## 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 containing a closed interval such that is contained in the from the maximal flow of . That is, assume that no matter where we start we can flow forward or backward by . But if we start at this means we can flow forward by to reach a point

But then by assumption we can flow forward again by to reach

And we can keep flowing forward to reach for arbitrarily large . 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 is compact there is. Indeed, at each point we can define the interval as the maximal open interval on which the integral curve based at is defined. Here, each and is a positive real number or . Since is compact, each of these functions has a minimum — and . And then the interval is contained in each , as asserted.

A little more analytically, let be an integral curve of a vector field , and suppose that there is a sequence increasing to for which . Then it can only make sense to define by defining for and , for continuity. But now if is the maximal integral curve of with then uniqueness tells us that for .

So what happens when is compact? In this case, any sequence of converging to gives rise to a sequence of converging to some common . Thus we can always extend any integral curve on a right-open interval to the closed interval , and then past 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 complete.

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 |