Character of $S_3$

Question

I am trying to learn about the characters of a group but I think I am missing something.

Consider $S_3$. This has three elements which fix one thing, two elements which fix nothing and one element which fixes everything. So its character should be $\chi=(1,1,1,0,0,3)$ since the trace is just equal to the number of fixed elements (using the standard representation of a permutation matrix).

Now I think this is an irreducible representation, so $\langle\chi,\chi\rangle$ should be 1. But it's $\frac{1}{6}(1+1+1+9)=2$.

So is the permutation matrix representation actually reducible? Or am I misunderstanding something?

Answer 1

The permutation representation is reducible. It has a subrepresentation spanned by the vector $(1,1,1)$. Hence, the permutation representation is the direct sum of the trivial representation and a $(n-1=2)$-dimensional irreducible representation.

Answer 2

For any $S_n$, the representation $\pi$ on ${\bf C}^n$ by permuting basis vectors is a product of the trivial representation $\pi\vert\{(x,x,\ldots,x)\vert x\in{\bf C}\}$ and the faithful representation $\pi\vert V$ with $V=\{(x_1,\ldots,x_n)\vert \sum x_j=0\}$, the latter being irreducible.

Answer 3

No the permutation representation is not irreducible.

The permutation representation contains the trivial representation. $\chi_U = (1,1,1)$. Here I choose to represent characters using conjugacy classes. The first position is the conjugacy class corresponding to $3 = 1 + 1 + 1$ and has 1 element. The second is the conjugacy classes corresponding $3 = 2 + 1$ and contains 3 elements. The third corresponds to $3 = 3$ and contains 2 elements.

The trace of the permutation representation is the number of fixed points. So as you mentioned $\chi_P = (3, 1, 0)$. The permutation representation always contains a copy of the trivial representation spaned by $e_1 + e_2 + e_3$, where $e_1, e_2, e_3$ are the basis for the vector space of the permutation representation.

Let $U$ denote the trivial representation. Then $\chi_P = \chi_U + \chi_V$ for some representation $V$. Since $\chi_U = (1, 1, 1)$, you must have that $\chi_V = (2,0, -1)$. By taking inner product of $\chi_V$ with itself, you see that $\langle \chi_V, \chi_V \rangle = 1$. Hence $V$ is also a irreducible representation. $V$ is called the standard representation of $S_3$.

Hence, $P = U \oplus V$.