I was reading "Linear Representations of Finite Groups" from Jean-Pierre Serre and in chapter 2.2 he says has the following:

If ϕ and ψ are functions on G, <ϕ,ψ> = $\frac{1}{g}$$\sum_{t∈G} $$\mathrm{ϕ(t}^{-1})$ψ(t)$ =$$\frac{1}{g}$ $\sum_{t∈G} ϕ(t)$$\mathrm{ψ(t}^{-1})$.

We have <ϕ,ψ> = <ψ,ϕ>.

I don't know why <ϕ,ψ> = <ψ,ϕ> is true.

I know that ${\displaystyle \langle ϕ,ψ\rangle ={\overline {\langle ψ,ϕ\rangle }}}$ But i don't know why ${\displaystyle \langle ψ,ϕ\rangle ={\overline {\langle ψ,ϕ\rangle }}}$.

I found out here https://en.wikipedia.org/wiki/Complex_conjugate that ${\displaystyle \varphi ^{2}=\operatorname {id} _{V}\,}$ therefore i have $\mathrm{ϕ^{-1}}$ = ${\overlineϕ}$

And then the result follows from that, but i don't know if this is right.

I'm sorry if my question is off-topic, but i don't know how to address it.

  • 1
    $\begingroup$ Inversion is a bijection, and your products commute because they occur in $\mathbb{C}$. $\endgroup$ – Randall Jul 31 at 16:33
  • $\begingroup$ Thank you so much! I feel so idiot for asking now. But is my argument right for the conjugate symmetry? $\endgroup$ – Vityôk Jul 31 at 16:36
  • $\begingroup$ What do you mean by $\varphi^2$ ? $\endgroup$ – darij grinberg Jul 31 at 17:38
  • $\begingroup$ @darijgrinberg i linked the wikipedia article that is saying what i meant. I don't know if i can use this argument, and this is also my question. $\endgroup$ – Vityôk Aug 1 at 0:00

The set $\{t\mid t\in G\}$ is equal to the set $\{t^{-1}\mid t\in G\}$ and therefore\begin{align}\langle\varphi,\psi\rangle&=\frac1{\#G}\sum_{t\in G}\varphi(t^{-1})\psi(t)\\&=\frac1{\#G}\sum_{t\in G}\varphi(t)\psi(t^{-1})\\&=\frac1{\#G}\sum_{t\in G}\psi(t^{-1})\varphi(t)\\&=\langle\psi,\varphi\rangle.\end{align}

  • $\begingroup$ You are using the fact that the product commute, right? $\endgroup$ – Vityôk Jul 31 at 16:38
  • 1
    $\begingroup$ Sure. It's the product of complex numbers and $(\mathbb C,\times)$ is commutative. $\endgroup$ – José Carlos Santos Jul 31 at 16:40

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