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If $X$ is a gradient-like vector field of a Morse function $f\colon M\to \mathbb{R}$, then the integral curve $c_p(t)$ starting at an arbitrary point $p$ approaches critical points as $t\to \pm \infty$.

This is a paragraph in Matsumoto's Morse theory and I'm having a bit of trouble understanding it. I assume that $M$ should be compact in this context, otherwise $f$ may not have any critical points or a global integral curve.

Can someone give me a hint on why this should be true? (I wonder if it is considered obvious by the author, or as a fact that he expects from the reader to take on faith.) Thanks!

EPS
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  • You're definitely right that compactness is required. I'd start by studying $f \circ c_p$ - away from critical points you should get a positive uniform lower bound on its growth rate, which limits the time the curve can spend in this region (since $f$ is bounded above). Thus the curve has to end up in the critical areas eventually, at which point the Morse lemma coordinates should make the behaviour easy to understand. – Anthony Carapetis Apr 11 '16 at 03:15
  • @AnthonyCarapetis This is exactly my intuition: for a small $\epsilon>0$ cover the finitely many critical points with $\epsilon$ balls whose union we denote by $U$. (I'm thinking of $M$ as sitting inside some Euclidean space.) Then $M\setminus U$ is compact and $Df$ being continuous takes a strictly positive lower bound $C>0$ in that region. Since $f\circ c_p$ is bounded, the mean value theorem implies that after a finite time, $c_p$ has to lie inside $U$. But since we can shrink $U$ to the critical points, it follows that $c_p$ should converge to one of them. This intuition should be flawed, – EPS Apr 12 '16 at 02:03
  • b/c it does not use the assumptions that $f$ is Morse (except for the fact that there are finitely many critical points) and that the vector field is gradient-like. If you could help with filling in holes in this proof, or revising a completely new proof, I would appreciate it. – EPS Apr 12 '16 at 02:04
  • You need to be a little more subtle - you haven't ruled out the possibility that $c_p$ enters and exits $U$ an infinite number of times such that the total time spent outside $U$ is finite. If you use the critical neighbourhoods provided in the definition of a gradient-like vector field rather than $\epsilon$-balls, my intuition tells me that these neighbourhoods cannot be re-entered - either $c_p$ visits once, never to return; or it stays inside forever (in which case it must converge to the critical point). The monotonicity of $f \circ c_p$ probably plays a large role here. – Anthony Carapetis Apr 12 '16 at 02:38
  • I meant to write "each of these neighbourhoods cannot be re-entered". If you can establish my claims then the finite number of critical points will complete the proof. – Anthony Carapetis Apr 12 '16 at 02:47
  • @AnthonyCarapetis I have a perhaps stupid question to ask, because I think the above argument is flawed in a different way. If my argument indeed would prove that "after a finite time $c_p$ would lie in $U$" as I had claimed, then $c_p$ would not have the possibility of entering and leaving $U$ indefinitely: after a finite time, it had to stay inside $U$. Does this make any sense at all? – EPS Apr 13 '16 at 01:56
  • your argument would really prove that "after a certain total time $T$ spent outside $U$, $c_p$ would lie in $U$", since the lower bound on $d/dt (f \circ c_p)$ applies only when $c_p$ is outside $U$. Thus it could re-enter as often as it liked if the length of the $n^{\rm th}$ visit to $U^c$ was e.g. $T/2^n$. – Anthony Carapetis Apr 13 '16 at 03:30
  • http://math.stackexchange.com/questions/501007/when-does-gradient-flow-not-converge – Thomas Rot Apr 24 '16 at 21:36

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