Metric associated to a Connection on a Vector Bundle - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T17:46:56Z http://mathoverflow.net/feeds/question/90707 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90707/metric-associated-to-a-connection-on-a-vector-bundle Metric associated to a Connection on a Vector Bundle Giovanni De Gaetano 2012-03-09T15:25:06Z 2012-03-20T02:03:42Z <p><strong>General question</strong>: Given a vector bundle $E \rightarrow M$ on a complex manifold $M$, and a connection $\nabla$ on $E$, is it possible to find an Hermitian structure on $E$ such that $\nabla$ is the associated metric connection (i.e. the unique connection compatible with both the metric and the complex structure)?</p> <p><strong>Specific question</strong>: Given a line bundle $L \rightarrow X$ on a compact Riemann surface $X$ equipped with a flat connection $\nabla$, is it possible to find an Hermitian structure on $L$ such that $\nabla$ is the associated metric connection? </p> <p><strong>Motivation</strong>: I´m trying to prove that a degree zero line bundle on a compact Riemann surface always admits an harmonic Hermitian metric. By a classical result of Weil and Atiyah every degree zero vector bundle admits a flat connection, moreover I think (though I still did not prove it) that the flatness condition on the connection could be translated (by computation on the vector fields $\partial z$ and $\partial \overline{z}$) into the harmonicity condition on the "associated" (in the sense of the question) Hermitian metric.</p> <p>The question is clearly related to the well discussed question: <a href="http://mathoverflow.net/questions/54434/when-can-a-connection-induce-a-riemannian-metric-for-which-it-is-the-levi-civita" rel="nofollow">When can a Connection Induce a Riemannian Metric for which it is the Levi-Civita Connection</a>. But I´m not able to adjust the proof given in the mentioned question to an answer for my own. I´m asking also a general version of the question because I´m just curious about the answer.</p> <p>Thank you for your time!</p> <p><strong>Edit</strong>: As pointed out by David Speyer the metric connection is constructed taking as input an Hermitian structure on the bundle and not an Hermitian metric on the manifold. I changed both the questions consequently.</p> http://mathoverflow.net/questions/90707/metric-associated-to-a-connection-on-a-vector-bundle/90718#90718 Answer by alvarezpaiva for Metric associated to a Connection on a Vector Bundle alvarezpaiva 2012-03-09T16:03:01Z 2012-03-09T16:03:01Z <p>I think the answer to the general question is bound to be no for the following reason: the holonomy group of metric connections is compact (a subgroup of the unitary group in your case), while this is not the case of a more general connection.</p> http://mathoverflow.net/questions/90707/metric-associated-to-a-connection-on-a-vector-bundle/91632#91632 Answer by YangMills for Metric associated to a Connection on a Vector Bundle YangMills 2012-03-19T16:38:39Z 2012-03-20T02:03:42Z <p>alvarezpaiva's answer shows that the answer to your question is no in general.</p> <p>I will answer your motivation question instead, for which the answer is positive.</p> <p>If $L$ is a holomorphic line bundle over a compact K&auml;hler manifold $(M^n,\omega)$ with $\omega$-degree zero, then there is a Hermitian metric on the fibers of $L$ with curvature $F$ and $\omega$-trace of the curvature equal to zero (what you call a harmonic Hermitian metric)</p> <p>$$\omega^{n-1}\wedge F=0.$$</p> <p>This is just the Hermitian-Yang-Mills equation, and I am saying that it can always be solved in the case of line bundles.</p> <p>This follows from simple Hodge theory. Start with any Hermitian metric on $L$, and call $F$ its curvature $(1,1)$-form. Since $L$ has $\omega$-degree zero, you have that $$\int_M \omega^{n-1}\wedge F=\int_M \Lambda F \omega^n=0.$$ Now solve the Poisson equation $$\Delta h=\Lambda F,$$ which can be done because of the integral of the RHS being zero. It follows that the new Hermitian metric on $L$ that you obtain by conformally rescaling the one you have by $e^h$, has curvature $$F_h=F-i\partial\overline{\partial}h$$ and by construction $$\omega^{n-1}\wedge F_h=0,$$ which is what you want.</p>