Countable groups are mapping class groups
A provocative statement due to Gromov is the dichotomy that any claim about all countable groups is either trivial or false. While this holds true in most cases, occasionally there are counterexamples.
Recall that a hyperbolic manifold is a connected orientable paracompact manifold equipped with a complete metric of constant sectional curvature -1. If $M$ is a hyperbolic 3-manifold, $\mathrm{Isom}(M)$ is the group of isometries of $M$, and $\mathcal{MCG}(M)$ is the group of isotopy classes of self-homeomorphisms of $M$, also known as the mapping class group of $M$. Recall that when $M$ is compact Mostow rigidity states that $\mathrm{Out(\pi_1}(M)) \cong \mathrm{Isom}(M)$ is a finite group, and every self-homeomorphism of $M$ is isotopic to an isometry.
My favourite counterexample to the Gromov dichotomy, which also generalises Mostow rigidity, is the following result due to Frigerio and Martelli:
Theorem 1: For every countable group $G$ there is a hyperbolic 3-manifold $M$ such that $$G \cong \mathcal{MCG}(M) \cong \mathrm{Out(\pi_1}(M)) \cong \mathrm{Isom}(M)$$
Corollary 2: There are uncountably many groups which arise as fundamental groups of hyperbolic 3-manifolds.
There are lots of restrictions on what groups can arise as the automorphisms of a group, e.g. it is an exercise to show that one cannot realise $\mathbb{Z}$ or any finite cyclic group as an automorphism group. The situation with outer automorphism groups is, in contrast, completely without restrictions. Minasyan showed that any countable group can be realised as the automorphism group of a 2-generated group with Kazhdan's property (T) using small cancellation techniques, and more recently Logan showed that for some fixed hyperbolic triangle group any group can be realised as the outer automorphism group of one of its HNN extensions. I find theorem 1 somewhat surprising probably because I grew up in the post-virtual fibring era when all the questions Thurston asked about hyperbolic 3-manifolds had already been solved, but that is somewhat misleading since there are still many interesting things which can be said about them.
The strategy to exhibit such manifolds will be to build 2-complexes with the right automorphism group and beef them up to get a 3-manifold in such a way that there are no extra symmetries introduced. I think of this as a generalisation of the exercise I was given as an undergraduate about how one can exhibit, for any finite group $G$, a finite polyhedron with $G$ as its isometry group.
Let's see how to do this. We will call a 2-dimensional polyhedron $P$ tame (the paper calls it special but since special is used so often elsewhere on this blog to refer to a specific class of cube complexes I thought I'd better use a different word) if each point $v \in P$ has a neighbourhood of one of the following forms:
take one piece of type $V$ for each vertex and $c(e)$ pieces of type $W$ for each edge $e$. The boundary of $V$ (resp. $W$) consists of 4 (resp. 2) $Y$-shaped graphs and 6 (resp. 3) arcs.
We will construct a piece $K$ with appropriate boundary so that whenever the tame polyhedron has a boundary we can just basically glue on $K$.
Consider the piece $V$: this is dual to a tetrahedron, which is combinatorially equivalent to an ideal regular hyperbolic octahedron $O$ with a checkerboard colouring of the faces. Define $K$ as the geometric object obtained by mirroring $O$ along its white faces. Since the dihedral angles of the regular ideal octahedron are right, $K$ is a complete hyperbolic manifold with non-compact geodesic boundary that consists of four geodesic thrice punctured spheres, each of which corresponds to a $Y$-shaped graph in $\partial V$. The six vertices of $O$ give rise to six annular cusps of $K$, which in turn intersect each component of $\partial K$ in neighbourhoods of the punctures.
It turns out, however, that if we only use $K$ then the orientation-reversing involution switching the ideal octahedra of this block would extend over the attempted thickening. To make our lives easier, we introduce the block $E$, which is chiral, i.e. has no orientation-reversing isometries.
Let $E_0$ be the manifold obtained by cutting the double of $K$ along one component of $\partial K$. By construction $E_0$ decomposes into two $K$-blocks, has 3 toric cusps, and 3 annular cusps. Moreover, $\partial E_0$ is given by two geodesic thrice-punctured spheres. This still has an orientation-reversing involution, which we now deal with.
Recall that a slope on a torus is an isotopy class of simple closed unoriented curves. Take three disjoint sections of the toric cusps: their boundaries are Euclidean tori whose slopes have a definite length. The group $\mathrm{Isom}^+(E_0)$ acts on the set of all slopes on the 3 toric cusps, and one can show that the orbit of any slope consists of 3 slopes of the same length, one in each of the 3 cusps. By a variant of Thurston's hyperbolic Dehn filling argument, there is a sufficiently long slope $s$ on a toric cusp such that:
- the slope is neither vertical nor horizontal, so the corrresponding triple is not invariant under orientation-reversing isometries;
- on the chosen horospherical sections of the cusps the slopes corresponding to $s$ have length $l(s)>12$;
- the manifold $E$ obtained via Dehn filling the 3 toric cusps by killing the 3 slopes corresponding to $s$, is hyperbolic with geodesic boundary;
- the shortest closed geodesics in $E$ are the cores of the added solid tori.
One can now show the following:
Proposition 3: $E$ and $K$ are complete, finite-volume hyperbolic 3-manifolds with totally geodesic boundary consisting of thrice punctured spheres such that:
- boundary components of $K$ and $E$ correspond to $Y$-shaped graphs in $\partial V$ and $\partial W$ respectively, with punctures corresponding to endpoints of the graphs;
- annular cusps of $K$ and $E$ correspond to arcs in the boundaries of $V$ and $W$ respectively;
- the intersections between the boundary components and the annular cusps of $K$ and $E$ correspond to intersections between the corresponding $Y-$shaped graphs and arcs in $\partial V$ and $\partial W$;
- denote by $\mathrm{Aut}(Z)$ the combinatorial automorphism group of a graph $Z$. The actions of the isometries on boundary components and cusps induce the isomorphisms: \[\mathrm{Isom}^+(K) \cong \mathrm{Aut}(\partial V) \cong S_4\] \[\mathrm{Isom}(E) \cong \mathrm{Isom}^+(E) \cong \mathrm{Aut}(\partial W) \cong S_3 \times \mathbb{Z}/2\]
- every geodesic thrice punctured sphere in $K$ and $E$ is contained in $\partial K$ and $\partial E$ respectively.
Now equip the blocks $K$ and $E$ with arbitrary orientations. Given $(P,c)$ decompose $P_0$ into pieces of type $V$ and $W$. Pick blocks of type $K$ and $E$ corresponding to pieces of type $V$ and $W$. For each $Y$-shaped graph we have a thrice-punctured sphere, whose punctures correspond to the endpoints of the graph. Given two geodesic thrice-punctured spheres, every bijection between their punctures is realized by a unique orientation-reversing isometry. The identifications of the $Y$-shaped graphs in $P_0$ therefore induce isometries between pairs of boundary punctured spheres, and we use such isometries for gluing all these pairs.
Since any symmetry of $V$ (resp.\ of $W$) translates into an orientation-preserving isometry of $K$ (resp.\ of $E$), there is no ambiguity in the construction, and the result is a complete oriented hyperbolic 3-manifold $M(P, c)$. Annular cusps glue up to toric cusps, which are in natural correspondence with the faces of $P$. Every vertex of $P$ corresponds to a $K$-block. Every edge $e$ of $P$ gives rise to $c(e)$ blocks of type $E$ and $c(e) + 1$ thrice-punctured spheres.
Remark 4: If $(P, c)$ is a decorated polyhedron, there are connected complete hyperbolic 3-submanifolds with geodesic boundary $M_i \subset M(P, c)$, $i \in \mathbb{N}$ with the following properties:
- $M_i$ is the union of a finite number of $K$-blocks and $E$-blocks, for all $i \in \mathbb{N}$;
- $M_i \subset \operatorname{int} M_{i+1}$ for all $i \in \mathbb{N}$, and $\bigcup_{i=0}^\infty M_i = M(P, c)$.
We say that a tame polyhedron is \emph{regular} if it has more than two vertices, and every edge in its 1-skeleton has distinct endpoints. A decorated polyhedron $(P, c)$ is \emph{big} if $P$ is regular and $c(e) > 4$ for every edge $e$ of $P$. We denote by $\operatorname{Aut}(P, c)$ the group of combinatorial automorphisms of $(P, c)$, i.e.\ the group of combinatorial automorphisms of $P$ preserving the decoration. As mentioned before we will exhibit tame polyhedra with arbitrary automorphism group and then thicken. First we exhibit polyhedra.
Proposition 5: Every countable group is the fundamental group of a regular tame polyhedron.
I think this proposition has a very funny proof.
Proof: A countable group has some presentation with countably many generators and relators. In general, the corresponding polyhedra are constructed by attaching a 1-cell for each generator and a 2-cell for each relator to some simply connected base-space (for instance, a point). Here, since cells need to be locally finite, the base-space needs to be non-compact.
Our base-space is the 2-skeleton of an infinite wall, i.e.\ of a strip $\mathbb{R}^2 \times [0, 1] \subset \mathbb{R}^3$ tessellated as in Fig.\ 5.
It is a simply connected tame polyhedron whose horizontal faces are hexagons and vertical faces are squares. Take an infinite line of consecutive bricks, a point inside the top-face of each brick, and add 1-cells to pairs of consecutive points, as in Fig.\ 5.
In our presentation, for each generator $g$, substitute its occurrences in the relators with new distinct generators $g_1, \ldots, g_k$, each occurring once, and add new relators $g_ig_{i+1}^{-1}$, so we can suppose that every generator occurs in exactly 3 relators, and once in each. Assign arbitrarily an orientation and a generator to each 1-cell, and attach for each relator a 2-cell running alternatively along a 1-cell (corresponding to a letter) and as a right-angled arc along the wall as in Fig.\ 6.
Each new two-dimensional region is a cell, and three distinct 2-cells run along each 1-cell. Therefore the resulting polyhedron is tame. It is easy to see that it is regular. $\blacksquare$
Proposition 6: Every countable group is the group of combinatorial automorphisms of a big decorated polyhedron.
Proof: In a decorated polyhedron, let us define the colour of a vertex as the 4-uple of colours of the adjacent edges. Let $G$ be a countable group and $P$ be a regular tame polyhedron with fundamental group $G$.
Let $e_0, e_1, \ldots, e_n, \ldots$ be an arbitrary ordering of the edges of $P$, and set $c(e_i) = i+4$ for all $i \in \mathbb{N}$. Since $P$ is regular, distinct vertices have distinct colours (because edges have distinct endpoints, and the singular locus of $P$ is not empty).
Let now $\widetilde{P}$ be the universal cover of $P$, and let $\widetilde{c}$ be the decoration of $\widetilde{P}$ induced by $c$. The fiber over a vertex or edge consists precisely of all vertices or edges of the same colour. Of course the group of deck transformations of $\widetilde{P}$ is isomorphic to $G$. By definition of $\widetilde{c}$, deck transformations are automorphisms. Conversely, every automorphism is a deck transformation, because it must preserve the colours of vertices and edges, and hence the fibers. $\blacksquare$
A little more work shows the following:
Proposition 7: Let $(P, c)$ be a big decorated polyhedron. Then
\[
\operatorname{Isom}(M(P, c)) = \operatorname{Isom}^+(M(P, c)) \cong \operatorname{Aut}(P, c).
\]
We will omit the proof since I think this feels very believable. To upgrade this to a result about outer automorphisms we need a bit more hyperbolic geometry background. Let $G$ be a finitely generated Kleinian group, i.e. a torsion-free discrete subgroup of $\mathrm{PSL}_2(\mathbb{C})$ and $M =\mathbb{H^3}/G$. Let $M_c$ be the compact core of $M$, i.e. a compact submanifold whose inclusion is a homotopy equivalence. It is a theorem of Scott that this always exists. One can show that the quantity $b(G)=-3\chi(M_c)$ is an invariant of the group and $G$ contains at most $b(G)$ conjugacy classes of rank-1 maximal parabolic subgroups. If this is in fact atained then $G$ is said to be maximally parabolic. We quote
Theorem 8: A finitely generated Kleinian group $G$ is maximally parabolic if and only if its convex core is a complete finite-volume hyperbolic 3-manifold with geodesic boundary, consisting of a finite number of totally geodesic thrice-punctured spheres.
Let $G$ and $G'$ be Kleinian groups and suppose $\phi \colon G \to G'$ is an isomorphism. We say that $\phi$ is type-preserving if $\phi$ sends parabolic elements to parabolic elements and loxodromic elements to loxodromic elements.
Theorem 9: Let $\phi \colon G \to G'$ be a type-preserving isomorphism between two Kleinian groups $G$ and $G'$. If $G$ is maximally parabolic, then there exists an element $h \in \operatorname{Isom}(\mathbb{H}^3)$ such that $\phi(g) = hgh^{-1}$ for all $g \in G$.
From isomorphisms of fundamental groups to isometries
Let $(P, c)$ be any decorated polyhedron, and let $\Gamma$ be the Kleinian group with $M(P, c) \cong \mathbb{H}^3/\Gamma$.
Theorem 10: Let $\phi \colon \Gamma \to \Gamma$ be an isomorphism of abstract groups. Then there exists an isometry $g \in \operatorname{Isom}(\mathbb{H}^3)$ with $\phi(\gamma) = g\gamma g^{-1}$ for all $\gamma \in \Gamma$.
Proof: If $P$ is compact then the conclusion follows from Mostow-Prasad's rigidity Theorem, so we concentrate here on the case when $P$ is non-compact.
The parabolic elements of $\Gamma$ can be characterized as those elements belonging to a $\mathbb{Z} \oplus \mathbb{Z}$ subgroup of $\Gamma$, so $\phi$ is type-preserving.
Let $M_i \subset M(P, c)$, $i \in \mathbb{N}$ be the finite-volume manifolds with geodesic boundary described in Remark 4, and take a basepoint $x_0 \in M_0$. The map $\pi_1(M_i, x_0) \to \pi_1(M(P, c), x_0)$ induced by the inclusion is injective, because $M_i$ has geodesic (and hence incompressible) boundary. Let $\Gamma_i < \Gamma$ be the subgroup corresponding to $\pi_1(M_i, x_0)$ under the identification $\pi_1(M(P, c), x_0) \cong \Gamma$. By Theorem 8, $\Gamma_i$ is maximally parabolic, so Theorem 9 implies that for every $i \in \mathbb{N}$ there exists $g_i \in \operatorname{Isom}(\mathbb{H}^3)$ such that $\phi(\gamma) = g_i\gamma g_i^{-1}$ for all $\gamma \in \Gamma_i$. It follows that for all $i \in \mathbb{N}$ the isometry $g_i g_0^{-1}$ commutes with all the elements in $\Gamma_0$. Since $\Gamma_0$ is non-elementary, this implies $g_0 = g_i$ for all $i \in \mathbb{N}$, whence $\phi(\gamma) = g_0\gamma g_0^{-1}$ for every $\gamma \in \bigcup_{i \in \mathbb{N}} \Gamma_i = \Gamma$. $\blacksquare$
We have denoted by $\operatorname{MCG}^{\text{hom}}(M(P, c))$ the group of homotopy classes of self-homeomorphisms of $M(P, c))$.
Corollary 11: We have:
\[
\operatorname{Out}(\pi_1(M(P, c))) \cong \operatorname{Isom}(M(P, c)) \cong \operatorname{MCG}^{\text{hom}}(M(P, c)).
\]
Proof: Since $\Gamma$ is not elementary, the natural map $\pi \colon \operatorname{Isom}(M(P, c)) \to \operatorname{Out}(\pi_1(M, c))$ is injective (in particular, there exists at most one isometry in every homotopy class of self-homeomorphisms of $M(P, c))$. By Theorem 16 it follows that $\pi$ is an isomorphism, so we are left to prove that any self-homeomorphism of $M(P, c))$ is homotopic to an isometry.
Let $f$ be a self-homeomorphism of $M(P, c))$. Then there exist an isomorphism $f_* \colon \Gamma \to \Gamma$ and a $f_*$-equivariant lift $\widetilde{f} \colon \mathbb{H}^3 \to \mathbb{H}^3$ of $f$ (i.e.\ a lift of $f$ with $\widetilde{f}(\gamma(x)) = f_*(\gamma)(\widetilde{f}(x))$ for all $\gamma \in \Gamma$, $x \in \mathbb{H}^3$). By Theorem 10, an element $g \in \operatorname{Isom}(\mathbb{H}^3)$ exists such that $\phi(\gamma) = g\gamma g^{-1}$ for all $\gamma \in \Gamma$. We now define $\widetilde{F} \colon \mathbb{H}^3 \times [0, 1] \to \mathbb{H}^3$ by setting $\widetilde{F}(x, t) = (1 - t) \cdot \widetilde{f}(x) + t \cdot g(x)$. Being $f_*$-equivariant, $\widetilde{F}$ projects onto a homotopy $F \colon M(P, c)) \to M(P, c))$ between $f$ and an isometry, whence the conclusion. $\blacksquare$
Finally, to get to the original mapping class group, we have to show that homotopic self-homeomorphisms of manifolds arising from our construction are isotopic.
Definition: Let $M$ be a non-compact manifold. We say that $M$ is end-reducible if there exist a compact set $W \subset M$ and a sequence $\{\lambda_n\}_{n \in \mathbb{N}}$ of simple loops in $M \setminus W$ with the following properties: any compact subset of $M$ intersects only a finite number of $\lambda_i$'s, and each $\lambda_i$ is homotopically trivial in $M$ and homotopically non-trivial in $M \setminus W$. A non-compact manifold is \emph{end-irreducible} if it is not end-reducible.
It has been shown that two homotopic self-homeomorphisms of an end-irreducible non-compmact manifold are isotopic, and the manifolds we built are end-irreducible basically because the boundaries are totally geodesic, so we are done.
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