Poisson homology and modular class

Say $\Om$ is a volume form on $M$.

$$\del_{\Pi}\Om = i_{\Pi}d\Om - di_{\Pi}\Om = -i_{\phi_{\Om}}\Om.$$

If $M$ is unimodular Poisson then there exists $\Om\in\Om^n(M)$ such that $\phi_{\Om}=0$, so $\del_{\Pi}\Om=0$ and thus $[\Om]\neq 0\in\rH_n^{\Pi}(M)$.

For this reason in quantization you can regard Connes axiom of having ''quantum'' homological dimension equal classical dimension as a condition of unimodularity of the underlying Poisson manifold.

Let us now consider the Poisson structure of example (). We want to compute its modular form starting from the standard volume form $\Omega=dx_1\wedge dx_2\wedge dx_3$. This means we want, for any $f\in\Coo(M)$

$$L_{X_f}\Omega=\phi(f)\Omega$$

Then, explicitely
$$\begin{align*} L_{X_{x_1}}\Omega &= d i_{g_3\partial_2-g_2\partial_3}\Omega \\ ... ... dx_2\right) \\ &= \left(\partial_2g_3-\partial_3 g_2\right)\Omega \end{align*}$$
And similar computations show that

$$\phi_\Omega(x_i)=\det\left(\begin{array}{cc}\partial_j &\partial_k \\ g_j & g_k\end{array}\right)$$

with $(i,j,k)=(1,2,3)$ or cyclic permutations. Therefore $\phi_\Omega=\nabla\times g$ (up to now we've never used the explicit form of $g$). Lastly, as remarked, $g$ is defined in such a way that $\nabla\times g=0$ and therefore such Poisson structure is unimodular. It is worth remarking that van den Bergh in its paper was commenting that this condition is exactly what makes computations of Poisson homology accesible through explicit formulas (unimodularity was at that time not recognized as an easily accesible, though very relevant, invariant of Poisson manifolds).