11
$\begingroup$

Let $M$ be a complete Riemannian manifold, with the property that $\exp_p\colon T_pM \to M$ is a diffeomorphism for every $p \in M$.

Can we say something about it's curvature? Is it true that its sectional curvature must be everywhere non-positive (or at least at some point)?

(If not, can we say something about the Ricci or scalar curvature?)


Note that it's clearly not true if we only assume $\exp$ is a diffeomorphism at a single point.

$\endgroup$
11
$\begingroup$

For complete Riemannian manifolds the exponential map is a covering map at all points if and only if the manifold has no conjugate points. In particular, the exponential map is a diffeomorphism at all points if and only if the manifold is simply-connected and the metric has no conjugate points.

Manifolds without conjugate points have been studied by many authors and they need not have everywhere nonpositive sectional curvature.

On the other hand it is still an open problem (as far as I know) whether every closed manifold without conjugate points admits a metric of nonpositive sectional curvature.

For one result relevant to your question is that if a complete Riemannian manifol has no conjugate points, then either it is flat (that is has zero sectional curvature everywhere) or it has negative Ricci curvature everywhere; see Geodesics without Conjugate Points and Curvatures at Infinity by SÉRGIO MENDONÇA and DETANG ZHOU. They also mention some related result of Leon Green involving scalar curvature.

EDIT: I misread the curvature condition in the paper linked above. Thanks to Sergei for correcting me!

$\endgroup$
  • 2
    $\begingroup$ Igor, a manifold without conjugate points can have some amount of positive Ricci curvature. There are 2-dimensional examples. The paper you cite claims that it should have negative Ricci curvature somewhere in a certain set. $\endgroup$ – Sergei Ivanov Jan 12 '16 at 0:21
3
$\begingroup$

Assuming that there are no conjugate points, one can conclude that the curvature is zero in the following special case. This is the case of a complete surface of nonnegative curvature. Namely, if the curvature is positive at a point $p$, a Jacobi field whose derivative vanishes at $p$ with have a convex graph (by the curvature assumption) and therefore will be forced to vanish eventually (just by the intermediate value theorem). Doing this in both directions along a geodesic will produce a pair of points which are conjugate to each other.

I am not sure if a similar argument will work in higher dimension. Perhaps Sergei can chip in?

Another relevant result is that of Burago and Ivanov from 1994, generalizing an older result of Hopf. Namely, a Riemannian metric on an $n$-torus without conjugate points must be flat.

$\endgroup$

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.