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?



Why is this a lattice?
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.


