Let $G$ be a finite group, $\mathbb{C}[G]$ the group algebra of $G$ over $\mathbb{C}$; $(W, \rho)$ an irreducible representation of $G$, ($\rho: G \longrightarrow GL(W)$) and $e_{\chi}$ the central idempotent attached to $\chi$:
\begin{equation} \mathbb{C}[G]=\bigoplus_{g \in G}\mathbb{C}\epsilon_g, \: \: \: \: \: e_{\chi}=\frac{\chi(1)}{|G|}\sum_{g \in G}\chi(g^{-1})\epsilon_g \end{equation}
Let $w \in W$, $\mu \in W^*$ (the dual of $W$) and $s$ the following sum: \begin{equation} s=\sum_{g \in G} \mu(g^{-1}w)\epsilon_g \end{equation}
Do you have an idea of what could be the more direct and simple proof as possible of the fact that $s \in \mathbb{C}[G]e_{\chi}$ ?
I thank you for any idea or suggestions.
Fix $\mu\in W^*.$ For an irreducible $\mathbb CG$ module $V$ and any given $v\in V,$ consider the map $\theta$ from $W$ to $V$ given by $\theta(w)=sv,$ where $s$ is given by your formula. Then (I am just writing $g$ instead of $\epsilon_g$ for the element of $\mathbb CG$), $$\theta(hw)=\sum_{g\in G}\mu((h^{-1}g)^{-1}w)gv=\sum_{g\in G}\mu(g^{-1}w)hgv=h\theta(w),$$ so $\theta$ is a homomorphism of $G$ modules. Hence $\theta=0$ for $V$ not isomorphic to $W,$ or in other words (since $v$ was arbitrary), $sV=0$ for all such $V.$ But $(1-e_\chi)\mathbb CG$ contains no copy of $W.$ (This is because the trace of $e_\chi$ on a $\mathbb CG$-module $X$ with character $\psi_X$ is $\chi(1)[\chi,\psi_X],$ but on the other hand $e_\chi(1-e_\chi)=0,$ so $X=(1-e_\chi)\mathbb CG$ has $[\chi,\psi_X]=0.$) Hence $s(1-e_\chi)=0,$ so $s=e_\chi s\in e_\chi\mathbb CG$ as required.