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 ![[-\delta,\delta]](http://s0.wp.com/latex.php?latex=%5B-%5Cdelta%2C%5Cdelta%5D&bg=e6e6e6&fg=333333&s=0 "[-\delta,\delta]")
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 ![\bar{\phi}:[\alpha,\beta]\to M](http://s0.wp.com/latex.php?latex=%5Cbar%7B%5Cphi%7D%3A%5B%5Calpha%2C%5Cbeta%5D%5Cto+M&bg=e6e6e6&fg=333333&s=0 "\bar{\phi}:[\alpha,\beta]\to M")
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 ![[\alpha,\beta]](http://s0.wp.com/latex.php?latex=%5B%5Calpha%2C%5Cbeta%5D&bg=e6e6e6&fg=333333&s=0 "[\alpha,\beta]")
, 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.
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June 1, 2011 -
Posted by John Armstrong |
Differential Topology, Topology
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
)