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Decompose the permutation into the direct sum of irreducible reps.

  1. May 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Note: I need help with part (c).

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

    a) Determine the character [itex]\chi_P : S_3 \rightarrow \mathbb{C}[/itex]

    b) Find all the irreducible representations of [itex]S_3[/itex].

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

    2. The attempt at a solution

    Here is my overall progress for the problem:

    I let [itex]e[/itex] to be the identity permutation, [itex]x = (1 \ 2 \ 3)[/itex] and [itex]y = (1 \ 2)[/itex]

    Then, I let [itex]P_x = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}[/itex], [itex]P_y = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}[/itex] and [itex]P_e = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}[/itex].

    The conjugacy classes are as followed:

    - [itex]\{e\}[/itex]

    - [itex]\{y, xy, yx\}[/itex]

    - [itex]\{x, x^2\}[/itex]

    The character table shows that:

    [itex]\chi (\{e\}) = 3[/itex]

    [itex]\chi (\{x, x^2\}) = 0[/itex]

    [itex]\chi (\{y, xy, yx\}) = 1[/itex]

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

    - trivial representation [itex]T[/itex]
    - sign representation [itex]\Sigma[/itex]
    - two-dimensional representation [itex]A[/itex], presented as the symmetries of equilateral triangle

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

    [itex]d_1^2 + d_2^2 + d_3^2 = |S_3| = 6[/itex]

    The only possibility for equality to hold is [itex]d_1 = d_2 = 1[/itex] and [itex]d_3 = 2[/itex].

    Another character table shows that:

    [itex]\chi_T (\{e\}) = 1[/itex]

    [itex]\chi_T (\{x, x^2\}) = 1[/itex]

    [itex]\chi_T (\{y, yx, xy\}) = 1[/itex]

    [itex]\chi_A (\{e\}) = 2[/itex]

    [itex]\chi_A (\{x, x^2\}) = -1[/itex]

    [itex]\chi_A (\{y, xy, yx\}) = 0[/itex]

    [itex]\chi_{\Sigma} (\{e\}) = 1[/itex]

    [itex]\chi_{\Sigma} (\{x, x^2\}) = 1[/itex]

    [itex]\chi_{\Sigma} (\{y, yx, xy\}) = -1[/itex]

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

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

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

    [itex]P_xv = w[/itex]

    [itex]P_yv = v[/itex]

    [itex]P_xw = -v[/itex]

    [itex]P_yw = -w[/itex]

    Here is what I have:

    [itex]v = \begin{bmatrix} a \\ b \\ c\end{bmatrix} \rightarrow w = \begin{bmatrix} c \\ a \\ b\end{bmatrix}[/itex]

    [itex]v = \begin{bmatrix} a \\ b \\ c\end{bmatrix} \rightarrow v = \begin{bmatrix} b \\ a \\ c\end{bmatrix}[/itex]

    [itex]w = \begin{bmatrix} c \\ a \\ b\end{bmatrix} \rightarrow -v = \begin{bmatrix} b \\ c \\ a\end{bmatrix}[/itex]

    [itex]w = \begin{bmatrix} c \\ a \\ b\end{bmatrix} \rightarrow -w = \begin{bmatrix} a \\ c \\ b\end{bmatrix}[/itex]

    From here, I can't determine what are the nonzero vectors for [itex]Q[/itex].

    Any advices or comments?
     
    Last edited: May 19, 2013
  2. jcsd
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