# Connection on line bundle on projective curve

Let $C$ be a smooth projective curve. It is known that a line bundle on $C$ is of degree 0, if we can impose a connection structure on it.

Now my question is: Given a line bundle $L$ of degree 0, if there exist connection structure on $L$.

When $C$ is projective line, the question is obvious, how about when the genus is greater than 0?

• What is a connection structure. Although I work in the field of Riemann surfaces and differential geometry, I have never seen that. Commented Apr 23, 2010 at 17:40
• I mean holomorphic connectioin. In my mind, a connection on $L$ is an operator $\nabla: L\mapsto \Omega \otimes L$, which satisfies Leibniz rule.
– JJH
Commented Apr 23, 2010 at 17:48

Yes (we're over C yes?)- the space of line bundles with connection forms a torsor for the cotangent bundle over the Jacobian. The class of this torsor (as an element in $H^1(Jac,\Omega^1)=H^{1,1}$) is the Chern class of the theta line bundle. In fact one can construct canonical connections on generic line bundles of degree zero on a curve (in fact line bundles outside the theta divisor), once you give yourself the choice of a theta characteristic on the curve -- this is the theory of the "prime form" or Szego kernel (it's much much older, but I think a fairly easy "modern" exposition of this and the nonabelian version is in here).

[EDIT: Nonalgebraically it is very easy to see this assertion from the Hodge theorem: the space of line bundles with a flat connection is $H^1(X,C^\times)$, which can be identified with the product $H^1(X,O^\times)^{\circ}\times H^0(X,\Omega^1)$ just by exponentiating the Hodge theorem for H^1. Note that ANY holomorphic connection on a Riemann surface/algebraic curve is flat for dimension reason - there are no holomorphic 2-forms that could serve as curvature.. For higher rank bundles the analogous result is the topic of the Corlette-Simpson nonabelian Hodge theorem.]

More generally Andre Weil proved that a rank n bundle on a curve /C admits a connection if and only if every indecomposable summand is a vector bundle of degree zero. There's a nice proof of this by Atiyah as an application of the Atiyah class (which is the canonical obstruction to the existence of a connection on a vector bundle). In particular over the moduli of stable degree zero bundles we again have a nontrivial torsor for the cotangent bundle parametrizing bundles with connection, and the class of this torsor is again the Chern class of the determinant line bundle.

• Since you mention dependence on the base field, could you maybe make some comments on the situation in positive characteristic or over non algebraically closed fields?
– Lars
Commented Apr 23, 2010 at 18:12
• I want to be careful so won't comment in total generality, but in char=p the same is true, and in fact things are much easier to understand.. see arxiv.org/abs/math/0602255 for a detailed analysis of the spaces of vector bundles with connection over a curve over an algebraically closed field in char p, which are almost the same thing as just vector bundles with endomorphism-valued one-forms.. Commented Apr 23, 2010 at 18:42
• David, I can't understand your argument in the first paragraph. Isn't the dimension of space of line bundles with connection the same as the dimension of cotangent bundle of Jacobian? How come one is the torsor of the other? I have understood the analytic situation, is there any easy algebraic proof?
– JJH
Commented Apr 23, 2010 at 22:39
• yes - the difference of any two connections on a given line bundle is a one-form, which is the same as a cotangent vector to the Jacobian, so connections form a torsor for the cotangent bundle. The former is a twisted form of the latter, in particular they have the same dimension.. Commented Apr 24, 2010 at 5:22

I also wanted to add a more elementary algebraic way of looking at this (compared to David's answer). Namely, let's write your line bundle as a divisor, so $L\simeq O(D)$. Outside of the support of $D$, $L$ is trivial, and so a connection on $L$ is the same as a connection on $O$, that is, a differential $1$-form $\omega$. But this holds only outside of the support of $D$: $\omega$ can have poles along $supp(D)$. A local calculation shows that for $\omega$ to give a regular connection on $L$, $\omega$ must have first-order poles at $x_i$ with residue $-k_i$, where $D=\sum k_ix_i$. So now the claim is reduced to the classical statement: a rational differential form with prescribed residues exists iff the sum of residues is zero. (Which may be viewed as an explicit calculation of $H^1(C,\Omega_C)$.)

Yes, you can: Every holomorphic line bundle of degree $0$ has a flat connection which is compatible with the holomorphic structure, i.e. $\nabla s=w_s s$ for local holomorphic sections, where $w_s\in\Gamma(M,K).$ But flatness implies $d w_s=0.$ Thus $\nabla$ is a holomorphic connection.

To see the existence of a flat connection compatible with the holomorphic structure, take any connection $\tilde\nabla$ compatible with the holomorphic structure. The curvature $F$ satisfies $\int_MF=0,$ so Serre duality implies the existence of a section $\eta\in\Gamma(M;K)$ with $d\eta=F.$ Then $\nabla=\tilde\nabla-\eta$ is flat and also compatible with the holomorphic structure.