1
$\begingroup$

(Cross-post from math.stackexchange, where it has received no attention.)

Orbifolds $\mathbb{C}^2/\mathbb{Z}_n$, given by the action $(x, y) \mapsto (\zeta x, \zeta^{-1} y)$ with $\zeta$ a primitive $n^\text{th}$ root of unity, admit smooth deformations. This is because they are hypersurface singularities; there are three independent invariants $u_1 = x^n$, $u_2 = y^n$, and $u_3 = xy$, and they satisfy one relation: $u_1 u_2 - u_3^n = 0$. Deforming to $u_1 u_2 - u_3^n + t = 0$ gives a flat family, and the fibres away from $t=0$ are smooth.

What about other orbifolds, like $(x, y) \mapsto (\zeta x, \zeta y)$? Is there an easy way to check whether they are smoothable?


(Some context: I know that not all orbifolds are smoothable, but don't really understand why. For example, it seems to be common knowledge that the orbifold $\mathbb{C}^3/\mathbb{Z}_3$, given by the action $(x, y, z) \to (\zeta x, \zeta y, \zeta z)$ where $\zeta = e^{2\pi i/3}$, is rigid. The invariant ring has 10 generators, satisfying 27 relations, and I have no idea how to go about showing that this admits no deformations.)

$\endgroup$
4
  • $\begingroup$ How do you deform the group action? the members of your flat family again orbit spaces? $\endgroup$ Apr 17, 2013 at 9:25
  • $\begingroup$ I'm just interested in deformations of these as varieties, so I'm not sure what you mean by 'deform the group action'. In the du Val examples, the smoothed varieties are not orbit spaces. $\endgroup$ Apr 17, 2013 at 9:44
  • 1
    $\begingroup$ You might find the paper by Schlessinger, "Rigidity of quotient singularities." Invent. Math. 14 (1971), 17–26 useful. In particular, it should explain the phenomenon described in your last paragraph. $\endgroup$
    – naf
    Apr 17, 2013 at 10:46
  • $\begingroup$ I get the impression that the OP writes "orbifold" to mean "geometric quotient of an orbifold". That is, a variety rather than a stack. In any case, it is true that every 2-dimensional quotient singularity is smoothable, because every 2-dimensional rational singularity is smoothable. See, for example, Artin's "An algebraic construction of Brieskorn's resolutions". $\endgroup$
    – inkspot
    Apr 17, 2013 at 13:51

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.