# Is the Hessian almost everywhere nondegenerate?

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 $B(o,R)$.

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 $T_xM$: $$b(x,x_1,...,x_n)=\Sigma_{i=1}^{n}\text{Hess}~d_{x_i}(x)$$.

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 $M^{n+1}$.

Any hints or references are warmly welcome! Many thanks!

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If in addition the convexity radius for all points in $B(o,R)$ is bigger that $2{\cdot}R$ then the $(\mathop{\rm Hess} d_p)(x)$ is nonnegative and it is zero only in the direction from $x$ to $p$. In this case your sum is strictly positive if not all the points $x_i$ and $x$ lie on one geodesic. Does it make you happy? –  Anton Petrunin May 4 '11 at 22:23
Thank you very much for your answer Anton! It does make me happy! What if we only assume that $R$ is not bigger than the convexity radius of $o$? A concrete example of this case, is that $B(o,R)$ is the open upper hemisphere. The main obstacle for me is that in this case the hessian is not necessarily positive definite and it can be negative definite, too. But I guess that for almost all $(x,x_1,...,x_n)$, the hessian is not degenerate. Have you some ideas for this? Thank you! –  ProbLe May 5 '11 at 7:54
I am sure that the answer is "YES" even for your original question. But the set where Hessian is degenerate can not be described that easely. I guess you have in mind some application; if you would explain it to me, I could think of a minimalistic solution. –  Anton Petrunin May 5 '11 at 15:35
I quite agree with you, the answer should be "YES". In fact, this is just a question from a conversation and I find it very interesting! Would you please tell me some references on the degeneracy of the Hessian? Many thanks! –  ProbLe May 6 '11 at 12:57