During one of my daily exercises, I was looking for properties of the elements of Cartan calculus. I stumbled on Wikipedia's page about interior products (here), and I've noticed a property that sounds very useful:
$$
\iota_{[X,Y]}\omega=[\mathcal L_X,\iota_Y]\omega.
$$
In Wikipedia's notations, $\iota$ is the interior product, $\mathcal L_X$ is the Lie derivative with respect to the vector field $X$ (and $X$ and $Y$ are vector fields). $\omega$ is a differential form on a manifold $M$. As there is no source given, I'm trying to prove this equality, and failing due to a sign. Maybe I'm doing some stupid error somewhere.
My attempt so far follows.
First: I note that the operator on the right has the property
$$
[\mathcal L_X,\iota _Y](\omega\wedge\eta)=[\mathcal L_X,\iota_Y]\omega\wedge\eta+(-1)^k\omega\wedge[\mathcal L_X,\iota_Y]\eta,
$$
where $\omega$ is assumed to be a $k-$form, and $\eta$ is an arbitrary form. This is the same interaction with wedge product as the left hand side. It follows that I can decide the value of the right hand side locally, where I can expand any form as tensor product of the basis forms. Hence, if I prove that the equality holds for $0-$ and $1-$forms, I'm done.
I start with $0-$forms: I take a function $f$ from $M$ to the field, and compute left and right hand side. Well, "compute": the left hand side is the application of an inner product to a function, that is zero by definition, while on the right hand side I either have a contraction first, annulling $f$, or I have a Lie derivative acting first. The Lie derivative of a function is a function, so it follows that $\iota_Y\mathcal L f=0$, and I'm done for this case.
For $1-$forms: let $\alpha$ be such an $1-$form. The left hand side is
$$
\iota_{[X,Y]}\alpha=\alpha([X,Y]).
$$
I split the calculation of the right hand side in two, and use Cartan's formula $\mathcal L_X=d\iota_X+\iota_Xd$ whenever necessary. Lie derivatives are ugly, all hail Cartan.
$$
\mathcal L_X\iota_Y\alpha=\mathcal L_X(\alpha (Y))=X(\alpha(Y)),\\
\iota_Y\mathcal L_X\alpha=\iota_Yd\iota_X\alpha+\iota_Y\iota_Xd\alpha=Y(\alpha(X))+\underline{d\alpha(Y,X)}.
$$
Underlined for your convenience is the step in which it is most likely I've done an error, but I fail to see why. I state that $\iota_Y\iota_Xd\alpha=d\alpha(Y,X)$ as I am applying $X$ first, and $Y$ second, and $\iota$ places vectors at the beginning of a string. Is it correct? Was I stupid here?
Continuing: I use
$$
d\alpha(Y,X)=Y(\alpha(X))-X(\alpha(Y))-\alpha([Y,X]).
$$
Here comes the failure. I'd expect stuff to cancel out, but that's not happening. The signs of $Y(\alpha(X))$ agree, so they do not cancel.
Where am I doing wrong? I strongly suspect that it is something in the underlined passage, but I need clarification about what I did wrong.
Thanks all in advance for your time.
(p.s.: in some other places of this site there is a proof of that relying on a property of $\mathcal L_X$ acting on differential forms. As I said, I despise $\mathcal L_X$. I'd prefer to see the error in this proof, as it is short and nice. Lie has done many wonderful things, and an orrible derivative)
Best Answer
Yes, you are off by a sign on the term you figured. Remember that (for a $2$-form $\phi$) $\iota_X\phi(\cdot) = \phi(X,\cdot)$, so $\iota_Y\big(\iota_X\phi\big)=\phi(X,Y)$.