Recently I was thinking about how local systems are the same thing as vector bundles with flat connection, and how representations of the fundamental group gave rise to vector bundles. This got me thinking about how we get a handle on line bundles in general, and made me suspect that I don't understand them as well as I should, so I thought I'd ask a novice question about them. In brief: one has two different ways of regarding line bundles on a smooth complex algebraic variety, as a set of transition functions and as an equivalence class of Weil divisors, and I want to see how the two relate.

Let's break it down and be very explicit: $E$ is an elliptic curve over $\mathbb{C}$ with (topological) fundamental group generated by $a, b$. I've chosen $E$ because I know all about its Picard group. A one-dimensional local system on $E$ corresponds to an assignation of nonzero complex numbers $\lambda, \mu$ to $a, b$, and by tensoring this setup with $\mathcal{O}$ I get the sheaf of sections of a certain line bundle, whose transition functions in an appropriate set of trivialisations should be $\lambda$ and $\mu$. I've convinced myself that different choices of $\lambda$ and $\mu$ give me non-isomorphic bundles. Now, if I fix a distinguished point $P_0$ of $E$, choosing $d \in \mathbb{Z}$ and $P \in E$ is the same thing as choosing an isomorphism class of line bundles on $E$, whose associated divisor will be $P + (d-1)P_0$.

So how are $\lambda$, $\mu$ and $P, d$ related? Is there a nice formula? What about for higher genus curves? Am I just confused?

  • 3
    $\begingroup$ As you mention, local systems are bundles equipped with flat connections. This means that, first of all, you only get line bundles of degree $d=0$, and secondly, it is not true that the assignment of a line bundle to $\lambda,\mu$ is one-to-one (because you forget the connection). For a curve of any genus, the variety of rank one local systems is isomorphic to $({\mathbb C}^\times)^{2g}$, and the variety of degree zero line bundles is a complex torus $J$ (its Jacobian). $J$ is identified with a quotient of $({\mathbb C}^\times){2g}$ by a subgroup isomorphic to ${\mathbb C}^g$. $\endgroup$
    – t3suji
    Feb 27, 2010 at 14:40

1 Answer 1


As was noted in the comment, $d$ must be $0$ (bundles with a flat connection can have only degree $0$), and different connections can lead to the same bundle. However, every line bundle of degree zero on an elliptic (or higher genus) curve has a unique flat unitary connection, i.e. in your example with $|\lambda|=|\mu|=1$. This identifies the Jacobian of any smooth projective complex curve with a product of $2g$ circles (once you fix a basis of $H_1(C,\Bbb Z)$). For an elliptic curve, this is the usual identification with the 2-torus.


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