orbits of linear algebraic group $G({\Bbb Q}_p)$ acting on subgroups of ${\Bbb Q}_p^n$ Let $G\subseteq GL(n)$ be a linear algebraic group, and let $G({\Bbb Q}_p)\subseteq GL(V)$ act on a ${\Bbb Q}_p$-vector space V of finite dimension.
Consider the action of $G$ on abelian subgroups $L\subseteq V$ such that $ L\otimes {\Bbb Q}_p=V$.
What is known about the orbits of this action ? Where can I read about it? I am also interested in the analogous question for adeles.
Specifically, I am interested in the following question.
Assume $G$ is reductive.
Assume $L={\Bbb Q}v_1+...+{\Bbb Q}v_n$ where $v_1,...,v_n$ is a ${\Bbb Q}_p$-basis of $V$.
Assume $G_L=\{ g \in G({\Bbb Q}_p) : g(L)=L\ setwise\ \}$ is dense in $G({\Bbb Q}_p)$ (in either Zariski or p-adic topology).
Is there finitely many orbits of the action of  $G({\Bbb Q}_p)$ on such subgroups $L$ ? or perhaps of some closely related group? Under what assumptions this is true?
UPDATE: I assume this is related to classifying discrete subgroups of $G({\Bbb Q}_p)$ up to conjugation. Thus related notions are probably Mostow/Margulis rigidity and arithmetics subgroups...Or perhaps another way to think is that $G_L$ is a ${\Bbb Q}$-form of $G({\Bbb Q}_p)$  and we are asking how many ${\Bbb Q}$-forms a linear algebraic reductive group may have up to conjugacy. I am not sure whether the latter can be made formal. 
 A: Probably the easiest case to understand is when $\mathbf{G}$ is an inner form that is adjoint. In this case, there will always be infinitely many orbits, unless $\mathbf{G}$ manages to be simply connected (even though it is adjoint), which means that every simple factor of $\mathbf{G}$ is of type $E_8$, $F_4$, or $G_2$.  (For a different case, see later in this answer for a discussion of all split groups.)
We are given a reductive group $\mathbf{G}$ over $\mathbb{Q}_p$, and a faithful representation $\rho \colon \mathbf{G} \to \mathbf{GL}_n$ that is defined over $\mathbb{Q}_p$. Assume, for now, that $\mathbf{G}$ is both inner and adjoint.
Suppose $\mathbf{H}$ is any inner $\mathbb{Q}$-form of $\mathbf{G}$, so we can think of $\rho$ as a representation of $\mathbf{H}$ that is defined over $\mathbb{Q}_p$. Since $\mathbf{H}$ is inner and adjoint, every representation of $\mathbf{H}$ can be realized over $\mathbb{Q}$. (A classic paper of Tits [J. Reine Angew. Math. 247 (1971) 196-220] describes the two obstructions to being able to realize a representation over the algebraic closure as a representation over the ground field. One obstruction comes from the $*$-action, which is trivial for inner forms, and the other comes from the center, which is trivial for adjoint groups.) Therefore, from the other answer, which relates orbits to $\mathbb{Q}$-forms, we just need to show that $\mathbf{G}$ has infinitely many different $\mathbb{Q}$-forms that are inner.
For each $\ell$ in a finite set $S$ of primes, choose an algebraic group $\mathbf{G}_\ell$ over $\mathbb{Q}_p$ that is isomorphic to $\mathbf{G}$ over an algebraic closure (and is inner). The proposition on page 525 of [Borel-Harder, J. Reine Angew. Math. 298 (1978) 53-64] implies that there is an inner $\mathbb{Q}$-form of $\mathbf{G}$ that is $\mathbb{Q}_\ell$-isomorphic to $\mathbf{G}_\ell$ for every $\ell$ in $S$. Assuming there is a simple factor of $\mathbf{G}$ that is not of type $E_8$, $F_4$, or $G_2$, then there is more than one possible choice of $\mathbf{G}_\ell$, for each prime $\ell$. Hence, there must be infinitely many different $\mathbb{Q}$-forms of $\mathbf{G}$ that are inner.

Proposition. Suppose that $\mathbf{G}$ is split (over $\mathbb{Q}_p$) and simply connected, and that $\rho$ is irreducible (and almost faithful). Let $\lambda$ be a highest weight of $\rho$. Then $\mathbf{G}(\mathbb{Q}_p) \times \mathbb{Q}^\times$ has finitely many orbits on the set of $\mathbb{Q}$-lattices for $\mathbf{G}$ if and only if $\rho$ is faithful and $\lambda$ is not fixed by any nontrivial diagram automorphism.
Proof. Since $\mathbf{G}$ is split, we know that $\rho$ is absolutely irreducible, not just irreducible, so $\mathbf{C}_\rho = \mathbb{Q}_p^\times$. (Also, the highest weight $\lambda$ is unique.) Therefore, the other answer shows that $\mathbf{G}(\mathbb{Q}_p) \times \mathbb{Q}^\times$ has finitely many orbits if and only if there are only finitely many $\mathbb{Q}$-forms of $\mathbf{G}$ such that $\rho$ can be realized as a representation over $\mathbb{Q}$.
Let $\mathbf{H}_0$ be the split $\mathbb{Q}$-form of $\mathbf{G}$.
($\Leftarrow$) Suppose $\mathbf{H}$ is $\mathbb{Q}$-form of $\mathbf{G}$ such that $\rho$ can be realized as a representation over $\mathbb{Q}$. Then Lemma 7.4 of the Tits paper tells us that $\lambda$ must be fixed by the $*$-action corresponding to $\mathbf{H}$.  However, by assumption, $\lambda$ is not fixed by any nontrivial diagram automorphisms, so this implies that the $*$-action is trivial, which means that $\mathbf{H}$ is an inner form.
Therefore, $\mathbf{H} = {}^\zeta \mathbf{H}_0$ is the Galois twist of $\mathbf{H}_0$ by some cocycle $\zeta \in H^1(\mathbb{Q}; \overline{\mathbf{H}_0})$, where $\overline{\mathbf{H}_0}$ is the adjoint group. Letting $Z$ be the center of $\mathbf{H}_0$, we obtain a cohomology class $c \in H^2(\mathbb{Q}; Z)$. Composing with $\lambda$ yields a cohomology class $\lambda \circ c \in H^2(\mathbb{Q}; \mu)$, where $\mu$ is the group of all $n$th roots of unity in $\overline{\mathbb{Q}}$.
Since the $*$-action is trivial, Corollary 3.5 of the paper of Tits states that $\rho$ can be realized over $\mathbb{Q}$ if and only if $\lambda \circ c$ is trivial (in the cohomology group $H^2(\mathbb{Q}; \mu)$). (This is one of the main results of the paper.)
Since $\lambda$ is faithful, and $\lambda \circ c$ is trivial, we conclude that $c$ is trivial (in the cohomology group $H^2(\mathbb{Q}; Z)$). This means that $\zeta$ lifts to a well-defined $1$-cocycle $\eta \in H^1(\mathbb{Q}; \mathbf{H}_0)$. 
Since $\mathbf{H}_0$ is simply connected (because $\mathbf{G}$ is simply connected), we know that $H^1(\mathbb{Q}_\ell ; \mathbf{H}_0) = 0$ for every prime $\ell$ [PR, Thm. 6.4,, p. 284. ([PR] = Platonov-Rapinchuk's book Algebraic Groups and Number Theory) Therefore, the natural map $H^1(\mathbb{Q} ; \mathbf{H}_0) \to H^1(\mathbb{R} ; \mathbf{H}_0)$ is finite-to-one [PR, Thm. 6.15, p.~316]. Since $H^1(\mathbb{R} ; \mathbf{H}_0)$ is finite [PR, Thm. 6.14, p.~316], we conclude that $H^1(\mathbb{Q} ; \mathbf{H}_0)$ is finite. Hence, there are only finitely many possibilities for $\eta$. So there are only finitely many possibilities for $\zeta$, which means there are only finitely many possibilities for the $\mathbb{Q}$-form $\mathbf{H} = {}^\zeta \mathbf{H}_0$.
($\Rightarrow$) Suppose $\lambda$ is fixed by the outer automorphism corresponding to some nontrivial automorphism $\varphi$ of the Dynkin diagram. Since $\varphi$ has order 2 or 3, I think it is easy to see that there are infinitely many different homomorphisms $\zeta$ from $\mathop{\mathrm{Gal}}(\overline{\mathbb{Q}}/\mathbb{Q})$ onto $\langle \varphi \rangle$, such that $\zeta$ is trivial on $\mathop{\mathrm{Gal}}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$. (For example, this is easier than Lemma 1.9 of [Borel-Harder].) Each $\zeta$ represents a different cohomology class in $H^1(\mathbb{Q}; \mathop{\mathrm{Aut}} \mathbf{H}_0)$, and therefore represents a different quasi-split $\mathbb{Q}$-form ${}^\zeta H_0$ of $\mathbf{G}$. (We are using the fact that $\zeta$ is trivial on $\mathop{\mathrm{Gal}}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$ to know that ${}^\zeta H_0$ is isomorphic to $\mathbf{G}$ over $\mathbb{Q}_p$.) 
By construction, the $*$-action for ${}^\zeta H_0$ is given by $\varphi$, so $\lambda$ is fixed by the $*$-action. Since ${}^\zeta H_0$ is quasi-split, Tits gives no additional obstruction, so the representation $\rho$ with highest weight $\lambda$ can be realized over $\mathbb{Q}$ with respect to each of the infinitely many $\mathbb{Q}$-forms ${}^\zeta H_0$
Suppose $\lambda$ is not faithful, so the kernel is some nontrivial subgroup $Z'$ of $Z = Z(\mathbf{H}_0)$. Then infinitely many cohomology classes in $H^1(\mathbb{Q}; \mathbf{H}_0/Z')$ have trivial restriction to $\mathop{\mathrm{Gal}}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$. (Indeed, since $H^1(\mathbb{Q}_\ell; \mathbf{H}_0/Z')$ is nontrivial for every prime $\ell$ [PR, Thm. 6.20, p. 326], this follows from the surjectivity result in Theorem 1.7 of [Borel-Harder].) Hence, twisting by elements $\zeta$ of $H^1(\mathbb{Q}; \mathbf{H}_0/Z')$ yields infinitely many different $\mathbb{Q}$-forms ${}^\zeta \mathbf{H}_0$ of $\mathbf{G}$. The corresponding element~$c$ of $H^2(\mathbb{Q}; Z)$ belongs to $H^2(\mathbb{Q}; Z')$. Since $Z' = ker \lambda$, then $\lambda \circ c$ is obviously trivial. So Tits tells us that $\rho$ can be realized over $\mathbb{Q}$ with respect to each of these infinitely many $\mathbb{Q}$-forms.
A: To turn your question into a problem about $\mathbb{Q}$-forms (as hinted in your update), you could include the action of the centralizer of $G$.
More precisely, for a representation $\rho \colon \mathbf{G} \to \mathbf{GL}_n$, let $\mathbf{C}_\rho$ be the centralizer of $\rho(\mathbf{G})$ in $\mathbf{GL}_n$. For convenience, I will call a subgroup $L$ satisfying your conditions a "$\mathbb{Q}$-lattice for $\mathbf{G}$." Given any $\mathbb{Q}$-lattice $L$ for $\mathbf{G}$ and any $c \in \mathbf{C}_\rho(\mathbb{Q}_p)$, it is easy to see that $c(L)$ is also a $\mathbb{Q}$-lattice for $\mathbf{G}$. So the group $\mathbf{G}(\mathbb{Q}_p) \times  \mathbf{C}_\rho(\mathbb{Q}_p)$ acts on the set of $\mathbb{Q}$-lattices for $\mathbf{G}$.
Assuming that $\rho$ is faithful, there are finitely many orbits of $\mathbf{G}(\mathbb{Q}_p) \times \mathbf{C}_\rho(\mathbb{Q}_p)$ on the set of $\mathbb{Q}$-lattices for $\mathbf{G}$ if and only if there are only finitely many $\mathbb{Q}$-forms of $\mathbf{G}$ (up to $\mathbb{Q}$-isomorphism), such that $\rho$ can be realized as a representation defined over $\mathbb{Q}$. (This question about $\mathbb{Q}$-forms can be translated into a problem of Galois cohomology, which should not be difficult to solve for any given representation $\rho$.)
The correspondence comes from the observation that any $\mathbb{Q}$-lattice $L$ for $\mathbf{G}$ determines a $\mathbb{Q}$-form $\mathbf{H}$ of $\mathbf{G}$, such that $\rho$ can be realized as a representation defined over $\mathbb{Q}$. Namely, $\mathbf{H}(\mathbb{Q}) = \{\, h \in G \mid \rho(h) L = L \,\}$.
Now, let $\mathbf{H}_1$ and $\mathbf{H}_2$ be the $\mathbb{Q}$-forms corresponding to two $\mathbb{Q}$-lattices for $\mathbf{G}$. Suppose $\rho(g)c(L_1) = L_2$ for some $g \in \mathbf{G}(\mathbb{Q}_p)$ and $c \in  \mathbf{C}_\rho(\mathbb{Q}_p)$. Then it is straightforward to verify that $g^{-1} \mathbf{H}_2(\mathbb{Q}) g = \mathbf{H}_1(\mathbb{Q})$. Therefore, conjugation by $g$ is a $\mathbb{Q}$-isomorphism from $\mathbf{H}_2$ to $\mathbf{H}_1$.
Conversely, suppose $\varphi$ is any $\mathbb{Q}$-isomorphism from $\mathbf{H}_2$ to $\mathbf{H}_1$. Over $\mathbb{Q}_p$, $\varphi$ is an automorphism of $\mathbf{G}$. The outer automorphism group of $\mathbf{G}$ is finite, so it causes no major harm to assume that $\varphi$ is inner. This means there is some $g \in \mathbf{G}$, such that $g^{-1} \mathbf{H}_2(\mathbb{Q}) g = \mathbf{H}_1(\mathbb{Q})$. Since $\mathrm{Ad} g$ is a $\mathbb{Q}_p$-point of the adjoint group, it is only another error of finite index to assume that $g \in \mathbf{G}(\mathbb{Q}_p)$. Then, by replacing $L_1$ with $\rho(g) L_1$, we may assume $g$ is trivial, so $\mathbf{H}_2 = \mathbf{H_1}$. This means that $L_1$ and $L_2$ are two $\mathbb{Q}$-lattices for $\mathbf{G}$ that are both invariant under $\mathbf{H}_1(\mathbb{Q})$. By letting $\rho_i(h)$ be the restriction $\rho(h)|_{L_i}$, we obtain two representations $\rho_i \colon \mathbf{H}_1(\mathbb{Q}) \to \mathrm{GL}(L_i)$ (on vector spaces over $\mathbb{Q}$). The representations are isomorphic when tensored with $\mathbb{Q}_p$ (since both of them are $\mathbb{Q}$-forms of $\rho$), so they are isomorphic over $\mathbb{Q}$. This means there is a $\mathbb{Q}$-linear isomorphism $c_{\mathbb{Q}} \colon L_1 \to L_2$, such that $c_{\mathbb{Q}} \rho_1(h) \ell = \rho_2(h) c_{\mathbb{Q}} \ell$, for all $\ell \in L_1$. Let $c$ be the $\mathbb{Q}_p$-linear extension of $c_{\mathbb{Q}}$ to $(\mathbb{Q}_p)^n$. Then, since $L_1$ spans $(\mathbb{Q}_p)^n$, and $\rho_i(h)$ is the restriction of $\rho(h)$, we have $c \rho(h) = \rho(h) c$, for all $h \in \mathbf{H}_1(\mathbb{Q})$. Since $\mathbf{H}_1(\mathbb{Q})$ is Zariski dense in $\mathbf{G}$, this equation must hold for all $h \in \mathbf{G}(\mathbb{Q}_p)$, so $c \in \mathbf{C}_\rho$. Hence, $L_2 = c(L_1)$ is in the orbit of $L_1$.
