3-manifolds with all minimal surfaces closed Question. Let the manifold $(M^3,g)$ be compact without boundary. Suppose that every complete, embedded minimal surface $\Sigma \subset M^3$ is closed. Must $M$ be diffeomorphic to $\mathbf{S}^3$ or $\mathbf{R}P^3$? If not, what if one strengthens the hypothesis to include also all immersed minimal surfaces?

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*Similar questions about geodesics are famous, but some of the tools used there—notably geodesic flow—have no immediate analogues in higher dimension.

*Not all three-manifolds satisfy the hypothesis: if $N^2$ is a compact surface that contains a non-closed geodesic, then one can take $M = N \times \mathbf{S}^1$; a concrete example is the three-torus. (Mind you I am not completely sure whether the result even holds in $\mathbf{S}^3$ and $\mathbf{R}P^3$.)

 A: Interesting question.  Note that all of the spherical space forms either have the property (all complete embedded minimal surfaces closed) or none do.
Does $S^3$ have a disjoint pair of closed, embedded, minimal surfaces?  If so, we might be able to use these to build a minimal lamination via a "spinning construction".

On the other hand, the three-sphere (and thus all spherical space forms) contain complete immersed minimal surfaces.  Namely, take a spherical tetrahedron $T$ with four dihedral angles being $\pi/2$ and the remaining two, non-adjacent, dihedral angles being irrational multiplies of $\pi$.  The medial square in $T$, between the two exceptional edges, will be a minimal surface.  Repeatedly reflecting $T$ across its faces, and gluing together the resulting medial squares, gives the desired minimal surface.
There is an easier argument giving immersed minimal (in fact, totally geodesic) surfaces in a closed hyperbolic three-manifolds.  This is because the universal cover is $\mathbb{H}^3$.  So we take an "irrational" geodesic plane, and project it down.
