Suppose we have a smooth compact manifold $M$ with boundary and a vector field $X$ on $M$. The maximal integral curves on $M$ are unique and hence their images give a partition of $M$.
Let $p\in M$ be a point in $M$, such that $X(p)=0$. The constant curve at $p$ fulfills the conditions on an integral curve and hence it is the unique maximal integral curve starting at $p$. This shows that the zero set of the vector field is in one-to-one correspondence to the constant maximal integral curves, right?
Now come my confusion: For sure there are vector fields $X$ with isolated zero-points, such that some inetgral curves of a point non zero $q$ runs in the direction of a zero point $p$. (Think of a gradient-like vector field for a morse function.) By the thoughts above, this curve cannot run into $p$, as the integral curve of the zero point is constant and unique, so my conclusion is, that the curve going to the $p$ has to stop right before $p$. But as $M$ is compact, integral curves are defined on the whole real line, so this tastes like a contradiction.
Where is my mistake?
Best Answer
The prototypical example for this situation is something like the vector field $-x \partial_x$ on a one-dimensional manifold. This has a zero at $x=0$, and indeed the flow at $x=0$ is constant; but this does not stop the flow starting at any other point from being maximally defined.
Your mistake is in confusing "defined on all of $\mathbb R$" with a statement about the image of the curve. In my example above, the integral curve starting at $x_0$ is simply $x(t) = x_0 e^{-t}$, which is defined on all of $\mathbb R$, but never hits $0$ - it just asymptotically approaches it. The fact that the image of the curve could be extended (i.e. the curve could be extended after a reparametrization) does not mean that the curve itself is not maximally defined.