Exhaustion of a noncompact manifold by domains with mean convex boundary

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.

• Could you define "mean convex"? Sep 17 '20 at 18:49
• 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. Sep 18 '20 at 19:57
• Take a flat cylinder. Sep 18 '20 at 23:51
• I think nonnegative Ricci curvature may imply what you want! Feb 20 '21 at 6:33
• In another direction, a simply connected non positively curved manifold (aka Hadamard manifold) satisfies this, you can even replace "mean convex" by "convex" and take balls around any point. Feb 20 '21 at 9:04

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.