The Moyal Product

Today I want to understand the Moyal product, as we will need to understand it in order to construct quantizations of symplectic quotients. (More precisely, to incorporate stability conditions.)



Let $A$ be the algebra of polynomial functions on $T^\ast \mathbb{C}^n$. This algebra has a natural Poisson bracket, given by
$$\{p_i, x_j\} = \delta_{ij}.$$
We would like to define a new associative product $\ast$ on $A((\hbar))$ satisfying:

1. $f  \ast g = fg + O(\hbar)$
2. $f \ast g - g \ast f = \hbar \{f, g\} + O(\hbar^2)$
3. $1 \ast f = f \ast 1 = f$
4. $(f \ast g)^\ast = -g^\ast \ast f^\ast$

In the last line, the map $(\cdot)^\ast$ takes $x_i \mapsto x_i$ and $p_i \mapsto -p_i$. To figure out what this new product should be, let's take $f,g \in A$ and expand $f \ast g$ in power series:

$$f \ast g = \sum_{n=0}^\infty c_n(f,g) \hbar^n$$

Now, equations (1) and (2) will be satisfied by taking $c_0(f,g) = fg$ and $c_1(f,g) = \{f,g\}/2$. Let $\sigma$ be the Poisson bivector defining the Poisson bracket. This defines a differential operator $\Pi$ on $A \otimes A$ by

$$\Pi = \sigma^{ij} (\partial_i \otimes \partial_j)$$

Let $B = \sum_{n=0}^\infty B_n \hbar^n$ and write the product as

$$f \ast g = m \circ B(f \otimes g).$$
Now, condition (2) tells us that $B(0) = 1$ and that
$$\left. \frac{dB}{d\hbar} \right|_{\hbar=0} = \frac{\Pi}{2}$$
So
$$B = 1 + \frac{\hbar \Pi}{2} + O(\hbar^2)$$
It is natural to guess that $B$ should be built out of powers of $\Pi$, and a natural guess is
$$B = \exp(\frac{\hbar \Pi}{2}),$$
which certainly reproduces the first two terms of our expansion. Let's see that this choice actually works, i.e. defines an associative $\ast$-product. Let $m: A \otimes A \to A$ be the multiplication, and
$m_{12}, m_{23}: A \otimes A \otimes A \to A \otimes A$, $m_{123}: A \otimes A \otimes A \to A$ the induced multiplication maps. Then
$$\begin{align} f \ast (g \ast h) &= m \circ(B( f \otimes m \circ B(g \otimes h) ) ) \\\ &= m \circ B( m_{23} \circ (1 \otimes B)(f \otimes g \otimes h) ) \\\ &= m_{123} (B \otimes 1)(1 \otimes B)(f \otimes g \otimes h) \end{align}$$
On the other hand, we have

$$\begin{align} (f \ast g) \ast h) &= m \circ(B( m \circ B(f \otimes g) \otimes h) ) ) \\\ &= m \circ B( m_{12} \circ (B \otimes 1)(f \otimes g \otimes h) ) \\\ &= m_{123} (1 \otimes B)(B \otimes 1)(f \otimes g \otimes h) \end{align}$$

Hence, associativity is the condition
$$m_{123} \circ [1\otimes B, B \otimes 1] = 0.$$

On $A \otimes A \otimes A$, write $\partial_i^1$ for the partial derivative acting on the first factor, $\partial_i^2$ on the second, etc. Then
$$1 \otimes B = \sum_n \frac {\hbar^n}{2^n n!}  \Pi^{i_1 j_1} \cdots \Pi^{i_n j_n} \partial^2_{i_1} \partial^3_{j_1} \cdots \partial^2_{i_n} \partial^3_{j_n}$$
and similarly for $B \otimes 1$. So we have
$$\begin{align} m_{123} (B\otimes 1)(1 \otimes B) &= \sum_n \sum_{k=0}^n \frac {\hbar^n}{2^n k! (n-k)!}  \Pi^{k_1 l_1} \cdots \Pi^{k_k l_k} \partial_{k_1} \partial_{l_1} \cdots \partial_{k_k} \partial_{l_k} \\\  & \ \times  \Pi^{i_1 j_1} \cdots \Pi^{i_{n-k} j_{n-k}} \partial_{i_1} \partial_{j_1} \cdots \partial_{i_{n-k}} \partial_{j_{n-k}} \\\ &= m_{123}(1 \otimes B)(B \otimes 1) \end{align}$$
Hence we obtain an associative $\ast$-product. This is called Moyal product.

Sheafifying the Construction

Now suppose that $U$ is a (Zariski) open subset of $X = T^\ast \mathbb{C}^n$. Then the star product induces a well-defined map
$$\ast: O_X(U)((\hbar)) \otimes_\mathbb{C} O_X(U)((\hbar)) \to O_X(U)((\hbar))$$
In this way we obtain a sheaf $\mathcal{D}$ of $O_X$ modules with a non-commutative $\ast$-product defined as above.

Define a $\mathbb{C}^\ast$ action on $T^\ast \mathbb{C}^n$ by acting on $x_i$ and $p_i$ with weight 1. Extend this to an action on $\mathcal{D}$ by acting on $hbar$ with weight -1.

Proposition: The algebra $C^\ast$-invariant global sections of $\mathcal{D}$ is naturally identified with the algebra of differential operators on $\mathbb{C}^n$.

Proof: The $\mathbb{C}^\ast$-invariant global sections are generated by $\hbar^{-1} x_i$ and $\hbar^{-1} p_i$. So define a map $\Gamma(\mathcal{D})^{\mathbb{C}^\ast} \to \mathbb{D}$ by
$$\hbar^{-1} x_i \mapsto x_i$$
$$\hbar^{-1} p_i \mapsto \partial_i$$
From the definition of the star product, it is clear that this is an algebra map, and that it is both injective and surjective.