Let $(M,g)$ be a complete noncompact Riemannian manifold. Can we find an exhaustion $M=\bigcup_{i \ge 1} U_i$ such that each $U_i$ is a bounded domain with smooth boundary $\partial U_i$ which is mean convex? Here, mean convex means the mean curvature of $\partial U_i$ with respect to the outer normal vector is positive everywhere.

If not, what condition can we impose on $(M,g)$ to guarantee this? For instance, if $M$ has positive sectional curvature, then the conclusion is true.

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    $\begingroup$ Could you define "mean convex"? $\endgroup$ – Igor Belegradek Sep 17 '20 at 18:49
  • $\begingroup$ In the context of the question, "mean convexity" means that the mean curvature vector $H_i$ of $\partial U_i$, dotted with the outward unit normal $\nu_i$ to $\partial U_i$ is positive. $\endgroup$ – Leo Moos Sep 18 '20 at 19:57
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    $\begingroup$ Take a flat cylinder. $\endgroup$ – Moishe Kohan Sep 18 '20 at 23:51

Unfortunately I don't have an answer to your question---I don't even know if this is an open problem---though perhaps I can suggest two places to start looking.

The same question was asked by Blaine Lawson, and is recorded as Problem 1.7 on page 459 in Proceedings of Symposia of Pure Mathematics Volume 44 (1986). Precisely he asks:

Characterize those complete Riemannian manifolds which admit an exhaustion by compact domains $\Omega_1 \subset \subset \Omega_2 \subset \subset \Omega_3 \subset \subset \cdots$ such that each boundary $\partial \Omega_k$ is smooth and of positive mean curvature with respect to the outer normal; e.g., concentric balls in $\mathbf{R}^n$. Some results in this direction are in [BR].

The paper that Lawson refers to here is Robert Brooks's `The fundamental group and the spectrum of the Laplacian', which appeared in Comment. Math. Helvetici in 1981. I must confess that I don't know the paper, but maybe somebody else does.


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