Is the Hessian almost everywhere nondegenerate? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T03:48:21Z http://mathoverflow.net/feeds/question/63949 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63949/is-the-hessian-almost-everywhere-nondegenerate Is the Hessian almost everywhere nondegenerate? ProbLe 2011-05-04T21:48:51Z 2011-05-04T21:58:06Z <p>Let $M$ be a complete Riemannian manifold. For a fixed point $p$ in $M$, the Riemannian distance to $p$ is denoted by $d_p$. Fix a strongly convex geodesic ball $B(o,R)$ in $M$ and some disjoint geodesic balls $B(a_0,r_0),...,B(a_n,r_n)$ in $B(o,R)$. </p> <p>Now for every $(x,x_1,...,x_n)\in B(a_0,r_0)\times...\times B(a_n,r_n)$, we consider a symmetric bilinear form on $T_xM$: $$b(x,x_1,...,x_n)=\Sigma_{i=1}^{n}\text{Hess}~d_{x_i}(x)$$.</p> <p>The question is that whether the bilinear form $b$ is nondegenerate for almost every point $(x,x_1,...,x_n)\in B(a_0,r_0)\times...\times B(a_n,r_n)$? Here the reference measure is the standard Lebesgue measure on $M^{n+1}$.</p> <p>Any hints or references are warmly welcome! Many thanks!</p>