Line bundles with complex connection - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:55:30Z http://mathoverflow.net/feeds/question/86355 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86355/line-bundles-with-complex-connection Line bundles with complex connection Blake 2012-01-22T05:25:05Z 2012-01-23T22:24:03Z <p>Suppose that we have a complex manifold $X$, and a line bundle $L$ over $X$. It is known that the line bundles over $X$ are parametrized by their Chern class, the Chern class being the cohomology class of the curvature of a connection on $L$, which must be integral. Thus $L$ admits a connection with curvature $\omega$ iff $[\omega]$ is an integral cohomology class.<br> My question is this: what if we are interested instead in complex connections on $L$ (connections that are $\mathbb{C}$-linear instead of $\mathbb{R}$-linear)? Given a complex (2,0)-form $\omega$ on $X$, under what circumstances will $L$ admit a complex connection with curvature $\omega$?</p> http://mathoverflow.net/questions/86355/line-bundles-with-complex-connection/86364#86364 Answer by Georges Elencwajg for Line bundles with complex connection Georges Elencwajg 2012-01-22T10:24:22Z 2012-01-22T13:53:05Z <p>I interpret your line bundle $L$ to be differentiable rather than holomorphic, else the statement that line bundles are parametrized by their Chern class would be completely false. </p> <p>There always exists a hermitian structure on $L$ and if you choose one, there exists a complex connexion $\nabla$ on $L$ compatible with the hermitian structure on $L$.<br> The curvature $\omega \in \mathcal E^{1,1}(X)$ of the connexion $\nabla$ is a $(1,1)$ form [and not a (2,0) form!] whose De Rham class computes the first Chern class of $L$ by the formula<br> $$c_1(L)=[\frac {i}{2\pi}\omega ] \in H^{1,1} (X,\mathbb C)\subset H^2_{DR} (X,\mathbb C)$$ </p> <p>That Chern class is independent of the chosen hermitian structure on $L$ and is integral in the sense that it lives in the image of the natural group morphism (where $H^2_{sing}$ is the topologists' singular homology) </p> <p>$$H^2_{sing}(X,\mathbb Z)\to H^2_{sing}(X,\mathbb C)=H^2_{DR} (X,\mathbb C)$$</p> <p><strong>NB</strong> My answer addresses the <em>cohomology class</em> $[ \omega]$ of the curvature. I don't know which $(1,1)$ actual forms $\omega$ you can reach by this procedure.<br> (You claim that you can reach any $\omega$ with integral class in the real case. I don't know why this is true). </p> http://mathoverflow.net/questions/86355/line-bundles-with-complex-connection/86491#86491 Answer by Xin Nie for Line bundles with complex connection Xin Nie 2012-01-23T22:24:03Z 2012-01-23T22:24:03Z <p>If I understood this right, the answer to your original question is somewhat trivial. You may check that if a complex connection $\nabla$ on $L$ has curvature $\omega$, then the cohomologous $2$-form $\omega+d\alpha$ is the curvature of the connection $\nabla+\alpha$. Hence any closed $2$-form representing integral class can be the curvature form of a complex connection on a line bundle (which just means a $C^\infty$ vector bundle, not holomorphic one).</p> <p>However, there is a version of your question which admits a non-trivial answer. </p> <p>If you take $L$ to be a <strong>holomorphic</strong> line bundle and endow it with a Hermitian metric then there is an unique Hermitian $(1,0)$-connection called the Chern connection, its curvature form $\omega$ is a real-valued (1,1)-form (real-valued means that $\omega=\bar{\omega}$) representing an integral class. </p> <p>A non-trivial result is that the converse is also true when $X$ is compact Kähler, i.e., any real-valued (1,1)-form representing an integral class is the curvature form of the Chern connection on a Hermitian holomorphic line bundle. This is a consequence of the $\partial\bar\partial$-lemma and the Lefschetz theorem on $(1,1)$-classes. </p>