Poisson bivector induced by a Poisson manifold?

A Poisson manifold is a pair $(M, \{\cdot, \cdot\}_M)$ where $M$ is a manifold and $$\{\cdot, \cdot\}: C^\infty(M)\times C^\infty(M)\longrightarrow C^\infty(M)$$ is a Lie bracket such that $$\{f, gh\}=g\{f, h\}+\{f, g\} h.$$ I read $\{\cdot, \cdot\}$ induces a Poisson bivector $\Pi$, that is, a section of the second exterior power of $TM$, that is, $\Pi\in \Gamma(\Lambda^2 TM)$, by: $$\Pi(df, dg):=\{f, g\}.$$ I know that $$\Gamma(\Lambda^2 TM)\simeq \textrm{Alt}^2_{C^\infty(M)}(\Omega^1(M), C^\infty(M)),$$ where $\Omega^1(M)$ is the space of $1$-forms on $M$ and $\textrm{Alt}^2_{C^\infty}$ is the space of alternating $C^\infty(M)$-bilinear maps.

Can anyone explain me the definition of $\Pi$. Why is $\Pi$ evaluated on $df$ and $dg$?

Thanks

• The Poisson bracket, by definition, is a derivation in each argument. You can use this to show that its value on $f, g$ only depends on the values of $df, dg$. – Qiaochu Yuan Feb 27 '17 at 20:09
• To elaborate on the previous comment, use the formula $\{f,g\} = \sum_{i,j} \frac{\partial f}{\partial x_i} \frac{\partial g}{\partial x_j} \cdot \{x_i,x_j\}$ where $x_1,\dots,x_n$ is a set of local coordinates. – Nick Feb 28 '17 at 4:35