Show that SL(2, Z) ⊂ SL(2, R) is a discrete subgroup

Question

I have already shown that it is a subgroup, and now I need only to check that it is a discrete one. So I need to prove that the identity is an isolated point. My questions is, what are the neighborhoods in these space?

With it I can try to construct the desired neighborhood. I sense that it has to do with the fact that $\mathbb{Z}$ is a discrete subspace in the subset of topology of $\mathbb{R}$

These spaces SL(2, Z), SL(2, R) are the set of all matrices with determinant equal to 1 and with integer and real entries, respectively.

Thanks.

Answer 1

It is usual to view $\mathrm{SL}(2, \Bbb{R})$ as a metric space under the metric it gets viewed as a subspace of $\Bbb{R}^4$ under the standard Euclidean metric. $\mathrm{SL}(2, \Bbb{Z})$ is then a subspace of $\Bbb{Z}^4$, which is a discrete subspace of $\Bbb{R}^4$.

Answer 2

The topology on $\mathrm{SL}(2,\mathbb R)$ is induced by a norm. Now consider if $A\neq B$ are matrices with interger entries, then how small can $$ ||A-B||$$ get? As $A\neq B$ at least one component differs by at least $1$. Thus $$ || A-B || \geq || C || > 0$$ where $C$ is matrix that is $1$ in one component and $0$ everywhere else. If we for example take $||.||$ the $1$-norm, then $$ || A-B || \geq 1$$ This implies that each point is isolated, so the induced Topology on $\mathrm {SL}(2,\mathbb Z)$ is discrete.