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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.

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Take two flat complex line bundles $L_i \to S_i$ whose Chern classes are generators. Then form $L_1 \times L_2$. Doesn't this answer your last question? –  Johannes Ebert Mar 24 '11 at 22:40
The holonomy of a flat bundle over $M$ factors through $\pi_1(M)$, so the classifying map $M\to BU(n)$ factors through $B(U(n)^d)$ ($U(n)$ with the discrete topology) so your question asks what is the map $H^*(BU(n))\to H^*(B(U(n)^d))$. The algebraic K-theory experts can probably say something, but I suspect the answer isnt completely known for $n>1$. –  Paul Mar 25 '11 at 2:40
@Paul: Maybe $GL(n,\mathbb C)$ instead of $U(n)$? A bundle admitting a flat connection might not admit an inner product compatible with a flat connection. –  Tom Goodwillie Mar 25 '11 at 2:52
Johannes - yes, thanks for pointing this out. I think I really meant to say something about classes on a product of 3 non-orientable surfaces. –  Dan Ramras Mar 25 '11 at 5:20
Paul: Morel recently proved the long-standing conjecture of Milnor that the homology of a Lie group with finite coefficients is the same if you use the discrete topology, so this yields no obstruction. To rule out an obstruction, you just need a surjection in cohomology, which is easy to see from the normalizer of a maximal torus. Also, we're probably in the stable range, which was dealt with by Suslin, decades ago. –  Ben Wieland Mar 25 '11 at 6:40
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2 Answers

Small correction: For a non-compact manifold $M$, the group $H_{2k-1}(M)$ might not be finitely generated. In this case Chern-Weil does not imply that the $k$th Chern class of a flat bundle on $M$ has finite order. Rather, it just implies that it belongs to the subgroup $Ext(H_{2k-1}(M),\mathbb Z)\subset H^{2k}(M)$.

Positive answer for first Chern class: Use the surjection $Hom(H_1(M),GL_1(\mathbb C))\to Ext(H_1(M),\mathbb Z)$ associated to the exponential exact sequence $0\to \mathbb Z\to \mathbb C\to GL_1(\mathbb C))\to 1$. An element of this $Hom$ group describes a flat complex line bundle on $M$ with prescribed Chern class in the $Ext$ part of $H^2(M)$.

Negative answer in general, for a pretty trivial reason: If $M$ is simply connected, then flat bundles on $M$ are necessarily trivial, but $M$ can still have torsion in $H^{2k}$ if $k>1$.

So a better question is, if $\Gamma$ is a group then can every element of (the $Ext$ part of ?) $H^{2k}(B\Gamma)$ be $c_k$ of a vector bundle arising from a homomorphism $\Gamma\to GL_r(\mathbb C)$ for some $r$? I don't know the answer if $k>1$.

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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 $Ext(Q_p/Z_p,Z)=0$.

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