Let $G$ be an abelian finite group act on $\mathbb C^n$, when the equivariant Hilbert scheme $\mathrm{Hilb}^{R}(\mathbb C^n)^G=\mathrm{Hilb}^{R}([\mathbb C^n/G])$ is connected? Now $R$ is a representation of $G$ which is in the K-group of $[\mathbb C^n/G]$. In particular, if we consider the $A_n$-type, that is, $G=\mathbb Z/(n+1)\mathbb Z$ act as $\zeta\cdot(x_1,...,x_n)=(\zeta x_1,\zeta^{-1}x_2,x_3,...,x_n)$, is $\mathrm{Hilb}^{R}([\mathbb C^n/(\mathbb Z/(n+1)\mathbb Z)])$ connected?
I find that $\mathrm{Hilb}^{R}([\mathbb C^2/G])$ connected in the paper Smooth and irreducible multigraded Hilbert schemes, see also 2.4 in the paper equivariant-hilbert, but I don't know whether this is right for $\mathrm{Hilb}^{R}([\mathbb C^n/(\mathbb Z/(n+1)\mathbb Z)])$ $n\geq3$ or not? These papers using the computational commutative algebra to show the rational chain connectedness.
For the ordinary connected variety $X$, the Hilbert scheme of points $\mathrm{Hilb}^n(X)$ is connected can be shown by the Fogarty's classical paper algebraic families on an algebraic surface using the Hilbert-Chow morphism and to show its fibres are connected, which are true since the hilbert scheme of points of finite dimensional local $\mathbb C$-algebra is connected. BUT it seems that this method can not generalized to the orbifolds.
Thank you for your help!
