Compactly supported smooth vectorfield on smooth manifold is complete

76 Views Asked by At

I have problems visualizing intuitively and understanding the proof of theorem 9.16 in John Lee's "Introduction to Smooth Manifolds", 2nd ed. The proof starts

"Suppose V is a compactly supported vector field on a smooth manifold M, and let $K= supp \, V$. For each $p \in K$, there is a neighborhood $U_p$ of p and a positive number $\epsilon_p$ such that the flow of V is defined at least on $ (-\epsilon_p,\epsilon_p) \times U_p$. ..."

  1. Does the author mean a neighborhood $U_p \subseteq K$ or a neighborhood $U_p \subseteq M$?

  2. I cannot figure out, why the second sentence is true. (The rest of the proof is clear to me.)

  3. Is there any geometric intuition possible in understanding why this theorem must be true? It would be particularly interesting because of the following corollary stating that on a compact smooth manifold every smooth vector field is complete.

Thanks for any help!

1

There are 1 best solutions below

0
On

The second sentence is true because of Lemma 9.15 (Uniform Time Lemma) on Lee.

The proof of this lemma is pretty intuitive. If integral curves exists for time $(-\varepsilon,\varepsilon)$ at every point, then by appropriately translating the integral curves, you automatically get existence at all time. Now, the uniform time lemma says that if a vector field fails to be complete, then the domains of definitions of the integral curves must taper off to $0$ (e.g. the area below the graph of $y=1/x$). This cannot happen on a compact manifold.