Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

1 Answer 1

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! – Hao Sun 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


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.