Discrete arithmetic subgroup

Question

Let $F=\mathbb{Q}(\sqrt{d})$ for some $d>0$ be a real quadratic field and $G=\rm{GL}_2$, say over $\mathbb{Z}$. For any embedding $F\hookrightarrow \mathbb{R}$, the group $G(\mathcal{O}_d)$ is not discrete in $G(\mathbb{R})$. But for the pair of the two embeddings of $F$ into $\mathbb{R}$, the group $G(\mathcal{O}_d)$ becomes discrete in $G(\mathbb{R})\times G(\mathbb{R})$. This is confusing me. Is $G(\mathcal{O}_d)$ now a discrete group in its own or not? Why is it discrete in one group, but not in the other. Are there different topologies or what is the matter?

Answer 1

You need to know the notion of induced topology.

A priori, there is no topology on the set $G(\mathcal O_d)$. But we have two embeddings $\iota_{1, 2}: G(\mathcal O_d) \hookrightarrow G(\Bbb R)$, given by the two real embeddings of $F$ into $\Bbb R$. They together give another embedding $\iota: G(\mathcal O_d) \hookrightarrow G(\Bbb R) \times G(\Bbb R)$ such that $\iota(x) = (\iota_1(x), \iota_2(x))$.

On the other hand, we have canonical topologies on $\Bbb G(\Bbb R)$ and on $G(\Bbb R) \times G(\Bbb R)$ (product topology). The three embeddings $\iota_1, \iota_2, \iota$ then induce three different topologies on $G(\mathcal O_d)$.

It turns out that, under the first two induced topologies, $G(\mathcal O_d)$ is not discrete; but under the third induced topology (via $\iota$), $G(\mathcal O_d)$ is discrete.

Is that clear now?