6
$\begingroup$

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 $M$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Ian Agol Jun 1 '15 at 3:31
  • $\begingroup$ 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? $\endgroup$ – Nyima Kao Jun 1 '15 at 3:42
  • $\begingroup$ 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. $\endgroup$ – Ian Agol Jun 1 '15 at 5:59
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Does the following condition help? Every closed geodesic on the minimal surface are closed geodesics in the ambient space M. I thought this is like some marked length spectral theorem, but I can't find a reference. $\endgroup$ – Nyima Kao Jun 1 '15 at 4:04
  • 1
    $\begingroup$ I am puzzled by this answer, because it does not answer the question (the closest it comes is that you think that the OP's conjecture might be true, since you can't think of a counterexample). $\endgroup$ – Igor Rivin Jun 1 '15 at 4:05
  • $\begingroup$ Igor, given the context, the OP was only assuming we had a minimal surface, not that all closed (intrinsic) geodesics were geodesics in the ambient space. In the latter case, I agree, it's likely that such a surface must be totally geodesic. It's late, but I'll try to muster a proof. $\endgroup$ – Andy Sanders Jun 1 '15 at 4:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.