# Why is this a lattice?

Why is $\text{SL}_3(\mathbf{Z}[1/2])$ a lattice in $\text{SL}_3(\mathbf{R})\times\text{SL}_3(\mathbf{Q}_2)$? Discreteness is pretty clear, but why finite covolume? I understand why $\text{SL}_3(\mathbf{Z})$ has finite covolume in $\text{SL}_3(\mathbf{R})$, but I'm having trouble seeing this extension.

For that matter, why does $\mathbf{Z}[1/2]$ have finite covolume in $\mathbf{R}\times\mathbf{Q}_2$?

Am I missing a simple argument?

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Take more and more primes and you will eventually end up with adeles;) $SL(n, \mathbb{Q})$ is a lattice in $SL(n, \mathbb{A})$! – Marc Palm Mar 22 '12 at 18:40
As to the second question, this is pretty elementary : use the "fractional part" in $\mathbf{Z}[1/2]$ to put any element of $\mathbf{R}\times\mathbf{Q}_2$ in $\mathbf{R}\times\mathbf{Z}_2$ (by subtraction), and the "remaining" $\mathbf{Z}$ to put the $\mathbf{R}$ component in $[0,1]$. This way you see that the quotient is a so-called solenoid, the compact quotient $[0,1]\times\mathbf{Z}_2/(1,x)\sim (0,x+1)$ (this is a compact connected topological group).
Informally, elements of $\mathbf{R}$ have dyadic expansion that is infinite to the right and finite to the left, and $2$-adic numbers have the opposite situation. Then $\mathbf{Z}[1/2]$ is their "intersection". The (diagonal) quotient just "blends" them.
As to the first question, you might first use that $\text{SL}_3(\mathbf{Z}[1/2])$ is dense in $\text{SL}_3(\mathbf{Q}_2)$ to put the $2$-adic component of any element of $\text{SL}_3(\mathbf{R})\times\text{SL}_3(\mathbf{Q}_2)$ in the open subgroup $\text{SL}_3(\mathbf{Z}_2)$, then $\text{SL}_3(\mathbf{Z})$ to put the $\text{SL}_3(\mathbf{R})$ component in a finite volume fundamental domain $D$. Then $D\times\text{SL}_3(\mathbf{Z}_2)$ is a finite volume fundamental domain.