conformally flat manifold with positive scalar curvature - MathOverflow

conformally flat manifold with positive scalar curvature

2011-06-04T12:48:46Z

In the paper "Conformal Deformation of a Riemannian metric to a constant scalar curvature" of Richard Schoen (see here: http://www.intlpress.com/JDG/archive/1984/20-2-479.pdf), in the first page, it says that "Note that the class of conformally flat manifolds of positive scalar curvature is closed under the operation of connected sum, and hence contains connected sums of spherical space forms with copies of $S^1\times S^{n-1}$."

My question is: Is there any other conformally flat manifold with positive scalar curvature which is not in these forms? Or the manifolds Schoen mentioned exhaust the list of all conformally flat manifolds of positive scalar curvature?

Answer by Ian Agol for conformally flat manifold with positive scalar curvature

2011-06-05T21:32:33Z

This is true in 3 and 4 dimensions (if you include manifolds finitely covered by $S^1\times S^{n-1}$ such as the non-orientable twisted product $S^1\tilde{\times}S^{n-1}$).

In three dimensions, this follows from the geometrization theorem (see section 6.1 of Perelman). A fortiori, he proves that any 3-manifold with positive scalar curvature is of this form (and therefore contains a conformally flat positive scalar curvature metric).

In four dimensions, this follows from the classification of manifolds with positive isotropic curvature due to Chen, Tang, and Zhu, completing an approach of Hamilton using ideas of Perelman. In particular, Hamilton pointed out that manifolds with positive scalar curvature which are conformally flat have positive isotropic curvature.

Another trivial observation is that a conformally flat manifold with finite fundamental group is a spherical space form. Its universal cover is a complete simply connected conformally flat manifold, and therefore must be $S^n$. Then by Liouville's theorem, the group of covering translations is conjugate into $O(n+1)$, so the manifold is a space form.

Answer by Misha for conformally flat manifold with positive scalar curvature

2012-03-31T23:46:45Z

Products $M^m \times S^{n-m}$ will be conformally flat, where $M^m$ is a compact manifold of curvature $-1$ and $S^{n-m}$ has curvature $1$. If $n>2m$ then the scalar curvature of the product will be positive (positive curvature of the sphere dominates the negative curvature of the hyperbolic manifold, so the scalar curvature of the product will be positive). You can also (frequently) deform these metrics a bit so that they no longer (locally) split as products and still have positive scalar curvature and remain conformally flat. Thus, you have counter-examples in all dimensions $n\ge 5$. See e.g. this paper by R.Mazzeo and N.Smale http://intlpress.com/JDG/archive/1991/34-3-581.pdf for further discussion.