MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathcal{O}$ be the ring of integers in an algebraic number field. Is $\text{SL}_2(\mathcal{O})$ generated by elementary matrices? If it isn't, is there any other natural generating set for it?

The usual argument shows that this is true for $\mathcal{O} = \mathbb{Z}$ (or, more generally, a Euclidean domain). However, I haven't been able to generalize this to other rings of integers.

share|cite|improve this question
1  
See uni-math.gwdg.de/nica/useiqr.pdf. – Guntram Aug 29 '12 at 16:53
3  
@Guntram : OK, reading that article it looks like it follows from results of Cohn and Vaserstein that it is so generated if and only if $\mathcal{O}$ is not a non-Euclidean ring of integers in an imaginary quadratic field. Why did you post this in the comments instead of as an answer? – Sue Aug 29 '12 at 16:55
4  
@Yves Cornulier : It is the right statement. Observe the phrase non-Euclidean in my comment. If you read Nica's survey even more carefully, you'd see that the imaginary quadratic fields for which there is elementary generation are exactly the Euclidean ones. – Sue Aug 30 '12 at 16:13
1  
@Sue: sorry, you're right. I got confused between $Z[\sqrt{-d}]$ and the ring of integers. Anyway, as an answer to your first comment: you can write an answer based on the Nica's survey (Guntram's link) and put it Community Wiki. (Such a short answer as "see (link)" is not perennial insofar as the link can disappear.) – YCor Aug 30 '12 at 19:48
1  
well, this is what happens when you are careful to properly put a theorem in its context: it makes the paper look like a survey :) well, the linked paper is not a survey, its main goal is to prove a theorem (thm 1.5). subsequently, and unfortunately, it turned out to be a theorem which could be deduced from other, more high-powered results in the literature (see comments following thm 1.5). – BN2 Sep 20 '13 at 10:55

L. N. Vaserstein's theorem [2] asserts that if $R$ is a Dedekind ring of arithmetic type with infinitely many units then $SL_2(R) = E_2(R)$ holds. In other words $R$ is a $GE_2$-ring in the sense of P. M. Cohn [1]. Here $E_2(R)$ denotes the subgroup of $SL_2(R)$ generated by the elementary matrices (also called transvections), i.e., the matrices of the form $\begin{pmatrix} 1 & r \\ 0 & 1\end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ r & 1\end{pmatrix}$ with $r \in R$. Since the class of Dedekind rings of arithmetic type comprises the class of rings of integers of algebraic number fields, Vaserstein’s theorem answers OP’s question in the positive when the unit group is infinite.

But you may object, like B. Liehl [3], that there is a gap in Vaserstein’s proof. Fortunately A. Leutbecher [Section 2, 3] corrected this gap and B. Liehl [3] subsequently extended Vaserstein’s theorem to orders of arithmetic type with infinitely many units.

By Dirichlet’s unit theorem, a ring of integers $R$ of an algebraic number field $K$ has finitely many units if and only if $K$ is either the field of rationals $\mathbb{Q}$ or an imaginary quadratic number field. Under this assumption P. M. Cohn proved that $SL_2(R) = E_2(R)$ if and only if $R$ is Euclidean with respect to complex modulus. Thus $\mathbb{Z}[\frac{1 + \sqrt{-19}}{2}]$ is a PID which is not a $GE_2$-ring. Remarkably, the subgroup $E_2(R)$ is non-normal in $SL_2(R)$ and has infinite index if $R$ is the ring of integers of an imaginary quadratic number field which is not Euclidean, see [Theorem 1.5, 6]. The proof of the latter exhibits infinitely many distinct coset representatives for the quotient $SL_2(R)/E_2(R)$, namely the so-called $S_{x, y}$ matrices, but not a transversal, a priori. The ring of integers $R$ of $\mathbb{Q}(\sqrt{-D})$ for $D = 5, 10$ and $14$ is scrutinized in [5] where a group presentation of $SL_2(R)$ is derived. I leave here the second part of OP's question, only partially answered.

These facts were already mentioned under the form of comments to OP’s question and are thoroughly discussed in B. Nica’s paper [6], which I warmly recommend. I put them in this answer because the question seems to be considered unsettled in a closely related MO post.

The following may help clear any doubt: If $R$ is an order in an algebraic number field $K$ which is not imaginary quadratic then $E_2(R)$ is a normal subgroup of $SL_2(R)$ of finite index. This is a result of [4], extracted under this form in [Theorem 1.6, 6]. The fact that the index of $E_2(R)$ in $SL_2(R)$ is actually $1$ if $R$ is moreover a maximal order, i.e., $R$ is the (full) ring of integers of $K$, may easily slip out of one’s mind.


[1] «On the structure of the $GL_2$ of a ring», P. M. Cohn, 1966.
[2] «On the group $SL_2$ over Dedekind rings of arithmetic type», L. N. Vaserstein, 1972.
[3] «Euklidischer Algorithmus und die Gruppe $GL_2$», A. Leutbecher, 1972.
[4] «On the group $SL_2$ over orders of arithmetic type», B. Liehl, 1981.
[5] "On the groups $SL_2(\mathbb{Z}[x])$ and $SL_2(k[x, y])$", F. Grunewald et al., 1994.
[6] «The Unreasonable Slightness of $E_2$ over Imaginary Quadratic Rings», B. Nica, 2013.

share|cite|improve this answer

If $\mathcal O$ be the ring of integers in an algebraic number field, whether $SL_2(\mathcal O)$ is generated by elementary matrices depends on the field $k$:

  • If $k = \Bbb Q$, or $k = \Bbb Q(\sqrt{-D}$ for $D\in\{1,2,3,7,11\}$, then $SL_2(\mathcal O)$ is generated by elementary matrices.

  • If $k = \Bbb Q(\sqrt{-D})$ for $D$ any other squarefree integer, then $SL_2(\mathcal O)$ is not generated by the elementary matrices. However, the index of the subgroup generated by elementary matrices in $SL_2(\mathcal O)$ is finite, so we can add a finite number of non-elementary matrices to the generating set, to ensure $SL_2(\mathcal O)$ will be generated by this set.

  • For all other $k$, $SL_2(\mathcal O)$ is not only generated by elementary matrices, but a bounded number of matrices is enough to generate any matrix. The bound depends on the field $k$.

share|cite|improve this answer
    
what do you mean "the ring generated by elementary matrices"? Do you mean the group? The survey Guntram posted says the index is infinite. – Will Sawin Aug 25 '15 at 2:07

No. If you look in Charles Frohman and Benjamin Fine, "Some Amalgam Structures for Bianchi Groups," 1988, Proceedings of the American Mathematical Society, Vol. 102, No. 2, pp. 221-229, we construct a splitting of $PSl_2(\mathcal{O})$ where we are in a $\mathbb{Q}[\sqrt{-d}]$ for d a positive square free integer that is big enough, and one of the factors is the elementary matrices. The fact that the elementary matrices do not generate was well known. I think I learned it from Morris Newman. Maybe Richard Swan proved it?

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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