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Let $V$ be an $\mathbb{R}$-vector space. Denote the space of all alternating $k$-linear forms from $V^k$ to $\mathbb{R}$ by ${\cal A}_k(V, \mathbb{R})$

Suppose $f\in{\cal A}_p(V, \mathbb{R})$ and $g\in{\cal A}_q(V, \mathbb{R})$. Munkres (Analysis on Manifolds) defines the wedge product of $f$ and $g$, $f\wedge g \in {\cal A}_{p+q}(V, \mathbb{R})$, as an alternating $(p+q)$-form given by:

$$ (f\wedge g)(\mathbf{x}) = \cfrac{1}{p!q!} \sum_{\sigma \in S_{p+q}} \epsilon(\sigma)f(x_{\sigma(1)}, x_{\sigma(2)}, \dots, x_{\sigma(p)})g(x_{\sigma(p+1)}, x_{\sigma(p+2)}, \dots, x_{\sigma(p+q)}) $$

where $x_i$ is the $i^{th}$ component of $\mathbf{x}$ and $\epsilon(\sigma)$ is the sign of the permutation.

In my differential geometry course, the instructor defined wedge product as:

$$(f\wedge g)(\mathbf{x}) = \sum_{\sigma\in S_{p,q}} \epsilon(\sigma)f(x_{\sigma(1)}, x_{\sigma(2)}, \dots, x_{\sigma(p)})g(x_{\sigma(p+1)}, x_{\sigma(p+2)}, \dots, x_{\sigma(p+q)})$$

where $S_{p,q} = \{ \sigma \in S_{p+q} : \sigma(1) < \sigma(2) < \dots < \sigma(p)$ and $\sigma(p+1) < \sigma(p+2) < \dots < \sigma(p+q) \}$.

How do I show the equivalence of these two definitions? Here is my attempt:

First of all, for convenience, we define the following subsets of $S_{p+q}$.

$ P = \{\sigma \in S_{p+q} : \sigma\ \ \text{fixes}\ \ p+1, p+2, \dots, p+q\}$

edit: $P$ is just a copy of $S_p$ in $S_{p+q}$

$ Q = \{\sigma \in S_{p+q} : \sigma\ \ \text{fixes}\ \ 1, 2, \dots, p\}$

edit: $Q$ is just a copy of $S_q$ in $S_{p+q}$

(By "$\sigma$ fixes $i$", I mean that $\sigma(i) = i$).

We know that, $|S_{p, q}| = {{p+q}\choose{p}}$.

Further, I want to claim that given any $\sigma \in S_{p+q}$, we can decompose, $\sigma = \phi \rho \tau$, where $\phi \in S_{p, q}$, $\rho \in P$ and $\tau \in Q$ (This is something that I believe to be true, but couldn't quite prove it).

Assuming this fact, we show the equivalence as follows: (The intermediate steps make use of the fact that $\rho$ and $\tau$ are disjoint and hence commute and also that $f$ and $g$ are alternating maps).

$$ \begin{align} &(f\wedge g)(\mathbf{x})\\ & = \cfrac{1}{p!q!} \sum_{\sigma \in S_{p+q}} \epsilon(\sigma)f(x_{\sigma(1)}, x_{\sigma(2)}, \dots, x_{\sigma(p)})g(x_{\sigma(p+1)}, x_{\sigma(p+2)}, \dots, x_{\sigma(p+q)})\\ & = \cfrac{1}{p!q!} \sum_{\phi \in S_{p, q}}\sum_{\rho \in P}\sum_{\tau \in Q} \epsilon(\phi\rho\tau)f(x_{\phi\rho\tau(1)}, x_{\phi\rho\tau(2)}, \dots, x_{\phi\rho\tau(p)})g(x_{\phi\rho\tau(p+1)}, x_{\phi\rho\tau(p+2)}, \dots, x_{\phi\rho\tau(p+q)})\\ & = \cfrac{1}{p!q!} \sum_{\phi \in S_{p, q}}\sum_{\rho \in P}\sum_{\tau \in Q} \epsilon(\phi\rho\tau)f(x_{\phi\tau(1)}, x_{\phi\tau(2)}, \dots, x_{\phi\tau(p)})g(x_{\phi\rho(p+1)}, x_{\phi\rho(p+2)}, \dots, x_{\phi\rho(p+q)})\\ & = \cfrac{1}{p!q!} \sum_{\phi \in S_{p, q}}\sum_{\rho \in P}\sum_{\tau \in Q} \epsilon(\phi\rho\tau) \epsilon(\tau) f(x_{\phi(1)}, x_{\phi(2)}, \dots, x_{\phi(p)})\epsilon(\rho)g(x_{\phi(p+1)}, x_{\phi(p+2)}, \dots, x_{\phi(p+q)})\\ & = \cfrac{1}{p!q!} \sum_{\phi \in S_{p, q}}\sum_{\rho \in P}\sum_{\tau \in Q} \epsilon(\phi) \epsilon(\tau)^2 \epsilon(\rho)^2 f(x_{\phi(1)}, x_{\phi(2)}, \dots, x_{\phi(p)}) g(x_{\phi(p+1)}, x_{\phi(p+2)}, \dots, x_{\phi(p+q)})\\ & = \cfrac{1}{p!q!} \sum_{\phi \in S_{p, q}}\sum_{\rho \in P}\sum_{\tau \in Q} \epsilon(\phi) f(x_{\phi(1)}, x_{\phi(2)}, \dots, x_{\phi(p)}) g(x_{\phi(p+1)}, x_{\phi(p+2)}, \dots, x_{\phi(p+q)})\\ & = \cfrac{1}{p!q!} \sum_{\phi \in S_{p, q}} p!q! \epsilon(\phi) f(x_{\phi(1)}, x_{\phi(2)}, \dots, x_{\phi(p)}) g(x_{\phi(p+1)}, x_{\phi(p+2)}, \dots, x_{\phi(p+q)})\\ & = \sum_{\phi \in S_{p, q}} \epsilon(\phi) f(x_{\phi(1)}, x_{\phi(2)}, \dots, x_{\phi(p)}) g(x_{\phi(p+1)}, x_{\phi(p+2)}, \dots, x_{\phi(p+q)}) \end{align} $$

which completes the proof.

Now, the only thing that remains to be shown is that the decomposition $\sigma = \phi\rho\tau$ is actually possible. But I am not sure how to do that. Any hints will be appreciated.

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$S_{p,q}$ is a subgroup of $S_{p+q}$. We show that the set of (right)-cosets of $S_{p,q}$ is precisely the collection of cosets of the form $S_{p,q}\rho \tau $, where $\rho \in P$ and $\tau \in Q$.

Firstly, we show that each coset in this collection is indeed distinct:

$S_{p,q}\rho_1\tau_1 = S_{p,q}\rho_2\tau_2 \implies \rho_1\tau_1(\rho_2\tau_2)^{-1} = \rho_1\tau_1\tau_2^{-1}\rho_2^{-1} \in S_{p,q}$

Now, $\rho_2^{-1}$ and $\tau_1\tau_2^{-1}$ are disjoint permutations, they commute. So, $\rho_1\tau_1\tau_2^{-1}\rho_2^{-1} = \rho_1\rho_2^{-1}\tau_1\tau_2^{-1} \in S_{p,q}$.

Since, $\rho_1\rho_2^{-1}$ is a permutation of first $p$ symbols and doesn't chage their ordering, $\rho_1\rho_2^{-1}$ is the trivial permutation similarly, so is, $\tau_1\tau_2^{-1}$. Thus, $\rho_1 = \rho_2$ and $\tau_1 = \tau_2$. So each element in the collection is indeed distinct.

Now, the index of $S_{p,q}$ in $S_{p+q}$ is $\cfrac{(p+q)!}{{p+q}\choose{p}} = p!q!$.

But, this is precisely the number of elements in our collection.This proves that each coset is of the form $S_{p,q} \rho\tau$.

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