14:48:30.410464
| 2020-04-25T14:14:33 |
358507
|
{
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"authors": [
"Andrés E. Caicedo",
"Dave L Renfro",
"David Roberts",
"Gerhard Paseman",
"Joe Silverman",
"LSpice",
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Stack Exchange
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Do mathematicians ignore mathematical works from non-mathematicians?
Is it true that mathematicians ignore and do not like to take a look at or comment on any mathematical work or manuscript from a person outside the field of mathematics (meaning is not a professional mathematician or does not have any prior publications/qualifications in mathematics)?
I am asking out of personal experience, over and over again, not from one but many.
Sorry If this is not the best place, but I dont have any other place.
Here are three: Academia, Math.StackExchange, and Workplace. I mention the last because the question phrased above generalizes to: why do professionals in an area whose job is to promote their profession not spend their time considering input from non professionals writing in that area? Phrased this way, you might come up with the answer: "It depends; sometimes one does consider such." This is the wrong forum for your question. Gerhard "Speaking As A Non-Professional Mathematician" Paseman, 2020.04.25.
Even if some one, say X, is not a professional mathematician, if X's work is endorsed by a professional mathematician, in the sense X acknowledges (in a research article) a professional mathematician, then there is a chance that people would like to spend some time.. This is only based on my experience.. Everyday professional mathematicians get emails from random person asking to check their idea (I heard it from a faculty whose area is (Algebraic) Number theory).. So, may be that spam emails is affecting a genuine question/response from some one who is not a professional mathematician..
Because, to begin with, much of mathematical work by professional mathematicians is ignored by other mathematicians, too. It is really difficult to do something in mathematics that would be recognized by mathematicians as relevant.
Sometimes, they do, in the case of work done by physicists. But, there has to be a specific reason for a mathematician to read any particular paper. Life is short and people are busy. Lastly, quality of math work written by non-mathematicians, on average, is atrocious, so, on average, reading such work amounts to waste of time.
I finally published my paper in a mathematics journal. I asked so many mathematicians for feedback, but none gave. One mathematicians said it was correct but nothing more than that. He did not even suggest any future directions also. I kept on improving it over my own criticism and it finally got accepted in a reasonable journal. Now I am writing my second paper in continuation, but finding it strange and demotivating to think that nobody in the world is going to read it except one or two who absolutely have to due to their profession.
A huge number of articles published in reputable mathematics journals are never cited by any other papers. This doesn't mean that they weren't read, but it indicates that no one was sufficiently interested to follow up on the contents. You can take two views. One is to get discouraged. The other (my preference) is to view doing mathematical research as something that I do for my own sake, pursuing problems that I find interesting. If perchance others find my work interesting, that's great. But if not, okay, it's still interesting to me.
"He did not even suggest any future directions also" I think expecting such suggestions is in general unreasonable. That said, congratulations on your publication!
Mathematicians spend a tremendous amount of time carefully reading and thoroughly commenting on work by people who have never published before: our PhD students.
Good illustrating example of opinion-based question (motivating this question on meta).
@RajeshDachiraju Only few have succeeded this way. You need to overcome the hidden bias that "Your work is unlikely to be impactful for them to spend time on it". How can you convince them that your work will help them as well?
This question seems to assume that the way you have been treated implies things about how mathematicians treat a wide group of people. Are you basing the claim in the title of your question solely on your own experience, or do you have other information/evidence?
I looked at your paper on arXiv (presumably, the one that was accepted for publication). It is quite far from my area of math, but looks like a solid piece of work. If it were ignored by mathematicians you sent the paper to, my guess is that this is not because you are a non-mathematician, but because they did not find it particularly interesting. They would have had the same reaction to a generic math paper posted on the arXiv. Keep working and maybe, eventually, you will find somebody interested in your work.
@MoisheKohan: Or even that they did find it interesting, but had other more interesting or pressing things on their mind. The number of papers I find interesting is much more than the number of papers I have time to read.
@NoahSnyder: Agreed.
@MoisheKohan Yes, probably they did not have time and energy to refocus and concentrate on my paper/topic which is not exactly related to their work. Also my experience is that say an expert in subfield $A$ would not like to give comments on something related to subfiled $B$ because they probably do not think they can give comment with same insight as they would have if it were topic $A$. So they may pass it on as it was someone else's job and not theirs. Thank you Andres Caicedo.
But they are very helpful, when I ask something about any mathematical sub problems I face, if it is generic. Infact I have thanked mathoverflow and math.SE communities in my paper. I am indebted to these communities, without which I wouldnt have arrived at right problem formulations and solutions.
@YemonChoi : Its based on my experience. I don't know if it is the general case and thats what I'd also like to know. "mathematicians treat a wide group of people" may not be correct but "mathematicians treat work from people outside math"
@JoeSilverman Thanks for your suggestion.
Dear Rajesh, perhaps it will comfort you to know that a mathematician may also face the same problem while trying to be taken seriously by professionals in other communities. I speak this as a professional mathematician who has been dabbling with statistics. A big, if not the biggest, problem I have is to understand what makes a problem "interesting" to people in that community. It helps to learn their language, and sometimes people would be interested if formulated in the right way.
I've rewritten the question in a less affirmative way (replacing "why do they ignore" with "do they ignore"), so as to make it a real question.
@MoisheKohan, also, the quality of math work written by mathematicians is often atrocious, at least from the point of view of writing rather than of mathematics.
This question is now unbelievably broad, and can be answered by a single counterexample: Ramanujan was taken seriously by Hardy. But the fact he was such an atypical case means this answer is useless. Here's another, more recent example that is not so outlandish: Aubrey de Grey, who a few years ago made progress on an open problem. What is really going on is mathematicians very rarely look at work that is badly presented and/or super niche and/or has a very high chance of being trivially wrong.
@LeonidPositselski : So as per your comment, there is rarely any team work in mathematics?
In my experience, looking at work by non-professionals, they seem to totally ignore the existing literature by mathematicians.
So, why should I as a professional look at someones work, where that person has not bothered with reading themselves?
Also, not learning to typeset stuff in LaTeX is just plain lazy - it is a standard nowadays, and not difficult to learn.
A person without patience to learn LaTeX, most likely do not have the patience to solve an interesting problem.
Edit: Vixra, (ViXra.org is an e-print archive set up as an alternative to the popular arXiv.org service owned by Cornell University.) is a good source of 'papers' written by mainly amateurs which gives quite a clear picture why professionals do not care.
Every third submission in the Number theory section is a 7-page paper claiming to solve the Riemann Hypothesis.
Moreover, it seems that all these different people completely ignore all the other 'proofs' of RH on vixra.
Papers on vixra, I feel, are fair game to criticize, as it claims to be an alternative to arxiv. One point of putting papers on arxiv, is to get input on the preprint, and allow the scientific community to give feedback on it.
The point is, an afternoon of browsing papers on vixra will answer the question why professional mathematicians rarely care about the works of amateurs.
What does latex has anything do with this question?
@PraphullaKoushik, a lot.
@PraphullaKoushik "any mathematical work or manuscript".
@PraphullaKoushik
For example, from vixra: https://vixra.org/pdf/2004.0556v1.pdf (proof that there are no odd almost perfect numbers)... and more than half of the references are the authors previous work...
@WlodAA but OP has not mentioned he do not know latex?
I do not know if it is ok to give some example of a badly written file (by a student in a school?) to answer this question... maybe you are assuming OP is not aware how to write in LaTeX...
@PraphullaKoushik that examle is from vixra. It is supposed to be an alternative to arxiv, and thus gives a good example on what 'papers' by amateurs do look like.
One may not be able to deduce a paper's validity from the non-use of LaTeX, but it's nonetheless a strong signal. See point (1) here: https://www.scottaaronson.com/blog/?p=304
@Neal of course, but a 'paper' lacking both LaTeX AND where there are no references to any paper or preprint written by a professional mathematician the last 10-20 years, is probably not something worth spending time reading.
@Per Alexandersson: lacking both LaTeX AND where there are no references to any paper or preprint written by a professional mathematician the last 10-20 years --- Interestingly, this manuscript that I wrote back in 2011 comes very close to qualifying, but I do agree with your general point. (There's a reason I was forced at the time to use MS Word, but I don't have room here to fully explain why.)
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