Exterior Product between an $m$-form and an $n$-form I am learning some differential geometry for GR. I want to prove the following:
Given an $m$-form $\omega$ and an $n$-form $\eta$, (the components of) the exterior product are given by:
$\tag 1
(\omega \wedge \eta)_{a_1a_2...a_mb_1b_2...b_n} = \frac{(m+n)!}{m!n!}\omega_{[a_1a_2...a_m} \eta_{b_1b_2...b_n]} $
I proceeded as follows. I first tried to take the tensor product between $\omega$ and $\eta$ and then antisymmetrise them. So the components of tensor product I can write as:
$\tag 2(\omega \otimes \eta)_{a_1a_2...a_mb_1b_2...b_n} = \omega_{a_1a_2...a_m}\eta_{b_1b_2...b_n} $
If I wanted to antisymmetrise the tensor on the left, I would get $(m+n)!$ permutations. If I wanted to antisymmetrise each component of $\omega$ and $\eta$ on the right, I would get $m!n!$
But this would give me:
$\tag 3 ((\omega \wedge \eta)_{[a_1a_2...a_mb_1b_2...b_n]} = \frac{(m+n)!}{m!n!}\omega_{[a_1a_2...a_m} \eta_{b_1b_2...b_n]} $
or
$\tag 4
(\omega \wedge \eta)_{a_1a_2...a_mb_1b_2...b_n} = \frac{(m+n)!}{m!n!}\omega_{a_1a_2...a_m} \eta_{b_1b_2...b_n} $
neither of which is what is required as per $(1)$
Alternatively, if I write down the tensor product in terms of the tensor product of the dual basis,
$ \tag 5\omega \otimes \eta = \omega_{a_1a_2...a_m}\eta_{b_1b_2...b_n} dx^{a_1} \otimes dx^{a_2}\otimes....\otimes dx^{a_m}\otimes dx^{b_1}\otimes.....\otimes dx^{b_n} $
I am confused if I should antisymmetrise the components alone, or if I should write it as the wedge products between the basis (and hence antisymmetrise the dual basis).
$ \tag 6\omega \otimes \eta = \frac{1}{(m+n)!}\omega_{a_1a_2...a_m}\eta_{b_1b_2...b_n} dx^{a_1}\wedge dx^{a_2}\wedge......dx^{b_n} $
I'm very confused between the whether I don't understand the notation, or the just the counting/definition of the wedge product.
 A: The expression $$ (\omega\wedge\eta)_{i_1...i_rj_1...j_s}=\frac{(r+s)!}{r!s!}\omega_{[i_1...i_r}\eta_{j_1...j_s]} $$is a definition or a convention. You cannot prove it. In fact some authors use the alternative convention in which there is no numerical factor in front of the antisymmetrization.
There are some ways of motivating it however. For example if $\omega_1,...,\omega_r$ are covectors then it is true that $$ (\omega_1\wedge...\wedge\omega_r)(v_1,...,v_r)=\det(\omega_i(v_j))_{i,j=1,...,n}, $$ while with the other convention it would give $1/r!\det(...)$.
Or another reasoning is that if we use the convention that if $\omega_{i_1...i_r}$ is an antisymmetric tensor, then the corresponding $r$-form is $$ \omega=\sum_{i_1<...<i_r}\omega_{i_1...i_r}dx^{i_1}\wedge...\wedge dx^{i_r}, $$ then in terms of an unordered sum and antisymmetric components, we can write $$ \omega=\frac{1}{r!}\omega_{i_1...i_r}dx^{i_1}\wedge...\wedge dx^{i_r}. $$
Now if we define the wedge product of two differential forms via multilinearity and associativity, we get $$ \omega\wedge\eta=\left(\frac{1}{r!}\omega_{i_1...i_r}dx^{i_1}\wedge...\wedge dx^{i_r}\right)\wedge\left(\frac{1}{s!}\eta_{j_1...j_s}dx^{j_1}\wedge...\wedge dx^{j_s}\right) \\ =\frac{1}{r!s!}\omega_{i_1...i_r}\eta_{j_1...j_s}dx^{i_1}\wedge...\wedge dx^{i_r}\wedge dx^{j_1}\wedge...\wedge dx^{j_s} \\ \frac{1}{r!s!}\omega_{[i_1...i_r}\eta_{j_1...j_s]}dx^{i_1}\wedge...\wedge dx^{i_r}\wedge dx^{j_1}\wedge...\wedge dx^{j_s}. $$Here we could antisymmetrize over the component indices "for free" because the differentials with which they are multiplied and contracted with are already antisymmetric.
The problem is that this is an $r+s$-form but the factor that appears is not $1/(r+s)!$ but $1/(r!s!)$. Thus, we both multiply and divide by $(r+s)!$: $$ \omega\wedge\eta=\frac{1}{(r+s)!}\frac{(r+s)!}{r!s!}\omega_{[i_1...i_r}\eta_{j_1...j_s]}dx^{i_1}\wedge...\wedge dx^{i_r}\wedge dx^{j_1}\wedge...\wedge dx^{j_s}. $$
We thus obtain that the components of $\omega\wedge\eta$ as a tensor are $$ (\omega\wedge\eta)_{i_1...i_rj_1...j_s}=\frac{(r+s)!}{r!s!}\omega_{[i_1...i_r}\eta_{j_1...j_s]}.  $$
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