I would compute the Schouten bracket of a bivector with itself. I am guessing the bivector is $ P = P^{ij}e_i \wedge e_j$ (repited indices add up), where $P^{ij}=P^{ij}(x)$ are $C^\infty$ functions defined over the manifold and $\{e_i\}_{i=1}^n$ is an holonomic base,
$$ [e_i,e_j] = 0 $$
for all $i,j=1,\dots, n$. So, the Schouten bracket of $P$ will be
$$ [P,P]_S = [P^{ij}e_i\wedge e_j,P^{kl}e_k \wedge e_l ]_S . $$
But, $P^{ij}$ aren't numbers, so they can't go out of the bracket. I try to apply the formula of excercise 4 (http://www.math.uiuc.edu/~ruiloja/Math595/homewrk2.pdf) but my result have been zero, thus I have a mistake.
So, can any body help me?
Thanks
If $$ P = \frac{1}{p!} P^{i_1\dots i_p} \partial_{i_1} \wedge \dots \wedge \partial_{i_p} $$ and $$ Q= \frac{1}{q!} Q^{j_1\dots j_q} \partial_{i_1} \wedge \dots \wedge \partial_{i_p} $$ are two multi-vector fields, then its Schouten-Nijenhuis bracket is expresed in coordinates by $$ [P,Q]^{k_1\dots k_{p+q-1}} = \frac{(-1)^p}{p!(q-1)!} \delta_{i_1\dots i_pj_2\dots j_q}^{k_1\dots k_{p+q-1}} Q^{mj_2\dots j_q} \frac{\partial P^{i_1\dots i_p}}{\partial x^m} + \\ + \frac{1}{(p-1)!q!} \delta_{i_2\dots i_p j_1\dots j_q}^{k_1\dots k_{p+q-1}} P^{mi_2\dots i_p} \frac{\partial Q^{j_1\dots j_q}}{\partial x^m} \, , $$ where $\delta^{i_1\dots i_p}_{j_1\dots j_p}$ is the generalized Kronecker delta.
This formula is given in page 10, equation (1.20) of Lectures on the Geometry of Poisson Manifolds, Izu Vaisman. Springer Bassel AG (1994).