4
$\begingroup$

Let $X$ be a normal scheme with quotient singularities and $Y\subset X$ its singular locus. The first order deformations of $X$ are parametrized by $\mathcal{E}xt^{1}(\Omega_{X},\mathcal{O}_{X})$. This sheaf is supported on $Y$.

I would like to relate $\mathcal{E}xt^{1}(\Omega_{X},\mathcal{O}_{X})$ and $\mathcal{E}xt^{1}(\Omega_{Y},\mathcal{O}_{Y})$ (here I am considering first order deformations of $Y$ as an abstract scheme not its embedded deformations as a subscheme of $X$). To do this I am thinking to consider the conormal exact sequence $$\mathcal{I}_{Y}/\mathcal{I}_{Y}^{2}\rightarrow \Omega_{X|Y}\rightarrow\Omega_{Y}\mapsto 0$$ Applying $\mathcal{H}om(-,\mathcal{O}_{Y})$ we get $$0\mapsto \mathcal{H}om(\Omega_{Y},\mathcal{O}_{Y})\rightarrow \mathcal{H}om(\Omega_{X|Y},\mathcal{O}_{Y})\rightarrow \mathcal{H}om(\mathcal{I}_{Y}/\mathcal{I}_{Y}^{2},\mathcal{O}_{Y})\rightarrow \mathcal{E}xt^{1}(\Omega_{Y},\mathcal{O}_{Y})\rightarrow \mathcal{E}xt^{1}(\Omega_{X|Y},\mathcal{O}_{Y})\rightarrow \mathcal{E}xt^{1}(\mathcal{I}_{Y}/\mathcal{I}_{Y}^{2},\mathcal{O}_{Y})\rightarrow ....$$ Is there any reason why one should have $\mathcal{E}xt^{1}(\mathcal{I}_{Y}/\mathcal{I}_{Y}^{2},\mathcal{O}_{Y}) = 0$ or $\mathcal{H}om(\mathcal{I}_{Y}/\mathcal{I}_{Y}^{2},\mathcal{O}_{Y}) = 0$ ?

$\endgroup$
6
  • $\begingroup$ What scheme structure are you giving the singular locus? $\endgroup$ Jan 13, 2014 at 14:25
  • $\begingroup$ Its reduced induced structure as a subscheme of $X$. $\endgroup$
    – Puzzled
    Jan 13, 2014 at 15:36
  • $\begingroup$ Ok, I would then say that you cannot expect that the $\mathcal{H}om(I_Y/I_Y^2, O_Y) = 0$. In fact, I think it's basically never zero (think about the generic point of $Y$). In terms of the $\mathcal{E}xt^1$, if $Y$ is Gorenstein, this might help as at least you could dualize the problem perhaps? But $X$ just having quotient singularities probably won't help you. Do you need general quotient singularities or do you have a particular variety in mind? Are there any examples worth trying (say even in a computer?) $\endgroup$ Jan 13, 2014 at 16:53
  • $\begingroup$ I need quotient singularities coming from the action of a finite group on a smooth scheme. Thank yuo very much. $\endgroup$
    – Puzzled
    Jan 18, 2014 at 14:53
  • $\begingroup$ Ok, what do you think the first interesting examples of such singularities might be for you? (ie, which ones have you already done). $\endgroup$ Jan 18, 2014 at 21:00

1 Answer 1

2
$\begingroup$

Here is a partial answer.

Let $Y$ be a smooth variety over a field $k$ of characteristic zero. Let $G$ be a finite group acting on $Y$, and let $X = Y/G$ be the quotient. Assume that the set of points where the isotropy is not trivial is in codimension greater or equal than three, that is the singular locus of $X$ is in codimension greater or equal than three. Then $Ext^1(\omega_X,\mathcal{O}_X) = 0$, that is $X$ is rigid.

One can find this in:

  • M. Schlessinger, "Rigidity of quotient singularities", Inventiones mathematicae, 1971, Volume 14, Issue 1, pp. 17-26.

This could fail is $X$ is singular in codimension two. For instance, we may consider the a singular point of type $\frac{1}{6}(2,4)$. Then, étale locally, in a neighborhood of $p$ the surface $X$ is isomorphic to $\mathbb{A}^{2}/\mu_{6}$ where the action is given by $$ \begin{array}{ccc} \mu_{6}\times\mathbb{A}^{2} & \longrightarrow & \mathbb{A}^{2}\\ (\epsilon,x_{1},x_{2}) & \longmapsto & (\epsilon^{2}x_{1},\epsilon^{4}x_{2}) \end{array} $$ The invariant polynomials with respect to this action are clearly $x_{1}^{3},x_{2}^{3},x_{1}x_{2}$. Therefore, étale locally, in a neighborhood of $p$ the surface $X$ is isomorphic to an étale neighborhood of singularity $$S = \{f(x,y,z) = z^{3}-xy = 0\}\subset\mathbb{A}^{3}.$$ Now, we have $$Ext^{1}(\Omega_{S},\mathcal{O}_{S})\cong K[x,y,z]/(f,\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}) = K[x,y,z]/(z^{3}-xy,-y,-x,3z^{2})\cong K[z]/(z^{2}).$$

Therefore $X$ is not rigid.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.