Let $(\pi,V)$ be a unitary finite dimensional unitary representation of a commutative Lie group $G$. I want to show that $\dim(V)=1$ if and only if $\pi$ is irreducible.
I think the following arguement works to show the $"\implies"$ direction.
Assume that $\dim(V)=1$. Then the only subspaces of $V$ are trivially $0$ and $V$ hence $\pi$ must be irreducible.
But I am not sure how to do the other direction. If we assume that $\pi$ is irreducible, then the only invariant subspaces of $V$ under $\pi$ are $0$ and $V$. I also know that schurs lemma gives $\text{End}(V)$ consists of $\lambda\cdot I $ with $\lambda \in \mathbb{C}$ but I'm not sure how this helps.
Best Answer
Take $g\in G$. Then the map$$\begin{array}{ccc}V&\longrightarrow&V\\v&\mapsto&\pi(g)(v)\end{array}$$is an endomorphism of $G$-modules (this is where the fact that $G$ is Abelian is used). So, by Schur's lemma, there is some $\lambda_g\in\mathbb C$ such that$$(\forall v\in V):\pi(g)(v)=\lambda_gv.$$Since $G$ acts by multiplication by scalars, every vector subspace of $V$ is a submodule. So, since this representation is irreducible, $\dim V=1$.