I have the following definition for Weinstein manifolds in my lecture:
A Weinstein manifold is an exact symplectic manifold $(W, \omega= d \lambda)$, whose associated Liouville vector field is complete (i.e. the flow exists for all times) and gradient-like for a smooth exhausting function $f:W \rightarrow \mathbb{R}$.
Now this confuses me, since $\omega = d \lambda$ means $0=[\omega] \in H^2(W)$, right? But for symplectic forms, it always holds that $[\omega]\neq 0$.
Also, it says that Stein manifolds give examples for Weinstein manifolds, but I can't see why, since Stein manifolds don't need to have a complex structure.
What is my mistake here?
(Answering just to get this off the unanswered page. Community wiki because I did nothing.)
As studiosus mentions in the comments, it is true for closed (compact, no boundary) symplectic manifolds that $[w]\in H^2(M;\mathbb{R})$ is non-zero. But in general, it can be the zero class. Consider, for example, the symplectic form $\omega = dx\wedge dy$ on $\mathbb{R}^2$.