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/>.



Is there an algebraic construction of the Quillen (determinant) Line Bundle?
For $G={\rm PGL}\_n({\mathbb C})$ there is no complex algebraic Quillen line bundle on the complex character variety $Y$ as the cohomology class $[\omega]=\alpha\_2\in H^2(Y;{\mathbb Q})$ is not pure by Proposition 4.1.8 in <http://arxiv.org/abs/math.AG/0612668>.


EDIT: It is instructive to think about the case $G={ GL}\_1({\mathbb C})$. Then the character variety is $M\_B=GL\_1^{2g}$ the complex torus while $M\_{DR}$ is an affine bundle over $Jac(C)$. Analytically $M\_B\cong M\_{DR}$ but the universal bundle $L$ on $M\_{DR}\times C$ which is just the pull back of the Poincare bundle from $Jac(C)\times C$ is not algebraic on $M\_B$, for example because the cohomology class $[\omega]$ of the symplectic form (which shows up in $c\_1(L)^2$) has weight $2$ in $H^2(M\_{DR};{\mathbb Q})$ but has weight $4$ (homogeneous weight $2$) in $H^2(M\_B;{\mathbb Q})$.
Thus in particular there is no complex algebraic Quillen bundle on $M\_B$ with first Chern class $[\omega]$. 


EDIT 2: To take David's example let $g=1$, then $M\_B={\mathbb C}^\* \times {\mathbb C}^\*$ which does not have non-trivial pure cohomology in $H^2(M\_B;{\mathbb Q})$ and even the Picard group is trivial (see an argument [here](https://mathoverflow.net/questions/68421/simplest-examples-of-nonisomorphic-complex-algebraic-varieties-with-isomorphic-an/68425#68425)); thus $M\_B$ does not support any non-trivial algebraic line bundle. 


