Poisson homology

Recall that

$$\del_{\Pi}=i_{\Pi}d - di_{\Pi}\: \Om^k M\to \Om^{k-1} M.$$

We showed that $\del_{\Pi}^2=0$ and defined Poisson homology as the homology of the complex $(\Om^{\bullet}, \del_{\Pi})$.





Note that $\partial_\pi(f_0 df_1)=\poiss{f_0}{f_1}$ and therefore The $0$-th Poisson homology is just given by: $\Coo(M)/ \poiss{\Coo(M)}{\Coo(M)}$. Thus it can be considered as the dual space to Poisson traces. This apparently easy definition does not mean that, even in very explicit examples, such invariant can be easily computed.





Poisson homology is functorial. Given a Poisson map $\vf\: M_1\to M_2$ there is a map $\vf^*\:\rH_k^{\Pi}(M_2, \Pi_2)\to \rH_k^{\Pi}(M_1, \Pi_1)$. In particular for any leaf $S$ of $M$

$$\xymatrix{ \rH_k^{\Pi}(S) \ar[d]_{\simeq} \ar[r]^{\vf^*} & \... ...r[d]_{\simeq} \\ \rH^{n-k}_{\mathrm{DR}}(S) \ar[uur]_{\vf*} }$$

Again deciding whether this map is injective or surjective is a difficult problem.

In the canonical double (mixed) complex you have $d\del_{\Pi}+\del_{\Pi}d=0$

$$\xymatrix{ \ar[d] & \ar[d] & \ar[d] \\ \Om^2(M) \ar[d]^{\d... ...^1(M) \ar[d]^{\del_{\Pi}} & \Om^0(M) \ar[l]_{d} \\ \Om^0(M) }$$

Starting from this you can define cyclic (negative, periodic) Poisson homology and a long exact sequence of Connes-type. Consider on $\bR^3$ the Poisson bracket
$$\begin{align*} \poiss{x_2}{x_3} &= 2p x_2x_3-q x_1^2 = g_1, \\ \poiss{x_1}{x_3}... ..._3-q x_2^2 = g_2, \\ \poiss{x_1}{x_2} &= 2p x_1x_2-q x_3^2 = g_3. \end{align*}$$
Check that

$$\phi=\frac{q}{3}(x_1^3+x_2^3+x_3^3)-2p x_1x_2x_3$$

is a Casimir element. Can you prove that there are no other functionally independent Casimirs?

Let

$$\nabla=(\del_{x_1}, \del_{x_2}, \del_{x_3}).$$

Verify that

$$\nabla\phi = (g_1, g_2, g_3),$$

and that

$$\nabla \times(g_1, g_2, g_3)=0.$$

(here we are denoting $\nabla\times$ to be the curl as in usal vector calculus). Then, again by direct computation you can verify that

$$\del_{\Pi}(x_1dx_1 + x_2dx_2 +x_3dx_3)=\nabla(x_1, x_2, x_3)\cdot\nabla\phi,$$

$$\del_{\Pi}(x_1dx_2\wedge dx_3 + x_2dx_3\wedge dx_1 + x_3dx_1... ...abla(x_1, x_2, x_3)d\phi - d[(x_1, x_2, x_3)\cdot \nabla\phi],$$

$$\del_{\Pi}(fdx_1\wedge dx_2\wedge dx_3)= - df\wedge d\phi.$$

These formulas are basically all one needs to thoroughly compute in an explicit manner the Poisson homology groups, as explained in [76].

The result of computation of Poisson homology is that $\rH_*^{\Pi}(\bR^3)$ is a free $\bR[\phi]$-module of rank 8, 8, 1, 1. $\rH_2^{\Pi}(\bR^3)$ is generated by $x_1dx_2dx_3$, $x_2dx_3dx_1$, $x_3dx_1dx_2$. $\rH_3^{\Pi}(\bR^3)$ is generated by $dx_1dx_2dx_3$.