{'fraction_non_alphanumeric': 0.05579556746014602, 'fraction_numerical': 0.010584524992439625, 'mean_word_length': 3.398156944708341, 'pattern_counts': {'":': 0, '<': 8, '<?xml version=': 0, '>': 3, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 56, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
Hyperbolic groups with 1-dimensional boundary

If a torsion-free hyperbolic group G has 1-dimensional boundary, then the boundary is a Menger curve or a Sierpinski carpet provided G does not split over a cyclic group. When the boundary of G is a Sierpinski carpet we show that G is a quasi-convex subgroup of a 3-dimensional hyperbolic Poincare duality group. We also construct a ``topologically rigid'' hyperbolic group G: any homeomorphism of the boundary of G is induced by an element of G.


Introduction
We recall that the boundary ∂ ∞ X of a locally compact Gromov hyperbolic space X is a compact metrizable topological space. Brian Bowditch observed that any compact metrizable space Z arises this way: view the unit ball B in Hilbert space as the Poincaré model of infinite dimensional hyperbolic space, topologically embed Z in the boundary of B, and then take the convex hull CH(Z) to get a locally compact Gromov hyperbolic space with ∂ ∞ CH(Z) = Z. On the other hand when X is the Cayley graph of a Gromov hyperbolic group G then the topology of ∂ ∞ X ≃ ∂ ∞ G is quite restricted. It is known that ∂ ∞ G is finite dimensional, and either perfect, empty, or a two element set (in the last two cases the group G is elementary). It was shown recently by Bowditch and Swarup [B2,Sw1] that if ∂ ∞ G is connected then it does not have global cut-points, and thus is locally connected according to [BM]. The boundary of G necessarily has a "large" group of homeomorphisms: if G is nonelementary then its action on ∂ ∞ G is minimal, and G acts on ∂ ∞ G as a discrete uniform convergence group. It turns out that the last property gives a dynamical characterization of boundaries of hyperbolic groups, according to a theorem of Bowditch [B3]: if Z is a compact metrizable space with |Z| ≥ 3 and G ⊂ Homeo(Z) is a discrete uniform convergence subgroup, then G is hyperbolic and Z is G-equivariantly homeomorphic to ∂ ∞ G.
There are two questions which arise naturally: If G ′ is a hyperbolic group whose boundary is homeomorphic to the boundary of a topologically rigid hyperbolic group G, then there is a finite normal subgroup N ⊳ G ′ so that G ′ /N embeds in G as a finite index subgroup. We construct a topologically rigid group in the section 7.
Most of the results of our paper concern hyperbolic groups with one-dimensional boundary.
Theorem 1 Let G be a hyperbolic group which does not split over a finite or virtually cyclic subgroup, and suppose ∂ ∞ G is 1-dimensional. Then one of the following holds (see section 2 for definitions): 1. ∂ ∞ G is a Menger curve. 2. ∂ ∞ G is a Sierpinski carpet. 3. ∂ ∞ G is homeomorphic to S 1 and G maps onto a Schwartz triangle group with finite kernel.
It is probably impossible to classify hyperbolic groups whose boundaries are homeomorphic to the Menger curve (since this is the "generic" case), however it appears that a meaningful study is possible in the case of hyperbolic groups whose boundaries are homeomorphic to the Sierpinski carpet. Recall that the Sierpinski carpet S has canonical collection of peripheral circles (see section 2).
Theorem 2 Suppose that ∂ ∞ G ∼ = S. Then: 1. There are only finitely many G-orbits of peripheral circles. 2. The stabilizer of each peripheral circle C is a quasi-convex virtually Fuchsian group which acts on C as a uniform convergence group. We call these subgroups peripheral subgroups of G.
3. If we "double" G along the collection of peripheral subgroups using amalgamated free product and iterated , then the result is a hyperbolic groupĜ which contains G as a quasiconvex subgroup.
Known examples are consistent with the following: Conjecture Let G be a hyperbolic group Sierpinski carpet boundary. Then G acts discretely, cocompactly, and isometrically on a convex subset of H 3 with nonempty totally geodesic boundary.
There is now some evidence for this conjecture. It would follow from a positive solution of Cannon's conjecture together with Theorem 2, see section 5. Alternately, in the torsion-free case, if one could show that (hyperbolic) 3-dimensional Poincaré duality groups are 3-manifold groups, then Thurston's Haken uniformization theorem could be applied to an irreducible 3-manifold with fundamental group isomorphic to the groupĜ produced in Theorem 2. Under extra conditions (such as coherence and the existence of a nontrivial splitting) one can show that a 3-dimensional Poincaré duality group is a 3-manifold group, [KK].
The conjecture above leads one to ask which hyperbolic groups have planar boundary. Concretely, one may ask if a torsion-free hyperbolic group with planar boundary has a finite index subgroup subgroup isomorphic to a discrete convex cocompact subgroup of Isom(H 3 ). Here is a cautionary example which shows that in general it is necessary to pass to a finite index subgroup: if one takes a surface of genus 1 with two boundary components and glues one boundary circle to the other by a degree 2 map, then the fundamental group G of the resulting complex enjoys the following properties: 1. G is torsion-free and hyperbolic. 2. G contains a finite index subgroup which isomorphic to a discrete, convex cocompact subgroup of Isom(H 3 ). In particular, the boundary of G is planar.
3. G is not a 3-manifold group.

Preliminaries
Properties of hyperbolic groups and spaces. For a proof of the following properties of hyperbolic groups, we refer the reader to [Gr,ABC+,GH,B3]. Let G be a nonelementary Gromov hyperbolic group, and suppose G acts discretely and cocompactly on a locally compact geodesic metric space X. Then the boundary of X is a compact metrizable space ∂ ∞ X on which Isom(X) acts by homeomorphisms. For any f ∈ Isom(X), we denote the corresponding homeomorphism of ∂ ∞ X by ∂ ∞ f . The action of G on ∂ ∞ X is minimal, i.e. the G-orbit of every point is dense in ∂ ∞ X. Let ∂ 2 ∞ X := ∂ ∞ X × ∂ ∞ X − Diag be the space of distinct pairs in ∂ ∞ X. Then the set of pairs of points (x, y) ∈ ∂ 2 ∞ X which are fixed by an infinite cyclic subgroup of G is dense in ∂ 2 ∞ X. We let∂ 2 ∞ X := ∂ 2 ∞ X/(x, y) ∼ (y, x). The group G acts cocompactly and properly discontinuously on ∂ 3 X := {(x, y, z) ∈ (∂ ∞ X) 3 | x, y, z distinct}. There is a natural topology on X ∪ ∂ ∞ X which is a Ginvariant compactification of X, and this is compatible with the topology on ∂ ∞ X.
Recall that a subset S of a geodesic metric space is C-quasi-convex if every geodesic segment with endpoints in S is contained in the C-tubular neighborhood of S. Quasi-convex subsets of δ-hyperbolic metric spaces satisfy a visibility property (cf. [EbOn]): then given any two unit speed geodesics γ 1 , γ 2 starting at x and ending in Y , and any t ∈ [0, R] we have d(γ 1 (t), Im(γ 2 )) < δ and d(γ 2 (t), Im(γ 1 )) < δ.
As a consequence of the visibility property, if Y k ⊂ X is a sequence of C-quasiconvex subsets of a δ-hyperbolic space X, and d(x, Y k ) → ∞ as k → ∞, then a subsequence of Y k 's converges to a single point ξ ∈ ∂ ∞ X.
Sierpinski carpets and Menger curves. The classical construction of a Sierpinski carpet is analogous to the construction of a Cantor set: start with the unit square in the plane, subdivide it into nine equal subsquares, remove the middle open square, and then repeat this procedure inductively on the remaining squares. If we take a sequence D i ⊂ S 2 of disjoint closed 2-disks whose union is dense in S 2 so that Diam(D i ) → 0 as i → ∞, then S 2 − ∪ i Interior(D i ) is a Sierpinski carpet; moreover any Sierpinski carpet embedded in S 2 is obtained in this way [W]. Sierpinski carpets can also be characterized as follows [W]: a compact, 1-dimensional, planar, connected, locally connected space with no local cut points is a Sierpinski carpet.
We will use a few topological properties of Sierpinski carpets S: 1. There is a unique embedding of S in S 2 up to post-composition with a homeomorphism of S 2 .
2. There is a countable collection C of "peripheral circles" in S, which are precisely the nonseparating topological circles in S.
3. Given any metric d on S and any number D > 0, there are only finitely many peripheral circles in S of diameter > D.
The Menger curve may be constructed as follows. Start with the unit cube I 3 in R 3 . Consider the orthogonal projections π ij : I 3 → F ij of the unit cube onto the ij coordinate square, and let S ij ⊂ F ij be the Sierpinski carpet as constructed above. The Menger curve is the intersection ∩ i<j π −1 ij (S ij ). The Menger curve is universal among all compact metrizable 1-dimensional spaces: any such space can topologically embedded in the Menger curve. By [A1, A2], a compact, metrizable, connected, locally connected, 1-dimensional space is a Menger curve provided it has no local cut points, and no nonempty open subset is planar.

Proof of Theorem 1
The fact that G does not split over a finite group implies [St] that G is one-ended, and ∂ ∞ G is connected. Recall that by the results of [BM,B2,Sw1], the boundary of a one-ended hyperbolic group is locally connected and has no global cut points; furthermore, if ∂ ∞ G has local cut points then G splits over a virtually infinite cyclic subgroup unless ∂ ∞ G ≃ S 1 and G maps onto a Schwarz triangle group with finite kernel. Therefore from now on we will assume that ∂ ∞ G has no local cut points.
A 1-dimensional, compact, metrizable, connected, locally connected space Z with no local cut points is a Menger curve provided no point z ∈ Z has a neighborhood which embeds in the plane (see section 2). Hence either ∂ ∞ G is a Menger curve or some ξ ∈ ∂ ∞ G has a planar neighborhood U; therefore we assume the latter holds.
Lemma 3 Let Γ ⊂ ∂ ∞ G be a subset homeomorphic to a finite graph. Then Γ is a planar graph.
Proof. Since the action of G on ∂ ∞ G is minimal, every G-orbit intersects the planar neighborhood U, and so every point of ∂ ∞ G has a planar neighborhood. Because ∂ ∞ G has no local cut points, we have ∂ ∞ G \ Γ = ∅. So we can find a hyperbolic element g ∈ G whose fixed point set {η 1 , η 2 } ⊂ ∂ ∞ G is disjoint from Γ (section 2). Hence for sufficiently large n, g n (Γ) is contained in a planar neighborhood of η 1 or η 2 .
We recall [C, M] that a compact, metrizable, connected, locally connected space X with no global cut points is planar as long as no nonplanar graph embeds in X. Therefore ∂ ∞ G is planar. Finally, by [W], ∂ ∞ G is Sierpinski carpet.

Groups with Sierpinski carpet boundary
Let M be a compact hyperbolic manifold with nonempty totally geodesic boundary and let G := π 1 (M) be its fundamental group. The universal coverM of M may be identified with a closed convex subset of H 3 which is bounded by a countable disjoint collection P of totally geodesic planes. Each P ∈ P bounds an open half-space disjoint fromM .M is obtained from H 3 by removing each of these open half-spaces, and ∂ ∞M ⊂ ∂ ∞ H 3 is obtained from ∂ ∞ H 3 ≃ S 2 by deleting the open disks corresponding to these half-spaces. The closures of these disks are disjoint since the distance between distinct elements of P is bounded away from zero. As ∂ ∞M has no interior points in S 2 , it is a Sierpinski carpet (see section 2). Note that the peripheral circles of ∂ ∞M are in 1-1 correspondence with elements of P, and therefore the conjugacy classes of G-stabilizers of peripheral circles are in 1-1 correspondence with P/G, the set of boundary components of M. The stabilizer of a peripheral circle is the same as the stabilizer of the corresponding element of P, so these stabilizers are quasi-convex in G.
The next theorem shows that similar conclusions hold for any hyperbolic group whose boundary is a Sierpinski carpet.
Theorem 4 Let G be a hyperbolic group with boundary homeomorphic to the Sierpinski carpet S. Then 1. There are finitely many G-orbits of peripheral circles in S. 2. The stabilizer of each peripheral circle C is a quasi-convex subgroup G whose boundary is C.
Proof. We recall that G acts cocompactly on the space ∂ 3 G := {(x, y, z) ∈ (∂ ∞ G) 3 | x, y, z distinct}. Therefore if C k ⊂ ∂ ∞ G is a sequence of peripheral circles, (x k , y k , z k ) ∈ ∂ 3 G and {x k , y k , z k } ⊂ C k , then after passing to a subsequence we may find a sequence g k ∈ G, (x ∞ , y ∞ , z ∞ ) ∈ ∂ 3 G so that (g k x k , g k y k , g k z k ) converges to (x ∞ , y ∞ , z ∞ ). But this means that Diam(g k (C k )) is bounded away from zero, so g k (C k ) belongs to a finite collection of peripheral circles, and hence g k (C k ) is eventually constant. We conclude that there are only finitely many G-orbits of peripheral circles, and the stabilizer of any C ∈ C acts cocompactly on the space of distinct triples in C. By [B2] Stab(C) is a quasi-convex subgroup of G, and ∂ ∞ Stab(C) = C. From now on we will refer to stabilizers of peripheral circles as peripheral subgroups. By [Ga, CJ, T] each peripheral subgroup is, modulo a finite normal subgroup, a cocompact Fuchsian group in Isom(H 2 ).

Doubling Sierpinski carpet groups along peripheral subgroups
In this section we prove Theorem 2.
Let G be a hyperbolic group with ∂ ∞ G ≃ S, and let H 1 , . . . , H k be a set of representatives of conjugacy classes of peripheral subgroups of G. We define a graph of groups G as follows. The underlying graph has two vertices and k edges (no loops). Each vertex is labelled by a copy of G, the i th edge is labelled by H i , and the edge homomorphisms H i → G are given by the inclusions. We letĜ be the fundamental group of G.
Next we construct a tree of spaces on which the groupĜ acts in a natural way. Let X 0 be a finite Cayley 2-complex for G, and let X i be a finite Cayley 2-complex for the group H i . The inclusion H i ֒→ G is induced by a cellular map h i : X i → X 0 between the 2-complexes. Let h : ∪X i → X 0 be the corresponding map from the disjoint union of the X i 's to X 0 , and let X denote the mapping cylinder of h.
Let DX be the double of X along the collection of subcomplexes X i , i = 1, ..., k. Consider now the universal cover DX of DX with the deck transformation groupĜ. Let Y be the 1-skeleton of DX. The 1-skeletons of the subcomplexes X i , i = 1, ..., k lift to disjoint edge subspaces of Y . The vertex subspaces of Y are closures of the connected components to the complement to edge spaces. Each vertex space is the cover of 1-skeleton of X. Let T be the graph dual to the decomposition of Y into vertex and edge subspaces: vertices v of T correspond to vertex spaces Y v ⊂ Y , the edges e correspond to the edge subspaces Y e ⊂ Y . An edge e is incident to a vertex v if and only if Y e is contained in Y v . It is standard that the graph T is actually a tree (compare [SW]). Let V and E denote the collections of vertices and edges in T respectively. If v ∈ T we let E v denote the collection of edges containing v.
Let σ : DX → DX be the natural involution of DX. A map τ : Y → Y is a reflection if it is a lift of σ and it fixes some point; each reflection fixes some edge space in Y , and each edge space Y e is the fixed point set of precisely one reflection r e . Let Γ be the group generated by the reflections in Y . The group Γ is normalized byĜ since conjugation of a reflection by an element ofĜ yields another reflection; likewisê G is normalized by Γ. Let v ∈ T be any vertex. Then Γ is the free product of order two subgroups of the form r e where e ∈ E v . The vertex space Y v is a fundamental domain for the action of Γ on Y . The group Γ preserves the tree structure of Y , so we have an induced action of Γ on T by tree automorphisms, each reflection r e acting on T as an inversion of the edge e. The action of Γ on T naturally induces an action of Γ on ∂ ∞ T . The space Y is a connected graph, and we give it the natural path-metric where each edge in Y has unit length.
Lemma 5 1. The space Y is Gromov-hyperbolic. 2. Edge and vertex spaces are all K-quasi-convex in Y for some K.
3. There is a function C(R) such that for every R, the intersection of R-neighborhoods of any two distinct vertex or edge spaces has diameter at most C(R) unless the spaces are incident.
We have a coarse Lipschitz projection p : to v for each v ∈ V , and maps each edge space to the midpoint of the corresponding edge of T . If γ : [0, ∞) → ∂ ∞ Y is a unit speed geodesic ray, then p • γ is a coarse Lipschitz path with the bounded backtracking property 1 by the quasi-convexity of vertex/edge spaces. Hence p • γ maps into a finite tube around a geodesic ray τ in T . If p • γ is unbounded in T , then the equivalence class of the ray τ is uniquely determined by γ and we label γ with the associated boundary point [τ ] ∈ ∂ ∞ T . By the quasi-convexity of edge spaces, if γ hits an edge space for an unbounded sequence of times, then it remains in a quasi-convex neighborhood of the edge space. In this case, we know that γ eventually remains in a bounded neighborhood of a unique edge space by property 3 in Lemma 5, and we label γ with this edge. If neither of the above two cases occurs, then for each edge e of the tree, we know that γ eventually lies in one of the two components of the complement of the edge space Y e , and we label the edge with an arrow pointing in the direction of the corresponding subtree of T . There must be some (and at most one) vertex v ∈ T such that all edges emanating from v have arrows pointing toward v; otherwise we could follow arrows and leave any bounded set. There must be an unbounded sequence of times t k such that γ(t k ) lies in the vertex space Y v (by the construction of the edge labelling); by quasi-convexity of Y v , this means that γ eventually lies in the R-neighborhood of Y v ; in this case we label γ by v. Equivalent geodesic rays are given the same label. We get a labelling map Label : We now examine the topology of ∂ ∞ Y . This space is metrizable and we fix a metric d on ∂ ∞ Y ; in what follows we will implicitly use d when discussing metric properties Sierpinski carpet. Similarly, the peripheral circles of the Sierpinski carpet ∂ ∞ Y v are in 1-1 correspondence with the boundaries of edge spaces Y e ⊂ Y v .
By the visibility property of the uniformly quasi-convex edge spaces, there is at most one boundary point of ∂ ∞ Y labelled by any ξ ∈ ∂ ∞ T . For each edge e in T , the set of points in ∂ ∞ Y labelled by e is the ideal boundary of the edge space Y e , i.e. a circle. For each vertex v ∈ T , the set of points labelled by v is i.e. the Sierpinski carpet ∂ ∞ Y v minus the union of its peripheral circles.
Our next goal is to describe the topology of ∂ ∞ Y using the tree T . Choose v ∈ T . Every edge e of T separates T into two subtrees, and we let T v,e ⊂ T be the subtree disjoint from v. We define the outward subset, Out v,e , for a pair (v, e) ∈ V × E to be the collection of points of ∂ ∞ Y labelled by elements of T v,e ∪ ∂ ∞ T v,e . The visibility property of Y implies that for a fixed v ∈ T and any ǫ > 0 there are only finitely many edges e ⊂ T so that the diameter of Out v,e exceeds ǫ. Outward subsets of ∂ ∞ Y are open since a geodesic ray γ with ∂ ∞ γ ∈ Out v,e will eventually leave any tubular neighborhood of the edge space Y e , and so nearby boundary points correspond to rays which eventually lie in the same component of the complement of Y e in Y . It follows that if ξ ∈ ∂ ∞ T , and e k is the sequence of edges occurring in the ray vξ, then the sequence of outward sets Out v,e k is a nested basis for the topology of ∂ ∞ Y at the point labelled by ξ.
In this case, for any subset E ⊆ E v containing all but finitely many elements of E v , the sequence ξ k eventually lies in 3. ξ ∞ is labelled by an edge e 0 . In this case, if v, w are the endpoints of e 0 then for any subset E ⊆ E v containing all but finitely many elements of E v , and any subset F ⊆ E w containing all but finitely many elements of E w , the sequence ξ k eventually lies in Proof. Case 1. If v is any arbitrary vertex of T , and e 1 , e 2 , . . . is the sequence of edges comprising the geodesic ray vξ ∞ ⊂ T , then Out v,e j ⊂ ∂ ∞ Y is a neighborhood basis for ξ ∞ . Therefore Label(ξ k ) converges to Label(ξ ∞ ) by the definition of the topology on T ∪ ∂ ∞ T .
Case 2. If this weren't the case, then a subsequence of ξ k would converge to an element of Out v,e = Out v,e ∪ ∂ ∞ Y e for some e / ∈ E. This contradicts the fact that ξ ∞ is labelled by v.
Proof. Let G ′ be the fundamental group of a compact hyperbolic 3-manifold M with nonempty totally geodesic boundary. Recall (see section 2) that ∂ ∞ G ′ is a Sierpinski carpet. Using the notation developed above (decorated with "primes"),Ĝ ′ is the fundamental group of the double of M, so ∂ ∞Ĝ ′ is homeomorphic to S 2 . We will construct a homeomorphism between ∂ ∞Ĝ ′ and ∂ ∞Ĝ .
Choose vertices v ∈ T and v ′ ∈ T ′ , and a bijection E v → E v ′ . This induces an isomorphism between Coxeter groups Γ → Γ ′ , which we will use to identify Γ with Γ ′ . There is a unique Γ-equivariant isomorphism T ∪ ∂ ∞ T → T ′ ∪ ∂ ∞ T ′ which induces the given bijection E v → E v ′ ; we will use primes to denote corresponding edges and vertices. Choose an enumeration By construction, f ∞ is compatible with label maps, i.e. the following diagram commutes: We claim that f ∞ extends continuously to a homeomorphism f : ∂ ∞ Y → ∂ ∞ Y ′ . In view of the naturality of our construction it is enough to show that f ∞ extends to a continuous map f : ∂ ∞ Y → ∂ ∞ Y ′ ≃ ∂ ∞Ĝ ′ ≃ S 2 , since the inverse map may be produced by exchanging the roles of G and G ′ . Pick a sequence ξ k ∈ ∂ ∞ Y which converges to some ξ ∈ ∂ ∞ Y . We will show that f ∞ (ξ k ) converges.
Case 1: ξ is labelled by some η ∈ ∂ ∞ T . In this case there is a unique ξ ′ ∈ ∂ ∞ Y ′ which is labelled by η ′ ∈ ∂ ∞ T ′ . We know that if e i (resp e ′ i ) is the sequence of edges of the ray vη (resp v ′ η ′ ), then the outward sets Out v,e i (resp. Out v ′ ,e ′ i ) form a basis for the topology of Since each e k occurs only finitely often, Case 3: ξ is labelled by an edge e 0 ∈ T . We leave this case to the reader, as it is similar to case 2.
Corollary 8 If G is torsion-free, then so isĜ, and in this caseĜ is a 3-dimensional Poincaré duality group by [BM], [Bes2]. By [DD], if one splits a P D(n) group over a P D(n−1) subgroup, then the vertex groups (together with the incident edge subgroups) define relative P D(n) pairs; therefore (G; H 1 , . . . , H k ) is a relative Poincaré duality pair. In particular χ(G) = 1 2 i χ(H i ) < 0.
Corollary 9 Let G be a torsion-free hyperbolic group with Sierpinski carpet boundary.

Suppose either
A. Cannon's conjecture is true or B. Every 3-dimensional Poincaré duality group with a nontrivial splitting is the fundamental group of a closed 3-manifold. Then G is the fundamental group of a compact hyperbolic 3-manifold with totally geodesic boundary.
Proof. Let H 1 , . . . , H k ,Ĝ, Γ, be as in the first part of this section. If A holds, thenĜ is the fundamental group of a closed hyperbolic 3-manifold M. SinceĜ splits nontrivially by its very definition, if B holds thenĜ = π 1 (M) where M is a closed irreducible 3-manifold. M is Haken since its fundamental group splits, and so Thurston's uniformization theorem implies that M admits a hyperbolic structure. In either case we haveĜ acting on H 3 discretely, cocompactly, and isometrically.
The reflection group Γ acts onĜ by conjugation, with each reflection centralizing a unique quasi-convex edge subgroup ofĜ. By Mostow rigidity, Γ acts isometrically on the universal cover of M normalizing the actionĜ H 3 . G ⊂Ĝ is a quasi-convex subgroup, and so it acts on H 3 as a convex cocompact subgroup. The limit set of G in ∂ ∞ H 3 is a Sierpinski carpet, and because every peripheral subgroup of G is centralized by a unique reflection in Γ ⊂ Isom(H 3 ), the peripheral circles are fixed by reflections in Γ. Thus each peripheral circle of the limit set of G is a round circle, and so the convex hull of the limit set is a convex subset bounded by disjoint totally geodesic hyperbolic planes. It follows that G is the fundamental group of a compact hyperbolic manifold with totally geodesic boundary.

Examples
We now use Theorems 1 and 2 to see that some classes of hyperbolic groups have Menger curve boundary.
We first remark that a torsion-free hyperbolic group with Sierpinski carpet boundary has negative Euler characteristic by Corollary 8. So if G is a torsion-free hyperbolic group with 1-dimensional boundary, G doesn't split over a trivial or cyclic group, and χ(G) ≥ 0, then ∂ ∞ G is a Menger curve.
Theorem 10 Let G be a torsion-free 2-dimensional hyperbolic group that does not split over trivial and cyclic subgroups and which fits into a short exact sequence: Proof. In view of Theorem 1, it is enough to show that ∂ ∞ G is not Sierpinski carpet. Suppose it is. Note that if F admits a finite Eilenberg-Maclane space, then it is easy to see that χ(G) = χ(F )χ(Z) = 0, so ∂ ∞ G cannot be a Sierpinski carpet by the remark above. However there are examples such that F is not a finitely presentable group (see [R]). We now consider the general case. Then (G; H 1 , . . . , H k ) is a relative Poincare duality pair. Let K 0 be a finite Eilenberg-Maclane space for the group G, let D be a disjoint union of finite Eilenberg-Maclane spaces for the groups H 1 , . . . , H k , and let K be the mapping cylinder for a map D → K 0 which induces the given maps H i ֒→ G. We view D as a subcomplex of K. Consider the finite cyclic coverings which are induced by the homomorphisms G → Z → Z n . Then each pair (K n , D n ) again satisfies relative Poincare duality in dimension 3, so H * (K n , D n ; Z/2) ∼ =H 3− * (K n ; Z/2) We will use the notation b j (L) to denote the rank of H j (L, Z/2). Thus lim n→∞ b 1 (D n ) = ∞ and b 1 (K n ) ≤ b 1 (F ) + 1 < ∞. Consider the exact sequence of the pair (K n , D n ): ... → H 1 (K n ; Z/2) → H 1 (D n ; Z/2) → H 2 (K n , D n ; Z/2) → ...
Since b 1 (K n ) is bounded by b 1 (F ) + 1 and lim n→∞ b 1 (D n ) = ∞, it follows that lim n→∞ Dim(H 2 (K n , D n ; Z/2)) = ∞. This contradicts the fact that H 2 (K n , D n ; Z/2) ∼ = H 1 (K n ; Z/2). Now let F be a finitely generated free group and φ : F → F be a hyperbolic automorphism (see [BF2] for the definition). Consider the extension 1 → F → G → Z → 1 induced by φ. The group G is hyperbolic by [BF2]. The cohomological dimension of G is 2 by the Mayer-Vietoris sequence, thus the boundary of G is 1-dimensional by [BM].
Proof. We will show that the group G does not split over a cyclic (possibly trivial) subgroup. Suppose that it does. Then we have the corresponding action of G on a minimal simplicial tree T with cyclic edge stabilizers. Consider the restriction of this action on the subgroup F . Let T ′ ⊂ T be the minimal F -invariant subtree, then T ′ is Z-invariant (since Z normalizes F ), thus T ′ = T . By Grushko's theorem (in the case of trivial edge stabilizers) and the generalized accessibility theorem [BF1] (in the case of infinite cyclic stabilizers), the quotient T /F is a finite graph Γ. The action of Z = z projects to action on Γ, after taking a finite iteration of φ (if necessary) we may assume that z acts trivially on Γ. Since G does not contain Z 2 -subgroups, the edge stabilizers for the action of F on T must be trivial. Thus we get a free product decomposition of F so that each factor is invariant under some iterate of z. This contradicts the assumption that the corresponding automorphism φ : F → F is hyperbolic.
Theorem 12 Let G be a finite graph of groups. Suppose 1. Each vertex group is a torsion-free hyperbolic group whose boundary is either a Menger curve or a Sierpinski carpet; and at least one vertex group has Menger curve boundary.
2. Each edge group is a finitely generated free group of rank at least 2, and includes as a quasi-convex subgroup of each of the corresponding vertex groups.
3. If T is the Bass-Serre tree for G, and e 1 , e 2 ⊂ T are two edges emanating from the same vertex v ∈ T , then their stabilizers intersect trivially. Then the fundamental group G of G is a hyperbolic group with Menger curve boundary.
Proof. Conditions 2 and 3 imply that G is hyperbolic by [BF2], and vertex groups are quasi-convex subgroup of G by [Mi,Sw2]. G is torsion-free since all vertex groups are torsion-free. G has cohomological dimension 2 by the Mayer-Vietoris sequence, so ∂ ∞ G has dimension 1 by [BM].
We claim that G does not split over trivial or infinite cyclic groups. To see this, let T be the Bass-Serre tree of G, and let S be the Bass-Serre tree of a splitting of G over trivial and/or cyclic groups. Consider two adjacent vertices v 1 , v 2 ∈ T , let G v i ⊂ G be their stabilizers, and let G e be the stabilizer of the edge joining them. Since G v i does not split over trivial or cyclic subgroups [B2], G v i has a nonempty fixed point set in S. If s i ∈ S is fixed by G v i , then the segment joining s 1 to s 2 will be fixed by G e . Since G e is free of rank at least 2, we see that s 1 = s 2 . Therefore by induction we find that G has a global fixed point in S, which is a contradiction.
If the stabilizer of v ∈ T has Menger curve boundary, then by the quasi-convexity of G v in G, the Menger curve embeds in ∂ ∞ G. This shows that ∂ ∞ G cannot be homeomorphic to S 1 or the Sierpinski carpet. By Theorem 1 ∂ ∞ G is a Menger curve.

Topologically rigid groups
In this section we will construct some examples of topologically rigid groups. Before proceeding, we first note a consequence of Theorem 1.
Corollary 13 Let G be a nonelementary hyperbolic group with Dim(∂ ∞ G) ≤ 1. Then G is not topologically rigid.
We will sketch a proof of the corollary, and leave the details to the reader.
Case I: G has more than one end. Then G splits as an amalgamated product or HNN extension over a finite group. Let G T be the action of G on the Bass-Serre tree associated to such a splitting, so there is only one edge orbit in T . Following along the same lines as in section 5, we construct a tree of spaces X, with vertex and edge spaces corresponding to vertices and edges in T . For each vertex v ∈ T , the vertex space X v ⊂ X is quasi-convex in X and as in section 5 we may label points in ∂ ∞ X with elements of T ∪ ∂ ∞ T . The outward sets (see section 5) are open and closed in ∂ ∞ X. If e 1 and e 2 are incident to a vertex v then they lie in the same G v -orbit (since G/T has only one edge). Out v,e 1 and Out v,e 2 are disjoint and homeomorphic, so we may define a homeomorphism of ∂ ∞ X by swapping them while holding everything else fixed. This construction yields a continuum of homeomorphisms of ∂ ∞ X, so G → Homeo(∂ ∞ X) cannot be surjective.
Case II: G is 1-ended. If ∂ ∞ G is homeomorphic to S 1 , the Sierpinski carpet, or the Menger curve then G cannot be topologically rigid since each of these spaces has uncountable homeomorphism group. Therefore by Theorem 1 we may assume that G splits as an amalgamated free product or HNN extension over a virtually cyclic group. Let G T be the action of G on the Bass-Serre tree associated with such a splitting. If e is an edge in T , e = v 1 v 2 , then Out v 1 ,e − ∂ ∞ X e and Out v 2 ,e − ∂ ∞ X e are open and closed in ∂ ∞ X − ∂ ∞ X e , and are preserved by G e . Take a g ∈ G e so that ∂ ∞ G fixes both points in ∂ ∞ G e , and define a homeomorphism f : ∂ ∞ X → ∂ ∞ X by f | Outv 1 ,e = ∂ ∞ g| Outv 1 ,e and f | Outv 2 ,e = id| Outv 2 ,e . This type of construction will give a continuum of homeomorphisms of ∂ ∞ X, so again G → Homeo(∂ ∞ X) cannot be surjective.
Our construction of topologically rigid groups is based on the idea (realized precisely in Proposition 16) that a homeomorphism of S 2 must be a Möbius transformation provided it preserves a sufficiently rich family of round circles. We begin with an analogous statement for homeomorphisms of S 1 .
Line configurations in H 2 . Let L be a locally finite collection of geodesics in H 2 so that the complementary regions of ∪ L∈L L are bounded, and we assume that there is a cocompact lattice Γ ⊂ Isom(H 2 ) stabilizing L. Let∂ 2 ∞ H 2 be the space of unordered distinct pairs in ∂ ∞ H 2 , and let ∂ ∞ L be the collection of pairs of endpoints ∂ ∞ L for L ∈ L, ∂ ∞ L := {∂ ∞ L | L ∈ L} ⊂∂ 2 ∞ H 2 . Note that if L 1 , L 2 ∈ L and ∂ ∞ L 1 ∩ ∂ ∞ L 2 = ∅ then L 1 = L 2 . Let Stab(∂ ∞ L) ⊂ Homeo(∂ ∞ H 2 ) be the group of homeomorphisms of ∂ ∞ H 2 which preserve ∂ ∞ L ⊂ ∂ 2 ∞ H 2 .
Lemma 14 1. If L 1 , L 2 ∈ L have nonempty intersection and g ∈ Stab(∂ ∞ L) fixes Proof. Our arguments essentially follow [CB, Proof of Theorem 2.7]. We will identify the space of geodesics in H 2 with ∂ 2 ∞ H 2 . (1) Suppose L 1 , L 2 ∈ L and g ∈ Stab(∂ ∞ L) fixes ∂ ∞ L 1 ∪∂ ∞ L 2 pointwise. If σ 1 , σ 2 are the connected components of ∂ ∞ H 2 − ∂ ∞ L 1 , then g(σ i ) = σ i since |∂ ∞ L 2 ∩ σ i | = 1 and ∂ ∞ L 2 is fixed by g. Observe that Σ i := {∂ ∞ L∩σ i | L ∈ L and |L ∩ L 1 | = 1} ⊂ σ i is a discrete subset of σ i with the order type (with respect to the ordering on σ i ≃ R) of the integers, and g(Σ i ) = Σ i . But g fixes the point ∂ ∞ L 2 ∩σ i ∈ Σ i and is orientation preserving, so g| Σ i = id Σ i . Therefore g fixes ∂ ∞ L for every L ∈ L with L ∩ L 1 = ∅. The incidence graph of L is connected, so we may apply this argument inductively to see that g fixes ∂ ∞ L for every L ∈ L. The set ∪ L∈L ∂ ∞ L is dense in ∂ ∞ H 2 , so g = id. This proves the first assertion of the lemma.
(2) We now show that every sequence g k ∈ Stab(∂ ∞ L) has a subsequence which is constant modulo Γ, which proves that [Stab(∂ ∞ L) : Γ] < ∞. Pick L 1 , L 2 ∈ L such that L 1 intersects L 2 in a point p. For each k let g k * L i ∈ L be the unique line with ∂ ∞ (g k * L i ) = g k (∂ ∞ L i ). Then (g k * L 1 ) ∩ (g k * L 2 ) = p k for some p k ∈ H 2 , and we may choose a sequence γ k ∈ Γ such that sup d(γ k (p k ), p) = R < ∞. Then the lines (γ k • g k ) * L i lie in the finite set {L ∈ L | L ∩ B(p, R) = ∅}, so after passing to a subsequence we may assume that (γ k • g k )| ∂∞L i independent of k for i = 1, 2. By the previous paragraph the sequence γ k • g k ∈ Homeo(∂ ∞ H 2 ) is constant.
Plane configurations in H 3 . Below we prove an analog of Lemma 14 for a collection H of totally geodesic hyperplanes in H 3 .
Let H be a locally finite collection of totally geodesic planes in H 3 , with stabilizer G := {g ∈ Isom(H 3 ) | g(H) ∈ H for every H ∈ H}. Let ∂ ∞ H := {∂ ∞ H | H ∈ H}. We assume that H satisfies the conditions: 1. G is a cocompact lattice in Isom(H 3 ). The local finiteness of H implies that there are finitely many G-orbits in H, and that the stabilizer of each H ∈ H acts cocompactly on H.
Definition 2 We will say that three circles Note that if the circles ∂ ∞ H 1 , ∂ ∞ H 2 , ∂ ∞ H 3 are in standard position and C 1 , C 2 , C 3 is another unordered triple of circles which bound elements of H, then C 1 , C 2 , C 3 is in standard position if and only if there is a homeomorphism f : ∂ ∞ H 1 ∪∂ ∞ H 2 ∪∂ ∞ H 3 → C 1 ∪ C 2 ∪ C 3 which carries elements of H to elements of H.
Let Stand denote the collection of unordered triples of circles in standard position. We will say that two elements of Stand are incident if they have exactly two circles in common.
Lemma 15 1. The incidence graph of Stand is connected.
2. If γ ⊂ ∂ ∞ H 3 is homeomorphic to S 1 , then either γ = ∂ ∞ H for some H ∈ H, or there is an H ∈ H so that ∂ ∞ H intersects both components of ∂ ∞ H 3 − γ.
Proof. The union ∪ H∈H H determines a polygonal subcomplex in H 3 with connected 1-skeleton. Therefore the assertion 1 follows.
To prove the assertion 2, let U and U ′ denote the connected components of ∂ ∞ H 3 − γ. We may find H, H ′ ∈ H so that ∂ ∞ H ⊂ U, ∂ ∞ H ′ ⊂ U ′ . Since the incidence graph for H is connected we can find a chain of planes H 0 = H, H 1 , ..., H n = H ′ in H so that consecutive planes intersect each other. We see that either γ = ∂ ∞ H j for some H j in this sequence or for some H j the circle ∂ ∞ H j intersects both U and U ′ .
since H 4 is the unique element of H whose boundary contains the 4 element set . Therefore by the previous paragraph we have f | ∂∞H 4 = id H 4 . Since the incidence graph of Stand is connected we see by induction that f | ∂∞H = id ∂∞H for all H ∈ H, and this forces f = id ∂∞H 3 .
Reasoning as in Lemma 14 we conclude that [Stab(∂ ∞ H) : G] < ∞. Let G ′ ⊂ G be a finite index normal subgroup of Stab(∂ ∞ H). Each f ∈ Stab(∂ ∞ H) normalizes the action G ′ ∂ ∞ H 3 , so by Mostow rigidity each f is a Möbius transformation. Therefore for every f ∈ Stab(∂ ∞ H) we have f = ∂ ∞ g for some g ∈ G.
Constructing topologically rigid groups. Let G ′ ⊂ G be a finite index torsionfree subgroup of G. Let {H 1 , . . . , H k } be a set of representatives of the G ′ -orbits in H, and let G i := Stab(H i ). For any 1 ≤ i ≤ k, the set of geodesics there is a finite collection Z i of conjugacy classes of maximal cyclic subgroups of G i with the property that for any g ∈ G ′ −G i , the intersection gG j g −1 ∩G i is an element of Z i . We now construct a double 2 G ′ along the collection of subgroups G i := Stab(H i ), 1 ≤ i ≤ k as follows: construct a graph of groups G with two vertices v 1 , v 2 and k edges e 1 , . . . , e k , where G v i is isomorphic to G ′ and G e i is isomorphic to G i . Identify G v i with G ′ . We choose the embeddings ι ij : G e i → G v j so that the image coincides with G i ⊂ G ′ , but so that the ι ij 's satisfying the following condition: LetĜ := π 1 (G), let T be the Bass-Serre tree associated with G, and let V and E denote the collections of vertices and edges in T respectively.Ĝ acts (discretely, cocompactly) on a tree of spaces X constructed as in section 5, with vertex spaces X v , v ∈ V and edge spaces X e , e ∈ E.
Lemma 17Ĝ is a hyperbolic group. All vertex and edge groups G x , x ∈ V ∪ T are quasi-convex subgroups ofĜ.
Proof. By [BF3], [Sw2], [Mi] it suffices to show that there is an upper bound on the length of essential annuli (see [BF3], section 1) in the graph of groups G. Or equivalently, we need to show that there is an upper bound on the length of any segment in T which is fixed by a nontrivial element g ∈Ĝ. We claim that if e 1 , e 2 , e 3 are 3 consecutive edges in the tree T , then G e 1 ∩ G e 2 ∩ G e 3 is trivial; for the twisting condition implies that the intersections G e 1 ∩ G e 2 and G e 2 ∩ G e 3 are cyclic subgroups of G e 2 with trivial intersection.
5. Pick e ∈ E, and let T 1 , T 2 ⊂ T be the two subtrees that one gets by removing the interior of the edge e. Then ∂ ∞ X − ∂ ∞ X e has two connected components, namely the closures of (∪ The proof of the lemma is similar to arguments from section 5, so we omit it.
Proof. We first claim that γ ⊂ ∂ ∞ X v for some v ∈ V . Otherwise by Alexander duality ∂ ∞ X v − γ is connected for every v ∈ V , and (∂ ∞ X v 1 ∪ ∂ ∞ X v 2 ) − γ is connected for any pair of adjacent vertices v 1 , v 2 ∈ V . By induction this implies that ∪ v∈V ∂ ∞ X v − γ is connected. By part 4 of Lemma 18 we conclude that ∂ ∞ X − γ is connected, a contradiction.
Hence we may assume that γ ⊂ ∂ ∞ X v for some v ∈ V . Suppose γ = ∂ ∞ X e for any e ∈ E adjacent to v. Then any point ξ ∈ ∂ ∞ X − γ lies in the same component of ∂ ∞ X − γ as one of the two components of ∂ ∞ X v − γ. By Lemma 15 we can find an edge e adjacent to v so that ∂ ∞ X e intersects both of the components U 1 , U 2 of ∂ ∞ X v − γ. So we may connect U 1 to U 2 within ∂ ∞ X w − γ where w is the other endpoint of e. This contradicts the assumption that γ separates ∂ ∞ X.
Let C denote the collection of unordered triples of circles C i = ∂ ∞ X e i , e i ∈ E, which are in standard position, i.e. there exists a triple H 1 , H 2 , H 3 ∈ H which are in standard position and a homeomorphism f : which carries each circle ∂ ∞ H i to one of the circles C j(i) . We define the incidence relation for elements of C the same way as before, let Γ(C) denote the associated incidence graph. Thus C contains the subsets S v where S v consists of triples of circles in standard position which are contained in ∂ ∞ X v . Then the incidence graph Γ(S v ) is isomorphic to the incidence graph of S, thus it is connected (see part 1 of Lemma 15). For each vertex v ∈ V the union of triples of circles Lemma 20 The subgraphs Γ(S v ) are the connected components of Γ(C).
Proof. It is enough to show that any {C 1 , C 2 , C 3 } ∈ C is contained in ∂ ∞ X v for some v ∈ T , since there is at most one ∂ ∞ X v containing any given pair of circles.
Pick {C 1 , C 2 , C 3 } ∈ C, with C i = ∂ ∞ X e i for e i ∈ E. Note that d(e i , e j ) ≤ 1 for 1 ≤ i, j ≤ 3 for otherwise we would have C i ∩ C j = ∅. Also, observe that if two of the circles lie in some ∂ ∞ X v , then the third one must too (because |∂ ∞ X e ∩ ∂ ∞ X v | ≤ 2 unless ∂ ∞ X e ⊂ ∂ ∞ X v ). Clearly this forces the edges e i to share a vertex.
Define the incidence graph with the vertex set {geoX v , v ∈ T }, where the vertices v, w are connected by an edge if and only if ∂ ∞ X v ∩ ∂ ∞ X w ≈ S 1 . Lemma 18 implies that this graph is isomorphic to the tree T .
Proposition 21 Any homeomorphism f : ∂ ∞ X → ∂ ∞ X preserves the collection of spheres {∂ ∞ X v , v ∈ V }. In particular, f induces an isomorphism of the tree T .
Proof. The homeomorphism f induces an automorphism f # of the graph Γ(C), thus it preserves its connected components. Therefore for each The paragraph preceeding Proposition implies that f induces an automorphism of the tree T .

Theorem 22
The homeomorphism group of ∂ ∞ X contains G as a subgroup of finite index. Therefore Homeo(∂ ∞ X) is a topologically rigid hyperbolic group.
Proof. For every v ∈ V , we identify ∂ ∞ X v with ∂ ∞ H 3 via a homeomorphism which carries the collection {∂ ∞ X e | e ∈ E, v ⊂ e} to ∂ ∞ H; this homeomorphism is unique up to a Möbius transformation by Proposition 16.
Suppose f ∈ Homeo(∂ ∞ X) and f | ∂∞Xv = id| ∂∞Xv for some v ∈ V . Then f fixes is a reflection. But condition 3 on H rules out the latter possibility. Therefore by induction we conclude that f fixes ∂ ∞ X w for every w ∈ V , and so f = id.
Pick v ∈ T , and consider the possibilities for f | ∂∞Xv where f ∈ Homeo(∂ ∞ X).
There are clearly only finitely many such possibilities up to post-composition with elements ofĜ; therefore by the preceding paragraphĜ has finite index in Homeo(∂ ∞ X).

An example of a plane configuration H.
We now construct a specific example of a plane configuration H satisfying the three required conditions. We start with the 3-dimensional hyperbolic polyhedron Φ described in Figure 1: the edges of the polyhedron are labelled with 2 and 3, they indicate that the corresponding dihedral angles of the polyhedron are π/2 and π/3 respectively. Such a polyhedron exists by Andreev's theorem [An]. Note that Φ has an order 3 isometry θ which is a rotation around the geodesic segment CE and reflection symmetries in each of three quadrilaterals, two of which are depicted in Figure 2.
The polyhedron Φ contains three squares which "bisect" Φ; one of them β 1 = P QRS which is indicated in Figure 1, the other two β 2 , β 3 are obtained from β 1 by applying the rotation θ. Lemma 23 The bisectors β 1 , β 2 , β 3 are realized by totally-geodesic 2-dimensional polygons in Φ which are orthogonal to the boundary of Φ. More precisely, for each 1 ≤ j ≤ 3 there is a totally geodesic plane H j ⊂ H 3 which intersects the same four edges of Φ as β j and H j intersects the faces of Φ orthogonally.
Proof. It is enough to prove the assertion for β 1 , the other two polygons are obtained via the rotation θ. The proof is similar to [Ka]: we first split open the cube Φ combinatorially along the bisector β 1 into two subcubes Φ + and Φ − . Each polyhedron Φ + , Φ − has a face F + , F − which corresponds to the bisector β 1 . We assign the label 2 to each edge of Φ ± is contained in F ± . Andreev's theorem again implies that Φ + and Φ − can be realized by polyhedra in H 3 (we retain the names Φ ± for these polyhedra). Our goal is to show that the homeomorphism F + → F − (which is given by identification with the bisector β 1 ) is isotopic (rel. vertices) to an isometry of the hyperbolic polygons. The polyhedron Φ admits a reflection symmetry which fixes the rectangle EJCA, and this symmetry also acts on the polyhedra Φ + , Φ − and quadrilaterals F ± so that the fixed point sets are the geodesic segments corresponding to P R. However it is clear that there exists a unique (up to vertex preserving isotopy) hyperbolic structure on quadrilateral P QRS so that the edges are geodesic, angles are π/2, π/3, π/2, π/3 and the quadrilateral has an order 2 isometry fixing P R. Thus we have a natural isometry F + → F − and we can glue Φ + to Φ − using this isometry. The result is a hyperbolic polyhedron Ψ which is combinatorially isomorphic to Φ this isomorphism preserves the angles. Thus by uniqueness part of Andreev's theorem (alternatively one can use Mostow rigidity theorem) the polyhedra Φ, Ψ are isometric.
On the other hand, the polyhedron Ψ contains totally geodesic 2-dimensional polygon F + = F − which is orthogonal to the boundary of Ψ. the dihedral angles between β j , j = 1, 2, 3 are all equal and are different from π/2 (otherwise the combinatorial cube P i which contains the vertex E would have all right angles which is impossible in hyperbolic space). Now we construct the collection of planes H as follows: let R ⊂ Isom(H 3 ) be the discrete group generated by reflections in the faces of Φ; the polyhedron Φ is a fundamental domain for R. The 2-dimensional hyperbolic polygons β j = H j ∩ Φ are orthogonal to ∂Φ, the plane H j is invariant under the subgroup R j of R generated by reflections in the faces of Φ which are incident to β j . The R-orbit of these hyperplanes is H. Note that (0) If H is a member of H and the intersection H ∩ Φ = ∅ then H ∩ Φ is equal to one of the bisectors β j . We next check that H satisfies the required properties: (1) The fundamental domain Φ for R is compact, hence the group R is a cocompact lattice.
(2) The complementary regions to H in H 3 are finite unions of the polyhedra P i , i = 1, ..., 8, thus they are bounded.
(3) Let ρ j be the reflection in the plane H j . Since the planes H j , 1 ≤ j ≤ 3 are not mutually orthogonal it follows that this reflection maps H i , i = j, to a plane which does not belong to H (see Property (0) above); it follows that ρ does not preserve the configuration H.