# Abstract Nonsense

## Irreducible Characters

Point of post: In this post we discuss what is arguably one of the most important tools in all of basic representation theory– the irreducible characters of a group.

Motivation

In our last few posts we’ve been developing the notion of a class function and the space of class functions under the hazy motivation that we will use them to ascertain that the cardinality of some set $X$ is the dimension of $\text{Cl}(G)$ (we, in our last post, showed that this was the number $k$ of conjugacy classes in $G$). In this set we shall see that this set $X$ for which we’d like to show has the quality $\#(X)=\dim\text{Cl}(G)$ are the irreducible characters of the group $G$. These are certain class functions which will occupy a fair amount of our efforts in the coming posts since, in a very real sense, they make up a vast portion of the substance in basic representation theory. But, for now we shall restrict our attention to defining the irreducible characters and showing that they form an orthonormal basis for $\text{Cl}(G)$. Of this we shall get the corollary that $\#\left(\widehat{G}\right)=k$.

Irreducible Characters

Let $G$ be a finite group, then making the necessary selections we produce the matrix representations $D^{(\alpha)}$ for each $\alpha\in\widehat{G}$. We then define the irreducible character associated with $\alpha$, denoted $\chi^{(\alpha)}$ by the rule

$\displaystyle \chi^{(\alpha)}(g)=\text{tr}\left(D^{(\alpha)}(g)\right)=\sum_{j=1}^{d_\alpha}D^{(\alpha)}_{j,j}(g)$

Some properties about characters are immediately evidently. Namely:

Theorem: Let $G$ be a finite group then for each $\alpha\in\widehat{G}$ and each $g\in G$ one has that:

\displaystyle \begin{aligned}&(1)\quad \chi^{(\alpha)}(e)=d_\alpha\\ &(2)\quad \chi^{(\alpha)}\left(g^{-1}\right)=\overline{\chi^{(\alpha)}(g)}\\ &(3)\quad \chi^{(\alpha)}\in\text{Cl}(G)\end{aligned}

Proof:

$(1)$: This follows immediately since $\chi^{(\alpha)}(e)=\text{tr}\left(D^{(\alpha)}(e)\right)=\text{tr}\left(I_{d_\alpha}\right)=d_\alpha$

$(2)$: We merely recall that the basis we’ve chosen for $\mathscr{V}^{(\alpha)}$ is such that $D^{(\alpha)}$ is unitary and so

\begin{aligned}\chi^{(\alpha)}(g^{-1}) &=\text{tr}\left(D^{(\alpha)}\left(g^{-1}\right)\right)\\ &=\text{tr}\left(D^{(\alpha)}(g)^{-1}\right)\\ &= \text{tr}\left(D^{(\alpha)}(g)^{\ast}\right)\\ &= \overline{\text{tr}\left(D^{(\alpha)}(g)\right)}\\ &=\overline{\chi^{(\alpha)}(g)}\end{aligned}

$(3)$: To see this we merely note that for any $x,y\in G$ one has that

\begin{aligned}\chi^{(\alpha)}\left(xyx^{-1}\right) &= \text{tr}\left(D^{(\alpha)}\left(xyx^{-1}\right)\right)\\ &= \text{tr}\left(D^{(\alpha)}(x)D^{(\alpha)}(y)D^{(\alpha)}(y)^{-1}\right)\\ &= \text{tr}\left(D^{(\alpha)}(y)\right)\\ &= \chi^{(\alpha)}(y)\end{aligned}

from where the  conclusion follows by the arbitrariness of $x,y$. $\blacksquare$

Orthonormality for Irreducible Characters

The real deal for the irreducible characters is that they form a basis for $\text{Cl}(G)$. We prove this in two steps

Theorem: The set $\left\{\chi^{(\alpha)}:\alpha\in\widehat{G}\right\}$ is orthonormal when thought of as elements of $\mathcal{A}(G)$.

Proof: Let $\alpha,\beta\in\widehat{G}$. Then one sees that

\begin{aligned}\left\langle \chi^{(\alpha)}(g),\chi^{(\beta)}\right\rangle &= \frac{1}{|G|}\sum_{g\in G}\chi^{(\alpha)}(g)\overline{\chi^{(\beta)}(g)}\\ &= \frac{1}{|G|}\sum_{g\in G}\sum_{i=1}^{d_\alpha}\sum_{j=1}^{d_\beta}D_{i,i}^{(\alpha)}(g)D^{(\beta)}_{j,j}(g)\\ &= \sum_{i=1}^{d_\alpha}\sum_{j=1}^{d_\beta}\frac{1}{|G|}\sum_{g\in G}D^{(\alpha)}_{i,i}(g)\overline{D^{(\beta)}_{j,j}(g)}\\ &= \sum_{i=1}^{d_\alpha}\sum_{j=1}^{d_\beta}\left\langle D^{(\alpha)}_{i,i},D^{(\beta)}_{j,j}\right\rangle\\ &= \sum_{i=1}^{d_\alpha}\sum_{j=1}^{d_\beta}\frac{1}{d_\alpha}\delta_{\alpha,\beta}\delta_{i,j}^2\\ &= \frac{1}{d_\alpha}\delta_{\alpha,\beta}\sum_{i,j=1}^{d_\alpha}\delta_{i,j}\\ &= \delta_{\alpha,\beta}\end{aligned}

where although it may appear there is a mistake with the indices we may assume that the indexes of the sums both range to $d_{\alpha}$ since otherwise $\alpha\ne\beta$ and the sum is zero anyways. Since $\alpha,\beta$ were arbitrary the conclusion follows. $\blacksquare$

References:

1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print

February 25, 2011 -

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