# The Unapologetic Mathematician

## Vector Fields on Compact Manifolds are Complete

It turns out that any vector field on a compact manifold is complete. That is, starting at any point we can follow the vector field on construct its integral curve as far forward or as far backward as we want. I’ll show this two ways.

First of all, let’s say that there is some open interval $I$ containing a closed interval $[-\delta,\delta]$ such that $I\times M$ is contained in the $W$ from the maximal flow of $X$. That is, assume that no matter where we start we can flow forward or backward by $\delta$. But if we start at $p_0$ this means we can flow forward by $\delta$ to reach a point

$\displaystyle p_1=\Phi_\delta(p_0)=\Phi(\delta,p_0)$

But then by assumption we can flow forward again by $\delta$ to reach

$\displaystyle p_2=\Phi_\delta(p_1)=\Phi_\delta(\Phi_\delta(p_0))=\Phi_{2\delta}(p_0)$

And we can keep flowing forward to reach $p_n=\Phi_{n\delta}(p_0)$ for arbitrarily large $n$. Similarly we can flow backward as far as we want.

So, is there such an interval? In general there doesn’t have to be, but if $M$ is compact there is. Indeed, at each point $p$ we can define the interval $I_p=\left(-a(p),b(p)\right)$ as the maximal open interval on which the integral curve based at $p$ is defined. Here, each $a(p)$ and $b(p)$ is a positive real number or $\infty$. Since $M$ is compact, each of these functions has a minimum — $a$ and $b$. And then the interval $(a,b)$ is contained in each $I_p$, as asserted.

A little more analytically, let $\phi:[\alpha,\beta)\to M$ be an integral curve of a vector field $X$, and suppose that there is a sequence $t_n$ increasing to $\beta$ for which $\lim\phi(t_n)\to p$. Then it can only make sense to define $\bar{\phi}:[\alpha,\beta]\to M$ by defining $\bar{\phi}(t)=\phi(t)$ for $\alpha\leq<\beta$ and $\bar{\phi}(\beta)=p$, for continuity. But now if $c:I\to M$ is the maximal integral curve of $X$ with $c(\beta)=p$ then uniqueness tells us that $c(t)=\bar{\phi}(t)$ for $\alpha\leq t\leq\beta$.

So what happens when $M$ is compact? In this case, any sequence of $t_n$ converging to $\beta$ gives rise to a sequence of $\phi(t_n)$ converging to some common $p$. Thus we can always extend any integral curve on a right-open interval $[\alpha,\beta)$ to the closed interval $[\alpha,\beta]$, and then past $\beta$ to a somewhat longer right-open interval, and so on as far as we want to go. And thus, again, every integral curve can be extended forever in either direction, which makes $X$ complete.

Advertisements

June 1, 2011 - Posted by | Differential Topology, Topology

## 1 Comment »

1. Hi,

I agree with the second proof you give, but the first one seems incomplete to me: For the functions a(p) and b(p) to have a minimum, you need that they are continuous. I don’t see any direct reason why this should be the case. It would certainly not be true for non-compact M: Tage e.g. M = R^2 – 0, and a constant vector field in x-direction. Then these functions will jump when we cross the x-axis.

Is there an argument why this cannot occur in the compact case? (Except the one that, because of the other proof, these functions are constantly infinity anyway đ )

Comment by Thiagor | November 15, 2011 | Reply