Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


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.