$\DeclareMathOperator\SU{SU}\DeclareMathOperator\SL{SL}$I'm working through some of the constructions in [*Introduction to Arithmetic Groups* by Dave Witte Morris][1], 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 $L/\mathbb{Q}$.


  [1]: https://arxiv.org/abs/math/0106063