# Are quotient varieties local complete intersections?

Let $$G$$ be a reductive group acting on the smooth affine variety $$X$$ such that the stabilizers are finite. Is it true that the quotient $$X/G$$ is a local complete intersection (LCI)? In particular, is the quotient of a smooth affine variety to the algebraic action of a finite group LCI? If no, is there any condition on the action that guarantees such a property?

• Did you consider $\mathbb{C}^m$ modulo $\pm 1$? My recollection (might be wrong) is that this is not lci for $m = 3,4...$. Sep 26, 2018 at 0:27
• The quotient described by Geordie Williamson is the determinantal variety $S_1$ of symmetric $m\times m$ matrices of rank $\leq 1$, with the quotient map given by $x\mapsto xx^{\rm T}$. Sep 26, 2018 at 4:22
• In other words, it is the (affine) cone over $\Bbb{P}^{m-1}$ embedded by the Veronese map. For $m\geq 3$ it is not Gorenstein, even less lci.
– abx
Sep 26, 2018 at 6:06
• I think that $\mathbb{C}^m$ mod $\pm 1$ is Gorenstein when $m$ is even; more generally, the quotient of $\mathbb{C}^m$ by a subgroup of $SL(m,\mathbb{C})$ is Gorenstein, see Thm. 1 here: ir.library.osaka-u.ac.jp/repo/ouka/all/12434/ojm11_01_01.pdf . But I agree that for all $m\geq 3$ it is not a complete intersection; see Theorem A here ams.org/journals/bull/1982-06-02/S0273-0979-1982-14989-8/… . Sep 27, 2018 at 2:28

As discussed in the comments, $$X/G$$ need not in general even be Gorenstein, let alone a local complete intersection.
Actually, if the ground field has positive characteristic, $$X/G$$ may not even be Cohen-Macaulay (let alone Gorenstein, let alone LCI). For example, if $$X = \mathbb{A}_k^4$$, where $$k = \overline{\mathbb{F}}_2$$, and $$G$$ is $$\mathbb{Z}/4$$, with the action given by cyclically permuting the coordinate axes, then $$X/G$$ is not Cohen-Macaulay.
If the stabilizers all have order prime to the characteristic, then $$X/G$$ is Cohen-Macaulay, though, by the Hochster-Eagon theorem (Proposition 13 here). Without this condition, even Cohen-Macaulayness is a very delicate question, so from this point on I will assume this condition (orders of all stabilizers are coprime to the ground field characteristic) is met. For example, perhaps you are working over a characteristic zero field.
In this situation, an overly strong sufficient condition is that the stabilizers are generated by elements acting as pseudoreflections (i.e. elements $$g$$ in a stabilizer $$G_x$$, $$x\in X$$ whose fixed-point sets have codimension $$1$$ in the neighborhood of $$x$$). By the Chevalley-Shephard-Todd theorem, this forces $$X/G$$ to be smooth.
I am not aware of a more precise sufficient condition. However, it seems to me a necessary condition is that the stabilizers be generated by bireflections, i.e. elements $$g$$ in a stabilizer $$G_x$$, $$x\in X$$ whose fixed point sets have codimension $$2$$ in a neighborhood of $$x$$. See this paper by Kac and Watanabe, which proves this in the situation of a linear action by a finite group on affine space, but it seems to me the argument works in general.
(Stabilizers-generated-by-bireflections is not sufficient. For examples of finite linear groups generated by bireflections whose action on affine 3-space over $$\mathbb{C}$$ has a quotient that is not a complete intersection, see here.)