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This question is inspired by a recent question about holomorphic bundles and factors of automorphy. Suppose $X$ is a compact, complex manifold whose universal cover $\widetilde{X}$ is Stein (the Stein condition is just to ensure all topologically trivial holomorphic vector bundles over $\widetilde{X}$ are holomorphically trivial). Then I've just recently come to appreciate the theorem that every holomorphic vector bundle (whose pullback to $\widetilde{X}$ is topologically trivial) \begin{align} E\rightarrow X \end{align} arises from a factor of automorphy, this is a holomorphic map: \begin{align} f:\pi_1(X) \times \widetilde{X}\rightarrow GL_n(\mathbb{C}) \end{align} satisfying the cocycle condition $f(\gamma\eta, x)=f(\gamma,\eta(x))f(\eta, x).$ The map $f$ defines a $\pi_1(X)$-action on the trivial bundle $\widetilde{X}\times \mathbb{C}^n\rightarrow \widetilde{X}$ and the quotient defines a holomorphic vector bundle over $X.$ Now suppose that the holomorphic structure on $E\rightarrow X$ is given as the $(0,1)$-part of a linear connection $\nabla$ on $E.$

If this connection is flat, then the factor of automorphy is nothing more than the monodromy representation $\rho:\pi_1(X)\rightarrow GL_n(\mathbb{C})$ of this flat vector bundle.

$\textbf{My question is:}$ if the connection is not flat, is there a gauge theoretic construction of the factor of automorphy defining the holomorphic structure on $E.$ Ideally this would be something involving parallel transport using the connection $\nabla,$ but in my first fumbling attempts I haven't found a way to do this.

My motivation for this question arises from something that has befuddled me for some time. I won't define everything in this section but a reference for anything I say is in Hitchin's paper "The self-duality equations on a Riemann surface." Let $X$ be a compact Riemann surface of genus at least 2. Let $\rho:\pi_1(X)\rightarrow SL_2(\mathbb{R})$ be a Fuchsian representation. Then via the non-abelian Hodge correspondence, the Higgs bundles $(E,\phi)$ corresponding to any Fuchsian representation all have the same underlying holomorphic bundle, namely \begin{align} E=K^{-\frac{1}{2}}\oplus K^{\frac{1}{2}} \end{align} where $K$ is the canonical bundle over $X.$ This has always been mysterious to me, starting from the flat bundle side; namely you form the flat bundle $E_{\rho}=\widetilde{X}\times_{\rho}\mathbb{C}^2$ and let's call the flat connection $B.$ Then by Hitchin, there exists a unique unitary connection $\nabla$ such that \begin{align} B=\nabla+\psi, \end{align} and $d_{\nabla}^{*}\psi=0.$ The holomorphic structure on $E_{\rho}$ in the Higgs bundle picture is then the one induced by the $(0,1)$-part of the unitary connection $\nabla,$ which is not the same as the one induced by the flat connection $B$ on $E_{\rho}.$ But, why should this always be $K^{-\frac{1}{2}}\oplus K^{\frac{1}{2}}.$ I've never reached a satisfying answer to this issue. I wonder if some insight into the question I asked above could clarify this situation.

Thank you for any help.

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    $\begingroup$ It is not true that all holomorphic vector bundles on a Stein manifold are trivial, this holds only for those which are topologically trivial. $\endgroup$ – abx Dec 25 '14 at 11:27
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    $\begingroup$ Of course you are right, I'll edit to make this clear. $\endgroup$ – Andy Sanders Dec 25 '14 at 19:37
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I will not answer your question, but tell you something about the holomorphic structure of a Fuchsian representation. I guess you are aware of the paper "Twisted harmonic maps..." of Donaldson (cited in Hitchin's paper). The thing is that if you start with a Fuchsian representation, your corresponding twisted harmonic map is just the developing map of you constant curvature metric into hyperbolic 2-space sitting in hyperbolic 3-space. This leads to some kind of symmetry of your connection $B$ and its decomposition into $B=\nabla+\psi$. Basically, $\nabla$ must be diagonal and $\psi$ off-diagonal. To work things out, you can also take a look into Hitchin's paper "Harmonic tori in $S^3$" (Proposition 1.9). There Hitchin also studied the case of harmonic maps into $S^2$ which led to the same kind of symmetry as in your case (but different signs...).

The next thing you need to understand is, why the entries of the diagonal part are a spin structure and its dual. To see this, look at the $(1,0)$ $\Psi$ part of $\psi$ which is a holomorphic off-diagonal 1-form with values in the endomorphism bundle:

If your Fuchsian representation is the one of your Riemann surface, than the harmonic map is conformal and therefore $\det \Psi=0.$ But the developing map has no branch point, hence $\Psi$ is nowhere vanishing. From this and trace-freeness you see that, with respect to your decomposition, one of the off-diagonal terms of $\Psi$ vanishes whereas the other is constant to $1.$ This implies that the holomorphic structure is $K^{1/2}\oplus K^{-1/2}.$

If you deform your Fuchsian representation slightly, everything depends real analytically on the deformation. In particular, there is locally one off-diagonal entry (the upper right one) of $\Psi$ which is not vanishing, but still holomorphic. So in a neighborhood of the Fuchsian representation of your Riemann surface $E=K^{1/2}\oplus K^{-1/2}.$

To deduce the general case observe, that the upper right entry of $\Psi$ either vanishes everywhere or nowhere (as it is constant), but the first case can only happen at the Fuchsian representation of the Riemann surface.

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