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?

  • $\begingroup$ This might help. $\endgroup$
    – Ivo Terek
    Dec 14, 2017 at 16:20
  • $\begingroup$ @IvoTerek Thanks. I had a look at that thread, but it didn't really help, as I'm unsure why the alternating sign appears in the first place? $\endgroup$
    – user35305
    Dec 14, 2017 at 16:23
  • $\begingroup$ The corresponding identity in geometric algebra is $X\,\lrcorner\,(\alpha\wedge\beta)=(X\,\lrcorner\,\alpha)\wedge\beta+(-1)^p\alpha\wedge(X\,\lrcorner\,\beta)$ . The sign comes from anticommuting the vector $X$ past all of the factors in the multivector $\alpha$. This is related to the double cross product/bac-cab formula by $a\,\lrcorner\,(b\wedge c)=-b(a\,\lrcorner\,c)+c(a\,\lrcorner\,b)$ $\endgroup$
    – mr_e_man
    Sep 11, 2018 at 0:42

1 Answer 1


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:

  1. 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.

  2. 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.

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


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