TITLE: Non-vanishing vector fields on non-compact manifolds
QUESTION [15 upvotes]: In several papers the following result is invoked:
Theorem. Every connected, non-compact, smooth manifold $M$ carries non-vanishing smooth vector fields $v$.
(we are assuming $M$ is $2$nd countable and Hausdorff. The case $\partial M \neq \emptyset$ reduces to the case $\partial M = \emptyset$, taking the double of $M$).
Questions:
1) where is this proved?
2) is there a "simple" proof of this result?
3) when $\partial M \neq \emptyset$, can we further require that
$\,$ 3.1) $v$  is tangent to $\partial M$
$\,$ 3.2) $v$ is transverse to $\partial M$?

REPLY [2 votes]: Here is a proof for 2. First of all, you can take a vector field $V_0$ in $M$ with isolated zeros as shown here. The idea of the proof is basically "pushing all the zeros to infinity".
Take a covering of the manifold $M$ by compacts $C_n$, $n\geq1$, with $C_n$ contained in the interior of $C_{n+1}$ $\forall n$. We´ll also need $M\setminus C_n$ not to have any component which is relatively compact in $M$ ($*$). You can achieve this by adjoining to $C_n$ all the relatively compact components of $M\setminus C_n$ (the resulting set is compact, see this).
Now we are going to inductively define a field $V_n$ that has no zeros in $C_n$. To do that, we take the vector field $V_n=(\phi_n)_*V_{n-1}$, where:

By induction hypothesis, $V_{n-1}$ has isolated zeros, and none of them are in $C_{n-1}$.

$\phi_n:M\to M$ is an diffeomorphism fixing $V_{n-1}$ and taking all the (finitely many) zeros of $V_{n-1}$ that were inside $C_n$ outside of it. (To make sure this exists you need the condition ($*$), which implies that every point outside $C_{n-1}$ is connected by a path in $M\setminus C_{n-1}$ to some point outside $C_n$, so you can take it outside $C_n$ by a diffeo fixing $C_{n-1}$).


Now as the sequence $V_n$ of vector fields is eventually constant in any compact subset of $M$, you can consider its limit $V$, which is equal to $V_n$ in $C_n$ so it can´t have any zeros in $C_n$ $\forall n$, meaning it has no zeros.
For 3.1 (or 3.2) to be true it is of course necessary that $\partial M$ has a tangent (transverse) vector field without zeros. This condition will also be sufficient if there is a vector field on $M$ which is tangent (transverse) to $\partial M$ and with isolated zeros outside of $\partial M$, because then we can apply the same proof as for $2)$. There surely is some version of the transversality theorem which implies the existence of such a vector field (if someone provides a reference that would be great). In that case, 3.1 only depends on the topology of $\partial M$ and 3.2 only depends on whether there is a non trivial section of the normal bundle of $\partial M$.