Aufgaben:Problem 7

Exercise

Compute the character table of \(S_4\)

Solution

Definitions

Let \(G\) be a finite group (in our case its \(S_4\)) and \( \left|G\right| \) the number of elements in the group (\(\left| S_4 \right| = 4! = 24 \))

A representation is irreducible if the following holds: \( (\chi^i , \chi^j)_G = \frac{1}{\left|G\right|} \sum_{K} \left| K \right| \chi_K^i \chi_K^{j*} = \delta_{ij} \) where \( \chi_K^i \) is the character of the i-th representation and the k-th conjugacy class and \( \left|K\right|\) is the number of elements of the conjugacy class.

For the definition of a character just look on Wikipedia: https://de.wikipedia.org/wiki/Charakter_(Mathematik)#Charaktere_von_Darstellungen

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:

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

Step 1:

\( \rho^{triv} = \chi^{triv}\) is the trivial map from \( S_4 \to GL(\mathbb{C}) \text{ where } \chi^{triv} \equiv 1\). \(\rho^{triv}\) is one dimensional and therefore irreducible.

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

Step 2:

We let act the the conjugacy class on \( (1,2,3,4) \in S_4\). That means for example we exchange two numbers for \(C_2\) or three numbers for \(C_3\). We then calculate the signum (for example by counting the inversion number -> You should know how this works from linear algebra). \(\rho^{sgn}\) is one dimensional and therefore irreducible. We see that it is not not isomorphic to \(\rho^{triv}\) since the character is different.

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

Step 3:

In Exercise 9 we proved that the representation of \(S_n\) on \(V= \mathbb{C}^n\) that permutes the basis vectors decomposes into a direct sum of the trivial rep. and an irreducible representation \(\rho^{std}\). Therefore

$$ \chi^V = \chi^{std} + \chi^{trv} \Rightarrow \chi^{std} = \chi^V -\chi^{trv}$$

We can easily evaluate \(\chi^{std}\) by looking at the matrices that interchange the basis vectors of \(\mathbb{C}^4\)

\( \chi^{V}(e) = tr \left( \begin{smallmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{smallmatrix} \right) = 4 \)

\( \chi^{V}(C_2) =\chi^{V}(12) = tr \left( \begin{smallmatrix}0 & 1 & 0 & 0\\1 & 0 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{smallmatrix} \right) = 2 \)

\( \chi^{V}(C_3) = \chi^{V}(123) = tr \left( \begin{smallmatrix}0 & 0 & 1 & 0\\1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 0 & 1\end{smallmatrix} \right) = 1 \)

\( \chi^{V}(C_4) = \chi^{V}(1234) = tr \left( \begin{smallmatrix}0 & 0 & 0 & 1\\1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\end{smallmatrix} \right) = 0 \)

\( \chi^{V}(C_{2,2}) = \chi^{V}(12)(34) = tr \left( \begin{smallmatrix}0 & 1 & 0 & 0\\1 & 0 & 0 & 0\\0 & 0 & 0 & 1\\0 & 0 & 1 & 0\end{smallmatrix} \right) = 0 \)

abusing notation:

\(\Rightarrow \chi^{std} = (4,2,1,0,0) - (1,1,1,1,1) = (3,1,0,-1,-1)\)

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

Alternative 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.

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

We check if that is irreducible:

\(\left(\chi^{std}, \chi^{std}\right)_G = \frac{1}{24} \cdot (1 \cdot 3 \cdot 3 + 6 \cdot 1 \cdot 1 + 8 \cdot 0 \cdot 0 + 6 \cdot -1 \cdot -1 + 3 \cdot -1 \cdot -1) = 1 \)

\(\left(\chi^{std}, \chi^{triv}\right)_G = \frac{1}{24} \cdot (1 \cdot 3 \cdot 1 + 6 \cdot 1 \cdot 1 + 8 \cdot 0 \cdot 1 + 6 \cdot -1 \cdot 1 + 3 \cdot -1 \cdot 1) = 0 \)

\(\left(\chi^{std}, sgn\right)_G = \frac{1}{24} \cdot (1 \cdot 3 \cdot 1 + 6 \cdot 1 \cdot -1 + 8 \cdot 0 \cdot 1 + 6 \cdot -1 \cdot -1 + 3 \cdot -1 \cdot 1) = 0 \)

Step 4:

The Tensor product of two representations is defined by: $$ (\rho_V \otimes \rho_W)(g) = \rho_V(g) \otimes \rho_W(g) \Rightarrow \chi^{V\otimes W} = \chi^V * \chi^W $$ (You might want to verify this, it follows from the definition of the tensor product)

(You should probably also check that this defines a representation: \((\rho_V \otimes \rho_W)(gh) = (\rho_V \otimes \rho_W)(g) (\rho_V \otimes \rho_W)(h) \)

Now we can calculate the inner tensor product \( \rho^{sgn} \otimes \rho^{std} \) (this is always a good idea to do in order to find representations).

\( \Rightarrow \chi^{sgn\otimes std } = \chi^{sgn} * \chi^{std } = (1,-1,1,-1,1) *(3,1,0,-1,-1) = (3,-1,0,1,-1)\)

We check if it's irreducible:

\(\left(\chi^{sgn\otimes std }, \chi^{sgn\otimes std }\right)_G = \frac{1}{24} \cdot (1 \cdot 3 \cdot 3 + 6 \cdot -1 \cdot -1 + 8 \cdot 0 \cdot 0 + 6 \cdot 1 \cdot 1 + 3 \cdot -1 \cdot -1) = 1 \)

\(\left(\chi^{sgn\otimes std }, \chi^{triv}\right)_G = \frac{1}{24} \cdot (1 \cdot 3 \cdot 1 + 6 \cdot -1 \cdot 1 + 8 \cdot 0 \cdot 1 + 6 \cdot 1 \cdot 1 + 3 \cdot -1 \cdot 1) = 0 \)

\(\left(\chi^{sgn\otimes std }, \chi^{sgn}\right)_G = \frac{1}{24} \cdot (1 \cdot 3 \cdot 1 + 6 \cdot -1 \cdot -1 + 8 \cdot 0 \cdot 1 + 6 \cdot 1 \cdot -1 + 3 \cdot -1 \cdot 1) = 0 \)

\(\left(\chi^{sgn\otimes std }, \chi^{std}\right)_G = \frac{1}{24} \cdot (1 \cdot 3 \cdot 3 + 6 \cdot -1 \cdot 1 + 8 \cdot 0 \cdot 0 + 6 \cdot 1 \cdot -1 + 3 \cdot -1 \cdot -1) = 0 \)

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

Step 5:

Now we can calculate the inner tensor product of \( \rho^{std} \) with itself and get \(\chi^{ std \otimes std } = (9,1,0,1,1)\) which isn't irreducible, but we know that every representation of \(G\) is isomorphic to a direct sum of irreducible representations of \(G\):

$$\Rightarrow \chi^{std \otimes std} = x_1*\chi^{triv} + x_2*\chi^{sgn} + x_3*\chi^{std} + x_4*\chi^{ sgn \otimes std } + x_5*\chi^{(5)}$$

Where \(\chi^{(5)}\) is the character of the missing irreducible representation \(\rho^{(5)}\). And the \(x\) are the number of times the representations are found in \(\rho^{ std \otimes std } \).

\(\left(\chi^{std \otimes std}, \chi^{triv}\right)_G = \frac{1}{24} \cdot (1 \cdot 9 \cdot 1 + 6 \cdot 1 \cdot 1 + 8 \cdot 0 \cdot 1 + 6 \cdot 1 \cdot 1 + 3 \cdot 1 \cdot 1) = 1 \)

\(\left(\chi^{std \otimes std}, \chi^{sgn}\right)_G = \frac{1}{24} \cdot (1 \cdot 9 \cdot 1 + 6 \cdot 1 \cdot -1 + 8 \cdot 0 \cdot 1 + 6 \cdot 1 \cdot -1 + 3 \cdot 1 \cdot 1) = 0 \)

\(\left(\chi^{std \otimes std},\chi^{std}\right)_G = \frac{1}{24} \cdot (1 \cdot 9 \cdot 3 + 6 \cdot 1 \cdot 1 + 8 \cdot 0 \cdot 0 + 6 \cdot 1 \cdot -1 + 3 \cdot 1 \cdot -1) = 1 \)

\(\left(\chi^{std \otimes std}, \chi^{sgn \otimes std }\right)_G = \frac{1}{24} \cdot (1 \cdot 9 \cdot 3 + 6 \cdot 1 \cdot -1 + 8 \cdot 0 \cdot 0 + 6 \cdot 1 \cdot 1 + 3 \cdot 1 \cdot -1) = 1 \)

Therfore \(\rho^{sgn}\) can't be part of the representation and \(\chi^{std \otimes std} = \chi^{triv} + \chi^{std} + \chi^{ sgn \otimes std } + x_5*\chi^{(5)}\)

But since \(\chi^{triv} + \chi^{std} + \chi^{ sgn \otimes std } = (7, 1, 1, 1, -1) \neq (9,1,0,1,1)\)

We know the representation \(\chi^{(5)}\) appears at least once in \(\chi^{ std \otimes std }\) and

\(x_5*\chi^{(5)} = \chi^{std \otimes std} - \chi^{triv} - \chi^{std} - \chi^{ sgn \otimes std } = (9,1,0,1,1)- (7, 1, 1, 1, -1) = (2,0,-1,0,2)\)

Now we just have to calculate the inner product of \(x_5* \chi^{(5)} \) with itself which looks like the following:

\(\left(x_5 *\chi^{(5)}, x_5*\chi^{(5)}\right)_G = \frac{1}{24} \cdot (1 \cdot 2 \cdot 2 + 6 \cdot 0 \cdot 0 + 8 \cdot -1 \cdot -1 + 6 \cdot 0 \cdot 0 + 3 \cdot 2 \cdot 2) = 1 \Rightarrow x_5 = 1 \)

This means that \(\rho^{(5)}\) is the last irreducible representation of \(S_4\) since we already checked the other 4 properties (by construction). And we are done.

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

Alternative Step 5:

This proof is based of the comment in the second link, that representation \((5)\) is the pullback (same idea as in general relativity) of the standard representation of \(S_3\). I will first sketh the proof and then prove the assumtion I made:

Let \(\rho^{std}\) be the standard rep. of \(S_3\) on \(V\). Let \(\phi: S_4 \rightarrow S_4/C_2^2 \) be a surjective group homomorphism, and \(\Omega: S_4/C_2^2 \rightarrow S_3 \) a group isomorphism.

Claim The pullback representation:

\begin{align} p: S_4 &\rightarrow GL(V)\\ g &\mapsto \rho^{std}(\Omega\circ \phi(g)) \end{align}

is an irreducible rep. of \(S_4\) on \(V\).

Proof: The homomorphic property of \(p\) follows directly from the homomorphic property of \(\rho^{std}\), \(\phi\) and \(\Omega\) therefore \(p\) is a representation on  \(V\). Let \(W\) be a non trivial invarinat subspace of \(V\) under \(p\):

$$p(g)w = \rho^{std}(\Omega\circ \phi(g)) w = \rho^{std}(h) w \in W$$

Because \(\Omega\) and \(\phi\) are surjective this holds for all \(h \in S_3\). Thus we get the contradiction that \(W\) is a non trivial invarinat subspace of \(V\) under \(\rho^{std}\).

To now prove the assumptions we calculate the Qutientgroup: \(S_4/C_2^2\) and we will see that it is isomorphic to \(S_3\) (The Qutient group is the group of conjugacy classes under the equivalence relation \(g_1 \sim g_2 \Leftrightarrow \exists h \in C_2^2, g_2 = g_1 h\) Where  \(g_1, g_2 \in S_4\)). After some calculation we find the Elements of \(S_4/C_2^2\):

\begin{align} [C_1] &= \{ (e), (12)(34), (13)(24), (14)(23) \}\\ [C_2] &= \{ (12), (34), (1324), (1423) \}\\ [C_3] &= \{ (13), (24), (1234), (1432) \}\\ [C_4] &= \{ (14), (23), (1243), (1342) \}\\ [C_5] &= \{ (123), (134), (243), (142) \}\\ [C_6] &= \{ (234), (124), (132), (143) \} \end{align}

the natural map \(\phi: S_4 \rightarrow S_4/C_2^2\) is obviously surjective and it is a group homomorphsim if \(S_4/C_2^2\) is again a group. We define the group operation \([g_1][g_2]:= [g_1 g_2]\) where \(g_1,g_2\in S_4\) and the bracket deontes the elment of \(S_4/C_2^2\) that \(g\) is apart of. \([\ ]\) is well defined, if the group operation is independent of the elements \(g_1,g_2\): let \(h_1, h_2 \in C_2^2\) then \(g_ih_i \in [g_i]\) and $$ [g_1h_1] [g_2h_2] = [g_1h_1g_2h_2] = [g_1g_2g_2^{-1}h_1g_2h_2] = [g_1g_2h_3h_2] = [g_1g_2]$$ where \(h_3 \in C_2^2\). Clousure follows from the fact that \(S_4\) is a group. Associativity follows from the Associativity of \(S_4\). The neutral element is \([C_1]\) and \([g]^{-1} = [g^{-1}]\).

Now we only have to show that \(S_4/C_2^2\) is isomorphic to \(S_3\). We define the map

\begin{align} \Omega : S_4/C_2^2 &\rightarrow S_3\\ [C_1] &\mapsto (e)\\ [C_2] &\mapsto (12)\\ [C_3] &\mapsto (13)\\ [C_4] &\mapsto (23)\\ [C_5] &\mapsto (123)\\ [C_6] &\mapsto (132) \end{align}

which is homomorphic (some more calculations) and obviously bijective thus a group isomorphism. This concludes the proof.

Now we can use the know character: \(\chi^{std} = (2,0,-1)\) to calculate \(\chi^p = (2,0,-1, 0, 2)\) since \(p(1234) = \rho^{std}(\Omega \circ \phi(1234)) = \rho^{std}((13))\) and \(p(12)(34) = \rho^{std}(\Omega \circ \phi(12)(34)) = \rho^{std}((e))\).

References

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

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