Interior multiplication of differential forms by bi-vector fields

Question

Suppose $M$ is a smooth manifold. If $X:M \to TM$ is a smooth vector field, then we can define a map $\iota_X: \Omega^k(M) \to \Omega^{k-1}(M) $ as a map which takes $\omega\in \Omega^k(M)$ to $\iota_X(\omega)\in \Omega^{k-1}(M)$ whose action is given by $$\iota_X(\omega)(X_1, \ldots, X_{k-1}) = \omega(X,X_1, \ldots, X_{k-1}).$$ Now if $ G:M \to \Lambda^2(TM)$ is a bivector field, how to define a map $\iota_G: \Omega^k(M) \to \Omega^{k-2}(M)$.

Answer 1

One choice is to define $$ \iota_{U\wedge V}(\omega)(X_1,\cdots,X_{k-2})=\omega(U,V,X_1,\cdots,X_{k-2}) $$ for vector fields $U,V$. Of course, one must show that $\iota$ is well defined and uniquely determined by this expression.