# A nice proof that the irreducible characters form a basis for class functions

Warning: This is a very shoddily organized post and can be vastly improved. The method of proof is still nice however.

Let $G$ be a finite group of size $g$ and $k$ an algebraically closed field such that $g \neq 0$ in it. Let $V_1,\dots, V_n$ be the irreducible representations of $G$ over $k$ and $\chi_1,\dots,\chi_n$ the corresponding characters.

A class function $f: G \to k$ is a set function such that $f$ is constant on conjugacy classes. That is, for all $g,h \in G, f(h^{-1}gh) = f(g)$. It is easy to see that any character is a class function by cyclicity of the trace function.

Let us denote by $H$ the vector space of class functions and $W$ the vector space of functions spanned by the $\chi_i$ within $H$. It is natural to wonder about the size of $W$ relative to $H$. In fact, the two vector spaces are equal!

To prove this, let us consider the algebra $k[G]$. By our hypothesis and the usual argument of Maschke’s theorem, this is a semisimple ring. That is, $k[G]$ is semisimple as a $G$ representation. Let $r_G$ be the character attached to this so called regular representation. A simple computation shows us that $r_1 = g$ and $r_G = 0$ otherwise.

Moreover, for any irreducible representation $V_k$, we see from last time that:

$Hom_G(V_k,k[G]) = \frac{1}{g}\sum_g r_G(g^{-1})\chi_k(g) = \chi_k(1) = \dim V_k.$

Therefore, by semisimplicity

$k[G] \cong \oplus_k V_k^{\dim V_k}$

as representations. Moreover, the orthogonality relations from last time also show that the $\chi_k$ are linearly independent. That is, $\dim W = \#$ irreducible characters.

Taking the endomorphism ring on both sides (as modules over $k[G]$), we obtain:

$k[G] \cong M_{\dim V_k}(k)$

as algebras. Considering dimensions, we have incidentally shown that $g = \sum_k \dim V_k^2$. Moreover, let us consider the central elements in $k[G]$. They are easily seen to exactly be the elements of the form $\sum_g f(g)g$ for $f:G \to k$ a class function.

On the other hand, the central elements in $M_n(k)$ are always just $k$. Therefore, the dimension of the center of $k[G]$ (= vector space of class functions) is equal to the number of irreducible representations. Therefore, $\dim W = \dim H$ and we are done.

The above map can be thought of as the representation of a class function in the basis defined by the functions $c_k\overline{\chi_k}$ for some appropriate constants $c_k$. This is because the element:

$\lambda_i = \sum_{g}\overline{\chi_i}g \in k[G]$

maps to a a diagonal element in $M_{\dim V_j}(k)$ that can be calculated by taking the trace. That is to say, the image of $\lambda_i$ in \$latex $M_{\dim V_j}(k)$ is given by:

$\frac{1}{\dim V_k}\sum_{g}\overline{\chi_i(g)}\chi_j(g) = \frac{g}{\dim V_k}\delta_{ij}$

by the orthogonality relations. We see that $c_k = \frac{\dim V_k}{g}$.