Characters and permutation matrices Suppose I represent the group $S_{\,n}$ using $n \times n$ permutation matrices. This is a valid group representation. Let $\chi$ be its character. Since $\chi(g)$ is complex and since $1=\chi(gg^{-1})=\chi(g)\chi(g^{-1})=\chi(g)\chi(g)^*=|\chi(g)|^2$, the value of $\chi(g)$ lies on the unit circle. However, for certain permutations $g$, the permutation matrix has 0's down the diagonal and the value of $\chi(g)$, being the trace of this matrix, is 0. Am I missing something here? 

The last part is correct: $\chi(g)$ is the number of fixed points of $g \in S_n$ acting on $\{1, \ldots, n\}$, e.g. $\chi(\operatorname{id}) = \operatorname{tr} I_n =  n$. It is true that the eigenvalues of the associated matrices must be roots of unity, since these matrices have finite order in $GL_n$. It could also be that you saw a formula like $\langle \chi, \chi\rangle = 1$ involving the inner product on class functions, where $\chi$ is the character of an irreducible representation. [But note that the permutation representation is not irreducible, since the span of $(1, \ldots, 1)$ is fixed.]
