An exercise in quantum Hamiltonian reduction

This post has been migrated from my old blog, the math-physics learning seminar.

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:

\[ \begin{aligned} {f_{ij}, f_{kl}} &= {p_j^\mu x_i^\mu, p_l^\nu x_k^\nu} \cr &= x_i^\mu p_l^\nu \delta_{jk} \delta^{\mu\nu} - p_j^\mu x_k^\nu \delta_{il} \delta^{\mu\nu} \cr &= f_{il} \delta_{jk} - f_{kj} \delta_{il}. \end{aligned} \]

So we see that the invariants form a Poisson subalgebra (as they should!). Let’s compute:

\[ \begin{aligned} \{Z_1, f_{ij}\} &= \sum_k \{ f_{kk}, f_{ij} \} \cr &= \sum_k \left( f_{kj} \delta_{ki} - f_{ik} \delta_{kj} \right) \cr &= f_{ij} - f_{ij} = 0. \end{aligned} \]

Hence \(Z_1\) is central with respect to the invariant functions \(f_{ij}\). Similarly,

\[ \begin{aligned} {Z_2, f_{kl}} &= \sum_{ij} {f_{ij} f_{ji}, f_{kl}} \cr &= \sum_{ij} f_{ij} \left(f_{jl} \delta_{ik} - f_{ki} \delta_{jl} \right) + f_{ji} \left(f_{il} \delta_{jk} - f_{kj} \delta_{il} \right) \cr &= \sum_j f_{kj} f_{jl} - \sum_i f_{il} f_{ki} + \sum_i f_{ki} f_{il} - \sum_j f_{jl} f_{kj} \cr &= 0. \end{aligned} \]

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.