Let's consider the moduli space of representations of $\pi=\pi_1(\Sigma)$ (a surface group) into $G$ (a lie group). Call this $X=\operatorname{Hom}(\pi,G)$, and let $Y=\operatorname{Hom}(\pi,G)/\\!/G$, where $G$ acts by conjugation on $X$ (and we take the GIT quotient). Let's denote the quotient map by $f:X\to Y$. Goldman has constructed a natural symplectic form $\omega$ on $Y$. By a construction of Quillen (the so-called determinant line bundle) there exists a line bundle $\mathcal L$ over $Y$ with curvature form equal to $\omega$ (or perhaps $\omega$ times some constant).

To construct this line bundle, though, one apparently needs to think of $\pi$ as the fundamental group of some specific Riemann surface (i.e. pick a holomorphic structure on $\Sigma$), and consider the entire (infinite-dimensional) moduli space of flat $G$-connections on $\Sigma$ (of which $Y$ is the quotient modulo the gauge group). We then need some infinite-dimensional analysis, and the notion of a Fredholm operator. I'm looking for a construction of $\mathcal L$ which stays in the algebraic world (and in particular avoids picking a holomorphic structure on $\Sigma$).

Some questions along those lines:

1. Is $f^\ast\omega$ exact? If so, then is there a natural choice of $1$-form $\gamma$ on $X$ such that $d\gamma=f^\ast\omega$? This would essentially answer my question, as taking the trivial bundle with connection form $\gamma$ (some might say $\exp\gamma$) would give $F^\ast\mathcal L$.

2. Is there another way to construct $\mathcal L$ staying in the algebraic category?

Perhaps a comment is relevant: Goldman proves that $\omega$ is closed by appealing to the infinite-dimensional picture of the moduli space of flat connections. This is apparently the standard proof, though I have found in the literature an entirely algebraic proof that $\omega$ is closed, via some direct (but coordinate independent, and thus not all that messy) calculations, see http://www.ams.org/journals/proc/1992-116-03/S0002-9939-1992-1112494-2/.

Let's consider the moduli space of representations of $\pi=\pi_1(\Sigma)$ (a surface group) into $G$ (a lie group). Call this $X=\operatorname{Hom}(\pi,G)$, and let $Y=\operatorname{Hom}(\pi,G)/\\!/G$, where $G$ acts by conjugation on $X$ (and we take the GIT quotient). Let's denote the quotient map by $f:X\to Y$. Goldman has constructed a natural symplectic form $\omega$ on $Y$. By a construction of Quillen (the so-called determinant line bundle) there exists a line bundle $\mathcal L$ over $Y$ with curvature form equal to $\omega$ (or perhaps $\omega$ times some constant).

To construct this line bundle, though, one apparently needs to think of $\pi$ as the fundamental group of some specific Riemann surface (i.e. pick a holomorphic structure on $\Sigma$), and consider the entire (infinite-dimensional) moduli space of flat $G$-connections on $\Sigma$ (of which $Y$ is the quotient modulo the gauge group). We then need some infinite-dimensional analysis, and the notion of a Fredholm operator. I'm looking for a construction of $\mathcal L$ which stays in the algebraic world (and in particular avoids picking a holomorphic structure on $\Sigma$).

Some questions along those lines:

1. Is $f^\ast\omega$ exact? If so, then is there a natural choice of $1$-form $\gamma$ on $X$ such that $d\gamma=f^\ast\omega$? This would essentially answer my question, as taking the trivial bundle with connection form $\gamma$ (some might say $\exp\gamma$) would give $f^\ast\mathcal L$.

2. Is there another way to construct $\mathcal L$ staying in the algebraic category?

Perhaps a comment is relevant: Goldman proves that $\omega$ is closed by appealing to the infinite-dimensional picture of the moduli space of flat connections. This is apparently the standard proof, though I have found in the literature an entirely algebraic proof that $\omega$ is closed, via some direct (but coordinate independent, and thus not all that messy) calculations, see http://www.ams.org/journals/proc/1992-116-03/S0002-9939-1992-1112494-2/.