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Answer by Michael_1812 for Generalization of Schur's Lemma

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...

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Answer by Glasby for Generalization of Schur's Lemma

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...

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Answer by Venkataramana for Generalization of Schur's Lemma

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...

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Generalization of Schur's Lemma

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...

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