We work over an algebraically closed field of characteristic 0. Let $\mathfrak{g}$ be a reductive Lie algebra and let $\mathfrak{p}\supset\mathfrak{m}$ be a parabolic subalgebra, respectively a Levi subalgebra. There is the adjoint action of $\mathfrak{m}$ on $\mathfrak{g}$ and I would like to know how it decomposes into simples. More precisely, what are the highest weights appearing? Does anyone know the answer or a reference? (Does it depend on the characteristic of the field?)

1$\begingroup$ Concerning the parenthetic question as the end, there is of course some problem in prime characteristic because representations need not be completely reducible. Anyway, the basic question here involves a very special case of the general branching problem (much studied in the literature, but not something I've actually done). Usually there are no closed formulas, just a recursive process. Basically you are restricting to a reductive subalgebra, or its semisimple derived algebra. Think about type $E_8$, where the process isn't trivial to carry out. $\endgroup$ – Jim Humphreys Jan 21 '14 at 18:32

$\begingroup$ P.S. Though it's wiser to stay away from prime characteristic at first, there is a characteristicfree aspect to your question: in terms of "Weyl modules" (obtained by reduction mod $p$), the Grothendieck group element you get by restriction in characteristic $p$ is the same as in characteristic 0. But Weyl modules can have many composition factors, typically still unknown, depending on $p$. $\endgroup$ – Jim Humphreys Jan 21 '14 at 21:31

$\begingroup$ Thank you James for the comment. Do you know some references where this branching for the adjoint rep is discussed? I'm interested only in the characteristic 0 case. $\endgroup$ – Dragos Fratila Jan 22 '14 at 10:13

$\begingroup$ As you might expect, most of the explicit results are in math physics papers, mainly for exceptional Lie algebras and their maximal semisimple subalgebras. See for instance Octonions and subalgebras of the exceptional algebras by F. Buccella, A. Della Selva, and A. Sciarrino, Journal of Mathematical Physics 30, 585 (1989). But $E_8$ has $2^8$ Levi subalgebras, up to conjugacy. $\endgroup$ – Jim Humphreys Jan 22 '14 at 16:36

$\begingroup$ P.S. It's useful to look at some special cases. The advantage of the adjoint representation is that all weights equal to roots occur with multiplicity 1. For example take type $A_2$, having simple roots $\alpha, \beta$, with a Levi subalgebra having derived algebra of rank 1 and simple root $\alpha$. The 8dimensional Lie algebra splits into summands with highest weights $\alpha, \alpha + \beta, \beta, 0$ (of dimensions $3, 2, 2, 1$). The number of times the trivial module occurs (here 1) is found recursively. $\endgroup$ – Jim Humphreys Jan 24 '14 at 0:10
Not sure that there would be a closed formula as Jim has pointed out. You can easily calculate it in examples because you can determine highest weight vectors in $g$ for $m$. Indeed, if $\alpha_1, \ldots \alpha_k$ are simple rooots for $m$, you just need to find those roots $\beta$ of $g$ such that no $\beta + \alpha_i$ is a root of $g$.
This will give you all nontrivial summands. The trivial summands sit in the Cartan and are easy to find as well.
And, if you are lazy or rich or both, just buy this book

$\begingroup$ great, thanks a lot! isn't it $\beta+\alpha_i$ not a root of $g$? $\endgroup$ – Dragos Fratila Jan 22 '14 at 13:16

$\begingroup$ It could be a root of $g$, then $[e_\beta, e_{\alpha_i}]$ is a mutliple of $e_{\beta+{\alpha_i}}$. But if it is not, $[e_\beta, e_{\alpha_i}]=0$, and that os what determines the highest weight vector... $\endgroup$ – Bugs Bunny Jan 22 '14 at 13:40

1$\begingroup$ P.S. For some reason I once acquired a more elaborate book of tables by W.G. McKay and J. Patera: Tables of dimensions, indices, and branching rules for representations of simple Lie algebras. Lecture Notes in Pure and Applied Mathematics, 69. Marcel Dekker, Inc., New York, 1981. v+317 pp. (Any offers?) Computers seem to have reolaced most tables. $\endgroup$ – Jim Humphreys Jan 22 '14 at 16:32

$\begingroup$ @Bugs Bunny: I read again your answer and I must had missed the "no" in front of $\beta+\alpha_i$; that was what I wanted to say in my comment. $\endgroup$ – Dragos Fratila Jan 22 '14 at 18:49

$\begingroup$ @BB: Why do you specify that $\beta$ should be a positive root? $\endgroup$ – Jim Humphreys Jan 23 '14 at 0:28