An Exercise in Quantum Hamiltonian Reduction

Semiclassical Setup

Let the group $GL(2)$ act on $V = \mathrm{Mat}_{2\times n}$ and consider the induced symplectic action on $T^\ast V$. If we use variables $(x,p)$ with $x$ a $2 \times n$ matrix and $p$ an $n \times 2$ matrix, then the classical moment map $\mu$ is given by
$$\mu(x,p) = xp$$
This is equivariant with respect to the adjoint action, so we can form the $GL(2)$-invariant functions
$$Z_1 = \mathrm{Tr} \mu$$
$$Z_2 = \mathrm{Tr} (\mu)^2$$
If we think of $x$ as being made of column vectors
$$x = ( x_1 \cdots x_n )$$
and similarly think of $p$ as being made of row vectors, then there are actually many more $GL(2)$ invariants, given by
$$f_{ij} = \mathrm{Tr} x_i p_j = p_j x_i$$
In terms of the invariants, the $Z$ functions are
$$Z_1 = \sum_k f_{kk}$$
$$Z_2 = \sum_{jk} f_{jk} f_{kj}$$
Let us compute Poisson brackets:
$$\begin{align}  \{f_{ij}, f_{kl}\} &= \{p_j^\mu x_i^\mu, p_l^\nu x_k^\nu\} \\\ &= x_i^\mu p_l^\nu \delta_{jk} \delta^{\mu\nu} - p_j^\mu x_k^\nu \delta_{il} \delta^{\mu\nu} \\\ &= f_{il} \delta_{jk} - f_{kj} \delta_{il}. \end{align}$$
So we see that the invariants form a Poisson subalgebra (as they should!). Let's compute:
$$\begin{align} \{Z_1, f_{ij} &= \sum_k \{ f_{kk}, f_{ij} \} \\\ &= \sum_k \left( f_{kj} \delta_{ki} - f_{ik} \delta_{kj} \right) \\\ &= f_{ij} - f_{ij} = 0. \end{align}$$
Hence $Z_1$ is central with respect to the invariant functions $f_{ij}$. Similarly,
$$\begin{align} \{Z_2, f_{kl}\} &= \sum_{ij} \{f_{ij} f_{ji}, f_{kl}\} \\\ &= \sum_{ij} f_{ij} \left(f_{jl} \delta_{ik} - f_{ki} \delta_{jl} \right) + f_{ji} \left(f_{il} \delta_{jk} - f_{kj} \delta_{il} \right) \\\ &= \sum_j f_{kj} f_{jl} - \sum_i f_{il} f_{ki} + \sum_i f_{ki} f_{il} - \sum_j f_{jl} f_{kj} \\\ &= 0. \end{align}$$
So we see that the $Z_i$ are in the center of the invariant algebra. In fact, they generate it, so we'll denote by $Z$ the algebra generated by $Z_1, Z_2$. Let $A$ be the algebra generated by the $f_{ij}$. The inclusion $Z \hookrightarrow A$ can be thought of as a purely algebraic version of the moment map. In particular, given any character $\lambda: Z \to \mathbb{C}$, we can define the Hamiltonian reduction of $A$ to be
$$A_\lambda := A / A\langle \ker \lambda \rangle$$
The corresponding space is of course $\mathrm{Spec} A$.

The Cartan Algebra and the Center

Define functions

$$h_1 = Z_1 = \sum_i f_{ii}$$
$$h_2 = Z_2 = \sum_{ij} f_{ij} f_{ji}$$
$$h_3 = \sum_{ijk} f_{ij} f_{jk} f_{ki}$$
$$h_k = \sum_{i_1, i_2, \ldots, i_k} f_{i_1 i_2} f_{i_2 i_3} \cdots f_{i_k i_1}$$

These are just the traces of various powers of the $n \times n$ matrix $px$. In particular, $h_k$ for $k>n$ may be expressed as a function of the $h_i$ for $i \leq n$. The algebra generated by the $H$ plays the role of a Cartan subalgebra. So we have inclusions
$$Z \subset H \subset A$$

Quantization

Now we wish to construct a quantization of $A$ and $A_\lambda$. The quantization of $A$ is obvious: we quantize $T^\ast V$ by taking the algebraic differential operators on $V$. Denote this algebra by $\mathbb{D}$. It is generated by $x_i$ and (\partial_i\) satisfying the relation
$$[\partial_i, x_j] = \delta_{ij}$$
Then we simply the subalgebra of $GL(2)$-invariant differential operators as our quantization of $A$. Call this subalgebra $U$. We can define Hamiltonian reduction analogously by taking central quotients. So we need to understand the center $Z(U)$, but this is just the subalgebra generated by quantizations of $Z_1$ and $Z_2$, i.e. the subalgebra of all elements whose associated graded lies in $Z(A)$.

More to come: stability conditions, $\mathbb{D}$-affineness, and maybe proofs of some of my claims.