Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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/.

share|improve this question

3 Answers 3

up vote 12 down vote accepted

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); thus $M_B$ does not support any non-trivial algebraic line bundle.

share|improve this answer

[Edit: Attempted to clarify arguments, and to explain why this doesn't answer the question!]

First I would like to claim that for any group, or more generally topological space/homotopy type X, we can make a "Quillen line bundle over the character variety" in the following sense. Let's consider all representations of the group (or local systems on X) with the condition that their Ext from the trivial representation (ie global cohomology/Ext from the trivial local system) is a perfect complex. This certainly applies to finite rank local systems on a compact manifold, as in the case in question. More generally given a group G and a representation V we can consider all G-local systems with the same condition on the associated local system of vector spaces in the representation V. (In our case we'll take the adjoint representation of an algebraic group and the corresponding local systems on Riemann surfaces).

Now for each such local system we get a canonically attached line, the determinant line of the cohomology (Ext) complex in question.

Next, there's a natural moduli stack of local systems (or G-local systems) of the above form -- for a compact manifold this is just the associated "character variety", or rather the underlying derived stack. In other words, there's a natural notion of algebraic family of local systems on X or representations of our group. (Concretely it's characterized I believe by asking traces of monodromies to be algebraic functions). Abstractly, in the G-case it's the stack $BG^X$, the derived mapping stack from X to the classifying space of $G$ --- i.e., families over a base $S$ are given by maps from $$S\times X\to BG$$ where it's crucial that we consider X as a homotopy type (simplicial set), not a variety!!

This is the Betti space of X. The claim is the above lines naturally form an algebraic line bundle on this moduli space.

[There's a claim I'm not too comfortable with that the choice of representaiton of our group $G$ doesn't affect the line bundle except up to a rational power, related to the ratio of Dynkin indices, though I don't think it's necessary for this discussion.]

Now the question is how does this compare (in the case of a Riemann surface X - again considered just as a topological space, in fact a $K(\pi,1)$) to the desired algebraic construction with curvature the symplectic form (which by Tamas' answer doesn't exist)? The answer is it doesn't at all..

By the Riemann-Hilbert correspondence the above moduli stack is identified with the moduli of flat G-bundles on the algebraic curve X (as derived ANALYTIC stacks), though not algebraically. However it seems clear that the line bundle we defined (hopefully) above agrees with the determinant line of de Rham cohomology of the universal connection on the moduli space times X. (i.e. Riemann-Hilbert preserves Ext from the trivial local system). Moreover the de Rham space carries a holomorphic symplectic form which agrees with the one defined on the Betti space by the intersection pairing on the curve. The symplectic form on the de Rham space can be described as the curvature of the line bundle, which is pulled back from the determinant line bundle (ie determinant of Dolbeault cohomology) on the moduli of G-bundles (the de Rham space is a twisted cotangent bundle of the latter moduli, twisted exactly by this bundle).

However this line bundle with desired curvature on the de Rham space is NOT the determinant line of de Rham cohomology, and in particular doesn't pass over to the Betti space.

EXAMPLE (following Tamas): For X=curve of genus one, the category of local systems is equivalent to that of coherent sheaves on $C^\times x C^\times$. The functor of Ext from the trivial local system becomes Ext from the skyscraper at the identity. In particular if we look at rank one local systems (which are identified with points in $C^\times x C^\times$) then we see that the determinant of de Rham cohomology is canonically trivialized on the complement of the identity - ie off of codimension two! So the corresponding line bundle is trivial, and certainly notidentified with the (analytic) Deligne line bundle (whose curvature is the canonical holomorphic symplectic two-form).

OK sorry for the wild goose chase!

share|improve this answer
Since I want to avoid thinking of $\Sigma=X$ as a variety, I'll replace $\det(R\pi_\ast E_V)$ with the determinant line of the group cohomology $H^\ast(\pi,\mathfrak g_{\operatorname{Ad}\phi})$ as the fiber over a point $\phi:\pi\to G$ in $\operatorname{Hom}(\pi,G)//G$. –  John Pardon Jun 19 '11 at 23:35
In fact you can think of $X$ as a plain homotopy type, which is all you need to define the variety (or rather stack) $M$ of local systems -- eg anything with the homotopy type of a finite CW complex will do for this construction. –  David Ben-Zvi Jun 19 '11 at 23:56
David, the universal bundle is only algebraic in the $M_{DR}$ algebraic structure - it is not algebraic in the character variety algebraic strucure, as its Chern classes are not pure. –  Tamas Hausel Jun 20 '11 at 13:59
I don't think (1) is true algebraically for the Betti (character variety) algebraic structure. –  Sam Gunningham Jun 20 '11 at 15:15
Sam - thanks, I meant 1 in de Rham, the point was to argue if the construction of the line bundle I gave is correct then indeed it should have the right curvature form. I think the problem is the construction itself, i.e., what kind of families of local systems we consider (if it's only isomonodromic ones then we can't compare with the character variety..) Very confused! –  David Ben-Zvi Jun 20 '11 at 15:55

I hope I'm not getting any details wrong, but I think the Narasimhan-Seshadri theorem asserts that the moduli space of $G=U(n)$ representations of $\pi_1$ is the same as the moduli space $M(n)$ of semistable rank $n$ vector bundles (i.e. principal $GL(n,\mathbb{C})$-bundles). (Probably this generalizes to any compact Lie group $G$?)

I believe $L$ is defined algebraically over $M(n)$ as $\operatorname{det} \mathbb{R}\pi_\ast U$, where $U$ is the universal bundle over $\Sigma \times M(n)$ and $\pi$ is the second projection.

share|improve this answer
I'd like to have $G$ be an algebraic group, e.g. $SL(n,\mathbb C)$ or $\GL(n,\mathbb C)$, which are clearly noncompact. Does the Narasimhan-Seshadri theorem apply in that case as well? –  John Pardon Jun 19 '11 at 19:49
The complex group version of Narasimhan-Seshadhri (sometimes called non-abelian Hodge theory) relates reps of $\pi_1$ to Higgs bundles. The moduli of Higgs bundles is (roughly, ignoring stability issues) the cotangent bundle to M(n), but the algberaic structure here is not the same as that on the character variety. –  Sam Gunningham Jun 20 '11 at 15:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.