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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.

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  • $\begingroup$ 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. $\endgroup$ Dec 2 '19 at 21:45
  • $\begingroup$ @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. $\endgroup$
    – balu
    Dec 21 '19 at 19:36
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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).

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  • $\begingroup$ 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? $\endgroup$
    – balu
    Dec 21 '19 at 19:34

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