{'fraction_non_alphanumeric': 0.08190699730228163, 'fraction_numerical': 0.011827285058907806, 'mean_word_length': 3.122910644715345, 'pattern_counts': {'":': 0, '<': 7, '<?xml version=': 0, '>': 23, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 53, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
Automorphism groups of right-angled buildings: simplicity and local splittings

We show that the group of type-preserving automorphisms of any irreducible semi-regular thick right-angled building is abstractly simple. When the building is locally finite, this gives a large family of compactly generated (abstractly) simple locally compact groups. Specializing to appropriate cases, we obtain examples of such simple groups that are locally indecomposable, but have locally normal subgroups decomposing non-trivially as direct products.


Introduction
Let (W, I) be a right-angled Coxeter system, i.e. a Coxeter system such that m i,j = 2 or m i,j = ∞ for all i = j. We assume that the generating set I is finite.
Haglund-Paulin have shown that for any tuple of (not necessarily finite) cardinalities (q i ) i∈I , there exists a right-angled building of type (W, I) with prescribed thicknesses (q i ) i∈I , in the sense that for each i ∈ I, all i-panels have thickness of the same cardinality q i . We refer to [Dav98,Th. 5.1] for a group-theoretic construction of that building. Moreover, such a building is unique up to isomorphism (see Proposition 1.2 in [HP03]). A right-angled building satisfying that condition on the panels is called semi-regular (this terminology is motivated by the case of trees). It is thick if q i > 2 for all i ∈ I.
The following shows that the automorphism groups of these buildings provide a large family of simple groups.
Theorem 1.1. Let X be a thick semi-regular building of right-angled type (W, I). Assume that (W, I) is irreducible non-spherical.
Then the group Aut(X) + of type-preserving automorphisms of X is abstractly simple, and acts strongly transitively on X.
Recall that strong transitivity means transitivity on pairs (c, A) consisting of a chamber c and an apartment A containing c (we implicitly refer to the complete apartment system). Haglund and Paulin [HP03, Prop. 1.2] have shown that Aut(X) + is chamber-transitive; in fact, the main tools in the proof of Theorem 1.1 rely on their work in an essential way.
If W is infinite dihedral, then a building X of type (W, I) with prescribed thicknesses (q i ) i∈I is nothing but a semi-regular tree. In that case the simplicity of the type-preserving automorphism group G = Aut(X) + is due to Tits [Tit70]. If (W, I) is a right-angled Fuchsian group (i.e. if I = {1, . . . , r} and m i,j = 2 if and only if |i − j| = 1 or r − 1), then a building X of type (W, I) is a Bourdon building, and the simplicity statement is due to Haglund-Paulin [HP98].
After this work was completed, K. Tent informed me that she had obtained independently a proof of bounded simplicity in the case right-angled buildings whose panels are of countable thickness; this stronger simplicity statement means that there is a uniform constant N such that the group can be written as a product of N copies of each of its non-trivial conjugacy classes. In the case of trees, bounded simplicity was proved without any restriction on the thickness by J. Gismatullin [Gis09]. Another related simplicity theorem was also obtained by N. Lazarovich [Laz]; it applies to a large family of groups acting on locally finite, finite-dimensional CAT(0) cube complexes. It is likely that the special case of Theorem 1.1 concerning locally finite right-angled buildings could also be deduced from [Laz], using the fact that right-angled buildings can be cubulated.
Notice that a building whose type-preserving automorphism group is chambertransitive, is necessarily semi-regular. The following is thus immediate from Theorem 1.1.
Corollary 1.2. Let X be an irreducible thick right-angled building of non-spherical type. If Aut(X) + is chamber-transitive, then it is strongly transitive and abstractly simple.
In the special case when X is locally finite, i.e. when q i < ∞ for all i ∈ I, the group G endowed with the compact-open topology is a second countable totally disconnected locally compact group. It is compactly generated since it acts chambertransitively on X. In particular Theorem 1.1 provides a large family of compactly generated simple locally compact groups. Our next goal is to describe their rich local structure.
A general study of the local structure of compactly generated, topologically simple, totally disconnected locally compact groups is initiated in [CRW13a] and [CRW13b]. The main objects of consideration in that study are the locally normal subgroups, namely the compact subgroups whose normaliser is open. The trivial subgroup, as well as the compact open subgroups, are obviously locally normal, considered as trivial. It is important to observe that a compactly generated, locally compact group can be topologically simple and nevertheless possess non-trivial locally normal subgroups. Basic examples of such groups are provided by the type-preserving automorphism group of semi-regular locally finite tree. It turns out that the group of type-preserving automorphisms of an arbitrary semi-regular locally finite tree always admits non-trivial locally normal subgroups; some of them even split nontrivially as direct products (see Lemma 9.1 below). The case of trees has however a special additional property: some compact open subgroups split as a direct product of infinite closed subgroups; the corresponding factors are a fortiori locally normal and non-trivial. It is thus natural to ask for which right-angled buildings that situation occurs, beyond the case of trees. The following provides a complete answer to this question, implying in particular that open subgroups admit non-trivial product decompositions only under very special circumstances.
Theorem 1.3. Let X be a building of right-angled type (W, I) and prescribed thicknesses (q i ) i∈I , with 2 < q i < ∞ for all i ∈ I. Assume that (W, I) is irreducible non-spherical.
Then the following assertions are equivalent: (i) All open subgroups of G = Aut(X) + are indecomposable. (ii) G is one-ended. (iii) W is one-ended.
By indecomposable, we mean the non-existence of a non-trivial direct product decomposition. The set of ends of a compactly generated locally compact group is defined with respect to compact generating sets in the same way as for discrete groups (see [Abe74]). Notice that Theorem 1.3 establishes a relation between the local structure of G (because the existence of an open subgroup splitting non-trivially as a product can be detected in arbitrarily small identity neighbourhoods) and its asymptotic properties.
The condition that W is one-ended can easily be read on the Coxeter diagram (see Theorem 9.2 for a precise formulation). H. Abels [Abe74] has shown that a natural analogue of Stallings' theorem holds for non-discrete locally compact groups. This ensures that G = Aut(X) + is one-ended if and only if it does not split non-trivially as an amalgamated free product over a compact open subgroup.
It follows from Theorem 1.3 that, if X is a Bourdon building, then compact open subgroups of Aut(X) + are indecomposable, but they have locally normal subgroups that split non-trivially as products, all of whose factors are themselves locally normal. With Theorem 1.3 at hand, one can construct buildings X of arbitrarily large dimension whose automorphism group has that property.
Acknowledgement. I thank Colin Reid and George Willis for numerous inspiring conversations; the main motivation for the present work was in fact provided by the common enterprise initiated in [CRW13b]. I am grateful to the anonymous referee for a very thorough reading of the paper; his/her numerous comments and suggestions helped much in correcting its inaccuracies and improving its readability.

Projections and parallel residues
Throughout the paper, we will mostly view a building X of type (W, I) as a Wmetric space; we refer to [AB08] for the basic concepts. Occasionally, geometric arguments will require to consider the Davis realisation of X, as defined in [Dav98] or [AB08,Ch. 12]. This point of view will be implicit when discussing configuration of walls in a given apartment. In order to avoid any confusion between these two viewpoints, we will avoid to identify a residue R with the chambers adjacent to it; instead the latter set of chambers is denoted by Ch(R).
A fundamental feature of buildings is the existence of combinatorial projections between residues. We briefly recall their basic properties, which will be frequently used in the sequel. All the properties which we do not prove in detail are established in [Tit74, §3.19].
Let X be a building of type (W, I). Given a chamber c ∈ Ch(X) and a residue σ in X, the projection of c on σ is the unique chamber of Ch(σ) that is closest to c. It is denoted by proj σ (c). For any chamber d ∈ Ch(σ), there is a minimal gallery from c to d passing through proj σ (c), and such that the subgallery from proj σ (c) to d is contained in Ch(σ). Moreover, any apartment containing c and meeting Ch(σ) also contains proj σ (c). An important property of proj is that it does not increase the numerical distance between chambers: for all c, c ′ ∈ Ch(X), the numerical distance from proj σ (c) to proj σ (c ′ ) is bounded above by the numerical distance from c to c ′ .
If σ and τ are two residues, then the set is the chamber-set of a residue contained in σ. That residue is denoted by proj σ (τ ). The rank of proj σ (τ ) is bounded above by the ranks of both σ and τ .
We shall often use the following crucial property of the projection map; we outsource its statement for the ease of reference.
Two residues σ and τ are called parallel if proj σ (τ ) = σ and proj τ (σ) = τ . In that case, the chamber sets of σ and τ are mutually in bijection under the respective projection maps. Since the projection map between residues does not increase the rank, it follows that two parallel residues have the same rank. A basic example of parallel residues is provided by two opposite residues in a spherical building. Another one is provided by the following.
Proof. Let Σ be an apartment containing x and y. By convexity, it also contains x ′ = proj τ (x) and y ′ = proj σ (y). Since σ and τ are parallel, every wall of Σ crossing σ ∩ Σ also crosses τ ∩ Σ and vice-versa. It follows that the respective stabilisers of σ ∩ Σ and τ ∩ Σ in the Weyl group W coincide. In particular the unique element w ∈ W mapping x to y ′ preserves both σ and τ . Since σ and τ are parallel, we have proj τ (y ′ ) = y so that w maps x ′ to y. Hence dist(x, τ ) = dist(x, x ′ ) = dist(y ′ , y) = dist(σ, y). The result follows.
The relation of parallelism plays a special role among panels. The following criterion will be used frequently.
If two chambers of σ ′ have distinct projections on σ, then σ and σ ′ are parallel.
Proof. If two chambers of σ ′ have distinct projections on σ, then proj σ (σ ′ ) is a panel, which is thus the whole of σ. Therefore, in an apartment intersecting both σ and σ ′ , we see that those panels lie on a common wall. It follows that proj σ ′ (σ) cannot be reduced to a single chamber. Hence proj σ ′ (σ) = σ ′ and the result follows.
The following result shows that two residues are parallel if and only if they share the same set of walls in every apartment intersecting them both. This useful criterion allows one to detect parallelism of residues by just looking at parallelism among panels.
Lemma 2.5. Let R and R ′ be two residues. Then R and R ′ are parallel if and only if for all panels σ of R and σ ′ of R ′ , the projections proj R ′ (σ) and proj R (σ ′ ) are both panels.
Proof. The 'only if' part is clear from the definition. Assume that R and R ′ are not parallel. Up to swapping the roles of R and R ′ , we may thus assume that proj R (R ′ ) is a proper residue of R. Let then c and d be a pair of adjacent chambers in R so that c is the projection of some chamber of R ′ and d is not.
If the latter two chambers coincide, then the projection on R ′ of the panel shared by c and d is a chamber and not a panel, and the desired condition holds. Otherwise the panel shared by c and d is parallel to the panel shared by c ′ and d ′ by Lemma 2.4. This implies that c = proj R (c ′ ) and d = proj R (d ′ ), contradicting the fact that d does not belong to the chamber-set of proj R (R ′ ).
Another useful fact is the following.
Lemma 2.6. Let R and R ′ be two residues.
Then proj R ′ (R) and proj R (R ′ ) are parallel.
Proof. Let σ be a panel contained in proj R (R ′ ). Then there is a panel σ ′ in R ′ such that σ = proj R (σ ′ ). It follows that σ and σ ′ are parallel. Therefore, we have where the second equality follows from Lemma 2.1. It follows that proj R ′ (σ) is a panel. Clearly, we have proj R ′ (σ) = proj proj R ′ (R) (σ). This shows that the projection of σ to proj R ′ (R) is a panel. By symmetry, the projection of any panel of proj R ′ (R) to proj R (R ′ ) is also a panel. By Lemma 2.5, we infer that proj R ′ (R) and proj R (R ′ ) are parallel.
We shall see that parallelism of residues has a very special behaviour in rightangled buildings. For instance, we have the following useful criterion.
Proposition 2.7. Let X be a right-angled building of type (W, I).
(i) Two parallel residues have the same type.
(ii) Given a residue R of type J, a residue R ′ is parallel to R if and only if R ′ is of type J and R and R ′ are both contained in a residue of type J ∪ J ⊥ .
We recall that J ⊥ is the subset of I defined by In the special case where J is a singleton, say J = {j}, it is customary to make a slight abuse of notation and write j = J and j ⊥ = J ⊥ when referring to the type of a residue; this should not cause any confusion.
Proof of Proposition 2.7. (i) In a right-angled building, any two panels lying on a common wall in some apartment have the same type. That two parallel residues have the same type is thus a consequence of Lemma 2.5.
(ii) Any two residues of type J in a building of type J ∪ J ⊥ are parallel. This implies that the 'if' part holds. Assume now that R and R ′ are parallel. Let c ∈ Ch(R) and c ′ = proj R ′ (c). We show by induction on dist(c, c ′ ) that the type of every panel crossed by a minimal gallery from c to c ′ belongs to J ⊥ . Let c = d 0 , d 1 , . . . , d n = c ′ be such a minimal gallery. Let also i be the type of the panel shared by c = d 0 and d 1 , and let σ i denote that i-panel. For any j ∈ J, let also σ j be the j-panel of c. By Lemma 2.5, the projection σ ′ j = proj R ′ (σ j ) is a panel. The panels σ j and σ ′ j lie therefore on a common wall in any apartment containing them both. If i and j did not commute, then the wall W i containing the panel σ i in such an apartment would be disjoint from the wall W j containing σ j . This implies c ′ is separated from W j by the wall W i , which prevents the panel σ ′ j from lying on W j . This shows that ij = ji. In other words, we have i ∈ J ⊥ . Let next R 1 be the J-residue containing d 1 , and let S be the (J ∪ {i})-residue containing c. Thus R and R 1 are both contained in S.
We claim that R 1 is parallel to R ′ . In order to establish the claim, we first notice that proj R ′ (S) = R ′ , since R ⊂ S. By Lemma 2.6, the residue R ′ = proj R ′ (S) is parallel to proj S (R ′ ). In particular proj S (R ′ ) is of type J by part (i). Since i ∈ J ⊥ , all J-residues in S contain exactly one chamber of σ i . Thus σ i is not contained in proj S (R ′ ), and it follows that all chambers of R ′ have the same projection on σ i ; that projection is the unique chamber of Ch(σ i ) ∩ Ch(proj S (R ′ )). By construction, we have proj σ i (c ′ ) = d 1 ; we deduce that d 1 belongs to Ch(proj S (R ′ )). This proves that proj S (R ′ ) is the J-residue of d 1 ; it coincides therefore with R 1 .
Thus we have shown that R ′ = proj R ′ (S) and that R 1 = proj S (R ′ ), and those residues are parallel by Lemma 2.6. The claim stands proven.
The claim implies by induction on n that R 1 and R ′ are contained in a common residue of type J ∪ J ⊥ . That residue must also contain R, since R and R 1 are contained in a common residue of type J ∪{i} ⊆ J ∪J ⊥ . This finishes the proof.
Corollary 2.8. Let X be a right-angled building.
Then parallelism of residues is an equivalence relation.
We emphasize that parallelism of panels is not an equivalence relation in general. In fact, we have the following characterization of right-angled buildings.
Proposition 2.9. Let X be a thick building.
Then parallelism of residues is an equivalence relation if and only if X is rightangled.
Proof. By Corollary 2.8, it suffices to show that if X is not right-angled, then parallelism of panels is not an equivalence relation. If X is not right-angled, then it contains a residue R which is an irreducible generalized polygon. Let σ and σ ′ be two distinct panels of the same type in R, at minimal distance from one another. It follows that σ and σ ′ are not opposite in R, and thus not parallel since they do not lie on a common wall in apartments containing σ and σ ′ . By [Tit74, 3.30], there is a panel τ in R which is opposite both σ and σ ′ . Thus σ is parallel to τ and τ is parallel to σ ′ . Parallelism is thus not a transitive relation.

Wall-residues and wings
Let X be a right-angled building of type (W, I).
Since parallelism of residues is an equivalence relation by Corollary 2.8, it is natural to ask what the equivalence classes are. The answer is in fact already provided by Proposition 2.7: the classes of parallel J-residues are the sets of Jresidues contained in a common residue of type J ∪ J ⊥ .
Given a residue R of type J, we will denote the unique residue of type J ∪ J ⊥ containing R by R. The special case of panels is the most important one. A residue of the form σ, with σ a panel, will be called a wall-residue.
In the case when (W, I) is a right-angled Fuchsian Coxeter group, wall-residues are what Marc Bourdon calls wall-trees, see [Bou97]. The terminology is motivated by the following observation: if the intersection of a wall-residue with an apartment is non-empty, then it is a wall in that apartment.
Our next step is to show how residues determine a partition of the chamber-set of the ambient building into convex pieces. To this end, we need some additional terminology and notation.
To any c ∈ Ch(X) and J ⊆ I, we associate the set wing is a J-wing for some J ⊆ I. The following results record some basic properties of wings.
Lemma 3.1. Let X be right-angled building of type (W, I), let J ⊆ I and c ∈ Ch(X). Then we have: Proof. (i) The inclusion ⊆ is clear. To check the reverse inclusion, let x be a chamber whose projection onto R = Res J (c) is different from c. Then there is a minimal gallery from x to c via x ′ = proj R (x). Let i be type of the last panel crossed by that gallery, and let σ be that panel. By construction we have proj σ (x) = c. Moreover, since x ′ = c, we have i ∈ J. This implies that x ∈ X i (c), thereby proving (i).
(ii) Let x ∈ Ch(X) and set y = proj Res J ∪J ⊥ (x). Let also R = Res J (c) and The desired equality thus follows using part (ii).
Proposition 3.2. In a right-angled building, wings are convex.
Proof. Let X be right-angled building of type (W, I). Fix c ∈ Ch(X) and J ⊆ I. By Lemma 3.1(i) it suffices to prove that a wing of the form Assume that the gallery is not entirely contained in X i (c). Let j be the minimal index such that d j+1 ∈ X i (c), and let j ′ be the maximal index such that d j ′ −1 ∈ X i (c). Thus j ′ > j.
By Lemma 2.4, the panel σ j shared by d j and d j+1 is parallel to σ. Similarly, so is the panel σ j ′ shared by d j ′ and d j ′ −1 . Therefore, by Proposition 2.7, the set For each k between j and j ′ , we now set a gallery that is strictly shorter than the given minimal gallery d j , d j+1 , . . . , d j ′ . This is absurd.
By definition of the projection, the set Ch(X) is the disjoint union of the wings X J (d) over all d ∈ Ch(Res J (c)). It thus follows from Proposition 3.2 that any residue containing q chambers yields a partition of the building into q convex subsets.
For the sake of future references, we record the following fact.
For any x ∈ X i (c) and x ′ ∈ X i (c), the gallery from x to x ′ obtained by concatenating a minimal gallery from x to proj σ (x), a minimal gallery from proj σ (x) to proj σ (x ′ ), and a minimal gallery from proj σ (x ′ ) to x ′ , is minimal.

Proof.
A gallery is minimal if and only if its length equals the numerical distance between its extremities. Therefore, it suffices to show that there is some minimal gallery from x to x ′ passing through proj σ (x) and proj σ (x ′ ). Let , the gallery γ must cross some panel which is parallel to σ. By Proposition 2.7, this implies that the gallery γ meets the residue σ.
Let j (resp. j ′ ) be the minimal (resp. maximal) index k such that the chamber x k of γ belongs to Ch(σ). Then there is a minimal gallery γ j from x to x j (resp. γ j ′ from x j ′ to x ′ ) passing through proj σ (x) (resp. proj σ (x ′ )). By concatenating γ j and γ j ′ with the gallery x j , x j+1 , . . . , x j ′ , we obtain a galleryγ, of the same length as γ, and joining x to x ′ . Thusγ is minimal. By construction, it passes through proj σ (x) and proj σ (x ′ ).
Remark that if Σ is an apartment of X containing a chamber c, then the intersection X i (c) ∩ Σ is a half-apartment. The set of half-apartments is partially ordered by inclusion; the following result shows that this order relation is reflected by the ordering of the wings in the ambient building. This will play a crucial role in the subsequent discussions.
Lemma 3.4. Let i, i ′ ∈ I and c, c ′ ∈ Ch(X). Suppose that at least one of the following conditions holds.
Proof. Assume first that (a) holds and let Σ be an apartment containing c and c ′ . Let W (resp. W ′ ) be the wall of Σ which bounds the half-apartment X i (c) ∩Σ (resp. X i ′ (c ′ ) ∩ Σ). The fact that i = i ′ or m i,i ′ = ∞ ensures that the walls W and W ′ have trivial intersection (the case W = W ′ is excluded in view of Lemma 3.1(ii)). Therefore the wall W is contained in the half-apartment The latter case is excluded, since it would imply that c ′ ∈ X i (c) ∩ Σ. This proves that (b) holds. Hence it suffices to prove the lemma under the hypothesis (b).
We may assume that By hypothesis (b), and since apartments are convex, both chambers belonging to Σ ∩ Ch(σ ′ ) have the same projection on σ, namely x. Therefore proj σ (σ ′ ) = x and, in particular, proj σ (c ′′ ) = x.
By assumption, we have d ∈ X i (c) and c ′ ∈ X i (c). Therefore, Lemma 3.3 implies that Moreover, by the claim, we also have So it suffices to show that dist(x, c ′′ ) = dist(x, c ′ ) + 1. But Lemma 3.3 applied to c and c ′′ also implies that We also need to analyse when a ball or a residue is contained in a given wing. This is the purpose of the next result, whose statements require the following notation. We denote by B(R, n) the ball of radius n around Ch(R), i.e. the collection of all chambers c such that dist(c, R) ≤ n.
Lemma 3.5. Let R be a residue, let i ∈ I and let σ be a residue of type i ∪ i ⊥ . Let , contradicting the hypothesis. The claim stands proven.
Choose a chamber y ∈ B(R, n + 1) − X i (c). Let x = proj R (y) and let x = x 0 , x 1 , . . . , x m = y be a minimal gallery. Hence m = dist(x, y) = dist(R, y) ≤ n + 1. By the claim above, we have x ∈ X i (c). On the other hand y ∈ X i (c) by assumption, so that it makes sense to define k 0 = min{ℓ | x ℓ ∈ X i (c)}. Thus k 0 > 0 and We next observe that the panel σ ′ shared by x k 0 −1 and x k 0 is parallel to σ = Res i (c) by Lemma 2.4, and is thus of type i by Proposition 2.7(i). Moreover x k 0 −1 and x k 0 both belong to Ch(σ) by Proposition 2.7(ii). In particular we have There is a minimal gallery from x to proj σ (x) passing through x ′ = proj R (proj σ (x))). The residues proj R (σ) and R ′ are parallel by Lemma 2.6. Therefore, we deduce from Lemma 2.3 that This implies that dist(x, proj σ (x)) ≥ n. From the sequence of inequalities above, we deduce that m = k 0 = n + 1. Part (i) follows.
. Thus the desired inclusion follows from Lemma 3.5.
However the type of R ′ is a subset of J, and therefore does not contain i by hypothesis. This shows that Ch(R ′ ) ⊆ X i (c). Applying Lemma 3.5(i) with n = 0, we obtain Ch(R) ⊆ X i (c), as required.

Extending local automorphisms
The following important result was shown by Haglund-Paulin.
In other words, this means that the canonical homomorphism Stab Aut(X) + (R) → Aut(R) + is surjective.
It will be important for our purposes to ensure that the extension constructed in Proposition 4.1 can be chosen to satisfy some additional constraints. In particular, we record the following.
Proposition 4.2. Let X be a semi-regular right-angled building of type (W, I). Let i ∈ I and σ be an i-panel.
It remains to check thatα is an automorphism. To this end, let x and y be any two chambers and denote by x ′ and y ′ their projections on σ.
Assume now that x ′ = y ′ . Let then x ′′ and y ′′ denote the projections of x and y on σ. By Lemma 3.3, it suffices to show thatα preserves the Weyl-distance from x to x ′′ , the Weyl-distance from x ′′ to y ′′ and the Weyl-distance from y ′′ to y. Since wings are convex by Proposition 3.2, and since the restriction ofα on each wing of σ preserves the Weyl-distance, it follows thatα preserves the Weyl-distance from x to x ′′ and from y ′′ to y. That the Weyl-distance from x ′′ to y ′′ is preserved is clear since the restriction ofα to Ch(σ) is the automorphism β.
This proves thatα preserves the Weyl-distance from x to y. Thusα is an automorphism.

Fixators of wings
As before, let X be a right-angled building of type (W, I).
The subsets X i (c) are analogues of half-trees in the case W is infinite dihedral. In view of this analogy, we shall consider the subgroups of Aut(X) + denoted by V i (c) and U i (c), consisting respectively of automorphisms supported on X i (c) and on its complement. In symbols, this yields Clearly U i (c) and V i (c) both fix c and stabilise σ. Moreover they commute and have trivial intersection, since their supports are disjoint. The following implies that they are both non-trivial.
Lemma 5.1. Assume that X is thick and semi-regular. Let i ∈ I be such that i ∪ i ⊥ = I. Then for all c ∈ Ch(X), the groups U i (c) and V i (c) are non-abelian.
Proof. By hypothesis, there exists j ∈ I not contained in i ∪ i ⊥ . Let x = c be a chamber j-adjacent to c. Then X j (x) ⊂ X i (c) by Lemma 3.4. This implies that U i (c) fixes pointwise X j (x) for all chambers x = c that are j-adjacent to but different from c.
In view of the symmetry between i and j, it only remains to show that U i (c) is not abelian.
Proposition 4.2 implies that U j (c) is non-trivial; so is thus V i (x) for all x ∈ Ch(X) in view of what we have just observed.
For each c ′ = c that is i-adjacent to c, the group V i (c ′ ) is contained in U i (c). Moreover, if c ′ , c ′′ are two distinct such chambers, the groups V i (c ′ ) and V i (c ′′ ) are different since they are non-trivial and have disjoint supports. By Proposition 4.2, there is u ∈ U i (c) mapping c ′ to c ′′ . We then have uV i (c ′ )u −1 = V i (c ′′ ) = V i (c ′ ). In particular u does not commute with V i (c ′ ), which proves that U i (c) is not abelian.
Given G ≤ Aut(X), the pointwise stabiliser of the chamber set Ch(R) of a residue R is denoted by Fix G (R). We shall next describe how the groups U i (c) and V i (c) provide convenient generating sets for the pointwise stabilisers of residues in X. We start with wall-residues; the case of spherical residues is postponed to Proposition 8.1 below.
Proposition 5.2. Let X be a right-angled building of type (W, I). Let c ∈ Ch(X) and i ∈ I, and let R = Res i∪i ⊥ (c) be the residue of type i ∪ i ⊥ of c.
Then we have We will use the following subsidiary fact.
Lemma 5.3. Let n > 0 be an integer, let C, W be sets and let δ : C n → W be a map. Let G denote the group of all permutations g ∈ Sym(C) such that δ(g(x 1 ), . . . , g(x n )) = δ(x 1 , . . . , x n ) for all (x 1 , . . . , x n ) ∈ C n . Let moreover (V i ) i∈I a collection of groups indexed by a set I, and for all i ∈ I, let ϕ i : V i → G be an injective homomorphism such that for all i = j, the subgroups ϕ i (V i ) and ϕ j (V j ) have disjoint supports. Then there is a unique injective homomorphism ϕ : The only relevant case for this paper is when C is the chamber set of a building X and δ : C × C → W is the Weyl-distance. In that case, the group G from Lemma 5.3 coincides with the group Aut(X) + of type-preserving automorphisms of X.
Proof of Lemma 5.3. The uniqueness of ϕ is clear; we focus on the existence proof. Set V = j∈I V j and let g = (g j ) j∈I ∈ V . Given x ∈ C, there is at most one index j ∈ I such that ϕ j (V j ) does not fix x, since the subgroups ϕ i (V i ) have disjoint supports. We set ϕ(g)(x) = ϕ j (g j )(x) if there exists such a j ∈ I, and ϕ(g)(x) = x otherwise. This defines a homomorphism ϕ : i∈I V i → Sym(C) such that ϕ•ι i = ϕ i for all i ∈ I. It is injective, since ϕ(g) = 1 implies that ϕ i (g i ) = 1 for all i, and hence g i = 1 for all i since all ϕ i are injective by hypohtesis. It remains to prove that ϕ(g) ∈ G. Given (x 1 , . . . , x n ) ∈ C n , let J ⊆ I be the (possibly empty) subset consisting all the indices j ∈ I such that ϕ j (g j ) does not fix all elements of {x 1 , . . . , x n }. Thus J is finite of cardinality ≤ n. Let g J denote the product of the elements ϕ j (g j ) ∈ G over all j ∈ J, in an arbitrary order; if J = ∅, we set g J = 1. Since two distinct subgroups ϕ i (V i ) and ϕ j (V j ) have disjoint supports, they commute, and it follows that the product g J is independent of the chosen order. Moreover, we have ϕ(g)(x i ) = g J (x i ) for all i ∈ {1, . . . , n}. Since g J ∈ G, we infer that δ(ϕ(g)(x 1 ), . . . , ϕ(g)(x n )) = δ(g J (x 1 ), . . . , g J (x n )) = δ(x 1 , . . . , x n ), as desired.
Proof of Proposition 5.2. Let d ∼ i c. Given any x ∈ Ch(R), we deduce from Lemma 2.2 that V i (d) fixes all chambers of the i-panel of x different from the projection of c. Hence V i (d) fixes all chambers of that panel. This proves that V i (d) is contained in Fix Aut(X) + (R).
Remark that for two different chambers d, d ′ that are i-adjacent to c, the groups V i (d) and V i (d ′ ) have disjoint supports. From Lemma 5.3, we deduce that the It remains to show that every element g ∈ Fix Aut(X) + (R) belongs to the product In order to see this, fix g ∈ Fix Aut(X) + (R) and d ∼ i c, and consider the permutation g d of Ch(X) defined by We claim that g d ∈ V i (d). To see this, let x, y ∈ Ch(X) and let δ : Ch(X)×Ch(X) → W denote the Weyl-distance. We need to show that δ(g d (x), g d (y)) = δ(x, y). By the definition of g d , it suffices to consider the case when x ∈ X i (d) and y ∈ X i (d) (or vice-versa). By Lemma 3.3, we have where x ′ = proj R (x), y ′ = proj R (y) and R = Res i∪i ⊥ (d). Moreover, the element g ∈ Aut(X) + fixes x ′ , y and y ′ and preserves R and X i (d). Thus we have proj R (g d (x)) = proj R (g(x)) = x ′ and, invoking Lemma 3.3 once more, we deduce as desired. Thus g d is a type-preserving automorphism of X. By construction, we have g d ∈ V i (d). Moreover the tuple (g d ) d∼ i c , which is an element of the direct product d∼ i c V i (d), coincides with g. Therefore g ∈ d∼ i c V i (d).

Strong transitivity
Proposition 6.1. Let X be a semi-regular right-angled building. Then the group Aut(X) + is strongly transitive on X.
We need the following basic consequence of Proposition 4.2.
Lemma 6.2. Let X be a semi-regular right-angled building of type (W, I). Let R be a residue and let n, t be non-negative integers. For all s ∈ {1, . . . , t}, let also: where R ′ s = proj σs (R). • π s be a permutation of Ch(σ s ) fixing c s , where σ s = Res is (c s ). Assume that the pairs (σ 1 , i 1 ), . . . , (σ t , i t ) are pairwise distinct, and that Ch(R ′ s ) ⊆ X is (c s ) for all s ∈ {1, . . . , t}.
Let now s ′ = s. If σ s ′ were parallel to σ s , we would have i s = i s ′ and σ s = σ s ′ by Proposition 2.7, contradicting our hypotheses. Therefore proj σs (σ s ′ ) is a single chamber (see Lemma 2.4). Moreover we have c s ′ ∈ Ch(σ s ′ ) ∩ B(R, n), and B(R, n) ⊆ X is (c s ) by Lemma 3.5. We infer that proj σs (σ s ′ ) = c s or, equivalently, that Ch(σ s ′ ) ⊆ X is (c s ). Therefore Ch(σ s ′ ) is pointwise fixed by g s . It follows that the product g = g 1 . . . g t enjoys the desired properties.
In order to facilitate future references, we state the following special case separately.
Lemma 6.3. Let X be a semi-regular right-angled building of type (W, I). Let x ∈ Ch(X) and let n, t be non-negative integers. For all s ∈ {1, . . . , t}, let also: • c s ∈ Ch(X) be such that dist(x, c s ) = n, • i s ∈ I be such that proj σs (x) = c s , where σ s = Res is∪i ⊥ s (c s ), • π s be a permutation of Ch(σ s ) fixing c s , where σ s = Res is (c s ). Assume that the pairs (c 1 , i 1 ), . . . , (c t , i t ) are pairwise distinct. Then there is g ∈ U is (c s ) | s = 1, . . . , t whose restriction to Ch(σ s ) is π s for all s. Moreover g fixes pointwise the set B(x, n + 1) − t s=1 Ch(σ s ). Proof. Since proj σs (x) = c s , we have dist(x, σ s ) = dist(x, c s ) = n. Clearly c s ∈ X is (c s ). Moreover, if (σ s , i s ) = (σ s ′ , i s ′ ), then (c s , i s ) = (c s ′ , i s ′ ) and hence s = s ′ by hypothesis. Thus the desired conclusion follows from Lemma 6.2.
Proof of Proposition 6.1. As observed by Haglund-Paulin [HP03], Proposition 4.1 readily implies that Aut(X) + is chamber transitive. We need to show that given a chamber c ∈ Ch(X) and two apartments A, A ′ containing c, there is an element g ∈ Aut(X) + fixing c and mapping A to A ′ .
Set g 0 = Id and let n > 0. We shall construct by induction on n an element g n ∈ Aut(X) + with the following properties: • g n fixes pointwise the ball of radius n − 1 around c; • g n g n−1 . . . g 0 (A) ∩ A ′ ⊇ B(c, n) ∩ A ′ , where B(c, n) is the ball of radius n around c. The first property ensures that the sequence (g n g n−1 . . . g 0 ) n≥0 pointwise converges to a well defined automorphism g ∞ ∈ Aut(X) + . The second property yields g ∞ (A) = A ′ , as desired.
Let n ≥ 0, and suppose that g 0 , g 1 , . . . , g n have already been constructed. Set A n = g n g n−1 . . . g 0 (A). Thus A n ∩ A ′ contains every chamber of A ′ at distance at most n from c.
We need to construct an automorphism g n+1 ∈ Aut(X) + fixing B(c, n) pointwise and such that g n+1 (A n ) ∩ A ′ contains B(c, n + 1) ∩ A ′ .
Let E be the set of those chambers in B(c, n + 1) ∩ A ′ that are not contained in A n . Notice that E is finite (since B(c, n + 1) ∩ A ′ is so) and that every chamber in E is at distance n + 1 from c.
If E is empty, then we set g n+1 = Id and we are done. Otherwise we enumerate E = {x ′ 1 , . . . , x ′ t } and consider s ∈ {1, . . . , t}. Let y s be the first chamber different from x ′ s on a minimal gallery from x ′ s to c. Thus dist(c, y s ) = n and y s ∈ B(c, n)∩A ′ , hence y s ∈ A n . Let σ s be the panel shared by x ′ s and y s and let i s ∈ I be its type. The pairs (y s , i s ) are pairwise distinct since (y s 1 , i s 1 ) = (y s 2 , i s 2 ) in the apartment A ′ implies that x ′ s 1 = x ′ s 2 and s 1 = s 2 . Finally, let x s ∈ A n be the unique chamber which is i s -adjacent to, but different from, y s . We claim that proj σs (c) = y s . In order to establish this, consider z s = proj σs (c). If z s = y s , then dist(c, z s ) < dist(c, y s ) = n. Therefore the unique chamber of z ′ s ∈ A ′ which i s -adjacent to, but different from, z s , also belong to A n . Since apartments are convex, we infer that proj σs (z ′ s ) ∈ A n ∩A ′ . On the other hand we have proj σs (z ′ s ) = x ′ s by Lemma 2.2. This contradicts the fact that x ′ s ∈ A n , and the claim stands proven. We are thus in a position to invoke Lemma 6.3. This yields an element g n+1 ∈ U is (y s ) | s = 1, . . . , t which maps x s to x ′ s for all s, and fixes B(c, n) pointwise. It follows that g n+1 has the requested properties, and we are done.
We are thus in a position to invoke Tits' transitivity lemma: Corollary 6.4. Let X be a thick semi-regular right-angled building of irreducible type.
Then every non-trivial normal subgroup of Aut(X) + is transitive on Ch(X).
In case when X is locally finite, the strong transitivity guaranteed by Proposition 6.1 is already enough to ensure that the intersection of all non-trivial closed normal subgroups of Aut(X) + is non-trivial, topologically simple and cocompact, see [CM11, Corollary 3.1]. This is of course a much weaker conclusion than Theorem 1.1.

Simplicity of the automorphism group
The following result is established by a similar argument as in Tits' commutator lemma (Lemma 4.3 in [Tit70] or Lemma 6.2 in [HP98]).
Lemma 7.1. Let X be a right-angled building of type (W, I). Let σ be a panel of type i ∈ I, let c, c ′ ∈ Ch(σ) be two distinct chambers, and let g ∈ Aut(X) + be such that g(c) is j-adjacent to c ′ , for some j ∈ I with m i,j = ∞.
Proof. Let V 0 = d∈Ch(σ)\{c,c ′ } V i (d), and remark that V 0 ≤ G by Proposition 5.2. For each n ≥ 0, we also set σ n = g n (σ), c n = g n (c), c ′ n = g n (c ′ ) and V n = g n V 0 g −n . For each n ≥ 0, the support of V n is contained in d∈Ch(σn)\{cn,c ′ n } X i (d). Given d ∈ Ch(σ n ) \ {c n , c ′ n } and m > n, we have d ∈ X i (c m ) and c m ∈ X i (d). Therefore X i (d) ⊂ X i (c m ) by Lemma 3.4. This implies that the sets d∈Ch(σn)\{cn,c ′ n } X i (d) and d∈Ch(σm)\{cm,c ′ m } X i (d) are disjoint. In other words, we have shown that for m > n ≥ 0, the subgroups V m and V n have disjoint support. By Lemma 5.3, it follows that the product V = n≥0 V n is a subgroup of G. Moreover, we have gV n g −1 = V n+1 for all n ≥ 0.
Given any h ∈ V 0 , we set x n = g n hg −n for all n ≥ 0. Then the tuple x = (x n ) n≥0 is an element of V ≤ G. So is thus the commutator [x, g]. Moreover, denoting by y n the n th component of an element y ∈ V according to the decomposition V = n≥0 V n , we have [x, g] n = x n (gx −1 g −1 ) n for all n ≥ 0. Hence [x, g] 0 = h and [x, g] n = x n gx −1 n−1 g −1 = x n x −1 n = 1 for all n > 0. Thus [x, g] = h, as required. We record the following consequence of Lemma 7.1, which is a crucial ingredient for the proof of Theorem 1.1.
Lemma 7.2. Let X be a right-angled building of type (W, I). Assume that the Coxeter system (W, I) is irreducible and that X is thick.
Then for any wall-residue R, every non-trivial normal subgroup of Aut(X) + contains Fix Aut(X) + (R).
Proof. Let N < Aut(X) + be a non-trivial normal subgroup.
Let σ be a panel of type i ∈ I and R = σ be the corresponding wall-residue. Choose two distinct chambers c, c ′ ∈ Ch(σ). Since (W, I) is irreducible and nonspherical, there exist j ∈ I such that m i,j = ∞. By Corollary 6.4, there is g ∈ N such that g(c) is j-adjacent to c ′ . In view of Lemma 7.1, we deduce that Since the latter holds for all pairs {c, c ′ } ⊂ Ch(σ) and since X is thick, we deduce that V i (c) and V i (c ′ ) are also contained in N. Therefore, so is Fix Aut(X) + (R) by Proposition 5.2.
We are now ready to complete the proof of simplicity.
Proof of Theorem 1.1. Let N = 1 be a non-trivial normal subgroup of G = Aut(X) + . By Corollary 6.4, the group N is transitive on Ch(X). Since G is strongly transitive on X, it is naturally endowed with a BN-pair. Therefore, if we show that N contains the full stabiliser Stab G (R) of some residue R, then it will follow from [Tit64, Proposition 2.2] that N itself is the stabiliser of some residue. The transitivity of N on Ch(X) forces that residue to be the whole building X, whence N = G as required. Therefore, the desired conclusion will follow provided we show that N contains Stab G (R) of some residue R. This is the final of the following series of claims.
Claim 1. For any residue R of irreducible type, the stabiliser Stab N (R) maps surjectively to Aut(R) + .
In order to prove the claim, we first observe that given two chambers c, c ′ ∈ Ch(R), any element of G mapping c to c ′ must stabilise R. Since N is chamber-transitive, it follows that for any residue R, the image of N ∩ Stab G (R) in Aut(R) + is non-trivial.
In case R is a proper residue of irreducible non-spherical type, we infer by induction on the rank that Aut(R) + is simple; notice that the base of the induction is provided by [Tit70], which settles the case of trees. Since moreover the homomorphism of Stab G (R) in Aut(R) + is surjective by Proposition 4.1, it follows that it remains surjective in restriction to N ∩ Stab G (R). In other words, we have shown that Stab N (R) maps surjectively to Aut(R) + for any proper irreducible non-spherical residue.
Assume now that R is spherical. Thus R is of rank one. Since (W, I) is irreducible, it follows that R is incident with a non-spherical residue R ′ of rank two. From the part of the claim which has already been proven, we deduce that Stab N (R ′ ) maps surjectively to Aut(R ′ ) + . Notice that R ′ , viewed as a building in its own right, is a semi-regular tree, in which the residue R corresponds to the set of edges emanating from a fixed vertex. It follows that the canonical map Stab Aut(R ′ ) + (R) → Aut(R) = Aut(R) + is surjective. Therefore, so is the map Stab N (R) → Aut(R) = Aut(R) + . The claim stands proven.
Claim 2. For any i ∈ I and any residue R of type i ∪ i ⊥ , the group Fix G (R) is contained in N.
This was established in Lemma 7.2.
Claim 3. Let J = J 0 ∪ J 1 ∪ · · · ∪ J s I be the disjoint union of pairwise commuting subsets such that (W J i , J i ) is irreducible non-spherical for all i > 0 and (W J 0 , J 0 ) is spherical (and possibly reducible or trivial). Let c ∈ Ch(X) and R = Res J (c) be its J-residue.
If Fix G (R) is contained in N, then so is Fix G (Res J 0 (c)).
Set P = Stab G (R) and U = Fix G (R). By Proposition 4.1, the quotient P/U is isomorphic to Aut(R) + .
For each i = 0, . . . , s, set R i = Res J i (c). By Lemma 2.2, we have a canonical decomposition Ch(R) ∼ = Ch(R 0 ) × · · · × Ch(R s ), which induces a corresponding product decomposition Aut(R) + ∼ = L 0 × · · · × L s , where L i = Aut(R i ) + . Let N ′ denote the image of N in Aut(R) + ∼ = L = L 0 × · · · × L s under the quotient map P → P/U. Let also L j denote the image of L j under the canonical embedding On the other hand, by Claim 1, the group Stab N (R j ) maps surjectively to L j . It follows that the projection π j : L → L j remains surjective in restriction to N ′ . Therefore, we have Since R j is of non-spherical type, we know that L j is simple by induction on the rank, whence L j = [L j , L j ] and N ′ ∩ L j = L j . In other words, we have L j ≤ N ′ . This holds for all j > 0; therefore {1} × L 1 × · · · × L s is also contained in N ′ .
Recalling that P fits in the short exact sequence and that N contains U by hypothesis, we deduce that N contains the preimage of {1} × L 1 × · · · × L s in P . This implies the claim, since the group coincides with the preimage in P of {1} × Stab L 1 (c) × · · · × Stab Ls (c).
Claim 4. N contains the full stabiliser Stab G (R) of some proper residue R.
Since (W, I) is irreducible non-spherical, we have i ∪ i ⊥ I for all i ∈ I. From Claims 2 and 3, we deduce that there exist spherical residues R 0 such that Fix G (R 0 ) is contained in N. Amongst all such residues, we pick one, say R, whose type J ⊆ I is of minimal possible cardinality.
If J = ∅, then R is a single chamber. Thus Stab G (R) = Fix G (R) is contained in N and we are done.
Assume next that J is not empty and let j ∈ J. Since (W, I) is irreducible, there exists i ∈ I − J such that m i,j = ∞. Now we distinguish two cases.
Assume first that J ∪ {i} is properly contained in I. Let R i be the unique residue of type J ∪ {i} incident with R. Then we have N ≥ Fix G (R) ≥ Fix G (R i ). Let R i = R 0 × Q 1 × · · · × Q s be the decomposition of R i into a maximal spherical factor R 0 and a number of irreducible non-spherical factors. By Claim 3, we have Fix G (R 0 ) ≤ N. By construction R i is not spherical and is incident to R. Therefore the type of R 0 is a proper subset of J. This contradicts the minimality property of R, hence the present case does not occur.
Assume finally that I = J ∪ {i}. Since (W, I) is irreducible, it follows that m i,j ′ = ∞ for all j ′ ∈ J. In other words, we have i ⊥ = ∅. Therefore, by Claim 2 we have Fix G (S) ≤ N for any i-residue S. It follows from the minimality assumption on R that J has cardinality 1 as well. Thus I = {i, j} and X is a tree, in which case the claim follows from the simplicity theorem in [Tit70].

Fixators of spherical residues
We now turn to fixators of spherical residues, i.e. residues whose type J ⊆ I generates a finite subgroup of W . We restrict ourselves to the case where the ambient building X is locally finite case. We endow the group Aut(X) with the compact open topology; the latter coincides with the topology of pointwise convergence on the discrete set Ch(X). The group Aut(X) is locally compact and totally disconnected.
Proposition 8.1. Let X be a semi-regular, locally finite, right-angled building of type (W, I). Let R be a residue of spherical type J ⊆ I. Then we have Specialising (i) to the case J = ∅, we obtain for any chamber c ∈ Ch(X).
Let c ∈ Ch(X) and i ∈ I. Set σ = Res i∪i ⊥ (c) and R ′ = proj σ (R). We say that the pair (c, i) is admissible of c ∈ Ch(R ′ ) and Ch(R ′ ) ⊆ X i (c). Now we set Notice that if c ∈ Ch(R), then (c, i) is admissible if and only if i ∈ J. Indeed, since c ∈ Ch(R), we have Ch(R ′ ) = Ch(R) ∩ Ch(σ); in particular c ∈ Ch(R ′ ). Now, if i ∈ J, then J ⊂ i ∪ i ⊥ . Therefore Ch(Res i (c)) ⊆ Ch(R) ⊂ Ch(σ) and R = R ′ ; in particular Ch(R ′ ) ⊆ X i (c). Conversely, if i ∈ J, then Ch(R ′ ) ⊆ Ch(R) ⊆ X i (c) by Corollary 3.7, so that (c, i) is indeed admissible.
This shows that U(0) = U i (c) | c ∈ Ch(R), i ∈ I − J . We need to show that G(0) = U(0). This is the last of the following series of claims.
Indeed, let (c, i) be an admissible pair with dist(c, R) = n. Then B(R, n) ⊆ X i (c) by Lemma 3.5(i). Thus U i (c) fixes B(R, n) pointwise, and hence U(n) ≤ G(n). The claim follows since G(n) is closed.
Let (c, i) be an admissible pair with dist(c, R) = n. We prove by induction on n that U i (c) ≤ U(0). The base case n = 0 is clear; we assume henceforth that n > 0. Let x = proj R (c). Let c ′ be the first chamber on a minimal gallery from c to x, and let j ∈ I be the type of the panel σ ′ shared by c and c ′ .
If m i,j = 2, then j ∈ i ⊥ and, by the definition of j, we have c ′ ∈ Res i∪i ⊥ (c). But c ′ is closer to x than c. Therefore n > dist(R, Res i∪i ⊥ (c)) = dist(R, proj Res i∪i ⊥ (c) (R)) = dist(R, z) for all z ∈ Ch(proj Res i∪i ⊥ (c) (R)) by Lemma 2.6 and Lemma 2.3. Therefore we have c ∈ Ch(proj Res i∪i ⊥ (c) (R)), in contradiction with the admissibility of (c, i). Thus m i,j = ∞, and hence we have Claim 3. For all n ≥ 0, we have G(n) ≤ U(n)G(n + 1).
Let h ∈ G(n). By the local finiteness of X, there are only finitely many panels σ 1 , . . . , σ r with Ch(σ s ) ⊂ B(R, n + 1) and that are not pointwise fixed by h. The set of panels σ 1 , . . . , σ r is partitioned according to the relation of parallelism. Upon reordering, we may assume {σ 1 , . . . , σ t } is a set of representatives of those classes such that for all s < s ′ ≤ t, the panels σ s and σ s ′ are not parallel.
Let i s ∈ I be the type of σ s . It follows from Proposition 2.7 that the pairs (σ 1 , i 1 ), . . . , (σ s , i s ) are pairwise distinct.
The projection proj σs (R) must be a single chamber, say c s , since h fixes Ch(R) pointwise but acts non-trivially on Ch(σ s ). In particular we have dist(c s , R) = n.
Let now R ′ s = proj σs (R) and pick z ∈ Ch(R ′ s ). Since h fixes pointwise B(R, n), it must also fix pointwise R ′ s . Since Res is (z) is parallel to σ s by Proposition 2.7(ii), it follows that h does not act trivially on Ch(Res is (z)). Therefore . This implies that c s ∈ Ch(R ′ s ). Moreover we have Ch(R ′ s ) ⊆ X is (c s ), since otherwise Ch(σ s ) would be contained in Ch(R ′ s ) by convexity, contradicting that h acts trivially on Ch(R ′ s ). Thus the pair (c s , i s ) is admissible. Now it follows from Lemma 6.2 that there is g ∈ U(n) such that gh fixes σ s pointwise for all s = 1, . . . , t. In particular gh fixes σ s pointwise for all s = 1, . . . , r.
By definition h fixes all chambers of B(x, n+1)− r s=1 Ch(σ s ). Moreover Lemma 6.3 ensures that g fixes all chambers of B(x, n + 1) − t s=1 z∈Ch(R ′ s ) Ch(Res is (z)). Let s ∈ {1, . . . , t} and z ∈ Ch(R ′ s ). By Lemma 2.6 and Lemma 2.3, we have dist(z, R) = dist(R ′ s , R) = dist(c s , R) = n. The panels σ s and Res is (z) are parallel by Proposition 2.7(ii). Thus h does not act trivially on Res is (z) and it follows that Res is (z) ∈ {σ 1 , . . . , σ r }. This implies that g fixes all chambers of B(x, n + 1) − r s=1 Ch(σ s ). Hence so does gh, so that gh ∈ G(n + 1) in view of the preceding paragraph. This proves the claim. Let g ∈ G(0). Invoking Claim 3 by induction on n ≥ 0, we find u n ∈ U(n) and g n ∈ G(n + 1) such that g = u 0 u 1 . . . u n g n for all n. By Claim 2 we have u n ∈ U(0) for all n. Since lim n→∞ g n = 1, we obtain g ∈ U(0). This proves that G(0) ≤ U(0). The reverse inclusion is provided by Claim 1.

Ends and local splittings
A locally normal subgroup of a locally compact group is a compact subgroup whose normaliser is open. We first record that the automorphism groups of rightangled buildings always admit many locally normal subgroups.
Lemma 9.1. Let X be a thick, semi-regular, locally finite, right-angled building of type (W, I). Assume that (W, I) is irreducible non-spherical.
Then Aut(X) + admits locally normal subgroups which decompose non-trivially as direct products, all of whose factors are themselves locally normal.
Proof. Given c ∈ Ch(C) and i ∈ I, the group V i (c) is closed by definition, compact because it fixes c, and non-trivial by Lemma 5.1. Let U = Fix G (Res i (c)). Since X is locally finite, the group U is a finite intersection of chamber stabilisers, and is thus open in G. Moreover, it normalises V i (c), which proves that V i (c) is a locally normal subgroup. The desired conclusion is thus provided by Proposition 5.2.
The following result is an extended version of Theorem 1.3 from the introduction.
Theorem 9.2. Let X be a thick, semi-regular, locally finite, right-angled building of type (W, I). Assume that (W, I) is irreducible non-spherical.
Then the following are equivalent.
(i) W is one-ended.
(ii) W does not split as a free amalgamated product over a finite subgroup.
(iii) There is no partition I = I 0 ∪ I 1 ∪ I 2 with I 1 , I 2 non-empty, m i,j = 2 for all i, j ∈ I 0 and m i,j = ∞ for all i ∈ I 1 and j ∈ I 2 . (iv) X is one-ended.
(v) G is one-ended. (vi) All compact open subgroups of G = Aut(X) + are indecomposable.
We shall need the following basic fact on right-angled Coxeter groups. Proof. The Davis complex of a right-angled Coxeter group (W, I) is a CAT(0) cube complex. We call it Σ. If (W, I) is irreducible, then for every wall W, we may find a wall W ′ disjoint from W (see [Hée93,Prop. 8.1 p. 309]). This implies that Σ is irreducible as a cube complex. Moreover, transforming W under the dihedral group generated by the reflections through W and W ′ , we find walls arbitrarily far from W in both of the half-spaces that it determines. This proves that W acts essentially on Σ in the sense of [CS11]. Moreover, since Σ is irreducible, it follows from [CS11,Th. 4.7] that W does not fix any point at infinity of Σ. The hypotheses of [CS11, Lem. 5.2] are thus fulfilled. The latter result ensures that at least one of the four sectors determined by the boundary walls of H and H ′ properly contains a half-space. Transforming that half-space by an appropriate element from the group generated by the reflections fixing H and H ′ , we find a half-space properly contained in H ∩ H ′ , as desired.
We also record an abstract group theoretic fact, where [g, V ] denotes the set of Proof. Given g ∈ Stab G (Y ), we define ϕ(g) ∈ Sym(C) by Then ϕ : Stab G (Y ) → Sym(C) is a homormorphism. Since V and aV a −1 have disjoint supports, they are both contained in Stab G (Y ). Moreover, Since [a, v] and [b, w] commute by hypothesis, so do their images under ϕ. Thus V is abelian, as claimed.
Proof of Theorem 9.2. The equivalences (i) ⇔ (ii) ⇔ (iii) are well-known, see [MT09]. The equivalence (iv) ⇔ (v) is clear since G acts properly and cocompactly on X, so that G and X are quasi-isometric.
(i) ⇒ (iv) By assumption all apartments are one-ended. Given x ∈ Ch(X), we need to prove that for all n ≥ 0, any two chambers c ′ , c ′′ at distance > n away from x can be connected by a gallery avoiding the ball B(x, n). We proceed by induction on n.
In the base case n = 0, either a minimal gallery from c ′ to c ′′ does not pass through x, and we are done, or every apartment containing c ′ and c ′′ also contains x, in which case we can find a gallery from c ′ to c ′′ avoiding x inside one of these apartments, since these are one-ended by hypothesis.
Let now n > 0 and assume that Ch(X) − B(x, n − 1) is gallery-connected. Let c ′ = c 0 , c 1 , . . . , c t = c ′′ be a gallery from c ′ to c ′′ which does not meet B(x, n − 1). Then for all i, if c i ∈ B(x, n) then dist(c i−1 , x) = dist(c i+1 , x) = n + 1. Therefore, it suffices to prove that if dist(c ′ , x) = dist(c ′′ , x) = n + 1 and c ′ , c ′′ are both adjacent to a common chamber d ∈ B(x, n), then there is a gallery from c ′ to c ′′ avoiding B(x, n). Let Σ be an apartment containing x and d. Let d ′ and d ′′ be the two chambers of Σ different from d and respectively sharing with d the common panel of d and c ′ , and of d and c ′′ . Let i ′ (resp. i ′′ ) be the type of the panel σ ′ (resp. σ ′′ ) shared by d, d ′ and c ′ (resp. d, d ′′ and c ′′ ). Clearly proj σ ′ (d) = d and proj σ ′′ (d) = d. Therefore, by Lemma 6.3, there is an element g ∈ G fixing X i ′ (d) ∩ X i ′′ (d) pointwise and such that g(d ′ ) = c ′ and g(d ′′ ) = c ′′ . Since x ∈ X i ′ (d) ∩ X i ′′ (d), it follows that g(Σ) is an apartment containing x, d, c ′ and c ′′ . Since apartments are one-ended, a gallery joining c ′ to c ′′ and avoiding B(x, n) can be found in the apartment g(Σ), and we are done.
(v) ⇒ (iii) Assume that (iii) fails and let I = I 0 ∪ I 1 ∪ I 2 be a partition with I 1 , I 2 non-empty, m i,j = 2 for all i, j ∈ I 0 and m i,j = ∞ for all i ∈ I 1 and j ∈ I 2 . Let T be the graph whose vertex set is the collection of residues of type I 0 ∪ I 1 and I 0 ∪ I 2 , and declare that two residues are adjacent if they contain a common residue of type I 0 . By Lemma 4.3 from [HP03] the graph T is a tree. Since I 0 is finite and since X is locally finite, the residues of type I 0 are finite and, hence, their stabilisers are compact open subgroups. In other words the edge stabilisers of the tree T are compact open subgroups. Since G is chamber-transitive, it acts edge-transitively on T . This yields a non-trivial decomposition of G as an amalgamated free product over a compact open subgroup. Hence G cannot be one-ended by [Abe74].
(vi) ⇒ (iii) Assume that (iii) fails and let I = I 0 ∪ I 1 ∪ I 2 be a partition with I 1 , I 2 non-empty, m i,j = 2 for all i, j ∈ I 0 and m i,j = ∞ for all i ∈ I 1 and j ∈ I 2 . Let R be a residue of type I 0 in X. Since I 0 is finite, the set Ch(R) is finite and hence Stab G (R) and Fix G (R) are both compact open subgroups of G. We shall prove that Fix G (R) splits non-trivially as a direct product.
For k = 1, 2, let U k = U i (c) | c ∈ Ch(R), i ∈ I k . Notice that U 1 and U 2 are both non-trivial by Lemma 5.1 since I 1 and I 2 are assumed non-empty.
We claim that U 1 and U 2 commute. Indeed, let c 1 , c 2 ∈ Ch(R), let i 1 ∈ I 1 , i 2 ∈ I 2 . It suffices to prove that U i 1 (c 1 ) and U i 2 (c 2 ) commute. This in turn will follow if one shows that they have disjoint supports.
By definition the support of U i 1 (c 1 ) is the union of the sets X i 1 (d) over all chambers d that are i 1 -adjacent to but different from c 1 . Let d be such a chamber. We claim that X i 1 (d) ⊂ X i 2 (c 2 ).
By Corollary 3.7, we have c 2 ∈ X i 1 (c 1 ) so that c 2 ∈ X i 1 (d). Similarly, Corollary 3.7, implies that c 1 ∈ X i 2 (c 2 ), which implies that d ∈ X i 2 (c 2 ), since otherwise a panel of type i 1 would be parallel to a panel of type i 2 by Lemma 2.4, which is impossible by Proposition 2.7(i). This proves that d ∈ X i 2 (c 2 ) and c 2 ∈ X i 1 (d). The claim then follows from Lemma 3.4.
The claim implies that the support of U i 1 (c 1 ) is pointwise fixed by U i 2 (c 2 ). By symmetry, the support of U i 2 (c 2 ) is pointwise fixed by U i 1 (c 1 ), so that U i 1 (c 1 ) and U i 2 (c 2 ) commute, as desired. This confirms that U 1 and U 2 commute.
By Proposition 8.1, we have Fix G (R) = U 1 ∪ U 2 . Since U 1 and U 2 commute, we have U 1 ∪ U 2 = U 1 U 2 . Moreover U 1 and U 2 are compact, since they are both closed subgroups of the compact group Stab G (R). Thus the product U 1 U 2 is closed, so that Fix G (R) = U 1 U 2 . In particular U 1 ∩ U 2 ≤ Z (Fix G (R)). Hence U 1 ∩ U 2 is contained in the quasi-centre of G, i.e. the collection of elements commuting with an open subgroup. By [BEW11,Theorem 4.8] the group G has trivial quasi-centre since G is compactly generated and simple. Thus Fix G (R) ∼ = U 1 × U 2 as desired.
(iv) ⇒ (vi). Assume finally that (iv) holds and let U ≤ G be a compact open subgroup with two commuting subgroups A, B such that U = AB. We shall prove that A or B is open. Since the closures A and B commute, we infer that B or A is in the quasi-centre of G, which is trivial by [BEW11, Theorem 4.8] since G is compactly generated and simple. Thus U = A or U = B and (vi) holds.
Therefore, all we need to show is that a compact open subgroup U = AB is the commuting product of two closed subgroups A and B, then A or B is open. To this end, it suffices to show that A or B is finite. This follows from the last of a series of claims which we shall now prove successively. Let x ∈ Ch(X). Upon replacing A and B by their respective intersections with the compact open subgroup Stab G (x) and then redefining U accordingly, we may assume that U fixes x. For all m ≥ 0, we set G(m) = Fix G (B(x, m)). Since U is open it contains G(n 0 ) for some n 0 ≥ 0. Without loss of generality, we may assume that n 0 > 1. We define In particular, if σ ∈ Π then dist(σ, x) ≥ n 0 .
Moreover, to each chamber c ∈ Ch(X), we associate two subsets of I defined as follows: I 0 (c) = {i ∈ I | proj Res i (c) (x) = c} and I Π (c) = {i ∈ I | Res i (c) ∈ Π}. Recall that a subset J ⊆ I is called spherical if it generates a finite subgroup of W . It is a classical fact that I 0 (c) is a spherical subset of I. Claim 1. Let c ∈ Ch(X) with dist(c, x) > n 0 and i ∈ I. Then i ∈ I Π (c) if and only if dist(x, Res i∪i ⊥ (c)) ≥ n 0 .
Assume conversely that dist(x, σ) < n 0 . Then the i-panel of c ′ lies entirely in B(x, n 0 ) and is thus pointwise fixed by G(n 0 ). Therefore Stab G(n 0 ) (σ) acts trivially on σ, and hence σ ∈ Π and i ∈ I Π (c).
Since (W, I) is right-angled, any collection of pairwise intersecting walls in an apartment is contained in the set of walls of a spherical residue. The cardinality of such a collection is bounded above by the largest cardinality of a spherical subset of I. In particular it is finite. In view of Ramsey's theorem, we infer that there is some n 1 > n 0 such that any set of more than n 1 walls contains a subset of more than n 0 + 1 pairwise non-intersecting walls.
Let now c ∈ Ch(X) be such that dist(c, x) > n 1 and Σ be an apartment containing c and x. By construction there is a set of more than n 0 + 1 pairwise non-intersecting walls in Σ that are crossed by any minimal gallery from c to x. In particular, at least one of these walls, say W, separates c from the ball B(x, n 0 + 1).
Since (W, I) is right-angled, no wall crossed by a shortest possible gallery from c to a chamber adjacent to W crosses W. Let W ′ be the first wall crossed by such a gallery. Thus W ′ is adjacent to c, and every chamber adjacent to W ′ is at distance > n 0 from x.
Let now k ∈ I be the type of the panel of c which belongs to W ′ . Since W ′ separates c from x, we have k ∈ I 0 (c). Notice that proj Res k∪k ⊥ (c) (x) belongs to Σ. Thus proj Res k∪k ⊥ (c) (x) is a chamber of Σ which is adjacent to the wall W ′ . This implies that dist(x, Res k∪k ⊥ (c)) > n 0 . Therefore k ∈ I Π (c) by Claim 1. Thus the sets I 0 (c) and I Π (c) have a non-empty intersection, as desired.
Claim 3. Let c ∈ Ch(X) and σ be a panel of c. If a(c) = c for some a ∈ Stab A (σ), then b(c) = c for all b ∈ Stab B (σ), and similarly with A and B interchanged.
Let i ∈ I be the type of σ. Notice that c 0 = proj σ (x) = c since a fixes x and stabilises σ. Let Σ be an apartment containing c and x. It also contains c 0 by convexity.
Since (W, I) is irreducible and non-spherical, there is j ∈ I such that m i,j = ∞. Let R = Res {i,j} (c). Let r be the reflection of Σ swapping c and c 0 and r ′ be the reflection of Σ through the j-panel of c. We set c ′ = (r ′ r) n 0 (c) and c ′ 0 = (r ′ r) n 0 (c 0 ). Thus c and c ′ are separated by 2n 0 walls of the residue R in Σ, and x lies on the same side as c of all those walls. Set y = proj Res i∪i ⊥ (c ′ ) (x). Then y belongs to Σ since apartments are convex. The chambers c ′ , c ′ 0 and y are all adjacent to the wall W = Σ ∩ Res i∪i ⊥ (c ′ ). Moreover the chambers x, y and c ′ 0 lie on the same side of W while c ′ lies on the opposite side. Thus X i (c ′ ) and X i (c ′ 0 ) are disjoint, and X i (c ′ 0 ) = X i (y) ⊇ B(x, n 0 ) by Lemma 3.1(ii) and Corollary 3.6. In particular X i (c ′ ) ∩ B(x, n 0 ) = ∅ so that V i (c ′ ) fixes B(x, n 0 ) pointwise, and is thus contained in U. By construction, we have X i (c ′ ) ∩ Σ ⊆ X i (c) ∩ Σ. Lemma 3.4 therefore ensures that X i (c ′ ) ⊂ X i (c), whence V i (c ′ ) ≤ V i (c). In particular the support of V i (c ′ ) and its image under a are disjoint.
Similarly, if b ∈ Stab B (σ) and b(c) = c, then the support of V i (c ′ ) and its image under b are disjoint. Since [a, V i (c ′ )] ≤ A and [b, V i (c ′ )] ≤ B, we deduce from Lemma 9.4 that V i (c ′ ) is abelian, in contradiction with Lemma 5.1. Therefore b(c) = c for all b ∈ Stab B (σ). Let σ ∈ Π. By definition there is u ∈ Stab G(n 0 ) (σ) and c ∈ Ch(σ) with u(c) = c. Write u = ab with a ∈ A and b ∈ B. Consider a gallery x 0 , x 1 , . . . , x k of minimal possible length joining a chamber in B(x, n 0 ) to a chamber in Ch(σ). Since u ∈ G(n 0 ), it fixes x 0 and by the minimality of the gallery, we have x k = proj σ (x 0 ), so that u fixes x k as well. Therefore u fixes x i for all i.
We claim that a and b both fix x i for all i. Otherwise there is some i such that a(x i ) = x i . Since u(x i ) = x i and u = ab, we must have b(x i ) = x i . If i the smallest such index, then a and b also fix x i−1 and thus both stabilise the panel shared by x i−1 and x i . This contradicts Claim 3.
It follows that a and b both fix x k and hence stabilise σ. As ab = u ∈ Fix G (σ), we have Stab A (σ) ≤ Fix G (σ) or Stab B (σ) ≤ Fix G (σ). It remains to show that these two possibilities are mutually exclusive. Let Ch A and Ch B be the subsets of Ch(σ) that are not fixed by A and B respectively. The previous claim guarantees that Ch A and Ch B are disjoint, and thus they are both stabilised by A and B and hence by U.
Let i ∈ I be the type of σ and c ′ = proj σ (x). Since σ ∈ Π, we have i ∈ I Π and U i (c ′ ) ≤ G(n 0 ) ≤ U by Claim 1 and Corollary 3.6. Consequently the group Stab U (σ) permutes arbitrarily the set Ch(σ) − {c ′ } by Proposition 4.2. Since Ch A and Ch B are disjoint and U-invariant, it follows that either Ch A or Ch B coincides with the whole of Ch(σ) − {c ′ }.
Claim 5. Let c ∈ Ch(X) and i, j ∈ I with m i,j = 2. Let σ i and σ j be the i-and j-panels of c respectively. If σ i and σ j belong to Π, then f (σ i ) = f (σ j ).
Suppose for a contradiction that f (σ i ) = A and f (σ j ) = B. Then there exist a ∈ A, b ∈ B stabilising respectively σ i and σ j , and such that a(c i ) = c i and b(c j ) = c j for some c i ∈ Ch(σ i ) and c j ∈ Ch(σ j ).
Let R be the {i, j}-residue of c and set c ′ = proj R (x). Let also σ ′ i and σ ′ j be the i-and j-panels of c ′ . Then a and b both fix c ′ and stabilise σ ′ i and σ ′ j . Moreover σ ′ i and σ ′ j are respectively parallel to σ i and σ j . Set c ′ i = proj σ ′ i (c i ) and c ′ j = proj σ ′ j (c j ). Then a(c ′ i ) = c ′ i and b(c ′ j ) = c ′ j . Therefore we have f (σ i ) = f (σ ′ i ) and f (σ j ) = f (σ ′ j ). Let Σ be an apartment containing x and c ′ . By Claim 1 and Corollary 3.6, the ball B(x, n 0 ) is contained in X i (c ′ ) ∩ X j (c ′ ). From Lemma 6.3, we deduce that there is some g ∈ G(n 0 ) ≤ U mapping Σ to an apartment containing c ′ i and c ′ j . Upon Since m i,j = 2, it follows that the i-panel of c ′ , say σ ′ i , is parallel to σ i since they are contained and opposite in the {i, j}-residue of c. Therefore, any element of G(n 0 ) ≤ U stabilises σ i and acts non-trivially on it if and only if it stabilises σ ′ i and acts non-trivially on it. Hence f (σ i ) = f (σ ′ i ) and therefore f (c) = f (c ′ ). Claim 9. We have A ∩ G(n 1 + 1) = 1 or B ∩ G(n 1 + 1) = 1.
By (iv) any two chambers at distance > n 1 from x can be joined by a gallery which does not meet the ball B(x, n 1 ). By the preceding claim, this implies that the map f is constant on Ch(X) − B(x, n 1 ). Upon exchanging A and B we may assume that this constant value is A. It follows that for all panels σ ∈ Π at distance > n 1 from x, we have Stab B (σ) ≤ Fix B (σ). An immediate induction now shows that for all m > n 1 , we have B ∩ G(m) ≤ G(m + 1). Therefore B ∩ G(n 1 + 1) is trivial.