# More on Characters

So it turns out that over at Rigorous Trivialities my idea for a set of posts on Grassmanians has already been done. So I’ll continue with characters for now.

The irreducible characters are the ones that come from irreducible representations. Denote $\chi_i$ the character afforded by $G\to GL(L_i)$ where $L_i$ is the i-th simple component of $kG$. So by breaking representations down to the simple components we get that every character is an (non-negative) integer combination of the irreducible characters.

Let’s quickly examine the regular character, or the character afforded by the $\mathbb{C}$-regular representation. Take as a basis for $\mathbb{C}G$ just $G$. If $g=1$, then $\chi(g)=|G|$ since the identity transformation just has trace the dimension of the vector space. But now if $g\neq 1$, then every basis element gets permuted, so every diagonal entry is 0. Thus $\chi(g)=0$. Hence $\chi(g)=\begin{cases} 0 \ if \ g\neq 1 \\ |G| \ if \ g=1\end{cases}$

I’d like to get to one really cool result today, so let’s start in that direction. A class function is a function (to some field, although I don’t think this is a necessary part of the definition) that is constant on conjugacy classes. We saw last time that every character is a class function. The set of all class functions, denoted $cf(G)$ is a vector space over the field that is your codomain. For ease, I’ll just start using $\mathbb{C}$ for everything.

Alright, now let’s note that every class function is a class sum, so by this post we have that $cf(G)= Z(\mathbb{C}G)$, and hence, since this has dimension the number of conjugacy classes which is the number of irreducible representations a natural conjecture is that the irreducible characters form a basis for $cf(G)$.

This is indeed the case. Alright. Well, all our work is basically done. We just need that the irreducible characters are linearly independent, since there are the right number of them. First, note that $\chi_i(u_j)=0$ when $u_j\in L_j$ and $i\neq j$. But also, if $e_i$ is the identity in $B_i=End_\mathbb{C}(L_i)$, then as we saw in the regular representation, $\chi_i(e_i)=n_i$ the dimension of that representation.

Suppose $\sum c_i\chi_i=0$. Then (as with most arguments of this form we will just carefully apply to the right things) for all j, $0=(\sum c_i\chi_i)(e_j)=c_j\chi_j(e_j)=c_jn_j$ and clearly $n_j\neq 0$, so each $c_j=0$. And we are done.

I think this is rather neat, since there is no immediate reason to believe that class functions in general have to do with characters, but in fact every class function must arise from the irreducible characters.

Next time we’ll expand this idea. It turns out that $cf(G)$ isn’t just a vector space, but a full blown inner product space, and the irreducible characters are not just a basis, but an orthonormal basis.