Let $ G $ be a simple linear algebraic group. Let $ G_\mathbb{R} $ be the real points of $ G $. Let $ G_\mathbb{Z} $ be the integer points of $ G $. Is $ G_\mathbb{Z} $ a maximal closed subgroup? In other words, are the only closed subgroups $ H $ of $ G_\mathbb{R} $ such that $$ G_\mathbb{Z} \subset H \subset G_\mathbb{R} $$ just $ G_\mathbb{R} $ and $ G_{\mathbb{Z}} $?

The answer is yes for for $\operatorname{SO}_3$. $ \operatorname{SO}_3(\mathbb{Z}) $ is the 24 element octohedral group (isomorphic to the symmetric group $ S_4 $) and this group is a maximal finite subgroup of rotations and indeed it turns out that adding any other rotation and taking the closure will generate all of $ \operatorname{SO}_3(\mathbb{R})$.

The group of integer points is also maximal in $ \operatorname{SL}_n(\mathbb R) $. See the answer by YCor given to Is $\operatorname{SL}_2(\mathbb Z)$ a maximal discrete subgroup in $\operatorname{SL}_2(\mathbb R)$?.

Although this only states that $ \operatorname{SL}_n(\mathbb{Z}) $ is maximal among discrete subgroups I'm guessing that it is in fact maximal among closed subgroups as well in this case.

  • 2
    $\begingroup$ Where by “no” you mean “yes”? $\endgroup$ Oct 14 at 1:22
  • $\begingroup$ haha oops I just fixed it with an edit $\endgroup$ Oct 14 at 2:29
  • $\begingroup$ By "Integer points are in dense for $\operatorname{SO}_3$", do you mean "the integer points form a maximal closed subgroup of the real points when $G = \operatorname{SO}_3$" or something? \\ Also, the hypothesis 'simple' should probably go in the title; otherwise the additive group $G$ is a counterexample. $\endgroup$
    – LSpice
    Oct 14 at 2:57
  • $\begingroup$ I just meant maximal closed. In my head I say "almost dense" instead of maximal closed because I think it's more interesting to frame it as "this subgroup is so close to being dense that if you add any single new element and take the closure of the group generated that way you get every element of $G_\mathbb{R}$". So I use "maximal closed subgroup" and "almost dense subgroup" interchangeably and that was supposed to say almost dense $\endgroup$ Oct 14 at 12:33

Your guess is correct: if $H$ is a closed subgroup containing $\operatorname{SL}_n({\mathbb Z})$,then it is a Lie subgroup. If $\mathfrak h$ is its Lie algebra, then $\mathfrak h$ is stable under the adjoint action of $SL_n(\mathbb Z)$ and hence under all of $\operatorname{SL}_n(\mathbb R)$ (since $\operatorname{SL}_n(\mathbb Z)$ is Zariski dense in $\operatorname{SL}_n(\mathbb R)$). If $\mathfrak h$ is non-zero, that means, by the simplicity of $\operatorname{SL}_n(\mathbb R)$, that $\mathfrak h$ is the Lie algebra of $\operatorname{SL}_n(\mathbb R)$ and hence $H$ is all of $\operatorname{SL}_n(\mathbb R)$.

If $\mathfrak h =0$, then $H$ is discrete and you have accepted the result in this case.

In the generality that you have asked, $G(\mathbb Z)$ need not be maximal in $G(\mathbb R)$; if $G$ is a simply connected semi-simple algebraic group over $\mathbb Q$ (and is $\mathbb Q$-simple), with $G(\mathbb R)$ non-compact, and $K_p$ is a maximal compact open subgroup of $G(\mathbb Q_p)$ for each prime $p$, then the intersection of $G(\mathbb Q)$ with all the $K_p$ may be "called" $G(\mathbb Z)$ and the same argument goes through to say that $G(\mathbb Z)$ is maximal among closed subgroups of $G(\mathbb R)$.

  • $\begingroup$ Wow this helps alot! For the second part of your answer could you give an example of a simple $ G$ with $ G_\mathbb{Z} $ not a maximal closed subgroup in $G_\mathbb{R} $? Can we take $ G$ to be an indefinite signature orthogonal group like $ SO_{4,1} $ perhaps? Can you also give a sense of what the group of integer points look like (is it a free product like modular group etc...) . I think one of the best ways to answer a question is with a specific example of groups! $\endgroup$ Oct 14 at 12:48
  • 1
    $\begingroup$ @IanGershonTeixeira Note that $G_\mathbf{Z}$ is well-defined only if you fix $G\subset\mathrm{GL}_n$ (or, alternatively, if you define it as scheme over $\mathbf{Z}$), but not starting from $G$ an abstract $\mathbf{R}$-group or even $\mathbf{Q}$-group. In particular, saying $G=\mathrm{SO}_{4,1}$ is highly ambiguous, since you have a lot of choice about the integral form. $\endgroup$
    – YCor
    Oct 14 at 12:54
  • $\begingroup$ Ok so there are many options for integral forms. Can you pick one and explain what the integer points look like and show they are not dense in the real points? I love specific examples of groups! $\endgroup$ Oct 14 at 13:07
  • $\begingroup$ @IanGershonTeixeira, are you saying dense when you mean maximal closed? $\endgroup$
    – LSpice
    Oct 14 at 13:14
  • $\begingroup$ Omg I'm so sorry I keep doing that I'll edit it. Oh wait its been too long since I posted I can't edit :( but yes same mental slip really sorry you guys ya I mean maximal closed or equivalently "almost dense" $\endgroup$ Oct 14 at 13:16

Regarding cases where $G(\mathbb{Z})$ is not maximal in $G(\mathbb{R})$: If I am just allowed to take $G$ to be any simple group scheme over $\mathbb{Z}$, then I take $$\left\{ (X,Y) : X \left[ \begin{smallmatrix} 2&0 \\ 0&1 \\ \end{smallmatrix} \right] = \left[ \begin{smallmatrix} 2&0 \\ 0&1 \\ \end{smallmatrix} \right] Y,\ \det X = \det Y = 1 \right\} \subset (SL_2)^2.$$ In other words, this is the functor which, to any commutative ring $R$, assigns the set of solutions to these equations in pairs of $2 \times 2$ matrices over $R$.

Over $\mathbb{Z}[1/2]$, this is just isomorphic to $SL_2$, since we can parameterize it as $$(X,\ \left[\begin{smallmatrix} 2&0 \\ 0&1 \\ \end{smallmatrix}\right] X \left[\begin{smallmatrix} 1/2&0 \\ 0&1 \\ \end{smallmatrix}\right] ).$$ However, the pair $$(\left[\begin{smallmatrix} a&b \\ c&d \end{smallmatrix}\right],\ \left[\begin{smallmatrix} a&2b \\ c/2&d \end{smallmatrix}\right])$$ is a $\mathbb{Z}$-valued point if and only if $c \equiv 0 \bmod 2$, so $G(\mathbb{Z})$ is the subgroup $\Gamma_0(2)$ of $SL_2(\mathbb{Z})$.

I do not know an example which does not seem artificial like this.

  • $\begingroup$ Wow this is a fabulous counterexample I feel like I'm learning alot! Thanks so much. (Also I think there is a tiny typo in the last line the second matrix in the pair should be [ 2a,b,c, d/2] not [a,2b,c/2,d]) $\endgroup$ Oct 14 at 13:35
  • $\begingroup$ Thanks for the correction. Actually, the typo was in a different place. See if this works. $\endgroup$ Oct 14 at 13:51
  • $\begingroup$ ah I see you are trying to arrange things to get $ \Gamma_0(2) $. I think you still have a typo but if you change your equation to $ X diag(1,2)=diag(1,2)Y $ then it should work out. $\endgroup$ Oct 14 at 19:14

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.