Aufgaben:Problem 7

Exercise

Compute the character table of \(S_4\)

Solution

Definitions

Let \(G\) be a finite group

Let \(g \in G\)

Let \(\psi \text{ and } \phi \in \mathbb{C}^G\)

There is an inner product defined of the character table defined as \( (\psi , \phi)_G = \frac{1}{\left|G\right|} \sum_{K} \left| K \right| \psi_K \phi_K^{*} \)

""We know that \( \psi_r , \psi_s \) are orthonormal if \( (\psi_r , \psi_s)_G = \delta_{rs} \)""

Step 0:

We know that \( S_4 \) is the symmetric group, that means the group of all permutations of a set with 4 elements, So we look at the Number 4 and ask ourself on how many ways we can combine smaller numbers or the 4 itself to get the value 4. We see that 1+1+1+1=4, 1+1+2=4, 1+3=4, 4=4 and 2+2=4 so we found 5 ways to do that this way. We know that \(S_4\) has \( n! = 4! = 24 \) elements which belong to 5 conjugacy classes which are represented by "e = identity", "(12) = permutation of two elements = \(C_2\)", "(123) = permutation of 3 elements = \(C_3\)", "(12)(34) = permutation of 2 times 2 elements = \(C_{2,2}\)" and "(1234) = permutation of 4 elements = \(C_4\)".

We can also count or calculate the number of elements in each conjugacy class and get the following: \( c[e]=1, c[C_2]=6, c[C_3]=8, c[C_4]=6, c[C_{2,2}]=3 \) which gives added together again 24.

The rows of the character table are the irreducible representations of the group and the columns are the conjugacy classes. Because the number of the irreducible representations is equal to the number of conjugacy classes, the table has to be quadratic. Now we can draw it like that:

\(S_4\)	\(e\)	\(6C_2\)	\(8C_3\)	\(6C_4\)	\(3C_{2,2}\)
\( \chi^{triv}\)
\( sgn \)
\( \chi^{std} \)
\( sgn \otimes \chi^{std }\)
\( \chi^{(5)} \)

Step 1:

\( \chi^{triv} \) is the trivial map from \( S_4 \to S^1 \text{ where } \chi^{triv} \equiv 1\) Therefore the table now looks like this:

\(S_4\)	\(e\)	\(6C_2\)	\(8C_3\)	\(6C_4\)	\(3C_{2,2}\)
\( \chi^{triv}\)	1	1	1	1	1
\( sgn \)
\( \chi^{std} \)
\( sgn \otimes \chi^{std }\)
\( \chi^{(5)} \)

Step 2:

We let act the the conjugacy class on \( (1,2,3,4) \in S_4\) and calculate the signum (for example by counting the inversion number). We use this as the second irreducible representation.

\(S_4\)	\(e\)	\(6C_2\)	\(8C_3\)	\(6C_4\)	\(3C_{2,2}\)
\( \chi^{triv}\)	1	1	1	1	1
\( sgn \)	1	-1	1	-1	1
\( \chi^{std} \)
\( sgn \otimes \chi^{std }\)
\( \chi^{(5)} \)

Step 3:

We now are looking for a 3-D representation of the Group. We find that a tetrahedron fulfills the conditions. To compute e we look at the identity mapping. In 3-D this is the 3x3 identity matrix. The trace of it is equal to 3. So we got our first value. We now give our corners the numers 1 to 4 and set the 1 and 2 on the x-axis on the ground. If we now use the permutation (1,2) we get the following transformation x->-x, y->y, z->z. So the transformation matrix is \( \left( \begin{smallmatrix}-1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{smallmatrix} \right) \). The Trace of this matrix ist given by 1. So we found our second value. Wueh. Imagine that we set the tetrahedron on the ground such that the corners 1,2,3 are on the xy-plane and the corner 4 is on the z-axis. We now look what happens if we take the permutation (1,2,3). ( The meaning of this is, that 3 -> 2, 2-> 1, 1-> 3) This is a rotation by 120° (2*pi/3). We see that the transformation matrix is given by \( \left( \begin{smallmatrix} \frac{-1}{2} & \frac{-\sqrt{3}}{2} & 0\\ \frac{\sqrt{3}}{2} & \frac{-1}{2} & 0\\0 & 0 & 1\end{smallmatrix} \right) \). You surely can imagine the trace of that. It's 0. Wuueeh. We got our third number. Still two to go. We now look at the permutation (1,2,3,4). It's not the easiest to imagine so we just get you the result. We first set all 4 points on one plane, then rotate by 90° and then make the point transformation v-> -v. We then get the transformation matrix \( \left( \begin{smallmatrix}0 & -1 & 0\\1 & 0 & 0\\0 & 0 & -1\end{smallmatrix} \right) \). The trace of that monster is -1. For the last one we look at the permutation (1,2)(3,4). We take again the corners 1,2,3 on the xy-plane and see that the transformation is given by x-> -x, y-> -y, z-> z. So we get the transformation matrix \( \left( \begin{smallmatrix}-1 & 0 & 0\\0 & -1 & 0\\0 & 0 & 1\end{smallmatrix} \right) \). The trace of that bastard is -1. So we got the whole third row. The conditions are fulfilled and we are happy.

\(S_4\)	\(e\)	\(6C_2\)	\(8C_3\)	\(6C_4\)	\(3C_{2,2}\)
\( \chi^{triv}\)	1	1	1	1	1
\( sgn \)	1	-1	1	-1	1
\( \chi^{std} \)	3	1	0	-1	-1
\( sgn \otimes \chi^{std }\)
\( \chi^{(5)} \)

Alternative Step 3:

Another way to find the third irreducible representation is to take the trace of every transformation matrix of all the different conjugacy classes. Then we get the representation (4, 2, 1, 0, 0). When we check if this is irreducible we see that it's not. It contains a copy of the trivial representation. We therefore have to subtract it (we then get (3, 1, 0, -1, -1) )and check again if it's irreducible and it is.

Step 4:

No we can calculate the inner tensor product of \( sgn \text{ and } \chi^{std}\) which gives us the new line in the table:

\(S_4\)	\(e\)	\(6C_2\)	\(8C_3\)	\(6C_4\)	\(3C_{2,2}\)
\( \chi^{triv}\)	1	1	1	1	1
\( sgn \)	1	-1	1	-1	1
\( \chi^{std} \)	3	1	0	-1	-1
\( sgn \otimes \chi^{std }\)	3	-1	0	1	-1
\( \chi^{(5)} \)

To check that this is an irreducible representation we have to calculate the wheightet inner product with all the other representations and with itself and should get a \( \delta_{ij} \)

Step 5:

No we can calculate the inner tensor product of \( \chi^{std} \) with itself and get (9,1,0,1,1) which isn't irreducible but we can calculat nner products with the existing irreducible characters and decompose it as \( \chi^{std} \otimes \chi^{std} = \chi^{triv} + sgn + \chi^{std} + sgn \otimes \chi^{std } + \chi^{(5)}\). This gives us the last line of the table which then looks like:

\(S_4\)	\(e\)	\(6C_2\)	\(8C_3\)	\(6C_4\)	\(3C_{2,2}\)
\( \chi^{triv}\)	1	1	1	1	1
\( sgn \)	1	-1	1	-1	1
\( \chi^{std} \)	3	1	0	-1	-1
\( sgn \otimes \chi^{std }\)	3	-1	0	1	-1
\( \chi^{(5)} \)	2	0	-1	0	2

Now we just have to calculate the inner product of \( \chi^{(5)} \) with itself and see that it is equal to 1 which means that it's irreducable.

References

https://unapologetic.wordpress.com/2010/11/08/the-character-table-of-s4/

https://www.itp.uni-hannover.de/~flohr/lectures/symm/handout2.pdf