# Problem proving Cartan's identity

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?

• I think the normalization factor should just be $1/(k! l!)$. For example Spivak has $\omega \wedge \eta = (k+l)!/k!l! \;{\rm Alt}(\omega \otimes \eta)$ but his ${\rm Alt}$ includes the normalization factor of $1/(k+l)!$. If you use this definition then you get the nice Leibniz formula for the interior product without the coefficients. Nov 2, 2015 at 15:15
• I got the coefficient off Lee, perhaps I didn't read what his Alt was :). Indeed, all problems come from that $(k+\ell)!$ in the normalization, so removing it fixes everything. Nov 2, 2015 at 16:24
• Indeed Lee does like Spivak with his Alt including the normalization factor. Nov 2, 2015 at 16:25
• Another method: math.stackexchange.com/questions/1480545/…
– user99914
Nov 2, 2015 at 21:20
• @JohnMa OMG!!! What is that?? No really, it looks so damn complex, much shorter to just fix a normalization and proceed my way :). Interesting reference anyway. Admitting one knows about homotopy -- and I fear path homotopy is not enough, and higher-dimension homotopies are total strangers to me :). Nov 2, 2015 at 21:30

As Anthony noted in his comment, correcting the normalization coefficient by eliminating the denominator $(k+\ell)!$ removes the problem and yields:
$$\iota_X(\alpha\wedge\beta)=\iota_X\alpha\wedge\beta+(-1)^k\alpha\wedge\iota_X\beta,$$