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More specifically, is it true that a representation of $\dim < p+1$ of the algebraic group $SL_2(\mathbb{F}_p)$ is always completely reducible? (of course above this dimension there are non completely reducible examples)

More general results that might help in this direction are also welcome.


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up vote 5 down vote accepted

The essential work in this direction was published from 1994 on by J.-P. Serre and J.C. Jantzen, concerning both algebraic groups and related finite groups of Lie type. Related papers by R. Guralnick and G.J. McNinch followed. There are uniform dimension bounds for complete reducibility, stricter in rank 1. For a finite group of simple type over a field of $q$ elements in characteristic $p$, Jantzen's upper bound is $p$ for rank at least 2 but $p-2$ in your case. The best I can do is list a few references:

MR1635685 (99g:20079) 20G05 (20G40) Jantzen, Jens Carsten (1-OR) Low-dimensional representations of reductive groups are semisimple. Algebraic groups and Lie groups, 255–266, Austral. Math. Soc. Lect. Ser., 9, Cambridge Univ. Press, Cambridge, 1997.

MR1753813 (2001k:20096) McNinch, George J.(1-NDM) Semisimple modules for finite groups of Lie type. J. London Math. Soc. (2) 60 (1999), no. 3, 771--792.

MR1717357 (2000m:20018) 20C20 Guralnick, Robert M. (1-SCA) Small representations are completely reducible. J. Algebra 220 (1999), no. 2, 531–541.

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Thank you very much! – Adam Gal Mar 15 '10 at 18:29

If I understand your question, then no. The finite group SL(2,5) has a projective, reducible, indecomposable representation of dimension 5 over the field of 5 elements, namely the projective cover of the principal module 5^1. It has composition series 5^1, 5^3, 5^1. This gives you an example of a non-completely-reducible module of dimension p-1 = 4 too if you'd like.

You probably only need to concern yourself with indecomposable modules, and since SL(2,p) has a cyclic Sylow p-subgroup, you can use the Brauer tree to write down the indecomposable modules.

The projective indecomposables for SL(2,p) are described on 48–49 of Alperin's Local representation theory book (MR 860771 google).

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I meant [i]algebraic[/i] representations of the [i]algebraic[/i] group, so you can think of it as the group over the algebraic closure, it's more or less the same thing. I'm not familiar with the notation you used but I do know that the finite groups have non-decomposable representations. My question was about the whole algebraic group. Is this representation you are talking about algebraic? – Adam Gal Mar 15 '10 at 17:14

Jim Humphreys gave already the answer, but I thought I'd try to clarify the question of "algebraic vs. non-algebraic" representations.

If G is any reductive algebraic group in char. p (say, over an alg. closure k of F_p), the result in the paper of Jantzen mentioned in Jim Humphreys' answer shows that any rational [=algebraic] SL_2-representation of dim <= p is semisimple.

If you consider G=SL_2 and look at the finite group of F_p-points H=SL_2(F_p), then the example given by Schmidt explains why Jantzen's result for this finite group has a poorer bound than p. As mentioned in Jim's answer, Jantzen's result for H only says that any kH-module of dimension <=p-2 is semisimple.

Incidentally, the two results just mentioned imply that a non-completely reducible kH representation of dimension p-1 is not the restriction to H of a rational [=algebraic] representation of SL_2. (Contrast this with the fact that every semisimple kH module is the restriction of a semisimple rational SL_2-representation).

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