# Space of representations of surface group into Lie groups

In the context of Goldman's paper The symplectic nature of fundamental groups of surfaces:

Consider a closed oriented surface $$S$$ with fundamental group $$\pi$$, and let $$G$$ be a connected Lie group. The space $$\operatorname{Hom}(\pi,G)$$ consisting of representations $$\pi\to G$$ is a real analytic variety a real analytic space (possibly singular). There is a canonical $$G$$-action on $$\operatorname{Hom}(\pi,G)$$ obtained by composing representations with inner automorphisms of $$G$$. When is the representation space $$\operatorname{Hom}(\pi,G)/G$$ a smooth manifold? I think that if the representation is irreducible, the space is smooth manifold nearby.

• Did you ask this previously? It reminds me of some similar question. – Ben McKay Apr 10 '20 at 20:10
• I would expect it's only a manifold when $G$ is trivial. But asking for a manifold structure is fairly peculiar on its own. Don't you want it to be related to the real analytic structure on $Hom(\pi, G)$? – Ryan Budney Apr 10 '20 at 22:32
• There are examples where Hom$(\pi,G)/G$ happens to be a topological manifold, somewhat by accident. For instance, if $S$ is the torus, then $\pi = \mathbb{Z}\times \mathbb{Z}$ and Hom$(\mathbb{Z}\times \mathbb{Z},SU(2))/SU(2)$ is homeomorphic to the 2-sphere $S^2$. This is discussed in various places in the literature, the oldest place I know of being Morgan, John W.; Shalen, Peter B. Valuations, trees, and degenerations of hyperbolic structures. I. Ann. of Math. (2) 120 (1984), no. 3, 401-476. – Dan Ramras Apr 11 '20 at 1:31
• Oddly, if you replace $SU(2)$ by $PSU(2)$ in the above example, the representation space Hom$(\mathbb{Z}^2, PSU(2))/PSU(2)$ is still homeomorphic to $S^2$, giving another example. – Dan Ramras Apr 11 '20 at 1:34
• I think the local condition to be a smooth manifold is that the representation is irreducible (i.e. not contained in any proper parabolic) and the centralizer of the image of pi is the center of G (second condition redundant for G=GL_n but not in general). – Will Sawin Apr 11 '20 at 1:41

The question of the structure of $$Hom(\pi, G)$$ and $$Hom(\pi, G)/G$$ is very complicated for a general connected Lie group $$G$$, though less so for a surface group than for a general finitely generated group. To see just how wild things can get, you should look at this paper of Kapovich and Millson http://www.math.umd.edu/~millson/papers/char19.pdf.

For a surface $$\Sigma$$, if $$G$$ is Abelian, then

$$Hom(\pi, G)\simeq Hom(\pi, G)/G\simeq H^{1}(\Sigma, G)\simeq G^{2\sigma}$$

where $$\sigma$$ is the genus of $$\Sigma.$$ In particular, in this case, it is naturally a smooth, even real analytic, manifold in a canonical way.

Meanwhile, as mentioned in the above comments, there are situations when $$Hom(\pi, G)$$ and/or $$Hom(\pi, G)/G$$ is homeomorphic to a manifold, although this is something of an accident.

Probably the cleanest statement available that lives in some level of generality, is that if $$G$$ is a connected reductive complex affine algebraic group (like $$GL_{n}(\mathbb{C})$$), then $$Hom(\pi, G)$$ is a complex affine variety with at most quadratic singularities, this is due to Goldman, Millson and Simpson (https://projecteuclid.org/euclid.bams/1183554530), and remarkably this result remains true as long as $$\pi$$ is the fundamental group of a compact Kahler manifold. But, even in this case, the quotient $$Hom(\pi, G)/G$$ is a total mess, in particular it is Hausdorff only when $$G$$ is Abelian.

This can be remedied (to some extent) in this case by studying the GIT quotient $$Hom(\pi, G)//G,$$ but since you ask about real Lie groups in general, GIT theory again is a bit of a mess since $$\mathbb{R}$$ is not algebraically closed.

As mentioned in the comments, if you take $$G$$ a connected semi-simple real Lie group, then the subset $$Hom^{\star}(\pi, G)$$ consisting of irreducible representations whose centralizer is equal to the center of $$G,$$ then this is a smooth (even real analytic) manifold upon which $$G$$ acts with constant stabilizer equal to the center, and properly, and therefore $$Hom^{\star}(\pi, G)/G$$ is a (real analytic) manifold.

Just as a final remark, there are many very good reasons why one might consider the space of conjugacy classes of all representations $$\pi\rightarrow G,$$ but as explained above, this is basically hopeless in any reasonable category of classical spaces such as: smooth manifolds, algebraic varieties, real analytic spaces...

One remedy to this hopelessness is to pass to a suitable category of stacks, but of course it heavily depends on what you want to do.

I'll expand a bit on @WillSawin's comment. Sikora (Corollary 50 of the article Character Varieties in Trans. AMS, 2012) proved that if $$G$$ is a complex linearly reductive algebraic group and $$\pi$$ is the fundamental group of a closed surface of genus greater than 1, then irreducible representations $$\pi\rightarrow G$$ whose centralizer is the center of $$G$$ do indeed lie in the smooth locus of the GIT quotient Hom$$(\pi, G)//G$$. (As Will said, irreducible here means the image is not contained in a proper parabolic subgroup.) I'm not sure if this always describes the entire smooth locus. (It would be nice to know!)

I don't know that there is any good understanding of these kinds of things for non-reductive Lie groups.

It seems natural to look at, even with $$G$$ non-reductive.

For instance, let $$G=G_m$$ be a $$2m+1$$-dimensional Heisenberg group: I write $$G=V\times K$$, where $$(V,\phi)$$ is a $$2m$$-dimensional symplectic space over $$K$$ (not of characteristic $$2$$) and law $$(X,T)\cdot (Y,U)=(X+Y,T+U+\phi(X,Y))$$. In particular (with suitable sign conventions), the group commutator $$[(X,T),(Y,U)]$$ is $$(0,2\phi(X,Y))$$.

Consider an $$2n$$-tuple in $$G^n$$, written as $$((x,t),(y,u))$$ with $$(x,t),(y,u)\in G^n$$ (so $$x,y\in V^n$$ and $$t,u\in K^n$$), $$x=(x_i)_{1\le i\le n}$$, $$y=(y_i)_{1\le i\le n}$$, with $$x_i,y_i\in V$$, and $$T=(t_i)_{1\le i\le n}$$, $$U=(u_i)_{1\le i\le n}$$, with $$t_i,u_i\in K$$.

The condition $$\prod_{i=1}^n[(x_i,t_i),(y_i,u_i)]=e$$ can be written as $$\sum_i\phi(x_i,y_i)=0$$. That is, $$\Phi(x,y)=0$$, where $$\Phi(x,y)$$ is defined as $$\sum_i\phi(x_i,y_i)$$, which makes $$(V^n,\Phi)$$ a $$2mn$$-dimensional symplectic space. Note that if we ignore the $$t_i,u_i$$, we obtain the hypersurface $$H_{mn}=\{(x,y)\in V^{2n}:\Phi(x,y)=0\}$$ in the $$4mn$$-dimensional space $$V^{2n}$$, which is non-singular outside zero (and the $$G$$-action by conjugation is trivial).

So if we consider the space $$\mathrm{Hom}(\Gamma_n,G_m)$$ (where $$\Gamma_n$$ is the genus $$n\ge 1$$ surface group), we have its description as $$H_{mn}\times K^{2n}$$, and it's non-singular outside $$\{0\}\times K^{2n}$$.

The conjugation action (after renormalization by $$2$$ and factoring the action through $$V$$) consists in letting $$V$$ act on this product as $$z\cdot (x,y,t,u)=(x,y,t+\psi(z,x),u+\psi(z,y)).$$ Here $$\psi(z,x)$$, for $$x\in V^n$$ and $$z\in V$$, is defined as $$(\phi(z,x_i))_{1\le i\le n}$$.

Since for $$(x,y)\neq (0,0)$$, the linear map $$z\mapsto\psi(z,x),u+\psi(z,y)$$ is injective, it's not hard to prove that the quotient remains smooth outside $$\bar{p}^{-1}(\{(0,0)\})$$, where $$p$$ is the projection $$\mathrm{Hom}(\Gamma_n,G_m)\to H_{mn}$$ and factors through a projection $$\bar{p}:\mathrm{Hom}(\Gamma_n,G_m)/G_m\to H_{mn}$$.

Note that fibers of $$p$$ are "translates" of $$K^{2n}$$; each fiber of $$\bar{p}$$ naturally carries a structure of affine space of dimension $$2n-1$$, except the fiber of $$(0,0)$$ which is still naturally identified to $$K^{2n}$$.

On generalizations:

Let $$G$$ be an arbitrary $$s$$-step nilpotent simply connected Lie group (that is, $$G^{s+1}=\{1\}$$ where $$(G^i)_{i\ge 1}$$ is the lower central series), and $$\Gamma$$ is an arbitrary finitely generated group. Let $$\Gamma(s)$$ be the real Malcev completion of the nilpotent quotient $$\Gamma/\Gamma^{s+1}$$ (this is a a simply connected $$s$$-step nilpotent Lie group in which $$\Gamma/\Gamma^{s+1}$$ modulo its finite torsion subgroup, sits as a lattice. Then we can identify $$\mathrm{Hom}(\Gamma,G)$$ to $$\mathrm{Hom}_{\mathbf{TopGrp}}(\Gamma(s),G)$$, and this identification commutes with the $$G$$-action. Furthermore, letting $$\Gamma[s]$$ be the Lie algebra of $$\Gamma(s)$$ and $$\mathfrak{g}$$ that of $$G$$, this can be identified to $$\mathrm{Hom}_{\mathbf{R}\text{-}\mathbf{LieAlg}}(\Gamma[s],\mathfrak{g})$$.

Hence all the study with nilpotent target reduces to that of spaces of homomorphisms between Lie algebras.

• Thanks for this answer: I've thought more than once that's there's probably a lot of rich geometry for non-reductive groups, but it's largely unexplored territory. – Andy Sanders Apr 12 '20 at 14:30