1
$\begingroup$

Let $X$ be a complex algebraic variety. Let $C\subset X$ be a compact (reduced) subvariety. Let $C^{(n)}$ denote the $n$th infinitesimal neighborhood of $C$ inside $X$. Let $Hilb(X)$ denote the Hilbert scheme of $X$. Thus $C$ defines the point $[C]\in Hilb(X)$.

In my situation one may assume that $X$ and $C$ are smooth varieties, and moreover $[C]$ is a smooth point of $Hilb(X)$.

I was told (here Basic questions on the Hilbert scheme/ Douady space) that $Hilb(C^{(n)})$ is a closed subscheme of $Hilb(X)$.

Question. Is it true that the $n$th infinitesimal neighborhood of the point $[C]$ in $Hilb(X)$ is equal to the connected component of $Hilb(C^{(n)})$ containing $[C]$ as a closed point?

I am particularly interested in a modification of this question when $X,C$ are complex analytic rather than algebraic manifolds; in that case Hilbert scheme is replaced by Douady space.

References are most welcome.

$\endgroup$
2
$\begingroup$

That already fails when $n$ equals $2$, $X$ equals $\mathbb{P}^3$ and $C$ is a line in $\mathbb{P}^3$. Choose homogeneous coordinates $[y_0,y_1,y_2,y_3]$ on $\mathbb{P}^3$ so that $C$ is $Z(y_2,y_3)$. Then an affine neighborhood $U$ of $[C]$ in $\text{Hilb}(X)$ is the affine $4$-space with affine coordinates $(a_{2,0},a_{2,1},a_{3,0},a_{3,1})$ associated to the closed subscheme $$Z(\widetilde{y}_2,\widetilde{y}_3), \ \ \widetilde{y}_2 := y_2-a_{2,0}y_0-a_{2,1}y_1, \ \ \widetilde{y}_3 := y_3-a_{3,0}y_0-a_{3,1}y_1.$$ The closed subscheme $C^{(2)}$ has homogeneous defining ideal $\langle y_2^2,y_2y_3,y_3^2\rangle$. Thus, the closed subscheme $\text{Hilb}(C^{(2)})\cap U$ of $U$ is the maximal closed subscheme such that each of the elements $$ y_2^2 = (\widetilde{y}_2+a_{2,0}y_0+a_{2,1}y_1)^2,y_2y_3 = (\widetilde{y}_2+a_{2,0}y_0+a_{2,1}y_1)(\widetilde{y}_3+a_{3,0}y_0+a_{3,1}y_1),y_3^2 =(\widetilde{y}_3+a_{3,0}y_0+a_{3,1}y_1)^2$$ is contained in the homogeneous ideal $\langle \widetilde{y}_2,\widetilde{y}_3 \rangle.$ Expanding out, this closed subscheme has defining ideal $$\langle a_{2,0}^2,a_{2,0}a_{2,1},a_{2,1}^2,a_{3,0}^2,a_{3,0}a_{3,1},a_{3,1}^2,a_{2,0}a_{3,0},a_{2,1}a_{3,1},a_{2,0}a_{3,1}+a_{2,1}a_{3,0} \rangle.$$ This ideal sits inside the full ideal $\langle a_{i,j} \rangle^2$, and the cokernel has length $1$. Thus, the component of $[C]$ in $\text{Hilb}(C^{(2)})$ contains the second order neighborhood of $[C]$ inside $\text{Hilb}(X)$, but the inclusion is strict.

$\endgroup$
2
  • $\begingroup$ Thanks a lot! I will have to think if I can modify the question. $\endgroup$
    – makt
    Aug 20 '14 at 14:13
  • $\begingroup$ It seems to me that $[C]^{(n)}$ is a closed subscheme of $Hilb(C^{(n)})$ in general. Moreover $[C]$ is the only closed point in its connected component in $Hilb(C^{(n)})$. Is that correct? Can one say anything more precise in general? $\endgroup$
    – makt
    Aug 20 '14 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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