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Alexandrov's Theorem says that a compact constant mean curvature hypersurface embedded in $\mathbb{R}^{n+1}$ must be a round sphere.

What happens when the mean curvature is small, or bounded? (For instance, we require that $\frac{\text{max}| \nabla \tau |}{\text{min} |\tau|},$ with $\tau$ the mean curvature, be small or bounded).

Does this condition place topological restrictions on compact surfaces?

(For context, I am led to this question in because of the Einstein constraint equations, where many people have studied solutions with "near constant" mean curvature.)

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  • $\begingroup$ Two side remarks which I am sure you are aware of: in the Einstein constraint equations the ambient manifold is Lorentzian and the topology of the ambient manifold is not necessarily $\mathbb{R}^{n+1}$. (This is not to say that there's something wrong with the question, but your motivation and your tag (general-relativity) are rather far from the question you actually asked.) $\endgroup$
    – Willie Wong
    Commented Oct 8, 2014 at 11:49
  • $\begingroup$ More precisely, by the well-known local existence theorem of Choquet-Bruhat, in the context of Einstein's equation there is absolutely no topological restrictions (beyond the initial thing being a smooth manifold) in the "general relativity" question. We can always change the topology of the solution to accommodate. $\endgroup$
    – Willie Wong
    Commented Oct 8, 2014 at 11:51
  • $\begingroup$ @WillieWong: Thanks for the comments. If I understand correctly, you point out that we can solve the IVP for any topology, because the constraint equation can be solved for any topology and Choquet-Bruhat's theorem then gives a solution to the evolution problem? However, I would add that if we impose other conditions on the data which must satisfy the constraints (for instance, that it be CMC), there may be restrictions on the topology. And showing that there are no restrictions can be hard (see for instance this article of Isenberg-Mazzeo-Pollack arxiv.org/abs/gr-qc/0206034) $\endgroup$
    – user142700
    Commented Oct 8, 2014 at 12:50
  • $\begingroup$ For compact manifolds, under CMC, the situation is entirely characterised (going back to the 80s); in the low regularity setting you can see this paper. In short, the fact that it is CMC does not constraint the topology: the topology does constrain, a little bit, which values the mean curvature can take (positive, negative, or zero). But since you only demand CMC and not the value of mean curvature, there is absolutely no restriction. $\endgroup$
    – Willie Wong
    Commented Oct 10, 2014 at 9:06
  • $\begingroup$ My point is mainly this: I think the question you ask is an interesting one in geometric topology (I even upvoted!). But the question you ask has pretty much absolutely nothing to do with general relativity: certainly you don't expect all initial data sets in GR embed in Euclidean $\mathbb{R}^{1+n}$ as an $n$ hypersurface with constant mean curvature, so your motivation is tenuous at best and entirely nonsensical at worst. So I think removing that tag and the parenthetical can only improve your question. $\endgroup$
    – Willie Wong
    Commented Oct 10, 2014 at 9:11

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