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 »

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