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?

  • $\begingroup$ There is no need to do the $2\pi_U$ trick (unless you want not linear maps but isomorphisms): you can simply take $T=\pi_W$. $\endgroup$ Commented Oct 28, 2013 at 7:02
  • $\begingroup$ You are completely right. $\endgroup$
    – pdfs
    Commented Oct 28, 2013 at 7:07
  • $\begingroup$ (In the algebra case, see mathoverflow.net/questions/2328/…) $\endgroup$ Commented Oct 28, 2013 at 7:12
  • $\begingroup$ 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. $\endgroup$ Commented Oct 28, 2013 at 7:17

1 Answer 1


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.


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