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?

3$\begingroup$ Did you consider $\mathbb{C}^m$ modulo $\pm 1$? My recollection (might be wrong) is that this is not lci for $m = 3,4...$. $\endgroup$– Geordie WilliamsonSep 26, 2018 at 0:27

3$\begingroup$ 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 $x\mapsto xx^{\rm T}$. $\endgroup$– Victor ProtsakSep 26, 2018 at 4:22

2$\begingroup$ In other words, it is the (affine) cone over $\Bbb{P}^{m1}$ embedded by the Veronese map. For $m\geq 3$ it is not Gorenstein, even less lci. $\endgroup$– abxSep 26, 2018 at 6:06

$\begingroup$ 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.osakau.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/19820602/S027309791982149898/… . $\endgroup$– benblumsmithSep 27, 2018 at 2:28
1 Answer
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 CohenMacaulay (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 CohenMacaulay.
If the stabilizers all have order prime to the characteristic, then $X/G$ is CohenMacaulay, though, by the HochsterEagon theorem (Proposition 13 here). Without this condition, even CohenMacaulayness 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 fixedpoint sets have codimension $1$ in the neighborhood of $x$). By the ChevalleyShephardTodd 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.
(Stabilizersgeneratedbybireflections is not sufficient. For examples of finite linear groups generated by bireflections whose action on affine 3space over $\mathbb{C}$ has a quotient that is not a complete intersection, see here.)