I suppose this question is probably elementary for experts, but I'd like to present my arguments, about which I have some doubts, and see if they are correct, or if corrections and improvements are possible.


The setting is as follows: $k$ is the base field of characteristic zero, $G$ a connected semisimple $k$-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 $H$ semisimple $k$-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* otherwise.


*Fact* For $H\subset G$ a semisimple $k$-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 semisimple $k$-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 semisimple $k$-subgroup, then $Z(H,G)$ is $k$-anisotropic.


*Sketch of 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 semisimple. 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 these arguments make sense. If so, is there any other elementary proof, essentially different (modulo the Tannakian duality). Moreover, what if one allows reductive $k$-subgroups? 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 fields (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.



When does an irreducible representation remain irreducible after restriction to a semi-simple subgroup?
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.


