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It is known that the cohomology ring of a Zoll manifold---a riemannian manifold all of whose geodesics are periodic with the same minimal period---must be the same as the cohomology ring of a compact rank one symmetric space (see Besse's book Manifolds all of whose geodesics are closed for references).

Is there a simple and elementary proof of the following much weaker property?

The first Betti number of a Zoll manifold is equal to zero.

Addendum. The comment by Thomas Richard got me thinking and here is something that should lead to a proof that the fundamental group of a Zoll manifold is either trivial or isomorphic to $\mathbb{Z}_2$:

Any two prime closed geodesics in a Zoll manifold are homotopic. Indeed, if $v_x$ is a unit vector tangent to a geodesic $\gamma$ and $w_y$ is a unit vectors tangent to a geodesic $\sigma$, then a continuous path on the unit tangent bundle joining these two unit vectors, taken as the initial conditions of prime closed geodesics, will define a homotopy between $\gamma$ and $\sigma$.

Note that there is at least one closed geodesic representing each non-trivial homotopy class of loops, but the geodesic doesn't have to be prime. Still ...

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  • $\begingroup$ I read the proof in Besse's book just today, and it wasn't very difficult to get the cohomology groups, if you are familiar with Gysin sequences (or look them up in McCleary's book). The ring structure is hard, and Besse just skips that. $\endgroup$
    – Ben McKay
    Commented Jul 25, 2013 at 21:11
  • $\begingroup$ Could one go further and show it is simply connected ? $\endgroup$
    – Thomas Richard
    Commented Jul 26, 2013 at 6:12
  • $\begingroup$ @ Ben: I had looked at Besse's book as well, but I really mean something elementary that maybe uses the full strength of the hypothesis (for example, the existence of a manifold of geodesics). Note that the Bott-Samelson theorem (or Theorem 7.37 in Besse) uses something much, much weaker. $\endgroup$
    – alvarezpaiva
    Commented Jul 26, 2013 at 7:03
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    $\begingroup$ @Thomas: It does not have to be simply-connected: the canonical metric on the projective plane is Zoll. However, the fundamental group has to be finite (Theorem 7.37 in Besse). $\endgroup$
    – alvarezpaiva
    Commented Jul 26, 2013 at 7:05
  • $\begingroup$ @Thomas: Now that I think of it, it would seem that the fundamental group of a Zoll manifold should not have more than two elements (i.e. that it has to be trivial or isomorphic to $\mathbb{Z}_2$). It's hard to reconcile two non-homotopic periodic geodesics with the existence of a manifold of geodesics. Maybe this is the idea of the elementary proof (?) $\endgroup$
    – alvarezpaiva
    Commented Jul 26, 2013 at 7:15

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As asked by @alvarezpaiva, I repost my remark as an answer.

After his addendum and answer showing that the fundamental group (if non-trivial) has one generator, given by a prime closed geodesic, you just have to observe that this geodesic is homotopic to itself with reversed orientation.

Hence the fundamental group has order at most $2$.

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  • $\begingroup$ Dear BS, I'm writing a paper in which I would like to use this result (that the fundamental group of a Zoll manifold has order at most two) and present the proof saying that it was worked out in a MO exchange with ... Well, I can't say "with BS" or an "with anonymous user of MO" although it would be quite funny. Please drop me a line. $\endgroup$
    – alvarezpaiva
    Commented Sep 10, 2013 at 15:03
  • $\begingroup$ A little comment: with this argument you actually prove that homology is Z/2Z or 0 and not the fundamental group.This gives the same result though. $\endgroup$
    – Olga
    Commented Jun 22, 2015 at 10:23
  • $\begingroup$ Sorry for the sloppiness. In fact I had in mind the $SC_l$ condition : all geodesics are closed, have the same minimal period $l$ and are simple. But if you mean condition $C_l$ (not necessarily simple) I think I have an argument. For $p\in M$ fixed, the minimum length in a non trivial class in $\pi_1(M,p)$ gives a geodedic loop $c$ with $c(0)=c(1)=p$ but a priori $c'(0)\neq c'(1)$. Then the length of $c$ is $l'<l$, but if one takes the minimum in the free homotopy class (conjugacy class in $\pi_1(M,p)$), it can only be less than $l'$. But it is $l$ by hypothesis, so $l'=l$ and $c'(0)=c'(1)$. $\endgroup$
    – BS.
    Commented Jun 22, 2015 at 20:36
  • $\begingroup$ I have some doubt about this argument, though, because Besse says (p 199) "there is no example of a $C$-manifold that is not an $SC$-manifold", and it would show quickly that they are all $SC_l$ indeed. Maybe they missed it ? Do you see a flaw ? Anyway, it would prove (with the above qualifications) that there can be at most one non-trivial element in $\pi_1(M,p)$ (if $n=\dim M>1$, of course), because $S^{n-1}$ is connected : all geodesics (they are closed) issued from $p$ are homotopic by rotating the initial velocity. BTW, I think we know each other ;-). $\endgroup$
    – BS.
    Commented Jun 22, 2015 at 21:08
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    $\begingroup$ I think I see why this doesn't prove $C\Rightarrow SC$ : it simply says nothing about simply connected manifolds. $\endgroup$
    – BS.
    Commented Jun 23, 2015 at 15:13
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I think I have it: The fundamental group of a Zoll manifold is a finite cyclic group.

Proof. In a compact riemannian manifold every non-trivial element of the fundamental group is represented by a closed geodesic. As was remarked in the addendum to the OP, all prime closed geodesic in a Zoll manifold are homotopic and, since every closed geodesic is an iterate of a prime geodesic, this implies that the fundamental group---if non-trivial---has just one generator.

In order to show that it is a finite group, consider the universal cover of the Zoll manifold, which is itself a Zoll manifold and hence compact. OK this part is still a bit fuzzy in my mind ... Is there a quick argument for proving that the universal cover of a Zoll manifold is compact? This was probably the crux of the matter from the beginning anyway.

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    $\begingroup$ I think you can use that a prime closed geodesic is homotopic to itself with reversed orientation, to conclude directly that the group has order at most $2$. $\endgroup$
    – BS.
    Commented Jul 27, 2013 at 13:40
  • $\begingroup$ @BS: that simple !! Wonderful! Could you please write the remark as an answer ? Just to follow MO protocol and to let people know the problem was solved. $\endgroup$
    – alvarezpaiva
    Commented Jul 27, 2013 at 15:26

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