Let $(M,\omega)$ be a symplectic manifold. Then we can define a Poisson bracket of $f$ and $g$ where $f,g\in C^{\infty}(M)$ by $$\{f,g\}=p(df\wedge dg)\text{ }(1)$$ where $p\in\Gamma(T^2M)$, $p=\omega^{-1}:T^*M\to TM$.
I want to connect/understand this definition with the definition that $\{f,g\}=\omega(X_f,X_g)\text{ }(2)$.
Since the symplectic form $\omega$ is non-degenerate, then we can see that we have the isomorphism $$\omega:TM\to T^*M$$ given by $X\to i_X\omega$. Then if we given a smooth function $f$, we can generate the vector field $X_f$ using the identity $$i_{X_f}\omega=df$$ i.e. we literally can define our vector field by $X_f=\omega^{-1}(df)$.
- Does the definition $(1)$ come before $(2)$ since we first need to find $df$ and $dg$ in order to obtain our vector fields $X_f$ and $X_g$?
- Why $p\in\Gamma(\bigwedge^2 TM)$ but $p$ is defined on $\bigwedge^2 T^*M$? Is the map defined by $$p(df\wedge dg)=p(df)\wedge p(dg)$$ where we have the following compositions of maps $$M\to T^*M\wedge T^*M\to TM\wedge TM?$$