Decompose $P$ into the direct sum of irreducible representations.

Question

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?

Answer 1

The permutation representation of $S_3$ on $\mathbb{C}^3$ (though everything works over the rationals in exactly the same way) decomposes as the direct sum of the trivial representation and the two dimensional reflection representation, which you have called $A$.

More explicitly, the subspace spanned by the vector $v=(1,1,1)^t$ is a copy of the trivial representation, with $S_3$-stable complement spanned by the vectors $\alpha_1=(1,-1,0)^t$ and $\alpha_2=(0,1,-1)^t$, isomorphic to $A$. The matrix $Q$ can therefore be taken to have these three vectors as its columns.

In case these vectors $\alpha_1$ and $\alpha_2$ seemed to come out of the blue, here is where they actually came from: the usual Hermitian inner product on $\mathbb{C}^3$ is $S_3$-invariant, so the orthogonal complement of $v=(1,1,1)^t$ is an $S_3$-stable complement to its span. But this orthogonal complement consists exactly of the vectors whose coordinates sum to $0$, and the $\alpha$'s are just what I consider the most obvious basis of this space.