Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is it possible to perform a connected sum of two Riemannian Manifolds or Orbifolds while keeping curvature bounded from below? More explicitly, If $M_1$ and $M_2$ are two Riemannian manifolds (or orbifolds) of (sectional or Ricci) curvature bounded from below, is it possible to obtain a Riemannian metric in $M_1\#M_2$ that also has curvature bounded below?

share|cite|improve this question
up vote 5 down vote accepted

It is possible to take the connected sum of manifolds of dimension at least 3 of positive scalar curvature, and get a metric of positive scalar curvature, by the work of Gromov and Lawson,

Gromov, Mikhael; Lawson, H. Blaine, Jr. Positive scalar curvature and the Dirac operator on complete Riemannian manifolds. Inst. Hautes Études Sci. Publ. Math. No. 58 (1983), 83–196 (1984).

For sectional and Ricci curvature this is not possible. Of course, you get some lower bound even in these cases by compactness. If you have other conditions in mind you need to make your question more specific (and in particular scale-invariant).

share|cite|improve this answer
I don-t mind that the bound changes, only that it stays bounded from below. – Sak Jul 30 '13 at 10:32
But it is always bounded from below by compactness. You still need to clarify your question. – Mikhail Katz Jul 30 '13 at 10:40
I don't quite get the compactness bit. Couldn't the connected sum be non compact if for example one of the manifolds is non compact? – Sak Jul 30 '13 at 12:23
Yes but the regions involved in constructing the connected sum can be taken to be compact, so behavior at infinity is irrelevant. – Mikhail Katz Jul 30 '13 at 12:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.