In this paper we show that given any 3-manifold $N$ and any non-fibered class
in H^1(N;Z) there exists a representation such that the corresponding twisted
Alexander polynomial is zero. This is obtained by extending earlier work of the
authors, together with results of Agol and Wise on separability of 3-manifold
groups. This result allows us to completely classify symplectic 4-manifolds
with a free circle action, and to determine their symplectic cones.
Let $M$ be a closed 4-manifold with a free circle action. If the orbit
manifold $N^3$ satisfies an appropriate fibering condition, then we show how to
represent a cone in $H^2(M;\R)$ by symplectic forms. This generalizes earlier
constructions by Thurston, Bouyakoub and Fern\'andez-Gray-Morgan. In the case
that $M$ is the product 4-manifold $S^1\times N$ our construction complements
the results of \cite{FV08} (arXiv:0805:1234 [math.GT]) and allows us to
completely determine the symplectic cone of such 4-manifolds.
In this paper we use the Lubotzky alternative for finitely generated linear
groups to determine which 4-manifolds admitting a free circle action can be
endowed with a symplectic structure with trivial canonical class. The content
of this paper partly overlaps with the content of the unpublished preprint
"Symplectic 4-manifolds with a free circle action" (arXiv:0801.1313 [math.GT]).
In a series of papers the authors proved that twisted Alexander polynomials
detect fibered 3-manifolds, and they showed that this implies that a closed
3-manifold N is fibered if and only if S^1 x N is symplectic. In this note we
summarize some of the key ideas of the proofs. We also give new evidence to the
conjecture that if $ is a symplectic 4-manifold with a free S^1-action, then
the orbit space is fibered.