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 $\mathbf{R}P^3$.)
  • $\begingroup$ 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. $\endgroup$
    – RBega2
    Nov 17, 2021 at 12:03

1 Answer 1


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.

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    $\begingroup$ 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. $\endgroup$
    – Leo Moos
    Nov 17, 2021 at 15:34
  • $\begingroup$ 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... $\endgroup$
    – Sam Nead
    Nov 19, 2021 at 16:25

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