Profinite rigidity and hyperbolic four-punctured sphere bundles over the circle

When $M$ (and $N$) are closed, Theorem 3.1 shows that Theorem 1.2 holds. Thus, we assume that $M$ is cusped and fix a fibration of $M$ over $S^{1}$ with fiber a punctured surface $\Sigma_{M}$ and monodromy $\varphi$. By Theorem 3.2, for a fixed isomorphism $\Phi:\widehat{\pi_{1}(M)}\to\widehat{\pi_{1}(N)}$, there is a punctured surface $\Sigma_{N}$ (with the same complexity as $\Sigma_{M}$) which is the fiber of a fibration of $N$ over $S^{1}$ (with monodromy $\psi$) and $\Phi(\widehat{\pi_{1}(\Sigma_{M}))}\cong\widehat{\pi_{1}(\Sigma_{N})}$. The goal is to show that $\Sigma_{M}$ is homeomorphic to $\Sigma_{N}$. The surface $\Sigma_{M}$ is a genus $g_{M}$ surface with $n_{M}$ punctures and the surface $\Sigma_{N}$ is a genus $g_{N}$ surface with $n_{N}$ punctures. Since $\chi(\Sigma_{M})=\chi(\Sigma_{N})$ by Theorem 3.2, it is sufficient to prove that $n_{N}=n_{M}$. To establish this we argue as follows:

By Theorem 3.1 and Theorem 3.2, we get the following diagram

where the $\times\mu:\hat{{\mathbb{Z}}}\to\hat{{\mathbb{Z}}}$ map is multiplication by a profinite unit $\mu$ in $\hat{{\mathbb{Z}}}$. Replace $\varphi$ and $\psi$ (as needed) with an appropriate power (say $k!$ for a large enough natural number $k$) such that both $\varphi^{k!}$ and $\psi^{k!}$ are pure mapping classes.

We first observe that $\hat{\varphi}^{k!}$ and $\hat{\psi}^{k!}$ have an equal number of fixed conjugacy classes in $\widehat{\pi_{1}(\Sigma_{M})}\cong\widehat{\pi_{1}(\Sigma_{N})}$ i.e. $\hat{n}_{\varphi^{k!}}=\hat{n}_{\psi^{k!}}$. To see this, set $t\in\widehat{\pi_{1}(M)}$ to be in the preimage of a unit in $\hat{{\mathbb{Z}}}$, the conjugation action of $t$ on $\widehat{\pi_{1}(\Sigma_{M})}$ is $\hat{\varphi}$. The conjugation action of $\Phi(t)$ on $\widehat{\pi_{1}(\Sigma_{N})}$ is $\hat{\psi}$ by the diagram above. The conjugation action on $\widehat{\pi_{1}(\Sigma_{M})}$ by $t^{k!}$ is $\hat{\varphi}^{k!}$ and the conjugation action on $\widehat{\pi_{1}(\Sigma_{N})}$ by $\Phi(t^{k!})$ is $\hat{\psi}^{k!}$. For every conjugacy class $\alpha$ of elements in $\widehat{\pi_{1}(\Sigma_{M})}$ fixed by $t^{k!}$-conjugation, there is a conjugacy class $\Phi(\alpha)$ of elements in $\widehat{\pi_{1}(\Sigma_{N})}$ fixed by $\Phi(t^{k!})$-conjugation, and vice versa. Thus, $\hat{n}_{\varphi^{k!}}=\hat{n}_{\psi^{k!}}$. By Proposition 4.1, $n_{M}=n_{\varphi^{k!}}=\hat{n}_{\varphi^{k!}}$ and $n_{N}=n_{\psi^{k!}}=\hat{n}_{\psi^{k!}}$. Thus, $n_{M}=n_{N}$, $g_{M}=g_{N}$, and $\Sigma_{M}$ is homeomorphic to $\Sigma_{N}$ as claimed. ∎

Let $\Sigma_{0,4}\hookrightarrow M\to S^{1}$ be a fixed fibration. By [WZ1], [JZ], and Theorem 3.2, a 3-manifold $N$ with $\widehat{\pi_{1}(N)}\cong\widehat{\pi_{1}(M)}$ will be hyperbolic and will fiber over $S^{1}$ with fiber a surface $S$ with Thurston norm 2. Fix an identification $\Phi:\widehat{\pi_{1}(M)}\to\widehat{\pi_{1}(N)}$. By Theorem 3.2, there is a connected fiber surface $\Sigma\hookrightarrow N$ with $\Phi(\widehat{\pi_{1}(\Sigma_{0,4}))}\cong\widehat{\pi_{1}(\Sigma)}$. By Theorem 1.2, $\Sigma$ is also homeomorphic to $\Sigma_{0,4}$, and therefore by Lemma 5.5 $M\cong N$. ∎

We denote the monodromies of $M$ and $N$ by $\varphi$ and $\psi$ respectively, and we will refer to the outer automorphisms of $F_{3}$ (the free group on three generators) induced by $\varphi,\psi$ by the same names. Let $\Sigma,\Sigma^{\prime}$ be the corresponding aligned fiber surfaces (both homeomorphic to $\Sigma_{0,4}$) for $M$ and $N$ respectively. We have the following exact sequences

where all the unlabelled vertical arrows are canonical inclusions of groups into their profinite completions.

The isomorphism $\Phi$ embeds $\pi_{1}(M)$ and $\pi_{1}(\Sigma)$ in $\widehat{\pi_{1}(N)}$ as dense subgroups of $\widehat{\pi_{1}(N)}$ and $\widehat{\pi_{1}(\Sigma^{\prime})}$ respectively. We consider $\Phi(\pi_{1}(\Sigma))<\Phi(\pi_{1}(M))$. Let $K_{i}$ be the intersection of all subgroups of $\Phi(\pi_{1}(\Sigma))$ of index $\leq i$. The tower $\{K_{i}\,|\,i\in\mathbb{N}\}$ is a cofinal tower of characteristic subgroups of $\Phi(\pi_{1}(\Sigma))$. Set $L_{i}=\widehat{K_{i}}\cap\pi_{1}(\Sigma^{\prime})$ to obtain a corresponding characteristic tower of $\pi_{1}(\Sigma^{\prime})$. Conjugation by elements of $\Phi(\pi_{1}(M))$ on $\Phi(\pi_{1}(\Sigma))$ induces outer automorphisms in $Out(\Phi(\pi_{1}(\Sigma))/K_{i})\cong Out(\widehat{\pi_{1}(\Sigma^{\prime})}/\widehat{L_{i}})$ for all $i$. Similarly, conjugation by elements of $\pi_{1}(N))$ on $\pi_{1}(\Sigma^{\prime})$ induces outer automorphisms in $Out(\pi_{1}(\Sigma^{\prime})/L_{i})\cong Out(\widehat{\pi_{1}(\Sigma^{\prime})}/\widehat{L_{i}})$ for all $i$.

The monodromy $\varphi$ is induced for the extension

by the conjugation action of an element $t\in\Phi(\pi_{1}(M))$ where the rightmost ${\mathbb{Z}}$ is a dense subgroup of $\hat{{\mathbb{Z}}}$ (and $t$ maps to a generator of this $\hat{{\mathbb{Z}}}$ under the fixed profinite epimorphism $\widehat{\pi_{1}(N)}\twoheadrightarrow\hat{{\mathbb{Z}}}$). Likewise, the monodromy $\psi$ is induced by the conjugation action of an element $s\in\pi_{1}(N)<\widehat{\pi_{1}(N)}$ that maps to a generator of ${\mathbb{Z}}$. Since the images of $t$ and $s$ under the fixed $\widehat{\pi_{1}(N)}\twoheadrightarrow\hat{{\mathbb{Z}}}$ both topologically generate $\hat{{\mathbb{Z}}}$, the actions of $t$ and $s$ (and therefore the actions of $\varphi$ and $\psi$) induce outer automorphisms $\varphi_{i}\in Out(\Phi(\pi_{1}(\Sigma))/K_{i})\cong Out(\widehat{\pi_{1}(\Sigma^{\prime})}/\widehat{L_{i}})$ and $\psi_{i}\in Out(\pi_{1}(\Sigma^{\prime})/L_{i})\cong Out(\widehat{\pi_{1}(\Sigma^{\prime})}/\widehat{L_{i}})$ such that $\varphi_{i}$ and $\psi_{i}$ generate conjugate cyclic subgroups $C_{i}<Out(\widehat{\pi_{1}(\Sigma^{\prime})}/\widehat{L_{i}})$.

Assume, for sake of contradiction, that $\varphi$ and $\psi$ are independent in $Mod(\Sigma_{0,4})$, i.e. that $\varphi$ and $\psi$ do not have conjugate powers. Since $Mod(\Sigma_{0,4})$ is omnipotent there is a finite quotient $q:Mod(\Sigma_{0,4})\twoheadrightarrow Q$ for which $o(q(\varphi))\neq o(q(\psi))$. Because $Mod(\Sigma_{0,4})$ has the Congruence Subgroup Property, $q$ factors through a principal congruence homomorphism

for some $i\in\mathbb{N}$. By construction, $\varphi_{i}=q_{i}(\varphi)$ and $\psi_{i}=q_{i}(\psi)$. By the previous paragraph, $\varphi_{i}$ and $\psi_{i}$ generate conjugate cyclic subgroups of $Out(\Phi(\pi_{1}(\Sigma)/K_{i})$, and this contradicts the claim that $o(q(\varphi))\neq o(q(\psi))$. Thus, $\varphi$ and $\psi$ are not independent. In particular, there are positive integers $m,n$ such that $\varphi^{m}$ and $\psi^{\pm n}$ are conjugate.

Following [BRW], apply omnipotence and CSP to the cyclic subgroup of $Mod(\Sigma_{0,4})$ generated by $\varphi$ to show that there is a congruence quotient $q:Mod(\Sigma_{0,4})\to Out(\pi_{1}(\Sigma)/{K_{i}})$ for which $mn$ divides $o(q(\varphi))=o(q(\psi))$. Since

it follows that $m=n$. Since $\varphi,\psi$ are pseudo-Anosovs in $Mod(\Sigma_{0,4})$ with a common conjugate power, there is a mapping class $\xi$ in $Mod(\Sigma_{0,4})$ with $(\xi\varphi\xi^{-1})^{m}=\xi\varphi^{m}\xi^{-1}=\psi^{m}$. Since $Mod(\Sigma_{0,4})$ is a (non-free) virtually free group, it splits as a finite graph of groups with finite vertex and edge groups (Theorem 1 [karrass_pietrowski_solitar_1973]) and so $Mod(\Sigma_{0,4})$ acts on a simplicial tree $T$. The elements $\xi\varphi\xi^{-1}$ and $\psi$ fix a common bi-infinite geodesic $\alpha$ in $T$. The stabilizer of $\alpha$ is cyclic (since all torsion in $Mod(\Sigma_{0,4})$ fixes a vertex in $T$) and so $\xi\varphi\xi^{-1}=\psi^{\pm}$.

∎

Let $\Sigma$ be a finite-type surface with negative Euler characteristic for which $Mod(\Sigma)$ is omnipotent and has the Congruence Subgroup Property. By Theorem 1.2 and [WZ1], any 3-manifold $N$ with the same profinite completion as a hyperbolic $\Sigma$-bundle $M$ over $S^{1}$ is a hyperbolic $\Sigma$-bundle over $S^{1}$. An analog of Lemma 5.5 shows that the corresponding monodromies have a common conjugate power. This common conjugate power is the monodromy of a common finite cyclic cover of $M$ and $N$. Thus $M$ and $N$ are cyclically commensurable. ∎

Let $\Sigma$ be a finite-type surface with negative Euler characteristic for which has the Congruence Subgroup Property. By Theorem 1.2 and [WZ1], any 3-manifold $N$ with the same profinite completion as a hyperbolic $\Sigma$-bundle $M$ over $S^{1}$ is a hyperbolic $\Sigma$-bundle over $S^{1}$. If all hyperbolic groups are residually finite, then Theorem 6.1 above can be combined with Lemma 5.5 to show that the corresponding monodromies will have a common conjugate power. This common conjugate power is the monodromy of a common degree finite cyclic cover of $M$ and $N$. Thus $M$ and $N$ are cyclically commensurable manifolds with the same volume. ∎