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 is residually if every element is mapped to a nontrivial element in a finite quotient of .
Examples. Finitely generated subgroups of (Selberg). Fundamental groups of compact 3-manifolds (Perelman, Hempel, Thurston).
Denote by the set of isomorphism classes of finite quotients of .
Question. To what extent does this set determine ?
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.
Terminology is influenced by the theory of quadratic forms over , where it is said that two quadratic forms have te same genus if they are equivalent locally, i.e. over and for all primes . In our context, finite quotients determine every finite portion of the Cayley graph of .
- For which does .
- If not, what can be said about groups in .
1.1. Examples of rigidity
. Indeed, if , then is abelian, all finite quotients are cyclic, this implies that .
Such a rigidity property also holds for all finitely generated groups. It implies that can be recovered from the genus of .
1.2. Examples of non-rigidity (Baumslag)
Lemma 2 (Stability Lemma)
- If , then .
- Let be finitely presented and , where has finite order in . Let be relatively prime to the order of . Then .
Baumslag used this in the following way. Let . Then admits an automorphism of order coprime to 1,2,3,4,6. This provides distinct groups which become isomorphic after a direct product with .
Hence to prove that , it suffices to show that for all ,
If , then . Remak-Krull-Schmidt’s Theorem implies that .
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 (Menny Aka). This is related to the Congruence Subgroup Property plus some number theory.
Open problem. What about .
Remark. 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.