# Character Table S4

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.

• This video might help you. By identity you mean trivial representation? For the other one dimensional, we have $\chi_{sgn}(\rho)=trace(sng(\rho))$ Jan 10, 2015 at 14:44
• You have to use the orthogonality relation between the columns. Jan 10, 2015 at 16:36

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 $$\chi_\mathrm{standard}$$:

• $$\chi_\mathrm{standard}(\mathrm{id})=3$$ (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 $$\rho_\mathrm{standard} : S_3 \to \mathrm{GL}(n,\mathbb{C})$$.