5
$\begingroup$

Zoll metric on a Riemannian manifold is a metric for which all geodesics are closed and have the same period. For sure, a standart metric on the sphere $S^2$ has this property: all its geodesics are great circles of period $2 \pi$. Projective space $RP^2$ as a factor of $S^2$ provided with the canonical metric also has all geodesics closed and of the same lenght $\pi$. There was lots of work done (Tannery, Zoll, Funk, Guillemin and others) studying Zoll surfaces. For example, a theorem of Green shows that there are no nontrivial Zoll metrics on $RP^2$. On the contrary, there is an abundance of such metrics on the sphere $S^2$, even without nontrivial isometries.

My question is why Zoll metrics exist only on the sphere and its factor $RP^2$? Of course, here I restrict myself to the $2$-dimensional case. The evidence that is true is mentioned in the book A.Besse "Manifolds all of whose geodesics are closed". The style of the book is very formal and the statement is proven in such a generality that it's impossible to understand. There should be some easy topological argument but I do not find it.

$\endgroup$
5
  • $\begingroup$ Just to clarify, do you have any evidence that it is actually true that Zoll metrics only exist on the sphere and the projective plane? $\endgroup$
    – Rasmus
    Oct 25 '13 at 7:37
  • $\begingroup$ Rasmus, I updated a post after your comment. Yes, of course, I want to prove a fact which really holds. It's mentioned in lots of places in the litterature, in particular, in the book that I mentioned above which is considered a main book on this topic. $\endgroup$
    – Olga
    Oct 25 '13 at 7:46
  • $\begingroup$ I see. It seems a bit strange that a book would make such a claim without a proof or a reference. $\endgroup$
    – Rasmus
    Oct 25 '13 at 7:52
  • $\begingroup$ Just a question: is it true that Zoll manifold has to be a symmetric space? $\endgroup$
    – user99914
    Oct 25 '13 at 7:57
  • $\begingroup$ @John: certainly not. There are many strange looking Zoll metrics. Are you asking whether a Zoll manifold must admit a metric making it a symmetric space? $\endgroup$ Dec 17 '13 at 11:17
6
$\begingroup$

I got the answer. One can prove that the fundamental group of Zoll surface has to be (if nontrivial) cyclic and, consequently, $\mathbb Z_2$. It follows from the fact that all geodesics are homotopic.

The proof is the following:

  1. In each class of homotopic equivalence in the fundamental group of the Zoll surface there is a geodesic (this is just the consequence of the fact that we can minimize the length in each homotopy class).

  2. Two geodesics on the Zoll surface are homotopically equivalent. Indeed, just take some point in the unitary tangent fiber corresponding to the first geodesic (a point lying on the geodesic and a tangent vector) and the analogous pair point-vector for the second one. Any path in the $T_1 S^2$ will give a homotopy.

  3. So, from $1.$ and $2.$ we have that $\pi_1$ of a Zoll surface is cyclic, let $a$ be its generator (if it is not trivial). Then, $a$ is homotopically equivalent to itself in another direction $a^{-1}$. So, $\pi_1$ is either trivial or $\mathbb{Z}_2$. So the surface is either $S^2$ or $\mathbb RP^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.