# Condition on local flows that imply completeness of vector field

Given any smooth vector field $$X$$ on a manifold $$M$$ we can cover $$M$$ by open sets $$\{U_\alpha\}_{\alpha\in A}$$ such that for each $$\alpha$$ there is an $$\epsilon_\alpha>0$$ such that if $$p \in U_\alpha$$ the local flow for $$X$$ at $$p$$ , say $$\phi^\alpha_t$$ is defined for $$t\in (-\epsilon_\alpha,\epsilon_\alpha)$$.

The $$\phi^\alpha_t$$ being local diffeomeorphisms we get a local family of local diffeomorphisms out of this i.e. $$\{U_\alpha, \phi^\alpha_t,\epsilon_\alpha\}_{\alpha\in A}$$

Now there is a result that if $$\inf_{\alpha\in A}\epsilon_\alpha>0$$ then the local family can be extended to a global one-parameter family of diffeomorphisms as follows:

Say $$\inf_{\alpha\in A}\epsilon_\alpha=\epsilon$$ . Then define

$$\phi_t(p)=\phi^\alpha_t(p)$$ where $$p\in U_\alpha$$ and $$|t|<\epsilon$$

else if $$n$$ is such that $$|\frac{t}{n}|<\epsilon$$ then $$\phi_t(p)=\underbrace{\phi^\alpha_{\frac{t}{n}}\circ\phi^\alpha_{\frac{t}{n}}\circ...\phi^\alpha_{\frac{t}{n}}}_{n\ \ \text{times}}(p)$$

The well-defined ness of the above definition etc needs to be checked here.

Now my problem here is that I don't see what problem arises when $$\inf_{\alpha\in A}\epsilon_\alpha=0$$.Given $$p$$ in a particular $$U_\alpha$$ the $$n$$ that we took above has to be adjusted so that $$|\frac{t}{n}|<\epsilon_\alpha$$ so we'd have to work with a variable $$n$$ in this case. But I don't see what harm this variable n causes to the well-definedness etc of $$\phi$$ (which has lead me to suspect that I have missed something vital that led to well-definedness in the result mentioned).

Please help see the necessity of $$\inf_{\alpha\in A}\epsilon_\alpha>0$$ in the result above.

• Hint: see what happens if $M = \mathbb{R}$ and $X(x) = x^2 \vec{e}$. Jan 16 at 11:46

It would be probably most instructive to study an example and see where the reasoning fails. Take $$M = \mathbb{R}$$ nad $$X(x) = x^2 \vec{e}$$ ($$\vec{e}$$ being the standard unit vector in $$\mathbb{R}$$). This corresponds to the ODE $$u'=u^2$$, which blows up in finite time. For example $$u(t) = \frac{1}{1-t}$$ is the solution starting from $$u(0)=1$$.
If we cover $$\mathbb{R}$$ by $$U_\alpha = (\alpha-1,\alpha+1)$$ ($$\alpha \in \mathbb{Z}$$), then you can see how $$\varepsilon_\alpha \to 0$$ with $$\alpha \to \infty$$. I would recommend tracing your reasoning on this concrete example first.
Given $$p$$ in a particular $$U_\alpha$$ the $$n$$ that we took above has to be adjusted so that $$|\frac{t}{n}|<\epsilon_\alpha$$ so we'd have to work with a variable $$n$$ in this case. But I don't see what harm this variable n causes to the well-definedness etc of $$\phi$$.
For $$\phi_t(p)=\underbrace{\phi^{\alpha_n}_{\frac{t}{n}}\circ\phi^{\alpha_2}_{\frac{t}{n}}\circ...\phi^{\alpha_1}_{\frac{t}{n}}}_{n\ \ \text{times}}(p)$$ to be well defined, you need the smallness condition to be satisfied for each $$\phi^{\alpha_k}_{\frac{t}{n}}$$ in the sequence: $$|\tfrac{t}{n}| < \varepsilon_{\alpha_1},\ldots,\varepsilon_{\alpha_n}.$$ The point here is that $$p' := \phi^\alpha_{\frac{t}{n}}(p)$$ is no longer in $$U_{\alpha}$$, but in some $$U_{\alpha_2}$$. If you want to apply $$\phi^{\alpha_{2}}_{\frac{t}{n}}$$ to it, you need another smallness condition.
• Yeah , my problem was $\underbrace{\phi^\alpha_{\frac{t}{n}}\circ\phi^\alpha_{\frac{t}{n}}\circ...\phi^\alpha_{\frac{t}{n}}}_{n\ \ \text{times}}(p)$ , the same $\alpha$ might not work all over . $\phi^\alpha_{\frac{t}{n}} (p)$ may not be in $U_\alpha$ . Jan 16 at 12:57