The Unapologetic Mathematician

Mathematics for the interested outsider

Symmetric Tensors

Wow, people are loving my zero-knowledge test. It got 1,743 views yesterday, thanks to someone redditing it. Anywho…

Today and tomorrow I want to take last Friday’s symmetrizer and antisymmetrizer and apply them to the tensor representations of \mathrm{GL}(V), which we know also carry symmetric group representations. Specifically, the nth tensor power V^{\otimes n} carries a representation of S_n by permuting the tensorands, and this representation commutes with the representation of \mathrm{GL}(V). Then since the symmetrizer and antisymmetrizer are elements of the group algebra \mathbb{F}[S_n], they define intertwiners from V^{\otimes n} to itself. The their images are not just subspaces on which the symmetric group acts nicely, but subrepresentations of symmetric and antisymmetric tensors — S^n(V) and A^n(V), respectively.

Now it’s important (even if it’s not quite clear why) to emphasize that we’ve defined these representations without ever talking about a basis for V. However, let’s try to get a better handle on what such a thing looks like by assuming V has finite dimension d and picking a basis \{e_i\}. Then we have bases for tensor powers: a basis element of the nth tensor power is given by an n-tuple of basis elements for V. We’ll write a general one like e_{i_1}\otimes e_{i_2}\otimes...\otimes e_{i_n}.

How does a permutation act on such a basis element? Well, basis elements are pure tensors, so the permutation \pi simply permutes these basis tensorands. That is:

\displaystyle\pi\left(e_{i_1}\otimes...\otimes e_{i_n}\right)=e_{i_{\pi(1)}}\otimes...\otimes e_{i_{\pi(n)}}

So the space of symmetric tensors S^n(V) is the image of V^{\otimes n} under the action of the symmetrizer. And so it’s going to be spanned by the images of a basis for V^{\otimes n}, which we can calculate now. The symmetrizer is an average of all the permutations in the symmetric group, so we find

\displaystyle\begin{aligned}S\left(e_{i_1}\otimes...\otimes e_{i_n}\right)=\frac{1}{n!}\sum\limits_{\pi\in S_n}\pi\left(e_{i_1}\otimes...\otimes e_{i_n}\right)\\=\frac{1}{n!}\sum\limits_{\pi\in S_n}e_{i_{\pi(1)}}\otimes...\otimes e_{i_{\pi(n)}}\end{aligned}

Now we notice something here: if two basic tensors are related by a permutation of their tensorands, then the symmetrizer will send them to the same symmetric tensor! This means that we can choose a preimage for each basic symmetric tensor. Just use whatever permutation we need to put the n-tuple of tensorands into order. That is, always select i_1\leq i_2\leq...\leq i_n. Given any basic tensor, there is a unique permutation of its tensorands which is in this order.

As an explicit example, let’s consider what happens when we symmetrize the tensor e_1\otimes e_2\otimes e_1. First of all, we toss it into the proper order, since this won’t change the symmetrization: e_1\otimes e_1\otimes e_2. Now we write out a sum of all the permutations of the three tensorands, with the normalizing factor out front

\displaystyle\begin{aligned}\frac{1}{3!}(e_1\otimes e_1\otimes e_2+e_1\otimes e_2\otimes e_1+e_1\otimes e_1\otimes e_2+\\e_1\otimes e_2\otimes e_1+e_2\otimes e_1\otimes e_1+e_2\otimes e_1\otimes e_1)\end{aligned}

Some of these terms are repeated, since we have two copies of e_1 in this tensor. So we collect these together and cancel off some of the normalizing factor to find

\displaystyle\frac{1}{3}e_1\otimes e_1\otimes e_2+\frac{1}{3}e_1\otimes e_2\otimes e_1+\frac{1}{3}e_2\otimes e_1\otimes e_1

Now no matter how we rearrange the tensorands we’ll get back this same tensor again.

Tomorrow we’ll apply the same sort of approach to the antisymmetrizer.

December 22, 2008 Posted by | Algebra, Linear Algebra, Representation Theory | 6 Comments