# the converse of Schur lemma

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?

• There is no need to do the $2\pi_U$ trick (unless you want not linear maps but isomorphisms): you can simply take $T=\pi_W$. Commented Oct 28, 2013 at 7:02
• You are completely right.
– pdfs
Commented Oct 28, 2013 at 7:07
• (In the algebra case, see mathoverflow.net/questions/2328/…) Commented Oct 28, 2013 at 7:12
• In the case of a reductive algebraic group in positive characteristic (which is analogous to the compact Lie group case), this fails for a large class of modules. More precisely, the modules induced from $1$-dimensional representations of the Borel have $1$-dimensional space of endomorphisms (due to Frobenius reciprocity), but they are generally not irreducible. Commented Oct 28, 2013 at 7:17

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.