# The Unapologetic Mathematician

## Maschke’s Theorem

Maschke’s theorem is a fundamental result that will make our project of understanding the representation theory of finite groups — and of symmetric groups in particular — far simpler. It tells us that every representation of a finite group is completely reducible.

We saw last time that in the presence of an invariant form, any reducible representation is decomposable, and so any representation with an invariant form is completely reducible. Maschke’s theorem works by showing that there is always an invariant form!

Let’s start by picking any form whatsoever. We know that we can do this by picking a basis $\{e_i\}$ of $V$ and declaring it to be orthonormal. We don’t anything fancy like Gram-Schmidt, which is used to find orthonormal bases for a given inner product. No, we just define our inner product by saying that $\langle e_i,e_j\rangle=\delta_{i,j}$ — the Kronecker delta, with value $1$ when its indices are the same and $0$ otherwise — and extend the only way we can. If we have $v=\sum v^ie_i$ and $w=\sum w^je_j$ then we find \displaystyle\begin{aligned}\langle v,w\rangle&=\left\langle\sum\limits_{i=1}^{\dim(V)}v^ie_i,\sum\limits_{j=1}^{\dim(V)}w^je_j\right\rangle\\&=\sum\limits_{i=1}^{\dim(V)}\sum\limits_{j=1}^{\dim(V)}\left\langle v^ie_i,w^je_j\right\rangle\\&=\sum\limits_{i=1}^{\dim(V)}\sum\limits_{j=1}^{\dim(V)}\overline{v^i}w^j\left\langle e_i,e_j\right\rangle\\&=\sum\limits_{i=1}^{\dim(V)}\sum\limits_{j=1}^{\dim(V)}\overline{v^i}w^j\delta_{i,j}\\&=\sum\limits_{i=1}^{\dim(V)}\overline{v^i}w^i\end{aligned}

so this does uniquely define an inner product. But there’s no reason at all to believe it’s $G$-invariant.

We will use this arbitrary form to build an invariant form by a process of averaging. For any vectors $v$ and $w$, define $\displaystyle\langle v,w\rangle_G=\sum\limits_{g\in G}\langle gv,gw\rangle$

Showing that this satisfies the definition of an inner product is a straightforward exercise. As for invariance, we want to show that for any $h\in G$ we have $\langle hv,hw\rangle_G=\langle v,w\rangle_G$. Indeed: \displaystyle\begin{aligned}\langle hv,hw\rangle_G&=\sum\limits_{g\in G}\langle ghv,ghw\rangle\\&=\sum\limits_{k\in G}\langle kv,kw\rangle\\&=\langle v,w\rangle_G\end{aligned}

where the essential second equality follows because as $g$ ranges over $G$, the product $k=gh$ ranges over $G$ as well, just in a different order.

And so we conclude that if $V$ is a representation of $G$ then we can take any inner product whatsoever on $V$ and “average” it to obtain an invariant form. Then with this invariant form in hand, we know that $V$ is completely reducible.

Why doesn’t this work for our counterexample representation of $\mathbb{Z}$? Because the group $\mathbb{Z}$ is infinite, and so the averaging process breaks down. This approach only works for finite groups, where the average over all $g\in G$ only involves a finite sum.

September 28, 2010