Difference between revisions of "Aufgaben:Problem 7"
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There is an inner product defined as \( (\psi , \phi)_G = \frac{1}{\left|G\right|} \sum_{y \in G} \psi (y) \phi (y)^{*} \) | 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 \( (\psi_r , \psi_s)_G = \delta_{rs} \) | + | ""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, and 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: \( c[e]=1, c[C_2]=6, c[C_3]=8, c[C_4]=6, c[C_{2,2}]=3 \) | ||
| + | |||
| + | The character table has to be quadratic where the rows are the irreducible representations of the group and the columns are the conjugacy classes of the group. So we can draw it like that: | ||
| + | {| class="wikitable" style="text-align:center" | ||
| + | ! style="width:40px" | \(S_4\) | ||
| + | ! style="width:40px" | \(e\) | ||
| + | ! style="width:40px" | \(6C_2\) | ||
| + | ! style="width:40px" | \(8C_3\) | ||
| + | ! style="width:40px" | \(6C_4\) | ||
| + | ! style="width:40px" | \(3C_{2,2}\) | ||
| + | |- | ||
| + | | \( \chi^{triv}\) || || || || || | ||
| + | |- | ||
| + | | \( sgn \) || || || || || | ||
| + | |- | ||
| + | | \( \chi^{std} \) || || || || || | ||
| + | |- | ||
| + | | \( sgn \otimes \chi^{std }\) || || || || || | ||
| + | |- | ||
| + | | \( \chi^{(5)} \) || || || || || | ||
| + | |} | ||
| + | |||
| + | '''Step 1: ''' | ||
| + | \( \chi^{triv} \) is the trivial map from \( S_4 \to S^1 \text{ where } \chi^{triv} \equiv 1\) | ||
| + | |||
| + | '''Step 2: ''' | ||
| Line 27: | Line 58: | ||
! style="width:40px" | \(3C_{2,2}\) | ! style="width:40px" | \(3C_{2,2}\) | ||
|- | |- | ||
| − | | \( | + | | \( \chi^{triv}\) || 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 |
|} | |} | ||
Revision as of 12:58, 15 June 2015
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, and 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: \( c[e]=1, c[C_2]=6, c[C_3]=8, c[C_4]=6, c[C_{2,2}]=3 \)
The character table has to be quadratic where the rows are the irreducible representations of the group and the columns are the conjugacy classes of the group. So 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 \( S_4 \to S^1 \text{ where } \chi^{triv} \equiv 1\)
Step 2:
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
