Wedge Product Convention In Wald’s General Relativity textbook he defines the wedge product as:
$$(w \wedge u)_{a_1 ... a_p b_1 ... b_q}= \frac{(p+q)!}{p!q!} w_{[a_1 ... a_p}u_{b_1 ... b_q]}$$
My question is relatively simple: why is he choosing to include the factor in the front? I believe that some authors use a different convention where they do not, so what is the advantage or intuition behind this choice? Note that the antisymmetrization symbol already includes a $1/n!$ factor, so his formula is really:
$$(w \wedge u)_{a_1 ... a_p b_1 ... b_q}=  \frac{1}{p!q!} \sum_{\text{permutations}}\text{sgn(perm)}w_{[a_1 ... a_p}u_{b_1 ... b_q]}$$
 A: Here is a reasoning. Define a vector space $V$ with basis $e^i$ ($i=1,...,n$). Define a unital associative algebra $\Lambda(V)$ generated by $V$ such that $v^2=0$ for all $v\in V$. If no additional relations are imposed then $\Lambda(V)$ is necessarily the exterior algebra over $V$. There are mathematically rigorous ways to construct this but for the purposes of this answer, it is not important.
Denote the multiplication in $\Lambda(V)$ by $\wedge$. The set $$ 1 \\ e^i \\ e^{i_1}\wedge e^{i_2} \quad i_1<i_2 \\ \vdots \\ e^{i_1}\wedge...\wedge e^{i_r}\quad i_1<...<i_r \\ \vdots \\ e^1\wedge...\wedge e^n $$ forms a linear basis for $\Lambda(V)$.
So far $\Lambda(V)$ is not a part of the tensor algebra $T(V)$ of $V$. In fact we have $$ \Lambda(V)=T(V)/I, $$ where $I$ is the two-sided ideal in $T(V)$ generated by elements of the form $v\otimes v$. But we can embed $\Lambda(V)$ into the tensor algebra such that homogenous elements of $\Lambda(V)$ are antisymmetric tensors in $T(V)$, however to do so we need to choose a convention for the wedge product.
A homogenous element of degree $0\le r\le n$ in $\Lambda(V)$ can be written as $$ w=\sum_{i_1<...<i_r}w_{i_1...i_r}e^{i_1}\wedge...\wedge e^{i_r}=\sum_{i_1...i_r}\frac{1}{r!}w_{i_1...i_r}e^{i_1}\wedge...\wedge e^{i_r}. $$
A natural convention would be to identify the unrestricted, antisymmetric coefficients $w_{i_1...i_r}$ that appear in the second term with the tensor components of $w$ (as a tensor), i.e. $$ w=\sum_{i_1...i_r}\frac{1}{r!}w_{i_1...i_r}e^{i_1}\wedge...\wedge e^{i_r}=\sum_{i_1...i_r}w_{i_1...i_r}e^{i_1}\otimes...\otimes e^{i_r}. $$ It follows that via this convention $$ e^{i_1}\wedge...\wedge e^{i_r}=r!e^{[i_1}\otimes...\otimes e^{i_r]}. $$ Now take two elements $$ v=\sum\frac{1}{r!}v_{i_1...i_r}e^{i_1}\wedge...\wedge e^{i_r},\quad w=\sum\frac{1}{s!}w_{j_1...j_s}e^{j_1}\wedge...\wedge e^{j_s}, $$(the sums are unrestricted) and calculate their product via associativity: $$ v\wedge w=\sum\frac{1}{r!s!}v_{i_1...i_r}w_{j_1...j_s}e^{i_1}\wedge...\wedge e^{i_r}\wedge e^{j_1}\wedge ... \wedge e^{j_s} \\=\sum\frac{1}{(r+s)!}\frac{(r+s)!}{r!s!}v_{[i_1...i_r}w_{j_1...j_s]}e^{i_1}\wedge...\wedge e^{i_r}\wedge e^{j_1}\wedge ... \wedge e^{j_s} \\ = \sum\frac{(r+s)!}{r!s!}v_{[i_1...i_r}w_{j_1...j_s]}e^{i_1}\otimes...\otimes e^{i_r}\otimes e^{j_1}\otimes ... \otimes e^{j_s}. $$
From this we get that in terms of tensor components $$ (v\wedge w)_{i_1...i_r j_1...j_s}= \frac{(r+s)!}{r!s!}v_{[i_1...i_r}w_{j_1...j_s]}. $$
