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A closed oriented Riemannian manifold with negative sectional curvatures has the property that all its geodesics have Morse index zero.

Is there a known counterexample to the "converse": if (M,g) is a closed oriented Riemannian manifold (Edit: assumed to be nondegenerate) all of whose geodesics have Morse index zero then M admits a (possibly different) metric g' with negative sectional curvatures?

Edit: Motivation for asking this (admittedly naive) question is that Viterbo/Eliashberg have proved that a manifold with a negatively curved metric cannot be embedded as a Lagrangian submanifold of a uniruled symplectic manifold. Actually their proof only seems to use the existence of a nondegenerate metric all of whose geodesics have Morse index zero. I wondered if that was known to be strictly weaker.

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2 Answers

As mentioned by Rbega the question should be amended to ask whether it's true that a closed manifold $M$ without conjugate points admits a metric of non-positive (rather than negative) curvature (otherwise a torus is an obvious counterexample). In that form this is a well-known open problem. The exponential map at any point is a universal covering of $M$ and the geodesics in $\tilde M$ are unique. This does show that $M$ is aspherical but that is a long way from admitting a metric of nonpositive curvature.

There are some partial results suggesting that fundamental groups of manifolds without conjugate points share some properties of fundamental groups of nonpositively curved manifolds. In particular, there is a result of Croke and Shroeder that if the metric is analytic then any abelian subgroup of $\pi_1(M)$ is embedded quasi-isometrically. By the following observation of Bruce Kleiner the analyticity condition can be removed: Croke and Schroeder show that even without assuming analyticity for any $\gamma\in\pi_1(M)$ its minimal displacement $d_\gamma$ satisfies $d_{\gamma^n}=nd_\gamma$ for any $n\ge1$. This then implies that $d_\gamma=\lim_{n\to\infty} d(\gamma^nx,x)/n$ for any $x\in\tilde M$. This in turn implies that the restriction of $d$ to an abelian subgroup $H \simeq \mathbb Z^n$ extends to a norm on $\mathbb R^n$. This implies that $H$ is quasi-isometrically embedded.

This result implies for example that nonflat nilmanifolds cannot admit metrics without conjugate points and more generally that every solvable subgroup of the fundamental group of a manifold without conjugate points is virtually abelian.

But it's unlikely that any such manifold admits a metric of non-positive curvature. It is more probable that its fundamental group must satisfy some weaker condition such as semi-hyperbolicity but even that is completely unclear. The natural bicombing on $\tilde M$ given by geodesics need not satisfy the fellow traveler property (at least there is no clear reason where it should come from).

So it might be worth trying to look for counterexamples and the first place I would look is among groups that are semi-hyperbolic but not $CAT(0)$. Specifically, any $CAT(0)$ group has the property that centralizers of non-torsion elements virtually split. This need not hold in a semi-hyperbolic group with the simplest example given by any nontrivial circle bundle over closed surfaces of genus $>1$. To be even more specific one can take the unit tangent bundle $T^1(S_g)$ to a hyperbolic surface. Note however that it's known that a closed homogenous manifold without conjugate points is flat so if there is a metric without conjugate points on $T^1(S_g)$ it can not be homogeneous. Edit: Actually, this last remark is irrelevant as $T^1(S_g)$ can not admit any homogeneous metrics at all.

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Your last comment implies that the answer is true in dimension 3 by geometrization and results of Leeb characterizing non-positively curved metrics. So the smallest example would have to be in dimension 4. –  Ian Agol Jan 29 '12 at 20:06
@Agol sorry, I don't follow. How does this imply that a unit tangent bundle to a higher genus surface can not admit a metric without conjugate points? I don't think that follows from any known results. –  Vitali Kapovitch Jan 29 '12 at 20:14
Oh, I misread what you wrote. So I guess the only open 3-dimensional case is for $\tilde{SL}_2R$ metrics? –  Ian Agol Jan 29 '12 at 20:34
I'm not sure - I think there are manifolds consisting of several geometric pieces that can not be ruled out. Specifically, I believe there are other examples of 3-manifolds with semi-hyperbolic groups like some graph manifolds which are known not to admit non-positively curved metrics but can not be ruled out from admitting metrics without conjugate points. –  Vitali Kapovitch Jan 29 '12 at 21:01
Sorry, I forgot what Leeb's results say. There's many graph-manifolds which do not admit non-positively curved metrics. So I suppose these are candidates for metrics which do not admit conjugate points. –  Ian Agol Jan 29 '12 at 22:12
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(This should be a comment)

What about the flat torus $\mathbb{S}^1\times \mathbb{S}^1$? I think you need to amend the question to ask for non-positive sectional curvature.

[Added after a little thought]

I should add that by infinite dimensional morse theory (for the energy functional on the loop space of $M$-which satisfies the Palais-Smale condition) you should (in principal) be able to conclude that each component of the loop space is contractible. In other words the homotopy groups vanish for $k>1$ that is, $\pi_k(M)=0$ for all $k>1$.

I'm not sure if that is enough to ensure the existence of a non-positively curved metric on $M$ but is certainly suggestive...

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Good point. What I tacitly had in mind was that g was nondegenerate (=> all geodesics isolated). I'll edit the question. –  Jonny Evans Jul 19 '11 at 6:44
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