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Let $K$ be a global field and set $O := \prod_{v\nmid \infty} O_v$ where $v$ runs over the finite places of $K$. Equip $\mathrm{GL}_n(O) = \prod_v \mathrm{GL}_n(O_v)$ with the product of the $v$-adic topologies and suppose that $H \le \mathrm{GL}_n(O)$ is a subgroup of finite index.

  1. Is $H$ necessarily open?
  2. Is $H$ necessarily a congruence subgroup? (I.e., does it contain $\mathrm{Ker}(\mathrm{GL}_n(O) \rightarrow \mathrm{GL}_n(O/NO))$ for some integer $N$?)

A positive answer to 2. would imply a positive answer to 1.

I will accept a complete answer to either 1. or 2. as an answer. Also, I would be especially happy if you discussed a generalization of 1. and 2. for arbitrary reductive group schemes (or even a more general class of group schemes) over the ring of integers of $K$.

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  • $\begingroup$ Can you give any example where you can make an $H$ that you know is finite index but for which you can't see the answers to either #1 or #2 immediately? That is: does this question have some motivation, or is it just idle curiosity? $\endgroup$
    – user27920
    Sep 6, 2014 at 0:57
  • $\begingroup$ The question arose by wondering whether 1.1 2) of arxiv.org/abs/math/0003131 is missing the condition that $K_{\varphi}$ be open. $\endgroup$
    – Question Mark
    Sep 7, 2014 at 6:11
  • $\begingroup$ Given the context, the author definitely meant to include openness; for such situations I'm not aware of any reason why one would care about such a finite-index subgroup $K_{\varphi}$ not assumed/known to be open). So in the end, your question is somewhat moot; it is always best to give motivation for questions asked. $\endgroup$
    – user27920
    Sep 9, 2014 at 13:50

2 Answers 2

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The answer to 1 is no. For example each $GL_n(O_v)$ has a continuous map to $\{ \pm 1\}$ so your product has a quotient isomorphic to $F_2^{\mathbf{N}}$. Zorn's lemma will give you a non continuous linear form on that, and the pullback of the quotient is a non closed subgroup of index $2$.

Edit: the pullback of the kernel, not of the "quotient"

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    $\begingroup$ And the answer to #2 is "yes" via congruence against nonzero ideals, not integers (if char. $>0$): if $S$ is a non-empty finite set of places containing the arch. places and $G$ is an affine $O_{K,S}$-group of finite type then congruence subgroups of $G(\mathbf{A}_K^S)$ constitute a base of opens around 1. Indeed, if $X$ is any affine $O_{K,S}$-scheme of finite type and ${O} := \prod_{v\not\in S} O_{K,v}$ then $X(O)$ is open in $X(\mathbf{A}_K^S)$ and a base of opens around $x\in X(O)$ is given by fibers over $x\bmod J$ under $X(O)\rightarrow X(O/JO)$ for nonzero ideals $J$ of $O_{K,S}$. $\endgroup$
    – user27920
    Sep 5, 2014 at 6:46
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In view of Laurent's answer, one may ask more generally for a criterion on a connected reductive $K$-group $G$ to ensure that for all compact open subgroups $U$ of $G(\mathbf{A}_K^S)$ (for a finite set $S$ of places of $K$ containing the arch. places) the finite-index subgroups of $U$ are necessarily open. If $G$ admits a non-trivial torus quotient then one typically gets an obstruction as in Laurent's answer. So assume $G$ is semisimple. Let $\widetilde{G} \rightarrow G$ be the simply connected central cover, with kernel $\mu$. If $\mu \ne 1$ then elementary "spreading out" arguments and Lang's theorem (on vanishing of degree-1 cohomology of smooth connected groups over finite fields) show that $U$ again admits a massive commutative quotient (related to Galois cohomology of $\mu$ over dvr's, roughly speaking), so problems arise again.

But what if $G$ is simply connected (such as ${\rm{SL}}_n$, not ${\rm{GL}}_n$)? Then there's no "obvious" commutative obstruction, and at least in characteristic 0 one might hope there is no obstruction. It comes down to a question like the following: is an index-N normal subgroup of $\prod_{p\nmid N} {\rm{SL}}_6(\mathbf{F}_p)$ necessarily open? (It is easy to show that such a subgroup contains ${\rm{SL}}_6(\mathbf{F}_p)$ for all but finitely many $p$, since such groups are generated by $\mathbf{F}_p$-points of unipotent radicals, a method which generalizes to all simply connected semisimple groups over finite fields of size at least 4, but that doesn't give access to the direct product.)

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