Let $f:\mathbb R^n\to\mathbb R$ be a Morse function with uniform nondegenerate Hessian at critical points, i.e. for some $\delta>0$ $$ \forall x \in \{\nabla f=0\}\;\forall \xi\in\mathbb R^n: \quad |\langle \xi, \nabla^2 H(x)\xi \rangle | \geq \delta |\xi|^2 . $$ Edit: Liviu Nicolaescu pointed out a condition to ensure the sublevel sets of $\{f\leq c\}$ to be diffeomorphic to the sphere for $c$ large enough. Therefore, we impose $f$ to have at least linear radial growth at infinity $$ \langle x , \nabla f(x) \rangle \geq A |x| - B > 0 ,\quad A,B>0 . $$ Especially, the above two conditions ensure that $f$ has only finitely many critical points.

Let w.l.o.g. $0$ be a local minima of $f$ and let $\Phi_s(x)$ be the negative gradient flow wrt. $f$, i.e. $$ \dot \Phi_s(x) = -\nabla f(\Phi_s(x)) \quad\text{and}\quad \Phi_0(x)= x . $$ In addtion $\Omega$ denotes the basin of attraction for $0$ or in other words just the stable manifold of $0$, i.e. $$ \Omega = \{ x : \Phi_s(x) \to 0 \text{ for } s\to \infty \} $$ Does $f$ satisfy Neumann boundary condition on $\partial\Omega$ in the sense that the following integration by parts hold $$ \int_\Omega (-\Delta f)\; g \; dx = \int_\Omega \nabla f \cdot \nabla g \; dx \quad\text{for all $g$ such that} \quad \int_\Omega |\nabla f| \; |\nabla g| \; dx < \infty\quad ? $$

Strategy so far:

  • If $f$ is Morse-Smale, then $\partial \Omega$ is the union of stable manifolds heteroclinic connected to $0$
  • for the integration by parts only the (n-1)-dimensional stable manifolds of saddles of index 1 are relevant.
  • hence $\mathit{H}^{n-1}$ almost all $x\in \partial \Omega$ lie on a stable manifold of a 1-saddle and there the proof follows by contradiction and the definition of $\Omega$. Hereby $H^{n-1}$ denotes the (n-1)-dimensional Hausdorff measure.

Is it necessary for $f$ to be Morse-Smale?
Is there some soft argument?
What are relevant references?

  • $\begingroup$ In your situation, is $f$ already given, and you look for conditions on $g\in C^1(\Omega)$ (that is, on its decay at $\partial \Omega$) in order that the formula holds? $\endgroup$ Feb 5, 2013 at 17:03
  • $\begingroup$ I tried to clarify the question. If you want I could give some more background. The focus is not on the noncompatness of $\mathbb R^n$, but more on the detailed structure of $\partial \Omega$ especially in the neighborhood of critical points of index larger 1. Here I expect the boundary of $\Omega$ to be nonsmooth with corners. $\endgroup$
    – Frederic
    Feb 5, 2013 at 19:52

1 Answer 1


What troubles me is the noncompactness of $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$. The boundary $\partial \Omega$ could be noncompact in a rather unpleasant way, difficult to control. For example, there might exists a sequence of critical points $p_\nu$ of index $1$ going to $\infty$ as $\nu\to \infty$ such that there exists a gradient flow trajectory from $p_\nu$ to $0$ for any $\nu$. In this case $\partial\Omega$ will have an $(n-1)$-dimensional stratum for each $p_\nu$ making it hard to predict.

On the other hand, in the compact case, i.e., gradient flows on compact manifolds there is a beautiful paper of Harvey and Lawson that describes one instance when an integration by parts holds.

Edit Another issue you might neet to confront on a noncompact manifold is that of the multiplicities of the top strata of $\partial \Omega$. If $p$ is a critical point of index $1$ and there are infinitely many gradient trajectories from $p$ to $0$ they you would be hard-pressed to give a mening of the multiplicity of the stratum of $\partial \Omega$ corresponding to $p$.

In the compact case this multiplicity is defined by assigning in a certain canonical fashion a sign $\pm 1$ to each such connecting trajectory and then adding up all these signs. Clearly, if there are infinitely many connecting trajectory this procedure does not make sense.

This is a serious problem that appears in symplectic Floer homology. As far as I know, this is dealt with by taking advantage of the peculiarities of each concrete situation.

  • $\begingroup$ I thought this case is ruled out, by the assumption of growth at infinity, i.e. $f(x)\to \infty$ whenever $|x|\to \infty$. Does this not ensure that there is not such a sequence of critical points going to infinity? $\endgroup$
    – Frederic
    Feb 5, 2013 at 16:21
  • 1
    $\begingroup$ well,no, think e.g. to $f(x):=\sqrt{x^2+1}/2+\sin x$. $\endgroup$ Feb 5, 2013 at 16:58
  • $\begingroup$ Also the function $x^2(2-\cos x)$. $\endgroup$ Feb 5, 2013 at 18:48
  • $\begingroup$ Sorry, you are right. But since I don't want to have trouble with noncompactness of $\mathbb R^n$ let's assume that $f$ has at least linear growth at infinity. I will adjust the question. $\endgroup$
    – Frederic
    Feb 5, 2013 at 19:33
  • $\begingroup$ The big problem is that you can still have infinitely many critical points. You need to impose a condition that prohibits this from happening. $\endgroup$ Feb 5, 2013 at 20:17

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.