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

Let $f$ be a polynomial in $k$ complex variables, and suppose the affine variety $V$ given by $f = 0$ has an isolated singularity at the origin, but is otherwise smooth. Now assume that some cyclic group $G \cong \mathbb{Z}_n$ acts on $\mathbb{C}^k$ in such a way that $V$ is invariant, and the origin is the unique fixed point. Are there simple conditions which determine whether $V/G$ can be smoothed?

An easy-to-read reference on smoothing singularities would also be appreciated.


Context:

I am interested in quotients of the threefold node, described in $\mathbb{C^4}$ by any non-degenerate homogeneous quadratic polynomial $f$, by cyclic subgroups of $GL(4,\mathbb{C}$). If $f$ is invariant, then by choice of coordinates it can be put in the form $f = x_1 x_4 - x_2 x_3$, $G$ acts on the whole family given by $f_t = f + t$, and this suffices to smooth the quotient. If $f$ is not invariant, then I'm not sure what happens, as there is generally no equivariant smoothing of the node (consider $(x_1, x_2, x_3, x_4) \to (i x_1, i x_2, i x_3, -i x_4)$ and $f \to -f$).


(Edit: No bites, a day later. Is the question unclear?)

share|improve this question
    
I'm a bit confused. In the second example you give, is it also a smoothing by itself as the quotient will identify $f_t$ and $f_{-t}$ as one smooth fiber for $t\neq 0$? –  CYXU Mar 9 '13 at 13:13
    
I see what you mean, but that seems strange: the smooth fibres are exactly the same as those for the deformations of the 'upstairs' geometry. –  Rhys Davies Mar 11 '13 at 16:27

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.