# Are indecomposable representations of a finite group of Lie type absolutely indecomposable?

Let $G = G(K)$ be a Chevalley group over an algebraically closed field $K$ of characteristic $p > 0$. Consider the finite group $G(q) = G(\mathbb{F}_q)$. (For example, if $G = \operatorname{SL}_n(K)$ then $G(q) = \operatorname{SL}_n(\mathbb{F}_q)$).

It was proven by Steinberg that every irreducible $\mathbb{F}_q[G(q)]$-module is absolutely irreducible, and correspond to $q$-restricted dominant weights (i.e. weights $\lambda$ such that $0 \leq \langle \lambda, \alpha \rangle < q$ for all simple roots $\alpha$).

Consider an indecomposable $\mathbb{F}_q[G(q)]$-module $V$. Is $V \otimes_{\mathbb{F}_q} K$ an indecomposable $K[G(q)]$-module?

EDIT: In case this is not true, I also would be interested if what I am asking for is true under some mild restrictions on $q$ and/or $G$.

• Sorry to have misunderstood at first what you were asking. As Jeremy Rickard points out, there is a problem when the finite group fails to have finite representation type. This is clear for groups of Lie type in the defining characteristic: see section 8.9 in my 2006 survey LMS Lecture Notes 326. Only SL(2,8) and PSL(2,p) for p>3 have cyclic Sylow p-subgroups. In other cases one typically doesn't know much about most of the indecomposable modules. – Jim Humphreys Mar 24 '18 at 20:20
• P.S. Note that, as usual when considering Sylow $p$-subgroups, it's convenient to consider only finite simple groups here. There are also some close relatives such as $\mathrm{SL}(2, p)$ for $p>3$ having finite representation type. – Jim Humphreys Mar 26 '18 at 17:55

This is rarely the case for all $V$. $KG(q)$ would need to have finite representation type, so $G(q)$ would need to have a cyclic Sylow $p$-subgroup. Otherwise a counterexample can be found by taking an indecomposable module over some field extension of $\mathbb{F}_q$, that is not defined over $\mathbb{F}_q$ and regarding it as a module for $\mathbb{F}_qG$.

Suppose $KG$ has infinite representation type, where $K$ is the algebraic closure of $\mathbb{F}_q$.

Then for some $d$ there are infinitely many isomorphism classes of $d$-dimensional indecomposable $KG$-modules. Since there are only finitely many $d$-dimensional $\mathbb{F}_qG$-modules up to isomorphism, there must be a $d$-dimensional $KG$-module $M$ that is not defined over $\mathbb{F}_q$: i.e., such that there is no $\mathbb{F}_qG$-module $N$ such that $M\cong N\otimes_{\mathbb{F}_q}K$.

The structure constants of $M$ only involve finitely many elements of $K$, and so $M$ is defined over some finite extension $k$ of $\mathbb{F}_q$: i.e., $M\cong L\otimes_kK$ for some $kG$-module $L$, where $L$ is indecomposable and can't be defined over $\mathbb{F}_q$.

Let $L'$ be $L$ regarded as an $\mathbb{F}_qG$-module (so $L'$ is $[k:\mathbb{F}_q]d$-dimensional).

Then $L'\otimes_{\mathbb{F}_q}k$ is a direct sum of $[k:\mathbb{F}_q]$ copies of $L$, so there can be no indecomposable direct summand $N$ of $L'$ with $N\otimes_{\mathbb{F}_q}k$ indecomposable.

• Thank you for the answer. One question, how does one show from infinite representation type that for some $d$ there are infinitely many indecomposables of dimension $d$? – spin Mar 24 '18 at 21:32
• @spin For general finite dimensional algebras that’s the “Second Brauer-Thrall Conjecture” (which was proved some time ago for algebras over algebraically closed fields, although it’s easier for group algebras). – Jeremy Rickard Mar 24 '18 at 21:44