Distinguishing 3-manifold geometries using finite quotients

This post documents my attempts to understand parts of the Wilton-Zalesski paper "Distinguishing geometries using finite quotients", which contains some of my favourites out of Wilton's results.

It is well-known that the question of whether two given groups are isomorphic is in general algorithmically undecidable, which makes it all the more interesting to find methods of showing two groups aren't undecidable when they fall in some special classes. One reasonably natural way of doing so is computing the list of finite quotients of the groups in question and comparing them: if the lists (of quotients of order at most $N$, say) differ, then this is a certificate that the groups can't be isomorphic. This list of finite quotients of a group $G$ can be assembled into a more compact object, known as the profinite completion $\hat{G}$ of $G$. One takes an inverse limit over the system of finite quotients of $G$.

More concretely, an inverse system consists of:

• A directed set $(I, \leq)$,
• An indexed family of finite groups $\{G_i : i \in I\}$, each having the discrete topology, and
•  A family of homomorphisms $\{f_i^j : G_j \to G_i \mid i,j \in I, i \leq j\}$ such that $f_i^i$ is the identity map on $G_i$, and the collection satisfies the composition property $f_i^j \circ f_j^k = f_i^k$ whenever $i \leq j \leq k$.

The inverse limit is the set:

\[\varprojlim G_i = \left\{(g_i)_{i \in I} \in \prod_{i \in I} G_i : f_i^j(g_j) = g_i \text{ for all } i \leq j\right\}\]

This is topologised as a subspace of the product with the product topology. When $G$ is residually finite, i.e. for any non-trivial element $g$ there is a finite quotient $Q$ and a map $\phi: G \to Q$ such that $\phi(g) \neq 1$, the profinite completion sees all of $G$ in the sense that the natural homomorphism $ \eta: G \to \hat{G}$ is injective.

The homomorphism $\eta$ is characterized by the following universal property: given any profinite group $H$ and any continuous group homomorphism $f \colon G \rightarrow H$ where $G$ is given the smallest topology compatible with group operations in which its normal subgroups of finite index are open, there exists a unique continuous group homomorphism $g \colon \widehat{G} \rightarrow H$ with $f = g\eta$.

Examples of profinite groups are finite groups, as well as the profinite completions of any residually finite group, e.g. the profinite completion of the integers is known as the procyclic group $\hat{\mathbb{Z}}$. It is a theorem that for any groups $G_1, G_2$ their profinite completions are isomorphic if and only if the groups themselves have the same set of finite quotients, but even with this result it is often very difficult to distinguish groups using this method.

Definition: A finitely generated residually finite group $G$ is said to be absolutely profinitely rigid if for any finitely generated residually finite group $H$, $\hat{G} \cong \hat{H} \implies G \cong H$. Equivalently, defining the genus to be the set of groups with the same profinite completion, $G$ is absolutely profinitely rigid if its genus consists only of itself.

From now on all groups will be assumed to be finitely generated and residually finite. Very few groups are known to be absolutely profinitely rigid: this is open for all fundamental groups of compact surfaces. The only examples of absolutely profinitely rigid groups we know are certain compact hyperbolic 3-manifold groups, certain arithmetic lattices, free metabelian groups, and soluble Baumslag-Solitar groups. Techniques from number theory are often used to show this.

Instead, one might ask this question for a restricted class $\mathcal{C}$ of groups.

Definition: Let $\mathcal{C}$ be a class of (residually finite) groups. A group $G \in \mathcal{C}$ is said to be profinitely rigid in the class $\mathcal{C}$ if for any $H \in \mathcal{C}$, $\hat{G} \cong \hat{H} \implies G \cong H$.

One can similarly define a $\mathcal{C}$ genus to be the set of groups in $\mathcal{C}$ with the same completion. Even here there are many hard questions: it remains open whether every group in the class $\mathcal{C}=$\{ fundamental groups of compact hyperbolic 3-manifolds\} is $\mathcal{C}$-profinitely rigid, although Liu has shown that the $\mathcal{C}$-genus of any such group is finite.

For more information on profinite groups, Gareth Wilkes' course notes and book are highly recommended.

Recall that a compact orientable 3-manifold admits a geometric decomposition: it can be cut up along incompressible tori so that each piece admits a geometric structure. The purpose of this and the next one or two blog posts is to discuss a series of results by Wilton and Zalesski that show that all of this geometry can be seen in the profinite completion: copies of $\hat{\mathbb{Z}^2}$ coming from the JSJ tori can be found in the profinite completion and if two geometric pieces have isomorphic profinite completions then they have the same geometric structure. Today we will aim to prove the following:

Theorem: Let $M,N$ be closed, orientable, irreducible 3-manifolds, and suppose that $\widehat{\pi_1M} \cong \widehat{\pi_1N}$. Then $M$ admits one of Thurston's eight geometric structures if and only if $N$ does, and in this case both $M$ and $N$ admit the same geometric structure.

Special cube complexes

The first ingredient we will need is special cube complexes, whose theory has been greatly developed by the work of Agol, Kahn-Markovic, Wise, and many others. Agol showed that a hyperbolic group acting geometrically on a CAT(0) cube complex has a finite index subgroup which acts on the same cube complex with special quotient. We won't define this and will simply state for later use some of the consequences of being special since ultimately the aim of the blog post is to cover the profinite aspects, but the reader should think of special cube complexes as ones without any silly pathologies. A consequence of this is that any compact hyperbolic 3-manifold virtually fibres, i.e. admits a finite cover which fibres over the circle (with fibre a surface). For cusped hyperbolic manifolds this was already known by work of Wise. This story, as well as more intuition for special cube complexes and their fundamental groups (which we call special groups for brevity), is covered by Calegari here so we will leave it to the interested reader to explore the more geometric and combinatorial aspects of cube complexes there. Here are some other results from the special cube complex theory that we will blackbox:

Definition: A subgroup $H \subseteq G$ is called malnormal if for any $\gamma \notin H$ $H^{\gamma} \cap H = 1$. If instead the intersection is finite, then the subgroup is called almost malnormal.

A finite family of subgroups $\{H_i\}$ is said to be malnormal if whenever $H_i^{\gamma} \cap H_j \neq 1$ then $i=j$ and $\gamma \in H_i$. An almost malnormal family of subgroups is defined analogously

Definition: The class of word-hyperbolic groups with a quasi-convex hierarchy is the smallest class of groups, closed under isomorphism, containing the trivial group and such that if

• $G=A*_C$ where $A$ has a qc hierarchy
•  $G=A*_CB$ where both $A,B$ have qc hierarchies

for $C$ quasi-convex in $G$, then $G$ also has a quasi-convex hierarchy. A malnormal quasi-convex hierarchy is defined analogously where every instance of quasi-convex is replaced by malnromal and quasi-convex.

The following important theorem is due to Haglund, Hsu, and Wise:

Theorem 1: Let $G$ be a word-hyperbolic group. TFAE:

• $G$ is virtually special; 
• $G$ virtually has a malnormal qc hierarchy;
• $G$ virtually has a qc hierarchy.

Theorem 2: Let $G$ be a hyperbolic group and the fundamental group of a compact, virtually special cube complex, and let $\mathcal{H} = \{H, K\}$ be a malnormal family of quasiconvex subgroups. For any finite-index subgroup $G_1$ of $G$ there exists a finite-index subgroup $G_0 \subseteq G_1$ and a retraction map $\rho \colon G_0 \to H \cap G_1$ with the following properties:

1. $\rho(H^\gamma \cap G_0) = 1$ unless $\gamma \in HG_1$; and
2. $\rho(K^\gamma \cap G_0) = 1$ for all $\gamma \in G$.

Definition: A group $G$ is said to be good in the sense of Serre or cohomologically separable if the natural homomorphism $G \to \hat{G}$ of the group in its profinite completion induces an isomorphism on cohomology with finite coefficients.

Proposition 3: If a group $G$ has a finite-index subgroup $G_0$ which is the fundamental group of a compact, special cube complex then $G$ is good. If $G$ is torsion-free then so is $\hat{G}$

We can now prove the main result we need about hyperbolic and cubulated groups:

Theorem 4: Let $G$ be word-hyperbolic and the fundamental group of a compact, virtually special cube complex. Let $\mathcal{H} = \{H_1,\ldots,H_n\}$ be a malnormal family of quasiconvex subgroups. Then the family $\widehat{\mathcal{H}} = \{\widehat{H}_1,\ldots,\widehat{H}_n\}$ is a malnormal family of subgroups of the profinite completion $\widehat{G}$.

Proof: It suffices to take $n = 2$. First, we prove that $\widehat{H} = \widehat{H}_1$ is a malnormal subgroup of $\widehat{G}$.

Let $\hat{\gamma} \in \widehat{G} \setminus \widehat{H}$, and suppose that $\hat{\delta} \in \widehat{H} \cap \widehat{H}^{\hat{\gamma}}$, so $\hat{\delta} = \hat{\epsilon}^{\hat{\gamma}}$ for $\hat{\epsilon} \in \widehat{H}$. Since $\widehat{H}$ is closed, there exists a finite quotient $q \colon G \to Q$ whose continuous extension $\hat{q} \colon \widehat{G} \to Q$ satisfies $\hat{q}(\hat{\gamma}) \notin q(H)$. Let $G_1 = \ker q$, and let $G_0$ be the finite-index subgroup guaranteed by Theorem 2. Let $n$ be such that $\hat{\delta}^n \in \widehat{G}_0$. If $\hat{\rho}$ is the continuous extension of $\rho$ to $\widehat{G}_0$ then

\[\hat{\delta}^n = \hat{\rho}(\hat{\delta}^n) = \hat{\rho}((\hat{\epsilon}^n)^{\hat{\gamma}})) = 1\]

where the final equality follows from item 1 of Theorem 2 by continuity using that the closure $HG_1$ is clopen in $\widehat{G}$. So $\hat{\delta}$ is torsion and therefore trivial, since $\widehat{G}$ is torsion-free by Proposition 3. This proves that $\widehat{H}_1$ is malnormal.

To complete the proof that $\widehat{\mathcal{H}}$ is a malnormal family, suppose that $\hat{\delta} = \hat{\epsilon}^{\hat{\gamma}}$, where $\hat{\delta} \in \widehat{H}_1$ and $\hat{\epsilon} \in \widehat{H}_2$. Let $G_1 = G$, and as before let $G_0$ be the finite-index subgroup guaranteed by Theorem 2 and let $n$ be such that $\hat{\delta}^n \in \widehat{G}_0$. Then, as before, we have that

\[\hat{\delta}^n = \hat{\rho}(\hat{\delta}^n) = \hat{\rho}((\hat{\epsilon}^n)^{\hat{\gamma}}) = 1\]

where the final inequality follows from item 2 of Theorem 2 by continuity. Again, since $\widehat{G}$ is torsion-free, we deduce that $\hat{\delta} = 1$, which proves the theorem. $\blacksquare$

Profinite trees

The second major ingredient needed for the proof is the notion of a profinite tree. I have to admit that I don't find the theory at all pretty, which would not be an appropriate remark to make in an academic paper. Luckily this isn't an academic paper but is my blog so I can be as unhinged as I like. I will content myself by saying that in the sequel I will be as brief as I can reasonably about the aspects of the proof pertaining to profinite tree. The reader is probably welcome to just pretend that profinite groups acting on profinite trees are whatever they think and take the one theorem (theorem 5) in this subsection on faith.

A graph $\Gamma$ is a disjoint union $E(\Gamma) \cup V(\Gamma)$ with two maps $d_0,d_1 \colon \Gamma \to V(\Gamma)$ that are the identity on the set of vertices $V(\Gamma)$. For an element $e$ of the set of edges $E(\Gamma)$, $d_0(e)$ is called the initial and $d_1(e)$ the terminal vertex of $e$.

Definition: A profinite graph $\Gamma$ is a graph such that:

•  $\Gamma$ is a profinite space (i.e., an inverse limit of finite discrete spaces); 
• $V(\Gamma)$ is closed; and
• the maps $d_0$ and $d_1$ are continuous.

It is a fact that every profinite graph $\Gamma$ is an inverse limit of finite quotient graphs of $\Gamma$. For a profinite space $X$ that is the inverse limit of finite discrete spaces $X_j$, $[[ \hat{Z} X ]]$ is the inverse limit of $[[ \hat{Z} X_j ]]$, where $[[ \hat{Z} X_j ]]$ is the free $\hat{Z}$-module with basis $X_j$. For a pointed profinite space $(X,*)$ that is the inverse limit of pointed finite discrete spaces $(X_j,*)$, $[[ \hat{Z} (X,*) ]]$ is the inverse limit of $[[ \hat{Z} (X_j,*) ]]$, where $[[ \hat{Z} (X_j,*) ]]$ is the $\hat{Z}$-vector space with basis $X_j \setminus \{*\}$.

For a profinite graph $\Gamma$, define the pointed space $(E^*(\Gamma),*)$ as $\Gamma/V(\Gamma)$ with the image of $V(\Gamma)$ as a distinguished point $*$, and denote the image of $e \in E(\Gamma)$ by $\overline{e}$. By definition, a profinite tree $\Gamma$ is a profinite graph with a short exact sequence

\[0 \to [[ \hat{Z} (E^*(\Gamma),*) ]] \xrightarrow{\delta} [[ \hat{Z} V(\Gamma) ]] \xrightarrow{\epsilon} \hat{Z} \to 0\]

where $\delta(\overline{e}) = d_1(e) - d_0(e)$ for every $e \in E(\Gamma)$ and $\epsilon(v) = 1$ for every $v \in V(\Gamma)$.

If $v$ and $w$ are elements of a profinite tree $T$, we denote by $[v,w]$ the smallest profinite subtree of $T$ containing $v$ and $w$ and call it geodesic.

By definition, a profinite group $G$ acts on a profinite graph $\Gamma$ if we have a continuous action of $G$ on the profinite space $\Gamma$ that commutes with the maps $d_0$ and $d_1$.

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 $C_n$.

Note that the group of automorphisms $\mathrm{Aut}(\mathbb{Z}pi)$ coincides with the group of units of the ring $\mathbb{Z}pi = \prod_{p\in\pi} \mathbb{Z}p$ and so is the direct product $\prod_{p\in\pi} \mathbb{Z}p^*$ of groups of units of $\mathbb{Z}p$. Remark also that $\mathbb{Z}p^* \cong \mathbb{Z}p \times C_{p-1}$ for $p \neq 2$ and $\mathbb{Z}_2^* \cong \mathbb{Z}_2 \times C_2$.

Profinite hyperbolisation

In this section we will figure out how to recognise hyperbolic manifolds from their profinite completions: the analogue of hyperbolisation holds and says that the manifold is hyperbolic if and only if there is no embedded $\widehat{\mathbb{Z}}^2$.

Definition: The action of a group $\Gamma$ on a tree $T$ is said to be \emph{$k$-acylindrical}, for $k$ a constant, if the set of fixed points of $\gamma$ has diameter at most $k$ whenever $\gamma \neq 1$.

Analogously, the action of a profinite group $\widehat{\Gamma}$ on a profinite tree $T$ is said to be \emph{$k$-acylindrical}, for $k$ a constant, if the set of fixed points of $\gamma$ has diameter at most $k$ whenever $\gamma \neq 1$.

A malnormal hierarchy for $\Gamma_0$ implies that $\Gamma_0$ has a $1$-acylindrical action on a tree. This result carries over to the profinite setting.

Lemma 6: Suppose that $\Pi$ is the (profinite) fundamental group of a graph of profinite groups $(\overline{G},\Delta)$ with one edge $e$, and suppose that the edge group $\Pi(e)$ is malnormal in $\Pi$. Then the action of $\Pi$ on the standard tree $S$ is $1$-acylindrical.

Recall that a projective group $G$ is a group of cohomological dimension at most 1. This is equivalent to every surjective morphism $H \to G$ having a section.

Theorem 7: Let $G$ be hyperbolic, virtually special group. If $H$ is a closed subgroup of $\widehat{G}$ that does not contain a free non-abelian pro-$p$ subgroup for any $p$ then $H$ is virtually isomorphic to $\mathbb{Z}^\pi \rtimes \mathbb{Z}^\rho$, where $\pi$ and $\rho$ are disjoint (possibly empty) sets of primes. If $G$ is torsion free, then 'virtually' can be omitted.

Proof: Put $H_0 = \widehat{\Gamma}_0 \cap H$, where $\Gamma_0$ is the subgroup from Theorem 1. By Proposition 3, $\widehat{\Gamma}_0$ is torsion free. We prove that $H_0 \cong \mathbb{Z}^\pi \rtimes \mathbb{Z}^\rho$, with $\pi \cap \rho = \emptyset$, by induction on the length of the malnormal hierarchy. By Theorem 4 and Lemma 6, $\widehat{\Gamma}_0$ acts $1$-acylindrically on a profinite tree and therefore so does $H_0$. By the inductive hypothesis, the vertex stabilizers in $H_0$ have the claimed structure. If $H_0$ stabilizes a vertex by the induction hypothesis we are done. 

Otherwise, 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. $\blacksquare$

Theorem 8: Let $M$ be a closed, orientable, aspherical $3$-manifold. Then $M$ is hyperbolic if and only if the profinite completion $\widehat{\pi_1M}$ does not contain a subgroup isomorphic to $\widehat{\mathbb{Z}}^2$.

Proof: Agol's theorem implies that if $M$ is a closed hyperbolic $3$-manifold then $\pi_1M$ is virtually special, and hence Theorem 7 applies to show that there are no $\widehat{\mathbb{Z}}^2$ subgroups. Conversely, if $\pi_1M$ contains no subgroups isomorphic to $\widehat{\mathbb{Z}}^2$ then, since it is known that abelian subgroups of $3$-manifold groups are separable, $\pi_1M$ contains no subgroups isomorphic to $\mathbb{Z}^2$ and so, by the Hyperbolization theorem, $M$ is hyperbolic. $\blacksquare$

Recognising Seifert-fibred geometry

We now briefly discuss how to recognise the Seifert-fibred fundamental groups.

Theorem 9: Let $M$ be a closed, orientable, aspherical 3-manifold. Then $M$ is Seifert fibred if and only if the profinite completion $\widehat{\pi_1M}$ has a non-trivial procyclic normal subgroup.

Proof sketch: For $M$ satisfying the conditions the fundamental group has a cyclic normal subgroup and all finitely generated subgroups are separable so this is visible in the profinite completion.

Conversely, if $M$ isn't Seifert-fibred and is a Sol-manifold, the $\mathbb{Z}^2 \rtimes \mathbb{Z}$ passes to the profinite completion. Since $\widehat{\mathbb{Z}}$ is as a profinite ring the direct product of all $p-$adic integers and $\widehat{\mathbb{Z^2}} \cong \widehat{\mathbb{Z}}^2 $ it will suffice to show that the image of $z$ generating the monodromy doesn't fix a non-trivial element of $\mathbb{Z}_p$. This implies that any procyclic subgroup of $\widehat{\pi_1(M)}$ intersects $\widehat{\mathbb{Z^2}}$ trivially so lives in the kernel of the map from $\widehat{<z>} \to \mathrm{GL_2}(\widehat{\mathbb{Z}})$. One now checks this on Sylow subgroups which is easy since everything is completely explicit.

For non-geometric or hyperbolic manifolds one can produce an acylindrical action on a profinite tree, but a procyclic normal subgroup would have to be in a vertex stabiliser, which contradicts acylindricity unless the subgroup is in fact trivial. $\blacksquare$

A compact Seifert fibred 3-manifold $M$ can admit any of six different geometries. We next explain how these geometric structures are also distinguished by the profinite completion of the fundamental group. Consider the corresponding short exact sequence

\[1 \to \mathbb{Z} \to \pi_1M \to \pi_1O \to 1\]

where $\mathbb{Z}$ is cyclic and $O$ is a compact, cone-type 2-orbifold. The geometric structure of $M$ is determined by two invariants: the geometric structure of $O$; and the `Euler number' $e(M)$.

We first note that the geometry of $O$ is detected by the profinite completion of $\pi_1M$.

Lemma 10: Let $M$ be a compact, Seifert fibred 3-manifold, as above. $O$ is spherical if and only if $\pi_1M$ is virtually procyclic, and $O$ is Euclidean if and only if $\pi_1M$ is virtually nilpotent.

Proof: This follows immediately from the following facts: $O$ is spherical if and only if $\pi_1M$ is virtually cyclic; $O$ is Euclidean if and only if $\pi_1M$ is virtually nilpotent; $\pi_1M$ is residually finite. $\blacksquare$

If $M$ is not spherical then $O$ is a good orbifold, and so is covered by a compact, orientable surface $\Sigma$. The short exact sequence for $\pi_1M$ then pulls back along the covering map $\Sigma \to O$ to a short exact sequence

\[1 \to \mathbb{Z} \to \pi_1N \to \pi_1\Sigma \to 1\]

where $N$ is a finite-sheeted covering space of $M$. We next explain how the Euler number of $M$ manifests itself in the structure of $\pi_1M$. The following lemma is standard in 3-manifold topology:

Lemma 11: Let $M$ be a compact, Seifert fibred 3-manifold. If $O$ is spherical then $e(M) = 0$ if and only $\pi_1M$ is infinite. If $O$ is not spherical then $e(M) = 0$ if and only if, for some finite-sheeted surface cover $\Sigma$ of $O$, the short exact sequence   \[1 \to \mathbb{Z} \to \pi_1N \to \pi_1\Sigma \to 1\]    splits.

Finally, we explain how the splitting of the short exact sequence

\[1 \to \mathbb{Z} \to \pi_1N \to \pi_1\Sigma \to 1\]

is detected by the profinite completion.

Lemma 12: Let $\Sigma$ be a compact, orientable surface. Then a short exact sequence

\[1 \to \mathbb{Z} \to G \to \pi_1\Sigma \to 1 \quad (1)\]

splits if and only if the induced exact sequence of profinite completions

\[1 \to \hat{Z} \to \widehat{G} \to \widehat{\pi_1\Sigma} \to 1 \quad (2)\]

splits.

Proof: If (1) splits then clearly so does (2). Suppose therefore that (1) does not split.

The exact sequence $1 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/n \to 1$ induces a long exact sequence in cohomology, part of which is

\[\cdots \to H^2(\pi_1\Sigma,\mathbb{Z}) \cong \mathbb{Z} \to H^2(\pi_1\Sigma,\mathbb{Z}/n) \to H^1(\pi_1\Sigma,\mathbb{Z}) \to \cdots\]

But $H^2(\pi_1\Sigma,\mathbb{Z}/n) \cong \mathbb{Z}/n$, whereas $H^1(\pi_1\Sigma,\mathbb{Z})$ is isomorphic to the abelianization of $\pi_1\Sigma$, which is torsion-free. Therefore, the induced map $H^2(\pi_1\Sigma,\mathbb{Z}) \to H^2(\pi_1\Sigma,\mathbb{Z}/n)$ is surjective. In particular, if (1) does not split then neither does the corresponding extension $1 \to \mathbb{Z}/n \to G \to \pi_1\Sigma \to 1$, for some $n$.

Since $\pi_1\Sigma$ is good, we have that $H^2(\widehat{\pi_1\Sigma},\mathbb{Z}/n) \to H^2(\pi_1\Sigma,\mathbb{Z}/n)$, and so the corresponding sequence $1 \to \mathbb{Z}/n \to \widehat{G} \to \widehat{\pi_1\Sigma} \to 1$ does not split. Therefore, (2) does not split, as required. $\blacksquare$

We can finally prove the theorem mentioned at the start: 

Theorem 13: Let $M,N$ be closed, orientable, irreducible 3-manifolds, and suppose that $\widehat{\pi_1M} \cong \widehat{\pi_1N}$. Then $M$ admits one of Thurston's eight geometric structures if and only if $N$ does, and in this case both $M$ and $N$ admit the same geometric structure.

Proof: We have show that $M$ is hyperbolic if and only if $N$ is. Since $3-$manifold groups are residually finite and admit Sol geometry if and only if they are soluble but not virtually nilpotent, $M$ admits Sol geometry if and only if $N$ does. The Seifert-fbred case follows from Theorem 9 and the last 3 lemmas. $\blacksquare$