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YCor
YCor
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I'm$\DeclareMathOperator\SU{SU}\DeclareMathOperator\SL{SL}$I'm working through some of the constructions in Introduction to Arithmetic Groups by Dave Witte Morris, and I'm confused by the construction of example 6.3.1 on page 121. For reference, here's the setup:

Take $a,b \in \mathbb{Q}^+$, yielding the totally real number field $L=Q(\sqrt a) \subset \mathbb{R}$ with ring of integers $\mathcal{O}$ commensurable to $\mathbb{Z}[\sqrt a]$. Let $\tau$ be the nontrivial element of $\operatorname{Gal}(L/\mathbb{Q})$, and let $A=(\begin{smallmatrix} b & 0 \\ 0 & -1 \end{smallmatrix})$. Define a ''unitary group'' with entries in $\mathcal O$ as $$G_{\mathcal O} = SU(A,\tau;\mathcal O) = \{g \in SL(2,\mathcal O) \mid \tau(g^T)Ag=A \} \subset SL(2,\mathbb{R}).$$$$G_{\mathcal O} = \SU(A,\tau;\mathcal O) = \{g \in \SL(2,\mathcal O) \mid \tau(g^T)Ag=A \} \subset \SL(2,\mathbb{R}).$$ The statement is that $G_\mathcal{O}$ is an arithmetic subgroup of $G=SL(2,\mathbb{R})$$G=\SL(2,\mathbb{R})$.

There are a few points in the provided construction that are confusing me, but I think I might be able to work them out if someone could explain one thing to me:

Why do we expect $G_{\mathcal O}$ to be an arithmetic subgroup of $G$?

From the avenue of restriction of scalars I would expect $G_\mathcal{O}$ to be arithmetic in $G \times G$, one factor for each of the Galois conjugates of the number field. I don't think either of the conjugates lands in a compact factor since the form $A$ that's being preserved doesn't change under automorphisms of $L/\mathbb{Q}$.

I'm working through some of the constructions in Introduction to Arithmetic Groups by Dave Witte Morris, and I'm confused by the construction of example 6.3.1 on page 121. For reference, here's the setup:

Take $a,b \in \mathbb{Q}^+$, yielding the totally real number field $L=Q(\sqrt a) \subset \mathbb{R}$ with ring of integers $\mathcal{O}$ commensurable to $\mathbb{Z}[\sqrt a]$. Let $\tau$ be the nontrivial element of $\operatorname{Gal}(L/\mathbb{Q})$, and let $A=(\begin{smallmatrix} b & 0 \\ 0 & -1 \end{smallmatrix})$. Define a ''unitary group'' with entries in $\mathcal O$ as $$G_{\mathcal O} = SU(A,\tau;\mathcal O) = \{g \in SL(2,\mathcal O) \mid \tau(g^T)Ag=A \} \subset SL(2,\mathbb{R}).$$ The statement is that $G_\mathcal{O}$ is an arithmetic subgroup of $G=SL(2,\mathbb{R})$.

There are a few points in the provided construction that are confusing me, but I think I might be able to work them out if someone could explain one thing to me:

Why do we expect $G_{\mathcal O}$ to be an arithmetic subgroup of $G$?

From the avenue of restriction of scalars I would expect $G_\mathcal{O}$ to be arithmetic in $G \times G$, one factor for each of the Galois conjugates of the number field. I don't think either of the conjugates lands in a compact factor since the form $A$ that's being preserved doesn't change under automorphisms of $L/\mathbb{Q}$.

$\DeclareMathOperator\SU{SU}\DeclareMathOperator\SL{SL}$I'm working through some of the constructions in Introduction to Arithmetic Groups by Dave Witte Morris, and I'm confused by the construction of example 6.3.1 on page 121. For reference, here's the setup:

Take $a,b \in \mathbb{Q}^+$, yielding the totally real number field $L=Q(\sqrt a) \subset \mathbb{R}$ with ring of integers $\mathcal{O}$ commensurable to $\mathbb{Z}[\sqrt a]$. Let $\tau$ be the nontrivial element of $\operatorname{Gal}(L/\mathbb{Q})$, and let $A=(\begin{smallmatrix} b & 0 \\ 0 & -1 \end{smallmatrix})$. Define a ''unitary group'' with entries in $\mathcal O$ as $$G_{\mathcal O} = \SU(A,\tau;\mathcal O) = \{g \in \SL(2,\mathcal O) \mid \tau(g^T)Ag=A \} \subset \SL(2,\mathbb{R}).$$ The statement is that $G_\mathcal{O}$ is an arithmetic subgroup of $G=\SL(2,\mathbb{R})$.

There are a few points in the provided construction that are confusing me, but I think I might be able to work them out if someone could explain one thing to me:

Why do we expect $G_{\mathcal O}$ to be an arithmetic subgroup of $G$?

From the avenue of restriction of scalars I would expect $G_\mathcal{O}$ to be arithmetic in $G \times G$, one factor for each of the Galois conjugates of the number field. I don't think either of the conjugates lands in a compact factor since the form $A$ that's being preserved doesn't change under automorphisms of $L/\mathbb{Q}$.

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Ethan Dlugie
Ethan Dlugie
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Clarification on arithmetic groups example

I'm working through some of the constructions in Introduction to Arithmetic Groups by Dave Witte Morris, and I'm confused by the construction of example 6.3.1 on page 121. For reference, here's the setup:

Take $a,b \in \mathbb{Q}^+$, yielding the totally real number field $L=Q(\sqrt a) \subset \mathbb{R}$ with ring of integers $\mathcal{O}$ commensurable to $\mathbb{Z}[\sqrt a]$. Let $\tau$ be the nontrivial element of $\operatorname{Gal}(L/\mathbb{Q})$, and let $A=(\begin{smallmatrix} b & 0 \\ 0 & -1 \end{smallmatrix})$. Define a ''unitary group'' with entries in $\mathcal O$ as $$G_{\mathcal O} = SU(A,\tau;\mathcal O) = \{g \in SL(2,\mathcal O) \mid \tau(g^T)Ag=A \} \subset SL(2,\mathbb{R}).$$ The statement is that $G_\mathcal{O}$ is an arithmetic subgroup of $G=SL(2,\mathbb{R})$.

There are a few points in the provided construction that are confusing me, but I think I might be able to work them out if someone could explain one thing to me:

Why do we expect $G_{\mathcal O}$ to be an arithmetic subgroup of $G$?

From the avenue of restriction of scalars I would expect $G_\mathcal{O}$ to be arithmetic in $G \times G$, one factor for each of the Galois conjugates of the number field. I don't think either of the conjugates lands in a compact factor since the form $A$ that's being preserved doesn't change under automorphisms of $L/\mathbb{Q}$.