# 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?

• 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 $$\mathbf{R}P^3$$.)
• Regarding your question about immersed surfaces: Nadirashvili's examples suggest every three manifold admits a complete minimally immersed disk (of course his argument uses the Weierstrass representation so would only hold if if the three manifold had a ball where the metric was flat. Nov 17, 2021 at 12:03
• @RBega2 That's a good thing to point out, thanks. This got me thinking: should one make a distinction between a minimally immersed surface, and one that would be minimal for 'outer variations'? By that I would mean something like: if $\Sigma$ has locally finite area in the open set $U \subset M$, then it is stationary in $U$ as a varifold. The Nadirashvili examples would only be minimal in the former sense, right, but not the latter? Nov 17, 2021 at 15:01

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.
• If I understood your question correctly, then the answer is negative: because a round $\mathbf{S}^3$ has positive Ricci curvature, any two closed minimal surfaces in it must intersect. (People usually call this the Frankel theorem.) Still, I am curious about the "spinning construction" you propose. With geodesics in a surface $M^2$, does it work if $M$ is a topological sphere? For example, say $M$ is a rotationally symmetric dumbbell - this would have the two equators disjoint, separated by a stable geodesic in the neck. Nov 17, 2021 at 15:34
• Hmm. I see how to "spin" in a plane with at least three mushrooms, say in a line. Let $\alpha$ and $\beta$ be simple closed curves that enclose (respectively) the left two and the right two mushrooms. So $\alpha$ and $\beta$ meet in two points. We can now "spin" - namely take one copy of $\alpha$ and $n$ parallel copies of $\beta$ and perform surgeries to get a single curve... Nov 19, 2021 at 16:25