Group action on the set of irreducible characters are orthogonal. I am studying representation theory and I came across the following problem:
Let $\mathbb{C}[G]$ be the set of $\mathbb{C}$-valued functions on a finite group $G$ and $S(\subset \mathbb{C}[G])$ be the set of irreducible characters of $G$. Let $\mathbb{C}_{\chi}[G]$ be the subspace of $\mathbb{C}[G]$ generated by the set $\{g* \chi \; : \; g \in G \} $ where $g*\chi$ is the action of $G$ on $\chi$ defined as $(g*\chi)(x) = \chi (g^{-1}x) \; \forall x \in G$.
Let us define the scalar product $<|>: \mathbb{C}[G] \times \mathbb{C}[G] \rightarrow \mathbb{C} $ s.t. $<\phi|\varphi> = \frac{1}{|G|} \sum_{g \in G} \phi(g) \overline{\varphi(g)}$.
Questions:

*

*How do we show that if $\chi_1 \neq \chi_2$, then $\mathbb{C}_{\chi_1}[G]$ and $\mathbb{C}_{\chi_2}[G]$ are othogonal w.r.t. above scalar product?


*Show that the subspaces $\mathbb{C}_{\chi}[G]$ are $\mathbb{C}[G \times G]$-submodules of $\mathbb{C}[G]$ irreducible by two non-isomporphs. (the action of $G \times G$ on $\mathbb{C}[G]$ is defined as $((g,h)*f)(x) = f (g^{-1}xh) $).
My attempt:

*

*We have the following:
\begin{align*}\label{key}
\mathbb{C}_{\chi_1}[G] &= \text{span} \{ g*  \chi_1 \; : \; g \in G \} \\[2pt]
\mathbb{C}_{\chi_2}[G] &= \text{span} \{ g*  \chi_2 \; : \; g \in G \}
\end{align*}
The dot product $ \langle \phi,\varphi \rangle $can be written as
\begin{align*}
\langle \phi,\varphi \rangle &= \langle \sum_{g \in G} \lambda_g \left( g* \pi \right) , \sum_{h \in G} \lambda_{h}^{'} \left( h* \pi'  \right)  \rangle \\[2pt]
\text{After simplifying, we get} \\
& =  \sum_{g \in G} \sum_{h \in G} \lambda_g \overline{\lambda_{h}^{'}} \langle  \left( g* \pi \right) , \left( h* \pi'  \right)  \rangle 
\end{align*}
If $g=h$, then the product will be $0$ but what if $g \neq h$?
(2) I don't even understand the statement.
 A: I am going to use the notation $\mathbb{C}^G$ rather than $\mathbb{C}[G]$.
(1) It suffices to check the basis elements $g\ast\chi$ of $\mathbb{C}^G_\chi$ are orthogonal to the basis elements $h\ast\psi$ of $\mathbb{C}_\psi^G$:
$$ \langle g\ast\chi,h\ast\psi\rangle=\frac{1}{|G|}\sum_{x\in G} \chi(g^{-1}x)\overline{\psi(h^{-1}x)}. \tag{$\circ$}$$
Suppose $\chi$ and $\psi$ are characters of $V$ and $W$. Note $\chi(g^{-1}x)\overline{\psi(h^{-1}x)}$ is then the trace of the linear map $T\mapsto (g^{-1}x)T(h^{-1}x)^{-1}$ on $V\otimes W^\ast \cong\hom(W,V)$. Then $(\circ)$ is the trace of
$$ T\mapsto g^{-1}\left(\frac{1}{|G|}\sum_{x\in G} xTx^{-1}\right)h, $$
on $\hom(W,V)$ but by Schur's lemma the expression in parentheses is zero (it is the projection of $T$ onto $\hom_G(W,V)$), so this is the trace of the zero map, which is zero.
(2) Saying subspaces are irreducible "by two non-isomporphs" does not seem like gramamatical English, so perhaps there is a typo or you transcribed/translated incorrectly.
But we can show $\mathbb{C}_\chi^G$ is an irrep of $G\times G$. The key, at least in the following, is in understanding how the Artin-Wedderburn theorem says the group algebra $\mathbb{C}[G]$ acts on irreps $V$. Recall a rep is an algebra homomorphism $\mathbb{C}[G]\to\mathrm{End}(V)$, and AW says $\mathbb{C}[G]\to\bigoplus\mathrm{End}(V)$ is an isomorphism. Keep in mind, also, that $\mathrm{End}(V)\cong V\otimes V^\ast$ as a $G\times G$-rep (where $(g,h)T:=gTh^{-1}$, which is an irrep since $V$ is. (Recall that if $V$ is a $G$-irrep and $W$ is an $H$-irrep then $V\otimes W$ is a $G\times H$-irrep.)
Suppose $\chi$ is the character of an irrep $\rho_V$. Then any $f\in\mathbb{C}^G_\chi$ is expressible as
$$ f_u(x)=\mathrm{tr}\rho(ux) $$
for some $u\in\mathbb{C}[G]$ (the group algebra). There is some redundancy in representing functions this way, though. Indeed the map $\mathbb{C}[G]\to\mathbb{C}^G_\chi$ given by $u\mapsto f_u$ annihilates every summand of $\mathbb{C}[G]$'s AW decomposition except $\mathrm{End}(V)$. Note also $(g,h)\ast f_u=f_{hug^{-1}}$, so in fact we have an onto $G$-morphism from $\mathrm{End}(V)$ to $\mathbb{C}^G_\chi$. (I am sure there is a better way to explain this, but this is how I think about the problem.)
