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Obtaining representations of the symmetric group

  1. Mar 5, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Consider the following permutation representations of three elements in ##S_3##: $$\Gamma((1,2)) = \begin{pmatrix} 0&1&0\\1&0&0\\0&0&1 \end{pmatrix}\,\,\,\,;\Gamma((1,3)) = \begin{pmatrix} 0&0&1\\0&1&0\\1&0&0 \end{pmatrix}\,\,\,\,\,; \Gamma((1,3,2)) = \begin{pmatrix} 0&1&0\\0&0&1\\1&0&0\end{pmatrix}$$

    The trivial representation is embedded in the subspace spanned by the vector ##\underline{r} = (1,1,1)##. Give ##x,y,\alpha, \beta## and then obtain the representation matrices ##\Gamma((1,2)), \Gamma((1,3,2))## and ##\Gamma((1,3))## in a basis where ##\underline{e}_1 = (x,y,y)## and ##\underline{e}_2 = (0,\alpha, \beta)##, with ##x,\beta > 0## and ##\underline{e}_i \cdot \underline{e}_j = \delta_{ij}\,\,\,, \underline{e}_i \cdot \underline{r} = 0##.

    There is a hint with the question that states that the matrix ##\Gamma((1,2))## satisfies, in the standard representation, $$\Gamma((1,2)) = \frac{1}{2}\begin{pmatrix} ..&\sqrt{3}\\..&1\end{pmatrix}$$

    2. Relevant equations
    $$\Gamma_{perm} = \Gamma_{triv} \oplus \Gamma_{stand} \Rightarrow \chi(\Gamma_{perm}) = \chi(\Gamma_{triv}) + \chi(\Gamma_{stand}) \Rightarrow \chi(\Gamma_{stand}) = 0\,\,\,\,\text{for}\,\,\,\Gamma((1,2))\,\,\,\,\text{and}\,\,\,\, \Gamma((1,3))$$
    3. The attempt at a solution
    I think ##(1,1,1)## is a common eigenvector to all the permutation representations. Using the given condition involving the Kronecker tensor, I obtained the conditions ##\alpha = -\beta## and ##x = -2y##. I am looking for a hint on how to compute these matrices in the desired basis. Using the equation in the relevant equations, I get that the left hand entry for the matrix in the hint is ##-1## and by orthonormality of the rows the entry ##(21)## is ##\sqrt{3}##. But I am not sure how to actually obtain the matrices in the basis.

    The RHS is a direct sum of the decomposition of two irreducible reps, which can be written like $$\begin{pmatrix}1&0&0\\0&a&b\\0&c&d \end{pmatrix},$$ but I don't see how to proceed.

    Thanks.
     
    Last edited: Mar 5, 2014
  2. jcsd
  3. Mar 5, 2014 #2

    vanhees71

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    It's easier to work with the bra-ket notation known from quantum theory, i.e., write [itex]|e_j \rangle[/itex] instead of [itex]\underline{e}_j[/itex] etc. Call the old basis [itex]|b_j \rangle[/itex]

    The first thing is to determine the new basis from the given properties. Then you get the unitary transformation matrix [itex]T_{jk}[/itex] such that
    [tex]|e_j \rangle=\sum_{k} T_{kj} |b_k \rangle.[/tex]
    Now for any linear operator [itex]\hat{\Gamma}[/itex] you have
    [tex]\Gamma_{jk}':=\langle e_j |\hat{\Gamma}|e_k \rangle.[/tex]
    In terms of the old basis that's given by
    [tex]\Gamma_{jk}'=\sum_{lm} \langle e_j|b_l \rangle \langle b_l \hat{\Gamma}|b_m \rangle \langle b_m|e_k \rangle=\sum_{lm}\langle e_j|b_l \rangle \Gamma_{lm} \langle b_m|e_k \rangle.[/tex]
    Now because of the orthonormality you have
    [tex]\langle b_m|e_k \rangle=T_{mk}, \quad \langle{e_j} \rangle_{b_l} = T_{lj}^*.[/tex]
    From this you get
    [tex]\Gamma_{jk}'=\sum_{lm} T_{lj}^* \Gamma_{lm} T_{mk}.[/tex]
    In matrix notation that means
    [tex]\hat{\Gamma}'=\hat{T}^{\dagger} \hat{\Gamma} \hat{T}.[/tex]
     
  4. Mar 5, 2014 #3

    CAF123

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    Thanks vanhees71,

    So, to obtain the permutation representation in the new basis I have to compute that transformation. Since ##T## is unitary, the result is the same as $$\Gamma' = T^{-1}\Gamma T \Rightarrow T\Gamma'T^{-1} = \Gamma$$ and hence the two representations are equivalent, related by this similarity transformation.

    But I don't quite see how to obtain the matrix ##T## in practice (which I think can be thought of as a rotation matrix that takes one basis where we have ##\Gamma## to the new basis where we have ##\Gamma'##)

    Using the orthonormality conditions, I can obtain a vector ##\underline{e}_3## by computing ##\underline{e}_1 \times \underline{e}_2## and thereby obtain $$\begin{pmatrix} e_1 \\e_2\\e_3 \end{pmatrix} = \begin{pmatrix} x&y&y\\0&\alpha&\beta\\(y\beta-y\alpha)& -x\beta&-2y^2 \end{pmatrix} \begin{pmatrix} b_1\\b_2\\b_3\end{pmatrix}$$ but I am not sure if this is what I need.
     
  5. Mar 5, 2014 #4

    CAF123

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    If ##T## is that 3x3 matrix above that I obtained involving ##x,y,\alpha, \beta##, then ##T^{\dagger} = T^T## (T transpose) since the question states ##x, \alpha > 0## (not β as I originally posted)(and since ##\beta = -\alpha## and ##x=-2y##, y < 0 and ##\beta < 0## ) and there is no hierarchy for the complex numbers, I inferred from this that all entries are real.

    I thought it might just be a case of subbing in for ##T^T, T ##and## \Gamma((1,2))##, evaluating ##T^T \Gamma((1,2))T## where $$\Gamma((1,2)) = \begin{pmatrix} 0&1&0\\1&0&0\\0&0&1\end{pmatrix}$$ and knowing 2 entries of the standard rep of ##\Gamma((1,2))## I could solve for ##x,y,\beta, \alpha## so that I could repeat the whole calculation and get the standard reps for the other elements. Does this seem reasonable? I tried this, but I keep end up with contradictions. Is there a more elegant/faster way?

    Thanks.
     
    Last edited: Mar 5, 2014
  6. Mar 5, 2014 #5

    vanhees71

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    Ok, you look for a orthonormal set of vectors such that
    [tex]|e_1 \rangle=x |b_1 \rangle + y |b_2 \rangle + y |b_3 \rangle, \quad |e_2 \rangle=\alpha |b_2 \rangle + \beta |b_3 \rangle, |e_3 \rangle = \frac{1}{\sqrt{3}} (|b_1 \rangle + |b_2 \rangle + b_3 \rangle).[/tex]
    Further you say, you want [itex]x,\alpha>0[/itex]. From
    [tex]\langle e_3 | e_1 \rangle=\frac{1}{\sqrt{3}}(x+2y)=0 \; \Rightarrow \; y=-\frac{x}{2}.[/tex]
    So [itex]x,y \in \mathbb{R}[/itex]. Then
    [tex]\langle e_1 | e_2 \rangle=y (\alpha+\beta)=0 \; \Rightarrow \; \beta=-\alpha <0.[/tex]
    Thus
    [tex]|e_1 \rangle=x (|b_1 \rangle-1/2 |b_2 \rangle -1/2 |b_3 \rangle).[/tex]
    From the normalization condition you get [tex]x=\sqrt{2/3}.[/tex]
    So you have
    [tex]|e_1 \rangle=\sqrt{\frac{2}{3}} \left ( |b_1 \rangle -\frac{1}{2} |b_2 \rangle -\frac{1}{2} |b_3 \rangle \right).[/tex]
    In the same way you find
    [tex]|e_2 \rangle = \sqrt{\frac{1}{2}}(|b_2 \rangle-|b_3 \rangle).[/tex]
    From this you can easily find [itex]\hat{T}[/itex] and then evaluate the representation of the [itex]\Gamma[/itex] matrices.
     
  7. Mar 5, 2014 #6

    CAF123

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    I see, I was way overthinking the problem.
    This gives $$\begin{pmatrix} e_1\\e_2\\e_3 \end{pmatrix} = \begin{pmatrix} \sqrt{\frac{2}{3}}&-\frac{1}{\sqrt{6}}&-\frac{1}{\sqrt{6}}\\0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}} \end{pmatrix} \begin{pmatrix} b_1\\b_2\\b_3 \end{pmatrix}$$

    T is the 3 x3 matrix there. Then I computed, via the transformation, ##\Gamma' = T^T \Gamma T## and this gave me $$\begin{pmatrix} 1/3& 1/\sqrt{3} + 1/3&-1/\sqrt{3}+1/3\\ 1/\sqrt{3}+1/3&-2/\sqrt{12}+1/3&1/3\\-1/\sqrt{3}+1/3&1/3&2/\sqrt{12}+1/3 \end{pmatrix}$$

    How do I obtain the block triangular form $$\begin{pmatrix} 1&0&0\\0&a&b\\0&c&d \end{pmatrix}?$$ so that I can then extract the standard representation knowing that a permutation representation decomposes into the trivial and the standard.
     
    Last edited: Mar 5, 2014
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