Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
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. –  fpqc Feb 2 '12 at 2:21
add comment

2 Answers

up vote 3 down vote accepted

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.

share|improve this answer
    
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
add comment

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.]

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.