In my differential geometry course we had the following

Theorem (Cartan-Hadamard): Let $M$ be a connected, simply connected, complete Riemannian manifold. Then the following are equivalent:

  1. $M$ has nonpositive curvature
  2. $|d\exp_p(v)\hat v| \ge |\hat v|$ for all $p \in M$, $v, \hat v \in T_pM$
  3. $d(\exp_p(v), \exp_p(w)) \ge |v - w|$ for all $p \in M$, $v, w \in T_pM$

In addition $\exp_p$ is a diffeomorphism if either of these statements holds.

Now, I understand the proof that was given, but I do not see where we have to use simple connectivity of $M$ (in the proof given by my professor the Cartan-Ambrose-Hicks Theorem was used). What's wrong with the following:

$1. \Leftrightarrow 2.$ didn't make use of simple connectivity. (at least I don't see where)

But now I think 2., completeness and connectedness imply that $\exp_p$ is a diffeo, and in particular $M$ is simply connected.

_ 2. implies 3.: Let $v, w \in T_pM$ be given, let $\gamma(t) = \exp_p(v(t))$ be a lenght minimizing geodesic connecting $\exp_p(v)$ and $\exp_p(w)$. Then

$ \begin{eqnarray} d(\exp_p(v), \exp_p(w)) &=& L(\gamma) \\ &=& \int_0^1 |\dot \gamma(t)| \, dt \\ &\ge& \int_0^1 |\dot v(t)| \, dt \\ &\ge& \left| \int_0^1 \dot v(t) \, dt \right| \\ &=& |v-w| \end{eqnarray} $

Therefore $\exp_p$ is injective. Completeness and connectedness imply that $\exp_p$ is surjective (Hopf-Rinow). 2 implies that $\exp_p$ is a local diffeo.

So in conclusion $\exp_p$ is seen to be a bijective local diffeo, hence a diffeo.

My question:

  • Is there anything wrong with this argument?
  • If yes: Could you give an example of a complete, connected manifold with nonpositive curvature, which is not simply connected?

Thanks a lot!


  • 2
    $\begingroup$ Dear S.L., a flat torus is a simple example of a complete, connected manifold with nonpositive (actually, zero) curvature that is not simply connected. (Any Riemannian manifold locally isometric to flat euclidean space has zero curvature.) $\endgroup$ Commented Jan 23, 2011 at 13:22
  • $\begingroup$ Ah, of course... That's a great counterexample, thanks. I'll have to go over the details again, I think. $\endgroup$
    – Sam
    Commented Jan 23, 2011 at 13:38
  • 1
    $\begingroup$ In your proof of 2 implies 3, to get that $\gamma(t) = \exp_p(v(t))$ requires simple connectivity. Let $V$ be the (geodesic) curve joining $p$ to $\exp_p(v)$, and $W$ similarly. Then to get that expression to hold you implicitly used that $V+\gamma$ and $W$ are homotopic. $\endgroup$ Commented Jan 23, 2011 at 13:59
  • $\begingroup$ Yeah, I've realized that one might have $\exp_p(v) = \exp_p(2v)$ (as is the case for a flat torus). Then of course one would not get the inequality in 3. I think this is the issue you are referring to, Willie. I guess in this case $\exp_p$ is a covering map, but only simple connectivity implies that it is actually a diffeo. $\endgroup$
    – Sam
    Commented Jan 23, 2011 at 14:08

1 Answer 1


Regarding your comment above: yes, if $M$ is complete and connected with non-postiive curvature, then the universal cover $\widetilde{M}$ of $M$ inherits a complete connected non-positive curvature metric (just pull-back the Riemannian metric from $M$), and so satisfies the conditions (and so also the conclusions!) of the Cartan--Hadamard theorem. One then finds that $\widetilde{M}$ is homeomorphic to the tangent space of any point $\widetilde{m}$ of $\widetilde{M}$ (via $\exp_{\widetilde{m}}$. If $m$ is the image of $\widetilde{m}$ in $M$, then the tangent space to $\widetilde{m}$ is naturally identified with the tangent space to $m$, and the exponential map $\exp_m$ is naturally identified with the composite of $\exp_{\widetilde{m}}$ and the projection $\widetilde{M} \to M$.

Thus indeed, one finds that $\exp_m$ is a covering map.

This circle of ideas is frequently applied, e.g. in the context of hyperbolic manifolds. If $M$ is a compact connected hyperbolic manifold, then one finds by Cartan--Hadamard that $\widetilde{M}$ is isometric to hyperbolic space of the appropriate dimension, and so $M$ is isometric to a quotient of hyperbolic space by a discrete cocompact group of isometries. (This is why hyperbolic manifold theory interacts so tightly with certain parts of group theory.)

Also, one concludes that if $M$ is compact and connected (and positive dimensional --- i.e. not a single point!) with non-positive curvature then $\pi_1(M)$ is infinite (because, writing $M$ as a quotient of $T_m M$ via $\exp_m$, we see that to get something compact, we have to quotient out by an infinite group of diffeos).

  • $\begingroup$ Thank you, Matt, for this answer. I always enjoy when people digress a bit on related topics and applications. I think that's great, so please keep it up! :-) $\endgroup$
    – Sam
    Commented Jan 23, 2011 at 16:39

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