Character table from a representation?

Question

Can anybody explain how to contruct a character table. A good explained example will be fantastic to me. For example, the character table of $S_4$.

I'm quite desperate about representation theory!!

Answer 1

I am guessing you have not yet learned about Schur Functors which give you a nice way to construct the irreducible $S_{n}$ representations. So I will do this by hand.

First, we figure out what the conjugacy classes are in $S_{4}.$ These are given by cycle types, so we can denote them with integer partitions of $4$ in nonincreasing order. We have the following conjugacy classes:

$$\begin{array}{rrrrrrr} (1, 1, 1, 1) & (2, 1, 1) & (2, 2) & (3, 1) & (4)\end{array}.$$

Thus, there are five conjugacy classes and hence there are five irreducible representations. We know two of them immediately: the standard and the alternating representation. This gives us: $$\begin{array}{rrrrrrrrrrr} & (1, 1, 1, 1) & (2, 1, 1) & (2, 2) & (3, 1) & (4)\\ \chi_{\text{triv}} & 1 & 1 & 1& 1 & 1 \\ \chi_{\text{sgn}} & 1 & -1 & 1 & 1 & -1\end{array}$$

We also have the permutation representation of $S_{4}$ on $\mathbb{C}^{4}.$ Note that this has a subrepresentation, called the standard representation, of dimension 3 given by those vectors whose coordinates add up to $0$. You can also think of this as the quotient of the permutation rep by the trivial rep (so its character is obtained by subtracting the latter from the former, which is just the number of fixed points i.e. 1's in the cycle type). Computing its character table shows us that $\langle \chi_{\text{std}}, \chi_{\text{std}} \rangle = 1$ and hence the standard representation is irreducible. So, we have

$$\begin{array}{rrrrrrrrrrr} & (1, 1, 1, 1) & (2, 1, 1) & (2, 2) & (3, 1) & (4)\\ \chi_{\text{triv}} & 1 & 1 & 1& 1 & 1 \\ \chi_{\text{sgn}} & 1 & -1 & 1 & 1 & -1\\ \chi_{\text{std}} & 3 & 1 & -1 & 0 & -1\end{array}$$

Note that the multiplication of the character of the sign representation and the standard representation is not any of the other representations. Since this corresponds to tensoring the standard representation with the one dimensional sign representation, this new representation is also irreducible. So, we now have

$$\begin{array}{rrrrrrrrrrr} & (1, 1, 1, 1) & (2, 1, 1) & (2, 2) & (3, 1) & (4)\\ \chi_{\text{triv}} & 1 & 1 & 1& 1 & 1 \\ \chi_{\text{sgn}} & 1 & -1 & 1 & 1 & -1\\ \chi_{\text{std}} & 3 & 1 & -1 & 0 & -1\\ \chi_{4} & 3 & -1 & -1 & 0 & 1\end{array}$$

Finally, we know that we have only one more irreducible character (since there are 5 conjugacy classes) which must normalized and orthogonal with respect to the other. This is easily computed by a computer. So the final character table is

$$\begin{array}{rrrrrrrrrrr} & (1, 1, 1, 1) & (2, 1, 1) & (2, 2) & (3, 1) & (4)\\ \chi_{\text{triv}} & 1 & 1 & 1& 1 & 1 \\ \chi_{\text{sgn}} & 1 & -1 & 1 & 1 & -1\\ \chi_{\text{std}} & 3 & 1 & -1 & 0 & -1\\ \chi_{4} & 3 & -1 & -1 & 0 & 1\\ \chi_{5} & 2 & 0 & 2 & -1 & 0 \end{array}$$

In general, computing characters can be hard. For $S_{n}$, I advise you look up Schur Functors. It turns out that the character computations have some wonderful combinatorics. Also, the $S_{n}$ representations give you knowledge about $GL_{n}(\mathbb{C})$ as well.