Totally geodesic submanifold of a hyperbolic 3-manifold If $M$ is a convex-cocompact hyperbolic 3-manifold, and $S$ is a closed surface with genus $\geq$ 2. Suppose $f:S\to M$ is a minimal immersion, and $f(S)$ is negatively curved. I know that all the closed geodesics in $f(S)$ are closed geodesics in $M$.  Can I conclude that $f(S)$ is totally geodesic in $M$?
 A: Reading the first paragraph of the introduction of the following paper:
http://homeweb.unifr.ch/parlierh/pub/BuserParlierOsaka.pdf
it seems to me that the set of unit tangent vectors $v_p$ to $S$ such that the $\gamma(t):=exp(tv_p)$ is a closed geodesic is dense. If I am not misunderstanding such a density then $\alpha(v_p,v_p) = 0$ for a dense set of unit vectors, where $\alpha$ is the second fundamental form of $f(S)$. Thus, $f(S)$ is totally geodesic in $M$ since $\alpha(v_p,v_p) \equiv 0 $ for all $v_p$ due to the density.
A: First, every immersed minimal surface has negative Gaussian curvature, which follows from the Gauss equation, which implies its Gauss curvature is at most $-1.$  Second, geodesics in $\Sigma$ (with the induced metric) are very unlikely to be geodesics in $M,$ I don't know an example where this happens with $\Sigma$ not totally geodesic, though I don't know a reason why it can't happen. 
Regarding your question, every convex, co-compact hyperbolic 3-manifold which contains a closed surface subgroup (this may be a tiny restriction, I'm not an expert) contains an immersed minimal surface, which follows from results of Hass.  So, the answer to your question is no, unless you get lucky and your minimal surface is totally geodesic.  As an example, ask the question for a quasi-Fuchsian hyperbolic 3-manifold.  Then, an immersed minimal surface is totally geodesic if and only if the manifold is Fuchsian.
