# First Chern class of a flat line bundle

A referee asked me to include a reference or proof for the following classical fact. It's not hard to prove, but I'd prefer to just give a reference -- does anyone know one?

Let $X$ be a nice space (eg a smooth manifold, or more generally a CW complex). The topological Picard group $Pic(X)$ is the set of isomorphism classes of $1$-dimensional complex vector bundles on $X$. The set $Pic(X)$ is an abelian group with group operation the fiberwise tensor product, and the first Chern class map

$$c_1 : Pic(X) \longrightarrow H^2(X;\mathbb{Z})$$

is an isomorphism of abelian groups.

Now make the assumption that $H_1(X;\mathbb{Z})$ is a finite abelian group. One nice construction of elements of $Pic(X)$ is as follows. Consider $\phi \in Hom(H_1(X;\mathbb{Z}),\mathbb{Q}/\mathbb{Z})$. Let $\tilde{X}$ be the universal cover, so $\pi_1(X)$ acts on $\tilde{X}$ and $X = \tilde{X} / \pi_1(X)$. Let $\psi : \pi_1(X) \rightarrow \mathbb{Q}/\mathbb{Z}$ be the composition of $\phi$ with the natural map $\pi_1(X) \rightarrow H_1(X;\mathbb{Z})$. Define an action of $\pi_1(X)$ on $\tilde{X} \times \mathbb{C}$ by the formula

$$g(p,z) = (g(p),e^{2 \pi i \psi(g)}z) \quad \quad \text{for g \in \pi_1(X) and (p,z) \in \tilde{X} \times \mathbb{C}}.$$

Observe that this makes sense since $\psi(g) \in \mathbb{Q} /\mathbb{Z}$. Define $E_\phi = (\tilde{X} \times \mathbb{C}) / \pi_1(X)$. The projection onto the first factor induces a map $E_{\phi} \rightarrow X$ which is easily seen to be a complex line bundle. The line bundle $E_{\phi}$ is known as the flat line bundle on $X$ with monodromy $\phi$.

Now, the universal coefficient theorem says that we have a short exact sequence

$$0 \longrightarrow Ext(H_1(X;\mathbb{Z}),\mathbb{Z}) \longrightarrow H^2(X;\mathbb{Z}) \longrightarrow Hom(H_2(X;\mathbb{Z}),\mathbb{Z}) \longrightarrow 0.$$

Since $H_1(X;\mathbb{Z})$ is a finite abelian group, there is a natural isomorphism $\rho : Hom(H_1(X;\mathbb{Z}),\mathbb{Q}/\mathbb{Z}) \rightarrow Ext(H_1(X;\mathbb{Z}),\mathbb{Z})$. We can finally state the fact for which I am looking for a reference :

$$c_1(E_{\phi}) = \rho(\phi).$$

• I'm hoping someone answers with a proof, and then you can give the referees this MO question as a reference. – Dylan Wilson Feb 20 '11 at 20:32
• @Mohan : (some more details) Flat just means that you can find a connection whose curvature vanishes. The Chern-Weil homomorphism will then be zero, but this only gives you an element of de Rham cohomology, so it doesn't see torsion phenomena. – Andy Putman Feb 20 '11 at 21:36
• @Andy:Thanks for clearing up my confusion. – Mohan Ramachandran Feb 20 '11 at 21:39
• You could just redefine $Pic(X)$ to be $H^1(X;\mathcal{C}^\times)$ (cohomology of the sheaf of continuous functions into $\mathbb{\C}^\times$). Then your construction is just the map on $H^1$ induced by $\mathbb{Q}/\mathbb{Z}\to \mathcal{C}^\times$, and the boundary map $H^1(X;\mathcal{C}^\times)\to H^2(X;\mathbb{Z})$ is $c_1$. – Charles Rezk Feb 20 '11 at 23:20
• Not a reference, but one proof goes a little like this: Think about $R/Z$ instead of $Q/Z$. Then $K(Z,2)=BU(1)=B(R/Z)$. Giving $R/Z$ the discrete topology gives a homomorphism of topological groups $Id:(R/Z)^d\to R/Z$ and hence a map on classifying spaces $K(R/Z,1)\to K(Z,2)$. This induces the Bockstein $H^1(-;R/Z)\to H^2(-;Z)$ whose image is the set of complex line bundles whose $c_2$ is sent to 0 in $H^2(-;R)$. Then to see the relation to your construction you just have to consider the "universal" case of the flat bundle over $K(R/Z,1)$. – Paul Feb 21 '11 at 0:31