Author: Eiko

Tags: Quiver Variety, Hilbert Scheme of Points, Nakajima Quiver Variety, Hilb^n(C^2)

We summarize why the Hilbert scheme of points ${\mathrm{Hilb}}^n({\mathbb{C}}^2)$ is a quiver variety here.

The Quiver

Consider the following quiver representation with dimension vector $v=(1,n)$, where the left vertex is a framing vertex and only the right vertex receives group actions.

The zero locus of the moment map ${\mu }^{-1}(0)$ is then given by

$$xy-yx+ij=0\in \mathrm{End}(k^n).$$

We can deduce that on ${\mu }^{-1}(0)$, we have

$$ji=\mathrm{tr}(ji)=\mathrm{tr}(ij)=-\mathrm{tr}(xy-yx)=0.$$

We know that if $j=0$, then the equivalence classes of $(x,y,i)$ in which $i$ give a cyclic vector is in bijection with ${\mathrm{Hilb}}^n({\mathbb{C}}^2)$. It turns out that this can be deduced considering only a subset of such representation space, i.e. by choosing a suitable stability parameter $\theta$ and consider the semistable locus.

Setting Stability Parameter Eliminates $j$

Let $\theta =(n,-1)$, then our representation $(x,y,i,j)$ is semi-stable iff it does not contain any subrepresentation of dimension $(1,m)$ for $m<n$, this means if image of $i$ is non-zero, it has to generate the entire $k^n$ together with $x$ and $y$.

With this generation requirement, $j$ will be forced to be $0$. We can see this by considering any closed path $jPi$ where $P$ is any (non-zero) number of products of $x,y$, we have

$$\begin{array}{rl}jPi& =\mathrm{tr}(jPi)\\ & =\mathrm{tr}(ijP)\\ & =-\mathrm{tr}([x,y]P)\\ & =0,\end{array}$$

since we can always put $x$, $y$ to be simultaneously upper triangular form, by a result in linear algebra and using the fact that $[x,y]=-ij$ has rank $1$.

This means $j=0$ since $i$ is cyclic vector and thus it is zero on the entire $k^n$.