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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.

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    $\begingroup$ Did you ask this previously? It reminds me of some similar question. $\endgroup$ – Ben McKay Apr 10 '20 at 20:10
  • $\begingroup$ 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 $Hom(\pi, G)$? $\endgroup$ – Ryan Budney Apr 10 '20 at 22:32
  • $\begingroup$ 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. $\endgroup$ – Dan Ramras Apr 11 '20 at 1:31
  • $\begingroup$ 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. $\endgroup$ – Dan Ramras Apr 11 '20 at 1:34
  • $\begingroup$ 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). $\endgroup$ – Will Sawin Apr 11 '20 at 1:41
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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.

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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Andy Sanders Apr 12 '20 at 14:30

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