Non-vanishing vector fields on non-compact manifolds

Question

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 $\partial M$?

Answer 1

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 $C_{n-1}$ 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 $\partial M$.

Answer 2

When the boundary is non-empty you can't put the transverse restriction (3.2) on, since the sum of the indices of the zeros of the vector field is the Euler characteristic, provided it is outward-pointing on the boundary. This is the Poincare-Hopf index theorem.

Similarly you can't demand (3.1), for example take the case 2-manifolds -- if the 2-manifold is oriented you could use a rotation of the vector field to satisfy (3.2).

The idea for non-compact manifolds is to take a handle decomposition of the manifold. You can define your vector field rather easily in a neighbourhood of the $n-1$-skeleton. So find a handle decomposition that does not use any $n$-cells. You can do this by a maximal tree construction in the dual 1-skeleton.