# The Unapologetic Mathematician

## The Character Table of a Group

Given a group $G$, Maschke’s theorem tells us that every $G$-module is completely reducible. That is, we can write any such module $V$ as the direct sum of irreducible representations:

$\displaystyle V=\bigoplus\limits_{i=1}^km_iV^{(i)}$

Thus the irreducible representations are the most important ones to understand. And so we’re particularly interested in their characters, which we call “irreducible characters”.

Of course an irreducible character — like all characters — is a class function. We can describe it by giving its values on each conjugacy class. And so we lay out the “character table”. This is an array whose rows are indexed by inequivalent irreducible representations, and whose columns are indexed by conjugacy classes $K\subseteq G$. The row indexed by $V^{(i)}$ describes the corresponding irreducible character $\chi^{(i)}$. If $k\in K$ is a representative of the conjugacy class, then the entry in the column indexed by $K$ is $\chi^{(i)}_K=\chi^{(i)}(k)$. That is, the character table looks like

$\displaystyle\begin{array}{c|ccc}&\cdots&K&\cdots\\\hline\vdots&&\vdots&\\V^{(i)}&\cdots&\chi^{(i)}_K&\cdots\\\vdots&&\vdots&\end{array}$

By convention, the first row corresponds to the trivial representation, and the first column corresponds to the conjugacy class $\{e\}$ of the identity element. We know that the trivial representation sends every group element to the $1\times 1$ identity matrix, whose trace is $1$. We also know that every character’s value on the identity element is the degree of the corresponding representation. We can slightly refine our first picture to sketch the character table like so:

$\displaystyle\begin{array}{c|cccc}&\{e\}&\cdots&K&\cdots\\\hline V^\mathrm{triv}&1&\cdots&1&\cdots\\\vdots&\vdots&&\vdots&\\V^{(i)}&\deg\left(V^{(i)}\right)&\cdots&\chi^{(i)}_K&\cdots\\\vdots&\vdots&&\vdots&\end{array}$

We have no reason to believe (yet) that the table is finite. Since $G$ is a finite group there can be only finitely many conjugacy classes, and thus only finitely many columns, but as far as we can tell there may be infinitely many inequivalent irreps, and thus infinitely many rows. Further, we have no reason to believe that the rows are all distinct. Indeed, we know that equivalent representations have equal characters — they’re related through conjugation by an invertible intertwinor — but we don’t know for sure that inequivalent representations must have distinct characters.

As an example, we can start writing down the character table of $S_3$. We know that conjugacy classes in symmetric groups correspond to cycle types, and so we can write down all three conjugacy classes easily:

\displaystyle\begin{aligned}K_1&=\left\{e\right\}\\K_2&=\left\{(1\,2),(1\,3),(2\,3)\right\}\\K_3&=\left\{(1\,2\,3),(1\,3\,2)\right\}\end{aligned}

We know of two irreps offhand — the trivial representation and the signum representation — and so we’ll start with those and leave the table incomplete below that:

$\displaystyle\begin{array}{c|ccc}&K_1&K_2&K_3\\\hline V^\mathrm{triv}&1&1&1\\V^\mathrm{sgn}&1&-1&1\\\vdots&\vdots&\vdots&\vdots\end{array}$

October 20, 2010 -

## 5 Comments »

1. [...] Products in the Character Table As we try to fill in the character table, it will help us to note another slight variation of our inner product [...]

Pingback by Inner Products in the Character Table « The Unapologetic Mathematician | October 21, 2010 | Reply

2. [...] this establishes what we suspected when setting up the character table: if and are inequivalent irreps then their characters and must be unequal. Indeed, since [...]

Pingback by Irreducible Characters are Orthogonal « The Unapologetic Mathematician | October 22, 2010 | Reply

3. [...] at most as many irreducible characters as there are conjugacy classes in . And so we know that the character table must have only finitely many rows. For instance, since has three conjugacy classes, it can have at [...]

Pingback by Consequences of Orthogonality « The Unapologetic Mathematician | October 25, 2010 | Reply

4. [...] we first defined the character table of a group, we closed by starting to write down the character table of [...]

Pingback by One Complete Character Table (part 1) « The Unapologetic Mathematician | October 26, 2010 | Reply

5. [...] Character Table is Square We’ve defined the character table of a group, and we’ve seen that it must be finite. Specifically, it cannot have any more rows [...]

Pingback by The Character Table is Square « The Unapologetic Mathematician | November 19, 2010 | Reply