# Interior product of a differential form with respect to a commutator of two vector fields

I've been working my way through Nakahara's book "Geometry, topology & physics" and I've reached chapter 5 section 4 in which he discusses the interior product and lie derivative of differential forms. There is a question in which he asks the reader, given $X,Y\in\mathscr{X}(M)$ (where $\mathscr{X}(M)$ is the set of all vector fields on $M$) and $\omega\in\Omega^{r}(M)$ (where $\Omega^{r}(M)$ is the set of all $r$-forms on $M$), to show that $$\iota_{[X,Y]}\omega=X(\iota_{Y}\omega)-Y(\iota_{X}\omega)\qquad [1]$$ where $\iota_{X}:\Omega^{r}(M)\rightarrow\Omega^{r-1}(M)$ is the interior product with respect to a vector field $X\in\mathscr{X}(M)$, and for $\omega\in\Omega^{r}(M)$, is defined by $$\iota_{X}\omega\left(X_{1},\ldots,X_{r-1}\right)\equiv\omega\left(X,X_{1},\ldots,X_{r-1}\right).$$

I've tried starting with the simple cases: clearly, fo $f\in\Omega^{0}(M)$ $[1]$ is trivially satisfied, as $\iota_{X}f=\iota_{Y}f=\iota_{[X,Y]}f=0$. However, when I move on to the next most simple case, i.e. $\omega\in\Omega^{1}(M)$, I already encounter issues. If I naively expand $\omega$ on a coordinate basis $\lbrace dx^{\mu}\rbrace$ such that $\omega =\omega_{\mu}dx^{\mu}$ it follows that $$\iota_{[X,Y]}\omega =[X,Y]^{\mu}\omega_{\mu}=\left(X^{\alpha}\partial_{\alpha}Y^{\mu}-Y^{\alpha}\partial_{\alpha}X^{\mu}\right)\omega_{\mu}=\left(X^{\alpha}\partial_{\alpha}Y^{\mu}\right)\omega_{\mu}-\left(Y^{\alpha}\partial_{\alpha}X^{\mu}\right)\omega_{\mu}\\ =X^{\alpha}\partial_{\alpha}\left(Y^{\mu}\omega_{\mu}\right)-Y^{\alpha}\partial_{\alpha}\left(X^{\mu}\omega_{\mu}\right)-X^{\alpha}Y^{\mu}\partial_{\alpha}\omega_{\mu}+Y^{\alpha}X^{\mu}\partial_{\alpha}\omega_{\mu}\qquad\;\;\\=X(\iota_{Y}\omega)-Y(\iota_{X}\omega)+\left(X^{\mu}Y^{\alpha}-Y^{\mu}X^{\alpha}\right)\partial_{\alpha}\omega_{\mu}\qquad\qquad\qquad\qquad\;\;\;\;\\ =X(\iota_{Y}\omega)-Y(\iota_{X}\omega)+d\omega (X,Y)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;$$

I'm probably being really daft, but I can't seem to make this last term vanish - what am I doing wrong?

My ultimate aim of this process is to prove that $\mathcal{L}_{[X,Y]}t=\mathcal{L}_{X}\mathcal{L}_{Y}t-\mathcal{L}_{Y}\mathcal{L}_{X}t$, for an arbitrary tensor field, and I think that I'll need that $\mathcal{L}_{X}\omega =d(\iota_{X}\omega)+\iota_{X}(d\omega)$ to do so.

• Are you sure you haven't missed an assumption? For $r=1$ it looks to me like this equation is equivalent to $d \omega = 0$. Sep 16, 2015 at 12:14
• @AnthonyCarapetis Not that I can tell from reading through the section in the book. Wouldn't $d\omega =0$ only in the case where $\omega$ is a top-form though?
– Will
Sep 16, 2015 at 12:49
• After looking at the book I'm pretty sure it's an error - compare the equation you arrived at with Nakahara's equation 5.70. Sep 16, 2015 at 12:59
• @AnthonyCarapetis Ah ok, so should it be $\iota_{[X,Y]}\omega=X[\iota_{Y}\omega]-Y[\iota_{X}\omega] -d\omega (X,Y)$ then?
– Will
Sep 16, 2015 at 13:15

Your derivation is right and equation [1] cannot be correct without further assumptions. In fact the exterior derivative of a 1-form can be shown to be $$\mathrm{d}\omega(X,Y)= X(\omega(Y)) - Y(\omega(X))-\omega([X,Y])$$ which is equivalent to what you wrote.
• There don't seem to be any other assumptions made in the book when the author poses the question - maybe it's a typo and the $d\omega$ term should be there? Otherwise, doesn't one have to assume that $dim\;M=r$ such that, in the case that I looked at (for $r=1$) we have $d\omega =0$?