# Non-compact manifolds of positive/non-negative Ricci curvature

Consider a non-compact complete Riemannian manifold $(M, g)$ with smooth compact boundary $\partial M$. Suppose also that $M \setminus \partial M$ has positive/non-negative Ricci curvature.

My question is, is it possible to remove $\partial M$, and replace it by another compact manifold $N$ with boundary $\partial N = \partial M$, and $M \cup N$ is a complete manifold (without boundary) of positive/non-negative Ricci curvature?

Let us assume the maximal volume growth on $M$, and let us also assume for convenience that $M \setminus K$, where $K$ is compact, has only one connected unbounded component. The naive mental picture I am having is "fitting a spherical cap to an end of a cylinder". But I am not sure how true such a heuristic is in general. Thanks in advance for any suggestions!

There are ways to glue while preserving positive Ricci curvature, but of course they require some geometric control near $\partial M$. Let me mention two relevant papers:

For example, Perelman shows how to glue two positive Ricci curvature manifolds with isometric boundaries to get a positively Ricci metric, which requires (roughly speaking) that the normal curvatures at one boundary is greater than the negative of the normal curvature at the other boundary when the normals are chosen correctly, like in filling a cylinder with a cap.

There are also obstructions to such gluings. For example in

[MR1216628, Greene, R. E., Wu, H. "Non-negatively curved manifolds which are flat outside a compact set". Differential geometry: Riemannian geometry (Los Angeles, CA, 1990), 327–335, Proc. Sympos. Pure Math., 54, Part 3, Amer. Math. Soc., Providence, RI, 1993]

one find the following: If $M$ is a complete Riemannian manifold of nonnegative Riccie curvature which is flat outside a compact set and is simply-connected at infinity , then $M$ is flat.

For example, $\mathbb R^n$ is simply-connected at infinity when $n>2$, so if you replace a round disk in $\mathbb R^n$ with a compact manifold with sphere boundary, and hope to have nonnegative Ricci curvature on the result, then the result must be flat, and in particular, be finitely covered by the product of a Euclidean space and a torus.

Note that a capped $2$-dimensional cylinder is not simply-connected at infinity, while a capped $n$-dimensional cylinder ($n>2$) is not flat outside the cap.

• Is there any criterion on $\partial M$ equipped with its second fundamental form that allows us to decide whether an $N$ with the required second fundamental form exists? The first obvious problem is that $\partial M$ must be zero bordant, but there could be geometric obstacles as well. – Sebastian Goette Jun 29 '17 at 13:07
• @SebastianGoette: I think there are no simple answers. The state of the art on these matters is given by the above-linked results of Perelman and Wraith and the announcements of Botvinnik, Wraith, and Walsh in Oberwolfach report of January 2017, see mfo.de/document/1702b/preliminary_OWR_2017_03.pdf. – Igor Belegradek Jun 29 '17 at 13:50

Take an n-sphere with its standard metric of positive curvature, cut out an n-ball and reglue it by a diffeomorphism of the bounding n-1-sphere. If the diffeomorphism is not isotopic to the identity, you will get an exotic sphere (this is called the twisted sphere construction) and you are asking whether the positively curved metric on the n-ball extends to a positively or non-negatively Ricci-curved metric on the exotic sphere.

However, Hitchin in his work on harmonic spinors has shown that certain exotic spheres do not admit a metric of positive scalar curvature, in particular not a metric of positive Ricci curvature. (You find more information in this survey.) Thus a constriction as you want can not work in general.