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

My question(s) relate(s) to pp51-52 of Local Representation Theory by JL Alperin -- the relevant pages are contained in the Google Books preview http://books.google.com/books?id=p7ylsZUmK3MC&printsec=frontcover&source=gbs_atb#v=onepage&q&f=false. In these pages he is dealing with representations of $SL(2,p)$ over a field $k$ of characteristic $p>0$. He has previously constructed the $p$ simple modules $V_1,\ldots,V_p$, and wants now to describe composition series for the corresponding projective indecomposable module $P_i$ when for $i$ less than $p$ (he deals with $P_p\cong V_p$ later, but assumes we know this isomorphism).

The book has a certain fluency with the notion of projective module which I haven't picked up yet. On p51, after having proved the Lemma that $V_2\otimes V_n\cong V_{n-1}\oplus V_{n+1}$ (if $n$ is less than $p$), he proceeds to show that $V_2\otimes V_p\cong P_{p-1}$.

His elegant proof just claims that $V_2\otimes V_p$ has a summand isomorphic with a submodule of the socle of $P_{p-1}$. I see how this proves what we want but I don't see how to show this simply.

On the next page, I can't even prove what he's claiming:

We examine $P_{p-2}$ by studying $V_2\otimes P_{p-1}$. Since $V_p$ is projective and since $V_{p-2}$ is a homomorphic image, we have that $V_2\otimes P_{p-1}$ has a direct summand isomorphic with $P_{p-2}\oplus V_p\oplus V_p$. Why is this?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

He claims that by the proof of Lemma 5 (which is not in google preview) it follows that $V_2\otimes V_p$ has a submodule isomorphic to $V_{p-1}$, but this is the socle of $P_{p-1}$ since $P_{p-1}$ is the projective cover of $V_{p-1}$ and a group algebra is always symmetric.

For the next question he first claims that we already know that $V_2\otimes P_{p-1}$ is projective. But then the universal property of projective shows, that $P_{p-2}$ is a direct summand (using the homomorphic image and the the projection onto the top as the maps). Since $V_p$ is projective, a direct application of the universal property leads the claim.

share|improve this answer
    
That makes sense. But why must $V_p$ appear as a direct summand twice? –  Clinton Boys Jul 5 '11 at 10:19
    
Because $V_2\otimes V_{p-1}$ appears twice in a series of submodules. And each of these has a direct summand $V_p$ according to $V_2\otimes V_{n}\cong V_{n-1}\oplus V_{n+1}$ for $n<p$. –  Julian Kuelshammer Jul 5 '11 at 11:16
    
And this follows because we know the explicit structure of $P_{p-1}$. It is uniserial and has a composition series, where $V_{p-1}$ is appearing twice and $V_2$ is appearing once. Now tensor with $V_2$. –  Julian Kuelshammer Jul 5 '11 at 11:20
    
Right, and the uniseriality and composition series are the same (i.e. the factors of the unique composition series for $V_2\otimes P_{p-1}$ are just the factors for $P{p−1}$, tensored with $V_2$) because $V_2$ is simple? I suspected this was true but he never proved it anywhere. –  Clinton Boys Jul 5 '11 at 11:57
    
No, you won't get a composition series of $V_2\otimes P_{p-1}$, and it won't be uniserial (the tensored are not simple anymore). But you will get a filtration because $\otimes$ is exact. –  Julian Kuelshammer Jul 5 '11 at 14:10

This is an extended comment, to put Julian's answer and comments in perspective. What Alperin does in his book is a direct but somewhat ad hoc treatment of one suggestive small case, which goes back to the pioneering work by Brauer and his student Nesbitt. The wider context provides the "correct" framework for understanding what is going on here, in terms of tensoring simple modules with the Steinberg module (here $P_{p-1}$). The starting point is the general fact (Curtis, Steinberg) that the simple modules are obtained directly from the "restricted" simple modules for the corresponding simple algebraic group, which in this case is SL$_2$. But the projective modules are more elusive (and in the general case remain so for small primes). Since the Steinberg module is simple as well as projective, one relies on the very general principle for finite dimensional Hopf algebras that tensoring any module with a projective one yields a projective module. Then the problem is to organize the direct summands of the latter and extract the one of interest.

For a group of rank 1 over the prime field, most of the tensor products you construct are already indecomposable. The one exception requires extra arguments, which Julian fills in. From the algebraic group point of view this is all fairly transparent, since you can start there with the classical Clebsch-Gordan formula adapted to prime characteristic and keep track of weights. (I worked out some of this program to get projective modules for $SL(2,q)$ in a 1973 J. Algebra paper, with $q$ a power of $p$, but after rank one it all gets much more complicated to make explicit. For a systematic exposition, look at my 2005 London Math. Society Lecture Note Series 326.)

The theoretical problem of most interest here, still unsolved for small primes, is to lift projectives for the restricted enveloping algebra of the Lie algebra to the ambient algebraic group and then analyze how the indecomposable ones restrict to the finite subgroups of Lie type: most such restrictions remain indecomposable, but some don't because the enveloping algebra has slightly larger dimension than the finite group algebra. All of this recurs in the study of modules for a related quantum group at a root of unity. So the rank one case treated by Alperin is just the tip of the iceberg.

share|improve this answer

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.