Obtaining representations of the symmetric group

[CAF123]

Homework Statement

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, $$\Gamma((1,2)) = \frac{1}{2}\begin{pmatrix} ..&\sqrt{3}\\..&1\end{pmatrix}$$

Homework Equations

$$\Gamma_{perm} = \Gamma_{triv} \oplus \Gamma_{stand} \Rightarrow \chi(\Gamma_{perm}) = \chi(\Gamma_{triv}) + \chi(\Gamma_{stand}) \Rightarrow \chi(\Gamma_{stand}) = 0\,\,\,\,\text{for}\,\,\,\Gamma((1,2))\,\,\,\,\text{and}\,\,\,\, \Gamma((1,3))$$

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.

 

[vanhees71]

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

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

 

[CAF123]

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.

 

[CAF123]

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.

 

[vanhees71]

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

 

[CAF123]

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.