# Decompose the permutation into the direct sum of irreducible reps.

1. May 19, 2013

### NasuSama

1. The problem statement, all variables and given/known data

Note: I need help with part (c).

Consider the representation $P: S_3 \rightarrow GL_3$ where $P_{\sigma}$ is the permutation matrix associated to $\sigma$.

a) Determine the character $\chi_P : S_3 \rightarrow \mathbb{C}$

b) Find all the irreducible representations of $S_3$.

c) Decompose $P$ into the direct sum of irreducible representations. That is, find a single matrix $Q$ so that $Q^{-1}P_{\sigma}Q$ is block diagonal where the blocks along the diagonal are either $T_{\sigma}$, $\Sigma_{\sigma}$ or $A_{\sigma}$

2. The attempt at a solution

Here is my overall progress for the problem:

I let $e$ to be the identity permutation, $x = (1 \ 2 \ 3)$ and $y = (1 \ 2)$

Then, I let $P_x = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$, $P_y = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $P_e = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

The conjugacy classes are as followed:

- $\{e\}$

- $\{y, xy, yx\}$

- $\{x, x^2\}$

The character table shows that:

$\chi (\{e\}) = 3$

$\chi (\{x, x^2\}) = 0$

$\chi (\{y, xy, yx\}) = 1$

By the theorem I have applied, I have found three representations, which is congruent to the number of conjugacy classes. They are:

- trivial representation $T$
- sign representation $\Sigma$
- two-dimensional representation $A$, presented as the symmetries of equilateral triangle

The sum of their dimensions corresponds to the theorem I applied:

$d_1^2 + d_2^2 + d_3^2 = |S_3| = 6$

The only possibility for equality to hold is $d_1 = d_2 = 1$ and $d_3 = 2$.

Another character table shows that:

$\chi_T (\{e\}) = 1$

$\chi_T (\{x, x^2\}) = 1$

$\chi_T (\{y, yx, xy\}) = 1$

$\chi_A (\{e\}) = 2$

$\chi_A (\{x, x^2\}) = -1$

$\chi_A (\{y, xy, yx\}) = 0$

$\chi_{\Sigma} (\{e\}) = 1$

$\chi_{\Sigma} (\{x, x^2\}) = 1$

$\chi_{\Sigma} (\{y, yx, xy\}) = -1$

Finally, I need to determine $Q$ for $Q^{-1}P_xQ = \begin{bmatrix}1 & 0 & 0 \\ 0 & -1/2 & -\sqrt{3}/2 \\ 0 & \sqrt{3}/2 & -1/2 \end{bmatrix}$ and $Q^{-1}P_yQ = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$ We need to find the basis for $\mathbb{R}^4$.

I know that $u = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$ is one of the vectors in the basis. That is because we want to find the vector $u$ such that $P_xu = u$ and $P_yu = u$.

Then, I need to find the remaining two vectors for the basis. Here is the hardest part. I attempt to find the vectors, say $v$ and $w$, but I can't find the nonzero basis. Here is what I have:

$P_xv = w$

$P_yv = v$

$P_xw = -v$

$P_yw = -w$

Here is what I have:

$v = \begin{bmatrix} a \\ b \\ c\end{bmatrix} \rightarrow w = \begin{bmatrix} c \\ a \\ b\end{bmatrix}$

$v = \begin{bmatrix} a \\ b \\ c\end{bmatrix} \rightarrow v = \begin{bmatrix} b \\ a \\ c\end{bmatrix}$

$w = \begin{bmatrix} c \\ a \\ b\end{bmatrix} \rightarrow -v = \begin{bmatrix} b \\ c \\ a\end{bmatrix}$

$w = \begin{bmatrix} c \\ a \\ b\end{bmatrix} \rightarrow -w = \begin{bmatrix} a \\ c \\ b\end{bmatrix}$

From here, I can't determine what are the nonzero vectors for $Q$.