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Let $d$ be a square-free positive integer, and let $\mathcal{O}_d$ be the ring of integers of the quadratic imaginary number field $\mathbb{Q}(\sqrt{-d})$. Consider the Bianchi group $\Gamma_d = \operatorname{SL}_2(\mathcal{O}_d)$. Is $\Gamma_d$ known to be residually solvable? I am also interested in the case of $\operatorname{PSL}_2(\mathcal{O}_d)$.

As mentioned in the comments, $\Gamma_d$ is never residually nilpotent.

(I might be overlooking something easy, sorry!)

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  • $\begingroup$ $\mathrm{SL}_2(\mathbf{Z})$ (and its overgroups) has elements of order 4 and 3 that don't commute, so it's not residually nilpotent. $\endgroup$
    – YCor
    Commented Dec 30, 2023 at 5:23
  • $\begingroup$ For $d=3$, an index 12 subgroup is the figure 8 knot group which is residually solvable since it is free-by-cyclic. So this group is residually solvable. $\endgroup$
    – Ian Agol
    Commented Dec 30, 2023 at 5:28
  • $\begingroup$ @IanAgol is this index 12 subgroup a normal subgroup? otherwise it is unclear how to conclude. $\endgroup$
    – YCor
    Commented Dec 30, 2023 at 5:48
  • $\begingroup$ In some papers (e.g. by Grunewald) the Bianchi group rather means the quotient PSL2 by $\pm 1$. Some little argument might be needed to show that residual solvability holds for SL2 iff it holds for PSL2. $\endgroup$
    – YCor
    Commented Dec 30, 2023 at 5:51
  • $\begingroup$ @YCor No, the figure eight knot subgroup mentioned by IanAgol is quite far from normal. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Dec 30, 2023 at 9:57

3 Answers 3

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This is true for $d=3$. Let $\zeta=\frac{1+\sqrt{-3}}{2}$, $\mathcal{O_3}=\mathbb{Z}[\zeta]$. The principal congruence subgroup of $PSL_2(\mathcal{O}_3)$ of level $1+\zeta$, which divides $3$, is a normal subgroup whose quotient is solvable (see picture of sequence of orbifold covers or prove algebraically).

enter image description here

Principal congruence subgroups are residually $p$, and hence residually nilpotent.

I think Chapter 1 of the thesis of Matthias Goerner should yield more examples.

This should also be true for $d=1$. There is a solvable cover corresponding to the 1-skeleton of an octahedron with edges labeled 2, which is index 2 in a right-angled Coxeter group. A right-angled Coxeter group is residually 2, hence residually nilpotent.

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(Inspired by Ian Agol's answer for d=3)

It seems to be true if $d$ is not $19$ modulo $24$. Indeed in this case $\mathcal{O}_d$ surjects as a ring onto either $\mathbf{Z}/2\mathbf{Z}$ or $\mathbf{Z}/3\mathbf{Z}$ (see below). Since $\mathrm{SL}_2(\mathbf{Z}/p\mathbf{Z})$ is solvable for $p=2,3$ and the principal congruence subgroup is residually nilpotent, we're done.

If $d$ is not $3$ modulo $4$, then $\mathcal{O}_d$ is reduced to $\mathbf{Z}[\sqrt{-d}]$, hence surjects as a ring onto $\mathbf{Z}/2\mathbf{Z}$.

Suppose henceforth that $d$ is $3$ modulo $4$. Then $\mathcal{O}_d$ is generated by $(1+\sqrt{-d})/2$ which has the minimal polynomial $X^2-X+(1+d)/4$.

If $d$ is $7$ modulo $8$, this minimal polynomial is $X^2+X$ modulo 2, so $\mathcal{O}_d$ surjects onto $\mathbf{Z}/2\mathbf{Z}$.

If $d$ is not $1$ modulo $3$ then modulo $3$ the above polynomial is $X^2-X$ or $X^2-X+1$ which has a root modulo $3$, so $\mathcal{O}_d$ surjects onto $\mathbf{Z}/3\mathbf{Z}$.

If $d$ is $19$ modulo $24$ (i.e., $3$ modulo $8$ and $1$ modulo $3$), the above minimal polynomial is $X^2+X+1$ modulo $2$ and $X^2-X-1$ modulo $3$, both irreducible, so $\mathcal{O}_d$ surjects onto neither $\mathbf{Z}/2\mathbf{Z}$ nor $\mathbf{Z}/3\mathbf{Z}$, and $\mathrm{SL}_2$ of larger finite fields are not solvable, so the argument cannot work. But this doesn't imply any conclusion. What about $d=19$?

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    $\begingroup$ It seems like $O_7$ surjects $Z/2$ since 2 splits. So I’m not sure your case analysis is complete. $\endgroup$
    – Ian Agol
    Commented Jan 2 at 16:11
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    $\begingroup$ @IanAgol thanks a lot: indeed this covers when $d$ is $7$ modulo $24$, which I indeed missed. I corrected the post: precisely $\mathcal{O}_d$ has no quotient field of cardinal $\le 3$ iff $d$ is $19$ mod $24$. $\endgroup$
    – YCor
    Commented Jan 2 at 16:23
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    $\begingroup$ This is very nice, thanks! Just one question: why is the principal congruence subgroup residually $p$? $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Jan 3 at 4:57
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    $\begingroup$ @Carl-FredrikNybergBrodda Let $G$ be $\mathrm{GL}_d(R)$ for $R$ a noetherian domain, and $M$ a maximal ideal in $R$. Let $G_n$ be the kernel of $\mathrm{GL}_d(R)\to\mathrm{GL}_d(R/M^nR)$. So $\mathrm{GL}_d(R)=G_0\supset G_1\supset G_2\dots$. Krull's intersection theorem ensures that $\bigcap_n M^n=0$, so $\bigcap_n G_n=1$. Then $G_n/G_{n+1}$ is naturally a finite-dim vector space over $R/M$. Hence if $R/M$ a finite field of characteristic $p$, then $G_n/G_{n+1}$ is a finite elementary abelian $p$-group, so $G_1/G_n$ is a finite $p$-group for all $n\ge 1$, and $G_1$ is residually-$p$. $\endgroup$
    – YCor
    Commented Jan 3 at 8:12
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    $\begingroup$ @YCor Thank you for spelling that out! (Implicit: every principal congruence subgroup is contained in one of level $q$ where $q$ is prime, and being residually $p$ is inherited by subgroups). $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Jan 3 at 9:13
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I am posting this answer (slightly edited) on behalf of Alan Reid:

$PSL(2,\mathcal{O}_{19})$ is not residually solvable. In fact it
doesn't have any non-abelian solvable quotients.

Here is the proof.

Let $G=PSL(2,\mathcal{O}_{19})$ which can be presented as

$<a,b,t,u | a^2, (ta)^3, b^3, (bt^{-1})^3, (ab)^2, (at^{-1}ubu^{-1})^2, [t,u]>$.

See p. 75 of

Swan, Richard G., Generators and relations for certain special linear groups, Adv. Math. 6, 1-77 (1971). ZBL0221.20060

which gives a presentation of $SL_2(\mathcal{O}_{19})$ (kill the center $J$ to get the above presentation of $PSL_2(\mathcal{O}_{19})$).

extract from Swan’s paper

It is easy to check that the abelianization of $G$ is $\mathbb{Z}$ (generated by
the image of $u$).

In fact setting $a=1$ (resp. $b=1$) gives the quotient map onto $\mathbb{Z}$.
Hence $K=[G,G]$ is generated by $G$-conjugates of $a$ (order 2) and $G$-conjugates $b$ (order 3), i.e. $K=\langle\langle a\rangle\rangle=\langle\langle b \rangle\rangle$, and so it follows that $K^{\rm ab}$ must be trivial, i.e. $K$ is perfect.

Thus if $S$ is solvable and $f:G\to S$ is an epimorphism then $f_{|K}$ has
solvable image hence trivial (since its perfect). Hence $S$ is cyclic.

Addendum: Alan found an orbifold description in a paper of Mark Baker Link Complements and the Bianchi Modular Groups

enter image description here

One sees that killing the order 2 or order 3 orbifold locus in the middle edges of the figure gives a bad orbifold whose fundamental group is $\mathbb{Z}$, giving a geometric proof of the above algebraic argument.

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  • $\begingroup$ That's beautifully simple - thank you (and say hi to Alan from me)! $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Jan 25 at 6:17
  • $\begingroup$ Nice. The next cases are thus $d=43$, $67$, $91$... are there available presentations then? $\endgroup$
    – YCor
    Commented Jan 25 at 8:50
  • $\begingroup$ @YCor it looks like Hatcher has worked these out for PGL from data of Riley (or one couple probably work directly with Riley’s data to get a presentation). pi.math.cornell.edu/~hatcher/Papers/Bianchi.pdf $\endgroup$
    – Ian Agol
    Commented Jan 25 at 13:35
  • $\begingroup$ I checked that a similar perfect subgroup exists (the normal closure of elements of order 2 and 3) when d=43. $\endgroup$
    – Ian Agol
    Commented Jan 26 at 15:32

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