# The Unapologetic Mathematician

## 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 $n$th 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 $n$th 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