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Let $F$ be a field and let $G$ be a group. Let $V$ be an irreducible $F$-linear representation of $G$.

If $F$ is algebraically closed, then $\dim_F \,(\operatorname{Hom}_G(V,V)) = 1$ by Schur's lemma.

I would like to know if there is a way to calculate $\dim_F \,(\operatorname{Hom}_G(V,V))$ when $F$ is not necessarily algebraically closed. If $V$ is absolutely irreducible, then Schur's lemma gives the answer, but I'm not sure what happens when this is not the case.

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1 Answer 1

The general form of Schur's lemma asserts that if $V$ is an irreducible representation and $\phi : V \to V$ an intertwining operator, then $\phi$ is either zero or an isomorphism; that is, invertible. So it follows that $\text{Hom}_G(V, V)$ is a division algebra over $F$. The usual form of Schur's lemma then follows from the proposition that a finite-dimensional division algebra over an algebraically closed field $F$ is necessarily itself $F$ (exercise).

The dimension of a division algebra may be arbitrary. For example, every field extension of $F$ is a division algebra over $F$. When $F = \mathbb{R}$, the Frobenius theorem asserts that the only possible division algebras are $\mathbb{R}, \mathbb{C}, \mathbb{H}$, with dimensions $1, 2, 4$, and all three examples occur among irreducible representations of finite groups. See also Frobenius-Schur indicator.

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