Aufgaben:Problem 7

Exercise

Compute the character table of \(S_4\)

Solution

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 as \( (\psi , \phi)_G = \frac{1}{\left|G\right|} \sum_{y \in G} \psi (y) \phi (y)^{*} \)

""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 (-1,0,0)(0,1,0)(0,0,1). 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 (-1/2,-sqrt(3)/2,0)(sqrt(3)/2,-1/2,0)(0,0,1). You surely can imagine the trace of that. It's 0. Wuueeh. We got our third number. Still two to go. We now look

Charactertable 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

References

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

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