The flux of a the negative gradient flow of a Morse-Bott function on a compact manifold converges to a critical value?

Question

Let $(M, g)$ be a compact Riemannian manifold and $f: M \rightarrow \mathbb{R}$ be a Morse-Bott function, i.e. the set a critical points of $f$, $Crit(f)$, has connected components which are smooth manifolds and which have as tangent spaces $T_x Crit(f) = \ker \nabla^2_x f$,

(where $\nabla^2_x f: T_x M \rightarrow T_xM$ is the linear operator obtained via $g$ from the hessian $f_{**,x} : T_x M \times T_x M \rightarrow \mathbb{R}$ defined as $f_{**,x}(v, w) = v(W(f))$ for $W \in \Gamma(TM)$ any extension of $w$ (this is well defined and symmetric at critical points) )

Let $\nabla f \in \Gamma(TM)$ be defined by $g(\nabla f, w) = w(f)$ and consider the flow of $-\nabla f$ denoted $\phi_t(y)$. I am trying to see why for any $y \in M$ it happens that $\lim\limits_{t \rightarrow \infty} \phi_t(y) \in Crit(f)$.

My attempt:

Since $M$ is compact, $\phi_t(y)$ is defined for all $t \in \mathbb{R}$. If the set $A_y:= \{ \phi_t(y) : t \in \mathbb{R} \}$ were closed, then, since $M$ is compact, this set would also be compact, and by Weierstrass $f$ would have to attain its minimum on it. Since moving along the flowlines of the negative gradient can only decrease $f$, this means that the minimum is attained at $x:= \lim\limits_{t \rightarrow \infty} \phi_t(y)$, so then $x$ would be a critical value for $f|_{A_y}$. But even ignoring the fact that I don't know why $A_y$ is necessarily closed, I don't see why if $x$ is a critical value for $f|_{A_y}$, then it is a critical value for $f$ as well.

I am thinking that this attempt not enough, as it doesn't use at all the fact that $f$ is a Morse-Bott function. But I don't see how to use this fact. I also know that $\nabla_x^2 f (v) = \nabla_V \nabla f$ for $x \in Crit(f)$ and $v \in T_xM$ and $V$ a vector field extending $v$, where $\nabla_V (\cdot)$ in the RHS is the Levi-Civita connexion of $g$, but I can't see how to use this either.

Answer 1

You can find the answer to this question in the appendix A of our preprint Symplectic circle actions on manifolds with contact type boundary. The proof was explained to us by Krzysztof Kurdyka and goes back to a result by Łojasiewicz.

The key point is to show that any gradient trajectory is of bounded length, and that paths of finite length converge at their end points.

It is not difficult to convince yourself that a gradient trajectory $\gamma$ accumulates at critical points of the Morse-Bott function $f$. Let us assume that there is a sequence $t_k$ with $t_k\to \infty$ such that $\gamma(t_k)$ converges to a critical point $q\in M$.

We can assume that $f(q) = 0$ without changing the gradient flow; simply replace $f$ by $f - f(q)$.

On a neighborhood~$U$ of $q$, we can assume that there is a constant $c>0$ such that $\| \nabla f\| \ge c\, \sqrt{|f|}$. For this it is enough to use a Morse-Bott chart around $q$, do the computations with respect to the Euclidean metric and use that all metrics are locally equivalent on a finite dimensional manifold.

If we consider now a gradient trajectory that comes arbitrarily close to $q$, then $f(\gamma(t))$ is for every $t\in \mathbb{R}$ strictly negative and strictly increasing.

The function $t\mapsto \sqrt{-f(\gamma(t))}$ is smooth, and we find $$ \frac{d}{dt} \sqrt{-f(\gamma(t))} = \frac{-df(\gamma'(t))}{2\sqrt{-f(\gamma(t))}} = \frac{-\bigl\langle \nabla f(\gamma(t)), \gamma'(t)\bigr\rangle}{2\sqrt{-f(\gamma(t))}} $$ Now use that $\gamma'$ and $\nabla f$ are parallel to write $$ \frac{d}{dt} \sqrt{-f(\gamma(t))} = \frac{-\bigl\|\nabla f(\gamma(t))\|\, \|\gamma'(t)\|}{2\sqrt{-f(\gamma(t))}} \le \frac{-c\,\sqrt{-f(\gamma(t))}\, \|\gamma'(t)\|}{2\sqrt{-f(\gamma(t))}} = -\frac{c}{2}\, \|\gamma'(t)\| \;. $$

This shows that $\|\gamma'(t)\|$ is bounded by $-\frac{2}{c}\, \frac{d}{dt}\sqrt{-f(\gamma(t))}$ in the neighborhood $U$ of $q$.

If we choose $C$ large enough, then length of $\gamma([C,\infty))$ for large $C$ is bounded by $+ \frac{2}{c}\, \sqrt{-f(\gamma(C))} -\frac{2}{c}\, \lim_{t\to \infty}\sqrt{-f(\gamma(t))} = \frac{2}{c}\, \sqrt{-f(\gamma(C))}$. Here you have to use that if $\gamma$ is sufficiently short, then it cannot get out of $U$ anymore.

The trajectory is thus of bounded length.