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In my textbook, W. Tu's An Introduction to Manifolds (page 26), the wedge product is defined to be

After the definition, the following explanation regarding the coefficient in the definition, namely $\displaystyle{\frac{1}{k!l!}}$, is given:

I am unable to follow this explanation: in my understanding, it states that there are as many groups, which have $k!$ elements, namely $\sigma\tau$, and have identical values when applied to the result of the tensor product, as the number of $\sigma\in S_{k+l}$. But in these groups, the elements overlap, since otherwise there will be $(k+l)!k!$ different permutations in $S_{k+l}$, which is not the case ($|S_{k+l}|=(k+l)!$). And I think this renders the division by $k!$ senseless,

I will be grateful for any kind of help. Thanks.

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For $f\in A_k(V)​$ ($f​$ is a $k​$-linear alternating function with domain $V^k​$ and codomain $\mathbb{R}​$) and $g\in A_l(V)​$, we define the wedge product as $$\displaystyle (f\wedge g)(v_1,\dots,v_k,v_{k+1},\dots,v_{k+l})=\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}(\mathrm{sgn}\,\sigma)f(v_{\sigma(1)},\dots,v_{\sigma(k)})g(v_{\sigma(k+1)},\dots,v_{\sigma(k+l)}).​$$

The justification of the existence of the coefficient $\displaystyle\frac{1}{k!l!}$ is that it is to compensate the repeating identical terms. First observe for $\displaystyle\frac{1}{k!}$; we proceed by repeating the second of the following step $\displaystyle\frac{(k+l)!}{k!}-1$ times to arrive at a classification of all the $\sigma$ into $\displaystyle\frac{(k+l)!}{k!}$ groups:

  • In step $1​$, pick an element $\sigma​$ from $S_{k+l}​$; find all permutations $\tau​$ that can be represented by matrix $\displaystyle\begin{pmatrix}1&\cdots&k&k+1&\cdots&k+l\\\tau(1)&\cdots&\tau(k)&k+1&\cdots&k+l\end{pmatrix}​$; then $\{\sigma\tau\,\big|\,\text{for all }k!\,\tau\}​$ forms a group, denoted by $G_1​$, under composition of permutation. Let $A_1=S_{k+l}\setminus\underline{G_1}​$ and proceed to the next step.
  • In step $n​$, pick an element $\sigma​$ from $A_{n-1}​$; find all permutations $\tau​$ that can be represented by matrix $\displaystyle\begin{pmatrix}1&\cdots&k&k+1&\cdots&k+l\\\tau(1)&\cdots&\tau(k)&k+1&\cdots&k+l\end{pmatrix}​$; then $\{\sigma\tau\,\big|\,\text{for all }k!\,\tau\}​$ forms a group, denoted by $G_i​$, under composition of permutation. Note that $\underline{G_i}​$ is disjoint from $A_{n-1}​$, since in each step we exclude all the permutations with one particular group of $\sigma(k+1),\dots,\sigma(k+l)​$ fixed from the pickée of the next step, thus each pick we will always have a new group of $\sigma(k+1),\dots,\sigma(k+l)​$ that is different from those picked and thus those in $\underline{G_i}​$ for $i<n​$ (since $\tau​$ does not permute $k+1,\dots,k+l​$). Let $A_n=A_{n-1}\setminus\underline{G_i}​$ and proceed to the next step.

SInce each pick and thus $\underline{G_i}$ is disjoint from those before, in each step we subtract $k!$ elements from the original $S_{k+l}$, which will result in exactly $\displaystyle\frac{(k+l)!}{k!}$ steps in total and that $A_{\frac{(k+l)!}{k!}}=\emptyset$.

Groups $G_1,\dots,G_{\frac{(k+l)!}{k!}}$ act on $ A_k(V)$, and we observe that all elements in the orbit of each of $G_i$ is identical: since $f$ is alternating, for each $\sigma\tau\in G_i$, $$\begin{align*}(\mathrm{sgn}\,\sigma\tau)f(v_{\sigma\tau(1)},\dots,v_{\sigma\tau(k)})&=(\mathrm{sgn}\,\sigma\tau)(\mathrm{sgn}\,\tau)f(v_{\sigma(1)},\dots,v_{\sigma(k)})\\&=(\mathrm{sgn}\,\sigma)f(v_{\sigma(1)},\dots,v_{\sigma(k)})\end{align*}.$$

Since each $G_i$ contains $k!$ elements, we divide the result by $k!$, and the repeated term will be eliminated.

A similar argument will then be applied to $\displaystyle\frac{1}{l!}$.

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