conformally flat manifold with positive scalar curvature - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:04:56Z http://mathoverflow.net/feeds/question/66886 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66886/conformally-flat-manifold-with-positive-scalar-curvature conformally flat manifold with positive scalar curvature Paul 2011-06-04T12:48:46Z 2012-03-31T23:46:45Z <p>In the paper "Conformal Deformation of a Riemannian metric to a constant scalar curvature" of Richard Schoen (see here: <a href="http://www.intlpress.com/JDG/archive/1984/20-2-479.pdf" rel="nofollow">http://www.intlpress.com/JDG/archive/1984/20-2-479.pdf</a>), 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 $S^1\times S^{n-1}$."</p> <p>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?</p> http://mathoverflow.net/questions/66886/conformally-flat-manifold-with-positive-scalar-curvature/66989#66989 Answer by Ian Agol for conformally flat manifold with positive scalar curvature Ian Agol 2011-06-05T21:32:33Z 2011-06-05T21:32:33Z <p>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 $S^1\tilde{\times}S^{n-1}$). </p> <p>In three dimensions, this follows from the geometrization theorem (see section 6.1 of <a href="http://front.math.ucdavis.edu/0303.5109" rel="nofollow">Perelman</a>). 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). </p> <p>In four dimensions, this follows from the classification of manifolds with positive isotropic curvature due to <a href="http://arxiv.org/abs/0810.1999" rel="nofollow">Chen, Tang, and Zhu</a>, completing an <a href="http://www.ams.org/mathscinet-getitem?mr=1456308" rel="nofollow">approach of Hamilton</a> using ideas of Perelman. In particular, Hamilton pointed out that manifolds with positive scalar curvature which are conformally flat have positive isotropic curvature. </p> <p>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 <a href="http://en.wikipedia.org/wiki/Liouville%2527s_theorem_%28conformal_mappings%29" rel="nofollow">Liouville's theorem</a>, the group of covering translations is conjugate into $O(n+1)$, so the manifold is a space form. </p> http://mathoverflow.net/questions/66886/conformally-flat-manifold-with-positive-scalar-curvature/92780#92780 Answer by Misha for conformally flat manifold with positive scalar curvature Misha 2012-03-31T23:46:45Z 2012-03-31T23:46:45Z <p>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 <a href="http://intlpress.com/JDG/archive/1991/34-3-581.pdf" rel="nofollow">http://intlpress.com/JDG/archive/1991/34-3-581.pdf</a> for further discussion. </p>