User atreyee - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T16:07:02Z http://mathoverflow.net/feeds/user/15654 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80452/diagonalizability-of-the-curvature-operator diagonalizability of the curvature operator atreyee 2011-11-09T06:12:29Z 2012-11-29T13:10:44Z <p>When can the curvature operator of a Riemannian manifold (M,g) be diagonalized by a basis of the following form</p> <p>'{\${E_i\wedge E_j }\$}' where '{\${E_i}\$}' is an orthonormal basis of the tangent space? If the manifold is three dimensional then it is always possible. But what about higher dimensional cases?</p> http://mathoverflow.net/questions/106325/curvature-of-the-cayley-projective-plane Curvature of the Cayley projective plane atreyee 2012-09-04T09:39:01Z 2012-09-04T12:31:20Z <p>The Cayley projective plane can be realized as the homogeneous space \$F_4/Spin(9)\$. In this way one can compute the curvature of this symmetric space in terms of a suitable orthonormal basis and the Lie brackets of the basic vectors. But is there any elegant expression for the curvature of \$CaP^2\$ which is independent of this homogeneous description due to the fact that \$CaP^2\$ is a rank one symmetric space?</p> http://mathoverflow.net/questions/67216/riemannian-manifolds-that-are-scalar-flat-but-not-ricci-flat Riemannian manifolds that are scalar flat but not ricci flat atreyee 2011-06-08T06:40:25Z 2011-06-20T02:27:12Z <p>What are the examples of Riemannian manifolds that have zero scalar curvature but non zero ricci curvature? Is there any sort of classification of such manifolds?</p> http://mathoverflow.net/questions/80452/diagonalizability-of-the-curvature-operator/80477#80477 Comment by atreyee atreyee 2011-11-10T05:08:41Z 2011-11-10T05:08:41Z Thanks.This result seems to be really useful.