Equivalent definitions of the exterior products

Question

In the Wikipedia article on alternating multilinear forms, two equivalent definitions of the exterior products are given.

Definition 1: $$ \omega \wedge \eta ={\frac {(k+m)!}{k!\,m!}}\operatorname {Alt} (\omega \otimes \eta ),\tag{1} $$ where $$ \operatorname{Alt}(\omega)(x_1,\ldots,x_k)=\frac{1}{k!}\sum_{\sigma\in S_k}\operatorname{sgn}(\sigma)\,\omega(x_{\sigma(1)},\ldots,x_{\sigma(k)});\tag{1.5} $$ Definition 2: $$ {\omega \wedge \eta(x_1,\ldots,x_{k+m})} = \sum_{\sigma \in Sh_{k,m}} \operatorname{sgn}(\sigma)\,\omega(x_{\sigma(1)}, \ldots, x_{\sigma(k)}) \eta(x_{\sigma(k+1)}, \ldots, x_{\sigma(k+m)}),\tag{2} $$ where here $Sh_{k,m} ⊂ S_{k+m}$ is the subset of $(k,m)$ shuffles: permutations $σ$ of the set $\{1, 2, ..., k + m\}$ such that $σ(1) < σ(2) < ... < σ(k)$, and $σ(k + 1) < σ(k + 2) < ... < σ(k + m)$.

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Note that (1) can be rewritten as $$ {\omega \wedge \eta(x_1,\ldots,x_{k+m})} = \frac{1}{k!m!}\sum_{\sigma \in S_{k+m}} \operatorname{sgn}(\sigma)\,\omega(x_{\sigma(1)}, \ldots, x_{\sigma(k)}) \eta(x_{\sigma(k+1)}, \ldots, x_{\sigma(k+m)})\tag{1'} $$ Thus to show that (1) and (2) are equivalent, it suffices to show that (1') are (2) are the same. Ignoring the $\operatorname{sgn}(\sigma)$ part, one can count the number ofterms in (1') and (2).

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Here is my question:

How can one show that (1') and (2) are the same?

Answer 1

Notations

I have used $l$ instead of $m$. Moreover, I have used $S_{(.,.)}$ instead of $Sh_{.,.}$.
I also use $v_i$ instead of $x_i$.

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Define the equivalence relation $\sim$ on $S_{k+l}$ by setting $\sigma \sim \sigma'$ iff $$\{\sigma(1), \ldots, \sigma(k)\} = \{\sigma'(1), \ldots, \sigma'(k)\}.$$ (Note that the equality is of sets.)
It is easily checked that $\sim$ is indeed an equivalence relation. Moreover, if $\sigma \sim \sigma',$ then we also have $$\{\sigma(k+1), \ldots, \sigma(k+l)\} = \{\sigma'(k+1), \ldots, \sigma'(k+l)\}.$$

Let $[\sigma]$ denote the equivalence class of $\sigma.$

We make the following simple observations:

1. Each equivalence class contains the same number of elements.
2. The above number is $k!l!.$
3. Every equivalence class contains exactly one $(k, l)$ shuffle.
4. Every $(k, l)$ shuffle is contained in some equivalence class.

Now, if we show that the quantity $\operatorname{sgn}(\sigma)\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)})$ is the same for all $\sigma$ belonging to a fixed equivalence class, then we would be done.

That is because, we could simply choose the shuffle present in the equivalence class as the representative of the class and then $(1')$ would reduce to $(2).$ To see this better, let $\Pi_1, \ldots \Pi_r$ denote the distinct equivalence classes and let $\sigma_i \in \Pi_i$ be the shuffle in that class. Then, we have $$S_{k+l} = \bigsqcup_{i=1}^r \Pi_i$$ and thus, \begin{align} & \dfrac{1}{k!l!}\sum_{\sigma \in S_{k+l}}\operatorname{sgn}(\sigma)\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)})\\ =&\;\dfrac{1}{k!l!}\sum_{i=1}^{r}\sum_{\sigma \in \Pi_i}\operatorname{sgn}(\sigma)\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)})\\ =&\;\dfrac{1}{k!l!}\sum_{i=1}^{r}\sum_{\sigma \in \Pi_i}\operatorname{sgn}(\sigma_i)\omega(v_{\sigma_i(1)}, \ldots, v_{\sigma_i(k)})\eta(v_{\sigma_i(k+1)}, \ldots, v_{\sigma_i(k+l)})\\ &\text{note that now the inner quantity is independent of $\sigma$}\\ =&\;\dfrac{1}{k!l!}\sum_{i=1}^{r}(k!l!)\operatorname{sgn}(\sigma_i)\omega(v_{\sigma_i(1)}, \ldots, v_{\sigma_i(k)})\eta(v_{\sigma_i(k+1)}, \ldots, v_{\sigma_i(k+l)})\\ =&\;\sum_{\sigma \in S_{(k, l)}}\operatorname{sgn}(\sigma)\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)}). \end{align}

Thus, now all we need to finish is the following claim.

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Claim. If $[\sigma] = [\sigma'],$ then \begin{align} \operatorname{sgn}(\sigma)&\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)})\\ =& \operatorname{sgn}(\sigma')\omega(v_{\sigma'(1)}, \ldots, v_{\sigma'(k)})\eta(v_{\sigma'(k+1)}, \ldots, v_{\sigma'(k+l)}). \end{align}

The claim is pretty simple and follows closely from what was written in the comments.

Proof. Since $\{\sigma(1), \ldots, \sigma(k)\} = \{\sigma'(1),\ldots, \sigma'(k)\},$ we can find a permutation $\tau \in S_{k+l}$ such that $$\tau\sigma(i) = \sigma'(i), \quad i = 1, \ldots, k$$ and $\tau$ acts as identity on $\{\sigma(k+1), \ldots, \sigma(k+l)\}.$

Similarly, we can find a permutation $\pi \in S_{k+l}$ such that $$\pi\sigma(i) = \sigma'(i), \quad i = k+1, \ldots, k+l$$ and $\pi$ acts as identity on $\{\sigma(1), \ldots, \sigma(k)\}.$

Thus, we actually get $$\pi\tau\sigma(i) = \sigma'(i), \quad i = 1, \ldots, k+l.$$

That is to say, $\sigma' = \pi\tau\sigma.$
In particular, we have $\operatorname{sgn}\sigma' = \operatorname{sgn}\pi\cdot\operatorname{sgn}\tau\cdot\operatorname{sgn}\sigma.$
This also gives us that $$\operatorname{sgn}\sigma'\cdot\operatorname{sgn}\pi\cdot\operatorname{sgn}\tau = \operatorname{sgn}\sigma.$$

With that in place, we prove the claim via the following calculation. \begin{align} & \operatorname{sgn}\sigma'\cdot\omega(v_{\sigma'(1)}, \ldots, v_{\sigma'(k)})\eta(v_{\sigma'(k+1)}, \ldots, v_{\sigma'(k+l)})\\ =&\;\operatorname{sgn}\sigma'\cdot\omega(v_{\tau\sigma(1)}, \ldots, v_{\tau\sigma(k)})\eta(v_{\pi\sigma(k+1)}, \ldots, v_{\pi\sigma(k+l)})\\ & \text{Now, we use that $\omega$ and $\eta$ are alternating}\\ =&\;\operatorname{sgn}\sigma'\cdot\operatorname{sgn}\tau\cdot\operatorname{sgn}\pi\cdot\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)})\\ =&\;\operatorname{sgn}\sigma\cdot\omega(v_{\sigma(1)}, \ldots, v_{\sigma(k)})\eta(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)}) & \blacksquare \end{align}