Connection on line bundle on projective curve - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T21:12:37Z http://mathoverflow.net/feeds/question/22371 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22371/connection-on-line-bundle-on-projective-curve Connection on line bundle on projective curve xiyu 2010-04-23T17:13:27Z 2010-04-23T18:36:01Z <p>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.</p> <p>Now my question is: Given a line bundle $L$ of degree 0, if there exist connection structure on $L$.</p> <p>When $C$ is projective line, the question is obvious, how about when the genus is greater than 0?</p> http://mathoverflow.net/questions/22371/connection-on-line-bundle-on-projective-curve/22379#22379 Answer by David Ben-Zvi for Connection on line bundle on projective curve David Ben-Zvi 2010-04-23T18:04:19Z 2010-04-23T18:36:01Z <p>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 <a href="http://arxiv.org/abs/math/0211441" rel="nofollow">here</a>).</p> <p>[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.]</p> <p>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.</p> http://mathoverflow.net/questions/22371/connection-on-line-bundle-on-projective-curve/22381#22381 Answer by Sebastian for Connection on line bundle on projective curve Sebastian 2010-04-23T18:04:47Z 2010-04-23T18:04:47Z <p>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. </p> <p>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. </p>