MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a complex smooth variety. We know for $Z\in Hilb^n(X)$, $T_Z(Hilb^n)=Hom(I_Z, O_Z)$. Hence there is a 1-1 correspodence between $f\in Hom(I_Z, O_Z)$ and the set of first-order deformations of Z. My question is can we express the correspodence explicitly. For example, let $X=A^2$ with coordinates x,y and $I_Z=(x,y)$. Take an element $f\in Hom(I_Z, O_Z)$. What is the corresponding first-order deformations of Z. Is it $(x+sf(x), y+sf(y))$?

share|cite|improve this question

You are right. And the same is true in general. Assume $X$ is affine. Choose a collection of generators $g_1,\dots,g_r$ of $I_Z$ --- then $I_Z = Im(O_X^r \stackrel{(g_1,\dots,g_r)}\to O_X)$. Given a map $f:I_Z \to O_Z$ consider the composition $O_X^r \to I_Z \to O_Z$ and choose its lift $\tilde{f} = (\tilde{f}_1,\dots,\tilde{f}_r):O_X^r \to O_X$. Then the first order deformation is given by the ideal $$ Im(O_X^r \stackrel{(g_1 + s\tilde{f}_1,\dots,g_r + s\tilde{f}_r)}\to O_X). $$

share|cite|improve this answer
    
Can you give us a proof? Thank you! – Diego Maradona Sep 21 '10 at 12:00
    
If you replace $X$ by $X \times Spec k[s]/s^2$ and define an ideal by the same formula, you will get a flat family over $Spec k[s]/s^2$ which is the same (since $Hilb$ represents a functor!) as a map $Spec k[s]/s^2 \to Hilb$, and this is the same as a point in the tangent space. Vice versa, any flat ideal in $X \times Spec k[s]/s^2$ can be written in this form. That's all. – Sasha Sep 21 '10 at 18:34

Your Answer

 
discard

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

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