Representation variety in $\mathrm{SU}(p,q)$

Question

$\DeclareMathOperator\SU{SU}$Let $\Gamma$ be a cocompact oriented Fuchsian group, and consider the representation variety $\textrm{Hom}(\Gamma, \SU(p,q))$. Consider a point $\rho : \Gamma \to \SU(p,q)$ with image $\rho(\Gamma) =\SU(p)\times \SU(q)$, a maximal compact subgroup of $\SU(p,q)$. How to decide whether $\rho$ is nonsingular or not? A result of André Weil gives a sufficient condition to decide whether a point, say $\lambda: \Gamma \to G$ with image $H$ of the representation variety is nonsingular or not, namely the centralizer of $H$ in $G$ has dimension zero. But, this criterion is not applicable here because the centralizer of $\SU(p)\times \SU(q)$ in $\SU(p,q)$ has dimension $1$.

Answer 1

I think it is always a singular point in $\mathrm{Hom}(\Gamma, \mathrm{SU}(p,q))$. Suppose $G$ is reductive and $\frak{g}$ its Lie algebra. Let $\rho \in \mathrm{Hom}(\Gamma, G)$. Then $\rho$ induces a $\Gamma$-module structure on $\frak{g}$ by the Adjoint action of $G$ on $\frak{g}$. Then the Zariski tangent space at $\rho$ is the group-cocycle space $T := Z^1(\Gamma, \frak{g})$. The dimension of $T$ is $$(2g - 1)\dim(G) + \dim(Z(\rho)),$$ where $g$ is the genus of the surface associated with $\Gamma$ and $Z(\rho)$ is the centralizer of the image of $\rho$.

In your case, $\dim(Z(\rho)) = 1$ while at a smooth point, $\dim(Z(\rho)) = 0$. So $\rho$ is singular.