Profinite Tits alternatives for 3-manifold groups

 This post documents my attempts to understand Theorem C of the Wilton-Zalesski paper "Distinguishing geometries using finite quotients"

Continuing our exploration of the Wilton-Zalesski paper, we aim to prove the following:

Theorem 14: If $M$ is any compact 3-manifold and $H$ is a closed subgroup of $\widehat{\pi_1M}$ that does not contain a free non-abelian pro-$p$ subgroup for any prime $p$ then $H$ is on the following list:

1.  $H$ is conjugate to the completion of a virtually soluble subgroup of $\pi_1M$; or
2. $H$ is isomorphic to $((\mathbb{Z}_{\sigma} \times \mathbb{Z}_{\pi}) \rtimes C)$, where $\pi$ and $\sigma$ are (possibly empty) sets of primes with $\pi \cap \sigma = \emptyset$ and $C$ is procyclic (possibly finite).

This is a strong structure theorem for profinite completions of 3-manifold groups and can be viewed as a sort of Tits alternative (recall that the Tits alternative states that a finitely generated linear group is either virtually soluble or contains a non-abelian free subgroup).

We will build on the techniques introduced in the previous blog post. Recall the following:

Theorem 5: Let $H$ be a profinite group acting on a profinite tree $T$. Suppose $H$ does not possess a non-abelian free pro-$p$ subgroup for every prime $p$. Then either $H$ stabilizes a vertex or there exists a unique infinite minimal $H$-invariant subtree $D$ of $T$ such that the quotient group $L = H/K$ modulo the kernel of the action on $D$ is soluble and isomorphic to one of the following groups:

1. $L \cong \mathbb{Z}_{\pi} \rtimes \mathbb{Z}rho$ where $\pi$ and $\rho$ are disjoint sets of primes;
2.  $L$ is a profinite dihedral group $\mathbb{Z}_{\pi} \rtimes C_2$;
3. $L$ is a profinite Frobenius group $\mathbb{Z}_{\pi}  \rtimes C_n$, i.e., the order of every prime divisor $p$ of $n$ divides $q - 1$ for some $q \in \pi$ and the centralizers of nonidentity elements of $C_n$ coincide with $C_n$.

Much of what we do will be making sure that the hypothesis hold in the appropriate setting so that this theorem can be applied. Note also the following in the proof of theorem 7:

Remark 15: If in the notation of Theorem 5 $H$ acts $k$-acylindrically for some natural number $k$ then $k = 1$ (why?). If in addition $H$ is torsion free we have just the first case of a projective soluble group $H \cong \mathbb{Z}_{pi} \rtimes \mathbb{Z}_{\rho}$, where $\pi$ and $\rho$ are disjoint sets of primes. In this case any torsion-free profinite group containing $H$ as an open subgroup has a similar structure.

We will use this fact again.

Relative malnormality

Since the theorem we want concerns all compact 3-manifold groups we will need to handle relatively hyperbolic groups (fundamental groups of cusped hyperbolic 3-manifold groups). In this section we will upgrade the techniques from the previous post to cover this case. Recall that more generally a toral relatively hyperbolic is a group that is torsion-free and hyperbolic relative to sets of finitely generated abelian subgroups.

Definition: Suppose that a group $G$ is hyperbolic relative to a collection of parabolic subgroups $\{P_1,\ldots,P_n\}$. A subgroup $H$ of $G$ is called relatively malnormal if, whenever an intersection of conjugates $H^\gamma \cap H$ is not conjugate into some $P_i$, we have $\gamma \in H$.

We will need the following result, which can be proved via the heavy machinery developed by Groves and Manning and a relatively hyperbolic version of the Malnormal Special Quotient Theorem:

Theorem 16: Suppose that $G$ is a virtually compact special group, which is also toral relatively hyperbolic with parabolic subgroups $P_1,\ldots,P_n$. Let $H$ be a subgroup which is relatively malnormal and relatively quasiconvex. Then $\widehat{H}$ is also a relatively malnormal subgroup of $\widehat{G}$, in the sense that $\widehat{H} \cap \widehat{H}^{\hat{\gamma}}$ is conjugate into $\widehat{P_i}$ (for some $i$) whenever $\hat{\gamma} \notin \widehat{H}$.

We will take this as a blackbox as the point of this post is to see the profinite aspects of 3-manifold groups rather than the theory of cube complexes and Dehn fillings, although these are very interesting in their own right.

It is an important fact that in cusped hyperbolic manifolds the peripheral subgroups form a malnormal family. We will see that an analogous fact is true in the profinite setting, but we will need to borrow a fact from number theory to prove it.

Lemma 17: Let $R$ be a finitely generated ring in a number field $k$, let $d$ be a non-zero element of $R$ that is not a root of unity, and let $x_l, x_2, ..., x_j$ be non-zero elements of $R$. Then there exists a positive integer $n$ with the following property:

 For each integer $m > n$, there exist a finite field $F$ and a ring homomorphism $\eta: R \to F$ such that the multiplicative order of $\eta(\delta)$ is equal to $m$ and for each $1 \leq i \leq j$, $\eta(x_i)\neq 0$.

 We will use lemma 17 to prove the following:

Lemma 18: Let $M = \mathbb{H}^3/\Gamma$ be a cusped hyperbolic 3-manifold of finite volume. Let $P$ and $Q$ be non-conjugate cusp subgroups of $\Gamma$. Then there exists a positive integer $n$ with the following property. For each integer $m \geq n$, there exist finite fields $\mathbb{F}_1$ and $\mathbb{F}_2$ and group homomorphisms $f_1 \colon \Gamma \to \mathrm{PSL}(2,\mathbb{F}_1)$ and $f_2 \colon \Gamma \to \mathrm{PSL}(2,\mathbb{F}_2)$ such that:

1.  the image of $P$ under $f_1 \times f_2$ is isomorphic to $\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$; 
2. for every element $p \in P$ and every $i$, if $f_i(p)$ is non-trivial, then $\mathrm{tr} f_i(p) = \pm 2$; 
3. for every element $q \in Q$ and every $i$, $\mathrm{tr} f_i(q) = \pm 2$.

In particular, if every $f_i(p)$ is conjugate into $f_i(Q)$, then $(f_1 \times f_2)(p)$ is trivial.

Proof: Since $P$ and $Q$ are non-conjugate, they correspond to distinct cusps of $M$. Let $T$ be the cusp of $M$ corresponding to $P$. The group $P$ is free abelian of rank 2. 

Thurston's hyperbolic Dehn surgery theorem states that for any cusped hyperbolic manifold there are finitely many exceptional slopes associated to each cusp such that if these are avoided, then the corresponding Dehn surgery yields a hyperbolic 3-manifold. This implies there exists a basis $\{p_1,p_2\}$ of $P$ and complete hyperbolic 3-manifolds $M_1$ and $M_2$, obtained by Dehn surgery on $M$ along $T$, such that if

\[

\varphi_1 \colon \Gamma \to \pi_1(M_1) \quad \text{and} \quad \varphi_2 \colon \Gamma \to \pi_1(M_2)

\]

are the homomorphisms induced by inclusion, then $\varphi_1(p_1)$ is a loxodromic isometry of $\pi_1(M_1)$, $\varphi_1(p_2)$ is trivial, $\varphi_2(p_1)$ is trivial and $\varphi_2(p_2)$ is a loxodromic isometry of $\pi_1(M_2)$.

Since $M_1$ has finite volume and $\varphi_1(p_1)$ is loxodromic, there exists a discrete, faithful representation

\[

\rho_1 \colon \pi_1(M_1) \to \mathrm{PSL}(2,\mathbb{C})

\]

such that $\rho_1(\pi_1(M_1)) \subset \mathrm{PSL}(2,L_1)$, for some number field $L_1$, and

\[

\rho_1(\varphi_1(p_1)) = \pm \begin{pmatrix}

\omega & 0 \\

0 & \omega^{-1}

\end{pmatrix}, \quad |\omega|=1.

\]

Let $R_1$ be the ring in $L_1$ generated by the coefficients of the generators of $\rho_1(\pi_1(M_1))$ over $\mathbb{Z}$. Then $\rho_1(\pi_1(M_1)) \subset \mathrm{PSL}(2,R_1) \subset \mathrm{PSL}(2,L_1)$. By Lemma 17, there exists a positive integer $n$ with the following property. For each integer $m \geq n$, there exist a finite field $\mathbb{F}_1$ and a ring homomorphism $\eta_1 \colon R_1 \to \mathbb{F}_1$ such that the multiplicative order of $\eta_1(\omega)$ is equal to $2m$. This ring homomorphism induces a group homomorphism

\[

\psi_1 \colon \rho_1(\pi_1(M_1)) \hookrightarrow \mathrm{PSL}(2,R_1) \to \mathrm{PSL}(2,\mathbb{F}_1).

\]

Let

\[

f_1 \colon \Gamma \to \mathrm{PSL}(2,\mathbb{F}_1)

\]

denote the composition $\psi_1 \circ \rho_1 \circ \varphi_1$. Then $f_1(p_1)$ has order $m$ and $f_1(p_2)$ is trivial. For every element $q \in Q$, $\rho_1 \circ \varphi_1(q)$ is parabolic. Therefore, $\mathrm{tr} f_1(q) = \pm 2$. For every element $p \in P$,

\[

\rho_1(\varphi_1(p)) = \pm \begin{pmatrix}

\omega^k & 0 \\

0 & \omega^{-k}

\end{pmatrix}, \quad \text{for some } k \in \mathbb{Z}.

\]

If $\eta_1(\omega^k + \omega^{-k}) = 2$, then $\eta_1(\omega^k) = \eta_1(\omega^{-k}) = 1$. If $\eta_1(\omega^k + \omega^{-k}) = -2$, then $\eta_1(\omega^k) = \eta_1(\omega^{-k}) = -1$. Therefore, if $f_1(p)$ is non-trivial, then $\mathrm{tr} f_1(p) = \pm 2$.

In a similar way, we can choose $n$ such that for each integer $m \geq n$ there exist a finite field $\mathbb{F}_2$ and a group homomorphism

\[

f_2 \colon \Gamma \to \mathrm{PSL}(2,\mathbb{F}_2)

\]

such that $f_2(p_1)$ is trivial, $f_2(p_2)$ has order $m$, for every $q \in Q$, $\mathrm{tr} f_2(q) = \pm 2$, and, for every $p \in P$, if $f_2(p)$ is non-trivial, then $\mathrm{tr} f_2(p) = \pm 2$. The homomorphisms $f_1$ and $f_2$ then satisfy the three conditions above. $\blacksquare$

Lemma 19: Let $\Gamma$ be the fundamental group of a cusped hyperbolic manifold and let $\{P_i\}$ be conjugacy representatives of the cusp subgroups. Then the set of their closures $\{\overline{P_i}\}$ in the profinite completion $\hat{\Gamma}$ forms a malnormal family: that is, if $\overline{P_i} \cap \overline{P_j}^{\hat{\gamma}}$ is non-trivial for some $\hat{\gamma} \in \hat{\Gamma}$ then $i = j$ and $\hat{\gamma} \in \overline{P_i}$.

Note that it is enough to consider the case of two cusps, so we adopt the notation $P = P_1$ and $Q = P_2$. We first consider the intersection $P \cap Q^{\hat{\gamma}}$.

Proof: Consider first the intersection $P \cap Q^{\hat{\gamma}}$. Let $\hat{p}$ be an element of the intersection $P \cap Q^{\hat{\gamma}}$. Let $f$ be an arbitrary homomorphism from $\Gamma$ to a finite group, and let $f_0$ be the restriction of $f$ to $P$. Choose $f_1,f_2$ as in Lemma 18 so that $f_0$ factors through $(f_1 \times f_2)|_P$, and extend them by continuity to homomorphisms $\hat{f}_i$ from the profinite completion. Choose $p \in P$ so that $f_i(p) = \hat{f}_i(\hat{p})$ for all $i$. Since $\hat{p}$ is conjugate into $Q$, item 3 implies that $\mathrm{tr} f_i(p) = \mathrm{tr} \hat{f}_i(\hat{p}) = 2$ for every $i$. Item 2 then implies that $f_i(p) = 1$ for every $i$, and so by item 1, $\hat{f}_0(\hat{p}) = f_0(p) = 1$. Thus, every finite quotient of $\Gamma$ kills $\hat{p}$, and so $\hat{p} = 1$ by the definition of the profinite completion. We have shown that the intersection $P \cap Q^{\hat{\gamma}}$ is trivial.

We now consider the intersection $P \cap P^{\hat{\gamma}}$, and suppose that $\hat{\gamma} \notin P$. Since $P$ is closed, there is a subgroup $\Gamma_0$ of finite index in $\Gamma$ that contains $P$ but such that $\hat{\Gamma}_0 \subseteq \hat{\Gamma}$ does not contain $\hat{\gamma}$. Let $\hat{\gamma} = \gamma \hat{\gamma}_0$, where $\gamma \in \Gamma \setminus \Gamma_0$ and $\hat{\gamma}_0 \in \hat{\Gamma}_0$. Then $\Gamma_0$ is the fundamental group of a closed hyperbolic manifold with non-conjugate cusp subgroups $P$ and $Q = P^\gamma \cap \Gamma_0$ (since $P$ is malnormal in $\Gamma$). By the argument of the previous paragraph applied to $\Gamma_0$, the intersection $P \cap Q^{\hat{\gamma}_0}$ is trivial. Since $Q$ is of finite index in $P^\gamma$, it follows that the intersection $P \cap P^{\hat{\gamma}}$ is finite. But 3-manifold groups are torsion-free and good, whence their profinite completions are also torsion-free. In particular, $P \cap P^{\hat{\gamma}}$ is trivial. $\blacksquare$

Proof of theorem 14

Our strategy to prove theorem 14 will be to prove it for each geometric piece and then assemble the results via the Grushko decomposition theorem. We start with the cusped hyperbolic case.

Theorem 20: A hyperbolic 3-manifold $M$ with cusps has a finite-sheeted covering space $N \to M$ that contains a disjoint family of connected, geometrically finite, incompressible subsurfaces $\{\Sigma_1,\ldots,\Sigma_n\}$ such that:

1. each cusp of $N$ contains a boundary component of some $\Sigma_i$;
2. each $\pi_1\Sigma_i \subseteq \pi_1N$ is relatively malnormal.

Proof: By the `Half Lives, Half Dies' Lemma (see here for the 3-manifold case or here for the more general case), there is a homomorphism $\pi_1M \to \mathbb{Z}$ such that each peripheral subgroup maps non-trivially. One can realize this by a smooth map $M \to S^1$, and the preimage of a generic point is a (possibly disconnected) surface that cuts every cusp, which can be compressed to obtain an incompressible surface with the same property. Passing to a finite-sheeted cover and applying an argument with the Thurston norm, one can further ensure that each component is geometrically finite.

Let $\Sigma$ be such a component. Geometrical finiteness implies that there are finitely many double cosets $\pi_1\Sigma g_i \pi_1\Sigma$ such that the intersection $\pi_1\Sigma^{g_i} \cap \pi_1\Sigma$ is non-peripheral. By subgroup separability, we can pass to a finite-sheeted covering space to which $\Sigma_i$ lifts but the elements $g_i$ do not. Doing this for every component, and passing to a deeper regular covering space (still of finite index), we obtain the required covering space.$\blacksquare$

 Cutting along the family of surfaces given by Theorem 20 produces a graph-of-groups decomposition $(G,\delta)$ for $\pi_1N$, with the property that every vertex and edge group is hyperbolic and virtually special, and the stabilizer of any infinite subtree of the Bass–Serre tree is cyclic. Note also that $(G,\delta)$ is efficient. Passing to the corresponding graph of profinite groups $( G,\delta)$, it follows from Theorem 16 that the stabilizer of any infinite subtree of the standard profinite tree is pro-cyclic (possibly trivial).

Theorem 21: Let $M$ be a cusped hyperbolic 3-manifold, $\pi_1M$ its fundamental group and $\widehat{\pi_1M}$ its profinite completion. If $H$ is a closed subgroup of $\widehat{\pi_1M}$ that does not contain a free non-abelian pro-$p$ subgroup for any $p$ then $H$ is isomorphic to $\mathbb{Z}_{\pi} \rtimes \mathbb{Z}_{\rho}$. Furthermore, if $H$ is not projective (i.e., $\pi \cap \rho = \emptyset$) then $H$ is conjugate into the closure of a cusp subgroup of $\widehat{\pi_1M}$.

Proof: Put $H_0 = \widehat{\pi_1N} \cap H$, where $N$ is the cover from Theorem 20. We first prove that $H_0$ has the claimed semi-direct product structure. Consider the action of $H_0$ on the standard profinite tree. By Theorem 7 the stabilizers of vertices in $H_0$ have the claimed structure. Since the stabilizers of infinite subtrees are pro-cyclic, the kernel $K$ of the action is torsion-free pro-cyclic.

Suppose first that $K$ is non-trivial. The closure of any peripheral subgroup $\widehat{P_i}$ is malnormal in $\widehat{\pi_1M}$ by Lemma 19, and since up to conjugation $K$ is in $\widehat{P_i}$, we deduce that $K$ is central in $H$ and $H$ is abelian. So $H$ is a subgroup of $\hat{\mathbb{Z}} \times \hat{\mathbb{Z}}$ in this case.

Alternatively, if $K$ is trivial then $H_0$ is isomorphic to a soluble projective group $\mathbb{Z}_{\pi} \rtimes \mathbb{Z}_{\rho}$ by Theorem 5. Observe now that $\pi_1M$ is torsion free, since $\Gamma$ is good. So $H$ is torsion free and hence, by Remark 15, a semi-direct product of the claimed form.

Finally, if $H$ is not projective then $K$ is necessarily non-trivial by Theorem 5, and so the above argument shows that $H$ is conjugate into the closure of some $\widehat{P_i}$ as claimed. $\blacksquare$

This does the hyperbolic case, which is often the most difficult one. Next we consider the Seifert-fibred case When the fundamental group itself is soluble we are done, so it remains to do the cases where the base orbifold is hyperbolic.

Proposition 22: Suppose that $M$ is any compact, Seifert fibred 3-manifold and that $\pi_1M$ is not virtually soluble. If $H$ is a closed subgroup of $\widehat{\pi_1M}$ that does not contain a free non-abelian pro-$p$ subgroup for any $p$ then $H$ is isomorphic to $(\mathbb{Z}_{\sigma} \times \mathbb{Z}_{\pi}) \rtimes \mathbb{Z}_{\rho}$, where $\pi \cap \rho = \emptyset$ and $\mathbb{Z}_{\rho}$ acts on $\mathbb{Z}_{\sigma}$ by inversion.

Proof: The fundamental group $G = \pi_1M$ fits into a short exact sequence

\[

1 \to \Z \to G \xrightarrow{p} \pi_1(O) \to 1

\]

where $\Z = \langle z \rangle$ is cyclic and $O$ is a cone-type 2-orbifold.

Since $G$ is not virtually soluble, $O$ is a hyperbolic orbifold, which is finitely covered by an orientable hyperbolic surface $\Sigma$; passing to a further cover if necessary, we may further assume that $\Sigma$ admits an essential simple-closed curve $\gamma$ which is not boundary parallel and that $\pi_1\Sigma$ is torsion-free.

The exact sequence above induces an exact sequence

\[

1 \to \hat{\mathbb{Z}} \to \widehat{G} \xrightarrow{\hat{p}} \widehat{\pi_1(O)} \to 1

\]

of profinite completions. Then $\widehat{\pi_1\Sigma}$ is naturally an open subgroup of $\widehat{\pi_1(O)}$, which pulls back to an open subgroup $\widehat{\pi_1N}$ of $\widehat{G}$. Cutting $\Sigma$ along $\gamma$ induces a splitting of $\pi_1\Sigma$ in which the edge group is malnormal; this in turn induces a graph of profinite groups for $\widehat{\pi_1\Sigma}$ with procyclic malnormal edge group.

Since $\widehat{\pi_1\Sigma}$ is torsion-free, by Theorem 5 $\hat{p}(H) \cong \mathbb{Z}_{\overline{\pi}} \rtimes \mathbb{Z}_{\overline{\rho}} $ is a projective soluble group with $\bar{\pi} \cap \bar{\rho} = \emptyset$. Denote by $\delta$ the set of primes dividing the orders of torsion elements of $\pi_1(O)$. Then $\hat{\mathbb{Z}} = \mathbb{Z}_{\sigma} \times \mathbb{Z}_{\delta}$ and $H/\mathbb{Z}_{\sigma}$ is a torsion-free soluble group containing a projective group $U$. Hence $H/\mathbb{Z}_{\sigma}$ is soluble projective (projective groups are groups of cohomological dimension 1 and analogously to the discrete group case, torsion-free overgroups of finite index preserve cohomological dimension. This requires an argument but we hope the reader will be happy to accept this is true.) and hence isomorphic to $\mathbb{Z}_{\pi} \rtimes \mathbb{Z}_{\rho}$ for some $\pi \cap \rho = \emptyset$ (this is another fact that we quote). Thus $H \cong \mathbb{Z}_{\sigma} \rtimes H/\mathbb{Z}_{\sigma} \cong \mathbb{Z}_{\sigma} \rtimes (\mathbb{Z}_{\pi} \rtimes \mathbb{Z}_{\rho})$. 

Since the action on $\hat{\mathbb{Z}}$ is induced from the action on $\mathbb{Z}$ which is either trivial or by inversion, moving the prime 2 from $\pi$ to $\rho$ if necessary we can rewrite $H$ as $H \cong (\mathbb{Z}_{\sigma} \times \mathbb{Z}_{\pi}) \rtimes \mathbb{Z}_{\rho}$. $\blacksquare$

Proposition 23: Let $M$ be a compact, orientable, irreducible 3-manifold whose torus decomposition is non-trivial. If $H$ is a closed subgroup of $\widehat{\pi_1M}$ that does not contain a free non-abelian pro-$p$ subgroup for any $p$ then $H$ is isomorphic to $(\mathbb{Z}_{\sigma} \times \mathbb{Z}_{\pi}) \rtimes \mathbb{Z}_{\rho}$, where $\mathbb{Z}_{\rho}$ acts on $\mathbb{Z}_{\sigma}$ by inversion and $\pi \cap \rho = \emptyset$.

Proof sketch: We quote previous results of Hamilton-Wilton-Zalesski which show that $\widehat{\pi_1M}$ acts 4-acylindrically on a profinite tree. Therefore so does $H$, and by Proposition 22 and Theorem 21 the stabilizers of vertices in $H$ have the claimed structure. So if $H$ stabilizes a vertex we are done. Otherwise, since the action is 4-acylindrical the kernel $K$ is trivial. Then $H$ is isomorphic to a soluble projective group $\mathbb{Z}_{\pi} \rtimes \mathbb{Z}_{\rho}$ by Remark 15 since $\pi_1M$ is torsion free.

Proof of Theorem 14: By doubling we may assume $M$ is closed, and the Grushko decomposition lets us split into factors $\pi_1(M) = (*A_i)*(B_i)*F$ with $A_i$ finite, $B_i$ geometric, and $F$ a free group. We consider the action of 

\[

\widehat{\pi_1M} = \widehat{A_1} \amalg \cdots \amalg \widehat{A_m} \amalg \widehat{B_1} \amalg \cdots \amalg \widehat{B_n} \amalg \widehat{F}

\]

on the standard profinite tree $S$ associated to the free profinite product decomposition of $\widehat{\pi_1M}$. If $H$ stabilizes a vertex then $H$ is conjugate to a subgroup of a free factor, so the structure of $H$ is described in the previous results. Otherwise, it follows from Theorem 5 combined with Remark 15 that $H$ is as in item 2 of Theorem 14. $\blacksquare$