Upper-triangular subgroup is not unimodular
$begingroup$
Consider the group $GL_n(mathbb{Q}_p)$ of $n times n$ invertible matrices over the $p$-adic field $mathbb{Q}_p$.
My goal is to prove that the subgroup $P_0$ of upper triangular matrices is not unimodular. This is often left as an exercise in scribed lecture notes.
Here is my so-far-unsuccessful attempt: fix a left-Haar measure $mu$, and normalize it so that $mu(A)=1$ for $A=P_0 cap GL_n(mathcal{O})$ (the subgroup of upper triangular matrices with integral entries). Now the modular function
$$Delta: P_0 to mathbb{R}_{>0}$$
is simply $Delta(x)=mu(Ax)=mu(x^{-1}Ax)$. Note that $x^{-1}Ax$ is a conjugate subgroup.
To start with, let us take $n=2$ and $x$ to be a diagonal matrix in $P_0$. In this case, $mu(Ax)$ is the (left) index of a congruence subgroup of $P_0$: that is, $P_0$ quotiented by a subgroup of matrices whose off-diagonal entry is a fixed power of $p$ (which is determined by the $x$).
But I am unable to proceed with this argument or use it to actually compute the modular character, even for the simple case of $n=2$ and $x$ being diagonal.
So how do we explicitly compute the modular character of the upper triangular subgroup?
abstract-algebra lie-groups topological-groups p-adic-number-theory haar-measure
asked Dec 4 '18 at 20:52
BharatRamBharatRam
905618
$endgroup$
add a comment |
$begingroup$
Consider the group $GL_n(mathbb{Q}_p)$ of $n times n$ invertible matrices over the $p$-adic field $mathbb{Q}_p$.
My goal is to prove that the subgroup $P_0$ of upper triangular matrices is not unimodular. This is often left as an exercise in scribed lecture notes.
Here is my so-far-unsuccessful attempt: fix a left-Haar measure $mu$, and normalize it so that $mu(A)=1$ for $A=P_0 cap GL_n(mathcal{O})$ (the subgroup of upper triangular matrices with integral entries). Now the modular function
$$Delta: P_0 to mathbb{R}_{>0}$$
is simply $Delta(x)=mu(Ax)=mu(x^{-1}Ax)$. Note that $x^{-1}Ax$ is a conjugate subgroup.
To start with, let us take $n=2$ and $x$ to be a diagonal matrix in $P_0$. In this case, $mu(Ax)$ is the (left) index of a congruence subgroup of $P_0$: that is, $P_0$ quotiented by a subgroup of matrices whose off-diagonal entry is a fixed power of $p$ (which is determined by the $x$).
But I am unable to proceed with this argument or use it to actually compute the modular character, even for the simple case of $n=2$ and $x$ being diagonal.
So how do we explicitly compute the modular character of the upper triangular subgroup?
abstract-algebra lie-groups topological-groups p-adic-number-theory haar-measure
asked Dec 4 '18 at 20:52
BharatRamBharatRam
905618
$endgroup$
add a comment |
$begingroup$
Consider the group $GL_n(mathbb{Q}_p)$ of $n times n$ invertible matrices over the $p$-adic field $mathbb{Q}_p$.
My goal is to prove that the subgroup $P_0$ of upper triangular matrices is not unimodular. This is often left as an exercise in scribed lecture notes.
Here is my so-far-unsuccessful attempt: fix a left-Haar measure $mu$, and normalize it so that $mu(A)=1$ for $A=P_0 cap GL_n(mathcal{O})$ (the subgroup of upper triangular matrices with integral entries). Now the modular function
$$Delta: P_0 to mathbb{R}_{>0}$$
is simply $Delta(x)=mu(Ax)=mu(x^{-1}Ax)$. Note that $x^{-1}Ax$ is a conjugate subgroup.
To start with, let us take $n=2$ and $x$ to be a diagonal matrix in $P_0$. In this case, $mu(Ax)$ is the (left) index of a congruence subgroup of $P_0$: that is, $P_0$ quotiented by a subgroup of matrices whose off-diagonal entry is a fixed power of $p$ (which is determined by the $x$).
But I am unable to proceed with this argument or use it to actually compute the modular character, even for the simple case of $n=2$ and $x$ being diagonal.
So how do we explicitly compute the modular character of the upper triangular subgroup?
abstract-algebra lie-groups topological-groups p-adic-number-theory haar-measure
asked Dec 4 '18 at 20:52
BharatRamBharatRam
905618
$endgroup$
Consider the group $GL_n(mathbb{Q}_p)$ of $n times n$ invertible matrices over the $p$-adic field $mathbb{Q}_p$.
My goal is to prove that the subgroup $P_0$ of upper triangular matrices is not unimodular. This is often left as an exercise in scribed lecture notes.
Here is my so-far-unsuccessful attempt: fix a left-Haar measure $mu$, and normalize it so that $mu(A)=1$ for $A=P_0 cap GL_n(mathcal{O})$ (the subgroup of upper triangular matrices with integral entries). Now the modular function
$$Delta: P_0 to mathbb{R}_{>0}$$
is simply $Delta(x)=mu(Ax)=mu(x^{-1}Ax)$. Note that $x^{-1}Ax$ is a conjugate subgroup.
To start with, let us take $n=2$ and $x$ to be a diagonal matrix in $P_0$. In this case, $mu(Ax)$ is the (left) index of a congruence subgroup of $P_0$: that is, $P_0$ quotiented by a subgroup of matrices whose off-diagonal entry is a fixed power of $p$ (which is determined by the $x$).
But I am unable to proceed with this argument or use it to actually compute the modular character, even for the simple case of $n=2$ and $x$ being diagonal.
So how do we explicitly compute the modular character of the upper triangular subgroup?
abstract-algebra lie-groups topological-groups p-adic-number-theory haar-measure
abstract-algebra lie-groups topological-groups p-adic-number-theory haar-measure
asked Dec 4 '18 at 20:52
BharatRamBharatRam
905618
asked Dec 4 '18 at 20:52
BharatRamBharatRam
905618
asked Dec 4 '18 at 20:52
BharatRamBharatRam
905618
asked Dec 4 '18 at 20:52
BharatRamBharatRam
905618
asked Dec 4 '18 at 20:52
BharatRamBharatRam
905618
905618
add a comment |
add a comment |
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