If $ T: V \to V $ is an operator such that for all $ g \in G $, $ T \Pi (g) = \Pi (g) T $ implies $ T = \lambda Id $ , then $ \Pi $ is irreducible.

Question

$Problem:$

Let $ \Pi $ be a representation ($ \mathbb {C} $ -linear) of a group of matrix Lie $ G $ on a normed vector space of finite dimension $ V $. Then:

If $ T: V \to V $ is an operator such that for all $ g \in G $, $ T \Pi (g) = \Pi (g) T $ implies $ T = \lambda Id $ , then $ \Pi $ is irreducible.

The following is definition of irreducible representation:

$ \textbf {DEF:} $ A representation $ (\Pi, V) $ of $ G $ is said $ \textbf {irreducible} $ if the only subspaces closed G-invariants are $ {\{0}\} $ and $ V $.

Can someone give me a suggestion to prove it?

Answer 1

I am not familiar with general representations of Lie groups. But assume that $G$ is any group and $\pi$ is a representation on a finite dimensional vector space $V$ with an inner product such that $\pi$ is a unitary representation (i.e. such that $\langle g v, g w \rangle = \langle v, w \rangle$). Note that whenever $G$ is finite (or compact) it is a restatement of Maschke's theorem that such an inner product exists.

If $W \subset V$ is $G$-invariant, then the orthogonal (w.r.t this inner product) projection $P$ onto $W$ satisfies $P \pi(g) = \pi(g) P$. Therefore $P = \lambda \text{Id}$, therefore $W = V$ or $W = \{0\}$.