General question: 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)?

Specific question: 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?

Motivation: 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.

The question is clearly related to the well discussed question: When can a Connection Induce a Riemannian Metric for which it is the Levi-Civita Connection. 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.

Thank you for your time!

Edit: 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.

  • $\begingroup$ The construction which I am aware of takes as input a complex manifold $M$, a holomorphic vector bundle $E$, and a smooth Hermitian structure on $E$, and returns a unique connection (the Chern connection) which is compatible with these. You are implying that there is a construction whose input is a a complex manifold $M$, a holomorphic vector bundle $E$, and a Hermitian metric on $M$. Did you mean to ask for the Hermitian structure to be on $E$, or is the "metric connection" something other than the Chern connection? $\endgroup$ Commented Mar 9, 2012 at 16:18
  • $\begingroup$ I´m really sorry for the delay in the response. You are totally right! Indeed I´m looking for an Hermitian structure on $L$ which is harmonic (i.e. the form $\Delta log \| s\|dxdy$ is identically zero for a rational sections $s$ of $L$, so for every rational section). I´m going to edit the question right now. $\endgroup$ Commented Mar 19, 2012 at 15:38

2 Answers 2


alvarezpaiva's answer shows that the answer to your question is no in general.

I will answer your motivation question instead, for which the answer is positive.

If $L$ is a holomorphic line bundle over a compact Kähler 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)

$$\omega^{n-1}\wedge F=0.$$

This is just the Hermitian-Yang-Mills equation, and I am saying that it can always be solved in the case of line bundles.

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.

  • $\begingroup$ @YangMills, Thank you very much for your very instructive answer, which really helped me so far. Unfortunately I don´t understand it completely. In specific may I assume that $\Lambda F$ is defined by the relation $\omega^{n-1}\wedge F= \Lambda F\omega^n$? I ask it for the following reason. Let the curvature of $\|\;\|$ defined as $i\partial \bar{\partial} log(\|\;\|)$ (correct?). Then in my situation (i.e. $M$ is a Riemann surface) $\partial \bar{\partial}\sim\Delta$, and in specific the equation $\Delta h =\Lambda F$ doesn´t need $deg(L)=0$ to be solved, so every line bundle has degree zero! $\endgroup$ Commented Mar 30, 2012 at 14:52
  • $\begingroup$ The last sentence is justified by the equation $\int_M \Delta log (\|\;\|)= cost \cdot deg(L)$. If you could help me in pointing out where my error lies I would very appreciate it because I´m badly stucked on it at the moment. $\endgroup$ Commented Mar 30, 2012 at 14:56
  • $\begingroup$ The function $log\|\ \|$ is not globally defined, it depends on the choice of local coordinate. On the other hand $i\partial\overline{\partial}\log\|\ \|$ is a globally defined $(1,1)$-form, which on a Riemann surface equals $\Lambda F$ up to some constant (perhaps $2\pi$) and indeed as you say its integral is the degree of $L$, which need not be zero. If it is zero, then $\Lambda F$ integrates to zero and then you can solve $\Delta h=\Lambda F$. Note that for any globally defined function $h$ you always have that $\Delta h$ integrates to zero. $\endgroup$
    – YangMills
    Commented Mar 30, 2012 at 19:35

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.

  • $\begingroup$ I apologize for the delay in the response, thank you for the answer! May I ask you a reference for the fact you use? $\endgroup$ Commented Mar 19, 2012 at 15:46
  • $\begingroup$ I think almost any books on connections will do. Parallel transport of a metric connection must preserve the metric, so the holonomy around a loop must be an orthogonal transformation and hence the holonomy group must be contained in the compact group of orthogonal transformations. In your case (because the metric is hermitian) the holonomy around a loop must be unitary. $\endgroup$ Commented Mar 19, 2012 at 18:21

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