## Notes of Alan Reid’s second Oxford lecture 23-03-2017

Profinite rigidity in low dimensions, II

1. Profinite completion

We organize finite quotients of a group ${\Gamma}$. The set ${\mathcal{N}}$ of normal finite index subgroups of ${\Gamma}$ is a directed set, where ${N_1\langle N_2}$ if ${N_1}$ contains ${N_2}$ (unfortunate conflict of notation).

The profinite completion ${\hat\Gamma}$ of ${\Gamma}$ is the inverse limite of finite quotients ${\Gamma/N}$, ${N\in\mathcal{N}}$. It can be viewed as a subset in the direct product ${\prod_{N\in\mathcal{N}} \Gamma/N}$. It is a compact topological group.

${\Gamma}$ maps to ${\hat\Gamma}$, its image is dense. This map is injective iff ${\Gamma}$ is residually finite.

Nikolov-Segal: finite index subgroups of ${\hat\Gamma}$ are open.

Theorem 1 (Correspondence theorem) There is a 1-1 correspondence between subgroups of finite index of ${\Gamma}$ and open subgroups of ${\hat\Gamma}$. Indices and normality are preserved. It follows that

$\displaystyle \begin{array}{rcl} \mathcal{C}(\Gamma)=\mathcal{C}(\hat\Gamma). \end{array}$

Theorem 2 Let ${\Gamma_1}$ and ${\Gamma_2}$ be finitiely generated residually finite groups. Then ${\mathcal{C}(\Gamma_1)=\mathcal{C}(\Gamma_2)}$ iff ${\hat\Gamma_1\simeq\hat\Gamma_2}$. Hence ${\hat\Gamma}$ can be recovered from the set of finite quotients of ${\Gamma}$.

Definition 3 Say ${\Gamma}$ is profinitely rigid if ${\mathcal{G}=\{\Gamma\}}$.

Conjecturally, free groups, surface groups, finite volume hyperbolic 3-manifold groups are profinitely rigid. Unclear for more general hyperbolic groups.

Not so much is known about the genus. Genus is finite for nilpotent groups. Also often for groups of arithmetic origin. It cost some efforts to find an example of a group with infinite genus (Bridson after Grunewald).

2. Crash course on 3-manifolds

Let ${M}$ be closed, orientable, 3 dimensional. Kneser-Milnor: ${M}$ decomposes as a connected sum of finitely manyprime manifolds (i.e. which cannot we decomposed further). From now on, stick to prime manifolds.

Either ${\pi_1}$ is finite. Then (Perelman) ${M}$ is covered by ${S^3}$. Otherwise, ${\tilde M\simeq{\mathbb R}^3}$ or ${S^2\times {\mathbb R}}$. In the latter case, ${\pi_1={\mathbb Z}}$. From now on, stick to prime manifolds with infinite ${\pi_1}$ different from ${{\mathbb Z}}$.

${M}$ either admits a geometric structure or admits a decomposition along incompressible tori. In the latter case, ${\pi_1}$ has a non-trivial JSJ decomposition. In the former case, ${M}$ is modelled on Euclidean space ${\mathbb{E}^3}$, ${\mathbb{H}^3}$, ${\mathbb{H}^2\times{\mathbb R}}$, Nil, Sol, ${\widetilde{Sl}_2}$. Manifolds modelled on ${\mathbb{E}^3}$, ${\mathbb{H}^2\times{\mathbb R}}$, Nil and ${\widetilde{Sl}_2}$ are Seifert fibered spaces. ${\mathbb{H}^2\times{\mathbb R}}$-manifolds are virtually products surface ${\times}$ circle. Manifolds modelled on ${\mathbb{E}^3}$, Nil and Sol are virtually torus bundles over the circle. ${\widetilde{Sl}_2}$-manifolds are never suface bundles.

Do hyperbolic manifolds virtually fiber over the circle? This has been open for decades. Agol came up with an idea which applied to a small class. Then Wise designed the theory of special cube complexes which fitted nicely with Agol’s initial idea. This has resulted into a proof that all hyperbolic manifolds virtually fiber.

Remark. Hempel has shown that there exist Seifert fibered bundles, which are not profinitely rigid.