Presentation of special linear group over localizations of the integers

Question

I am looking (for $n,k\in{\mathbb Z}$) for a presentation (in the best of all worlds concretely, as a list of relators) for the group ${\rm SL}_n(R)$ for $R={\mathbb Z}[\frac{1}{k}]=\{\frac{a}{k^l}\mid a\in {\mathbb Z}\}$.

A search in MathSciNet found a paper of Behr and Mennicke(A presentation of the groups PSL(2,p), Can. J. Math. 20, 1432-1438 (1968)) that gives a presentation for the special case of $n=k=2$; (add the diagonal matrix $(2,\frac{1}{2})$ as extra generator and describe its conjugation action on suitable generators of ${\rm SL}_n({\mathbb Z})$); but a reference search did not yield a further result.

Similarly mathOverflow carried the same question question for other rings, but not $R$.

(A follow-up question would be the same question for ${\rm Sp}$)

Answer 1

The question appears to be rather difficult. I'll give some literature references and discuss how it can be approached, and where the difficulties lie.

First, the results mentioned by Luc Guyot in the comments can also be found in Section II.1.4 of Serre's book "Trees". More precisely, there is an amalgam decomposition of $SL_2(\mathbb{Z}[1/p])$ on p.80, and after that a description how to successively write $SL_2(\mathbb{Z}[1/(p_1\cdots p_k)])$ as amalgams of congruence subgroups of $SL_2(\mathbb{Z})$. Then he gives the presentations for $SL_2(\mathbb{Z}[1/2])$ and $SL_2(\mathbb{Z}[1/3])$ and leaves the case $SL_2(\mathbb{Z}[1/6])$ as an exercise.

The general procedure to obtain a presentation of a group $SL_n(\mathbb{Z}[1/(p_1\cdots p_k)])$ would be the following: the group acts on a symmetric space which is a product of $SL_n(\mathbb{R})/SO(n)$ and a copy of the Bruhat-Tits building for $SL_n(\mathbb{Q}_{p_i})$ for each prime $p_i$ that is being inverted. If one can determine a fundamental domain for this action, together with presentations for the stabilizer groups, then one can put all these data together to a presentation of the group. (Of course it's more complicated. The case $SL_2$, where the corresponding buildings are trees is discussed in detail in Serre's book. I'd have to dig for good references outlining the general case...) Alternatively, one can add each of the primes to be inverted successively, then only the action of the group on a single building has to be considered, but then the stabilizers are more complicated.

So, where are the difficulties. First, determining the fundamental domain plus all the stabilizer groups isn't easy. Next, knowing presentations for the stabilizer subgroups isn't easy. Finally, the group $SL_n(\mathbb{Z}[1/(p_1\cdots p_k)])$ isn't usually the amalgam of the stabilizer subgroups because the quotient of the symmetric space might have nontrivial topology. For instance, in the tree case, the quotient might contain nontrivial loops, in which case we get HNN-extensions instead of simple amalagams. All these questions have to be solved to get a presentation; I guess this is one of the reasons why there appear to be no general results on this question in the literature.

Something more can be said, though. The technique of using the action on the symmetric space, computing fundamental domains and stabilizers and so forth is also used to compute group homology. There is this paper of Williams and Wisner which contains partial information on the abelianization of $SL_2(\mathbb{Z}[1/k])$ for general $k$:

• F. Williams and R.J. Wisner. Cohomology of certain congruence subgroups of the modular group. Proc. Amer. Math. Soc. 126 (1998), 1331--1336. (link to paper on journal website)

In particular, Corollary 4.4 of their paper states that the abelianization of $PSL_2(\mathbb{Z}[1/n])$ is a subgroup of an explicit torsion group. Note that nontriviality of the abelianization would already tell us something about necessary generators for the group. Unfortunately, they don't determine the subgroup or its order precisely, because that would require identification of a lot of higher differentials in the spectral sequence they are considering...

There is also an explicit GAP-implementation of an algorithm constructing resolutions for $SL_2(\mathbb{Z}[1/m])$ to compute homology:

• B.A. Tuan and G. Ellis. The homology of $SL_2(\mathbb{Z}[1/m])$ for small m. J. Algebra 408 (2014), 102–108. (link to paper on journal website)

The paper contains a table listing the abelianizations for small $m$. Maybe it's possible to extract some preliminary information on the shape of the presentation from the resolution constructed by their algorithm. Note that they also use the technique suggested by Serre: write the group as amalgam and then use that to produce the resolution. The size and representatives for elements in $H_2$ might also (from the point of view of Steinberg groups and K-theory) say something about relations in the presentation.

Answer 2

In my recent paper (arXiv:2401.08146), I give a new presentation for $\operatorname{SL}_2(\mathbf{Z}[\frac{1}{2}])$. This group is generated by the two matrices $$ A = \begin{pmatrix}1 & 0 \\ 1 & 1\end{pmatrix}, \quad Q = \begin{pmatrix} 1 & -1/2 \\ 0 & 1 \end{pmatrix} $$ and if one maps $A \mapsto x$ and $Q \mapsto y$, then one obtains the presentation $$ \operatorname{SL}_2(\mathbf{Z}[\frac{1}{2}]) \cong \langle x, y \mid x^2yx^2 = yx^2y, \: y^2xy^2 = xy^2x, \: (xy^2x)^4 = 1\rangle. $$ Replacing all exponents which are $2$ by $m$ for $m \geq 3$ does not give an isomorphism, however (but there is a surjection from the resulting group; see the article).

Note that the abelianizations mentioned by @MatthiasWendt in his answer are also completely determined in my article, namely: $$ \operatorname{SL}_2(\mathbf{Z}[\frac{1}{m}])^{\operatorname{ab}} \cong \begin{cases} 1 & \text{if } 6 \mid m, \\ \mathbf{Z} / 3 \mathbf{Z} & \text{if } 2 \mid m \text{ and } \gcd(3,m)=1,\\ \mathbf{Z} / 4 \mathbf{Z} & \text{if } 3 \mid m \text{ and } \gcd(2,m)=1, \\ \mathbf{Z} / 12 \mathbf{Z} & \text{if } \gcd(m,6)=1. \end{cases} $$ for all $m \geq 1$. (The same result was obtained independently - the very previous day! - by Mirzaii & Torres Pérez at arXiv:2401.06330).