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Section 2.10 An example: \(S_3\)

Let us now study representations of our favourite symmetric group \(S_3\) using all that we have seen about characters. This should make the various theorems and constructions more explicit.

Subsection 2.10.1 The character table

First, recall that \(S_3\) is the group of permutations of three objects. It has six elements. They are split into three conjugacy classes, corresponding to the three possible cycle structures. There is one permutation with three one-cycles (the identity permutation); 3 permutations with one two-cycle and one one-cycle (the transpositions); and 2 permutations with a three-cycle (the cyclic permutations of length 3). Let us denote the corresponding conjugacy classes as \(C_1\text{,}\) \(C_2\)and \(C_3\text{,}\) with number of elements respectively \(n_1=1\text{,}\) \(n_2=3\) and \(n_3=2\text{.}\)

Since the number of inequivalent irreducible representations is equal to the number of conjugacy classes (Theorem 2.8.4), we know that \(S_3\) has three inequivalent irreducible representations. Let us denote those by \(T^{(\alpha)}\) with \(\alpha=1,2,3.\) We have already found two of them: the identity representation (\(T^{(1)}\)) and the parity representation (\(T^{(2)}\)), which are both one-dimensional. We do not know yet the third one.

From Theorem 2.8.6, we know that the dimensions of the three irreducible representations must satisfy the equation

\begin{equation*} \sum_{\alpha=1}^3 d_\alpha^2 = |S_3| = 6. \end{equation*}

The only possibility is that two of the representations are one-dimensional (we know them already), and the third one is two-dimensional.

Even if we do not know the third two-dimensional irreducible representation explicitly, we can still construct the character table of \(S_3\) using orthogonality of characters. The character table should look like:

\(C_1\) \(C_2\) \(C_3\)
\(T^{(1)}\) \(1\) \(1\) \(1\)
\(T^{(2)}\) \(1\) \(-1\) \(1\)
\(T^{(3)}\) \(2\) \(a\) \(b\)
Table 2.10.1. Constructing the character table for \(S_3\)

We have already filled many entries, and left two unknowns (\(a\) and \(b\)), which we will determine shortly. The first column is given by the trace of the matrices corresponding to the identity element of the group, which are just identity matrices. So these entries are just the dimension of the representations. This is true in general: the first column is always given by the dimensions of the representations.

The first row corresponds to the identity representation, which assigns 1 to all group elements. So the characters are always 1.

The second row corresponds to the parity representation, which assigns 1 to even permutations and -1 to odd permutations. The permutations in the conjugacy classes \(C_1\) and \(C_3\) are even, while those in \(C_2\) are odd, so we can fill the second row appropriately.

There are two entries left, \(a\) and \(b\text{.}\) To find them, we use orthogonality of characters. We will use orthogonality of columns, which is just the statement of the second orthogonality relation Theorem 2.8.3. This relation says that

\begin{equation*} \sum_{\alpha=1}^3 (\chi_i^{(\alpha)} )^* \chi_j^{(\alpha)} = \frac{|G|}{n_i} \delta_{ij}. \end{equation*}

Applying this to the first and second conjugacy classes (the first and second columns), that is, setting \(i=1\) and \(j=2\text{,}\) the right-hand-side becomes zero, and we get:

\begin{equation*} (1)(1) + (1)(-1) + (2) (a) = 0. \end{equation*}

Therefore \(a=0\text{.}\) Similarly, using the first and third columns (\(i=1\) and \(j=3\)), we get:

\begin{equation*} (1)(1) + (1)(1) + (2)(b) = 0, \end{equation*}

that is \(b = -1\text{.}\) We have thus completed the character table of \(S_3\text{:}\)

\(C_1\) \(C_2\) \(C_3\)
\(T^{(1)}\) \(1\) \(1\) \(1\)
\(T^{(2)}\) \(1\) \(-1\) \(1\)
\(T^{(3)}\) \(2\) \(0\) \(-1\)
Table 2.10.2. The character table for \(S_3\)

For fun we can check that the other orthogonality relations are also satisfied. For instance, sticking with the second orthogonality relation Theorem 2.8.3, but applying it to the third column (\(i=j=3\)), we get:

\begin{equation*} (1)(1) + (1)(1) + (-1)(-1) = \frac{|G|}{2} = \frac{6}{2} = 3, \end{equation*}

which is indeed true.

The first orthogonality relation Theorem 2.8.1 correspond to orthogonality between rows. Recall that it says that:

\begin{equation*} \sum_{i=1}^3 n_i (\chi_i^{(\alpha)} )^* \chi_i^{(\beta)} = |G| \delta_{\alpha \beta}. \end{equation*}

For instance, using it for the second and third irreducible representations (\(\alpha=2\) and \(\beta=3\)), we get:

\begin{equation*} 1 (1)(2) + 3 (1)(0) + 2(1)(-1) = 0, \end{equation*}

which is indeed correct. Applying to the third row (\(\alpha=\beta=3\)), we get:

\begin{equation*} 1 (2)(2) + 3 (0)(0) + 2(-1)(-1) = |G| = 6, \end{equation*}

which is again correct. Orthogonality relations are indeed satisfied!

We can also check that the reducibility criterion Theorem 2.9.1 is satisfied for the irreducible representations. For instance, looking at the two-dimensional irreducible representation \(T^{(3)}\text{,}\) we get:

\begin{equation*} \sum_{i=1}^3 n_i |\chi_i^{(3)}|^2 = 1 (2)^2 + 3 (0)^2 + 2 (-1)^2 = 6, \end{equation*}

which is indeed equal to \(|G|=6\text{,}\) therefore the representation is irreducible, as we know. In fact, we know this two-dimensional irreducible representation of \(S^3\) very well: it is the two-dimensional representation constructed from symmetries of the equilateral triangle in Example 2.1.6. Indeed, one can check that the characters of this representation match with the third row of the character table for \(S^3\text{.}\)

Subsection 2.10.2 The decomposition theorem

Let us now study the reducibility criterion and decomposition theorem. Let us consider the three-dimensional permutation representation \(T\) of \(S_3\) from Example 2.1.4. Writing down the \(3 \times 3\) matrices explicitly, it is easy to calculate the characters by taking the traces. The trace of the identity matrix is 3; the trace of the three matrices corresponding to the transpositions is 1; and the trace of the two matrices corresponding to the cyclic permutations are 0. Thus, the characters of \(T\) are:

\begin{equation*} \chi_1^{(T)} = 3, \qquad \chi_2^{(T)} = 1, \qquad \chi_3^{(T)} = 0. \end{equation*}

Let us first show that this representation is reducible (we already know that since \(S_3\) has only three irreducible representations of dimensions 1, 1 and 2.) We use the reducibility criterion Theorem 2.9.1. We calculate:

\begin{equation*} \sum_{i=1}^3 n_i |\chi_i^{(T)}|^2 = 1 (3)^2 + 3 (1)^2 + 2 (0)^2 = 12, \end{equation*}

which is certainly greater than \(|G| = 6\text{.}\) Thus the representation is reducible, as expected.

We now want to find its direct sum decomposition. Recall the decomposition theorem Theorem 2.9.2. If we write the direct sum decomposition as

\begin{equation*} T = \bigoplus_{\alpha=1}^c m_\alpha T^{(\alpha)}, \end{equation*}

then the coefficients are given by

\begin{equation*} m_\alpha = \frac{1}{|G|} \sum_{i=1}^c n_i \chi_i^{(T)} (\chi_i^{(\alpha)})^*. \end{equation*}

Let us calculate these coefficients using the character table Table 2.10.2. We get:

\begin{align*} m_1 =\amp \frac{1}{6} \left( 1 (3)(1) + 3 (1)(1) + 2 (0)(1) \right) = 1,\\ m_2 =\amp \frac{1}{6} \left( 1 (3)(1) + 3 (1)(-1) + 2 (0)(1) \right) = 0,\\ m_3 =\amp \frac{1}{6} \left(1 (3)(2) + 3 (1)(0) + 2 (0)(-1) \right) = 1. \end{align*}

We thus obtain the direct sum decomposition:

\begin{equation*} T = T^{(1)} \oplus T^{(3)}. \end{equation*}

What this means is that the three-dimensional permutation representation of \(S_3\) is equivalent to the direct sum of the trivial one-dimensional representation \(T^{(1)}\text{,}\) and the two-dimensional irreducible representation \(T^{(3)}\text{.}\) This is something that you could show explicitly by finding a similarity transformation that brings all six \(3 \times 3\) matrices in block diagonal form. But you see how much faster this was using characters! No need to find similarity transformations, only simple algebraic calculations are needed! Characters are really awesome!

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