User atreyee - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-20T16:07:02Zhttp://mathoverflow.net/feeds/user/15654http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/80452/diagonalizability-of-the-curvature-operatordiagonalizability of the curvature operator atreyee2011-11-09T06:12:29Z2012-11-29T13:10:44Z
<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-planeCurvature of the Cayley projective planeatreyee2012-09-04T09:39:01Z2012-09-04T12:31:20Z
<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-flatRiemannian manifolds that are scalar flat but not ricci flatatreyee2011-06-08T06:40:25Z2011-06-20T02:27:12Z
<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#80477Comment by atreyeeatreyee2011-11-10T05:08:41Z2011-11-10T05:08:41ZThanks.This result seems to be really useful.