# Interior multiplication of differential forms by bi-vector fields

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 $$\iota_G: \Omega^k(M) \to \Omega^{k-2}(M)$$.

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.
• What is the domain of $U$ and $V$, is it whole of $M$ or some open subsets of $M$? May 4 at 10:20