TITLE: Which torsion classes in integral cohomology are Chern classes of flat bundles?
QUESTION [13 upvotes]: Chern-Weil theory tells us that the integral Chern classes of a flat bundle over a compact manifold (i.e. a bundle admitting a flat connection) are all torsion.  Given a compact manifold $M$ whose integral cohomology contains torsion, one can then ask which (even-dimensional) torsion classes appear as the Chern classes of flat bundles.  What is known about this question?  I would be interested both in statements about specific manifolds and about general (non)-realizability results.
One specific thing that I know: if $S$ is a non-orientable surface, then there is a flat bundle $E\to S$ whose first Chern class is the generator of $H^2 (S; \mathbb{Z}) = \mathbb{Z}/2$.  This shows up, for example, in papers of C.-C. Melissa Liu and Nan-Kuo Ho.  As Johannes pointed out in the comments, this also shows that the fundamental class of a product of surfaces can be realized by a flat bundle.
However, I suspect that for a product of 3 Klein bottles, not all the 4-dimensional torsion classes can be realized as second Chern classes of flat bundles.  In fact, I think I know a proof of this if one restricts to unitary flat connections: the space of unitary representations has too few connected components.

REPLY [3 votes]: The answer to Tom's formulation is no. It's possible if you restrict to finitely generated groups that my argument falls apart, but I doubt this is essential.
Take a group $\Gamma$ so that $B\Gamma^+=K(Q/Z,2n-1)$, ie, $H^k(\Gamma;Z)=H^k(K(Q/Z,2n-1);Z)$. Since $Ext(Q/Z,Z)=\hat Z$, there lots of interesting classes in $H^{2n}(\Gamma;Z)$. If we could lift them to flat bundles over $B\Gamma$, then after applying the plus construction and profinite completion, we would have split $K(\hat Z;2n)$ off of $BU^{\hat{}}$. But the torsion homology of the Eilenberg-MacLane space cannot be a retract of the torsion-free homology of $BU$.
I wanted to work one prime at a time, but $Ext(Q_p/Z_p,Z)=0$.