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Let $X$ be a smooth quasi projective variety over $\mathbb{C}$. Let $G$ be a finite abelian group acting via automorphisms on $X$.

Denote by $G$-$\text{Hilb}(X)$ the subscheme of the Hilbert scheme of $X$ which parametrises $G$-invariant length $\#G$ subschemes $P$ of $X$ such that $O_P$ is isomorphic to $\mathbb{C}[G]$ as a $G$-module.

Assume there is a subgroup $H\subset G$ with the property that $H$ acts freely on $X$, then do we have the following isomorphism:

$G$-$\text{Hilb}(X)\cong$ $G/H$-$\text{Hilb}(H$-$\text{Hilb}(X))$?

Since $H$ acts freely we just have $\text{Hilb}(H$-$\text{Hilb}(X))\cong X/H$. So the question of describing $G$-$\text{Hilb}(X)$ reduces to describing $G/H$-$\text{Hilb}(X/H)$.

Or is this not true? Do we need some more assumptions for this to be true?

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    $\begingroup$ Yes, you do have that isomorphism. You can check this directly, but you can also use the fact that $G-\text{Hilb}(X)$ equals the Hilbert scheme of the Deligne-Mumford stack $[X/G]$ with the "Hilbert polynomial" $p(t)$ equal to a constant polynomial corresponding to a copy of the group ring $\mathbb{Z}[G]$ as an element in $K^0(G)[t]$, cf. Olsson-Starr. For the quotient algebraic space $Y=X/H$ and for $\Gamma=G/H$ with its induced action on $Y$, the Deligne-Mumford stack $[X/G]$ is naturally isomorphic to $[Y/\Gamma]$. Thus, the Hilbert schemes of these stacks are naturally isomorphic. $\endgroup$
    – Jason Starr
    Commented Oct 17, 2016 at 14:15
  • $\begingroup$ @Jason Starr: Thanks a lot. This answer really enlightens the situtaion. $\endgroup$
    – Bernie
    Commented Oct 18, 2016 at 7:11

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