Schouten–Nijenhuis bracket

Remarks

For vector fields $X_1, \dots, X_p$ and $Y_1, \dots, Y_q$ with $p, q \ge 1$, the bracket is $$[X_1 \wedge \dots \wedge X_p,\ Y_1 \wedge \dots \wedge Y_q] = \sum_{i=1}^{p} \sum_{j=1}^{q} (-1)^{i+j} [X_i, Y_j] \wedge X_1 \wedge \dots \widehat{X_i} \dots \wedge X_p \wedge Y_1 \wedge \dots \widehat{Y_j} \dots \wedge Y_q,$$ where a hat marks an omitted factor. The bracket also satisfies a graded Jacobi identity, so $\mathfrak{X}^{\bullet + 1}(M)$, with degrees shifted by one, is a graded Lie algebra.

A bivector field $\pi \in \mathfrak{X}^2(M)$ defines the bracket $\{f, g\} = \pi(df, dg)$ on functions. This bracket satisfies the Jacobi identity, so that $\pi$ is a Poisson structure, exactly when $[\pi, \pi] = 0$.

Example: on $\R^2$, let $\pi = \partial_x \wedge \partial_y$. All brackets of the coordinate vector fields vanish, so the formula gives $[\pi, \pi] = 0$, and $\{f, g\} = f_x g_y - f_y g_x$ is the standard Poisson bracket.

Non-example: the wedge product $(P, Q) \mapsto P \wedge Q$ is bilinear but is not this bracket. It fails property 2: for a vector field $X$ and a function $f$, it gives $X \wedge f = fX$, while property 2 requires the function $X(f)$.