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:
- $M$ has nonpositive curvature
- $|d\exp_p(v)\hat v| \ge |\hat v|$ for all $p \in M$, $v, \hat v \in T_pM$
- $d(\exp_p(v), \exp_p(w)) \ge |v - w|$ 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!
S.L.