Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a simply connected semi-simple algebraic group over an algebraically closed field of positive characteristic. The Steinberg tensor product theorem gives a tensor product decomposition of an irreducible rational $G$-module $S(\lambda)$ with heighest weight $\lambda$ according to the $p$-adic expansion of $\lambda$.

I am trying to understand the proof of this theorem as given by Cline, Parshall, Scott in Journal of Algebra 63, 264-267 (1980). I have two major problems with this proof:

  1. In theorem 1 where it is proven that any irreducible $\mathcal{U}$-module over the restricted universal enveloping algebra extends uniquely to a rational $G$-module, how do I get this representation $\rho:G \rightarrow PGL(V)$? I was able to see that I have a morphism $G \rightarrow \mathrm{Aut}_{_kAlg}(\mathrm{End}_k(V))$. But what next? It seems to be well-known that the group of algebra-automorphisms of the endomorphism ring is just $PGL(V)$, but why? Moreover, why do I get a morphism of algebraic groups?

  2. In the proof of theorem 2, why acts the Lie algebra $L(G)$ trivially on $\mathrm{Hom}_{L(G)}(S _1, S)$? (Here $S$ is some irreducible $G$-module, $S _1$ is an irreducible $L(G)$-submodule of $S$ and $S _1$ also denotes the unique extension to a rational $G$-module). This is not explicitly mentioned, but I think it is used in b). This must have something to do with theorem 1!?

Any hints and ideas are welcome.

share|improve this question
add comment

3 Answers

up vote 5 down vote accepted

The 1980 CPS paper is short but not easy to read without enough background. They gave the first conceptual alternative to Steinberg's somewhat opaque and computational proof of the tensor product theorem in 1963 (which built on the 1950s work of Curtis on "restricted" Lie algebra representations coming from the algebraic group plus the older work of Steinberg's teacher Richard Brauer on rank 1). Steinberg relied quite a bit on working with covering groups and projective representations.

CPS already realized the importance of getting beyond the Lie algebra by using Frobenius kernels. The best modern source is the large but well-organized 1987 book by J.C. Jantzen, Representations of Algebraic Groups (expanded AMS edition in 2003). Here the foundations are worked out thoroughly and the CPS proof is given an efficient treatment in part II, 3.16-3.17. While CPS had in mind the analogy with Clifford theory for finite groups, Jantzen gives a self-contained treatment avoiding use of projective representations or Skolem-Noether.

Apart from sources, the essential goal is to single out the finitely many "restricted" simple modules for the Lie algebra among the infinitely many simple (rational) modules for the ambient algebraic group, then realize the latter modules as twisted tensor products of the former. This requires a notion of Frobenius morphism for each power of the prime, Frobenius kernels being infinitesimal group schemes. The Lie algebra just plays the role of first Frobenius kernel (a normal subgroup scheme), so the Clifford theory analogue developed by Ballard and CPS makes sense here.

share|improve this answer
    
Thanks for the background information. Except for one little problem I could finally understand all arguments. –  user717 Mar 13 '10 at 9:55
add comment

For #1, there's a standard argument that the automorphisms of End(V) are PGL(V). (I'll take for granted that the conjugation map is injective, and we just have to show surjectivity). If you have an automorphism of End(V), you get a new representation $V^\sharp$ on the same underlying vector space of End(V) given by applying the automorphism, and then acting on V. Since End(V) only has one irreducible rep, this must be isomorphic to V as a representation. This isomorphism is only unique up to scalar multiplication, so it's well defined as an element of PGL(V). The original automorphism must have come from this guy.

To see why this is an isomorphism of algebraic groups, just note that this works over any ring. The only part that's harder is the uniqueness of V, but you just have to say "$\mathrm{End}(V)\otimes R$ has a unique irreducible representation which is a free R-module of the right rank."

share|improve this answer
    
Thanks! I think I just found another way (but your's might be better, I don't know): According to Skolem-Noether Aut(End_k(V)) = Inn(End_k(V)) and this is is the image of the adjoint representation of GL(V) which has kernel Z(GL(V)). With the inverse of the induced isomorphism (of algebraic groups, I hope) I also get into PGL(V). –  user717 Jan 6 '10 at 18:14
add comment

[This was intended to be comment to Ben's reply but I exceeded the allowable limit for comments.] Actually it doesn't work over any ring. Just take any ring $R$ for which $GL(V)(R) \to PGL(V)(R)$ is not surjective. This exists if $R$ has a non-trivial rank $1$ projective $L$ such that $L^n$ is isomorphic to $R^n$, $n:=\dim V$. Then $V\bigotimes L$ is an $End(V)\bigotimes R$ module which is not isomorphic to $V$ though they are both indecomposable projective $End(V)\bigotimes R$-modules (rarely irreducible though). As $R$-module $V\bigotimes L \cong L^n\cong R^n\cong V\bigotimes R$ so that we get a different action of $End(V)\bigotimes R$, i.e., an automorphism $End(V)\bigotimes R \to End(V)\bigotimes R$ which is not given by conjugation by some element of $GL(V\bigotimes R)$. For $R$ a local ring such an $L$ can not exists as all projective modules are free and there one does indeed get that every automorphism of $End(V)\bigotimes R$ is given by conjugation by an element of $GL(V)(R)$. This is enough to show that $PGL(V)$ is the automorphism group scheme of $End(V)$.

share|improve this answer
add comment

Your Answer

 
discard

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.