## Zariski tangent spaces to representation varieties

In Bill Goldman's paper "The Symplectic Nature of the Fundamental Groups of Surfaces" (Advances, 54, 200-225, '84) it is stated that the "Zariski tangent space" to a representation space Hom$(\pi, G)/G$ at a representation $\rho$ is the cohomology group $H^1 (\pi; Ad\ \rho)$. Here G is a Lie group, and Ad $\rho$ is the representation of $\pi$ on the Lie algebra of G induced by the adjoint action of G. In the context of Goldman's paper, maybe this is just meant to refer to the case when $\pi = \pi_1 S$ with $S$ a Riemann surface.

My question is: is there some sense in which this is true for other discrete groups $\Gamma$? In general, $Hom(\pi, G)/G$ is only a semi-algebraic set, and not a variety, so maybe the question is not really meaningful. But I'd like to know whether these first cohomology groups behave like tangent spaces in some useful way.

Here are two specific questions. I'm most interested in the case when G = U(n).

Edit: Let me emphasize that I'm really talking about the topological quotient Hom(G, U(n))/U(n), which is a reasonably nice space. Since U(n) is compact, general nonsense implies that this space is Hausdorff, and even better, it's a semi-algebraic set. So in particular, it's homeomorphic to a simplicial complex.

1. Say $[\rho]\in$ Hom$\left(\Gamma, U(n)\right)/U(n)$ has an open neighborhood homeomorphic to $\mathbb{R}^m$ for some $m$. Is it then true that dim $H^1 (\Gamma; Ad\ \rho) = m$? In other words, does the cohomology group give the "topological" dimension at "smooth" points in Hom$\left(\Gamma, U(n)\right)/U(n)$? When $H^1 (\Gamma; Ad\ \rho) = 0$, a theorem of Weil (Ann. of Math. (2) 80 1964 149--157) says that $[\rho]$ is an isolated point in Hom$\left(\Gamma, U(n)\right)/U(n)$. This is the converse statement for m=0.

2. If $[\rho]$ is not a smooth point in the above sense, then in any triangulation of Hom$\left(\Gamma, U(n)\right)/U(n)$, we see that $[\rho]$ must not lie in the interior of a maximal simplex. If $\sigma$ is a maximal simplex of dimension m containing $[\rho]$ (in its boundary), is it true that dim $H^1 (\Gamma; Ad\ \rho) > m$? In other words, does the dimension of the "tangent space" jump up at non-smooth points?

Any ideas, references, examples, or counterexamples would be welcomed!

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 Surface groups are exceptionally nice. I expect 1. is false for arbitrary $\Gamma$ (the representation "scheme" can be pretty wild). See a paper by Lubotzky and Magid in the Memoirs. If no one else answers, I'll try to think of something more specific tomorrow. – Donu Arapura Jun 3 2010 at 22:36 I actually just checked out Lubotzky and Magid before posting this. They seem to focus on the case of general linear representations. There the space of conjugacy classes of semi-simple representations does have the structure of a variety. Looking some more, I see that they show that these cohomology groups are the Zariski tangent spaces. So this seems to answer my questions positively in this setting, but I'm really interested in unitary representations. – Dan Ramras Jun 3 2010 at 23:01 If $\Gamma$ is the fundamental group of a compact Kaehler manifold, then by Donaldson ..., your space can be identified with tangent space to the moduli scheme of polystable bundles with trivial Chern classes. And I'm convinced this can be bad i.e nonreduced, but I don't have a specific example in mind. Perhaps you can find something in Goldman and Millson's paper in IHES (1988). I guess I'm out of ideas for now. – Donu Arapura Jun 4 2010 at 1:57

There are some people around who know much more about this subject than I do, but since none of them has responded so far, let me make a few remarks. By an old theorem of Narasimhan and Seshadri (1965), moduli of flat unitary bundles with irreducible holonomy and of holomorphic stable bundles on a Riemann surface $\Sigma$ are isomorphic. The former space is basically the representation variety you've described and has a canonical symplectic structure; the latter space is a Mumford GIT quotient, so it's an algebraic variety. There is a big machine which implies such a relation between the Kahler/symplectic reduction and a GIT quotient for finite-dimensional group actions, except that the group in this situation is the gauge group, so some care needs to be taken.
Irreducibility of the representation $\rho$ assures that the holonomy of the corresponding connection has trivial centralizer (scalar matrices) and that the corresponding point $[\rho]$ in the moduli space is regular (non-singular and non-orbifold). Without it, you have to be careful about defining the quotient. The first cohomology group
$$\text{H}^1(\Sigma, \operatorname{End}\rho)=\text{H}^1(\pi, \operatorname{End}\rho)$$ classifies $G$-conjugacy classes of infinitesimal deformations of $\rho:\pi\to G$ by abstract Kodaira-Spencer theory, so the answer to Q1 is affirmative: if the point $[\rho]$ is smooth, the dimension of the moduli space at $[\rho]$ is the dimension of the first cohomology. This statement generalizes to arbitrary finitely generated discrete groups $\Gamma$. You can let $\Sigma=\text{K}(\pi,1)$ and consider topological cohomology of $\Sigma$ instead of group cohomology of $\pi$. The case studied by Weil is when $\Sigma$ is a hyperbolic $n$-manifold, so that the moduli space no longer has complex or algebraic structure for $n\geq 3.$ One difficulty in addressing Q2 is that if $\rho$ is reducible, you get an orbifold point, so it doesn't quite make sense to speak about triangulations. Another difficulty is that if the cohomological dimension of $\pi$ is greater than 2 (i.e. $\dim \Sigma\geq 3$), it becomes harder to keep track of the dimensions of higher cohomology groups by means of the Euler characteristic. Representation varieties of 3-manifold groups have been extensively studied (e.g. look at papers of Shalen).
 I actually insist on not being careful in forming the quotient! (I promise I have my reasons...) What I mean is, I'm really interested in the topological quotient space Hom$(\Gamma, U(n))/U(n)$, which is a reasonably nice space. General nonsense tells you it's Hausdorff (because U(n) is compact), and it's even semi-algebraic. In particular, it's homeomorphic to an ordinary simplicial complex. Maybe Kodairi-Spencer theory is telling me I need to figure out whether my topological notion of dimension - triangulate and look at the maximal simplices - agrees with some sort of stacky notion? – Dan Ramras Jun 4 2010 at 4:21 I don't have a good feel about what happens at singular points. Do take a look at 3-manifolds people's papers, they've likely answered nearly any question like that you may have. – Victor Protsak Jun 4 2010 at 4:52