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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.

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

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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.


But here is what went wrong:

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.

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  • $\begingroup$ 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 $U_\alpha$ . $\endgroup$
    – aritracb
    Jan 16 at 12:57

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