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Let $X$ be a variety over $\mathbb{C}$ with finite quotient singularities, i.e. every point has a Zariski-open neighbourhood isomorphic to $U/H$ where $U$ is a smooth variety and $H$ is a finite group acting on it.

Now assume we have a connected reductive group $G$ acting on $X$ with finite stabilizers, let $\mathcal L$ be an ample $G$-linearized line bundle on $X$ and let $Y = X^{ss}/G$ be the corresponding GIT quotient. If necessary we may assume that $X^{ss} = X^s$.

Is it true that $Y$ again has finite quotient singularities?

If $X$ is already smooth, then this should be true (see Mumford's GIT, Chapter 8, §4).

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  • $\begingroup$ You certainly need to restrict to the properly stable locus $X^s$. Already the cone over a smooth quadric surface, considered as a quotient of $\mathbb{A}^4$ by a $\mathbb{G}_m$-action, shows that the result cannot hold on the semistable locus. $\endgroup$ – Jason Starr Feb 3 '16 at 19:50
  • $\begingroup$ Thank you for your example. However, I think the problem there is that the action of $\mathbb{G}_m$ on $\mathbb{A}^4$ has nontrivial stabilizer at the origin, which is excluded in the question. I have posted an answer to the question using stronger assumptions on $X$ and $G$ below and would be happy for comments. $\endgroup$ – JoS Feb 4 '16 at 10:11
  • $\begingroup$ I will look at your answer below. But in my example, I can always replace $\mathbb{A}^4$ by $\mathbb{A}^4\setminus \{(0,0,0,0)\}$. That does not change the GIT quotient. $\endgroup$ – Jason Starr Feb 4 '16 at 10:57
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    $\begingroup$ I don't see why the quotient map from $\mathbb{A}^4$ to the quadratic cone, restricted to $\mathbb{A}^4 \setminus \{0\}$ is still a categorical quotient. By Theorem 1.10 in GIT, it should then still be affine, however the target remains affine but $\mathbb{A}^4 \setminus \{0\}$ is not. $\endgroup$ – JoS Feb 4 '16 at 11:28
  • $\begingroup$ You are correct, I was wrong. If you remove the origin, then the semistable locus of the new (quasi-affine) scheme equals the properly stable locus of the original scheme. The quotient stack of the entire quasi-affine scheme is a Deligne-Mumford stack whose coarse moduli space is a non-separated scheme: the quadric cone with doubled vertex. So my example above is wrong. $\endgroup$ – Jason Starr Feb 4 '16 at 11:58
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Since I asked the question, I have come to realize that in my application, I know more about $X$ and the action of $G$. For this application, $X$ has finite quotient singularities as it is the coarse moduli space of a smooth DM stack and the action of $G$ comes from an action of the group on this stack in the sense Romagny's paper. Using this, I think I am able to show the desired conclusion. Let me sketch the proof here:

Proposition Let $\mathcal{M}$ be an orbifold, that is a smooth separated Deligne-Mumford stack with connected coarse moduli space and containing a non-empty open substack which is a scheme. Let the smooth group scheme $G$ act on $\mathcal{M}$ with finite, reduced stabilizers at geometric points. Then the quotient $\mathcal{M}/G$ is again a smooth DM stack.

Proof Consider the frame bundle $\mathcal{F} = \text{Fr}\left( T_\mathcal{M} \right)$ of the tangent bundle $T_\mathcal{M}$ of $\mathcal{M}$. We know $\mathcal{F}$ is an algebraic space and for $n=\dim(\mathcal{M})$, the group $\text{GL}_n$ acts on $\mathcal{F}$ on the right by $$ (v_1, \ldots, v_n) . (a_{ij})_{i,j=1}^n = (\sum_{i=1}^n a_{i1} v_i, \ldots, \sum_{i=1}^n a_{in} v_i) $$ and we have that $\mathcal{M} = [\mathcal{F}/\text{GL}_n]$. On the other hand, the action of $G$ on $\mathcal{M}$ induces an action of $G$ on $\mathcal{F}$ by $$ g . (v_1, \ldots, v_n) = (g_* v_1, \ldots, g_* v_n), $$ where $g_*$ denotes the pushforward under the map $p \mapsto gp$ on $\mathcal{M}$. One shows that both actions are strict actions on the stack $\mathcal{F}$ and that they commute (as pushforward is $\mathbb{C}$-linear) and hence induce an action $$G \times \text{GL}_n \times \mathcal{F} \to \mathcal{F}$$ with finite reduced stabilizers.

With these preparations done, we simply note that $$\mathcal{M}/G = [\mathcal{F}/\text{GL}_n]/G = (\mathcal{F}/\text{GL}_n)/G = \mathcal{F}/(\text{GL}_n \times G).$$ The second isomorphism is a consequence of Theorem 4.1 and Remark 2.4 of Romagny's paper. But now $\mathcal{F}$ is a smooth algebraic space and the action of $G \times \text{GL}_n$ has finite, reduced stabilizers at geometric points, so the quotient $\mathcal{F}/(\text{GL}_n \times G)$ is again a smooth Deligne-Mumford stack. qed

Hence instead of first taking the coarse moduli space $X$ of $\mathcal{M}$ and then performing the quotient, one can first take the quotient stack, which is smooth DM, and then take the coarse moduli space, which thus has finite quotient singularities.

I don't know whether this can be used to answer the original question. In Proposition 2.8 of a paper by Vistoli it is shown that any normal complex variety with quotient singularities is the coarse moduli space of a smooth stack. However, to apply this to the question above one would need a way to lift the action of $G$ to a strict action on this stack.

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