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Are there simple conditions on a Riemannian metric on the two-sphere that imply that a geodesic loop of minimal length is actually a periodic geodesic?

For example, is this true for small perturbations of the round metric, for metrics with suitably pinched curvature, ... ?

Addendum (16/09/2013). When I asked this question I thought it might have an easy or well-known answer, but apparently that is not so. It may be helpful then if I provide some context. It will also be helpful for me since I'm now writing these things up in a paper with Balacheff.

An open question in Riemannian geometry that I learned from Florent Balacheff asks whether every Riemannian metric on the two-sphere that is sufficiently close to the round metric and whose area is $4\pi$ carries a closed geodesic of length at most $2 \pi$. What really caught my imagination is that this conjectured inequality is really local: there are metrics far from the round metric for which it is not true. However, it was a bit disappointing that the round metric or Zoll metrics did not come out to be the extremal metrics for this problem and I started to look for a similar inequality where it would be at least reasonable to conjecture that the extremal metrics are Zoll. I found the following variation, which I call the "figure 8 conjecture":

Figure 8 conjecture. A Riemannian metric on the two-sphere with area equal to $4\pi$ carries a closed geodesic that is regular homotopic to a figure 8 and whose length is at most $4\pi$. Moreover, if equality is attained at a (smooth) metric, the metric is Zoll.

Note that this result would imply that a Riemannian metric on the two-sphere with area equal to $4\pi$ carries a geodesic loop whose length is at most $2\pi$. The OP inquires when this implies that there is a periodic geodesic of length at most $2\pi$.

The reason the figure 8 conjecture may just be true is that it follows from a natural version of the Viterbo conjecture:

Let $\alpha$ be a contact form on the three-dimensional sphere $S^3$. If ${\rm Ker} \alpha$ is the standard (tight) contact structure on $S^3$, then the volume of $(S^3,\alpha)$, defined by $$ {\rm vol}(S^3,\alpha) = \int_{S^3} \alpha \wedge d\alpha , $$ is no less than the square of the smallest period of a closed Reeb orbit (i.e., a periodic integral curve of the vector field defined by $\alpha(X) = 1$, $d\alpha(X,\cdot) = 0$).

Michael Hutchings also arrived at this conjecture from his work on contact homology (see his blog) and actually at some point I thought I had found a counterexample and he kindly re-set me on the right track. I believe the only partial answers that are known are as follows (and this is joint work with Florent Balacheff):

1. If among all the smooth tight contact forms on the three sphere there is one for which the ratio (volume/square of the smallest period of a periodic Reeb orbit) is minimal, then the conjecture is true. In other words, the existence of a smooth minimizer for this ratio settles the conjecture.

2. If $\alpha_t$ is a smooth constant-volume deformation of the standard contact form on $S^3$ that is not formally trivial, then for each sufficiently small $t$ different from zero $(S^3, \alpha_t)$ carries a periodic Reeb orbit with period less than $\pi$.

A defomation $\alpha_t$ is formally trivial if for every positive integer $k$ there exists an isotopy $\phi_t$ (depending on $k$) such that $\alpha_t$ and $\phi_t^* \alpha$ agree to order $k$ at $t = 0$.

3. Let $K \subset \mathbb{R}^4$ be a convex body and let us identify $\mathbb{R}^4$ with the space of quaternions. If $K$ is invariant under left multiplication by the quaternions $i$, $j$, and $k$, then the boundary of $K$ carries a closed Reeb orbits whose period is less than the square root of its (contact) volume.

This last result follows from Sergei Ivanov's Finsler generalization of Pu's isosystolic inequality (a truly neat result, which has many beautiful corollaries including Mahler's inequality for the volume product of symmetric convex bodies on the plane).

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Do you mean minimality given a fixed base-point, or over all possible base-points? –  Benoît Kloeckner Aug 22 '13 at 20:52
    
@Benoît: All possible base points. –  alvarezpaiva Aug 23 '13 at 10:14

1 Answer 1

I am not sure this could be an answer or not, but it is too long to be a comment...

It has been known that if a loop $L$ realizes the injectivity radius at some point $p$, then it must be smooth at the antipodal point $q$ of $p$, i.e., the nearest cut point of $p$. Otherwise we can deform the corner to shorten the loop (cf. Cheeger-Ebin's book.) The only possible non-smooth corner of $L$ is $p$ itself. However, if $p$ is a point where the injectivity radius achieves it minimum, that is $inj(p)=\min_{M} inj(x)$, then we can take $q$ as a base point and consider all loops passing through $q$. Among these loop, $L$ must be the shortest one (because it realizes the minimal injectivity radius.) Hence it must be smooth at the antipodal point of $q$, that is, $p$. So the conclusion is: the loop which realizes the minimum of injectivity radius must be smooth. (We don't need to deal with conjugate points because the minimal loop collapses to a segment there.)

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@Chi-Wei Chen : Thanks! Let me think about this for a while. –  alvarezpaiva Apr 14 at 18:11

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