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Let $\rho : G \to GL(V)$ be an irreducible representation of a finite group. Schur's lemma says if $\pi:GL(V) \to GL(V)$ intertwines with $\rho$, that is, $\pi \rho(g) = \rho(g) \pi$ for every $g\in G$, then $\pi = \lambda I$ for some $\lambda \in \mathbb{C}$.

Is there a similar lemma for $\rho = m_1 \rho_1 \oplus \ldots \oplus m_k \rho_k$, where $\rho_1, \ldots, \rho_k$ are different irreducible representations? The question is, given such $\rho$, if $\pi \rho = \rho \pi$, then what is the structure of $\pi$?

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3 Answers 3

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This is an easy exercise. If $\rho _i$ are "different" (i.e. inequivalent) then by Schur's lemma, $Hom _G(\rho _i,\rho _i)={\mathbb C}I$ and $Hom _G(\rho _i, \rho _j)=0$. Hence the commutant of $G$ in $End(\rho)$ is easily seen to be the product

$$M_{m_1}({\mathbb C})\times \cdots \times M_{m_k}({\mathbb C}),$$ where $M_n({\mathbb C})$ is the algebra of $n\times n$ matrices with complex entries.

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Schur's lemma has a different generalization when the coefficient field $F$ is not algebraically closed. Then you get $M_{m_1}(D_1)\times\cdots\times M_{m_k}(D_k)$ where $D_i:={\rm Hom}_{FG}(\rho_i,\rho_i)$ is a division algebra over $F$ by Schur's lemma. If $F$ is infinite, noncommutative $D_i$ can arise. For example, the quaternion group of order 8 has a 4-dimensional irreducible representation $\rho$ over the rationals, where ${\rm Hom}_{\mathbb{Q}G}(\rho,\rho)$ is the rational quaternions.

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In this lemma, the group G does not need to be finite.

The representation space V of an infinite group G may be either finite-dimensional or countable-dimensional (or, better to say, of dimension less than continuum). When the dimensionality is continuum, this lemma fails.

The proofs of these facts are surprisingly elementary, see the discussion at

Dixmier's lemma as a generalisation of Schur's first lemma

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