[[Idempotent of the complex group ring]]
# Idempotent primitivity criterion

An [[Idempotent of the complex group ring|idempotent]] $e_{\mu} \in \mathbb{C}[G]$ is primitive iff for every $q \in \mathbb{C}[G]$ there exists a scalar $\lambda_{q} \in \mathbb{C}$ such that $e_{\mu}* q * e_{\mu} = \lambda_{q} e_{\mu}$. #m/thm/rep

[^kep]: 2023, [[@keppelerGroupsRepresentations2023|Groups and representations]], p. 58

> [!check]- Proof
> Let $L^{\mu\alpha}$ be the minimal left ideäl generated by primitive idempotent $e_{\mu\alpha}$.
> Since for all $q,s \in \mathbb{C}[G]$
> $$
> \Rho_{\mathbb{C}[G]}(e_{\mu\alpha} * q * e_{\mu\alpha}) \Lambda_{\mathbb{C}[G]}(s)
> = \Lambda_{\mathbb{C}[G]}(s) \Rho_{\mathbb{C}[G]}(e_{\mu\alpha} * q * e_{\mu\alpha})
> $$
> and $L^{\mu\alpha}$ transforms in an irrep $\Gamma^{\mu}$, in particular
> $$
> \Rho_{\mathbb{C}[G]}(e_{\mu\alpha}*q*e_{\mu\alpha}) \Gamma^{\mu\alpha}(g) = \Gamma^{\mu\alpha}(g) \Rho_{\mathbb{C}[G]}(e_{\mu\alpha}*q*e_{\mu\alpha})
> $$
> for all $g \in G$ and thus by [[Schur's lemma]]  $\Rho_{\mathbb{C}[G]}(e_{\mu\alpha}*q*e_{\mu\alpha})$ is $\lambda_{q} \mathbf{I}$ in $L^{\mu\alpha}$ and zero everywhere else,
> i.e. $e_{\mu\alpha}*q*e_{\mu\alpha} = \lambda_{q}e_{\mu\alpha}$.
> 
> For the converse, assume $e_{\mu}$ is non-primitive and for every $q \in \mathbb{C}[G]$ there exists a scalar $\lambda_{q} \in \mathbb{C}$ such that $e_{\mu}* q * e_{\mu} = \lambda_{q} e_{\mu}$.
> From non-primitivity $e_{\mu} = e_{1} + e_{2}$ for nonzero idempotents with $e_{1}*e_{2}=0=e_{2}*e_{1}$.
> Then on the one hand
> $$
> \begin{align*}
> e_{\mu} * e_{1} * e_{\mu} = (e_{1}+e_{2})*e_{1} * (e_{1} + e_{2}) = e_{1}
> \end{align*}
> $$
> but on the other hand
> $$
> \begin{align*}
> e_{\mu} * e_{1} * e_{\mu} = \lambda e_{\mu}
> \end{align*}
> $$
> so $\lambda e_{\mu} = e_{1}$.
> But this is a contradiction, since it implies 
> $$
> \begin{align*}
> e_{1}*e_{1}=\lambda^2 e_{\mu} * e_{\mu} = \lambda^2e_{\mu } = e_{1} = \lambda e_{\mu}
> \end{align*}
> $$
> and thus $\lambda=\lambda^2$ and $\lambda \neq 0$.
> Hence the assumption is false.[^kep]
> <span class="QED"/>

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