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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.

Proof:

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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ (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
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    $\begingroup$ (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
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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. $\Box$

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  • $\begingroup$ 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
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    $\begingroup$ @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

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