Alternating sign in the interior product If $X$ is a vector field on a manifold $M$, then it induces an interior product $\iota_{X}:\Omega^{p}(M)\rightarrow\Omega^{p-1}(M)$, where $\Omega^{p}(M)$ is the set of all $p$-forms on $M$. For a 1-form, $\omega$, we therefore have that $\iota_{X}\omega=\omega(X).$
Given this, if $\alpha$ is a $p$-form and $\beta$ is a $q$-form, then $$\iota_{X}(\alpha\wedge\beta)=(\iota_{X}\alpha)\wedge\beta +(-1)^{p}\,\alpha\wedge (\iota_{X}\beta)$$
My question is (and apologies if it is a naive one): What is the reason for the alternating sign? Is it possible to show why this must be the case?
 A: NB by definition the appearance of the factor $(-1)^p$ in the second term precisely makes $\iota_X$ an antiderivation, rather than a derivation (for which that factor does not appear). It's not clear what "why" means here exactly, but here are a few relevant facts that give explanations of various sorts:


*

*The interior product can be (and usually is) defined explicitly by
$$(\iota_X \alpha)(Y_1, \ldots, Y_{p - 1}) := \alpha(X, Y_1, \ldots, Y_{p - 1}),$$
and checking directly using the definition of wedge product shows that it is indeed an antiderivation and not a derivation.

*Any derivation $D : \Omega^p(M) \to \Omega^{p - 1}(M)$ of degree $-1$ on $\Omega^{\bullet}(M)$) is necessarily zero on $1$-forms: For any such operation and any $1$-form $\alpha$, we have
$$0 = D(\alpha \wedge \alpha) = D(\alpha) \wedge \alpha + \alpha \wedge D(\alpha) = 2 D(\alpha) \alpha .$$
In particular, no such map can satisfy the property $\iota_X \alpha = \alpha(X)$ that we require of interior multiplication.

*The interior multiplication map satisfies a certain duality with the exterior derivative $d$---which, like any interior multiplication operator $\iota_X$, is an antiderivation---
$$\mathcal L_X \omega = \iota_X d\omega + d(\iota_X \omega) .$$
