I am interested in the converse of the following form of Schur's lemma:
Lemma. (Schur) A group $G$, a $\mathbb{C}$-vector space $V$ and a homomorphism $D : G \rightarrow \operatorname{GL}(V)$ are given. Suppose that $D$ is a irreducible reprsentation. If a linear map $T : V \rightarrow V$ commutes with $D(g)$ for all $g \in G$, then $T$ is some scalar multiple of the identity operator.
Edit) What I want to prove or disprove is the converse of the above:
Suppose that the representation $D : G \rightarrow \operatorname{GL}(V)$ is reducible. Then there is a linear map $T : V \rightarrow V$ which is not a scalar multiple of the identity such that $T D(g) = D(g) T$, $\forall g\in G$.
When $G$ is a finite group or a compact Lie group, this trivally holds, because every reducible representation of $G$ is decomposable (i.e. fully reducible); if $W \leq V$ is a subspace invariant under $D : G \rightarrow \operatorname{GL}(V)$, then we can find an invariant subspace $U \leq V$ satisfying $V = W \oplus U$, and $T=\pi_W+2\pi_U$ commutes with every $D(g)$ but is not a scalar multiple of the identity. ($\pi$: projection operators)
QUESTION. But what if $G$ is a general group, which admits reducible but indecomposable representations? Does the converse of Schur lemma still hold?
Best Answer
Let $A=\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right)$ and $B=\left( \begin{array}{cc} 2 & 0 \\ 0 & 1 \\ \end{array} \right)$, let $F$ be the free group on generators $a$ and $b$, and consider the representation of $F$ on $\mathbb C^2$ which maps $a$ and $b$ to $A$ and $B$. This has trivial endomorphism ring, so in particular it is indecomposable, yet it is not simple: the vector $(1,0)$ spans a proper submodule.