I have the following, maybe naive question:
Given a smooth vector field $X$ on a smooth manifold $M$ with $\dim M \geq 1$. Then this defines a linear operator $$ D: C^\infty(M,\mathbb C) \to C^\infty(M,\mathbb C) ; f \mapsto X.f\,. $$ What I would like to know is: What is known about the eigenvalues of this linear map $D$ ?
It is clear to me that constant maps are always mapped to $0$, so $0$ is always an eigenvalue. It is also clear that if $X=0$, then $D=0$ and there is no other eigenvalue.
But assume, $X$ is not the constant $0$-vector field. Does this imply that $D$ has a complex non-zero eigenvalue ? Does this depend on whether $M$ is compact or not?
EDIT: It looks to me as if $X$ being not constantly zero is not enough to conclude the existence of a non-zero eigenvalue. Motivated by the comment of Ben Grossmann I looked at the real line $\mathbb R$ and the circle $\mathbb R/\mathbb Z$ and it seems that $X(x)=\cos^2(x)$ should be an example of a vector field where $D$ has only the eigenvalue $0$ -- if my calculus was correct.
So, maybe the correct question is: Assume that $X$ is everywhere non-zero, does this imply that we have a non-zero eigenvalue ?
The comment of Ben Grossmann answeres this in the affirmative for $M=\mathbb R$ and I think a very similar argument should work for the circle $M=\mathbb R/\mathbb Z$. The comment of Eric solves this in the affirmative locally but it is not clear how the global geomety of $M$ comes into play here.