# Inner product from Jean-Pierre Serre representation theory book

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.

• Inversion is a bijection, and your products commute because they occur in $\mathbb{C}$. – Randall Jul 31 at 16:33
• Thank you so much! I feel so idiot for asking now. But is my argument right for the conjugate symmetry? – Vityôk Jul 31 at 16:36
• What do you mean by $\varphi^2$ ? – darij grinberg Jul 31 at 17:38
• @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. – 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}
• Sure. It's the product of complex numbers and $(\mathbb C,\times)$ is commutative. – José Carlos Santos Jul 31 at 16:40