# The Unapologetic Mathematician

## The Symmetrizer and Antisymmetrizer

Today we’ll introduce two elements of the group algebra $\mathbb{F}[S_n]$ of the symmetric group $S_n$ which have some interesting properties. Each one, given a representation of the symmetric group, will give us a subrepresentation of that representation.

The symmetrizer gets its name from the way that it takes an arbitrary vector and turns it into one that the symmetric group will act trivially on. Specifically, we use the element

$\displaystyle S=\frac{1}{n!}\sum\limits_{\pi\in S_n}\pi$

That is, we take all $n!$ permutations in the symmetric group and — in a sense — average them out. If we compose this with any permutation $\hat{\pi}$ we find

$\displaystyle\begin{aligned}\hat{\pi}S=\hat{\pi}\frac{1}{n!}\sum\limits_{\pi\in S_n}\pi\\=\frac{1}{n!}\sum\limits_{\pi\in S_n}\hat{\pi}\pi\end{aligned}$

But as $\pi$ runs over all the permutations in the group, $\hat{\pi}\pi$ does as well. So this is just the symmetrizer $S$ back again. The upshot is that if we have a representation $\rho:S_n\rightarrow\mathrm{GL}(V)$ we find that

$\displaystyle\begin{aligned}\left[\rho(\hat{\pi})\right]\left(\left[\rho(S)\right](v)\right)=\left[\rho(\hat{\pi}S)\right](v)\\=\left[\rho(S)\right](v)\end{aligned}$

Thus every vector in the image of $\rho(S)$ is left unchanged by the action of any permutation. That is, $\mathrm{Im}\left(\rho(S)\right)\subseteq V$ is a subrepresentation on which $S_n$ acts trivially.

The antisymmetrizer, on the other hand, on the other hand, will give us vectors on which the symmetric group acts by the signum representation. We use the group algebra element

$\displaystyle A=\frac{1}{n!}\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\pi$

Now if we compose this with any permutation $\hat{\pi}$ we find

$\displaystyle\begin{aligned}\hat{\pi}A=\hat{\pi}\frac{1}{n!}\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\pi\\=\frac{1}{n!}\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\hat{\pi}\pi\\=\frac{1}{n!}\sum\limits_{\pi\in S_n}\mathrm{sgn}(\hat{\pi})\mathrm{sgn}(\hat{\pi}\pi)\hat{\pi}\pi\\=\mathrm{sgn}(\hat{\pi})\frac{1}{n!}\sum\limits_{\pi\in S_n}\mathrm{sgn}(\hat{\pi}\pi)\hat{\pi}\pi\\=\mathrm{sgn}(\hat{\pi})\hat{\pi}A\end{aligned}$

Now given a representation $\rho:S_n\rightarrow\mathrm{GL}(V)$ we find that

$\displaystyle\begin{aligned}\left[\rho(\hat{\pi})\right]\left(\left[\rho(A)\right](v)\right)=\left[\rho(\hat{\pi}A)\right](v)\\=\left[\rho(\mathrm{sgn}(\hat{\pi})A)\right](v)\\=\mathrm{sgn}(\hat{\pi})\left[\rho(A)\right](v)\end{aligned}$

Thus every vector in the image of $\rho(A)$ is multiplied by the signum of any permutation. That is, $\mathrm{Im}\left(\rho(A)\right)\subseteq V$ is a subrepresentation on which $S_n$ acts by the signum representation.

Now, one thing to be careful about: I haven’t said that the subrepresentations are nontrivial. That is, when we (anti)symmetrize a representation, the subrepresentation we get may be zero — maybe no vectors in the representation transform trivially or by the signum representation. In fact, let’s check what happens when we multiply the symmetrizer and antisymmetrizer:

$\displaystyle\begin{aligned}SA=\left(\frac{1}{n!}\sum\limits_{\pi\in S_n}\pi\right)\left(\frac{1}{n!}\sum\limits_{\hat{\pi}\in S_n}\mathrm{sgn}(\hat{\pi})\hat{\pi}\right)\\=\frac{1}{n!}\frac{1}{n!}\sum\limits_{\substack{\pi\in S_n\\\hat{\pi}\in S_n}}\mathrm{sgn}(\hat{\pi})\pi\hat{\pi}\\=\frac{1}{n!}\frac{1}{n!}\sum\limits_{\substack{\pi\in S_n\\\hat{\pi}\in S_n}}\mathrm{sgn}(\pi)\mathrm{sgn}(\pi\hat{\pi})\pi\hat{\pi}\\=\frac{1}{n!}\frac{1}{n!}\left(\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\right)\left(\sum\limits_{\pi\hat{\pi}\in S_n}\mathrm{sgn}(\pi\hat{\pi})\pi\hat{\pi}\right)\\=\frac{1}{n!}\frac{1}{n!}\left(\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\right)A\\=0\end{aligned}$

Where the sum comes to zero because we’re just adding up $\frac{n!}{2}$ terms where $\mathrm{sgn}(\pi)=1$ and $\frac{n!}{2}$ where $\mathrm{sgn}(\pi)=-1$, and so everything cancels out. That is, the symmetric part of an antisymmetrized representation is trivial. Similarly, the antisymmetric part of a symmetrized representation is trivial.

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Posted by John Armstrong | Algebra, Representation Theory

## 5 Comments	»

1. [...] and tomorrow I want to take last Friday’s symmetrizer and antisymmetrizer and apply them to the tensor representations of , which we know also carry symmetric group [...]

Pingback by | December 22, 2008 | Reply

2. [...] of , which actions commute with each other. Our antisymmetric tensors are the image of a certain action from the symmetric group, which is an intertwiner of the action. Thus we have a one-dimensional representation of , which [...]

Pingback by | December 31, 2008 | Reply

3. [...] if its value is unchanged when we swap two of its inputs. Equivalently, it commutes with the symmetrizer. That is, it must kill everything that the symmetrizer kills, and so must really define a linear [...]

Pingback by | October 22, 2009 | Reply

4. [...] may well not be symmetric itself. Instead, we will take the tensor product of and , and then symmetrize it, to give . This will be bilinear, and it will work with our choice of grading, but will it be [...]

Pingback by | October 26, 2009 | Reply

5. [...] may not be antisymmetric itself. Instead, we will take the tensor product of and , and then antisymmetrize it, to give . This will be bilinear, but will it be [...]

Pingback by | October 27, 2009 | Reply

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