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

I guess the question is somehow elementary to experts, but I'd like to put down my arguments, which appear doubtful, and see if they are correct and if corrections and improvements are possible.

The setting is as follows: $k$ is the base field of characteristic zero, $G$ a connected semi-simple $k$-group, and $Rep(G)$ the Tannakian category of finite-dimensional algebraic $k$-representations of $G$, with the canonical fiber functor in $k$-vector spaces, whose objects are called $k$-representations for short. Unless otherwise stated, reductive $k$-groups are connected.

The motivation is as follows: for $H$ semi-simple $k$-subgroup of $G$, one has the restriction functor: $Rep(G)\rightarrow Rep(H)$ sending a $k$-representation $V$ of $G$ to the restriction $V$ as a $k$-representation of $H$. What kind of irreducible $k$-representation of $G$ remains irreducible viewed in $Rep(H)$?

Recall that a reductive $k$-group is $k$-isotropic if it contains a $k$-split $k$-torus, and $k$-anisotropic if otherwise.

fact: for $H\subset G$ a semi-simple $k$-subgroup, $H$ extends to a parabolic $k$-subgroup $H\subset P\subsetneq G$ if and only if $Z(H,G)$ the centralizer of $H$ in $G$ is $k$-isotropic. (from this one also sees that if $L$ is the Levi $k$-subgroup of a $k$-parabolic $P$, then its connected center $C(L)$ is $k$-isotropic.)

claim: let $H$ be a semi-simple $k$-subgroup of $G$ as above, such that $Z(H,G)$ is $k$-anisotropic, then for any irreducible $k$-representation $(\rho,V)$ of $G$, its restriction to $H$ is irreducible as an algebraic $k$-representation of $H$. Conversely, if the restriction functor $Rep(G)\rightarrow Rep(H)$ respects irreducibility, with $H\subset G$ a semi-simple $k$-subgroup, then $Z(H,G)$ is $k$-anisotropic.

Sketch of the proof: To prove the first part, assume that for some irreducible $(\rho,V)$, the restriction to $H$ is not irreducible. Then in $Rep(H)$ one has a non-trivial splitting $V=V_1\oplus V_2$. Define $F_0(V)=V$, $F_1=V_1$, $F_2=0$ etc, one gets a non-trivial decreasing filtration on $V$. $V$ generates a full Tannakian subcategory, which is of the form $Rep(G')$, equipped with the non-trivial filtration generated by $F(V)$. By Tannaka duality, $Rep(G')\rightarrow Rep(G)$ corresponds to an epimorphism $G\rightarrow G'$. $G'$ is thus semi-simple. The non-trivial filtration on $Rep(G')$ corresponds to a cocharacter defined over $k$, which is equivalently characterized by $k$-parabolic $P'$ of $G'$, and $P'$ lifts to a $k$-parabolic $P$ of $G$. One checks easily that $P$ contains $H$, because $H$ preserves the filtration generated by $F(V)$. This shows that $Z(H,G)$ is $k$-isotropic.

Conversely, when $Z(H,G)$ is $k$-isotropic, $H$ extends to a non-trivial $k$-parabolic $H\subset P\subsetneq G$. This gives a filtration on $Rep(G)$, preserved by $P$ and $H$. In particular, there exists at least one irreducible $k$-representation $(\rho,V)$ of $G$ on which $F(V)$ is non-trivial, and then the restriction of $\rho$ to $H$ splits non-trivially.

Here I use the notion of filtration on $Rep(G)$, which means for each $V\in Rep(G)$ one has a finite separated exhaustive decreasing filtration $F(V)$, moving functorially: it respects the tensor products and direct sums in the filtered sense, and is strict with respect to all exact sequences in $Rep(G)$. To see a filtration on $Rep(G')$ extends to a filtration on $Rep(G)$ for an epimorphism $G\rightarrow G'$ as above, it suffices to transfer to the Lie algebra side: $LieG=LieG'\oplus Lie G''$ for some semi-simple $k$-subgroup $G''$ of $G$, then use the fact that $Rep(LieG)$ equals the "exterior tensor product" of $Rep(LieG')$ with $Rep(LieG'')$, and pass equivalently to the $k$-group side, as $k$ is of characteristic zero. In this way the filtration on $Rep(G')$, together with the trivial filtration on $Rep(G'')$, gives a filtration on $Rep(G)$ by tensorial construction.

I would like to know if the above arguments makes sense. If it is, is there any other elementary proof, essentially different (modulo the Tannakian duality). Moreover, what if one allows reductive $k$-subgroup? Does that imply the claim that over $\mathbb{R}$, if one takes a pair of compact groups, say $SO_3\subset SO_4$, every irreducible representation of $SO_4$ remains irreducible when restricted to $SO_3$? and does it have anything to do with the branching rule? I would be grateful if further references, like expository articles, are mentioned concerning branching rules for reductive $k$-groups, even in the case of non-algebraically base field (I guess one might do something from the algebraically closed case through Galois descent, but I'm quite lost when doing this for reductive $k$-groups.)

Thanks a lot.

share|improve this question
2  
I don't see how every irreducible representation of $G$ could restrict to an irreducible representation of $H$. Could you give an example? –  Bruce Westbury Sep 15 '10 at 8:32
    
I don't understand your comment (in the last paragraph) about $SO_3 \subseteq SO_4$; certainly the standard representation of $SO_4$ on $\mathbb{R}^4$ is not irreducible when restricted to $SO_3$. The next question about the branching rule doesn't seem to make much sense either: of course, if every irreducible for $G$ remains irreducible when restricted to $H$ then the branching rule is not very interesting. Can you clarify? –  Sheikraisinrollbank Sep 15 '10 at 10:13
1  
The claim is false. To amplify on the comments, consider $G=SL_n, H=SO_n$ for $n\geq 3$ (for any form of the orthogonal group), then $Z(H,G)$ is 0-dim and hence anisotropic over any field. A simple $G$-module with highest weight $\lambda$ remains simple upon restriction to $H$ is and only if $\lambda$ is a fundamental weight (so that the module is an exterior power of the defining module of $SL_n$). Certainly the restriction of $S^2$ to $H$ contains an invariant vector. A similar example is $G=SL_{2n}, H=Sp_{2n}:$ only symmetric powers of the defining module or its dual remain simple. –  Victor Protsak Sep 15 '10 at 14:41
    
I suggest that you re-post your comment from 13:29 as a new question. It is very clear and does not contain several paragraphs of heavy notation interspersed with wrong claims to wade through (and MO is not the best place to check proofs). –  Victor Protsak Sep 15 '10 at 14:46
add comment

1 Answer

up vote 1 down vote accepted

One method for computing branching rules in favorable situations is to use the Littelmann path model---this has a wiki page

http://en.wikipedia.org/wiki/Littelmann_path_model

In this situation (of semisimple $G$ and with $H$ the Levi subgroup of a parabolic) irreducibles essentially never remain irreducible.

Edit in response to the comment below: Another situation that sometimes occurs (e.g. $SO_n \subseteq SL_n$) is that $H$ is the fixed point set of an involution $\phi$ of $G$. In this case, given any irreducible $G$ module $V$ we can construct another $G$-module $V^\phi$ by twisting by $\phi$. If $V$ and $V^\phi$ are isomorphic as $G$-modules, then we obtain an involution of $V$ as an $H$-module whose $\pm 1$ eigenspaces are $H$-submodules. This allows one to prove that certain $G$-modules are not irreducible as $H$-modules.

share|improve this answer
    
Thank you for the reference. As I found it, branching rules are mainly concerned with behavior of irreducible representations restricted to Levi subgroups. The case $SO_3\subset SO_4$ should finally fall into this situation. And what about $SO_3\subset SL_3$? $SO_3$ is its own normalizer in $SL_3$, with trivial centralizer. In view of mathoverflow.net/questions/18764, it is not associated to a Dykin subdiagram for that of $SL_3$. I expect the situation to be different in this case. –  genshin Sep 15 '10 at 13:04
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.