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I am trying to understand how to build a character table of S4. I've already read many articles about it but I am stuck at one point.

S4 has 5 conjugacy classes and therefore 5 irreducible representation. One of them is the identity which is one-dimensional.

The dimensions of all representations is determined by the sum of dimensions squares, which must equal 24, because 4! = 24. From this, the representations have dimensions 1, 1, 2, 3, 3.

And now I am stuck. I know that one row will be filled with 1 (identity, dimension 1), but what about the rest? The other 1-dimensional representation is the signum of permutations, but I honestly don't know why (mostly because signum is not present in the character table of A4).

I'll appreciate any help.

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    $\begingroup$ This video might help you. By identity you mean trivial representation? For the other one dimensional, we have $\chi_{sgn}(\rho)=trace(sng(\rho))$ $\endgroup$ – Math137 Jan 10 '15 at 14:44
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    $\begingroup$ You have to use the orthogonality relation between the columns. $\endgroup$ – ArthurStuart Jan 10 '15 at 16:36
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Two 1-dimensional representations for $S_4$ are immediate: as you said, the trivial representation $\rho_\mathrm{trivial}$ and the sign representation $\rho_\mathrm{sign}$. $A_4$, however, do not have a distinct sign representation because it is indeed trivial.

Another immediate representation is the one in $\mathbb{C}^4$ sending each permutation to its associated permutation matrix. This representation is reducible, with an obvious invariant subspace the span of $e_1+e_2+e_3+e_4$. As in the case of $S_3$ we may study its orthogonal complement of this 1-dimensional invariant space, i.e., the subspace spanned by $\{e_2-e_1,e_3-e_1,e_4-e_1\}$, which is again invariant. This sub-representation is the standard representation $\rho_\mathrm{standard}$. Using this particular basis, for example, we may compute the character $\chi_\mathrm{trivial}$:

  • $\chi_\mathrm{standard}(\mathrm{id})=3$ (size 1);
  • $\chi_\mathrm{standard}((1 \ 2))=1$ (size 6);
  • $\chi_\mathrm{standard}((1 \ 2 \ 3))=0$ (size 8);
  • $\chi_\mathrm{standard}((1 \ 2 \ 3 \ 4))=-1$ (size 6);
  • $\chi_\mathrm{standard}((1 \ 2) (3 \ 4))=-1$ (size 3).

Since $\langle \chi_\mathrm{standard}, \chi_\mathrm{standard} \rangle = 1$, it follows that $\rho_\mathrm{standard}$ is irreducible.

To cook up the other 3-dimensional irreducible representation, we may simply multiply $\rho_\mathrm{sign}$ and $\rho_\mathrm{standard}$. We may simply compute the character and verify that it has norm 1 to show that it is irreducible.

Finally we only have 1 row left in the character table, which corresponds to a 2-dimensional representation. Using orthogonality and the three other known irreducible representations, (say start with the class function that is 2 on identity element and 0 otherwise, and subtract the orthogonal projections, then normalize). It turns out the last character of irreducible representation $\chi$ is

  • $\chi(\mathrm{id})=2$ (size 1);
  • $\chi((1 \ 2))=0$ (size 6);
  • $\chi((1 \ 2 \ 3))=-1$ (size 8);
  • $\chi((1 \ 2 \ 3 \ 4))=0$ (size 6);
  • $\chi((1 \ 2) (3 \ 4))=2$ (size 3).

In case you want to know, the corresponding representation is the composition of the canonical surjection $S_4 \to S_4/V \cong S_3$, where $V$ is the subgroup generated by the double transpositions, and the standard 2-dimensional representation $\rho_\mathrm{standard} : S_3 \to \mathrm{GL}(n,\mathbb{C})$.

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