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How would you prove the interior product formula? Namely for $\omega \in \Omega^k (X), \mu\in \Omega^l(X)$, where $X$ is smooth manifold with vector field $v$ we have

$$i(v)(\omega \wedge\mu)=i(v)\omega \wedge\mu+(-1)^k\omega\wedge i(v)\mu.$$

I am trying to expand everything from definition but formula gets complicated and I do not know how to proceed.

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up vote 4 down vote accepted

Because the lhs and the rhs are both multilinear, it is sufficient to prove the "interior product formula" when $\omega$ and $\mu$ are decomposable, i.e. wedge products of $1$-forms.

Working in such a reduced context our formula will be an instance of $$\iota_X(\omega_1\wedge\ldots\wedge\omega_p)=\sum_{i=1}^p(-1)^{i-1}(\iota_X\omega_i)\omega_1\wedge \ldots\wedge\widehat{\omega_i}\wedge\ldots\wedge\omega_p.\tag{1}$$ (Where $\widehat{\phantom{a}}$ denotes an omitted factor.)
The formula $(1)$ holds if and only if its two sides assume the same value on any $(X_2,\ldots,X_p).$
So, using the definition of interior product and denoting $X=X_1,$ we have to show that $$(\omega_1\wedge\ldots\wedge\omega_p)(X_1,\ldots,X_p)=\\=\sum_{i=1}^p(-1)^{i-1}(\iota_X\omega_i)(\omega_1\wedge \ldots\wedge\widehat{\omega_i}\wedge\ldots\wedge\omega_p)(X_2,\ldots,X_p)\tag{2}$$ If $A$ is the matrix $(\omega_i(X_j))_{i,j=1,\ldots,p},$ then the lhs of $(2)$ is $\det(A)$ while the rhs of $(2)$ is the Laplace expansion of $\det(A)$ w.r.t. the first column of $A$.

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