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Parabolic Induction in Stages
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$DeclareMathOperatorIndInd$Let $G$ be a connected, reductive group over a local field $k$.
Let $P_ast = M_astN_ast subseteq P = MN$ be parabolic subgroups, standard with respect to a given minimal parabolic, with standard Levi subgroups $M_ast subseteq M$.
Then $M_ast(N_ast cap M)$ is a parabolic subgroup of $M$ with unipotent radical $N_ast cap M$. If $(pi,V)$ is a representation of $M_ast$, then we can regard $pi$ as a representation of $P_ast$ by making it trivial on $N_ast$, and form the induced representation $Ind_P_ast^G pi$. On the other hand, we can regard $pi$ as a representation of $M_ast(N_ast cap M)$ by making it trivial on $N_ast cap M$, and form the induced representation
$$sigma = Ind_M_ast(N_ast cap M)^M pi$$
Then we can extend $sigma$ to $P$ by making it trivial on $N$, and form
$$Ind_P^G sigma$$
I expect that we should be able to identify the representations
$$Ind_P^GInd_M_ast(N_ast cap M)^M pi = Ind_P_ast^G pi$$
This initially appears to be an application of the transitivity of induction, but this principle cannot be immediately implied. Are these two representations of $G$ the same? What if we do these by normalized induction? (Taking into account the modulus characters)
representation-theory algebraic-groups reductive-groups
asked Jul 27 at 4:52
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$DeclareMathOperatorIndInd$Let $G$ be a connected, reductive group over a local field $k$.
Let $P_ast = M_astN_ast subseteq P = MN$ be parabolic subgroups, standard with respect to a given minimal parabolic, with standard Levi subgroups $M_ast subseteq M$.
Then $M_ast(N_ast cap M)$ is a parabolic subgroup of $M$ with unipotent radical $N_ast cap M$. If $(pi,V)$ is a representation of $M_ast$, then we can regard $pi$ as a representation of $P_ast$ by making it trivial on $N_ast$, and form the induced representation $Ind_P_ast^G pi$. On the other hand, we can regard $pi$ as a representation of $M_ast(N_ast cap M)$ by making it trivial on $N_ast cap M$, and form the induced representation
$$sigma = Ind_M_ast(N_ast cap M)^M pi$$
Then we can extend $sigma$ to $P$ by making it trivial on $N$, and form
$$Ind_P^G sigma$$
I expect that we should be able to identify the representations
$$Ind_P^GInd_M_ast(N_ast cap M)^M pi = Ind_P_ast^G pi$$
This initially appears to be an application of the transitivity of induction, but this principle cannot be immediately implied. Are these two representations of $G$ the same? What if we do these by normalized induction? (Taking into account the modulus characters)
representation-theory algebraic-groups reductive-groups
asked Jul 27 at 4:52
D_S
12.8k51550
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$DeclareMathOperatorIndInd$Let $G$ be a connected, reductive group over a local field $k$.
Let $P_ast = M_astN_ast subseteq P = MN$ be parabolic subgroups, standard with respect to a given minimal parabolic, with standard Levi subgroups $M_ast subseteq M$.
Then $M_ast(N_ast cap M)$ is a parabolic subgroup of $M$ with unipotent radical $N_ast cap M$. If $(pi,V)$ is a representation of $M_ast$, then we can regard $pi$ as a representation of $P_ast$ by making it trivial on $N_ast$, and form the induced representation $Ind_P_ast^G pi$. On the other hand, we can regard $pi$ as a representation of $M_ast(N_ast cap M)$ by making it trivial on $N_ast cap M$, and form the induced representation
$$sigma = Ind_M_ast(N_ast cap M)^M pi$$
Then we can extend $sigma$ to $P$ by making it trivial on $N$, and form
$$Ind_P^G sigma$$
I expect that we should be able to identify the representations
$$Ind_P^GInd_M_ast(N_ast cap M)^M pi = Ind_P_ast^G pi$$
This initially appears to be an application of the transitivity of induction, but this principle cannot be immediately implied. Are these two representations of $G$ the same? What if we do these by normalized induction? (Taking into account the modulus characters)
representation-theory algebraic-groups reductive-groups
asked Jul 27 at 4:52
D_S
12.8k51550
$DeclareMathOperatorIndInd$Let $G$ be a connected, reductive group over a local field $k$.
Let $P_ast = M_astN_ast subseteq P = MN$ be parabolic subgroups, standard with respect to a given minimal parabolic, with standard Levi subgroups $M_ast subseteq M$.
Then $M_ast(N_ast cap M)$ is a parabolic subgroup of $M$ with unipotent radical $N_ast cap M$. If $(pi,V)$ is a representation of $M_ast$, then we can regard $pi$ as a representation of $P_ast$ by making it trivial on $N_ast$, and form the induced representation $Ind_P_ast^G pi$. On the other hand, we can regard $pi$ as a representation of $M_ast(N_ast cap M)$ by making it trivial on $N_ast cap M$, and form the induced representation
$$sigma = Ind_M_ast(N_ast cap M)^M pi$$
Then we can extend $sigma$ to $P$ by making it trivial on $N$, and form
$$Ind_P^G sigma$$
I expect that we should be able to identify the representations
$$Ind_P^GInd_M_ast(N_ast cap M)^M pi = Ind_P_ast^G pi$$
This initially appears to be an application of the transitivity of induction, but this principle cannot be immediately implied. Are these two representations of $G$ the same? What if we do these by normalized induction? (Taking into account the modulus characters)
representation-theory algebraic-groups reductive-groups
asked Jul 27 at 4:52
D_S
12.8k51550
asked Jul 27 at 4:52
D_S
12.8k51550
asked Jul 27 at 4:52
D_S
12.8k51550
asked Jul 27 at 4:52
D_S
12.8k51550
12.8k51550
add a comment |Â
add a comment |Â
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$DeclareMathOperatorHomHom$I think that this follows from a version of Frobenius reciprocity: if $(tau,W)$ is a representation of $G$, and $W_N = W/W(N)$ is the Jacquet module of $P$ (where $W(N)$ is the linear span of $w - tau(n)w : w in W, n in N$), then we have
$$Hom_G(W, Ind_P^Gpi) = Hom_M(W_N,pi)$$
Here $W_N$ inherits the structure of a representation of $M$. We can further restrict this to a representation of the parabolic subgroup $M_ast(N_ast cap M)$ of $M$ and take the Jacquet module again.
Lemma: The Jacquet module $W_N/[W_N(N_ast cap M)]$ coincides with $W_N_ast$ as a representation of $M_ast$.
Proof: By the third isomorphism theorem, this comes down to showing that the subspace $W_N(N_ast cap M)$ of $W_N = W/W(N)$ is equal to
$$W_N(N_ast cap M) = fracW(N_ast)W(N)$$
It is clear that the left hand side is contained in the right. Now consider a generator $w - tau(n)w + W(N)$ of the right hand side, for some $w in W$ and $n in N_ast$. Since $N_ast$ is the semidirect product of $N$ by $N_ast cap M$, we can write $n = n_1n_2$ for $n_1 in N_ast cap M$ and $n_2 in N$. Then
$$w - tau(n)w + W(N) = w - tau(n_2)w + tau(n_2)w - tau(n)w+W(N) = tau(n_2)w - tau(n)w + W(N)$$
with $tau(n_2)w - tau(n)w = tau(n_2)w - tau(n_1)tau(n_2)w$. So this is in the left hand side. $blacksquare$
Now we have natural bijections in both variables
$$beginequation*
beginsplit
Hom_G(W,Ind_P^G Ind_M_ast(N_ast cap M)^M pi) & = Hom_M(W_N,Ind_M_ast(N_ast cap M)^M pi) \ & = Hom_M_ast(W_N_ast,pi) \
& = Hom_G(W,Ind_P_ast^Gpi)
endsplit
endequation*$$
so the functors $Ind_P_ast$ and $Ind_M_ast(N_ast cap M)^M$ must be naturally isomorphic.
We get the same result if we replace induction by normalized induction. This is on account of the fact that if $delta_1$ is the modulus character for $M_ast$ on $(N_ast cap M)$, and $delta_2$ is the modulus character for $M$ on $N$, then $delta_1 delta_2|_M_ast$ is the modulus character for $M_ast$ on $N_ast$.
answered Jul 27 at 5:53
D_S
12.8k51550
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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up vote
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$DeclareMathOperatorHomHom$I think that this follows from a version of Frobenius reciprocity: if $(tau,W)$ is a representation of $G$, and $W_N = W/W(N)$ is the Jacquet module of $P$ (where $W(N)$ is the linear span of $w - tau(n)w : w in W, n in N$), then we have
$$Hom_G(W, Ind_P^Gpi) = Hom_M(W_N,pi)$$
Here $W_N$ inherits the structure of a representation of $M$. We can further restrict this to a representation of the parabolic subgroup $M_ast(N_ast cap M)$ of $M$ and take the Jacquet module again.
Lemma: The Jacquet module $W_N/[W_N(N_ast cap M)]$ coincides with $W_N_ast$ as a representation of $M_ast$.
Proof: By the third isomorphism theorem, this comes down to showing that the subspace $W_N(N_ast cap M)$ of $W_N = W/W(N)$ is equal to
$$W_N(N_ast cap M) = fracW(N_ast)W(N)$$
It is clear that the left hand side is contained in the right. Now consider a generator $w - tau(n)w + W(N)$ of the right hand side, for some $w in W$ and $n in N_ast$. Since $N_ast$ is the semidirect product of $N$ by $N_ast cap M$, we can write $n = n_1n_2$ for $n_1 in N_ast cap M$ and $n_2 in N$. Then
$$w - tau(n)w + W(N) = w - tau(n_2)w + tau(n_2)w - tau(n)w+W(N) = tau(n_2)w - tau(n)w + W(N)$$
with $tau(n_2)w - tau(n)w = tau(n_2)w - tau(n_1)tau(n_2)w$. So this is in the left hand side. $blacksquare$
Now we have natural bijections in both variables
$$beginequation*
beginsplit
Hom_G(W,Ind_P^G Ind_M_ast(N_ast cap M)^M pi) & = Hom_M(W_N,Ind_M_ast(N_ast cap M)^M pi) \ & = Hom_M_ast(W_N_ast,pi) \
& = Hom_G(W,Ind_P_ast^Gpi)
endsplit
endequation*$$
so the functors $Ind_P_ast$ and $Ind_M_ast(N_ast cap M)^M$ must be naturally isomorphic.
We get the same result if we replace induction by normalized induction. This is on account of the fact that if $delta_1$ is the modulus character for $M_ast$ on $(N_ast cap M)$, and $delta_2$ is the modulus character for $M$ on $N$, then $delta_1 delta_2|_M_ast$ is the modulus character for $M_ast$ on $N_ast$.
answered Jul 27 at 5:53
D_S
12.8k51550
add a comment |Â
up vote
0
down vote
$DeclareMathOperatorHomHom$I think that this follows from a version of Frobenius reciprocity: if $(tau,W)$ is a representation of $G$, and $W_N = W/W(N)$ is the Jacquet module of $P$ (where $W(N)$ is the linear span of $w - tau(n)w : w in W, n in N$), then we have
$$Hom_G(W, Ind_P^Gpi) = Hom_M(W_N,pi)$$
Here $W_N$ inherits the structure of a representation of $M$. We can further restrict this to a representation of the parabolic subgroup $M_ast(N_ast cap M)$ of $M$ and take the Jacquet module again.
Lemma: The Jacquet module $W_N/[W_N(N_ast cap M)]$ coincides with $W_N_ast$ as a representation of $M_ast$.
Proof: By the third isomorphism theorem, this comes down to showing that the subspace $W_N(N_ast cap M)$ of $W_N = W/W(N)$ is equal to
$$W_N(N_ast cap M) = fracW(N_ast)W(N)$$
It is clear that the left hand side is contained in the right. Now consider a generator $w - tau(n)w + W(N)$ of the right hand side, for some $w in W$ and $n in N_ast$. Since $N_ast$ is the semidirect product of $N$ by $N_ast cap M$, we can write $n = n_1n_2$ for $n_1 in N_ast cap M$ and $n_2 in N$. Then
$$w - tau(n)w + W(N) = w - tau(n_2)w + tau(n_2)w - tau(n)w+W(N) = tau(n_2)w - tau(n)w + W(N)$$
with $tau(n_2)w - tau(n)w = tau(n_2)w - tau(n_1)tau(n_2)w$. So this is in the left hand side. $blacksquare$
Now we have natural bijections in both variables
$$beginequation*
beginsplit
Hom_G(W,Ind_P^G Ind_M_ast(N_ast cap M)^M pi) & = Hom_M(W_N,Ind_M_ast(N_ast cap M)^M pi) \ & = Hom_M_ast(W_N_ast,pi) \
& = Hom_G(W,Ind_P_ast^Gpi)
endsplit
endequation*$$
so the functors $Ind_P_ast$ and $Ind_M_ast(N_ast cap M)^M$ must be naturally isomorphic.
We get the same result if we replace induction by normalized induction. This is on account of the fact that if $delta_1$ is the modulus character for $M_ast$ on $(N_ast cap M)$, and $delta_2$ is the modulus character for $M$ on $N$, then $delta_1 delta_2|_M_ast$ is the modulus character for $M_ast$ on $N_ast$.
answered Jul 27 at 5:53
D_S
12.8k51550
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$DeclareMathOperatorHomHom$I think that this follows from a version of Frobenius reciprocity: if $(tau,W)$ is a representation of $G$, and $W_N = W/W(N)$ is the Jacquet module of $P$ (where $W(N)$ is the linear span of $w - tau(n)w : w in W, n in N$), then we have
$$Hom_G(W, Ind_P^Gpi) = Hom_M(W_N,pi)$$
Here $W_N$ inherits the structure of a representation of $M$. We can further restrict this to a representation of the parabolic subgroup $M_ast(N_ast cap M)$ of $M$ and take the Jacquet module again.
Lemma: The Jacquet module $W_N/[W_N(N_ast cap M)]$ coincides with $W_N_ast$ as a representation of $M_ast$.
Proof: By the third isomorphism theorem, this comes down to showing that the subspace $W_N(N_ast cap M)$ of $W_N = W/W(N)$ is equal to
$$W_N(N_ast cap M) = fracW(N_ast)W(N)$$
It is clear that the left hand side is contained in the right. Now consider a generator $w - tau(n)w + W(N)$ of the right hand side, for some $w in W$ and $n in N_ast$. Since $N_ast$ is the semidirect product of $N$ by $N_ast cap M$, we can write $n = n_1n_2$ for $n_1 in N_ast cap M$ and $n_2 in N$. Then
$$w - tau(n)w + W(N) = w - tau(n_2)w + tau(n_2)w - tau(n)w+W(N) = tau(n_2)w - tau(n)w + W(N)$$
with $tau(n_2)w - tau(n)w = tau(n_2)w - tau(n_1)tau(n_2)w$. So this is in the left hand side. $blacksquare$
Now we have natural bijections in both variables
$$beginequation*
beginsplit
Hom_G(W,Ind_P^G Ind_M_ast(N_ast cap M)^M pi) & = Hom_M(W_N,Ind_M_ast(N_ast cap M)^M pi) \ & = Hom_M_ast(W_N_ast,pi) \
& = Hom_G(W,Ind_P_ast^Gpi)
endsplit
endequation*$$
so the functors $Ind_P_ast$ and $Ind_M_ast(N_ast cap M)^M$ must be naturally isomorphic.
We get the same result if we replace induction by normalized induction. This is on account of the fact that if $delta_1$ is the modulus character for $M_ast$ on $(N_ast cap M)$, and $delta_2$ is the modulus character for $M$ on $N$, then $delta_1 delta_2|_M_ast$ is the modulus character for $M_ast$ on $N_ast$.
answered Jul 27 at 5:53
D_S
12.8k51550
$DeclareMathOperatorHomHom$I think that this follows from a version of Frobenius reciprocity: if $(tau,W)$ is a representation of $G$, and $W_N = W/W(N)$ is the Jacquet module of $P$ (where $W(N)$ is the linear span of $w - tau(n)w : w in W, n in N$), then we have
$$Hom_G(W, Ind_P^Gpi) = Hom_M(W_N,pi)$$
Here $W_N$ inherits the structure of a representation of $M$. We can further restrict this to a representation of the parabolic subgroup $M_ast(N_ast cap M)$ of $M$ and take the Jacquet module again.
Lemma: The Jacquet module $W_N/[W_N(N_ast cap M)]$ coincides with $W_N_ast$ as a representation of $M_ast$.
Proof: By the third isomorphism theorem, this comes down to showing that the subspace $W_N(N_ast cap M)$ of $W_N = W/W(N)$ is equal to
$$W_N(N_ast cap M) = fracW(N_ast)W(N)$$
It is clear that the left hand side is contained in the right. Now consider a generator $w - tau(n)w + W(N)$ of the right hand side, for some $w in W$ and $n in N_ast$. Since $N_ast$ is the semidirect product of $N$ by $N_ast cap M$, we can write $n = n_1n_2$ for $n_1 in N_ast cap M$ and $n_2 in N$. Then
$$w - tau(n)w + W(N) = w - tau(n_2)w + tau(n_2)w - tau(n)w+W(N) = tau(n_2)w - tau(n)w + W(N)$$
with $tau(n_2)w - tau(n)w = tau(n_2)w - tau(n_1)tau(n_2)w$. So this is in the left hand side. $blacksquare$
Now we have natural bijections in both variables
$$beginequation*
beginsplit
Hom_G(W,Ind_P^G Ind_M_ast(N_ast cap M)^M pi) & = Hom_M(W_N,Ind_M_ast(N_ast cap M)^M pi) \ & = Hom_M_ast(W_N_ast,pi) \
& = Hom_G(W,Ind_P_ast^Gpi)
endsplit
endequation*$$
so the functors $Ind_P_ast$ and $Ind_M_ast(N_ast cap M)^M$ must be naturally isomorphic.
We get the same result if we replace induction by normalized induction. This is on account of the fact that if $delta_1$ is the modulus character for $M_ast$ on $(N_ast cap M)$, and $delta_2$ is the modulus character for $M$ on $N$, then $delta_1 delta_2|_M_ast$ is the modulus character for $M_ast$ on $N_ast$.
answered Jul 27 at 5:53
D_S
12.8k51550
answered Jul 27 at 5:53
D_S
12.8k51550
answered Jul 27 at 5:53
D_S
12.8k51550
answered Jul 27 at 5:53
D_S
12.8k51550
12.8k51550
add a comment |Â
add a comment |Â
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