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}.


  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.


About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See
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