# Absolute irreducibility of p-restricted highest weight modules

Let $K$ be a finite field of characteristic $p$, $G/K$ be a connected, reductive, split algebraic group. Fix some maximal split torus $T$ and a system of positive roots $\Phi^+ \subset \Phi (G,T)$.

Consider a $p$-restricted weight $\lambda \in X_*(T)$: this is a (dominant) weight such that $0 \le \langle \lambda, \alpha \rangle < p$ for all simple roots $\alpha$.

Just like for any other dominant weight $\lambda$, we have an associated highest weight module $L(\lambda)$: this is a rational, finite-dimensional $K$-representation of $G$ characterized by having highest weight $\lambda$.

Question: if $\lambda$ is $p$-restricted, is it true that $L(\lambda)$ is absolutely irreducible? If so, I'd be grateful to anyone who provides a reference.

## 2 Answers

Yes, it's absolutely irreducible. The standard reference is R. Steinberg's 1963 paper here (which is freely available online). His 1967-68 Yale lectures (see $\S13$), now published by AMS in a LaTeX version, may be a good alternative source.

It should be emphasized, however, that our knowledge of these irreducibles is indirect and incomplete. This is somewhat parallel to the infinite dimensional theory in characteristic 0, but far more complicated to resolve.

ADDED: For a treatment in somewhat more modern language, see also my LMS Lecture Note volume 326 (2006), section 5.2. As in Steinberg's early work, the finite groups of Lie type are the main theme; but I've tried to make the splitting field notion more explicit. (Revisions are posted on my webpage here.)

One other comment is that for the algebraic groups (or the finite groups over arbitrary finite fields), Steinberg's tensor product theorem shows that all irreducibles are then absolutely irreducible when constructed over appropriate finite fields.

Jim Humphreys already gave several good references, but another is J. C. Jantzen's text "Representations of Algebraic Groups", Corollary II.2.9. In Jantzen's setting, $L(\lambda)$ is constructed as the socle of the "standard representation" $H^0(\lambda)$, and it follows more-or-less from the construction that $\operatorname{End}_G(H^0(\lambda)) = \operatorname{End}_G(L) = k.$

As Jim points out, the absolute irreducibility of $L$ is true whether or not $\lambda$ is restricted.

While it isn't what you asked about, let me point out that if $G$ is not split reductive over $k$, then in general its irreducible representations (as algebraic group over $k$) are not absolutely irreducible. The best reference I know concerning irreducibles when $G$ is not split is: [J. Tits, "Représentations linéaires irréductibles d'un groupe réductif sur un corps quelconque", J. Reine Angew. Math. 247 1971 196–220.]