Iwahori versus Bruhat decompositions
$begingroup$
I am faced with the following issue that I do not understand but seems contradictory, coming from the book of Roberts and Schmidt about $GSp(4)$. Consider a local non-archimedean field $F$, let $p$ be its maximal ideal, $mathcal{O}$ its ring of integers and $G=GSp(4, F)$. We are interested in the following Klingen congruence subgroup
$$K =
left(
begin{array}{cccc}
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
p & mathcal{O} & mathcal{O} & mathcal{O} \
p & mathcal{O} & mathcal{O} & mathcal{O} \
p & p & p & mathcal{O}
end{array}
right)
$$
(from now on all the subgroups written in this matrix-entries form is meant to be their intersection with $GSp(4, F)$. I am interested in computing the index of this subgroup in the maximal compact subgroup $K_0$ (where all the entries are integers).
Iwahori decomposition
We have
$$
K =
left(
begin{array}{cccc}
1 & & & \
p & 1 & & \
p & & 1 & \
p & p & p & 1
end{array}
right)
left(
begin{array}{cccc}
mathcal{O}^times & & & \
& mathcal{O} & mathcal{O} & \
& mathcal{O} & mathcal{O} & \
& & & mathcal{O}^times
end{array}
right)
left(
begin{array}{cccc}
1 & mathcal{O} & mathcal{O} & mathcal{O} \
& 1 & & mathcal{O} \
& & 1 & mathcal{O} \
& & & 1
end{array}
right)
$$
so that in particular by decomposing the left subgroup we should obtain
$$
left(
begin{array}{cccc}
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O}
end{array}
right)
=
bigsqcup_{a, b, c in mathcal{O}/p}
left(
begin{array}{cccc}
1 & & & \
a & 1 & & \
b & & 1 & \
c & b & -a & 1
end{array}
right)
K
$$
(where the fact that the entries on the right are this way comes from the conditions of belonging to $GSp(4)$). So that in particular the index should be, writting $N(p)$ for the norm of $p$,
$$[K_0:K] =N(p)^3$$
Bruhat decomposition
On the other hand, introducing the subgroup
$$
Q =
left(
begin{array}{cccc}
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
& mathcal{O} & mathcal{O} & mathcal{O} \
& mathcal{O} & mathcal{O} & mathcal{O} \
& & & mathcal{O}
end{array}
right)
$$
the Bruhat decomposition yields that for any field $k$,
$$
GSp(4, k) = Q
sqcup Qx
left(
begin{array}{cccc}
1 & k & & \
& 1 & & \
& & 1 & k \
& & & 1
end{array}
right)
sqcup
Qxy
left(
begin{array}{cccc}
1 & & k & \
& 1 & k & k \
& & 1 & \
& & & 1
end{array}
right)
sqcup
Qxyx
left(
begin{array}{cccc}
1 & k & k & k\
& 1 & & k\
& & 1 & k \
& & & 1
end{array}
right)
$$
where the transformations $x$ and $y$ are defined by
$$x=
left(
begin{array}{cccc}
& 1& & \
1 & & & \
& & & 1 \
& &1 &
end{array}
right)
$$
$$y =
left(
begin{array}{cccc}
1 & & &\
& & 1 & \
& -1 & & \
& & & 1
end{array}
right)
$$
In particular if $k$ is the finite field with $N(p)$ elements, the index we search for is exactly the cardinality of $GSp(4,k) / Q$, and this one is $(1+N(p))(1+N(p)^2)$, so that we should say
$$[K_0:K] =(1+N(p))(1+N(p)^2)$$
Here is the question following from this discussion:
Both results are different even if asymptotically equivalent, what is
happening?
group-theory number-theory proof-verification
$endgroup$
add a comment |
$begingroup$
I am faced with the following issue that I do not understand but seems contradictory, coming from the book of Roberts and Schmidt about $GSp(4)$. Consider a local non-archimedean field $F$, let $p$ be its maximal ideal, $mathcal{O}$ its ring of integers and $G=GSp(4, F)$. We are interested in the following Klingen congruence subgroup
$$K =
left(
begin{array}{cccc}
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
p & mathcal{O} & mathcal{O} & mathcal{O} \
p & mathcal{O} & mathcal{O} & mathcal{O} \
p & p & p & mathcal{O}
end{array}
right)
$$
(from now on all the subgroups written in this matrix-entries form is meant to be their intersection with $GSp(4, F)$. I am interested in computing the index of this subgroup in the maximal compact subgroup $K_0$ (where all the entries are integers).
Iwahori decomposition
We have
$$
K =
left(
begin{array}{cccc}
1 & & & \
p & 1 & & \
p & & 1 & \
p & p & p & 1
end{array}
right)
left(
begin{array}{cccc}
mathcal{O}^times & & & \
& mathcal{O} & mathcal{O} & \
& mathcal{O} & mathcal{O} & \
& & & mathcal{O}^times
end{array}
right)
left(
begin{array}{cccc}
1 & mathcal{O} & mathcal{O} & mathcal{O} \
& 1 & & mathcal{O} \
& & 1 & mathcal{O} \
& & & 1
end{array}
right)
$$
so that in particular by decomposing the left subgroup we should obtain
$$
left(
begin{array}{cccc}
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O}
end{array}
right)
=
bigsqcup_{a, b, c in mathcal{O}/p}
left(
begin{array}{cccc}
1 & & & \
a & 1 & & \
b & & 1 & \
c & b & -a & 1
end{array}
right)
K
$$
(where the fact that the entries on the right are this way comes from the conditions of belonging to $GSp(4)$). So that in particular the index should be, writting $N(p)$ for the norm of $p$,
$$[K_0:K] =N(p)^3$$
Bruhat decomposition
On the other hand, introducing the subgroup
$$
Q =
left(
begin{array}{cccc}
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
& mathcal{O} & mathcal{O} & mathcal{O} \
& mathcal{O} & mathcal{O} & mathcal{O} \
& & & mathcal{O}
end{array}
right)
$$
the Bruhat decomposition yields that for any field $k$,
$$
GSp(4, k) = Q
sqcup Qx
left(
begin{array}{cccc}
1 & k & & \
& 1 & & \
& & 1 & k \
& & & 1
end{array}
right)
sqcup
Qxy
left(
begin{array}{cccc}
1 & & k & \
& 1 & k & k \
& & 1 & \
& & & 1
end{array}
right)
sqcup
Qxyx
left(
begin{array}{cccc}
1 & k & k & k\
& 1 & & k\
& & 1 & k \
& & & 1
end{array}
right)
$$
where the transformations $x$ and $y$ are defined by
$$x=
left(
begin{array}{cccc}
& 1& & \
1 & & & \
& & & 1 \
& &1 &
end{array}
right)
$$
$$y =
left(
begin{array}{cccc}
1 & & &\
& & 1 & \
& -1 & & \
& & & 1
end{array}
right)
$$
In particular if $k$ is the finite field with $N(p)$ elements, the index we search for is exactly the cardinality of $GSp(4,k) / Q$, and this one is $(1+N(p))(1+N(p)^2)$, so that we should say
$$[K_0:K] =(1+N(p))(1+N(p)^2)$$
Here is the question following from this discussion:
Both results are different even if asymptotically equivalent, what is
happening?
group-theory number-theory proof-verification
$endgroup$
add a comment |
$begingroup$
I am faced with the following issue that I do not understand but seems contradictory, coming from the book of Roberts and Schmidt about $GSp(4)$. Consider a local non-archimedean field $F$, let $p$ be its maximal ideal, $mathcal{O}$ its ring of integers and $G=GSp(4, F)$. We are interested in the following Klingen congruence subgroup
$$K =
left(
begin{array}{cccc}
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
p & mathcal{O} & mathcal{O} & mathcal{O} \
p & mathcal{O} & mathcal{O} & mathcal{O} \
p & p & p & mathcal{O}
end{array}
right)
$$
(from now on all the subgroups written in this matrix-entries form is meant to be their intersection with $GSp(4, F)$. I am interested in computing the index of this subgroup in the maximal compact subgroup $K_0$ (where all the entries are integers).
Iwahori decomposition
We have
$$
K =
left(
begin{array}{cccc}
1 & & & \
p & 1 & & \
p & & 1 & \
p & p & p & 1
end{array}
right)
left(
begin{array}{cccc}
mathcal{O}^times & & & \
& mathcal{O} & mathcal{O} & \
& mathcal{O} & mathcal{O} & \
& & & mathcal{O}^times
end{array}
right)
left(
begin{array}{cccc}
1 & mathcal{O} & mathcal{O} & mathcal{O} \
& 1 & & mathcal{O} \
& & 1 & mathcal{O} \
& & & 1
end{array}
right)
$$
so that in particular by decomposing the left subgroup we should obtain
$$
left(
begin{array}{cccc}
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O}
end{array}
right)
=
bigsqcup_{a, b, c in mathcal{O}/p}
left(
begin{array}{cccc}
1 & & & \
a & 1 & & \
b & & 1 & \
c & b & -a & 1
end{array}
right)
K
$$
(where the fact that the entries on the right are this way comes from the conditions of belonging to $GSp(4)$). So that in particular the index should be, writting $N(p)$ for the norm of $p$,
$$[K_0:K] =N(p)^3$$
Bruhat decomposition
On the other hand, introducing the subgroup
$$
Q =
left(
begin{array}{cccc}
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
& mathcal{O} & mathcal{O} & mathcal{O} \
& mathcal{O} & mathcal{O} & mathcal{O} \
& & & mathcal{O}
end{array}
right)
$$
the Bruhat decomposition yields that for any field $k$,
$$
GSp(4, k) = Q
sqcup Qx
left(
begin{array}{cccc}
1 & k & & \
& 1 & & \
& & 1 & k \
& & & 1
end{array}
right)
sqcup
Qxy
left(
begin{array}{cccc}
1 & & k & \
& 1 & k & k \
& & 1 & \
& & & 1
end{array}
right)
sqcup
Qxyx
left(
begin{array}{cccc}
1 & k & k & k\
& 1 & & k\
& & 1 & k \
& & & 1
end{array}
right)
$$
where the transformations $x$ and $y$ are defined by
$$x=
left(
begin{array}{cccc}
& 1& & \
1 & & & \
& & & 1 \
& &1 &
end{array}
right)
$$
$$y =
left(
begin{array}{cccc}
1 & & &\
& & 1 & \
& -1 & & \
& & & 1
end{array}
right)
$$
In particular if $k$ is the finite field with $N(p)$ elements, the index we search for is exactly the cardinality of $GSp(4,k) / Q$, and this one is $(1+N(p))(1+N(p)^2)$, so that we should say
$$[K_0:K] =(1+N(p))(1+N(p)^2)$$
Here is the question following from this discussion:
Both results are different even if asymptotically equivalent, what is
happening?
group-theory number-theory proof-verification
$endgroup$
I am faced with the following issue that I do not understand but seems contradictory, coming from the book of Roberts and Schmidt about $GSp(4)$. Consider a local non-archimedean field $F$, let $p$ be its maximal ideal, $mathcal{O}$ its ring of integers and $G=GSp(4, F)$. We are interested in the following Klingen congruence subgroup
$$K =
left(
begin{array}{cccc}
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
p & mathcal{O} & mathcal{O} & mathcal{O} \
p & mathcal{O} & mathcal{O} & mathcal{O} \
p & p & p & mathcal{O}
end{array}
right)
$$
(from now on all the subgroups written in this matrix-entries form is meant to be their intersection with $GSp(4, F)$. I am interested in computing the index of this subgroup in the maximal compact subgroup $K_0$ (where all the entries are integers).
Iwahori decomposition
We have
$$
K =
left(
begin{array}{cccc}
1 & & & \
p & 1 & & \
p & & 1 & \
p & p & p & 1
end{array}
right)
left(
begin{array}{cccc}
mathcal{O}^times & & & \
& mathcal{O} & mathcal{O} & \
& mathcal{O} & mathcal{O} & \
& & & mathcal{O}^times
end{array}
right)
left(
begin{array}{cccc}
1 & mathcal{O} & mathcal{O} & mathcal{O} \
& 1 & & mathcal{O} \
& & 1 & mathcal{O} \
& & & 1
end{array}
right)
$$
so that in particular by decomposing the left subgroup we should obtain
$$
left(
begin{array}{cccc}
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O}
end{array}
right)
=
bigsqcup_{a, b, c in mathcal{O}/p}
left(
begin{array}{cccc}
1 & & & \
a & 1 & & \
b & & 1 & \
c & b & -a & 1
end{array}
right)
K
$$
(where the fact that the entries on the right are this way comes from the conditions of belonging to $GSp(4)$). So that in particular the index should be, writting $N(p)$ for the norm of $p$,
$$[K_0:K] =N(p)^3$$
Bruhat decomposition
On the other hand, introducing the subgroup
$$
Q =
left(
begin{array}{cccc}
mathcal{O} & mathcal{O} & mathcal{O} & mathcal{O} \
& mathcal{O} & mathcal{O} & mathcal{O} \
& mathcal{O} & mathcal{O} & mathcal{O} \
& & & mathcal{O}
end{array}
right)
$$
the Bruhat decomposition yields that for any field $k$,
$$
GSp(4, k) = Q
sqcup Qx
left(
begin{array}{cccc}
1 & k & & \
& 1 & & \
& & 1 & k \
& & & 1
end{array}
right)
sqcup
Qxy
left(
begin{array}{cccc}
1 & & k & \
& 1 & k & k \
& & 1 & \
& & & 1
end{array}
right)
sqcup
Qxyx
left(
begin{array}{cccc}
1 & k & k & k\
& 1 & & k\
& & 1 & k \
& & & 1
end{array}
right)
$$
where the transformations $x$ and $y$ are defined by
$$x=
left(
begin{array}{cccc}
& 1& & \
1 & & & \
& & & 1 \
& &1 &
end{array}
right)
$$
$$y =
left(
begin{array}{cccc}
1 & & &\
& & 1 & \
& -1 & & \
& & & 1
end{array}
right)
$$
In particular if $k$ is the finite field with $N(p)$ elements, the index we search for is exactly the cardinality of $GSp(4,k) / Q$, and this one is $(1+N(p))(1+N(p)^2)$, so that we should say
$$[K_0:K] =(1+N(p))(1+N(p)^2)$$
Here is the question following from this discussion:
Both results are different even if asymptotically equivalent, what is
happening?
group-theory number-theory proof-verification
group-theory number-theory proof-verification
edited Jan 13 at 9:40
TheStudent
asked Jan 10 at 9:21
TheStudentTheStudent
3239
3239
add a comment |
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