Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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 $A$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.