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

• It is true that compact Riemannian manifolds which are not simply connected have a closed geodesic. I think (however, I am not entirely sure) that complete Riemannian manifolds which are not simply connected also have a closed geodesic. Therefore, if $M^k$ was not simply connected, then it would have a closed geodesic. Being a totally geodesic submanifold of $H^n$, this closed geodesic would also be a geodesic of $H^n$, a contradiction. I would be glad if someone confirmed (or refuted) what I said. – Aloizio Macedo Nov 2 '16 at 0:15
• nice result @Aloizio, for sure. I am expecting that someone provides the confirmation for your claim. This would solve the problem. Indeed, I did not know this fact from manifolds, could please tell me a possible reference? – L.F. Cavenaghi Nov 2 '16 at 0:18
• I don't remember if doCarmo has this fact. Klingenberg (Lectures on Closed Geodesics) has, however his proof is very involved in the theory of the free loop space of a compact manifold. Nevertheless, he gives four references of different proofs, which proceed through more elementary means. – Aloizio Macedo Nov 2 '16 at 0:30

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.

• Could you remember me why the Fiber fas more than two points If the base is not simply connected? – L.F. Cavenaghi Nov 3 '16 at 13:25
• @frusciante14 More than one, it can have two (for example, $S^2 \to \mathbb{R}P^2$). To see this, you can use the fact that the number of sheets of a universal cover is equal to the cardinality of the fundamental group of the base space (more generally, the number of sheets is equal to the index of the image of the $\pi_1$ of the cover on the $\pi_1$ of the base space by the induced map $p_{\#}$ (induced by the projection $p$) on the fundamental group). See Bredon (section "The action of $\pi_1$ on the Fiber" for a reference). – Aloizio Macedo Nov 3 '16 at 14:17
• Yes! Sure! I totallt forgot this. Thank you! – L.F. Cavenaghi Nov 3 '16 at 14:32
• One of the standard examples of complete connected but not simply connected hyperbolic manifolds without nonconstant closed geodesics is the quotient of the hyperbolic plane (upper half plane) by the group of integer horizontal translations. – Moishe Kohan Nov 3 '16 at 18:12