Are curvature forms in complex line bundles symplectic

Question

I know that the curvature form $F_\nabla$ of a connection $\nabla$ in a complex line bundle $L \to B$ is presymplectic (i.e. antisymmetric and closed). Does it also have to be non-degenerate, i.e symplectic (I am excluding the obvious case $F_\nabla = 0$)?

Answer 1

As in your previous question you can make the curvature form anything you want, provided it's closed and lives in the right cohomology class. Most 2-forms are not nondegenerate. In particular, there are manifolds that support no symplectic form (take $S^4$, say...) or cohomology classes $c_1$ such that $c_1^n$ is zero in cohomology, hence cannot be symplectic on a closed manifold.