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Let $(M,g)$ be a compact smooth orientable riemannian manifold, and let $f: M \to \mathbb R$ be a Morse function. All functions here are assumed to be smooth. We will be considering the integral curves and the flow $\varphi$ of the gradient vector field of $f$. That is \begin{align} & \varphi : \mathbb R \times M \to M, \\ & (t,x) \mapsto \varphi^t(x), \end{align} where $\varphi^t(x)$ is the solution of the ODE $\dot{u} = -\nabla(f(u))$ at time $t$, with initial value $x$. It is known that $\varphi^t(x)$ converges to critical points of $f$, as $t \to \pm \infty$. Given $p,q$ critical points of $f$, we set \begin{align} \mathcal M(p,q) := \{u: \mathbb R \to M : \dot{u} = -\nabla(f(u)), \lim_{t \to -\infty}u(t) = p, \lim_{t \to +\infty}u(t) = q \}. \end{align} Next, we assume that $f$ is Morse-Smale, so that $\mathcal M(p,q)$ is a smooth manifold (it essentially corresponds to the transverse intersection of a stable manifold with an unstable manifold). We consider on $\mathcal M(p,q)$ the $C^\infty_{loc}$-convergence, that is, the uniform convergence on every compacta of the derivatives of all orders. There is a $\mathbb R$-action of $\mathcal M(p,q)$, defined by translation: \begin{equation} (\tau \cdot u)(s) = \varphi^\tau(u(s)) = u(s+\tau). \end{equation} The quotient $\widehat{\mathcal M}(p,q) = \mathcal M(p,q) / \mathbb R$ is the space of gradient flow lines from $p$ to $q$.

I'd like an answer or a reference for the following question:

How do you prove that the above action is proper and free, so that the quotient $\widehat{\mathcal M}(p,q) = \mathcal M(p,q) / \mathbb R$ is indeed a smooth manifold?

This question is possibly related to this one.

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2 Answers 2

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You can find an explicit proof of this in Schwarz, Morse Homology, Proposition 2.31.

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Edit: I rewrote this since I did not understand my previous argument.

Let me consider $M(p,q)$ as a submanifold of $M$, it is the intersection of the unstable and stable manifolds of $p$ and $q$. (There is a map of this $M(p,q)$ to your $\mathcal M(p,q)$ by sending a point to the solution curve through this point.)

Anyway: The action is free on this space: The function $f$ is a strict Lyapunov function: if $s>0$ then $f(\phi_0(x))=f(x)>f(\phi_s(x))$. This implies that there can't be a time $s>0$ such that $u(0)=u(s)$. By symmetry this also works for $s<0$. This argument also shows that $f:M(p,q)\rightarrow \mathbb{R}$ is a submersion everywhere. Let $c=(f(p)-f(q))/2$. Since $f$ is submersive, $Y=f^{-1}(c)$ is a smooth codimension $1$ manifold of $M(p,q)$. On $Y\times \mathbb{R}$ there is an action $\psi_t(x,s)=\phi(x,s+t)$.

Consider now the map

$$ h:Y\times \mathbb{R}\rightarrow M(p,q) $$ defined by $h(x,s)=\phi_s(x)$. This map is a diffeomorphism and intertwines the action of $\phi$ and $\psi$. The conclusion is that we can identify $M(p,q)$ with the flow action with $Y\times \mathbb{R}$.

So you can also prove that the translation action $\psi$ is proper on $Y\times \mathbb R$. But this is not too hard.

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  • $\begingroup$ I did not quite understand the "proper because it is basically precomposing the flow". So we have a map $\mathbb{R} \times \mathcal{M}(p,q) \rightarrow \mathcal{M}(p,q) \times \mathcal{M}(p,q) $ which is $(t,a) \rightarrow (\gamma_t (a),a)$ , now what is the diffeomorphism you're talking about ? $\endgroup$
    – Soham
    Commented May 27, 2018 at 15:06
  • $\begingroup$ @LucyferZedd: My neither. I have edited the answer. $\endgroup$
    – Thomas Rot
    Commented May 29, 2018 at 8:20

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