Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

(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.)

share|improve this question
    
How do you deform the group action? the members of your flat family again orbit spaces? –  Peter Michor Apr 17 '13 at 9:25
    
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. –  Rhys Davies Apr 17 '13 at 9:44
1  
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. –  ulrich Apr 17 '13 at 10:46
    
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". –  inkspot Apr 17 '13 at 13:51
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.