# Abstract Nonsense

## Character Table of S_3 By Finding the Irreducible Representations

Point of post: In this post we construct the first of a few character tables, namely we construct the character table for $S_3$.

Motivation

We now start off nice and easy and construct the classic character table for $S_3$ using the techniques from the last post. $S_3$ may be perhaps the easiest charater table to construct, but it will give us a good start to stretch our proverbial legs. In this post though, we find the character table using minimal machinery by actually constructing the irreducible characters of $S_3$ instead of using the techniques in our previous post. This method is, in my opinion, for the purpose of  character table construction, not preferable. Indeed, one must actually come up with representatives from each equivalency class of irreps of $S_3$. This post shall be useful to illustrate how beautifully simple the construction of character tables is made by the theory we’ve developed.

Character Table for $S_3$ By Finding the Irreducible Characters

We being by determining the degrees of the irreps for $S_3$. To do this, we recall that $\displaystyle \sum_{\alpha\in\widehat{S_3}}d_\alpha^2=\left|S_3\right|=6$. But, we have that (as for any group) $S_3$ admits the trivial irrep $\tau:S_3\to\mathbb{C}:g\mapsto 1$ which clearly has degree $1$. Thus, we have that

$\displaystyle 1+\sum d_\alpha^2=6\quad\quad\bold{(1)}$

where the sum runs over all $\alpha\in\widehat{S_3}$ with $\tau\notin\alpha$. We recall next that for every $\alpha\in\widehat{S_3}$ we must have that $d_\alpha\mid \left|S_3\right|=6$ so that $d_\alpha=1,2,3$ but since $d_\alpha^2\leqslant 6$ this implies that $d_\alpha=1,2$. Thus, with these possible choices of $d_\alpha$ it follows from a simple case analysis of $\mathbf{(1)}$ that the only solutions are that $\widehat{S_3}=\left\{\alpha_1,\alpha_2,\alpha_3\right\}$  with $d_{\alpha_1}=d_{\alpha_2}=1$ and $d_{\alpha_3}=2$ or $\widehat{S_3}=\left\{\alpha_k:k\in[6]\right\}$ and $d_{\alpha_k}=1$ for $k=1,\cdots,6$. But, since we know that the degrees of all the irreps of a group are $1$ if and only if the group is abelian we may exlcude this second possibility and conclude that the first must be true. This also tells us that the number of conjugacy classes of $S_3$ is $3$. Thus, we may begin setting up our character table as

$\begin{array}{c|ccc} & e & \mathcal{C}_1 & \mathcal{C}_2\\ \hline \chi^{\text{triv}} & 1 & 1 & 1\\ \chi^{(\alpha_2)} & 1 & a & b\\ \chi^{(\alpha_3)} & 2 & c & d\end{array}$

where we’ve denoted the character of the trivial irrep $\tau$ as $\chi^{\text{triv}}$, the equivalence class containing $e$ as itself, and the characters corresponding to the two other non-trivial classes in $\widehat{S_3}$ as $\chi^{(\alpha_2)},\chi^{(\alpha_3)}$. Moreover, we were able to fill in the first row since the nature of $\chi^{\text{triv}}$ is known and apparent and the first column since $\chi^{(\alpha_k)}(e)=d_{\alpha_k}\;\; k=2,3$. We next seek to obtain representatives for $\mathcal{C}_1,\mathcal{C}_2$. To do this we do what’s natural. We pick some element of $S_3$, say $(1,2)$ and since this conjugacyclass is distinct from the one represented by $e$ we may designate it as representative for $\mathcal{C}_2$. Next, we pick another element, say $(1,2,3)$ and since $(1,2,3)^2\ne e$ we know that the order of $(1,2,3)$ is not two and thus cannot be conjugate to $(1,2)$ and since it is clearly not conjugate to $e$ we may conclude that $(1,2,3)$ may serve as a representative for $\mathcal{C}_3$. Thus, our table can be rewritten as

$\begin{array}{c|ccc} & e & (1,2) & (1,2,3)\\ \hline \chi^{\text{triv}} & 1 & 1 & 1\\ \chi^{(\alpha_2)} & 1 & a & b\\ \chi^{(\alpha_3)} & 2 & c & d\end{array}$

We now wish to find the actual characters $\chi^{(\alpha_2)}$ and $\chi^{(\alpha_3)}$ by constructing them. Now, for $\chi^{(\alpha_2)}$ we’re looking for homomorphisms $\rho:G\to\mathcal{U}\left(\mathbb{C}\right)$. Now, the first one that may come to mind is the classic homomorhpism $\text{sgn}:S_3\to\{-1,1\}\leqslant\mathcal{U}\left(\mathbb{C}\right)$. Now, the only question is whether $\text{sgn}$ is irreducible. To do this, it suffices to check our alternative characterization of irreducibility. Indeed:

\displaystyle \begin{aligned}\left\langle \chi_{\text{sgn}},\chi_{\text{sgn}}\right\rangle &= \frac{1}{\left|S_3\right|}\sum_{g\in S_6}\left|\chi_{\text{sgn}}(g)\right|^2\\ &= \frac{1}{6}\sum_{g\in S_3}\left|\pm 1\right|^2\\ &= 1\end{aligned}

from where the irreducibility of $\text{sgn}$ follows. Thus, we may conclude that $\chi_{\text{sgn}}$ is what we have called $\chi^{(\alpha_2)}$. Thus, since $\chi_{\text{sgn}}((1,2))=-1$ and $\chi_{\text{sgn}}((1,2,3))=1$ we may fill in our table accordingly to get

$\begin{array}{c|ccc} & e & (1,2) & (1,2,3)\\ \hline \chi^{\text{triv}} & 1 & 1 & 1\\ \chi_{\text{sgn}} & 1 & -1 & 1\\ \chi^{(\alpha_3)} & 2 & c & d\end{array}$

Thus, it remains to find a character that corresponds to a degree two representation. To do this we recall that $S_3\cong D_3$ and interpret $D_3$ as the symmetries of a planar triangle. We can then correspond $(2,3)$ to the matrix $\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$ which gives a flip accross the $y$-axis and $(1,2,3)$ to the matrix $\displaystyle \begin{pmatrix}\frac{1}{2} & \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix}$ and extend this to a homomorphism $\rho:S_3\to\mathcal{U}\left(\mathbb{C}^2\right)$ since $\left\{(1,2),(1,2,3)\right\}$ is a generating set for $S_3$. Explicitly

$\begin{array}{ccc}e\mapsto \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} & (1,2)\mapsto \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix} & (2,3)\mapsto \begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\\ & &\\ (1,3)\mapsto\begin{pmatrix}-\frac{1}{2} & -\frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix} & (1,2,3)\mapsto\begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix} & (1,3,2)\mapsto \begin{pmatrix}-\frac{1}{2} & \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}\end{array}$

Now, to check that this is actually an irrep we use our alternative characterization again. Indeed:

\displaystyle \begin{aligned}\left\langle\chi_\rho,\chi_\rho\right\rangle &= \frac{1}{|S_3|}\sum_{g\in S_6}\left|\chi_\rho(g)\right|^2\\ &= \frac{1}{6}\left(2^2+0^2+0^2+0^2+(-1)^2+(-1)^2\right)\\ &=1\end{aligned}

From where the irreducibility of $\rho$ follows. It follows then that $\chi_\rho=\chi^{(\alpha_3)}$. Thus, we may complete our character table as:

$\begin{array}{c|ccc} & e & (1,2) & (1,2,3)\\ \hline \chi^{\text{triv}} & 1 & 1 & 1\\ \chi_{\text{sgn}} & 1 & -1 & 1\\ \chi_{\rho} & 2 & 0 & -1\end{array}$

References:

1. Isaacs, I. Martin. Character Theory of Finite Groups. New York: Academic, 1976. Print.

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