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.
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.
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 $k$.
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?
