I would like to know if my proof of the following statement is correct

If $M$ is a compact manifold, then every vector field $X$ over $M$ is complete.


I take $p\in M$ and $(\gamma_p,I_p)$ a maximal integral curve through $p$. $I_p$ is open and non empty so if $I_p$ is closed, it is equal to $\mathbb{R}$.

So take $s\in \text{Adh}(I_p)$. There exists $t_i\rightarrow s, t_i\in I_p$. Define $q$ as $\text{lim}(\gamma_p(t_i))$ and take a maximal integral curve passing by $q$ as $(\gamma_q,I_q)$. Then since the vector fields are $C^\infty$, $\gamma_q(I_q)\cap\gamma_p(I_p)$ is nonempty (intuitively, this is clear but don't really know how to show it).

On the intersection, $\gamma_p$ and $\gamma_q$ coincide. So we can extend the maximal curve $\gamma_p$ to include $\gamma_q$. This implies that $q\in\gamma_p(I_p)$ and is actually equal to $\gamma_p(s)$

I think the main idea is there even though it probably lacks a bit of rigour.

  • $\begingroup$ Your title is not a correct reflection of the question. To say that a manifold is complete means that it is endowed with a complete Riemannian metric. Your question is about showing that a vector field is complete. $\endgroup$ – Jack Lee Dec 29 '17 at 22:39
  • 1
    $\begingroup$ A couple of questions about your proof: (1) How do you know that the sequence $\gamma_p(t_i)$ converges? (2) How do you show that the two integral curves coincide where both are defined? $\endgroup$ – Jack Lee Dec 29 '17 at 22:41
  • 1
    $\begingroup$ (1) by continuity of $\gamma_p$? (2) In my course we have a result that says that if two integral curves pass by the same point, then they coincide in their intersections (by showing that it is non empty, open and closed) $\endgroup$ – tomak Dec 30 '17 at 9:16
  • 2
    $\begingroup$ (1) Continuity at what point? If you knew $\gamma_p$ were continuous at $t$, then you could conclude $\gamma_p(t_i)\to \gamma_p(t)$ as $t_i\to t$. But you don't even know it's defined at $t$ -- that's what you're trying to prove. (2) If two integral curves pass through the same point, then one is a reparametrization of the other, but they're not necessarily equal. What point do they both pass through? $\endgroup$ – Jack Lee Dec 30 '17 at 16:31

As Jack Lee is already addressing the flaws of your proof, I thought I would drop a complete proof of the statement, so it is at hand for everyone having a question on it.

First, let us work out a general statement:

Proposition. Let $M$ be a smooth manifold and $X$ be a vector field on $M$ with a local flow given by $(\phi_t)_t$. Assume that there exists $\varepsilon>0$ such that $\phi$ is defined on $]-2\varepsilon,2\varepsilon[\times M$, then $X$ is complete.

Proof. For all $t\in\mathbb{R}$, let $k(t)$ be the integer part of $t/\varepsilon$, then one has: $$t-k(t)\varepsilon\in[-\varepsilon,0]\subseteq]-2\varepsilon,2\varepsilon[,$$ so that one can define the following diffeomorphism of $M$: $$\psi_t:={\phi_{\varepsilon}}^{k(t)}\circ\phi_{t-k(t)\varepsilon}.$$ For all $x\in M$, since $k(0)=0$, one has: $$\psi_0(x)=\phi_0(x)=x.$$ Furthermore, for all $s\in\mathbb{R}$, one has the following equality: $$\begin{align}\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert t=s}\psi_t(x)&=\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert t=s}{\phi_{\varepsilon}}^{k(s)}\circ\phi_{t-k(s)\varepsilon}(x),\\&=\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert t=s-k(s)\varepsilon}{\phi_{\varepsilon}}^{k(s)}\circ\phi_t(x),\\&=T_{\phi_{s-k(s)\varepsilon}(x)}{\phi_{\varepsilon}}^{k(s)}\left(\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert t=s-k(s)\varepsilon}\phi_t(x)\right),\\&=T_{\phi_{s-k(s)\varepsilon}(x)}{\phi_{\varepsilon}}^{k(s)}(X(\phi_{s-k(s)\varepsilon}(x))),\\&=T_{\phi_{\varepsilon}(\phi_{s-k(s)\varepsilon}(x))}{\phi_{\varepsilon}}^{k(s)-1}(T_{\phi_{s-k(s)\varepsilon}(x)}\phi_{\varepsilon}(X(\phi_{s-k(s)\varepsilon}(x)))),\\&=T_{\phi_{\varepsilon}(\phi_{s-k(s)\varepsilon}(x))}{\phi_{\varepsilon}}^{k(s)-1}(X(\phi_{\varepsilon}(\phi_{s-k(s)\varepsilon}(x)))),\\&=T_{{\phi_{\varepsilon}}^{k(s)}(\phi_{s-k(s)\varepsilon}(x))}{\phi_{\varepsilon}}^0(X({\phi_{\varepsilon}}^{k(s)}(\phi_{s-k(s)\varepsilon}(x)))),&\\&=X(\psi_s(x)).\end{align}$$ Therefore, by the unicity part of Picard-Lindelöf theorem, $\phi=\psi$ and $\phi$ is in fact defined on $\mathbb{R}\times M$. Whence the result. $\Box$

Remark. The key point of these computations is that $\phi$ preserves $X$, for all $t$ such that $\phi_t$ exists and $x\in M$: $$T_x\phi_t(X(x))=X(\phi_t(x)).$$ Which is almost tautological, since by the very definition of the flow, one has: $$X(\phi_t(x))=\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert s=t}\phi_s(x)=\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert s=0}\phi_t\circ\phi_s(x)=T_{\phi_0(x)}\phi_t\left(\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert s=0}\phi_s(x)\right)=T_x\phi_t(X(x)).$$ Another thing to notice is that for all $t$ sufficiently close to $s$, $k(t)=k(s)$.

From there it is easy to derive the desired result.

Corollary. Let $M$ be a compact smooth manifold and $X$ be a vector field on $M$, then $X$ is complete.

Proof. Let $p\in M$, using the existence part of Picard-Lindelöf theorem, there exists $\varepsilon_p>0$ and $U_p$ an open neighborhood of $p$ in $M$ such that $\phi$ the flow of $X$ is defined on $]-\varepsilon_p,\varepsilon_p[\times U_p$. By construction, $\{U_p\}_{p\in M}$ is an open cover of $M$, which is compact, hence there exists $p_1,\ldots,p_k$ in $M$ such that $\{U_{p_i}\}_{1\leqslant i\leqslant k}$ is still a cover of $M$. Let then define the following existence time: $$\varepsilon:=\min_{1\leqslant i\leqslant k}\varepsilon_{p_i}>0,$$ by construction, for all $i\in\{1,\ldots,k\}$, $\phi$ is defined on $]-\varepsilon,\varepsilon[\times U_i$, therefore on the whole $]-\varepsilon,\varepsilon[\times M$. Whence the result using the above proposition. $\Box$

  • $\begingroup$ I don't understand the definition $\psi_t:=\phi_\epsilon^{k(t)}\circ \phi_{t-k(t)\epsilon}$. By the property $\phi_t\circ \phi_s=\phi_{t+s}$, this definition just says $\psi_t=\phi_{k(t)\epsilon +t-k(t)\epsilon}=\phi_t$, right? What am I missing? $\endgroup$ – rmdmc89 Apr 16 '18 at 16:56
  • 1
    $\begingroup$ @AguirreK The equality $\phi_t\circ\phi_s=\phi_{t+s}$ means nothing if $s,t$ and $s+t$ are not in the domain of definition of the flow! The aim of this discussion is to prove that $\phi$ is indeed a global flow and not only a local flow. The idea for defining $(\psi_t)_t$ comes from the well-known formula you mentioned. $\endgroup$ – C. Falcon Apr 16 '18 at 18:09
  • $\begingroup$ Oh, I see. Last question: the functions $k(t)$ and $t-k(t)\epsilon$ are not smooth, so how do we gurantee $\psi_t$ is smooth? $\endgroup$ – rmdmc89 Apr 17 '18 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.