[Math] To show $X$ is a complete vector field on $M$

differential-geometry

Well, I have solved myself the problem : every smooth vector field on a compact manifold is complete.

Now I have got this problem which I am not able to progress:

let $X$ is a vector field on $M$, suppose $\exists \epsilon >0\ni (-\epsilon,\epsilon) \subsetneq (a(m),b(m))\forall m\in M$, Then show that $X$ is a complete vector field i.e $(a(m), b(m))=\mathbb{R}$

thank you for any help.

where $(a(m),b(m))$ is the domain of maximal integral curve, for each $m\in M$ we will get this interval $(a(m),b(m))$

here I asked some realted things

A series of Lemmas about $C^{\infty}$ vector fields

Best Answer

Let $\gamma(t)$ be the maximal integral curve with initial condition $\gamma(0) = m$, and let's suppose $b(m) < \infty$.

If we denote $\gamma'(t)$ the maximal integral curve with initial condition $\gamma'(0) = \gamma(b(m) -\epsilon/2)$ then $$ \gamma''(t) = \begin{cases} \gamma(t) & \text{if } t\in (a(m), b(m) - \epsilon/2)\\ \gamma'(t - b(m) + \epsilon/2)& \text{if } t\in [b(m) - \epsilon/2, b(m) + \epsilon/2) \end{cases} $$ prolongs $\gamma(t)$. That contradicts the assertion that $\gamma(t)$ is maximal.

A similar reasoning can be used to show $a(m) = -\infty$.

Edit - Some clarifications

$\gamma''$ prolongs $\gamma$ means that $\gamma''$ has an interval of existence, $(a(m), b(m) + \epsilon/2)$, that is a proper superset of the interval of existence, $(a(m), b(m))$, of $\gamma$ and the two curves are equal on the common domain, $(a(m), b(m))$.

The "uniformity" of $\epsilon$ ensures that the integral curve with initial condition $$ \gamma'(0) = \gamma(b(m) - \epsilon/2) $$ exists on the interval $[0, \epsilon)$. Without "uniformity" it could happen that the maximal right interval of existence of $\gamma'$ is $[0, \epsilon/2)$. In such a case, a curve $\gamma''$, constructed as above, would coincide with $\gamma$: it would not prolong $\gamma$.

The initial condition of $\gamma'$ is chosen in such a way to allow us to smoothly join $\gamma'$ to $\gamma$ in order to form a new curve, $\gamma''$, that prolongs $\gamma$.

The contradiction arises because we constructed an integral curve that prolongs a maximal integral curve, which, as such, cannot be prolonged by definition.