Decompose the permutation into the direct sum of irreducible reps.

In summary, the conversation discusses finding the character and decomposing a representation of a permutation matrix into irreducible representations. The character table is used to find the representations and the remaining two vectors for the basis can be determined using the character table, the orthogonality of the vectors, and the determinant of the change of basis matrix.
  • #1
NasuSama
326
3

Homework Statement



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:
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  • #2


Hi there, great job on your progress so far! I can offer a few suggestions for part (c):

1. Try using the character table to determine the remaining two vectors for the basis. Since the character table gives the values of the characters for each conjugacy class, you can use this information to determine the remaining two vectors.

2. Remember that the vectors for the basis must be orthogonal to each other. This means that their dot product must be equal to 0. You can use this information to help determine the remaining two vectors.

3. You can also use the fact that the determinant of Q must be equal to 1 (since it is a change of basis matrix). This can help narrow down the possible values for the remaining two vectors.

I hope these suggestions help you make progress on part (c). Keep up the good work!
 

1. What is a permutation?

A permutation is a rearrangement of a set of objects or elements in a specific order. It is often represented as a sequence of numbers or symbols, and each permutation has a unique identity.

2. What is a direct sum of irreducible reps?

A direct sum of irreducible reps is a way of breaking down a larger representation into smaller, irreducible representations. This allows us to understand the structure and behavior of the larger representation in a more manageable way.

3. Why is it important to decompose a permutation into irreducible reps?

Decomposing a permutation into irreducible reps helps us understand the underlying symmetry and structure of the permutation. It also allows us to simplify complex problems and make them easier to solve.

4. How is a permutation decomposed into the direct sum of irreducible reps?

The process of decomposing a permutation into the direct sum of irreducible reps involves finding the eigenvalues and eigenvectors of the permutation. These eigenvectors form the basis for the irreducible representations, and the eigenvalues determine the size of the irreducible representations.

5. What are some real-world applications of decomposing a permutation into irreducible reps?

Decomposing a permutation into irreducible reps has many applications in various fields, such as physics, chemistry, and computer science. For example, it is used in quantum mechanics to study the symmetry of atomic and molecular systems, and in computer science to analyze algorithms and data structures.

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