Setup and notation:
Let $f:M\to \mathbb{R}$ be a Morse-function on the compact $m$-dimensional manifold $M$ and let $X$ be a gradient-like vector field for the function $f$. Denote the unstable respectively stable manifold by $W^u_f(p)=\{x\in M\, |\, \lim_{t\to -\infty}\varphi^t(x)=p\}$ resp. $W^s_f(p)=\{x\in M\, |\, \lim_{t\to \infty}\varphi^t(x)=p\}$ where $\varphi^t$ is the flow of the vector field $X$ and $p\in Crit(f)$ is some critical point of $f$.
We have a free $\mathbb{R}$-action (at least if $p\neq q$) on the intersection $\mathcal{M}(p,q):=W^u_f(p)\cap W^s_f(q)$ by $s\cdot x = \varphi^s(x)$.
I would like to check whether the vector field $X$ is Morse-Smale, i.e. whether $W^u_f(p)\pitchfork W^s_f(q)$ for all $p,q\in Crit(f)$.

Question: Suppose $p\neq q$ and $\mathcal{M}(p,q)\neq \emptyset$ and that the intersection is transverse at $r\in \mathcal{M}(p,q)$ (i.e. if $T_rM=T_rW^u_f(p)+T_rW^s_f(q)$). Does this imply that the intersection is transverse at $\varphi^s(r)$ for every $s\in \mathbb{R}$?

In other words, to check whether $X$ is Morse-Smale, is it enough to check that the intersection is transverse for any single point on each flow line connecting $p$ and $q$?

Relevant references are also much appreciated.


1 Answer 1


I'd say that the answer to your question is yes. Note that $\varphi^s$ preserves stable (resp. unstable) manifolds: $$x\in W_f^s(p)\Leftrightarrow\lim_{t\rightarrow\infty} \varphi^t(x)=p\Rightarrow\lim_{t\rightarrow\infty} \varphi^t(\varphi^s(x))=p\Leftrightarrow\varphi^s(x)\in W_f^s(p).$$ Now, since $\varphi^s$ is a diffeomorphism we have the first equality of the following line: $$T_{\varphi^s(r)}M=D_r\varphi^s(T_rM)=D_r\varphi^s(T_rW_f^u(p)+T_rW_f^s(q))=D_r\varphi^s(T_rW_f^u(p))+D_r\varphi^s(T_rW_f^s(q)).$$ Finally, since $\varphi^s$ preserves (un)stable manifolds the last sum is contained in $T_{\varphi^s(r)}W_f^u(p)+T_{\varphi^s(r)}W_f^s(q)$. The other inclusion being obvious one gets the equality we were looking for.

  • $\begingroup$ That's a nice argument, thanks a lot! $\endgroup$
    – Dave
    Commented Apr 11, 2013 at 15:09
  • $\begingroup$ @Dave: You're welcome! $\endgroup$ Commented Apr 11, 2013 at 15:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .