In the paper Conformal Deformation of a Riemannian metric to a constant scalar curvature of Richard Schoen (J. Differential Geom. 20(2) (1984) 479-495, doi:10.4310/jdg/1214439291), 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?


2 Answers 2


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.


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's Ricci flow with surgery on three-manifolds). 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 (Complete classification of compact four-manifolds with positive isotropic curvature), completing an approach of Hamilton (Four-manifolds with positive isotropic curvature) 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.

  • $\begingroup$ What does "must be $S^n$" mean? Diffeomorphic, conformally diffeomorphic, isometric? $\endgroup$ Apr 25, 2021 at 4:04
  • $\begingroup$ Conformally equivalent $\endgroup$
    – Ian Agol
    Apr 25, 2021 at 6:14

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