Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\pi$?

share|improve this question

2 Answers 2

up vote 8 down vote accepted

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.