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Let $G$ be a simple Lie group over ${\mathbb C}$, $P \subset G$ a parabolic subgroup, and $L \subset P$ its Levi subgroup. Let $\lambda$ be a $G$-dominant weight and $V_G^\lambda$ an irreducible representation of $G$ with highest weight $\lambda$. I am interested in restrictions on highest weights of irreducible components of the restriction $$ (V_G^\lambda)_{|L}. $$

In the simplest case, when $P = B$ is the Borel subgroup, $L = T$ is the maximal torus, and there is a well-known restriction --- the weights of $V_G^\lambda$ all lie in $$ Conv(\lbrace w\lambda \rbrace_{w \in W}), $$ where $W$ is the Weyl group. I would be happy to know something of the same sort for arbitrary parabolic subgroup.

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Note that not all weights in that convex hull occur -- only the ones that differ from $\lambda$ by an element of the root lattice. This gets worse for other branching problems, though maybe not yours. –  Allen Knutson Aug 17 '10 at 1:59
    
I guess in type A one can use Gelfand-Tsetlin rules to deduce the restriction. For example, if $G=GL_3, L=GL_2\times C^*$ (the maximal parabolic corresponding to the root $\alpha_2$) these rules show that highest weights of $(V_G^\lambda)_{|L}$ all lie in $Conv(\lambda,s_2\lambda,s_2s_1\lambda,\lambda+s_2s_1\lambda - s_2\lambda)$. I would like to have something similar in general - a universal collection of elements in $\mathbb{Z}[W]$, so that every highest weight of $(V_G^\lambda)_{|L}$ is contained in the convex envelope of these elements applied to $\lambda$. –Sasha 2010-08-17 03:56:43Z UTC –  Victor Protsak Aug 17 '10 at 5:30
    
General case may be considerably more involved than what happens in type A. For your modified question, Guillemin-Sjamaar would be a natural place to check. However, I've spent enough time on your problem with not even symbolic gain and I don't have any incentive to think about it further. – Victor Protsak 2010-08-17 04:36:22Z UTC –  Victor Protsak Aug 17 '10 at 5:32
    
Sasha, I do not know if you are still interested in the question, but this [www2.math.umd.edu/~tjh/HKMm.pdf] might be helpful. –  Misha Dec 21 '12 at 5:33
    
@Misha, Thanks a lot! In fact I already managed to circumvent the problem with branching that I had at that time, but I am still interested in the question. Can you give me a reference to a precise statement about the branching in the paper? I looked the paper through but haven't noticed "branching" mentioned in the text. –  Sasha Dec 23 '12 at 9:54

2 Answers 2

The standard classical question concerns multiplicities of the irreducible representations of $L$ (or its derived group) in the restriction: these are given by branching rules. This is complicated to work out in detail but is treated in many textbooks and other sources. I'm not sure exactly how much information you are asking for. The original $W$-invariant set of weights relative to $T \subset L$ is unchanged, but the original representation decomposes into a direct sum of irreducibles for $L$, with the subgroup $W_L \subset W$ acting on the separate weight diagrams within the convex region you describe. (It's easy to picture this in restricting from rank 2 to rank 1, for example.)

By the way, your $G$-dominant weight means a weight of $T$ which is dominant relative to a fixed Borel subgroup $B$ for which $T \subset B \subset P$.

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I do understand that the branching rules is a rather complicated question, but the point is that I ask about a simpler thing. This is analogous to the following - if we consider the tensor product $V_G^\lambda \otimes V_G^\mu$ then the multiplicities of irreducible $G$-modules are complicated. However, there is a simple restriction saying that all the highest weights are contained in $Conv(\lbrace\lambda + w\mu\rbrace_{w \in W})$. I ask about a similar restriction in this branching problem. –  Sasha Aug 17 '10 at 3:17

This is not a complete answer, but maybe it will be of use.

You're asking which $L$-high weights $\mu$ occur in the $G$-irrep $V_\lambda$. Let me say that $\mu$ occurs classically if for some $N>0$, $N\mu$ occurs in $V_{N\lambda}$.

  1. The set of such $\mu$ form a rational polytope lying inside $L$'s positive Weyl chamber.

  2. The vertices of this polytope strictly inside $L$'s chamber ("regular vertices") are exactly those of the form $w\cdot \lambda$ that are lucky enough to be in there.

  3. The vertices of this polytope lying on $L$'s Weyl walls are very likely to be very complicated. In particular they may not be integral weights of $L$. As I recall this already happens for $GL(3) \supset GL(2)\times GL(1)$.

Parts 1 & 3 apply to any branching problem (and much further). Part 2 is special to your case that $L$ has the same rank as $G$ (I'm not actually using that it's a Levi).

If all you want is an upper bound, as your comment to Jim suggests, then that's easy: the $L$-high weights that can occur are a subset of the $T$-weights that occur, which you already described. Probably you want something better than that though. In principle it wouldn't be too hard to figure out the local structure of your polytope nearby the regular vertices, but I expect that not all facets contain regular vertices.

Littelmann describes (in the case of a Levi) the highest weights that occur and their multiplicities: one looks at all the Littelmann paths for the irrep $V_\lambda$ that lie entirely inside the closed $L$-chamber.

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Allen, thanks a lot! You convinced me that the problem is more complicated than I thought. Although it seems slightly strange for me - I somehow believed that the branching problem is very similar to the problem of decomposing a tensor product (and in type A even equivalent, is it true?). Now, for tensor product we have a very nice restriction, so why don't we have something similar for branching? –  Sasha Aug 23 '10 at 20:37
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I think the simplest answer is that even for tensor product questions, we don't have a reasonable handle on the vertices of the resulting polytope, just the facets. If you want the list of facets for a general branching problem, check out Ressayre's article arxiv.org/abs/0903.1187 I haven't thought about how that list might simplify for a Levi. –  Allen Knutson Aug 24 '10 at 2:04

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