I have seen around the internet that this should hold:


where $X$ is a vector field, $\alpha$ a $k$-form, $\beta$ an $\ell$-form, $\iota_X$ is the interior product (i.e. $\iota_X\alpha(v_1,\dotsc,v_{k-1})=\alpha(X,v_1,\dotsc,v_{k-1})$), and $\wedge$ is the exterior product. Now I define the exterior product as:

$$\alpha\wedge\beta(v_1,\dotsc,v_k,v_{k+1},\dotsc,v_{k+\ell})=\sum_{\sigma\in S_{k+\ell}}\operatorname{sgn}\sigma\alpha(v_{\sigma(1)},\dotsc,v_{\sigma(k)})\beta(v_{\sigma(k+1)},\dotsc,v_{\sigma(k+\ell)}),$$

where some others define it with a coefficient in front of it involving factorials of $\ell$ and $k$. I tried all I could to prove the above identity. I reduced it to proving the case $\alpha=df$. And I'm simply stuck on that case. No matter what, there are sign problems. So could you help me figure this out? Are there coefficients missing with my definition of wedge btw?

  • 1
    $\begingroup$ It suffices to prove it pointwise for forms of the... form $dx^I$, and you may as well assume the vector is $\partial/\partial x_j$. $\endgroup$
    – user98602
    Nov 1 '15 at 21:30
  • $\begingroup$ I'm doing the following: 1) Case $f\wedge\omega$: done; 2) Case $df\wedge\omega$: stuck!; 3) Case of forms of that form, proving general formula for interior product of wedge of many $dx_i$ by induction on their number using case 2) in the middle; done, assuming 2); 4) Generalize. $\endgroup$
    – MickG
    Nov 1 '15 at 21:34
  • $\begingroup$ It should be easy enough to prove algebraically without such a reduction. The sign $(-1)^k$ comes from the number of slots you have to move $X$ to get it into the first slot of $\beta$ (in order to match up with the definition of $\iota_X \beta$). $\endgroup$ Nov 2 '15 at 0:20
  • $\begingroup$ I don't think that this can work out without the normalizing factor in the wedge product. (In particular, it is clear that it cannot work out with and without the normalizing factor, and it does work out with the normalization.) This normalizing factor is not a trivial issue. If you define a wedge product of one-forms via the alternation (which is needed to see how $dx^{i_1}\wedge\dots\wedge dx^{i_k}$ acts) then the normalizing factor $\frac1{k!\ell!}$ is needed to make the wedge product associative. $\endgroup$ Nov 3 '15 at 8:17
  • $\begingroup$ Indeed @AndreasCap, with the correct normalization (see this) it works, and as the answer here shows there are extra coefficients without the normalization. $\endgroup$
    – MickG
    Nov 3 '15 at 8:35

If we let brackets denote unnormalized antisymmetrization to match your wedge convention, then in index notation we have (choosing a basis $e_{i_k}$ with $e_1 = X$)

$$\begin{align*} \iota_X (\alpha \wedge \beta)_{i_2 \ldots i_{k+l}} &= \alpha_{[i_1 \ldots i_{k}} \beta_{i_{k+1} \ldots i_{k+l}]}. \end{align*}$$

Now let us expand just the $i_1$ in the antisymmetrization:

$$\begin{align*} \iota_X (\alpha \wedge \beta)_{i_2 \ldots i_{k+l}} &= \alpha_{i_1 [i_2\ldots i_{k}} \beta_{i_{k+1} \ldots i_{k+l}]} \\ &- \alpha_{[i_2|i_1|i_3\ldots i_{k}} \beta_{i_{k+1} \ldots i_{k+l}]} \\ &+ \alpha_{[i_2 i_3|i_1|i_4\ldots i_{k}} \beta_{i_{k+1} \ldots i_{k+l}]} \\ &\;\vdots\\ &+(-1)^k \alpha_{[i_2 i_3 i_4\ldots i_{k+1}} \beta_{|i_1 |i_{k+2} \ldots i_{k+l}]} \\ &-(-1)^k \alpha_{[i_2 i_3 i_4\ldots i_{k+1}} \beta_{i_{k+2}|i_1 |i_{k+3} \ldots i_{k+l}]}\\ &\;\vdots \end{align*}$$

Now separating this in to two sums, note that by shifting $i_1$ to the first slot of whichever form it appears in we eliminate the alternating signs, leaving us with

$$ \iota_X (\alpha \wedge \beta)_{i_2 \ldots i_{k+l}} = k\alpha_{i_1 [i_2 \ldots i_k} \beta_{i_{k+1}\ldots i_{k+l}]} + l(-1)^k \alpha_{[i_2 \ldots i_{k+1}} \beta_{|i_1|i_{k+2} \ldots i_{k+l}]}\\ = k(\iota_X \alpha)_{[i_2 \ldots i_k} \beta_{i_{k+1}\ldots i_{k+l}]} + l(-1)^k \alpha_{[i_2 \ldots i_{k+1}} (\iota_X \beta)_{i_{k+2} \ldots i_{k+l}]} $$

and so it seems the correct formula for your definition of the wedge product is in fact

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

  • $\begingroup$ Problem solved. But then I know the Lie derivative of a wedge is the sum of Lie(1) wedge form 2 and form 1 wedge Lie(2), so I'm having problems with extra coefficients in Cartan... $\endgroup$
    – MickG
    Nov 2 '15 at 11:36
  • $\begingroup$ More precisely, my proof of Cartan starts from functions, then goes by induction on the rank of the derived form. Suppose it holds for $k-1$-forms, then 1) prove it for exact $k$-forms; 2) prove it for $f\wedge\omega$ where $\omega$ is exact and $f$ a function; 3) conclude. 2) is problematic because $d\iota_X(f\wedge\omega)=d(kf\wedge\iota_X\omega)=kf\wedge d\iota_X\omega=kf\wedge\mathcal{L}_X\omega$, and that $k$ is not supposed to be there! $\endgroup$
    – MickG
    Nov 2 '15 at 11:46
  • $\begingroup$ That's why normalizing is better, because the formula above with normalized wedge has $k+\ell$ on both members, which fixes Cartan. $\endgroup$
    – MickG
    Nov 2 '15 at 11:49
  • $\begingroup$ @MickG: The case $k=0$ definitely brings out a more obvious downside of your definition: the wedge $f \wedge \omega$ is not equal to $f \omega$! $\endgroup$ Nov 2 '15 at 12:00
  • $\begingroup$ I know, I know. Remarked that in my thesis as a downside just now :). But nowhere as bad as no Cartan, don't you think :)? $\endgroup$
    – MickG
    Nov 2 '15 at 12:09


For the sake of practice I've done normed version. Normalization means that I use factorials and sum over all permutations in definition of wedge product, i.e.:

$$(\alpha\wedge\beta)(v_1,\dots,v_{k+l})=\frac{(k+l)!}{k!l!}\sum_{\sigma\in S_{k+l}}\sgn(\sigma)\alpha(v_{\sigma(1)},\dots,v_{\sigma(k)})\beta(v_{\sigma(k+1)},\dots,v_{\sigma(k+l)})$$

So we fix base and use Ricci calculus. This means that we can work only with coefficients.

Given coefficients $\alpha_{[i_1\dots i_k]}$ and $\beta_{[j_{1}\dots j_{l}]}$ of covariant tensors $\alpha$ and $\beta,$ the coefficients of wedge product are given by formula $$(\alpha\wedge\beta)_{i_1\dots i_{k+l}}=(\alpha\wedge\beta)_{[i_1\dots i_{k+l}]}=\frac{(k+l)!}{k!l!}\alpha_{[i_1\dots i_k}\beta_{i_{k+1}\dots i_{k+l}]}.$$ Given coefficients $\omega_{[i_1\dots i_s]}$ of covariant tensor $\omega$ and coefficients $X^i$ of contravariant one tensor (i.e. vector) $X,$ the coefficients of interior product are given by formula $$(\iota_X\omega)_{[i_1\dots i_{s-1}]}=X^i\omega_{[ii_1\dots i_{s-1}]}.$$

In order to prove $$\iota_X(\alpha\wedge\beta)=\overbrace{\iota_X\alpha\wedge\beta}^{(1)}+\overbrace{(-1)^k\alpha\wedge\iota_X\beta}^{(2)}\hspace{20pt}(\star)$$ we will compare coefficients on left and right hand side of $(\star)$

LHS: $$[\iota_X(\alpha\wedge\beta)]_{[i_1\dots i_{k+l-1}]}=X^i(\alpha\wedge\beta)_{[ii_1\dots i_{k+l-1}]}=\frac{(k+l)!}{k!l!}X^i\alpha_{[ii_1\dots i_{k-1}}\beta_{i_k\dots i_{k+l-1}]}$$ RHS (1): $$[(\iota_X\alpha)\wedge\beta]_{[i_1\dots i_{k+l-1}]}=\frac{(k+l-1)!}{(k-1)!l!}(\iota_X\alpha)_{[i_1\dots i_{k-1}}\beta_{i_k\dots i_{k+l-1}]}=\frac{(k+l-1)!}{(k-1)!l!}X^i\alpha_{[ii_1\dots i_{k-1}}\beta_{i_k\dots i_{k+l-1}]}.$$ RHS (2): $$(-1)^k[\alpha\wedge(\iota_X\beta)]_{[i_1\dots i_{k+l-1}]}=(-1)^k\frac{(k+l-1)!}{k!(l-1)!}\alpha_{[i_1\dots i_k}(\iota_X\beta)_{i_{k+1}\dots i_{k+l-1}]}=(-1)^k\frac{(k+l-1)!}{k!(l-1)!}\alpha_{[i_1\dots i_k}X^i\beta_{ii_{k+1}\dots i_{k+l-1}]}=(-1)^k\frac{(k+l-1)!}{k!(l-1)!}X^i\alpha_{[i_1\dots i_k}\beta_{ii_{k+1}\dots i_{k+l-1}]}=(-1)^k\frac{(k+l-1)!}{k!(l-1)!}X^i(-1)^k \alpha_{[ii_1\dots i_{k-1}}\beta_{i_k\dots i_{k+l-1}]}=\frac{(k+l-1)!}{k!(l-1)!}X^i\alpha_{[ii_1\dots i_{k-1}}\beta_{i_k\dots i_{k+l-1}]}$$ Adding factorials from RHS (1) and RHS (2) we get $$\frac{(k+l-1)!}{(k-1)!l!}+\frac{(k+l-1)!}{k!(l-1)!}=\frac{(k+l-1)!k}{k!l!}+\frac{(k+l-1)!l}{k!l!}=\frac{(k+l-1)!(k+l)}{k!l!}=\frac{(k+l)!}{k!l!}.$$ What finishes the proof.


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