Vtyjkyuk

Let $Gamma$ be an arithmetic lattice in a linear algebraic $mathbbQ$-group $mathbfG$, that is, $Gamma$ is a subgroup of $mathbfG(mathbbQ)$ that is commensurable with $mathbfG(mathbbZ)$.

For a prime $p$, we can consider $Gamma$ as a subspace of $mathbfG(mathbbQ_p)$. My question is:

What does the closure of $Gamma$ in $mathbfG(mathbbQ_p)$ with respect to the $p$-adic topology look like?

The closure of $mathbbZ$ in $mathbbQ_p$ is the ring of $p$-adic integers $mathbbZ_p$. So it seems plausible to me that, for example, the closure of the lattice $Gamma = SL(n,mathbbZ)$ in the $p$-adic topology of $SL(n,mathbbQ_p)$ would be $SL(n,mathbbZ_p)$. Is this correct?

Also, what about other lattice, for example, what is the closure of a congruence subgroup $$Gamma(c) := g in SL(n,mathbbZ) : g - I_n equiv 0 ;text mod c, subset SL(n,mathbbZ)$$ in the $p$-adic topology of $SL(n,mathbbQ_p)$?

Suppose $G$ is $mathbb Q$ simple (i.e. has no connected normal algebraic subgroups which are defined over $mathbb Q$) and is simply connected (i.e. $G(mathbb C)$ is simply connected). Assume also that $G(mathbb R)$ is not compact. With these assumptions, the closure of an arithmetic lattice in $G(mathbb Z_p)$ is an open subgroup. This statement is known as strong approximation.

More generally, if $G(mathbb Z)$ is Zariski dense in $G$, and $G(mathbb C)$ is connected and simply connected, then the closure of a finite index subgroup of $G(mathbb Z)$ is open in $G(mathbb Z _p)$.

Examples are $G=SL_n$ and $Sp_2n$. But not $PGL_n$ (this is not simply connected).

In your example, the closure of $SL(n,mathbb Z)$ is indeed $SL(n,mathbb Z _p)$; this can be proved by using the fact that $SL(n,mathbb Z), SL(n,mathbb Z _p)$ are generated by unipotent elements.