# Generating congruence subgroups of SL_n over totally imaginary number rings

Fix some $n \geq 3$. Let $k$ be an algebraic number field with ring of integers $\mathcal{O}$ and let $\alpha$ be an ideal of $\mathcal{O}$. Define $\text{SL}_n(\mathcal{O},\alpha)$ to be the congruence subgroup defined by $\alpha$, i.e. the kernel of the natural map $\text{SL}_n(\mathcal{O}) \rightarrow \text{SL}_n(\mathcal{O}/\alpha)$. I am interested in finding generators for $\text{SL}_n(\mathcal{O},\alpha)$.

This was almost'' done by Bass-Milnor-Serre, who proved the following. Letting $\text{E}_n(\mathcal{O},\alpha)$ be the normal subgroup of $\text{SL}_n(\mathcal{O},\alpha)$ generated by elementary matrices whose off-diagonal entries lie in $\alpha$, we have

1. The group $\text{Q}_n(\mathcal{O},\alpha) := \text{SL}_n(\mathcal{O},\alpha) / E_n(\mathcal{O},\alpha)$ is a finite cyclic group which is isomorphic to a subgroup of the roots of unity lying in $k$.
2. If $k$ has a real embedding, then $\text{Q}_n(\mathcal{O},\alpha)=0$.

The only remaining problem, then, is when $k$ does not have a real embedding, so when $k$ is totally imaginary. My question is then in that case, does anyone know an explicit matrix (or matrices) in $\mathcal{SL}_n(\mathcal{O},\alpha)$ which projects to an element (or elements) which generates $\text{Q}_n(\mathcal{O},\alpha)$? Bass-Milnor-Serre also show that we may take said matrices to lie in $\text{SL}_2(\mathcal{O})$, but I do not demand this.

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I don't know the answer (I haven't looked at Bass-Milnor-Serre), but clearly $Q_n(\mathcal{O},\alpha)$ is a quotient of the abelianization $H_1(G)=G/[G,G]$, where $G=\mathcal{SL_n(O,\alpha)}$. It's a fact that $SL_n(\mathcal{O},\alpha)/SL_n(\mathcal{O},\alpha^2)$ is abelian (basically the $\alpha$-adic expansion converts products to sums), so I would look for quotients of this group (I'm not sure what $G/[G,G]$ is in this case). – Ian Agol Oct 11 '13 at 3:44
@Ian Agol : Bass-Milnor-Serre actually say which finite cyclic group it is, though the answer is a little complicated, and they give an explicit description of the isomorphism between $Q_n(\mathcal{O},\alpha)$ in terms of power residue symbols. But I can't extract explicit matrices from their description. I also don't know what the abelianization of $G$ is in this case; for the rings of integers where I know how to calculate the abelianization of $G$, that calculation depends on the CSP holding (so it doesn't work in the totally imaginary case). – Frank Oct 11 '13 at 3:51
(in particular, I think that it is definitely false in the totally imaginary case that the commutator subgroup of $\text{SL}_n(\mathcal{O},\alpha)$ is $\text{SL}_n(\mathcal{O},\alpha^2)$). – Frank Oct 11 '13 at 4:00