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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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These lecture notes by Brian Conrad might be helpful: math.stanford.edu/~conrad/248APage/handouts/adelelattice.pdf –  Tom Church Mar 22 '12 at 17:03
Take more and more primes and you will eventually end up with adeles;) $SL(n, \mathbb{Q})$ is a lattice in $SL(n, \mathbb{A})$! –  plusepsilon.de Mar 22 '12 at 18:40
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1 Answer 1

up vote 7 down vote accepted

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.

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Thanks very much. I worked out the first half of your answer shortly after asking the question, but the second part is very helpful. –  Sean Eberhard Mar 22 '12 at 20:36
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