Every irrep of $G L 𝑛 (ℂ)$ is an irrep of $U (𝑛)$ and $S U (𝑛)$

Let $Γ : G L 𝑛 (ℂ) \to G L (𝑉)$ be a finite-dimensional irrep of $G L 𝑛 (ℂ)$ . Then the restrictions $Γ ↾ U (𝑛)$ and $Γ ↾ S U (𝑛)$ are also irreps of $U (𝑛)$ and $S U (𝑛)$ respectively. #m/thm/rep/lie

[!chec\Spanroof (sketch) A representation of a Lie group is reducible iff its infinitesimal representation is reducible. We can give a basis to the Lie algebra of each group as follows
$𝔰 𝔲 (𝑛) = s p a n {𝐽 𝑗} 2 𝑛 - 1 𝑗 = 1 𝔲 (𝑛) = s p a n {𝐽 𝑗} 2 𝑛 - 1 𝑗 = 0, 𝐽 0 = 𝟙 𝔤 𝔩 𝑛 (ℂ) = s p a n {𝐽 𝑗, 𝑖 𝐽 𝑗} 2 𝑛 - 1 𝑗 = 1$
hence clearly a block diagonalization of a representation of $𝔤 𝔩 𝑛 (ℂ)$ yields a block diagonalization of the restrictions to $𝔰 𝔲 (𝑛)$ and $𝔲 (𝑛)$ .

#state/tidy | #lang/en | #SemBr

Every irrep of GL𝑛(ℂ) is an irrep of U(𝑛) and SU(𝑛)

Every irrep of $G L 𝑛 (ℂ)$ is an irrep of $U (𝑛)$ and $S U (𝑛)$