# Question on the coefficient in the definition of wedge product

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.

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 $$i (since $$\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 $$\displaystyle\frac{1}{l!}$$.