The Moyal product for a constant Poisson tensor

Since ${\mathrm{\Pi }}^{ij}$ is constant, every derivative in [F2] is zero, so its Jacobiator vanishes. Thus [F1] gives a Poisson bracket for every constant skew matrix, with no rank assumption.

On ${\mathbb{R}}^{2n}$, $\omega ={\sum }_{i}d{q}^i\wedge d{p}_i$ is closed because $d\omega ={\sum }_{i}d(d{q}^i\wedge d{p}_i)={\sum }_i(d^2{q}^i\wedge d{p}_i-d{q}^i\wedge d^2{p}_i)=0.$ For $X={\sum }_i(a_i{\mathrm{\partial }}_{q^i}+b_i{\mathrm{\partial }}_{p_i})$, ${\iota }_X\omega ={\sum }_i(a_{i}d{p}_i-b_{i}d{q}^i)$; if this vanishes then every $a_i=b_i=0$, so it is pointwise nondegenerate and is symplectic by [F7]. Equating ${\iota }_{X_H}\omega$ with $dH={\sum }_i(H_{q^i}d{q}^i+H_{p_i}d{p}_i)$, [F5] gives $X_H={\sum }_i(H_{p_i}{\mathrm{\partial }}_{q^i}-H_{q^i}{\mathrm{\partial }}_{p_i})$. Therefore $X_G(F)={\sum }_i(F_{q^i}{G}_{p_i}-F_{p_i}{G}_{q^i})$ by [F6], exactly the bracket of [F1] for $\mathrm{\Pi }={\sum }_i{\mathrm{\partial }}_{q^i}\wedge {\mathrm{\partial }}_{p_i}$.

The zeroth exponential term is $fg$, so $B_0=m$. For $r\ge 1$, expanding the formal exponential gives $B_r(f,g)=\frac{1}{2^{r}r!}{\sum }_{i_1,j_1,\dots ,i_r,j_r}{\mathrm{\Pi }}^{i_1{j}_1}\cdots {\mathrm{\Pi }}^{i_r{j}_r}({\mathrm{\partial }}_{i_1}\cdots {\mathrm{\partial }}_{i_r}f)({\mathrm{\partial }}_{j_1}\cdots {\mathrm{\partial }}_{j_r}g)\mathrm{.}$ Each coefficient is a finite bidifferential expression. If either input is $1$, every positive-order term vanishes, so the unit is preserved; the series is well-defined coefficientwise by [F3].

On three independent tensor factors define $P_{ab}={\sum }_{i,j}{\mathrm{\Pi }}^{ij}{\mathrm{\partial }}_{x_a^i}\otimes {\mathrm{\partial }}_{x_b^j}$ for $1\le a<b\le 3$, with each operator acting as the identity on the remaining factor, and define ${\mu }_3(f\otimes g\otimes h)=fgh$. These operators commute because they are sums of products of commuting partial derivatives with constant coefficients. The product rule says that differentiation of the first composite factor is ${\mathrm{\partial }}^{(1)}+{\mathrm{\partial }}^{(2)}$, so $(f\star g)\star h={\mu }_3{e}^{\frac{\mathrm{\hslash }}{2}(P_{13}+P_{23})}{e}^{\frac{\mathrm{\hslash }}{2}{P}_{12}}(f\otimes g\otimes h),$ while differentiating the second composite factor gives $f\star (g\star h)={\mu }_3{e}^{\frac{\mathrm{\hslash }}{2}(P_{12}+P_{13})}{e}^{\frac{\mathrm{\hslash }}{2}{P}_{23}}(f\otimes g\otimes h)\mathrm{.}$ For commuting operators $U,V$, the coefficient of ${\mathrm{\hslash }}^n$ in $e^{\mathrm{\hslash }U\mathrm{/}2}{e}^{\mathrm{\hslash }V\mathrm{/}2}$ is $2^{-n}{\sum }_{r+s=n}{U}^r{V}^s\mathrm{/}(r!s!)=2^{-n}(U+V{)}^n\mathrm{/}n!$ by the binomial theorem. This is a finite identity at each order, so it applies coefficientwise to these formal exponentials. Commutativity of the three operators makes both expressions ${\mu }_3{e}^{\frac{\mathrm{\hslash }}{2}(P_{12}+P_{13}+P_{23})}(f\otimes g\otimes h)$, proving associativity coefficientwise.

The first coefficient is $B_1(f,g)=\frac{1}{2}{\sum }_{i,j}{\mathrm{\Pi }}^{ij}{f}_i{g}_j$. Skew-symmetry of $\mathrm{\Pi }$ gives $B_1(f,g)-B_1(g,f)={\sum }_{i,j}{\mathrm{\Pi }}^{ij}{f}_i{g}_j=\{f,g{\}}_{\mathrm{\Pi }}$. Hence the product in this example satisfies the first-order normalization in [F4]. When $\mathrm{\Pi }=0$, every $P_{ab}=0$ and the product is the ordinary commutative product; no rank or nondegeneracy assumption is needed.