User jean-marc schlenker - MathOverflow

Permutations of $(Z/pZ)^*$

2013-05-20T15:08:11Z

Let $p$ be a prime integer, and let $(\mathbb Z/p\mathbb Z)^*$ be the set of non-zero elements of $\mathbb Z/p \mathbb Z$. Denote by $S((\mathbb Z/p \mathbb Z)^*)$ the group of permutations of $(\mathbb Z/p \mathbb Z)^*$.

Say that a map $a:(\mathbb Z/p \mathbb Z)^*\to S((\mathbb Z/p \mathbb Z)^*)$ satisfies condition (A) if, for any two distinct elements $i,j\in (\mathbb Z/p \mathbb Z)^*$ , $a(i)-a(j)\in S((\mathbb Z/p \mathbb Z)^*)$.

For example, let $a(i)(k) = ik.$ This satisfies condition (A). The same is true if we permute the functions $a'(i) = a(c(i))$, or relabel the objects $a''(i)(k) = b^{-1}(i \cdot b(k))$, or both. Are these modifications of $a(i)(k) = ik$ the only ways to get a map satisfying condition (A)?

If $a$ satisfies (A), are there $b,c\in S((\mathbb Z/p \mathbb Z)^*)$ such that, for all $i\in (\mathbb Z/p \mathbb Z)^*$ and all $k\in (\mathbb Z/p \mathbb Z)^*$, $a(i)(k)=b^{-1}(c(i)\cdot b(k))$, where the dot is multiplication in $\mathbb Z/p \mathbb Z$?

Note: it would probably be sufficient to prove that, if $a$ satisfies (A), then, for all $i,j\in (\mathbb Z/p \mathbb Z)^*$, $a(i)$ and $a(j)$ commute.

Answer by Jean-Marc Schlenker for Poincare Metric on Hyperbolic Plane

2011-11-25T12:30:03Z

The question is not obvious because it is stated in the upper half-plane, while it is much easier if you translate it in terms of the Klein projective model of the hyperbolic plane (in a ball of radius $1$). Then I believe that the formula is equivalent to the one given by the Hilbert metric of the disk: $$ d_H(U,V)=-\frac{1}2 \log[U,V;A,B]~, $$ where $A$ and $B$ are the intersections with the boundary of the ball of the line through $U$ and $V$, and $[U,V;A,B]$ is the cross-ratio of the four points. (The minus sign might depend on the convention for the cross-ratio).

Now in this form there is a beautiful and quite simple proof of the triangle inequality, originally due to Hilbert but which can be found in "Metric spaces, convexity and nonpositive curvature" by Papadopoulos, pp153-154, here: http://books.google.com/books?id=JrwzXZB0YrIC&lpg=PA153&ots=V5xkvJE6rO&dq=hilbert%20metric%20triangle%20inequality%20convex&pg=PA153#v=onepage&q=hilbert%20metric%20triangle%20inequality%20convex&f=false

In other terms: the harder part is to check that the distance you wrote is the same as the expression given by Hilbert in the ball, then you can use his simple and nice proof.

Answer by Jean-Marc Schlenker for Isoperimetric-like inequality for non-convex sets

2011-10-11T13:59:51Z

I believe that the answer is positive. If $A$ is connected, then it has the same mean shadow as its convex hull $CH(A)$ so the isoperimetric inequality for $CH(A)$ shows that the mean shadow of $A$ is larger than the mean shadow of $B$.

If $A$ is not connected I believe the same inequality holds. I'll sketch a proof when $A$ has finitely many connected components $A_1, \cdots, A_n$, the general case then follows by an approximation argument. Choose a point $x\in CH(A)$ and define a 1-parameter family of deformations of $A$ by making a parallel translate of each connected component $A_i$ so that its barycenter moves towards $x$, say at constant speed to reach it at time $t=1$. Stop this 1-parameter family of deformation as soon as a contact occurs between two connected components, then merge those two connected components and repeat.

The point is that the area of $A$ does not change under this deformation, however the mean shadow is non-increasing -- actually the size of the shadow is non-increasing in every direction. At the end of this deformation one obtains a connected set to which the usual isoperimetric inequality can be applied, and the same inequality then also applies to $A$.

Answer by Jean-Marc Schlenker for Questions on Thurston's earthquake flow

2011-10-04T04:56:46Z

On Q1 I would guess that you can find explicit deformations corresponding to an earthquake path on a non-simple lamination in the quite special case of the punctured torus. You might find some help for instance in

MR0697067 (85d:32047) Waterman, Peter; Wolpert, Scott Earthquakes and tessellations of Teichmüller space. Trans. Amer. Math. Soc. 278 (1983), no. 1, 157–167.

The answer to Q2 is yes, the earthquake flow is the Hamiltonian flow of the length function. I'm not sure where this was first proved but you could check in Kerckhoff's paper on the Nielsen realization problem.

On Q3 I'm convinced that the earthquake path $\phi^t_\alpha(h)$ limits to $\alpha$, I believe you could prove it by understanding the asymptotic behavior of the length of closed curves on this 1-parameter family of metrics, using the tools in the Kerckhoff paper mentioned above.

As for your remark, as Igor Rivin mentiond, I don't believe the Bonahon paper that you cite states that the earthquake flow extends to the boundary as nicely.

Answer by Jean-Marc Schlenker for Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature

2011-10-02T06:06:37Z

There is a satisfactory answer to Q1 if you restrict to convex surfaces: one way to state the question is then as the Minkowski problem. That is, you choose a positive function $k$ on the sphere and look for a surface $S$ in $R^3$ with Gauss curvature $k(n)$ at the point where the unit normal vector is $n$. This problem was solved in the early 50', see the Math Review of MR0058265 (15,347b) Nirenberg, Louis The Weyl and Minkowski problems in differential geometry in the large. Comm. Pure Appl. Math. 6, (1953). 337–394. In higher dimensions you can still play the same game, for convex hypersurfaces (or sometimes using a weaker form of convexity) and find one with prescribed "curvature", where the curvature can be a symmetric function of the eigenvalues of the shape operator. For instance the determinant of the shape operator, which corresponds to the Minkowski problem in higher dimension, which was also solved in the early 50', but it's not exactly what you're asking for in Q2.

Answer by Jean-Marc Schlenker for Geodesics in $\mathbb{R^2} \times \mathbb{S^1}$ under "segment" metric

2011-09-23T15:43:01Z

A useful keyword for this problem is the Wasserstein distance, see wikipedia. I believe that this Wasserstein distance, for $p=1$, provides a variant of the distance you're considering but for unoriented segments. There is a well-developed theory here, in particular concerning the geodesics and their behavior. Incidentally things my turn out to be easier if you take the means of the squares of the distances, rather than of the distances.

Elements of unit modulus in ring generated by root of unity

2011-09-18T14:47:24Z

When thinking of an apparently unrelated problem I stumbled upon the following question, which is certainly elementary to many readers of this site. Let $\omega_l=e^{i2\pi/l}$, and let $z\in Z[\omega_l]$ have $|z|=1$. Can we conclude that $z=\omega_l^k$ for some integer $k$? Thanks in advance for any answer!

Answer by Jean-Marc Schlenker for About Jacobi fields on nonpositive curvature

2011-08-27T10:28:38Z

I think that the answer to quetion 2 is yes. The basic property of Jacobi fields defined along a geodesic $\gamma$ (parameterized at speed $1$) is that they satisfy the equation $$ \eta''(t)=\pm R(\gamma',\eta)\gamma'~, $$ where $R$ is the curvature operator, $\eta''$ is with respect to the covariant derivative along $\gamma$, and the sign depends on the convention used.

Note that $H:x\mapsto \pm R(\gamma', x)\gamma'$ is self-adjoint, and positive semi-definite if the sectional curvature is non-positive. Moreover if $M$ is a symmetric space then its curvature operator is parallel, so $H$ is also parallel. So if you bring everything back at a point by parallel transport, the Jacobi field is solution of an equation of the form $\eta''=H\eta$ with $H$ constant and positive semi-definite. So your positivity condition should hold.

Answer by Jean-Marc Schlenker for estimating lattice sums of concave functions

2011-08-18T14:23:05Z

It looks like the error is in $O(1/n^2)$, with a precise and optimal bound $C/n^2$ if you have a fixed bound on (1) the second derivative of the function (2) the radius of the region where it is non-negative.

As the question is stated there are two sources for the error term:

the error in each square, centered at a point of the lattice, on which the function is strictly positive. This terms is controled by the second derivative of the function (it clearly vanishes for a linear function) at it is bounded by $O(1/n^4)$, since the number of squares is $O(n^2)$ the estimate on this whole term is $O(1/n^2)$,
the error term in the boundary squares, those on which the function takes both a $>0$ and a zero value. On those squares the error is $O(1/n^3)$ and the number or such boundary squares is $O(n)$ so we get again a bound $O(1/n^2)$.

(Note that a complete argument has to be more precise because the function $f$ could have zero derivative at the points where it vanishes, then the number of boundary squares is $O(n^2)$ but I think the result does not change).

To check that this estimate is optimal you can think of a function which is invariant under a rotation of angle $\pi/2$ and equal to say $N-x$ on $y>0, -y+u\leq x\leq y-u$ for some small $u>0$. Then the first error term can be made smaller than the second, while the second "boundary" error term is indeed of the order of $1/n^2$ (the boundary errors all sum up).

Answer by Jean-Marc Schlenker for minimal $L^2$ norm with $L^1$ norm fixed to one

2011-08-17T10:13:02Z

There are lots of others -- think of the function equal to $1/|\Omega|$ on one half of $\Omega$, and to $-1/|\Omega|$ on the other half. However if you restrict to non-negative functions then it is the only minimizer, as the equality case in Cauchy-Schwarz shows.

Answer by Jean-Marc Schlenker for harmonic function on surface

2011-08-11T05:43:20Z

Two remarks:

the notion of harmonic function on a surface is conformally invariant, so if your question is local the it is about harmonic function on $C$.
on a surface, a harmonic function is the same as the real part of a holomorphic function, see wikipedia. So you can reformulate your question on zeros as a question on the inverse image of the real line by a holomorphic function defined on (a subset of) $C$.

Answer by Jean-Marc Schlenker for Shortest "painting" of the sphere

2011-07-20T09:19:10Z

This question is somewhat related to this recent one. More precisely, the comment by Gjergji Zaimi in the earlier question gives a painting of length $2\sqrt{2}\pi$ for $d=\pi/4$, which, as explained in another comment there, is optimal for a path at constant distance from the sphere. So for $d=\pi/4$ the optimal length should be $2\sqrt{2}\pi$.

Answer by Jean-Marc Schlenker for Quasiconformal harmonic extension of a quasi-symmetric map on $S^1$

2011-07-19T20:10:15Z

The answer to the question depends on the Riemannian metric you choose in the target space (but only on the conformal class of the domain). One possibility is to choose the Euclidean metric, as in David's answer, but a more natural one is to consider the hyperbolic plane, as mentioned by Deane Yang. In this second case the question is still open. It was conjectured by Schoen that any quasisymmetric homeomorphism of the circle has a unique quasiconformal harmonic extension to the hyperbolic plane, see Richard M. Schoen, The role of harmonic mappings in rigidity and deformation problems, Complex geometry (Osaka, 1990), Lecture Notes in Pure and Appl. Math., vol. 143, Dekker, New York, 1993, pp. 179–200. MR MR1201611. The uniqueness is known, as for the existence there are newer results than those mentioned in other answers, see Vladimir Markovic, Harmonic diffeomorphisms of noncompact surfaces and Teichm¨uller spaces, J. London Math. Soc. (2) 65 (2002), no. 1, 103–114. MR MR1875138, but no complete answer.

Answer by Jean-Marc Schlenker for Random manifolds

2011-07-19T07:29:41Z

In a slightly different direction there are natural (but probably not uniquely defined notions of random surfaces, and also of random 3-manifolds, as in this paper by Dunfield and Thurston, based on Heegaard splittings. For those notions of randomness no curvature hypothesis is needed. But it does not seem obvious how to generalize them in higher dimension.

Answer by Jean-Marc Schlenker for Is a smooth closed surface in Euclidean 3-space rigid?

2011-07-16T05:31:41Z

Apparently the question is still open for smooth enough surfaces and deformations (that is, at least $C^2$).

Mike Anderson wrote a preprint claiming to prove local rigidity of smooth enough surfaces, but it was later withdrawn.

Idjad Sabitov and his collaborators have been working on this question, developing for instance a theory of higher-order isometric deformations, see e.g. Sabitov, I. Kh. Local theory of bendings of surfaces [MR1039820 (91c:53004)]. Geometry, III, 179–256, Encyclopaedia Math. Sci., 48, Springer, Berlin, 1992. He conjectures that local rigidity holds for analytic surfaces.

Answer by Jean-Marc Schlenker for Curvature formula

2011-07-09T10:57:56Z

There is basic formula in Riemannian geometry that, used twice, gives fairly direct answer. Let $M$ be a Riemannian manifold, let $N$ be a codimension $1$ submanifold of $M$, and let $f$ be a function defined on $M$. Call $f$ also the restriction of $f$ to $N$. Then the Hessian of $f$ on $N$ and the restriction to $TN$ of the Hessian of $f$ on $M$ are related by $$ Hess_N(f) = Hess_M(f)+df(n) II~, $$ where $n$ is the oriented unit normal of $N$ in $M$ and $II$ is the second fundamental form of $N$ in $M$.

You can apply this first for $S^3$ considered as a hypersurface in $R^4$, and obtain that $$ Hess_{S^3}(f) = Hess_{R^4}(f)+(\partial_rf) g~, $$ where $g$ is induced metric on $S^3$. Then you can apply it to $X$ considered as a hypersurface of $S^3$ to obtain that $$ 0 = Hess_X(f) = Hess_{S^3}(f) + df(n) II~, $$ where $n$ is the unit normal to $X$ in $S^3$ and $II$ its second fundamental form.

Putting all this together yields a formula $$ II = -(Hess_{R^4}(f) + (\partial_rf) g)/df(n)~. $$ The curvature of $X$ the follows from the Gauss formula, $K=1+det(II)$.

Answer by Jean-Marc Schlenker for A comparison question for non-positively curved disks

2011-07-01T19:08:29Z

You could check a recent paper by Burago and Ivanov, arXiv:1011.1570: Area minimizers and boundary rigidity of almost hyperbolic metrics. They prove the statement you want when $A$ is close to being hyperbolic (but in any dimension). See their Thm 1.6. They proved earlier the same statement if A is close to being Euclidean. Actually they prove a stronger rigidity statement: if the boundary distances are equal then the interior metrics are equal, not only the areas. (Somehow this stronger rigidity statement follows from the strict minimality of the area.)

Their proof is quite involved, but in the 2d case it's possible that some elementary arguments can work, for instance using the Santalo formula which they recall in their introduction.

Answer by Jean-Marc Schlenker for Parametrizing the realization space of a polyhedron by its edges

2011-06-27T08:44:55Z

In cases like this it might be practical to use the square of the lengths of the edges rather than the lengths. If $e_i$ is the edge between vertices $v_j$ and $v_k$ then the derivative of its square length can be written as $2\langle v_j-v_k, v'_j-v'_k\rangle$, which is quite simple.

Your problem is nice but probably difficult. If you consider only the dihedral angles and only convex polyhedra, you might want to look at a recent result by Mazzeo and Montcouquiol, http://arxiv.org/abs/0908.2981 They prove a rigidity result relevant to your question but their proof uses some real analysis.

Comment by Jean-Marc Schlenker

2013-05-21T10:36:52Z

Peter Mueller: are you sure that maps satisfying (A) correspond bijectively to latin squares? According to wikipedia, "a Latin square is an n × n array filled with n different symbols, each occurring exactly once in each row and exactly once in each column." So it implies that $a(i)-a(j)$ is non-zero for $i\neq j$. But it looks like (A) is more restrictive?

Comment by Jean-Marc Schlenker

2013-05-21T09:11:46Z

Douglas Zare: thanks for the additional explanations, I should have included this when submitting the question...

Comment by Jean-Marc Schlenker

2013-05-20T19:10:20Z

Joël: the question comes up rather naturally in a "recreational" but still serious project we're working on with Teo Banica and Ion Nechita, on complex Hadamard matrices. It looks like it might involves some ideas from basic number theory, things that I don't really know but that should be obvious to someone like you, so perhaps MO can help here... BTW, thanks for the editing.

Comment by Jean-Marc Schlenker

2013-05-20T16:08:21Z

P Vanchinathan: yes, the - sign in condition (A) is the substraction in the ring $Z/pZ$ and your interpretation is correct. No, we do not assume that $a$ is a group homomorphism, just a map.

Comment by Jean-Marc Schlenker

2011-10-11T15:34:26Z

You're right -- I think that the mean shadow probably decreases, but the argument I gave is not correct. I'll look into this later, unless some else can make it work...

Comment by Jean-Marc Schlenker

2011-10-10T10:03:41Z

I don't quite understand the question. You can always compare an open set to its convex hull, it will have the same mean shadow, but a smaller area. So the isoperimetric inequality for open subsets should follow from the convex case?

Comment by Jean-Marc Schlenker

2011-10-05T10:29:14Z

@Joseph: well technically it's a sort of answer to your Q1, although probably not what you had in mind.

Comment by Jean-Marc Schlenker

2011-10-02T18:45:29Z

As pointed our by Deane Yang above, this type of answer differs from the motivating result concerning curves in $R^2$ since it's global rather than local.

Comment by Jean-Marc Schlenker

2011-09-29T07:25:06Z

I'm pretty sure there is such a formula. If the 2d case is as you say, it should follow that the Weingarten operator is $B^{-1}=\pm Hess(h)+hI$, and it should be possible to prove this by considering what happens in 2-planes containing the eigenvalues of the shape operator.

Comment by Jean-Marc Schlenker

2011-09-28T18:42:10Z

@ Joseph: abolutely -- he's hard to read but he was a great geometer and a great analyst, too. And there is a long story of not quoting his work sufficiently!

Comment by Jean-Marc Schlenker

2011-09-27T16:31:36Z

@ Joseph: this is included in the definition of a convex cap that Pogorelov uses (I should have made this precise, sorry). See his uniqueness thm in the book mentioned above, on p. 78: books.google.com/… So I still believe that the existence and uniquess thms of Pogorelov, together, seem to be very close to the Hong thm you cite. (for some reason I can't make the google books link go directly to p. 78...)

Comment by Jean-Marc Schlenker

2011-09-27T11:31:24Z

I believe that the theorem you cite on isometric realization of convex caps is very close to one obtained by Pogorelov a while ago, see thm 4 on p. 104 of "Extrinsic geometry of convex surfaces" books.google.com/…

Comment by Jean-Marc Schlenker

2011-09-24T14:07:43Z

Thanks Joseph. Actually I'm not sure it's that useful, because the Wasserstein metric geodesics will not respect the structure of the segments -- you'll have segments at the beginning and at the end, but not in between. In other terms the space of segments is a kind of submanifold in the space of measures, but it's not geodesic. Back to your original question, it's probably necessary to do a computation to obtain the ODE describing geodesics, as mentioned by Sergei.

Comment by Jean-Marc Schlenker

2011-09-18T17:09:48Z

Thanks! This is exactly what I needed. MathOverflow is great, too!

Comment by Jean-Marc Schlenker

2011-08-31T15:13:26Z

There is no canonical geodesic representative of a homotopy class of curves in a Riemannian manifold in general (there is one if the curvature is negative). Moreover the notion of homotopy class of simple closed curves does not make sense in dimension larger than 2.