# The Unapologetic Mathematician

## Exterior Algebras

Let’s continue yesterday’s discussion of algebras we can construct from a vector space. Today, we consider the “exterior algebra” on $V$, which consists of the direct sum of all the spaces of antisymmetric tensors $\displaystyle \Lambda(V)=\bigoplus\limits_{n=0}^\infty A^n(V)$

Yes, that’s a capital $\lambda$, not an $A$. This is just standard notation, probably related to the symbol for its multiplication we’ll soon come to.

Again, despite the fact that each $A^n(V)$ is a subspace of the tensor space $T^{\otimes n}$, this isn’t a subalgebra of $T(V)$, because the tensor product of two antisymmetric tensors may not be antisymmetric itself. Instead, we will take the tensor product of $\mu\in S^m(V)$ and $\nu\in A^n(V)$, and then antisymmetrize it, to give $\mu\wedge\nu\in A^{m+n}(V)$. This will be bilinear, but will it be associative?

Our proof parallels the one we ran through yesterday, writing the symmetric group as the disjoint union of cosets indexed by a set $C$ of representatives $\displaystyle S_{l+m+n}=\biguplus\limits_{\gamma\in\Gamma}\gamma S_{l+m}$

and rewriting the symmetrizer in just the right way. But now we’ve got the signs of our permutations to be careful with. Still, let’s dive in with the antisymmetrizers \displaystyle\begin{aligned}\left(\frac{1}{(l+m+n)!}\sum\limits_{\pi\in S_{l+m+n}}\mathrm{sgn}(\pi)\pi\right)\left(\frac{1}{(l+m)!}\sum\limits_{\hat{\pi}\in S_{l+m}}\mathrm{sgn}(\hat{\pi})\hat{\pi}\right)&=\left(\frac{1}{(l+m+n)!}\sum\limits_{\gamma\in\Gamma}\sum\limits_{\pi\in\gamma S_{l+m}}\mathrm{sgn}(\pi)\pi\right)\left(\frac{1}{(l+m)!}\sum\limits_{\hat{\pi}\in S_{l+m}}\mathrm{sgn}(\hat{\pi})\hat{\pi}\right)\\&=\left(\frac{1}{(l+m+n)!}\sum\limits_{\gamma\in\Gamma}\sum\limits_{\pi\in S_{l+m}}\mathrm{sgn}(\gamma\pi)\gamma\pi\right)\left(\frac{1}{(l+m)!}\sum\limits_{\hat{\pi}\in S_{l+m}}\mathrm{sgn}(\hat{\pi})\hat{\pi}\right)\\&=\left(\frac{1}{(l+m+n)!}\left(\sum\limits_{\gamma\in\Gamma}\mathrm{sgn}(\gamma)\gamma\right)\left(\sum\limits_{\pi\in S_{l+m}}\mathrm{sgn}(\pi)\pi\right)\right)\left(\frac{1}{(l+m)!}\sum\limits_{\hat{\pi}\in S_{l+m}}\mathrm{sgn}(\hat{\pi})\hat{\pi}\right)\\&=\left(\frac{1}{(l+m+n)!}\sum\limits_{\gamma\in\Gamma}\mathrm{sgn}(\gamma)\gamma\right)\left(\frac{1}{(l+m)!}\sum\limits_{\pi\in S_{l+m}}\sum\limits_{\hat{\pi}\in S_{l+m}}\mathrm{sgn}(\pi)\mathrm{sgn}(\hat{\pi})\pi\hat{\pi}\right)\\&=\frac{1}{(l+m+n)!}\left(\sum\limits_{\gamma\in\Gamma}\mathrm{sgn}(\gamma)\gamma\right)\left(\sum\limits_{\pi\in S_{l+m}}\mathrm{sgn}(\pi)\pi\right)\\&=\frac{1}{(l+m+n)!}\sum\limits_{\pi\in S_{l+m+n}}\mathrm{sgn}(\pi)\pi\end{aligned}

Where throughout we’ve used the fact that $\mathrm{sgn}$ is a representation, and so the signum of the product of two group elements is the product of their signa. We also make the crucial combination of the double sum over $S_{l+m}$ into a single sum by noting that each group element shows up exactly $(l+m)!$ times, and each time it shows up with the exact same sign, which lets us factor out $(l+m)!$ from the sum and cancel the normalizing factor.

Now this multiplication is not commutative. Instead, it’s graded-commutative. If $\mu\in\Lambda^m(V)$ and $\nu\in\Lambda^n(V)$ are elements of the exterior algebra, then we find $\displaystyle\mu\wedge\nu=(-1)^{mn}\nu\wedge\mu$

That is, elements of odd degree anticommute with each other, while elements of even degree commute with everything.

Indeed, given $\mu\in S^m(V)$ and $\nu\in S^n(V)$, we can let $\tau_{m,n}$ be the permutation which moves the last $n$ slots to the beginning of the term and the first $m$ slots to the end. We can construct $\tau_{m,n}$ by moving each of the last $n$ slots one-by-one past the first $m$, taking $m$ swaps for each one. That gives a total of $mn$ swaps, so $\mathrm{sgn}(\tau_{m,n})=(-1)^{mn}$. Then we write \displaystyle\begin{aligned}\mu\wedge\nu&=\left(\frac{1}{(m+n)!}\sum\limits_{\pi\in S_{m+n}}\mathrm{sgn}(\pi)\pi\right)(\mu\otimes\nu)\\&=\left(\frac{1}{(m+n)!}\sum\limits_{\pi\in S_{m+n}}\mathrm{sgn}(\pi\tau_{m,n})\pi\tau_{m,n}\right)(\mu\otimes\nu)\\&=\mathrm{sgn}(\tau_{m,n})\left(\frac{1}{(m+n)!}\sum\limits_{\pi\in S_{m+n}}\mathrm{sgn}(\pi)\pi\right)\left(\tau_{m,n}(\mu\otimes\nu)\right)\\&=(-1)^{mn}\left(\frac{1}{(m+n)!}\sum\limits_{\pi\in S_{m+n}}\mathrm{sgn}(\pi)\pi\right)(\nu\otimes\mu)\\&=(-1)^{mn}\nu\wedge\mu\end{aligned}

as asserted.

The dual to the exterior algebra $\Lambda(V^*)$ is the algebra of all alternating multilinear functionals on $V$, providing a counterpart to the algebra of polynomial functions on $V$. But where the variables in polynomial functions commute with each other, the basic covectors — analogous to variables reading off components of a vector — anticommute with each other in this algebra.

October 27, 2009 - Posted by | Algebra, Linear Algebra

## 6 Comments »

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2. […] to see what this does for our tensor algebras. Again, I’ll be mostly interested in the exterior algebra , so I’ll stick to talking about that […]

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4. […] interpretation of what’s going on that I learned from Todd Trimble, which is that “the exterior algebra is the symmetric algebra of a purely odd supervector […]

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6. […] the dual space: . And of course we can take the direct sum of these spaces over all to get the exterior algebra […]

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