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 such that $T\_x Crit(f) = \ker \nabla^2\_x f$,


(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 $g(\nabla f, w) = w(f)$. Consider the flow of $-\nabla f$ denoted $\phi\_t$.


How to prove that $\phi\_t(y)$ converges when $t \rightarrow + \infty $ and it converges in the critical set of $f$?


Ps: This question appeared under bounty for 500 points here <https://math.stackexchange.com/questions/4265927/the-flux-of-a-the-negative-gradient-flow-of-a-morse-bott-function-on-a-compact-m?noredirect=1#comment9393403_4265927> but didn't receive any answer so far.



The negative gradient flow of a Morse-Bott function on a compact manifold converges to a critical point?
I will assume that "converges in the critical set of $f$" is asking that if $f$ is MB then $\phi\_t(y)$ (the flowlines) converge as $t\to\infty$ to $y\_\infty$ a critical point (of course, depending on $y$).




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For any function $f$ on $(M,g)$ a closed Riemannian manifold, note that
$$
\tag{\*} \frac{d}{dt} f(\phi\_t(y)) = - |\nabla f|^2(\phi\_t(y).
$$
Since $f$ is bounded, we can integrate this from $0$ to $\infty$ to find that
$$
\int\_0^\infty |\nabla f|^2(\phi\_s(y)) ds < \infty.
$$



> 
> Claim: $|\nabla f|^2(\phi\_s(y)) \to 0$ as $s\to\infty$.
> 
> 
> 


One way to see this is to note that
$$
\frac{d}{ds} |\nabla f|^2(\phi\_s(y)) = 2 D^2 f(\nabla f,\nabla f)|\_{\phi\_s(y)} 
$$
is bounded and to prove that if $u \in L^2([0,\infty))$ has $|u'|\leq C$ then $u(s)\to 0$ as $s\to\infty$ (if $u$ is $>\epsilon$ far out at infinity, it will be $>\epsilon/2$ on a definite interval by the derivative estimate, this will contribute too much to the $L^2$ norm). This proves the claim.


In particular, for any sequence $t\_i\to\infty$, we can pass to a subsequence (compactness of $M$) so that $\phi\_{t\_i}(y) \to y\_\infty$. By the above claim, $|\nabla f|(y\_\infty) = 0$.


Of course, well-known examples (the goat tracks) show that $y\_\infty$ need not be the UNIQUE limit (without some further assumption on $f$).


The usual assumptions that would suffice are either Morse, Morse Bott, or real analytic.




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Now, let us assume that $f$ is MB. There are several ways to conclude that $y\_\infty$ is the limit of $\phi\_t(y)$. A good tool is the MB lemma, which says that you can find coordinates where $f$ is exactly a quadratic form, with $+1,-1,0$ eigenvalues. (Note that the metric need not be Euclidean in these coordinates, so you can't conclude that the gradient of $f$ is as simple as one might expect).


There are various ways to argue, I will describe one below. My proof is based on the "Lojasiewicz method" (but is much simpler due to the Morse Bott assumption + the MB lemma). I am sure that it is possible to argue in a more direct manner, but it's a bit annoying to handle the various cases (e.g., $x\_1\gg x\_2$, etc) and this way is a bit smoother (although somewhat indirect).


For simplicity, let me assume that $n=3$ and in some neighborhood $U$ of $y\_\infty$, we can choose coordinates $x\_1,x\_2,x\_3$ so that $y\_\infty$ corresponds to $(0,0,0)$
$$
f(x) = -x\_1^2 + x\_2^2.
$$
Then, locally the critical set of $f$ is $x\_1=x\_2=0$. (The general case is essentially identical.)


In the coordinates with respect to the MB neighborhood, let me choose some constant $C$ so that the Riemannian metric satisfies
$$
C^{-1}\delta \leq g \leq C \delta
$$
for $\delta$ the Euclidean metric in these coordinates.


Then, we find that $|\nabla f|\_g \geq C^{-1} |\nabla f|\_\delta$, so indeed
$$
|\nabla f|\_g^2 \geq c(x\_1^2+x\_2^2).
$$
Thus, we find that
$$
\tag{\*\*} |\nabla f|\_g^2 \geq c|f|.
$$


I claim that if $f$ satisfies (\*\*) then the limit $y\_\infty$ is unique. (Morse Bott will not be used again).


We compute
$$
\frac{d}{dt} f(\phi\_t(y)) = - |\nabla f|^2(\phi\_t(y)) \leq - c f(\phi\_t(y))
$$
(This is valid as long as $\phi\_t(y)$ remains in $U$, the neighborhood where (\*\*) holds.) We can assume that $\phi\_0(y) \in U$ (just shift time). Below, we will assume that $\phi\_0(y)$ is very close to $y\_\infty$ (by a final shift in time).


Integrating the above ODE we find
$$
f(\phi\_t(y)) \leq C e^{-ct}
$$
as long as $\phi\_t(y)$ remains in $U$.


We can also compute differently:
$$
\frac{d}{dt} f(\phi\_t(y))^{1/2} = - \frac 12 f^{-1/2}|\nabla f|^2(\phi\_t(y)) \leq -c|\nabla f|(\phi\_t(y)).
$$
In particular,
$$
\textrm{length}(\phi\_t(y)|\_{[0,T]}) = \int\_0^T |(\phi\_t(y)'|\_g = \int\_0^T |\nabla f|\_g(\phi\_t) \leq - c \int\_0^T \frac{d}{dt} f(\phi\_t(y))^{1/2} \leq C (f(\phi\_0(y))^{1/2} - f(\phi\_T(y))^{1/2}) \leq C (f(\phi\_0(y))^{1/2}.
$$
This shows that as long as $f(\phi\_0(y)$ is sufficiently close to $y\_\infty=0$ (which we can arrange), $\phi\_t(y)$ cannot travel far enough to leave the neighborhood $U$. Thus, the above analysis is valid for all times.


Repeating the same calculation, we find
$$
\textrm{length}(\phi\_t(y)|\_{[t,\infty]}) \leq C f(\phi\_t(y))^{1/2} \leq C e^{-ct/2}.
$$
This proves that $y\_\infty$ is the unique limit point. Indeed, assume that $y\_\infty'$ is also a limit point. Then, the curve $\phi\_t(y)|\_{[t,\infty]}$ will have to pass infinitely many times between points close to $y\_\infty$ and points close to $y'\_\infty$. This will force the curve to have infinite length (but we proved it has finite length).


