3
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ @Carnahan, Thanks! That was a mistake on my part. I will delete my comment. $\endgroup$ Feb 17 '15 at 14:14
  • $\begingroup$ $G({\mathbb Z}_p)$ is compact, hence the only discrete subgroups are finite. On the other hand, $G({\mathbb Q}_p)$ doe have discrete subgroups, which, if the ${\mathbb Q}_p$ rank is high, are arithmetic, if they are of finite covolume. $\endgroup$ Feb 18 '15 at 5:42
  • $\begingroup$ @Venkataramana, thank you, I corrected that. It does seem that the answer might be implied by the theory of arithmetic subgroups, but it feels the question is standard enough to be somewhere in the literature... $\endgroup$
    – mmm
    Feb 24 '15 at 9:15
  • $\begingroup$ 1) It would be safe to write the algebraic group as $\mathbf{G}$ and $G=\mathbf{G}(\mathbb{Q}_p)$. With this notation, $\mathbf{G}$ doesn't act on subgroups of $V$, $G$ does act. 2) By dense in $G$ do you mean in the Zariski or ordinary topology? 3) "$g(L)=L$ setwise": I don't guess any other meaning of $g(L)=L$ (nobody would understand that $g$ fixes $L$ pointwise) 4) "Is there finitely many orbits on such subgroups $L$": I don't quite understand (do you mean "on" or "of"?) 5) $L\otimes\mathbb{Q}_p=V$ would be better stated as: $L$ spans $V$ as $\mathbb{Q}_p$-vector space. $\endgroup$
    – YCor
    Feb 24 '15 at 10:26
  • $\begingroup$ Thanks, I've made a few correction. Yes, I consider the action of the group of p-adic points on the subgroups L with these properties. Concerning being dense in either Zariski or p-adic topology, it is unclear what the right question is. $\endgroup$
    – mmm
    Feb 24 '15 at 11:01
5
+50
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ Thanks! And what about $G=GL_n$, should it not be an exception as well? What I ask is true for $G=GL_n$... $\endgroup$
    – mmm
    Feb 25 '15 at 7:29
  • $\begingroup$ Thanks for pointing this out. I overlooked that example, because of my assumption that $\mathbf{G}$ is adjoint. However, suppose $\rho$ is a representation of $\mathbf{G} = \mathbf{GL}_n$ that is not faithful on $\mathbf{SL}_n$. Then infinitely many elements of the Galois cohomology set $H^1 \bigl( \mathbb{Q};\ \mathbf{G}/\mathrm{ker}(\rho) \bigr)$ are trivial at $\mathbb{Q}_p$. The $\mathbb{Q}$ forms corresponding to these cohomology classes produce non-conjugate subgroups $L$. So, even for $\mathbf{GL}_n$, many representations have infinitely many orbits. $\endgroup$ Feb 25 '15 at 9:24
  • $\begingroup$ More generally, I think $\mathbf{G}(\mathbb{Q}_p) \times \mathbb{Q}_p^\times$ has only finitely many orbits on the set of vector-space lattices whenever $\mathbf{G}$ is split, semisimple, and simply connected, and $\rho$ is faithful and irreducible. So, for non-adjoint groups, it was not accurate for me to say that there will almost always be infinitely many orbits. $\endgroup$ Feb 26 '15 at 3:31
  • $\begingroup$ Dear Dave, thank you! It is interesting and shall take some time for me to digest; but it does solve (probably) my question so I "accepted" your answer. $\endgroup$
    – mmm
    Feb 26 '15 at 9:24
  • $\begingroup$ Dear Dave, oh. It seems the following simple argument works. $L=A({\Bbb Q}{\Bbb Q}\times ... {\Bbb Q})$ is vector-space lattice iff $A^{-1} G({\Bbb Q}_p)A=G({\Bbb Q}_p)$, i.e. $A\in GL({\Bbb Q}^n)$ is in the normaliser of $G({\Bbb Q}_p)$. $<=$ is easy; $=>$: $ A^{-1}Stab(L)A=G({\Bbb Q}_p)\cap GL({\Bbb Q}^n)$; pass to the closures and use $Stab(L)$ is dense in $G(({\Bbb Q}_p)$: $A^{-1}G(({\Bbb Q}_p)A=cl(A^{-1}Stab(L)A)=cl(G({\Bbb Q}_p)\cap GL({\Bbb Q}^n))\subseteq G({\Bbb Q}_p)$. $\endgroup$
    – mmm
    Feb 26 '15 at 18:43
3
$\begingroup$

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 $L_1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.