# Decomposition of permutations and wedge products.

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.

$$S_{p,q}$$ 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 $$S_{p,q} \rho\tau$$.