User alvarezpaiva - MathOverflow

Answer by alvarezpaiva for Reference request: Riemannian manifold of linear isometries from $\mathbb{C}^n$ into $\mathbb{C}^m$

2013-05-17T07:29:29Z

At first sight, it seems to me you talking about partial isometries. If so, this paper by Andruchow and Corach (and/or its references) may help.

A question on the theorem of Minkowski-Hlawka

2013-05-16T19:53:30Z

The Minkowski-Hlawka theorem (as stated by Gruber in his lovely book Convex and Discrete Geometry) says that if $S$ is a Jordan measurable set in $\mathbb{R}^n$ with volume < 1. Then there is a lattice $L \subset \mathbb{R}^n$ with determinant $1$ which contains no point of $S$, with the possible exception of the origin.

In the special case where the set $S$ is a star-shaped body (with respect to the origin), this inequality can be written as $$ 1 < \text{vol}(S)/\Delta(S) , $$ where $\Delta(S)$ is the critical determinant of $S$ (i.e., the infimum of the determinants of all lattices that intersect $S$ only at the origin).

Question. Does the the linear-invariant functional $S \mapsto \text{vol}(S)/\Delta(S)$ attain its greatest lower bound when defined on (1) the space of star-shaped (compact) bodies and (2) the space of convex bodies that contain the origin as an interior point.

I'm really just interested in the question of existence of minima and specially interested in the case of convex bodies. In any case, even in the "easy" case of $0$-symmetric convex bodies in $\mathbb{R}^2$, where the existence of minima is obvious, the minimum value is still just a conjecture (the Reinhardt conjecture).

Answer by alvarezpaiva for Algebraic topology in low regularity

2013-05-09T12:43:37Z

This does not answer your specific questions, but degree theory has been extended to certain classes of non-continuous functions (including some Sobolev spaces) by Brezis and Nirenberg. See

Brezis, H., & Nirenberg, L. (1995). Degree theory and BMO; part I: Compact manifolds without boundaries. Selecta Mathematica, New Series, 1(2), 197-263.

Brezis, H., & Nirenberg, L. (1996). Degree theory and BMO; part II: Compact manifolds with boundaries. Selecta Mathematica, New Series, 2(3), 309-368.

Answer by alvarezpaiva for Good book on Calculus of Variations

2013-04-29T17:24:57Z

The book by Gelfand and Fomin is quite good (and its Dover ...). Another one I like a great deal are those of Giaquinta and Hildebrandt (specially volume 1), but those are not Dover: check them out from the library!

A question on the Mahler conjecture

2013-04-29T14:18:03Z

In its asymmetric version, the Mahler conjecture states that if $K \subset \mathbb{R^n}$ is a convex body containing the origin as an interior point and $$ K^* := \{y \in \mathbb{R}^n : \langle y, x \rangle \leq 1 \mbox{ for all } x \in K \} $$ is its polar body, then the product of volumes $|K| |K^*|$ is bounded below by $(n+1)^{n+1}/(n!)^2$. Equality is conjectured to hold only for simplexes.

Is it known whether there is a unique minimum of $K \mapsto |K| |K^*|$ modulo linear equivalence?

Is there some dimension dependent bound on the number of minima (modulo linear equivalence)?

For a given dimension is it least known that the number of minima (modulo linear equivalence) is finite?

This last would follow if it were known that the minima or local minima are isolated points in the space of linear equivalence classes of convex bodies containing the origin as an interior point. I guess that's only known for the explicit case of the simplex.

I know that determining whether the minima are polytopes is still open (and that in the symmetric case there are different linearly inequivalent classes of conjectured minima), but I don't remember having seen the problem of uniqueness discussed in the asymmetric case.

A spectral inequality for positive-definite matrices

2013-04-27T16:49:03Z

Question. Given a positive-definite $n \times n$ matrix $A = (a_{ij})$ with eigenvalues $$ \lambda_1 \leq \cdots \leq \lambda_n , $$ is there a sharp upper bound for the product $\lambda_2 \cdots \lambda_n$ in terms of the quantity $$ \|A\|_\infty := \max_{1 \leq i, j \leq n} |a_{ij}| ? $$

A classic inequality due to A. Hirsch states that the modulus of an eigenvalue of an $n \times n$ complex matrix $A$ is less than $n \|A\|_\infty$, which implies that $$ |\lambda_2 \cdots \lambda_n| \leq n^{n-1}\|A\|_\infty^{n-1} . $$ However, this seems like a rather rough estimate for positive-definite matrices. I'm interested in any estimate that is substantially better than this.

Motivation. Using Hirsch's inequality one can improve Lemma 4 in page 43 of Siegel's Lectures on Quadratic Forms to yield the following result:

Theorem. If $A = (a_{ij})$ is a positive-definite $n \times n$ matrix, then for every $x \in \mathbb{R}^n$ we have that $$ \frac{\det(A)}{n^{n-1}a_{11} a_{22} \cdots a_{nn}} \sum a_{ii} x_i^2 \leq \sum a_{ij} x_i x_j \leq n \sum a_{ii} x_i^2 . $$

The inequality on the left would be greatly improved if we had the sharp upper bound required in the question. This in turn would yield a better answer to this enclosure problem (see my answer to that question).

Addendum. If we apply the estimate in Suvrit's answer in the proof of the theorem above, the inequality is indeed improved to:

$$ \frac{\det(A)}{2^{n-1}a_{11} a_{22} \cdots a_{nn}} \sum a_{ii} x_i^2 \leq \sum a_{ij} x_i x_j \leq n \sum a_{ii} x_i^2 . $$

In fact, in the proof the estimate for $\lambda_2 \cdots \lambda_n$ is applied to an auxiliary matrix $B = (b_{ij})$ whose diagonal entries are all $1$ and for which $|b_{ij}| < 1$ if $i \neq j$.

In turn this yields the following improved bound for the enclosure problem:

Theorem. Let $E \subset \mathbb{R^n}$ be an $n$-dimensional ellipsoid centered at the origin and containing no other integer point. There exists a transformation $T \in GL(n,\mathbb{Z})$ such that $T(E)$ is contained in the ball of radius $$ \left(\frac{3}{2}\right)^{(n-1)(n-2)/2} \frac{2^n}{\epsilon_n}\sqrt{2^{n-1}} $$ centered at the origin.

Here $\epsilon_n$ is the volume of the unit ball of dimension $n$.

Does anyone know a better bound?

Ellipsoids and lattices: an enclosure problem.

2012-02-08T20:12:58Z

$E \subset {\mathbb R}^2$ is an ellipse of area $1$ centered at the origin that contains no other point with integer coordinates. Is there a matrix $A \in SL(2,{\mathbb Z})$ such that the ellipse $A(E)$ is contained in a disc of radius 10?

Hopefully, this is really easy and it is only my ignorance in "reduction theory" (or other similar techniques) that is at the heart of my difficulties.

More generally, given an ellipsoid $E$ of unit volume in ${\mathbb R}^n$ centered at the origin and containing no other point with integer coordinates, I'm interested in a good upper estimate of the size of the ball in which I can enclose $A(E)$ for some matrix $A \in SL(n,{\mathbb Z})$.

Answer by alvarezpaiva for Ellipsoids and lattices: an enclosure problem.

2013-04-26T18:41:17Z

This is an addition to Noam's answer. In higher dimensions the same type of reasoning---using explicit conditions on the coefficients of reduced quadratic forms---is not feasible. Nevertheless, it is possible to prove the following result:

Here $\epsilon_n$ is the volume of the unit ball of dimension $n$.

This theorem follows easily from the following two results:

Theorem. Given a positive-definite $n \times n$ matrix $A = (a_{ij})$, there exists a unimodular integer matrix $U$ such that the matrix $A' = (a'_{ij})$ defined by $A' := U^t A U$ satisfies the following inequality: $$ \left(\frac{2}{3}\right)^{(n-1)(n-2)} \left(\frac{\epsilon_n}{2^n}\right)^2 \leq \frac{\det(A')}{a'_{11} a'_{22} \cdots a'_{nn}} . $$

The first of these two results is an important theorem of Minkowski on the reduction of positive quadratic forms (see Theorem 3 in page 69 of Lekkerkerker), while the second is an improved version of Lemma 4 in page 43 of Siegel's Lectures on Quadratic Forms.

Answer by alvarezpaiva for Intuition for Levi-Civita connection via Hamiltonian flows

2013-04-12T18:07:27Z

The intuition is that the Levi-Civita connection corresponds to the linearization of the geodesic flow plus a simple projective-geometric construction:

Let $c(t)$ be an orbit of the geodesic flow (projecting down to a geodesic), consider the vertical subspaces $V(t)$ along $c(t)$ and bring them back to the tangent space of the cotangent bundle over the point c(0) by using the differential of the flow. You get a family of (Lagrangian) subspaces $l(t) := D\phi_{-t}(V(t))$ that is "fanning" or "regular".

Now forget you ever had a geodesic flow: all that you need is the curve of subspaces. A bit of differential projective geometry shows that you also get a second curve $h(t)$ of (Lagrangian subspaces) in $T_{c(0)}(T^*M)$that is transversal to $l(t)$. The subspace h(0) is the horizontal subspace of the connection and $T_{c(0)}(T^*M) = l(0) \oplus h(0)$ is the decomposition into vertical and horizontal subspaces.

I think this is nicely written in this paper ;-)

However, this is classical: it's just the geometry behind the Schwartzian derivative. There are plenty of references in the paper if you're interested in this.

Addendum on the geometry of the Schwartzian derivative.

Since the Schwartzian derivative has many "standard" geometric interpretations, for completeness sake I'll sketch the (new?) one that I'm refering to. I will do it in a way just a bit more conceptual than was done by Duran and myself in the paper. Actually, it's just a "high brow" version of what was done in the paper, but I will do it here just for curves on the projective line. The reader may amuse himself/herself at extending this to curves on the Grassmannian of $n$-planes in $\mathbb{R}^{2n}$ and checking against the paper.

Consider the action of the linear group $GL(2;\mathbb{R})$ on the projective line and lift it to an action on its cotangent bundle. The moment map of this action takes values on the set of nilpotent matrices. In fact, if you take away the zero section, the moment map takes value in the coadjoint orbit formed by $2 \times 2$ nilpotent matrices of rank one. In higher dimensions, one needs to take away a bit more than the zero section and "rank one" is replanced by "the kernel and the image of the matrix (seen as linear transformation) are the same".

We will also need a neat thing about the geometry of the tangent space of the projective line: there is a canonical way to identify non-zero tangent vectors to non-zero covectors. In fact, the tangent space of the projective line at a line $\ell$ is the space of linear maps between $\ell$ and the quotient space $\mathbb{R}^2/\ell$, which is the tensor product $(\mathbb{R}^2/\ell) \otimes \ell^* $. If the map is invertible, its inverse is a map from $\mathbb{R}^2/\ell$ to $\ell$ and, therefore, an element of $ \ell \otimes (\mathbb{R}^2/\ell)^*$, which is the cotangent space at $\ell$.

Now consider a curve $\gamma(t)$ on the projective line whose derivative never vanishes. We lift the curve to the curve $\dot{\gamma}(t)$ on the tangent bundle and use the isomorphism described above to obtain a curve $\Gamma(t)$ on the cotangent of the projective line. Use the moment map to obtain a curve $F(t)$ of nilpotent matrices. Note that everything we have done is projective-equivariant.

Finally we come to the little miracle: the time derivative of $F(t)$ is a curve of reflections $\dot{F}(t)$ (i.e., $\dot{F}(t)^2 = I$) whose -1 eigenspace is the curve of lines $\gamma(t)$ and whose $1$-eigenspace defines a "horizontal curve" $h(t)$ equivariantly attached to $\gamma(t)$. This is the construction that yields the Levi-Civita connection (and what is behind the formalisms of Grifone and Foulon for connections of second order ODE's on manifolds).

Differentiate $F(t)$ a second time to find the Schwartzian derivative. Geometrically, it just describes how the curve $h(t)$ moves with respect to $\gamma(t)$. For comparison, recall that the curvature of a connection is obtained by differentiating (i.e., bracketing) horizonal vector fields and projecting onto the vertical bundle.

Monotonicity of Loewner ellipsoid?

2013-04-10T16:56:01Z

Given two $0$-symmetric convex bodies $K \subset L \subset \mathbb{R}^n$, is it true that the Loewner ellipsoid of $K$ is contained in the Loewner ellipsoid of $L$?

I have just finished a proving a lemma stating that the Loewner ellipsoid depends continuously on parameters and the proof is a bit more elaborate than I first expected. I suddenly realized that I was unconsciously assuming that the Loewner ellipsoid is not monotone (otherwise the lemma would be trivially true), but that I did not have a ready example showing that this was the case.

I profit to ask a second question: is there a reference for the fact that the Loewner ellipsoids of a continuous family of convex bodies form a continuous family?

Rephrase in terms of normed or Finsler bundles and Euclidean structures if you want to be very rigourous.

My proof of this fact involves "looking under the hood" at the proof of uniqueness of the Loewner ellipsoid and using Berge's maximum theorem for set-valued maps. It's natural (after all this is just a problem in mathematical programming), but I was expecting a triviality.

Answer by alvarezpaiva for Fubini-Study metric for an infinite dimensional Hilbert space

2013-03-31T12:15:29Z

Maybe it's easier to see that the definition extends if you don't use a formula:

The unit sphere inherits a Riemannian metric from the Hilbert space in the standard manner and since it is invariant under the circular symmetry $(e^{i\theta},x) \mapsto e^{i\theta} x$, it will project down to a Riemannian metric on the complex projective space.

A question of compactness in the geometry of numbers

2013-03-25T14:21:40Z

Given a star body $S \subset \mathbb{R}^n$ with the origin as interior point, the critical determinant of $S$---usually denoted as $\Delta(S)$---is the infimum of the determinants of all lattices that intersect $S$ only at the origin.

The quantity ${\rm vol}(S)/\Delta(S)$ is a linear invariant of compact, star bodies. If we restrict its domain to the space $\mathcal{K}_0^n$ of convex bodies in $\mathbb{R}^n$ that contain the origin in their interior, then it is a continuous linear invariant.

The question: Is there a nice little proof or a reference for the statement that the the sublevel sets $$ \{ K \in \mathcal{K}_0^n : {\rm vol}(K)/\Delta(K) \leq C \} $$ are compact and hence the functional must attain a global minimum?

I think this is true (because of Macbeath's compactness theorem and the fact that $\Delta(K)^{-1}$ blows up if the origin is near the boundary of $K$), but have not really checked thoroughly thinking that is must be widely known although I cannot find the explicit statement in Cassel or Lekkerkerker.

Addendum. Sergei has given an excellent reason why this is not to be found in Cassel or Lekkerkerker ... However, much work has gone into giving lower bounds of the functional $K \mapsto {\rm vol}(K)/\Delta(K)$. For example, the Minkowski-Hlawka theorem says it is bounded below by $1$ (or $\zeta(n)$ to be more accurate). Is it known whether this functional attains a (global) minimum on $\mathcal{K}_0^n$ ?

Reversibility vs geodesic reversibility for Finsler metrics on the two-sphere

2013-01-11T21:19:07Z

Problem. To give a concrete example of a geodesically reversible Finsler metric on the two-sphere that is not just the sum of a reversible Finsler metric and an exact $1$-form.

Some background may be useful since the problem is more of a problem in variational calculus than a problem in geometry.

If $M$ is a smooth manifold and $L: TM \rightarrow \mathbb{R}$ is a (Lagrangian) fuction that is smooth outside the zero section and homogeneous of degree one in the velocities (i.e. $L(x,tv) = tL(x,v)$ for every $t > 0$), the variational problem $$ \gamma \mapsto \int_\gamma L $$ is invariant under orientation-preserving reparametrization of the curve $\gamma$. The Lagrangian $L$ is geodesically reversible if changing the orientation of any of its extremals yields another extremal. If the Lagrangian is reversible ($L(x,-v) = L(x,v)$), then it is geodesically reversible, but the converse is not true. For example, an asymmetric norm on $\mathbb{R}^n$ is geodesically reversible, but it is reversible if and only if the norm is symmetric. Another example can be constructed by taking a reversible Lagrangian and adding to it a closed $1$-form considered as a function on the tangent bundle that is linear in the velocities. This last example is somehow "trivial" and I would like to find many examples of geodesically reversible Lagrangians on compact manifolds that are not of this form. On the torus they are easy to construct because we can always compactify the example with the asymmetric norm, but what happens on the sphere?

I can point to a non-solution: If all the geodesics of a geodesically reversible Finsler metric on the $n$-sphere are closed and of the same length, then the metric is the sum of a reversible metric and an exact $1$-form.

Answer by alvarezpaiva for Characterizing Hessians among symmetric bilinear tensors

2013-03-05T18:01:17Z

This is more of a comment/thinking aloud than an answer, but perhaps the symplectic viewpoint helps:

Think of the Hessian with respect to a Riemannian metric as follows: through the Legendre transfom, the Levi-Civita connection can be seen as a splitting of the tangent bundle of the cotangent bundle of $M$ into vertical and horizontal bundles: $T(T^*M) = V(T^*M) \oplus H(T^*M)$. At each point $p_x \in T^*M$, the subspace $V_{p_x}(T^*M) = V_{p_x}$ and $H_{p_x}(T^*M) = H_{p_x}$ are Lagrangian subspaces.

If $f$ is a function on $M$, the graph of its differential is a Lagrangian submanifold $L \subset T^*M$. The tangent space of $L$ at the point $df(x)$ is a Lagrangian subspace of $T_{df(x)} T^*M$ which is transverse to the vertical space $V_{df(x)}$ and, therefore, we can think of $T_{df(x)}L$ as the graph of a linear transformation from $H_{df(x)}$ to $V_{df(x)}$. Modulo some basic identifications, this linear map is the Hessian of $f$ at $x$.

A nice thing about this description is that it works on Finsler spaces since the splitting $T(T^*M) = V(T^*M) \oplus H(T^*M)$ just comes from the linearlization of the geodesic flow and the convexity of the Hamiltonian.

If you start with a Hessian, this description suggests the following procedure to construct a function $f$: given a point $p_x \in T^*M$, use the Hessian of $f$ at $x$, the identification of $V_{p_x}$ and $H_{p_x}$ with $T_xM$ and the decomposition $T_{p_x}T^*M$ to construct a Lagrangian subspace in $T_{p_x}T^*M$. One ends up with a field of Lagrangian subspaces in $TT^*M$. We have to find an integral manifold $L$. Such a manifold will be Lagrangian by construction and, given that the constructed Lagrangian subspaces are transverse to the vertical subspaces, it will also project in a locally-diffeomorphic way onto $M$. It's a good candidate for the graph of $df$.

Answer by alvarezpaiva for determining a convex set by mixed volumes

2013-02-09T16:53:00Z

Dear Karl, if by $K$ convex you mean centrally-symmetric convex body, the answer is yes, the map is injective. There must be a neat proof somewhere, but on the spur of the moment I came up with this one which works in $n$-dimensions if you use the mixed volume $$ MV(A,K) := \lim_{h \rightarrow +0} (V(A + hK) - V(A))/h . $$ Define a norm $\psi_K$ in the space of $(n-1)$-vectors in $\mathbb{R^n}$ by setting $$ \psi(v_1 \wedge \cdots \wedge v_{n-1}) = MV([v_1,\cdots,v_{n-1}],K), $$ where $[v_1,\cdots,v_{n-1}]$ is the parallelotope defined by the vectors $v_1$,$\ldots$,$v_{n-1}$.

Assume you have two centrally symmetric bodies $K$ and $K'$ giving rise to the same norm in $\Lambda^{n-1} \mathbb{R}^n$, then they must be the same because both bodies can be reconstructed from this norm (up to a constant multiplcative factor depending on dimensions, which I've been lazy to specify) by the Wulff construction: Let $B \subset \Lambda^{n-1} \mathbb{R}^n$ be the unit ball of the norm, and let $B^* \subset \Lambda^{n-1} \mathbb{R}^{n*}$ be its polar body. Note that the volume form $\Omega$ in $\mathbb{R}^n$ defines an isomorphism between $\mathbb{R^n}$ and $\Lambda^{n-1} \mathbb{R}^{n*}$ by the contraction map $$ v \mapsto i_v\Omega = \Omega(v\wedge \cdot). $$ Up to a dilation that depends only on the dimension of the space, your convex body $K$ (and $K'$) are the images of $B^*$ under this isomorphism.

As I said before, I think there is simpler than this (and I have not given thought to the non-symmetric case). But if you want more details on this see section 6 of my paper with Thompson Volumes on normed and Finsler spaces

Pencils of circles and Liouville's theorem

2013-02-08T17:57:01Z

Is there any relation (maybe implicit) between the conformal geometry in the space of circles and spheres and the study of harmonic functions?

In the original question I was musing whether the following argument could be made into rigorous proof of Lioville's theorem (any bounded harmonic function on the plane (or $\mathbb{R}^n$) is constant.)?

Geometer "proof" (i.e. just as bad as a physicist's "proof"):

By the mean value theorem, the average of a harmonic function over any two concentric circles is the same. Notice that (1) a pencil of concentric circles is just an elliptic pencil of circles where one of the limit points is infinity, and that (2) all elliptic pencils are equivalent by some Moebius transformation. Since Moebius transformations take harmonic functions to harmonic functions, the mean value theorem actually says that the average of a harmonic function over any two circles in an elliptic pencil is the same. To see that the value of the function at any two points is the same, apply this remark to an elliptic pencil of circles having these two points as limit points.

Obviously this is not a proof : those Moebius transformations are moving infinity all over the place and some of the circles in the elliptic pencils are actually straight lines. However, I never used the fact that the harmonic function is bounded either ! Can one use the hypothesis that the function is bounded to make these ideas rigorous?

Disclaimer: this is just for fun.

Noam Elkies pointed out (see his comment) that there is the additional difficulty that the way we measure averages on circles depends also on the point at infinity and is not preserved by Moebius transformations.

Answer by alvarezpaiva for Submersions from compact flat manifold

2013-01-31T18:43:38Z

If the submersion is Riemannian, the answer is yes for $N$: Any Riemannian submersion with complete flat total space and compact base must have a flat base space. This was proved by Luis Guijarro and Peter Petersen in Annales Scientifiques de l’École Normale Supérieure Volume 30, Issue 5, 1997, Pages 595–603.

However, this uses more hypotheses than you have (and gets a stronger result than you wish for).

Answer by alvarezpaiva for Research trends in geometry of numbers?

2013-01-30T12:06:18Z

I can't say that what I'll relate is fundamental, but it does fit into the new ideas category. Since I and (my collaborator) Florent Balacheff have given talks on the subject and the paper will be in the ArXiv in a few days I feel free to comment on it. This post is an annoucement of joint work with Florent Balacheff and Kroum Tzanev.

As you comment, the basic result in the geometry of numbers is Minkowski's (first) theorem: If the volume of a $0$-symmetric convex body $K \subset \mathbb{R}^n$ is at least $2^n$, then $K$ contains a non-zero integer point.

But what happens when the body is not $0$-symmetric? It is easy to see that Minkowski's theorem fails completely, but that's because one is not thinking symplectically. By using some Hamiltonian dynamics of the sort Balacheff and I used to study isosystolic inequalities in this paper, we guessed that the "right" result should be the following:

Conjecture. If a convex body in $\mathbb{R}^n$ contains no integer point other than the origin, then the volume of its dual body with respect to the origin is at least (n+1)/n!

In other words, one should have a sort of uncertainty principle: if the origin is localized as the unique integer point inside a convex body, the dual body cannot be too small. In fact, its volume is bounded below by $(n+1)/n!$. Another formulation of the conjecture that seems more elementary goes as follows:

If every hyperplane $m_1x_1 + \cdots m_nx_n = 1$, where the $m_i$ are integers not all equal to zero, intersects a convex body $K \subset \mathbb{R}^n$, then the volume of $K$ is at least $(n+1)/n!$

We proved the conjecture in the case $n = 2$ and the asymptotic version:

Theorem. There exists a (universal) constant $C \leq 1$ such that if a convex body $K \subset \mathbb{R}^n$ contains no integer point other than the origin, then the volume of $K^*$ is at least $C^n(n+1)/n!$.

In fact, this result is equivalent to Bourgain-Milman. Moreover, it easily implies the asymptotic version of a conjecture of Ehrhart:

Theorem. There exists a universal constant $c \geq 1$ such that if $K \subset \mathbb{R}^n$ is a convex body with barycenter at the origin and containing no other integer point, then the volume of $K$ is at most $c^n (n+1)^n/n!$.

However, what is really interesting for us is that at least in the case $n=2$ the result trascends the geometry of numbers and is really a result in Hamiltonian dynamics. I just need a definition:

Definition. A hypersurface in the cotangent bundle of a manifold $M$ is said to be optical if its intersection with every cotangent space is a convex hypersurface enclosing the origin.

To an optical hypersurface in the cotangent of a compact manifold we can associate two numbers: the symplectic volume of the region enclosed by $\Sigma$ and the least action of its periodic characteristics.

Theorem. An optical hypersurface $\Sigma$ in the cotangent space of the two-torus carries a periodic characteristic whose action is less than or equal to the square root of two-thirds the symplectic volume enclosed by $\Sigma$.

The inequality is sharp.

Finsler geometers will be happier if I translate: If the Holmes-Thompson volume of a (non-reversible) Finsler $2$-torus $(T^2,F)$ is $3/2\pi$, then $(T^2,F)$ carries a (non-contractible) periodic geodesic of length at most $1$.

In other words, this is the (non-reversible) Finsler version of Loewner's systolic inequality. The reversible Finsler version (replace $3/2\pi$ by $2/\pi$) is due to Stéphane Sabourau and can be found here.

Symmetric matrices and Hilbert's fourth problem

2013-01-23T15:53:43Z

From the analytic viewpoint, the Busemann-Pogorelov solution of Hilbert's fourth problem is summarized in the following result:

Theorem. All straight lines are extremals of the variational problem $$ \gamma \mapsto \int_\gamma L \ , $$ where the Lagrangian function $L : T\mathbb{R}^n \rightarrow \mathbb{R}$ is smooth outside the zero section and absolutely homogeneous of degree one (i.e., $L(x,\lambda v) = |\lambda|L(x,v)$), if and only if there exists a smooth function $f: \mathbb{R} \times S^{n-1} \rightarrow \mathbb{R}$ for which $$ L(x,v) = \int_{\xi \in S^{n-1}} |\xi\cdot v|f(\xi\cdot x,\xi)dA(\xi) . $$

In trying to embellish this preprint for submission, I started thinking in how to give nice, (even more) explicit formulas of Finsler metrics and Lagrangians on the tangent space of the $n$-sphere or projective $n$-space for which all great circles or projective lines are extremals. I came up with this construction:

Given a positive definite $n\times n$ symmetric matrix $A$ and a non-zero tangent vector $(x,v)$ to the $(n-1)$-sphere in Euclidean $n$-space, let $\sigma(A;x,v)$ denote the absolute value of the sine of the angle formed by the vectors $Ax$ and $Av$.

Proposition. If $\mu$ is a smooth (signed) measure of compact support on the space $PS(n)$ of symmetric positive-definite $n\times n$ matrices, then all great circles are extremals of the Lagrangian $$ L(x,v) := \int_{A \in PS(n)} \frac{\|Av\|}{\|Ax\|} \sigma(A;x,v) \ d\mu(A) $$ defined on the tangent space of the $(n-1)$-sphere $x_1^2 + \cdots + x_n^2 = 1$.

Proof. A simple computation will show that the integrand is nothing but the pullback of the arc-length element for the standard metric on the sphere under the map $x \mapsto Ax/\|Ax\|$. Geodesics for this Lagrangian are great circles. The set of all $1$-homogeneous Lagrangians on a manifold sharing the same extremals is a linear space and so adding or integrating such Lagrangians yields another of the same kind. Q.E.D.

In other words, the whole point is that in the sphere there are lots of collineations that are not isometries and this can be exploited.

Finally:

Question 1. What are the natural conditions on the measure $\mu$ for which the construction works? I gave "smooth" and "compact support" as conditions, but these are not necessary: use a delta function and the construction gives a nice Riemannian metric.

Question 2. What conditions of the measure will guarantee that $L$ is a Finsler metric (i.e. that $L(x,\cdot)$ be a norm on each tangent space $T_x S^{n-1}$)? Requiring $\mu$ to be non-negative and non-zero will do the trick, but this maybe too strong for $n > 3$.

In fact, this construction hits the same snag as the Busemann-Pogorelov construction: if one uses positive measures, one ends up with hypermetric metrics on the sphere (because the Minkowski sum of ellipsoids---the little something hidden behind the construction---is a zonoid). This brings me to

Question 3. Is every hypermetric Finsler solution of Hilbert's fourth problem on the sphere obtained by the above construction by choosing an appropriate measure $\mu$.

Convex bodies with symmetric shadows.

2013-01-22T14:13:51Z

Theorem. If all orthogonal projections of a convex body $K \subset \mathbb{R}^n$ onto $2$-dimensional subspaces have a center of symmetry, then $K$ has a center of symmetry.

This is a classic result of Blaschke and Hessenberg (that I just learned thanks to Guillaume's comment.). A short simple proof of it can be found in Bonnesen and Fenchel.

I wonder if it is necessary to know what happens for every orthogonal projection or whether we can get by with less:

Question 1. Let $K \subset \mathbb{C}^{n}$ be a convex body. Assume all orthogonal projections of $K$ onto complex lines have a center of symmetry. Does it follow that $K$ must also have a center of symmetry?

Note. The center of symmetry of the shadows may depend on the subspace containing it.

A similar question is:

Question 2. Let $K \subset \mathbb{C}^{n}$ be a convex body. Assume all orthogonal projections of $K$ onto Lagrangian subspaces have a center of symmetry. Does it follow that $K$ must also have a center of symmetry?

Can a simple Riemannian metric on the disc be extended to a Zoll metric on the sphere?

2013-01-19T14:03:03Z

Given a simple Riemannian metric $(D,g)$ on the two-disc---its geodesics have no conjugate point and the boundary of the disc is strictly convex---, is it possible to embed $(D,g)$ isometrically into a Zoll two-sphere?

I'm also interested in the same question on the $n$-disc. A related question is whether one can distinguish a Zoll metric locally (or at least say: this metric cannot be Zoll because in the neighborhood of this point it does not behave in such and such way). I'm guessing the answer to this last question is no and this prompted the first question.

Answer by alvarezpaiva for Known size invariant for Riemannian manifolds?

2013-01-12T20:58:02Z

Remark. Now that Cadoi has disclosed the origin of the invariant (which was absent from the original formulation of the problem) my musings in trying to guess the symplectic geometry behind it seem a bit silly. I'll leave the post anyway, since it maybe useful in understanding the question. However, maybe there is something to be said for including motivation and sources (right from the start) in MO questions.

Dear Cadoi,

This is just a comment on your question, but since it is a bit long and I can see what I'm TeXing I'll post it as an answer. I myself have not seen this invariant before, however it has a symplectic look about it and may turn out to be some capacity in disguise. Note that if you define a Riemannian invariant by taking a symplectic capacity and evaluating it on unit codisc bundles, then you get a size invariant. This is because isometries lift to symplectic transformations that preserve the unit codisc bundle and because capacities are monotone.

The reason I say your invariant looks symplectic is because, seen geometrically, the differential of $f$ defines a Lagrangian submanifold of the cotangent bundle and the condition that the norm of the gradient of $f$ be at least $1$ on $K$ means this Lagrangian submanifold does not intersect the unit co-disc bundle over $K$. The quantity $\|f\|$ is a relative symplectic invariant of the pair of Lagrangians given by the zero section and the differential of $f$. I don't exactly remember Viterbo's notation for this, but it is in his paper Symplectic topology as the geometry of generating functions.

Just for the fun of it, and without implying that this will give something non-trivial, I'll attempt a "symplectization" of your invariant, at least to the extent that it might give size invariant of Riemannian or Finsler metrics that gives the same value on two metrics that have symplectomorphic codisc bundles:

Let $D^*M$ be the unit codisc bundle of your metric and let $K \subset D^*M$ be a compact subset. Consider the set $P_K$ of pairs of Lagrangian submanifolds $(E,F)$ such that (1) $E$ is contained in $D^*M$, (2) $F$ does not intersect $K$, and (3) both $E$ and $F$ are Hamiltonian isotopic to the zero section.

Define $S(D^*M,K)$ as the infimum of the Viterbo capacities of all pairs of Lagrangians $(E,F) \in P_K$ and now define $S(D^*M)$ as the supremum of the quantities $S(D^*M,K)$ as $K$ ranges over all compact subsets of $D^*M$.

If you restrict $E$ to always be the zero section, $F$ to the the graph of the differential of a function, and $K$ to the the codisc bundle over some compact subset of $M$ this is exactly your invariant.

Caveat emptor, this definition may require some fidgeting (like restricting the class of compact sets $K$) to make it non-obviously trivial.

Answer by alvarezpaiva for Reversibility vs geodesic reversibility for Finsler metrics on the two-sphere

2013-01-14T09:30:16Z

After a few days thinking I understand my own question ... . This is not to say I've solved it, but I think I understood why it may be very hard. In fact, geodesically-reversible Finsler metrics that are not of the form reversible metric + closed 1-form are very rare and there may actually be none in the two-sphere nor in any other compact surface besides the torus and the Klein bottle, both of which admit flat non-revesible Finsler metrics.

Instead of going through some relatively easy partial results (e.g., if the flow of a geodesically reversible Finsler metric is ergodic, then the metric is the sum of a reversible metric and a closed 1-form) I'll explain by an integral-geometric analogy, which is loosely related to the problem, why the problem is probably very hard.

The injectivity of the X-ray transform on the projective plane is not particularly hard, but I don't think it can be called trivial at all. However, the existing proofs use the symmetries of the transform very strongly and, as far as I know, given a Zoll reversible Finsler metric on the projective plane (and there are tons of these)---including one for which all geodesics are projective lines such as solutions of Hilbert's fourth problem---it is not known whether the associated $X$-ray transform is injective. However, in the case of solutions of Hilbert's fourth problem, some of the work of Quinto points in this direction.

Assume for a second you have happily proved this, but someone (maybe yourself) comes along and asks the following

Question. Let $ds$ be the arc-length element of a Riemannian or reversible Finsler metric on the projective plane and suppose a smooth function $f$ on $\mathbb{RP}^2$ is such that the integral of $fds$ over all closed geodesics of $\mathbb{RP}^2$ is equal to zero. Is it true that $f$ must be zero?

Since there are no counter-examples to the affirmative answer of this question, your theorem suddenly looks incredibly small. However, the answer to the new question seems impossibly hard, specially since all results in this direction concern metrics that are negatively curved or Anosov (via Livsic's theorem).

Warning: this note is slightly autobiographical: see here

Forms satisfying the zero-energy condition on the projective plane

2013-01-02T19:48:13Z

Theorem (Michel). A $1$-form on the projective plane is exact if and only if its integral over any projective line is equal to zero.

Is there a simple proof of this result due, I think, to R. Michel ?

I'm guessing there must be a representation theoretic proof (everything is $SL(3;\mathbb{R})$ equivariant) and a complex-geometric proof (where $\mathbb{R}P^2$ is complexified to $\mathbb{C}P^2$ ) and I would appreciate reference for these, but I'm really interested in a simple proof that one could teach to seniors or first year grads.

P.S. I have not been able to get a hold of Michel's papers. Perhaps everything is there and in that case I apologize before hand.

Answer by alvarezpaiva for Conley index for isolated invariant sets with no exit points

2013-01-10T23:40:43Z

The Conley index is a generalization of the Morse index and it must give basically "the same" answer when the Morse index is defined: when the invariant set is an isolated non-degenerate critical point and the isolating neighborhood is a small neighborhood where the Morse lemma gives you a normal form for the function. If you work out the dictionary between Morse index and Conley index, you will see why in the case of an isolated minimim the convention comes up.

Characterizing maximal powers of closed 2-forms in odd-dimensional manifolds

2012-12-31T12:29:52Z

Given a nowhere-zero, closed $2n$-form $\Omega$ in a manifold of dimension $2n +1$, how do we know if there exists a closed $2$-form $\omega$ such that $\Omega = \omega^n$?

Remark. This question started off as a question on multilinear algebra because I thought that perhaps there was an algebraic point-wise condition on $\Omega$ that was non-trivial, but that is not the case: every element of $\Lambda^{2n}(\mathbb{R}^{2n+1})$ is the $n$-th power of an element of $\Lambda^{2}(\mathbb{R}^{2n+1})$.

Background. The odd dimensional manifolds in which I'm really interested are spherized tangent bundles (i.e. $STM := (TM \setminus 0)/\mathbb{R}^+$. The reason for this is that this question comes up in the study of inverse problems in the calculus of variations.

Answer by alvarezpaiva for New grand projects in contemporary math

2012-12-31T11:10:25Z

The simultaneous study of a space $X$ and its observables $F(X)$ (real, complex, or operator-valued functions on $X$) is an old topic, but with quantum groups and non-commutative geometry it has been the source of much modern mathematics. The introductory paper by Connes does a really nice job at explaining this.

In the paper Connes underlines the pioneering work of I.M. Gelfand in this area. However he misses one little thing. Gelfand's work on integral geometry was also motivated by this philosophy. The idea is to consider the incidence relation as a special type of multivalued map between the two spaces and to consider how functions, forms, densities, and other functional objects correspond under the map.

Answer by alvarezpaiva for Old books still used

2012-12-28T19:31:02Z

In metric geometry Busemann's "The Geometry of Geodesics" (1955) is still wonderful reading. This book is now published by Dover.

Extreme rays in the cone of (semi)metrics

2012-12-28T10:58:41Z

How many extreme rays are there on the polytopal cone formed by all semimetrics on a set with $n$ elements?

Some background. Given a set $X$ with $n$ elements, the set of all semimetrics $d:X \times X \rightarrow [0,\infty)$ can be seen as the cone of symmetric matrices $(d_{i,j})$ with zeroes on the diagonal and satifying the system of linear inequalities $d_{i,j} \geq 0$, $d_{i,j} + d_{j,k} - d_{i,k} \geq 0$. The polytopal cone defined by this finite system of inequalities can also be described (at least in principle) by exhibiting its extreme rays. Some extreme semimetrics are easy to describe. Here is a class of examples: if $Y$ is a subset of $X$, define the cut semimetric $d_Y$ by setting the distance between two points to be equal to $1$ if one of the points is in $Y$ and the other in its complement, otherwise the distance between the two points is zero. Another example, given by Avis, of an extreme metric is the length metric of the graph $K_{3,2}$ on the cone of semimetrics on a set with $5$ elements.

More elaborate question. It may be hard to know exactly how many extremal rays there are on the semimetric cone, but I'm interested in a good estimate (some asymptotic estimate would also be nice). I'm also interested in knowing about classes of examples of extremal semimetrics other than the cut semimetrics and the examples given by Avis in his 1980 paper On the extreme rays of the metric cone.

Answer by alvarezpaiva for Liouville's theorem with your bare hands

2012-12-20T20:44:24Z

I beg to differ with Andy and propose Nelson's proof as THE book proof that a bounded analytic function is constant. In fact Nelson's little gem is for harmonic functions, but the proof is incredibly beautiful and, of course, applies to analytic functions by considering their real and imaginary parts.

There are just two ingredients:

1. The mean value theorem: the value of a harmonic function in the plane (or n-space) at some point is the average of the function over any disc centered at that point. We can even take that as definition of a harmonic function if we have a family of "discs" and a measure with which to average.

2. A geometric property of metric discs on the plane: Given any two points $x$ and $y$, for sufficiently large radii $R$, the symmetric difference of the disc of radius $R$ centered at $x$ and the disc of the same radius centered at $y$ has negligible area compared with the area of the discs.

Now the proof goes as follows: take any two points $x$ and $y$ on the plane and choose discs of a very large radius $R$ centered at each of these points. If the harmonic function is bounded, its average over the two discs, that by (2) basically coincide, has to be almost the same. Let $R$ go to infinity and you're done.

Remark/Question Clearly the proof makes sense for a class of metric measure spaces. Has this sort of spaces (symmetric difference between large metric balls having small relative measure) been studied on its own? Normed spaces are in this category.

Comment by alvarezpaiva

2013-05-18T06:39:16Z

@Deane, well, for me that's the magic of the Pythagorean theorem: it gives you a way to compute the distance between two points $x$ and $z$ if you know their distances to a third point $y$ at which the geodesics $xy$ and $yz$ are perpendicular (something you can see "at" the point $y$). It's not even clear there should exist a geodesic metric space where one can do this ! Oh, and it was also a bit of a joke ...

Comment by alvarezpaiva

2013-05-18T06:32:57Z

The recent book "Nonlinear Perron-Frobenius Theory" by B. Lemmens and R. Nussbaum covers a great many aspects of Hilbert geometries (and does so beautifully, I may add). That would be a good place to look for this if you haven't done so already.

Comment by alvarezpaiva

2013-05-17T14:29:13Z

Here is a way to see that the theorem is not trivial ;-) : assume you have a complete Riemannian manifold such that any two distinct points are joined by a unique geodesic. Assume that any triple of distict points $x$, $y$ and $z$ such that the geodesics $xy$ and $yz$ form a right angle at $y$ satisfy $d(x,y)^2 + d(y,z)^2 = d(x,z)^2$. Prove the manifold is flat.

Comment by alvarezpaiva

2013-05-17T08:12:23Z

Yeah, that too! :-) Although I think that if this is coming from "quantum detection theory" he could do worse than to look at more general partial isometries as well.

Comment by alvarezpaiva

2013-05-17T08:01:16Z

I don't think it matters that $m > n$. The kernel of the maps has dimension zero and the map is an isometry on the space perpendicular to the kernel. From another viewpoint: if $A$ is your map $A = AA^*A$, which characterizes partial isometries.

Comment by alvarezpaiva

2013-05-17T07:37:30Z

Oh, "unexpectedly" because of the title.

Comment by alvarezpaiva

2013-05-17T07:36:32Z

There is a very nice book where this is (somewhat unexpectedly) REALLY nicely done: Schrödinger Operators: With Applications to Quantum Mechanics and Global Geometry by Barry Simon et al. As a bonus, they have a good exposition of the relation between Hodge theory and Morse theory that was first uncovered by Witten.

Comment by alvarezpaiva

2013-05-17T07:17:09Z

@Yoav: Thanks for your input. I don't see why the min should be achieved either and I'm starting to suspect it is not. What puzzles me is that no one seems to mention this.

Comment by alvarezpaiva

2013-05-15T22:06:34Z

Do you get the equality case this way?

Comment by alvarezpaiva

2013-05-14T12:24:48Z

@Misha: Actually, I think you may be right and that the trick is to use the Loewner-John ellipsoid (ellipse in this case), but one needs to use the sharp bound for quotient between the area of the ellipse and the area of the body (instead of using John's theorem).

Comment by alvarezpaiva

2013-05-13T20:15:35Z

@Misha: this will give a similar bound, but I don't think it will be sharp. Behrend's bound is sharp with equality if and only if $F$ is a triangle. But of course one should check ...

Comment by alvarezpaiva

2013-05-13T19:24:50Z

Funny, I'm also looking at this reference and I can't even find the statement of this result. It is supposed to be in pages 745 and 746.

Comment by alvarezpaiva

2013-05-06T19:20:27Z

I gather you're mostly interested in the Nikodym distance of $K$ to a ball. Do you expect the Nikodym distance between $K$ and the ball to be at most $c_n \epsilon$, where $c_n$ depends only on the dimension of the space?

Comment by alvarezpaiva

2013-05-06T18:14:57Z

I think one just needs to add that every rotation is a composition of reflections and so the fact that the body $K$ or its gauge function is almost reflection invariant implies that is is almost invariant under rotations.

Comment by alvarezpaiva

2013-05-05T18:44:29Z

Nice reference indeed! Thanks.