# Removing the boundary from a manifold-with-boundary while preserving non-negativity of scalar curvature & completeness

Let $$(M, g)$$ be a complete $$n$$-dimensional Riemannian manifold ($$n \geq 3$$) with non-negative scalar curvature and compact boundary. Is there a canonical way to "close" the manifold at the boundary while preserving the sign of the scalar curvature and the number of asymptotic ends?

That is, I would like to "remove" the $$(n-1)$$-dimensional boundary $$\partial M$$ without losing completeness and, to this end, glue some new $$n$$-dimensional piece to $$M$$ along the boundary $$\partial M$$, so that the result is a Riemannian $$n$$-manifold $$(\tilde{M}, \tilde{g})$$ which:

• is complete
• has no boundary
• contains $$\operatorname{int}(M)$$ as an open Riemannian submanifold
• has the same number of topological ends as $$M$$, i.e. no additional ends compared to $$M$$
• has scalar curvature $$R \geq 0$$

If it helps, take $$M$$ to be asymptotically flat with one or more ends, i.e. let there be some compact set $$K \subset M$$ s.t. $$M \setminus K$$ is a union of open sets each of which be diffeomorphic to $$\mathbb{R}^n$$ without a closed ball and such that, at infinity, the metric falls off uniformly and sufficiently fast to the Euclidean metric.

Without the scalar-curvature condition the solution is trivial: Namely take the double of $$M$$ and for each new asymptotic end introduced in this way, "close" this end by doing a 1-point compactification.

• You should define what you mean by completeness of a Riemannian manifold with boundary (metric completeness, I assume) and, more importantly, what do you mean by nonnegative sectional curvature: What boundary conditions do you impose (product neighborhood?, convex/concave boundary, etc). Without such conditions, the answer will be negative already when $n=3$ and $M$ is compact. Dec 2 '19 at 21:45
• @MoisheKohan Yes, sorry, I should have been more explicit – I indeed meant metric completeness. As for additional assumptions, please see my comment to your answer below.
– balu
Dec 21 '19 at 19:36

Here is an example which shows that without further conditions on $$\partial M$$ the answer is negative. Start with a compact connected smooth orientable 3-dimensional manifold $$M$$ whose boundary is diffeomorphic to $$S^2$$. I will assume that $$M$$ is obtained from, say, $$\Sigma\times S^1$$ by removing an open ball, where $$\Sigma$$ is a compact surface of genus $$\ge 2$$.

Let $$N$$ be the open manifold obtained by attaching a collar to $$\partial M$$. Since $$N$$ is open and orientable, it admits an immersion in $$S^3$$ (this was proven by Whitehead and, independently, by Hirsch in 1959-1960).

J. H. C. Whitehead, The Immersion of an Open 3‐Manifold in Euclidean 3‐Space, Proc. LMS, 11 (1961) 81-90.

Taking pull-back of the spherical metric on $$S^3$$ yields a metric of positive sectional (hence, scalar) curvature on $$N$$ and, hence, by restriction, on $$M$$. Now, suppose that $$X$$ is a compact 3-manifold with boundary diffeomorphic to $$S^2$$ such that gluing $$M$$ and $$X$$ along their boundary spheres results in a closed 3-manifold $$Y$$. Thus, $$Y$$ is diffeomorphic to the connected sum $$\Sigma\times S^1 \# Z$$, where $$Z$$ is obtained by attaching $$B^3$$ to $$\partial X$$.

I claim that $$Y$$ cannot admit a Riemannian metric of nonnegative scalar curvature. Suppose, to the contrary, that it does. Then, by a theorem of Gromov and Lawson, $$Y$$ cannot have aspherical connected summands, except when $$Y$$ admits a flat metric, i.e. is finitely covered by the 3-torus $$T^3$$.

M. Gromov, H.B. Lawson, Positive scalar curvature and the Dirac operator on complete Riemannian manifolds, Inst. Hautes Etudes Sci. Publ. Math. No. 58 (1983), 83–196

In our case, $$Y$$ has the aspherical manifold $$\Sigma\times S^1$$ as its connected summand and, clearly, $$Y$$ is not covered by $$T^3$$ (this follows, for instance, from the fact that $$\pi_1(Y)$$ contains a free nonabelian subgroup).

• Sorry for the huge delay – I got knocked out by the flu – and thank you so much for your help! I have to admit that parts of your answer go a bit over my head but I'll read up on those! As for your comment/question regarding boundary conditions, what would change if we assumed the boundary to be mean convex w.r.t. the inward-pointing normal?
– balu
Dec 21 '19 at 19:34