Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:41:22Z http://mathoverflow.net/feeds/question/36777 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36777/congruence-subgroups-as-open-subgroups-of-the-modular-group-under-the-right-topol Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology David Corwin 2010-08-26T16:36:52Z 2012-04-02T16:08:19Z <p>It occurred to me that a subgroup of the modular group $\Gamma$ is a congruence subgroup iff it contains a subgroup of the form $\Gamma(N)$, while a subgroup of a general topological group is open iff it contains an open subgroup. This suggests making a topology on the modular group $\Gamma$ with the subgroups $\Gamma(N)$ as a basis of open neighborhoods of the origin so that $\Gamma$ becomes a topological group. It would then follow that a subgroup of $\Gamma$ is a congruence subgroup iff it is open.</p> <p>Furthermore, for any $\gamma \in \Gamma$ not equal to the identity, there exists $N$ such that $\gamma \notin \Gamma(N)$, so this topology is Hausdorff, even totally disconnected.</p> <p>I was inspired in part by <a href="http://mathoverflow.net/questions/20929/distinguishing-congruence-subgroups-of-the-modular-group" rel="nofollow">this thread</a> and looked at <a href="http://www.ams.org/journals/proc/1996-124-05/S0002-9939-96-03496-X/S0002-9939-96-03496-X.pdf" rel="nofollow">this paper</a> but could not find anything about this idea.</p> <p>Has anyone considered this topology? Does it provide insight into the problem of determining whether a group is a congruence subgroup?</p> http://mathoverflow.net/questions/36777/congruence-subgroups-as-open-subgroups-of-the-modular-group-under-the-right-topol/36781#36781 Answer by HW for Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology HW 2010-08-26T16:53:35Z 2010-08-26T17:03:21Z <p>It's called the congruence topology, and is (obviously) always at least as coarse as the profinite topology. If your group has the Congruence Subgroup Property (the modular group doesn't, but $SL_n(\mathbb{Z})$ does for $n>2$) then it's the same as the profinite topology.</p> <p>A google search found, for instance, <a href="http://books.google.com/books?id=1D6crOEoRFEC&amp;pg=PA200&amp;lpg=PA200&amp;dq=congruence+topology&amp;source=bl&amp;ots=d-mel58Uo0&amp;sig=Lxim14zcfsKZdMxgO2gU9otjPGY&amp;hl=en&amp;ei=TZx2TN3sGIW-sQPjktCgDQ&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=11&amp;ved=0CFEQ6AEwCg#v=onepage&amp;q=congruence%2520topology&amp;f=false" rel="nofollow">Section 7.3</a> of <em>Algebraic theory of the Bianchi groups</em> by Benjamin Fine.</p> http://mathoverflow.net/questions/36777/congruence-subgroups-as-open-subgroups-of-the-modular-group-under-the-right-topol/36782#36782 Answer by Jim Humphreys for Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology Jim Humphreys 2010-08-26T17:06:27Z 2010-08-26T17:12:06Z <p>To expand Henry Wilton's concise answer, the Congruence Subgroup Problem has a distinguished history including important work by Serre and a number of others (exploiting effectively the congruence topology). See for example: MR0272790 (42 #7671) 14.50, Serre, Jean-Pierre, Le probl`eme des groupes de congruence pour SL2. (French) Ann. of Math. (2) 92 1970 489–527. </p> <p>This sort of topology on a group originates earlier, but the application here is highly original. </p> <p>ADDED: Like many other journal articles, the one mentioned here by Serre is available in PDF format but only through JSTOR (or other library resource). There is a lot of literature, including my 1980 Springer Lecture Notes 789 <em>Arithmetic Groups</em> which cover some of the background as well as an expository account of Matsumoto's thesis. </p> http://mathoverflow.net/questions/36777/congruence-subgroups-as-open-subgroups-of-the-modular-group-under-the-right-topol/91830#91830 Answer by Marc Palm for Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology Marc Palm 2012-03-21T15:47:57Z 2012-04-02T16:08:19Z <p>You can take the profinite completion $\widehat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$ of $\mathbb{Z}$, then open subgroups of $G( \widehat{\mathbb{Z}})$ correspond to congruence subgroups in $G(\mathbb{Z})$.</p> <p>The identification is easy:</p> <p>$$\Gamma \; congruence \leftrightarrow K \; open$$</p> <p>"$\rightarrow$": Assume you have a congruence subgroup $\Gamma \subset \Gamma(N)$, then we can consider $$\Gamma / \Gamma(N) \subset SL_2(\mathbb{Z} / N) = \prod\limits_{p^k || N} SL_2(\mathbb{Z} / p^k ).$$ Define $K = K(\Gamma)$ as the pullback of $\Gamma / \Gamma(N)$ along the surjection <code>$$p: \prod\limits_{p} SL_2(\mathbb{Z}_p) \rightarrow \prod\limits_{p^k || N} SL_2(\mathbb{Z} / p^k )$$</code></p> <p>"$\leftarrow$": Pick $\Gamma = K \cap SL_2(\mathbb{Q})$, where $SL_2( \mathbb{Q})$ is diagonal subgroup $\prod\limits_p SL_2(\mathbb{Q}_{p})$.</p>