The Unapologetic Mathematician

Mathematics for the interested outsider

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.

About these ads

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

6 Comments »

  1. […] The three constructions we’ve just shown — the tensor, symmetric tensor, and exterior algebras — were all asserted to be the “free” constructions. This makes them […]

    Pingback by Functoriality of Tensor Algebras « The Unapologetic Mathematician | October 28, 2009 | Reply

  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 […]

    Pingback by Tensor Algebras and Inner Products I « The Unapologetic Mathematician | October 29, 2009 | Reply

  3. […] alternating multilinear functional, with the sides as variables, and so it lives somewhere in the exterior algebra . We’ll have to work out which particular functional gives us a good notion of volume as we […]

    Pingback by Parallelepipeds and Volumes I « The Unapologetic Mathematician | November 2, 2009 | Reply

  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 […]

    Pingback by Set-multiset duality and supervector spaces « Annoying Precision | November 7, 2009 | Reply

  5. […] Now that we’ve used exterior algebras to come to terms with parallelepipeds and their transformations, let’s come back to apply […]

    Pingback by The Jacobian « The Unapologetic Mathematician | November 11, 2009 | Reply

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

    Pingback by Tensor Bundles « The Unapologetic Mathematician | July 6, 2011 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 392 other followers

%d bloggers like this: