4
$\begingroup$

Suppose $M$ is a Riemannian manifold and $f:M\to\mathbb{R}$ a differentiable function. I can form the Hessian $H$ of $f$ (with respect to the Levi-Civita connection); this is a symmetric bilinear form in the tangent bundle of $M$. The eigenspaces of $H$ fit together into distributions on $M$. My question is, under what conditions are these distributions integrable?

More precisely, assume that $\lambda\in\mathbb{R}$ is nowhere on $M$ an eigenvalue of $H$, then at every point of $M$ the tangent space splits into a direct sum of $T^{<\lambda}$ and $T^{>\lambda}$, where $T^{<\lambda}$ is the sum of the eigenspaces of $H$ whose corresponding eigenvalue is $<\lambda$ (and similarly for $T^{>\lambda}$). Under what conditions are $T^{<\lambda}$ and $T^{>\lambda}$ integrable distributions?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ I think you should clarify your question along the lines of "Under what conditions on ?? are $T^{<\lambda}$ and $T^{>\lambda}$ (?always?) integrable distributions?" Thus, for example, one might ask "Under what conditions on the Riemannian metric are $T^{<\lambda}$ and $T^{>\lambda}$ integrable distributions for every $f$?", and the answer to this is "Only when the dimension of $M$ is $2$ or when one of the bundles is trivial". $\endgroup$
    – Robert Bryant
    Jun 26, 2014 at 13:12
  • $\begingroup$ @euklid345 Are you assuming that $f$ is a Morse function? According to your first statement, could you please more explain about this distribution? Is the dimension of this distribution fixed(along M)? $\endgroup$
    – Ali Taghavi
    Jun 27, 2014 at 22:09
  • $\begingroup$ Thanks Robert, for the comment. In my situation, the function f is fixed and I'm hoping to find a metric that would make my statement true. It would be fine if the metric depended on the function. (The function has a big critical locus, though.) $\endgroup$
    – euklid345
    Jul 7, 2014 at 1:55

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.