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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?

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  • $\begingroup$ the vector of all 1's is an eigenvalue for all the permutation matrices, which should reduce to a 2-d problem. you probably get $A_{\sigma}$ if the determinant is -1. $\endgroup$ – mike May 18 '13 at 22:35
  • $\begingroup$ @mike I know that I will get $A_{\sigma}$ as the block of the matrix, but I can't seem to go one from here. I attempted to find the eigenvalues and eigenvectors of the matrix and find the way to obtain the diagonalization of the matrix, but this doesn't seem to work. Mind if you show me some work? $\endgroup$ – NasuSama May 19 '13 at 14:45
  • $\begingroup$ I attempt to work out the two dimensional subspace of $\mathbb{R}^3$ in which $P$ acts as $A$, but I can't find the nonzero vectors. $\endgroup$ – NasuSama May 19 '13 at 21:17
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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.

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  • $\begingroup$ Do you have another way to work out the problem instead of abstract reasoning? I would like some computation. The purpose of asking this is that I want to see how invariant subspaces are taken in place for each of the representation. I have the invariant subspace spanned by the vector $(1 \ 1 \ 1)$ correct, but I can't get other two correct. $\endgroup$ – NasuSama May 21 '13 at 2:02
  • $\begingroup$ @NasuSama, The other subspace is spanned by the two vectors $\alpha_1,\alpha_2$ I mentioned. Is that not concrete enough? The matrix $Q$ is now completely explicit (it has columns $v,\alpha_1,\alpha_2$) and you can write your representation in block diagonal form, with irreducible blocks. $\endgroup$ – Stephen May 21 '13 at 2:07
  • $\begingroup$ So the matrix I have now is $\begin{bmatrix}1 & 1 & 0 \\ 1 & -1 & 1 \\ 1 & 0 & -1\end{bmatrix}$. If I compute $Q^{-1}P_xQ$, then I don't get $\begin{bmatrix}1 & 0 & 0 \\ 0 & -1/2 & -\sqrt{3}/2 \\ 0 & \sqrt{3}/2 & -1/2 \end{bmatrix}$ $\endgroup$ – NasuSama May 21 '13 at 2:10
  • $\begingroup$ Why do you expect to get that matrix? $\endgroup$ – Stephen May 21 '13 at 2:10
  • $\begingroup$ I expect to get that matrix because of the representations. The top left 1 by 1 block has to be 1 (probably because of the common eigenvalue, which is 1) while the 2 by 2 block on the bottom right must be the 2 by 2 matrix $A_x$, which is the one I found for the two dimensional representation. $\endgroup$ – NasuSama May 21 '13 at 2:12

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