Manifold with all geodesics of Morse index zero but no negatively curved metric? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T22:37:13Z http://mathoverflow.net/feeds/question/70635 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70635/manifold-with-all-geodesics-of-morse-index-zero-but-no-negatively-curved-metric Manifold with all geodesics of Morse index zero but no negatively curved metric? Jonny Evans 2011-07-18T15:50:44Z 2012-01-29T21:07:46Z <p>A closed oriented Riemannian manifold with negative sectional curvatures has the property that all its geodesics have Morse index zero.</p> <p>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?</p> <p>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.</p> http://mathoverflow.net/questions/70635/manifold-with-all-geodesics-of-morse-index-zero-but-no-negatively-curved-metric/70641#70641 Answer by Rbega for Manifold with all geodesics of Morse index zero but no negatively curved metric? Rbega 2011-07-18T16:32:43Z 2011-07-18T16:48:43Z <p>(This should be a comment)</p> <p>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. </p> <p><strong>[Added after a little thought]</strong></p> <p>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 $k>1$.</p> <p>I'm not sure if that is enough to ensure the existence of a non-positively curved metric on $M$ but is certainly suggestive...</p> http://mathoverflow.net/questions/70635/manifold-with-all-geodesics-of-morse-index-zero-but-no-negatively-curved-metric/86975#86975 Answer by Vitali Kapovitch for Manifold with all geodesics of Morse index zero but no negatively curved metric? Vitali Kapovitch 2012-01-29T19:28:35Z 2012-01-29T21:07:46Z <p>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 <strong>non-positive</strong> (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.</p> <p>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 <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=MR&amp;s1=0847526" rel="nofollow">result of Croke and Shroeder</a> 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.</p> <p>This result implies for example that nonflat nilmanifolds can not 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.</p> <p>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 <em>semi-hyperbolicity</em> but even that is completely unclear. The natural <em>bicombing</em> 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).</p> <p>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.*</p>