Problem proving Cartan's identity

Question

There is a famous identity stating that, if $X$ is a field and $\omega$ a form, then:

$$\mathcal{L}_X\omega=\iota_Xd\omega+d\iota_X\omega.$$

I'm trying to prove it. Thanks to Anthony Carapetis, I know that:

$$\iota_X(\alpha\wedge\beta)=(k+\ell)(\iota_X\alpha\wedge\beta+(-1)^k\alpha\wedge\iota_X\beta),$$

where $\alpha$ is a $k$-form and $\beta$ an $\ell$-form. Now I start on Cartan. I first suppose $\omega=d\alpha$ for $\alpha$ a $(k-1)$-form. Then:

$$\mathcal{L}_Xd\alpha=d\mathcal{L}_X\alpha=d(d\iota_X\alpha+\iota_Xd\alpha)=d\iota_Xd\alpha=d\iota_X\omega,$$

and being $\omega$ exact, it is closed, so $\iota_Xd\omega=\iota_X0=0$. Now consider $f\omega$ for any smooth function $f$. $f$ will then be a 0-form. If I try induction, I get stuck with:

$$\mathcal{L}_X(f\omega)=f\mathcal{L}_X\omega+\mathcal{L}_Xf\cdot\omega=fd\iota_X\omega+\iota_Xdf\cdot\omega,$$

whereas the RHS of Cartan would be:

\begin{align*} d\iota_X(f\omega)+\iota_Xd(f\omega)={}&d(kf\iota_X\omega)+\iota_X(df\wedge\omega)={} \\ {}={}&kdf\wedge\iota_X\omega+kfd\iota_X\omega+(k+1)\iota_Xdf\wedge\omega-(k+1)df\wedge\iota_X\omega, \end{align*}

and those coefficients get in the way, because the two sides only equate for $k=1$. Am I doing something wrong? Have I used the linked question too hastily to deduce a formula for the normalized wedge?

Details

Anthony proved, in his answer, that, if $\overline\wedge$ denotes the unnormalized antisimmetrization, then:

$$\iota_X(\alpha\overline\wedge\beta)=k\iota_X\alpha\overline\wedge\beta+(-1)^k\ell\beta\overline\wedge\iota_X\beta.$$

The relationship between $\wedge$ and $\overline\wedge$ is that, if $\alpha$ is a $k$-form and $\beta$ an $\ell$-form:

$$\alpha\wedge\beta=\frac{(k+\ell)!}{k!\ell!}\alpha\overline\wedge\beta.$$

The above identity then becomes:

$$\iota_X\left(\frac{k!\ell!}{(k+\ell)!}\alpha\wedge\beta\right)=\frac{(k-1)!\ell!}{(k+\ell-1)!}k\iota_X\alpha\wedge\beta+(-1)^k\frac{k!(\ell-1)!}{(k+\ell-1)!}\ell\alpha\wedge\iota_X\beta,$$

which with due simplifications done after extracting the fraction from the parenthesis on the left yields:

$$\frac{1}{k+\ell}\iota_X(\alpha\wedge\beta)=\iota_X\alpha\wedge\beta+(-1)^k\alpha\wedge\iota_X\beta,$$

which is almost identical to what I said at the start of the question after the link to my previous question. Am I doing something wrong here?

Answer 1

As Anthony noted in his comment, correcting the normalization coefficient by eliminating the denominator $(k+\ell)!$ removes the problem and yields:

$$\iota_X(\alpha\wedge\beta)=\iota_X\alpha\wedge\beta+(-1)^k\alpha\wedge\iota_X\beta,$$

which allows for Cartan to be proved my way.

Another way is this, which John Ma posted in his comment, but which is out of my way because it uses homotopy of chains and I have left homotopy as path homotopy.

I'm posting this to get this question answered.