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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?

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Although the representation is a homomorphism, $\chi$ is not in general a homomorphism, so the second of your equalities is not valid in general. The first equality is also not valid in general, because the trace of the identity matrix $I_d$ is $d$, not $1$. –  user16299 Feb 2 '12 at 2:20
I think you can only do that when the representation is one dimensional. –  user38268 Feb 2 '12 at 2:21
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Call the representation $\rho$, then $\chi$ is the trace of $\rho$. The trace is a homomorphism of the additive group, not the multiplicative; your $\chi(gg^{-1})=\chi(g)\chi(g^{-1})$ is false. Only characters of degree 1 are multiplicative.

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So you are saying that it should be $\chi(gg^{-1})=\chi(g)+\chi(g^{-1})$? And what do you mean by "characters of degree 1"? Aren't all characters one-dimensional representations? –  echoone Feb 2 '12 at 15:59
@echoone By "degree" I think he's referring to the dimension of the representation from which $\chi$ arises. You'll often see degree $1$ characters referred to as "one-dimensional characters". This is all admittedly confusing. –  Dylan Moreland Feb 2 '12 at 16:30
Dylan has interpreted me correctly. If the degree of $\rho$ is greater than 1 then there is, in general, no simple expression for $\chi(gh)$ in terms of $\chi(g)$ and $\chi(h)$. On the level of matrices, if $A$ and $B$ are $n\times n$ matrices with $n\gt1$, there is in general no simple relation between the trace of $AB$ and the traces of $A$ and $B$. –  Gerry Myerson Feb 2 '12 at 23:48
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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.]

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Are saying that because the permutation representation is reducible, $\chi$ is not a bonafide "character"? –  echoone Feb 2 '12 at 16:09
@echoone Ah, no. It's definitely a character. But $\langle\chi, \chi\rangle \Leftrightarrow \chi$ is irreducible. –  Dylan Moreland Feb 2 '12 at 16:28
What Dylan meant was, $\langle\chi,\chi\rangle=1$ if and only if $\chi$ is irreducible. –  Gerry Myerson Feb 2 '12 at 23:49
@Gerry Ah, thanks, I didn't catch that. I wish I could edit comments. –  Dylan Moreland Feb 2 '12 at 23:52
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