21

There's a lot of work about the existence, number and other properties of closed geodesics on a Riemannian manifold (belonging to some specific class of manifolds).

In the case of geodesics representing some non trivial homotopy class of closed curves on a (not simply connected) manifold, I understand the importance/usefullness of knowing that such a homotopy class (element of the fundamental group) can be represented by a geodesic. But what about simply connected manifolds? In dimension 2 the simply connected surfaces are $S^2, \mathbb{R}^2$ and $\mathbb{H}^2$; according to Lusternik-Fet, $S^2$, being compact, admits non trivial closed geodesics (whereas the other two do not), but they are all homotopically trivial; I don't get what these closed geodesic tell about the topology of $S^2$; my feeling is that maybe the whole situation is quite trivial in dimension 2, but I'm not much familiar with manifolds of dimension 3 or higher (spheres, euclidean and projective spaces apart)...so any interesting example is welcome!

I know that existence results are always of fundamental importance on their own, but here I can't figure out which are the implications of the existence of closed geodesic (especially in the simply connected case, as pointed out above).

To sum up, I'm interested in the following questions:

1) what if a manifold has a closed geodesic?

2) what if a manifold has no closed geodesic? (can I say that then the manifold is simply connected?...the punctured euclidean plane seems a counterexample...what if I restrict my question to complete manifolds?)

3) what if a manifold has more than one closed geodesic? or even infinite? (here I mean "distinct" geodesics, in some sense)

4) what if every geodesic is closed?

I would appreciate explicit examples as well as theorems or references on the subject.

Henry
  • 169,616
Lor
  • 5,738

5 Answers5

14

I think closed geodesics are interesting because they place powerful constraints on the geometry and topology of a Riemannian manifold.

For instance, it is possible to place bounds on the volume and diameter of a manifold if you know the lengths of the longest or shortest closed geodesics. In fact there is sort of an entire subfield of Riemannian geometry called "systolic geometry" based on this principle; you might check out the wikipedia page for further discussion.

Moreover, there is a very nice theorem of Cartan which asserts that every free homotopy class of loops in a compact Riemannian manifold has a geodesic. This result answers your second question in the compact case, and there are counterexamples among complete manifolds (for instance, take an infinite cylinder whose radius monotonically decreases). It's useful because it can be used to relate topology and geometry; for instance this is the key ingredient in the proof of a theorem of Preissman which asserts that every nontrivial abelian subgroup of the fundamental group of a compact Riemannian manifold with negative curvature is infinite cyclic. This helps rule out metrics of negative curvature on a lot of manifolds.

M Turgeon
  • 10,785
Paul Siegel
  • 9,467
  • Thanks for mentioning systolic geometry: it's a nice interplay between metric and topological properties. But let me focus on the rest of your answer, as I'd like to understand what's going on here:
    1. Cartan's theorem implies that if I have no closed geodesic on a compact manifold, then it's simply connected. But Lusternik-Fet's theorem says I always have a closed geodesic on a compact manifold. So the point is: is that closed geodesic homotopically trivial? Am I right?

    2. Why doesn't your cylinder have closed geodesics? I can see lots of them (though not minimizing)..what am I missing?

    – Lor May 23 '12 at 14:05
  • In 2) I'm referring to the various circles at fixed radius. They are all freely homotopical and non homotopically trivial geodesics, arent't they? – Lor May 23 '12 at 14:06
  • 1
  • The sphere is a good example to have in mind: it is simply connected, but it has lots of closed geodesics. Note that you can introduce nullhomotopic closed geodesics to any manifold by attaching a spherical handle to it, so I guess I would interpret Cartan's theorem as saying that if a manifold doesn't have very many closed geodesics then its fundamental group is small.
  • – Paul Siegel May 24 '12 at 02:30
  • 1
  • Consider the intersection of the standard unit sphere in $\mathbb{R}$ with the plane $x = 1/2$. The intersection is a circle and it might feel like it's a geodesic, but it's not: given two nearby points on the circle it's a little bit cheaper to move toward the point $(1,0,0)$ than it is to move along the circle. The same is true of your circles on the sloped cylinder.
  • – Paul Siegel May 24 '12 at 02:36
  • In the case of the cylinder, do you mean something like this? http://postimage.org/image/ucw687gzf/ What does the geodesic do away from this? will the two branches go away for ever never meeting again? or will the geodesic selfintersect infinitely many times, but never close (like on an elliptic paraboloid)? On the other hand, if I let the radius increase (i.e. take a one-sheeted hyperboloid), the "obvious shortest geodesic" is indeed a shortest geodesi, right? Has this something to do with the sign of curvature? Thanks a lot! – Lor May 24 '12 at 13:16