I am working on a problem and I was very tempted to take for granted that for any complex character $\chi$ of a finite group $G$, the norm $$\langle \chi, \chi \rangle = \frac{1}{|G|}\sum_{g\in G} \chi(g)\chi( g^{-1})$$ Is an integer. I have pondered this for awhile and I don't see a reason why it is necessarily true, but nontheless I have never seen any inner product of characters be anything but an integer.

My initial thought was that any character is the integral sum of irreducible characters, so the norm will be a sum of square integers, but that isn't sufficient

Am I missing something obvious here? If this is not true, can someone provide an explicit example for my repertoire of examples?

  • $\begingroup$ @Timkinsella by complex character I mean the class function given by the trace of a representation $\rho: G \rightarrow GL_n(\mathbb{C})$ $\endgroup$ Jul 24, 2021 at 3:27

1 Answer 1


Let $G$ be finite, and restrict our attention to $G$-representations of finite degree over $\Bbb C$.

As you observe, to a $G$-representation $V$ we have a decomposition: $$V\cong V_1^{d_1}\oplus \dots \oplus V_h^{d_h},$$ where the $V_i$ are the irreducible representations corresponding to the irreduciblecharacters $\chi_i$ of $G$, i.e. the character $\chi$ of $V$ is $d_1\chi_1+\dots + d_h\chi_h$, and the orthogonality condition on the irreducible characters implies that $$(\chi|\chi) = \sum_{i,j=1}^h (d_i\chi_i|d_j\chi_j) = \sum_{i=1}^h (d_i\chi_i|d_i\chi_i)=\sum_{i=1}^h d_i^2(\chi_i|\chi_i) = \sum_{i=1}^h d_i^2,$$ which is, as you say, a sum of integers, and hence an integer. I am not sure why you say this is insufficient.

  • $\begingroup$ I am concerned about the factor of 1/|G|, which you have not addressed here. $\endgroup$ Jul 24, 2021 at 5:49
  • $\begingroup$ @VanishingPhilosopher Do you already know that $(\chi_i|\chi_j) = \delta_{ij}$ for $\chi_i$ irreducible $\Bbb C$-characters? $\endgroup$
    – Horizon
    Jul 24, 2021 at 5:52
  • 2
    $\begingroup$ Ahh, I see. So the factor of 1/|G| is brought in to the inner product and thus goes away with the orthonormal inner products. That was what I missed, thank you $\endgroup$ Jul 24, 2021 at 5:56

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