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 $M$?

• You need more hypotheses likely, since one has immersed minimal planes which have no closed geodesics, so satisfy the hypothesis vacuously. Similar for immersed minimal annuli which contain a single closed (primitive) geodesic. – Ian Agol Jun 1 '15 at 3:31
• Thanks. Let say $\Sigma=f(S)$ is a minimal immersed image of a closed genus $\geq$ 2 surface $S$, and $f$ is $\pi_1$ injective. Does this help? – Nyima Kao Jun 1 '15 at 3:42
• Ok, that's a reasonable assumption, although then $\pi_1$-injectivity is redundant, since this follows from the assumption that closed (immersed) geodesic curves on the surface are geodesics in the manifold. – Ian Agol Jun 1 '15 at 5:59

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.
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.