# Equivalent definitions of the exterior products

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 $σ(k + 1) < σ(k + 2) < ... < σ(k + m)$.

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).

Here is my question:

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

• ...by inserting the definition of $\otimes$ in equation 1, expanding out equation 1, via 1.5 and then gathering terms in a way that matches equation 1'. In particular, when you apply equation 1.5 to equation 1, you're taking Alt of a $k+m$ form, so the sum is over $S_{k+m}$, which is the same index set used in equation 1'. May 6, 2017 at 19:55
• @JohnHughes: I have just added the tag (1.5) to match the labels in your comment. (I was confused..)
– user9464
May 6, 2017 at 20:01
• Yeah ... I figured you could work it out. :) May 6, 2017 at 20:02
• OK, so I lied about the index sets being the same. But the $k,m$ shuffles do form a subset of the $k+m$ permutations. And if you remove the order restrictions on the shuffles, then for each order-restricted shuffle of $k,m$ items, you get $k!m!$ order-unrestricted shuffles (by permuting the elements of each pile), and looks like a nice correspondence that can account for a factorial or two. To be honest...I'd try this for $\omega\wedge \eta$ for a couple of 2-forms, and I'll bet that the correspondence suddenly gets really obvious. Since it's only 24 terms, it shouldn't be too tough. May 6, 2017 at 20:06
• I tried the case $k=1,m=2$ to see what is really going on. I might come back later with my own answer. As I see, the key point is that for any given permutation $\sigma\in S_{k+m}$, it can be written as $$\sigma=\pi^{-1}\sigma_2\pi\sigma_1\tau$$ where $\tau\in Sh_{k,m}$, $(\pi(1),\cdots,\pi(k+m))=(k+1,\cdots,k+m,1,\cdots,k)$, and $\sigma_1\in S_k$, $\sigma_2\in S_m$. It follows that $$\hbox{sgn}(\sigma)=\hbox{sgn}(\sigma_2)\hbox{sgn}(\sigma_1)\hbox{sgn}(\tau).$$ Combining with the alternating properties of $\omega$ and $\eta$, one gets the desired results.
– user9464
May 7, 2017 at 19:54

Notations

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

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.

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}