Author: Eiko

Time: 2024-12-14 11:37:09 - 2024-12-14 11:37:09 (UTC)

References: Lectures on the Hilbert Schemes of Points on Surfaces by Hiraku Nakajima

Recall On $({\mathbb{A}}^2{)}^{[n]}$

Recall that the simplest Hilbert scheme $({\mathbb{A}}^2{)}^{[n]}$ can be expressed as the space of $n$-dimensional cyclic $\mathbb{C}[x,y]$-module together with a cyclic vector. Equivalently,

$$({\mathbb{A}}^2{)}^{[n]}=\{(B_1,B_2,i):[B_1,B_2]=0,\text{no submodule containing }i\}/{\mathrm{GL}}_n(\mathbb{C})$$

It is also easy to pass the module to ideals, by just taking the annihilator of the cyclic vector, $I=\{f\in \mathbb{C}[x,y]:f(B_1,B_2)i=0\}$.

Nakajima’s Proof, With Extra Details

• First Nakajima shows that the differential of the map $(B_1,B_2,i)\mapsto [B_1,B_2]$ has constant rank. Start by looking at the cokernel of the differential $\mathrm{d}\phi :V\to W$, which is $W/\mathrm{Im}f$. Think about the dual space of the cokernel is the kernel of $W^{\ast }\to (\mathrm{Im}f{)}^{\ast }$, and since there is a non-degenerate inner product, the trace form on $W$, such subspace can be identified back to a subspace of $W$. Thus we identify the cokernel in $W$ as the subspace

  $$\begin{array}{rl}\{w:\mathrm{tr}(w\mathrm{d}f)=0\}& =\{w:\mathrm{tr}(w([\mathrm{d}{B}_1,B_2]+[B_1,\mathrm{d}{B}_2]))=0,\mathrm{\forall }\mathrm{d}{B}_1,\mathrm{d}{B}_2\}\\ & =\{w:\mathrm{tr}(\mathrm{d}{B}_1[B_2,w])+\mathrm{tr}(\mathrm{d}{B}_2[w,B_1])=0,\mathrm{\forall }\mathrm{d}{B}_1,\mathrm{d}{B}_2\}\\ & =\{w:[w,B_1]=[w,B_2]=0\}.\end{array}$$

  Here we used an elementary fact that $\mathrm{tr}(A[B,C])$ is a sign representation of $S_3$ and is a cyclic invariant in $A,B,C$.

Luna’s Slice Theorem

Remarks On $({\mathbb{A}}^2{)}^{[n]}$

Symplectic $({\mathbb{C}}^2{)}^{[n]}$

The matrix description can be understood as a holomorphic symplectic quotient, and we have a holomorphic symplectic form $\omega \in {\mathrm{\Omega }}_{({\mathbb{A}}^2{)}^{[n]}}^2$ that is everywhere non-degenerate. In fact

$$\text{Symplectic Structure }X\Rightarrow \text{Symplectic Structure }{X}^{[n]}.$$

Parallel Translation on $({\mathbb{C}}^2{)}^{[n]}$

We can use the affine translation structure on ${\mathbb{C}}^2$ to move every point such that the average of $n$ points is the origin. This gives us a decomposition

$$({\mathbb{C}}^2{)}^{[n]}={\mathbb{C}}^2\times (({\mathbb{C}}^2{)}^{[n]}/{\mathbb{C}}^2).$$

In terms of matrices, this sends

$$(B_1,B_2,i)\mapsto ((\mathrm{tr}{B}_1,\mathrm{tr}{B}_2),(B_1-\mathrm{tr}{B}_1,B_2-\mathrm{tr}{B}_2,i)).$$

And the set $({\mathbb{C}}^2{)}^{[n]}/{\mathbb{C}}^2$ is identifiable with the subset of $({\mathbb{C}}^2{)}^{[n]}$ consisting of traceless matrices.

The Dimension Estimation of ${\pi }^{-1}(n[0])$

We will see the existence of symplectic structure helps us estimate the dimension of fibres of the Hilbert-Chow morphism $\pi :({\mathbb{C}}^2{)}^{[n]}\to {\mathrm{Sym}}^n({\mathbb{C}}^2)$.

• The subvariety ${\pi }^{-1}(n[0])$ lies in the subset $({\mathbb{C}}^2{)}^{[n]}/{\mathbb{C}}^2$ and is isotropic with respect to the symplectic form on $({\mathbb{C}}^2{)}^{[n]}/{\mathbb{C}}^2$. This tells us we must have

  $$\dim {\pi }^{-1}(n[0])\le \frac{\dim ({\mathbb{C}}^2{)}^{[n]}/{\mathbb{C}}^2}{2}=n-1.$$

• Moreover there exists at least one component of ${\pi }^{-1}(n[0])$ of dimension $n-1$.

Proof.

• There is a torus action of $({\mathbb{C}}^{\ast }{)}^2$ acting on ${\mathbb{C}}^2$ given by

$${\mathrm{\Phi }}_{t_1,t_2}(z_1,z_2)=(t_1{z}_1,t_2{z}_2).$$

Lifting to $({\mathbb{C}}^2{)}^{[n]}$ this is the action of multiplying matrices by scalars, or changing the module structure by multiplying scalars.

Hilbert Scheme Of Points On A Surface