# The Unapologetic Mathematician

## The Inner Product of Characters

When we’re dealing with characters, there’s something we can do to rework our expression for the inner product on the space of class functions.

Let’s take a $G$-module $(W,\sigma)$, with character $\psi$. Before, we’ve used Maschke’s theorem to tell us that all $G$-modules are completely reducible, but remember what it really tells us that there is some $G$invariant inner product on $W$ (we’ll have to keep straight the two inner products by which vector space they apply to). With respect to the inner product on $W$, every transformation $\sigma(g)$ with $g\in G$ is unitary, and if we pick an orthonormal basis to get a matrix representation $Y$ each of the matrices $Y(g)$ will be unitary. That is:

$\displaystyle Y\left(g^{-1}\right)=Y(g)^{-1}=\overline{Y(g)}^\top$

So what does this mean for the character $\psi$? We can calculate

\displaystyle\begin{aligned}\overline{\psi(g)}&=\overline{\mathrm{Tr}\left(Y(g)\right)}\\&=\mathrm{Tr}\left(\overline{Y(g)}\right)\\&=\mathrm{Tr}\left(Y\left(g^{-1}\right)^\top\right)\\&=\mathrm{Tr}\left(Y\left(g^{-1}\right)\right)\\&=\psi\left(g^{-1}\right)\end{aligned}

And so we can rewrite our inner product

\displaystyle\begin{aligned}\langle\chi,\psi\rangle&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}\overline{\chi(g)}\psi(g)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}\chi\left(g^{-1}\right)\psi(g)\end{aligned}

The nice thing about this formula is that it doesn’t depend on complex conjugation, and so it’s useful for any base field (if we were using other base fields).

The catch is that for class functions in general we have no reason to believe that this is an inner product. Indeed, if $g\in G$ is some element that isn’t conjugate to its inverse then we can define a class function $f$ that takes the value $1$ on the class $K_g$ of $g$, $-1$ on the class $K_{g^{-1}}$ of $g^{-1}$, and $0$ elsewhere. Our new formula gives

$\displaystyle\langle f,f\rangle=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}f\left(g^{-1}\right)f(g)=-\lvert K_{g^{-1}}\rvert\lvert K_g\rvert$

so this bilinear form isn’t positive-definite.

October 18, 2010