Poisson manifold

Remarks

Clauses 1 to 3 say that $(C^\infty(M), \{\cdot, \cdot\})$ is a Lie algebra over $\R$. By clause 4, there is a unique bivector field $\pi$, a section of $\Lambda^2 TM$, with $\{f, g\} = \pi(df, dg)$; the Jacobi identity is then the condition $[\pi, \pi] = 0$ for the Schouten bracket.

Example: every symplectic manifold with its Poisson bracket is a Poisson manifold. Another example is the dual $\mathfrak{g}^*$ of a finite-dimensional Lie algebra $\mathfrak{g}$, with $\{f, g\}(\xi) = \langle \xi, [df_\xi, dg_\xi] \rangle$, where $df_\xi \in (\mathfrak{g}^*)^* = \mathfrak{g}$. On $\mathfrak{g}^*$ for $\mathfrak{g} = \mathfrak{so}(3)$ with suitable coordinates, $\{x, y\} = z$, $\{y, z\} = x$ and $\{z, x\} = y$.

Non-example: on $\R^3$, define a bracket by clauses 1, 2, 4 and $\{x, y\} = 1$, $\{z, x\} = x$, $\{y, z\} = 0$. Then $\{x, \{y, z\}\} + \{y, \{z, x\}\} + \{z, \{x, y\}\} = 0 + \{y, x\} + \{z, 1\} = -1$, so the Jacobi identity fails.