Every closed and totally geodesic submanifold $M^k$ of $H^n$ is isometric to $H^k$, $k \le n.$ Let $H^n$ be the hyperbolic space. Then the claim is: Every closed and totally geodesic submanifold $M^k$ of $H^n$ is isometric to $H^k$, $k \le n.$
This problem comes from do Carmo's book of Riemannian geometry. It supposed to be easy under the assumption that $M^k$ is simply connected. The result would come from the combination of Hadamard theorem, implying that the exponential map is a diffeo and from the fact that since $M^k$ is closed, it is also complete (since $H^n$ is) and once $M^k$ is totally geodesic, its curvature is $-1$.
We know yet that the universal covering of $M^k$ is isometric to $H^k$ and we can also speculate that $M^k$ cannot be bounded, since if it was, this would imply compactness. So if we stated that $M^k$ is simply connected it would be a contradiction.
How to solve it?
Thanks a lot
 A: I said in the comments that I suspected that every complete manifold which is not simply connected has a closed geodesic. While I don't know if that is true (actually, I think that is open, as is implied by this article that I found after I searched for references), we don't need the statement in this full generality. It suffices to get a geodesic which self-intersects, and we can get this easily by doing the following: (*)
It suffices to get a geodesic which self-intersects, and we can get this easily by doing the following:
Suppose your complete manifold $M$ is not simply connected. Go to the universal cover $\widetilde{M}$. Since $M$ is not simply connected, $\widetilde{M}$ has more than one point in a fiber. Take two such (different) points and join by a geodesic (which you can do since $\widetilde{M}$ is also complete). Project this geodesic. We get a geodesic which self-intersects.
Note that the above argument does not provide a closed geodesic, since a closed geodesic must be a $C^{\infty}$ map on the circle (equivalently, it must meet its intersection with the same derivative), while the geodesic constructed above may meet itself transversally.
However, it suffices, since this geodesic would be a geodesic on $H^n$ since $M$ is totally geodesic, and no geodesic on $H^n$ self-intersects.
(*) It is NOT true that every complete Riemannian manifold which is not simply connected has a closed geodesic. See here, more specifically section 2. I'll leave my comments on the question in order to preserve coherence of the conversation.
