Character of $S_3$ 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?

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, $P = U \oplus V$. 
