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In [Zeghib: Laminations et hypersurfaces géodésiques des variétés hyperboliques, Annales scientifiques de l'ENS, 1991] it is shown, that in a compact manifold of negative curvature, there exists only a finite number of totally geodesic hypersurfaces (codim=1) without self-intersections.

Now my question: Is there an example of such a compact manifold M (with dim(M) > 2) , that allows even one totally geodesique hypersurface? Or a result that proofs the existence?

In manifolds of variable curvature it is not clear, that there exists even one totally geodesic hypersurface.

In the case of constant curvature it is clear in the universal covering (hyperbolic space), bur for a compact manifold?

Balou
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  • The theorem (if I understand correctly) can be illustrated by a compact hyperbolic surfaces. There are at least $2g$ many simple closed geodesics (which are hypersurfaces) –  Jun 22 '15 at 10:24
  • Yes, you are right. I forgot the assumption, that the dimension of M is bigger or equal than 3. I added it to the question. – Balou Jun 22 '15 at 11:57
  • Thanks for your posts. I would accept both solutions if it was possible... – Balou Jun 24 '15 at 06:53
  • Any ideas for dimension greater than 3, or the case of finite volume? – Balou Jun 24 '15 at 07:01

1 Answers1

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Examples in dimension $3$ follow from hyperbolization theorems for 3-manifolds combined with Mostow rigidity. Suppose that you have a compact 3-manifold $M$ with connected boundary $\partial M$ of genus $\ge 2$, such that $M$ is irreducible, atoroidal, and acylindrical and $\partial M$ is incompressible --- in brief this says $M$ has no "badly embedded" spheres, discs, annuli, or tori.

Now produce a closed 3-manifold $DM$, the "double" of $M$, by taking the quotient of two copies of $M$ identifying their boundaries by the "identity map". It follows that $DM$ is irreducible, atoroidal, and has infinite fundamental group.

The closed 3-manifold $DM$ has a hyperbolic structure; you can use Thurston's original hyperbolization theorem for this, because $\partial M$ becomes an incompressible surface in $DM$.

By the Mostow rigidity theorem, the "reflection" map from $DM$ to itself, interchanging the two copies of $M$ and being the identity on $\partial M$ itself, is an isometry. Therefore, $\partial M$ is totally geodesic in $DM$.

Lee Mosher
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  • Is there a method for dimension greater than $3$, perhaps by considering algebraic properties of lattices in $SL(n,\Bbb{R})$? – Neal Jun 23 '15 at 00:30
  • @Neal: Sorry, I cannot say. I do not know much at all about dimension greater than 3. – Lee Mosher Jun 23 '15 at 14:43
  • Is it possible to do such a construction, so that the manifold DM is not compact but has finite volume? I think the problem is the existence of the hyperbolic structure, but I don't know much about this. – Balou Jun 24 '15 at 07:00
  • @Balou: Yes, it works just the same way. Thurston's hyperbolization theorem applies in this context, and one uses Prasad's version of Mostow rigidity which applies to the finite volume context. – Lee Mosher Jun 24 '15 at 18:22