# Obtaining representations of the symmetric group

• CAF123
Is there a quick way to read this off from the matrix without having to multiply it out and then read it off?In summary, the given permutation representations of three elements in ##S_3## can be obtained in a new basis by using the unitary transformation matrix ##T## and then evaluating the transformed representation matrices ##\Gamma' = T^T \Gamma T##. The resulting matrices can be written in a block triangular form, with the trivial representation in the first block and the non-trivial representation in the remaining block. The entries in the non-trivial block can then be used to obtain the standard representation.
CAF123
Gold Member

## 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.

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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
$$\hat{\Gamma}'=\hat{T}^{\dagger} \hat{\Gamma} \hat{T}.$$

1 person
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.

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:
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.

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.

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## 1. What is the symmetric group?

The symmetric group, denoted as Sn, is a group that consists of all the possible permutations of n distinct elements. In other words, it is the set of all possible ways to arrange n objects.

## 2. Why is it important to obtain representations of the symmetric group?

Obtaining representations of the symmetric group is important because it allows us to study and understand the group's structure and properties. It also helps in solving problems in various fields such as algebra, geometry, and physics.

## 3. How do you obtain representations of the symmetric group?

There are several methods to obtain representations of the symmetric group, including the Young tableaux method, the character theory method, and the representation theory method. Each method uses different techniques to construct representations of the symmetric group.

## 4. What are the applications of representations of the symmetric group?

Representations of the symmetric group have various applications in different areas of mathematics, such as group theory, algebraic geometry, and number theory. They are also used in physics to study the symmetry of physical systems and in computer science for data compression and error correction techniques.

## 5. Can representations of the symmetric group be generalized to other groups?

Yes, the concept of obtaining representations of a group can be generalized to other groups. In fact, representation theory is a fundamental tool in the study of abstract algebra and is used to understand the structure and properties of different groups, including the symmetric group.

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