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


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


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:


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:


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?


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


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


The above identity then becomes:


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


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?

  • 1
    $\begingroup$ 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. $\endgroup$ Nov 2, 2015 at 15:15
  • $\begingroup$ 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. $\endgroup$
    – MickG
    Nov 2, 2015 at 16:24
  • $\begingroup$ Indeed Lee does like Spivak with his Alt including the normalization factor. $\endgroup$
    – MickG
    Nov 2, 2015 at 16:25
  • 1
    $\begingroup$ Another method: math.stackexchange.com/questions/1480545/… $\endgroup$
    – user99914
    Nov 2, 2015 at 21:20
  • 1
    $\begingroup$ @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 :). $\endgroup$
    – MickG
    Nov 2, 2015 at 21:30

1 Answer 1


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


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .