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Let $M$ be a smooth manifold. I would like to understand why the moduli space of flat $U(1)$-connections modulo gauge equivalence is the torus $$ H^1(M;\mathbb{R})/H^1(M;\mathbb{Z}). $$ How should I see this?

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Sorry, I'm not sure what you mean by the moduli space of flat $\text{U}(1)$-connections. Are these connections on the trivial bundle? –  Qiaochu Yuan Sep 1 '12 at 5:47
    
@QiaochuYuan The result holds for the moduli space of flat connections on any principal $\mathrm{U}(1)$-bundle, not just the trivial bundle. Maybe in a day or two I'll take the time to write up the details in an answer here. –  Henry T. Horton Sep 2 '12 at 5:12

1 Answer 1

I'm not sure about arbitrary manifolds, but there is an easy answer for compact Riemann surfaces.

Let $\Sigma\times G\to \Sigma$ be the trivial $G$-bundle over a compact Riemann surface $\Sigma$. There is a bijective correspondence between gauge equivalence classes of flat connections and representations of the fundamental group (in the gauge group) $\mathrm{Hom}(\pi_{1}(\Sigma),G)$, modulo conjugation by $G$.

When $G = U(1)$ you don't have much choice for a representation. The Seifert-Van Kampen theorem gives the following presentation of $\pi_{1}(\Sigma) = \langle \alpha_{1},\beta_{1},\dots,\alpha_{g},\beta_{g} \,\vert\, \prod_{i} [\alpha_{i},\beta_{i}] = 1 \rangle$. How can you describe a representation $\rho: \pi_{1}(\Sigma) \to G$? Just pick a target for each of the generators while ensuring that the relations are still satisfied. In this case, $U(1)$ is abelian so there is nothing to be done. A representation is just a selection of $2g$ points in a circle - i.e. we have that:

$\mathcal{A}_{fl} / \mathcal{G} = U(1)^{2g} = H^{1}(M;\mathbb{R})/H^{1}(M;\mathbb{Z})$

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