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 $A_{\sigma}$
My Attempt
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:
$\chi_T (\{e\}) = 1$
$\chi_T (\{x, x^2\}) = 1$
$\chi_T (\{y, yx, xy\}) = 1$
$\chi_A (\{e\}) = 2$
$\chi_A (\{x, x^2\}) = -1$
$\chi_A (\{y, xy, yx\}) = 0$
$\chi_{\Sigma} (\{e\}) = 1$
$\chi_{\Sigma} (\{x, x^2\}) = 1$
$\chi_{\Sigma} (\{y, yx, xy\}) = -1$
Now, I am stuck in determining what is the matrix $Q$ for $Q^{-1}P_{\sigma}Q$.
I know that I need to do "change of basis" and work out the vectors and stuff like this, but I can't seem to find the thorough approach.
EDIT:
Here is what I currently have:
For the trivial representation, I have the vector $(1 , 1 , 1)$ spanning the invariant subspace.
For the two-dimensional representation, I need to find two vectors $v$ and $w$ such that:
$P_x v = -v/2 + \sqrt{3}w/2$
$P_x w = -v/2 - \sqrt{3}w/2$
$P_y v = v$
$P_y w = -w$
I found the $Q$ matrix, which is:
$$\begin{bmatrix} 1 & 1 & -\frac{(1 + \sqrt{3})}{2} \\ 1 & 1 & \frac{(1 + \sqrt{3})}{2} \\ 1 & -2 & 0 \end{bmatrix}$$
But it is wrong.
Any advices or comments you have?