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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?

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    $\begingroup$ 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". $\endgroup$ Commented 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)? $\endgroup$ Commented 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
    Commented Jul 7, 2014 at 1:55

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