show/hide this revision's text 4 added 151 characters in body

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.
show/hide this revision's text 3 Spelling corrected

when When does an irreducible representation remains ireducible after restriction to a semi-simple subgroup?

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 $k$-vecor 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 parablic 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 $k$-represnetation k$-representation of $H$. Conversely, if the retriction 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 $k$-represnetation 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 direc direct sums in the filtred 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.

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 represenation 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.
show/hide this revision's text 2 added 14 characters in body

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 $k$-vecor 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 parablic $k$-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 $k$-represnetation of $H$. Conversely, if the retriction 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 $k$-represnetation $(\rho,V)$ 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 direc sums in the filtred 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.

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 represenation 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.
show/hide this revision's text 1