## Notes of Alan Reid’s first Oxford lecture 21-03-2017

Profinite rigidity in low dimensions

The third lecture will report on joint work with Martin Bridson and Henry Wilton. This deals with profinite rigidity in the context of 3-manifolds. I will give a crash coarse on profinite completion, and then on 3-manifold topology after so many recent progresses.

1. Residually finite groups

Say a group ${\Gamma}$ is residually if every element is mapped to a nontrivial element in a finite quotient of ${\Gamma}$.

Examples. Finitely generated subgroups of ${Gl(n,{\mathbb C})}$ (Selberg). Fundamental groups of compact 3-manifolds (Perelman, Hempel, Thurston).

Denote by ${\mathcal{C}(\Gamma)}$ the set of isomorphism classes of finite quotients of ${\Gamma}$.

Question. To what extent does this set determine ${\Gamma}$?

This is a rather classical path. For instance, to guess wether a presentation is nontrivial, produce a finite quotient. Also, 3-manifold topology has turned a lot around finite covers.

Definition 1

$\displaystyle \begin{array}{rcl} \mathcal{G}(\Gamma)=\{\Delta\,;\,\mathcal{C}(\Delta)<\mathcal{C}(\Gamma)\}. \end{array}$

Terminology is influenced by the theory of quadratic forms over ${{\mathbb Z}}$, where it is said that two quadratic forms have te same genus if they are equivalent locally, i.e. over ${{\mathbb R}}$ and ${{\mathbb Z}_p}$ for all primes ${p}$. In our context, finite quotients determine every finite portion of the Cayley graph of ${\Gamma}$.

Questions

1. For which ${\Gamma}$ does ${\mathcal{G}(\Gamma)=\{\Gamma\}}$.
2. If not, what can be said about groups in ${\mathcal{G}(\Gamma)}$.

1.1. Examples of rigidity

${\mathcal{G}({\mathbb Z})=\{{\mathbb Z}\}}$. Indeed, if ${\mathcal{C}(\Delta)<\mathcal{C}(\Gamma)}$, then ${\Delta}$ is abelian, all finite quotients are cyclic, this implies that ${\Delta={\mathbb Z}}$.

Such a rigidity property also holds for all finitely generated groups. It implies that ${H_1(\Gamma,{\mathbb Z})}$ can be recovered from the genus of ${\Gamma}$.

1.2. Examples of non-rigidity (Baumslag)

Lemma 2 (Stability Lemma)

1. If ${G\times{\mathbb Z}=H\times{\mathbb Z}}$, then ${\mathcal{C}(G)=\mathcal{C}(H)}$.
2. Let ${N}$ be finitely presented and ${G_\phi=N\times_\phi{\mathbb Z}}$, where ${\phi}$ has finite order in ${Out(N)}$. Let ${k}$ be relatively prime to the order of ${\phi}$. Then ${G_{\phi^k}\times{\mathbb Z}=G_\phi\times{\mathbb Z}}$.

Baumslag used this in the following way. Let ${N={\mathbb Z}/11{\mathbb Z}}$. Then ${N}$ admits an automorphism of order coprime to 1,2,3,4,6. This provides distinct groups which become isomorphic after a direct product with ${{\mathbb Z}}$.

Comment. Let

$\displaystyle \begin{array}{rcl} \Gamma(n)=\bigcap\{H<\Gamma\,;\,[\gamma:N]\leq n\}. \end{array}$

Hence to prove that ${\mathcal{C}(G)=\mathcal{C}(H)}$, it suffices to show that for all ${n}$,

$\displaystyle \begin{array}{rcl} G/G(n)=H/H(n). \end{array}$

If ${G\times{\mathbb Z}=H\times{\mathbb Z}}$, then ${G/G(n)\times{\mathbb Z}/{\mathbb Z}(n)=H/H(n)\times{\mathbb Z}/{\mathbb Z}(n)}$. Remak-Krull-Schmidt’s Theorem implies that ${G/G(n)=H/H(n)}$.

Pickel anwered above questions for nilpotent groups. The solvable case was solved by Grunewald-Segal and Pickel.

There exist constructions of lattices in semi-simple Lie groups with ${|\mathcal{G}(\Gamma)|>1}$ (Menny Aka). This is related to the Congruence Subgroup Property plus some number theory.

Open problem. What about ${Sl_3({\mathbb Z})}$.

Remark. ${|\mathcal{G}(\Gamma)|}$ can be infinite (Bridson).

In these lectures, we shall focus on surface groups and 3-manifold groups.

Question (Remeslennikov). What about finitely generated free groups?

Exercise. Recognize a surface group from a free group. Hint: Surface group does not have odd rank finite index subgroups, this leads to a finite nilpotent group which cannot be a quotient of a surface group.