# Irreducibility of a character if and only if inner product equals 1

I am working through the following representation theory notes: Notes

I am having some trouble understanding why the Corollary to Theorem 15 is true.

Theorem 15 Suppose $G$ is a finite group with irreducible characters $\chi_{1},\dots \chi_{k}$ over $\mathbb{C}$ If $\chi$ is any character, expressible as a sum of irreducible characters by $\chi = \Sigma m_{i} \chi_{i}$ then:

1. For each i , $m_{i}=\langle \chi_{i} | \chi_{i} \rangle$
2. $\langle \chi | \chi \rangle = \Sigma m_{i}^{2}$

Corollary A character $\chi$ is irreducible if and only if $\langle \chi | \chi \rangle=1$

I understand that if $\chi$ is irreducible then it must be one of $\chi_{1},\dots \chi_{k}$ and then $\langle \chi | \chi \rangle=1$

I am having trouble understanding why the converse must be true. For example if I had $\chi =\frac{3}{5}\chi_{1}+\frac{4}{5}\chi_{2}$ then $\langle \chi | \chi \rangle = \Sigma m_{i}^{2}=1$ but this is reducible.

I would appreciate some clarification on what is going on and a 'proof' of the corollary. Thank you.

You are not allowed to have rational numbers in the linear combination of the characters. The $m_i$'s are natural numbers. Theorem 10 in your notes says that

Every character is a sum of irreducible characters.


So you can't have half of a character.

So then, if $\chi = m_1\chi_1 + \dots + m_k\chi_k$, you see that $\chi$ is irreducible if and only if exactly one $m_i =1$ and the rest zero.

• Why the downvote? – Thomas Sep 9 '15 at 17:01
• @Ravi: Nice catch! lol – Thomas Jul 8 '17 at 13:01

Any arbitrary (non-virtual) character is a linear sum $\chi = \sum m_i \chi_i$ with the $\chi_i$ irreducible characters and the $m_i$ positive integers. In terms of the underlying representations, any (complex) representation $\rho$ is the direct sum $\rho = \rho_1^{\oplus n_1} \oplus \cdots \oplus \rho_k^{\oplus n_k}$ with each $\rho_i$ an irrep and the $n_i$ positive integers; taking traces gives $\chi_{\rho} = n_1\chi_{\rho_1} + \cdots + n_k \chi_{\rho_k}$ for the corresponding characters. Given that, any character $\chi = \sum m_i \chi_i$ with $\langle{\chi, \chi}\rangle = 1$ must have $\sum m_i^2 = 1$ and thus exactly one nonzero $m_i = 1$.

A character is specifically the trace of a representation; not every class function on the group $G$ is a character. This does become true after tensoring with $\mathbb{C}$ (that is, the set of irreducible characters form a basis for the vector space of class functions on $G$) but an arbitrary sum of characters is not necessarily itself a character. For example, the values of characters are algebraic integers.