Convex cores for actions on finite-rank median algebras
aa r X i v : . [ m a t h . G T ] J a n CONVEX CORES FOR ACTIONS ON FINITE-RANK MEDIAN ALGEBRAS
ELIA FIORAVANTI
Abstract.
We show that every action of a finitely generated group on a finite-rank median algebraadmits a nonempty “convex core”, even when no metric or topology is given. We then use this todeduce an analogue of the flat torus theorem for actions on connected finite-rank median spaces. Wealso prove that isometries of connected finite-rank median spaces are either elliptic or loxodromic.
Contents
1. Introduction. 12. Preliminaries. 42.1. Some classical facts. 42.2. Restriction quotients. 62.3. The fundamental lemma. 72.4. Products. 83. Automorphisms of finite-rank median algebras. 93.1. Dynamics of halfspaces. 93.2. Some technical results. 103.3. Convex cores. 123.4. Semisimple automorphisms. 163.5. Non-transverse automorphisms. 184. Compatible metrics. 225. Actions of polycyclic groups. 245.1. Lineal median algebras. 255.2. Lineal median spaces. 255.3. Actions on lineal median spaces. 26References 271.
Introduction.
A metric space X is median if, for all x , x , x ∈ X , there exists a unique m = m ( x , x , x ) ∈ X with the property that( ∗ ) d ( x i , x j ) = d ( x i , m ) + d ( m, x j ) for all ≤ i < j ≤ . We refer to median metric spaces simply as median spaces . The rank of aconnected median space is the supremum of the topological dimensions of its compact subsets.The simplest examples of finite-rank median spaces are provided by R –trees and finite-dimensional CAT(0) cube complexes with the ℓ metric. The Guirardel core of a pair of actions on R –trees isalso a median space of rank ≤ [Gui05]. On the other hand, for every non-discrete measure space (Ω , µ ) , the Banach space L (Ω , µ ) is an infinite-rank median space [CDH10]. Universität Bonn, Bonn, Germany
E-mail address : [email protected] . ost importantly, finite-rank median spaces arise as ultralimits of sequences of CAT(0) cubecomplexes of uniformly bounded dimension, or, in fact, any sequence of spaces that can be suitablyapproximated by such cube complexes. Thus, the asymptotic cones of any coarse median group canbe given a natural structure of finite-rank median space [Bow13, Zei16]. Such groups include allcocompactly cubulated groups and mapping class groups of compact surfaces [Bow13, BHS17].There is a sharp divide between the properties of median spaces of finite rank , which closelytrack those of
CAT(0) cube complexes, and the properties of infinite-rank median spaces, such as L ([0 , . For this reason, we restrict our attention to finite-rank spaces in this work.Our first goal is to provide a proof of the following two facts (Corollaries A and B), which followfrom our more general results on group actions on median algebras.To begin with, isometries of finite-rank median spaces are semisimple : Corollary A.
Let X be a connected, finite-rank median space. For every g ∈ Isom X :(1) the translation length of g is realised by some point of X ;(2) if X is geodesic and g does not fix a point, then g admits an axis : a bi-infinite h g i –invariantgeodesic along which g translates nontrivially. When X is a CAT(0) cube complex, Corollary A is due to Haglund [Hag07] and also holds when X is infinite-dimensional . In median spaces, however, the finite-rank assumption is essential forCorollary A to hold (see for instance Example 2.6 below).Haglund’s proof is combinatorial in nature and does not carry over to the general median setting.We will develop an alternative approach, based on the study of “ convex cores ” for general actions G y X . These are certain G –invariant convex subsets of X naturally attached to the G –action.Our approach also easily yields the following version of the flat torus theorem. Similar resultswhen X is a CAT(0) cube complex were obtained in [WW17, Woo17, Gen20].
Corollary B.
Let A be a finitely generated, virtually abelian group. Let A y X be an isometricaction on a connected, finite-rank median space. Then A stabilises a nonempty convex subset C ⊆ X isometric to a subset of ( R n , d ℓ ) with n ≤ rk X . Some version of Corollaries A and B was certainly to be expected if one subscribes to the viewthat “anything that holds in
CAT(0) cube complexes and R –trees should also hold in finite-rankmedian spaces”. What I found surprising is that these results do not require the metric on X to be complete , which might appear necessary since both involve finding a fixed point of sorts.In fact, most of the results in this paper hold for general automorphisms of finite-rank medianalgebras, even when no metric or topology is present.As an example of the breadth of this extension, consider the case when X = R . While, up toindex , every isometry of R is a translation, the automorphism group of the underlying medianalgebra is the entire homeomorphism group of R .Another interesting example is provided by the Baumslag–Solitar groups BS (1 , n ) with n ≥ .By [Fio18, Theorem A], these groups do not admit any free (or proper) actions by isometries onfinite-rank median spaces. However, BS (1 , n ) admits a properly discontinuous, cocompact actionby median-preserving homeomorphisms on the rank– median algebra T n × R , where T n is thestandard Bass–Serre tree (Example 2.7).This level of generality will prove particularly useful in our upcoming work [Fio20a], where wegeneralise to all fundamental groups of compact special cube complexes the classical fact that everyouter automorphism of a hyperbolic group G can be realised as a homothety of a small G –tree Haglund’s theorem is normally stated with the additional assumption that g act stably without inversions . Thisis only necessary if we want the axis to intersect the –skeleton of the cube complex. In our setting, the fact that themedian space is connected guarantees that there cannot be any inversions (see Remark 4.3). Pau97, BH92, GJLL98, GLL98]. Actions on R –trees will have to be replaced by actions on higher-dimensional median spaces. It will then be important that many results can be expressed purely inmedian-algebra terms, leaving us free to modify the median metric to suit our needs.In the rest of the introduction, we state our results in the general context of median algebras.This inevitably requires giving a few important definitions first. Statement of results on finite-rank median algebras. A median algebra is a pair ( M, m ) ,where M is a set and m : M → M is a map satisfying:(1) m ( x, x, y ) = y , for all x, y ∈ M ;(2) m ( x , x , x ) = m ( x σ (1) , x σ (2) , x σ (3) ) , for all x , x , x ∈ M and σ ∈ Sym(3) ;(3) m ( m ( x, y, z ) , y, w ) = m ( x, y, m ( z, y, w )) .Median algebras were originally introduced in order theory as a common generalisation of dendritesand lattices. They have been extensively studied in relation to semi-lattices (see e.g. [Sho52, Isb80,BH83]) and, more recently, in more geometrical terms because of their connections to the geometry of CAT(0) cube complexes and mapping class groups (for instance in [Rol98, CDH10, Bow13, Bow14]).A map ϕ : M → N between median algebras is a median morphism if it satisfies the equality ϕ ( m ( x, y, z )) = m ( ϕ ( x ) , ϕ ( y ) , ϕ ( z )) for all x, y, z ∈ M . We denote by Aut M the group of medianautomorphisms of M .If X is a median space and m : X → X is the map given by ( ∗ ), then the pair ( X, m ) is a medianalgebra. Every isometric action G y X gives an action by median automorphisms on ( X, m ) , butthe converse does not hold (as already mentioned, every homeomorphism of R preserves its usualmedian-algebra structure).The rank rk M , is the supremum of k ∈ N for which M contains a median subalgebra isomorphicto { , } k (the vertex set of a k –cube). For connected median spaces, this is equivalent to thedefinition of rank given earlier in this introduction (see Remark 2.1).A subset C ⊆ M is convex if we have m ( x, y, z ) ∈ C for all x, y ∈ C and z ∈ M . If M is theunderlying median algebra of a geodesic median space X , then a subset C is convex exactly whenit contains all geodesics joining two of its points.A halfspace is a nonempty convex subset h ⊆ M such that its complement h ∗ := M \ h isalso convex and nonempty. A wall is an unordered pair { h , h ∗ } , where h is a halfspace. A wall w = { h , h ∗ } separates two points x, y ∈ M if h ∩ { x, y } is a singleton. We write W ( x | y ) for the setof walls that separate x and y , which is nonempty as soon as x = y [Rol98, Theorem 2.7]. Definition 1.1.
Let M be a median algebra. Consider g ∈ Aut M .(1) The minimal set Min( g ) is the set of points x ∈ M such that the sets W ( g n x | g n +1 x ) arepairwise disjoint for n ∈ Z .(2) We say that g is semisimple if Min( g ) = ∅ .When X is a geodesic median space and g ∈ Isom X , a point x ∈ X lies in Min( g ) if and only ifeither gx = x or x lies on an axis of g . Definition 1.2.
Let M be a median algebra. An action by median automorphisms G y M is:(1) essential if, for every halfspace h ⊆ M , there exists g ∈ G such that g h ( h ;(2) minimal if there exists no proper, G –invariant, convex subset of M .(3) without wall inversions if there do not exist h ∈ H ( M ) and g ∈ G such that g h = h ∗ .Perhaps counterintuitively, any action G y M that originates from an isometric G –action on a connected finite-rank median space is automatically without wall inversions (see Remark 4.3).We are now ready to state the main result of this paper. For a more canonical (though moretechnical) result that is roughly equivalent to Theorem C, we refer the reader to Theorem 3.17. heorem C. Let M be a finite-rank median algebra. Let G be a finitely generated group acting on M by median automorphisms without wall inversions. Then:(1) there exists a nonempty, G –invariant, convex subset C ⊆ M such that G y C is essential;(2) G y M is essential if and only if G y M is minimal. When M is the vertex set of a CAT(0) cube complex X , the subset C is G –equivariantly isomor-phic to the essential core of Caprace and Sageev [CS11, Subsection 3.3]. Thus, Theorem C impliesthat the essential core embeds into X in this case (cf. Propositions 3.5 and 3.12 in [CS11]).The set C in part (1) of Theorem C is allowed to be a single point, in which case the action G y M has a global fixed point. When G ≃ Z , the set C has a very special structure (it is aninterval with endpoints in the zero-completion of M , an analogue of the Roller boundary of a cubecomplex). From this, one can deduce the following result, which, in turn, implies Corollary A.
Corollary D.
Let M be a finite-rank median algebra. If, for g ∈ Aut M , the action h g i y M iswithout wall inversions, then g is semisimple. We will also deduce the following result, which, along with Theorem C, implies Corollary B.
Corollary E.
Let G y X be a minimal, isometric action without wall inversions on a finite-rankmedian space. Let H ≤ G be a commensurated polycyclic subgroup. Then X has a G –invariantproduct splitting C = F × P such that: • H y P fixes a point and H y F factors through a virtually abelian group; • F isometrically embeds in ( R r , d ℓ ) for r = rk X . A very similar result for actions on (possibly infinite-dimensional)
CAT(0) cube complexes wasrecently obtained by Genevois (see Theorem 1.1 and Theorem 1.2 in [Gen20]).It would be interesting to see if Corollary E admits an analogue for general actions by medianautomorphisms on finite-rank median algebras. There are significant complications that arise inthis setting (above all, the failure of Corollary 4.6), but I was not able to find a counterexample.
Paper Outline.
Section 2 contains some basic facts on median algebras. We also obtain a novelextension of Helly’s lemma to infinite families of halfspaces (Lemma 2.9), which will prove hugelyuseful for many of the results in this paper.Section 3 constitutes the heart of this work. After some preliminary facts, we introduce the convex core in Subsection 3.3. This will be our main tool throughout the paper. Theorem 3.17,proving that cores are nonempty, can also be viewed as our main result, having Theorem C as oneof its consequences. The two parts of Theorem C are proved as Proposition 3.24 and Corollary 3.26at the end of Subsection 3.3. Corollary D is obtained in Subsection 3.4.In Subsection 3.5, we study non-transverse automorphisms of median algebras. The main resultsare Proposition 3.36 and Proposition 3.40, which will prove useful in [Fio20a].In Section 4, we prove Corollary A and refine a few results in the presence of a median metric(see in particular Corollary 4.6 and Proposition 4.9). Finally, Section 5 is devoted to the proof ofCorollaries B and E.
Acknowledgments.
I would like to thank Anthony Genevois for conversations related to Ex-ample 2.7, and Ric Wade for pointing me to [GL95, Example II.6]. I am grateful to Max PlanckInstitute for Mathematics in Bonn for its hospitality and financial support while part of this workwas being completed. 2.
Preliminaries.
Some classical facts.
Several fundamental notions in the study of median algebras werealready defined in the Introduction. In this subsection, we collect some additional terminology and otation, along with a few basic facts. For a more detailed introduction to median algebras, see e.g.[CDH10, Sections 2–4], [Bow13, Sections 4–6] and [Fio20b, Section 2].Let M be a median algebra. Given two subsets A, B ⊆ M , we use the notation: H ( A | B ) := { h ∈ H ( M ) | A ⊆ h ∗ , B ⊆ h } , W ( A | B ) := {{ h , h ∗ } ∈ W ( M ) | h ∈ H ( A | B ) ∪ H ( B | A ) } , H A ( M ) := { h ∈ H ( M ) | h ∩ A = ∅ , h ∗ ∩ A = ∅} . Given a subset
H ⊆ H ( M ) , we write H ∗ := { h ∗ | h ∈ H} .If w = { h , h ∗ } is a wall, we say that h and h ∗ are its two sides and that w bounds h and h ∗ .Halfspaces h and k are transverse if each of the four intersections h ∩ k , h ∗ ∩ k , h ∩ k ∗ , h ∗ ∩ k ∗ isnonempty. Two walls are transverse if they bound transverse halfspaces. If U and V are sets ofhalfspaces or walls, we say that U and V are transverse if every element of U is transverse to everyelement of V . Remark 2.1 (Definitions of rank) . In the introduction, we defined rk M as the largest k such that M has a median subalgebra isomorphic to { , } k . When rk M is finite, this coincides with themaximal cardinality of a set of pairwise-transverse walls of M [Bow13, Proposition 6.2].If X is a connected median space, then the rank of the underlying median algebra also coincideswith the supremum of the topological dimensions of the compact subsets of X . One inequality followsfrom Theorem 2.2 and Lemma 7.6 in [Bow13], while the other from [Bow16, Proposition 5.6].A pocset is a triple ( P, (cid:22) , ∗ ) , where ( P, (cid:22) ) is a poset and ∗ is an order-reversing involution. Givenpocsets P, P ′ , a map f : P → P ′ is a morphism of pocsets if, for all x, y ∈ P with x ≤ y , we have f ( x ) ≤ f ( y ) and f ( x ) ∗ = f ( x ∗ ) . For every median algebra M , the triple ( H ( M ) , ⊆ , ∗ ) is a pocset.An ultrafilter is a subset σ ⊆ H ( M ) such that the halfspaces in σ pairwise intersect and σ ∩ { h , h ∗ } is a singleton for every h ∈ H ( M ) . For every x ∈ M , the set of all halfspaces containing x is anultrafilter, which we denote by σ x .Convex subsets were defined in the introduction. In this paragraph, we recall two importantproperties that they enjoy. First, Helly’s lemma: every finite family of pairwise-intersecting convexsubsets has nonempty intersection [Rol98, Theorem 2.2]. Second, if C , C ⊆ M are convex subsets,then W ( C | C ) = ∅ [Rol98, Theorem 2.8]. Remark 2.2.
Consider median algebras M and N , and a median morphism φ : M → N . If C ⊆ M and C ⊆ N are convex, then φ − ( C ) in M , and φ ( C ) is convex in φ ( N ) .When φ is surjective, we obtain a well-defined map φ ∗ : H ( N ) → H ( M ) given by φ ∗ ( h ) = φ − ( h ) (surjectivity is needed to avoid empty preimages). The map φ ∗ is an injective morphism of pocsetsand it preserves transversality. The image of φ ∗ is exactly the set of those halfspaces h ∈ H ( M ) for which φ ( h ) ∩ φ ( h ∗ ) = ∅ .A subset C ⊆ M is gate-convex if there exists a gate-projection π : M → C , i.e. a map thatsatisfies m ( x, π ( x ) , y ) = π ( x ) for all x ∈ M and y ∈ C . When they exist, gate-projections areunique and they are median morphisms. Moreover, H ( x | π ( x )) = H ( x | C ) for every x ∈ M .Every gate-convex subset is convex, and every convex subset is a median subalgebra. Given x, y ∈ M , the set I ( x, y ) := { z ∈ M | m ( x, y, z ) = z } is called the interval with endpoints x and y . This is a gate-convex subset of M with gate-projectiongiven by the map z m ( x, y, z ) . Nonempty intersections of intervals are again intervals. Remark 2.3. (1) If S ⊆ M is a subalgebra, we have a map res S : H S ( M ) → H ( S ) given by res S ( h ) = h ∩ S .This is a morphism of pocsets and, by [Bow13, Lemma 6.5], it is a surjection.
2) If C ⊆ M is convex, then res C is injective and it preserves transversality. In this case, thesets H ( C ) and H C ( M ) are naturally identified.In order to see this, observe that, if h , k ∈ H C ( M ) satisfy h ∩ k = ∅ , then Helly’s lemmayields ( h ∩ C ) ∩ ( k ∩ C ) = ∅ . Note that h = k if and only if h ∩ k ∗ and h ∗ ∩ k are empty.(3) If C ⊆ M is gate-convex with gate-projection π C , we have, in the notation of Remark 2.2: res C ◦ π ∗ C = id H ( C ) and π ∗ C ◦ res C = id H C ( M ) . Remark 2.4.
Let G y M be an action by median automorphisms with no wall inversions. If C ⊆ M is a G –invariant convex subset, then G y C has no wall inversions. This is because, asobserved in Remark 2.3, every halfspace of C is of the form h ∩ C for a unique h ∈ H ( M ) . Thus,if g h ∩ C = h ∗ ∩ C for some g ∈ G , we must have g h = h ∗ . Lemma 2.5.
Let G y M be an action by median automorphisms with a finite orbit. Then thereexists a G –invariant median subalgebra C ⊆ M such that C ≃ { , } k for some ≤ k < + ∞ .Proof. Let x ∈ M be a point such that G · x is finite. The median subalgebra of M generated by G · x is G –invariant and, by [Bow13, Lemma 4.2], it is finite. It was observed in [Bow16, Section 3]that every finite median algebra is isomorphic to the –skeleton of a finite CAT(0) cube complex.If X is a finite CAT(0) cube complex with X (0) isomorphic to h G · x i , the induced action G y X (0) preserves the median of X , hence it extends to an action by cubical automorphisms on the entire X .Since X is a compact CAT(0) space, it has a unique barycentre p ∈ X . Since the G –action preservesthe CAT(0) metric on X , the point p is fixed by G . Thus G leaves invariant the –skeleton of theunique open cube of X containing p . This provides the required median subalgebra C ⊆ M . (cid:3) The following examples were mentioned in the Introduction.
Example 2.6.
There are isometries of X = L ( R ) whose translation distance is not realised. Notethat X is a complete, geodesic, infinite-rank median space with the metric induced by its norm.As far as I know, the following is the quickest way to see this. There exist a standard Borel space Ω and an infinite, σ –finite, atomless measure ν such that H embeds in Y = L (Ω , ν ) isometricallyand equivariantly with respect to an embedding ι : Isom H ֒ → Isom Y (see e.g. Example 3.7 andProposition 3.14 in [CDH10]). Since Ω is standard, the space L (Ω , ν ) is isometric to L ( R ) .If g is a parabolic isometry of H , then h g i acts with unbounded orbits on H , so ι ( g ) acts withunbounded orbits on Y . In particular, ι ( g ) does not fix a point of Y . However, since the translationlength of g in H is zero, there exist points of Y that are moved arbitrarily little by ι ( g ) . Example 2.7.
Let G be the group BS (1 , n ) = h a, b | aba − = b n i with n ≥ . Let G y T be theBass–Serre tree corresponding to the HNN splitting evident from this presentation. Consider theaction G y R where b is a translation of length and a is a homothety of factor n fixing the origin.Thus, G y R is an action by homotheties and G y T is an action by isometries. The product T × R is a rank– median space, and the diagonal action G y T × R is by homeomorphisms that arealso automorphisms of the median-algebra structure. It is straightforward to check that the latteraction is cocompact, free, and metrically proper (though not uniformly proper).As mentioned in the introduction, [Fio18, Theorem A] implies that G does not admit any free iso-metric actions on finite-rank median spaces. However, G does act properly, freely and by isometrieson infinite-rank median spaces, since G has the Haagerup property [CDH10, Theorem 1.2].2.2. Restriction quotients.
Let M be a median algebra with an action G y M by medianautomorphisms. In the special case when M is the vertex set of a CAT(0) cube complex, the notionof restriction quotient was introduced in [CS11, Subsection 2.3].Consider a G –invariant subset U ⊆ W ( M ) . Given x, y ∈ M , we write x ∼ U y if W ( x | y ) ∩ U = ∅ .Observe that, for all x, y, z ∈ M and x ′ , y ′ , z ′ ∈ M , we have: W ( x | y ) ⊆ W ( x | z ) ∪ W ( z | y ) , W ( m ( x, y, z ) | m ( x ′ , y ′ , z ′ )) ⊆ W ( x | x ′ ) ∪ W ( y | y ′ ) ∪ W ( z | z ′ ) . e conclude that ∼ U is an equivalence relation and that the ∼ U –equivalence class of m ( x, y, z ) only depends on the ∼ U –equivalence classes of x, y, z . Thus, the quotient M/ ∼ U naturally inheritsa structure of median algebra, which we denote by M ( U ) . Since U is G –invariant, so is ∼ U , and M ( U ) is again endowed with a G –action by median automorphisms. Definition 2.8.
We say that M ( U ) is the restriction quotient of M associated to U .The quotient projection π U : M → M ( U ) is a G –equivariant, surjective median morphism. Re-mark 2.2 shows that the projection π U induces an injective, G –equivariant morphism of pocsets π ∗U : H ( M ( U )) → H ( M ) . Since π ∗U preserves transversality, this shows that rk M ( U ) ≤ rk M .We say that a subset A ⊆ M is ∼ U –saturated if it is a union of ∼ U –equivalence classes; equiva-lently A = π − U ( π U ( A )) . The image of π ∗U is precisely the set of ∼ U –saturated halfspaces. In general,this will be larger than the set of halfspaces bounded by the walls in U .2.3. The fundamental lemma.
Let M be a median algebra of rank r < + ∞ .Given a subset H ⊆ H ( M ) , we denote by T H the intersection of all halfspaces that it contains.We say that a subset C ⊆ H ( M ) is a chain of halfspaces if it is totally ordered by inclusion.The following simple fact is a key ingredient of many results in this paper. It can be viewed asan extension of Helly’s lemma to infinite sets of halfspaces. Lemma 2.9.
Let
H ⊆ H ( M ) be a set of pairwise-intersecting halfspaces such that any chain in H admits a lower bound in H . Then the intersection of all halfspaces in H is nonempty.Proof. Pick any point x ∈ M and let H x ⊆ H be the subset of those halfspaces that do not contain x . We begin by finding a point y ∈ M that lies in every element of H x .Given h , k ∈ H x , we have h ∩ k = ∅ by our hypothesis on H , and x ∈ h ∗ ∩ k ∗ by definition of H x .Hence either h ⊆ k , or k ⊆ h , or h and k are transverse. Since M has finite rank r , Dilworth’s lemmaallows us to partition H x = C ⊔ · · · ⊔ C k , where k ≤ r and each C i is totally ordered by inclusion.Let h , . . . , h k ∈ H be lower bounds for C , . . . , C k . Helly’s lemma implies that h ∩ · · · ∩ h k = ∅ ,and we pick the point y in this intersection.Now, let H x,y ⊆ H be the subset of halfspaces that contain exactly one point from the set { x, y } .Note that H x,y ⊆ H ( x | y ) ⊔ H ( y | x ) . Hence, again by Dilworth’s lemma, we can partition H x,y intofinitely many chains and there exists a point z ∈ T H x,y . Set w = m ( x, y, z ) .Suppose for the sake of contradiction that w ∈ h ∗ for some h ∈ H . Then at least two of the threepoints x, y, z must lie in h ∗ . By our choice of y , there does not exist h ∈ H with { x, y } ⊆ h ∗ . Byour choice of z , there does not exist h ∈ H with h ∈ H ( x, z | y ) ⊔ H ( y, z | x ) . We have reached thedesired contradiction. (cid:3) Remark 2.10. If C is a chain of halfspaces and T C = ∅ , then T C is a halfspace. Indeed, both T C and its complement are convex (ascending unions of convex sets are convex).If C is a chain of halfspaces, we say that a subset C ′ ⊆ C is cofinal if every halfspace in C contains a halfspace in C ′ . In this case, we have T C ′ = T C . Lemma 2.11.
A convex subset C ⊆ M is gate-convex if and only if there does not exist a chain C ⊆ H C ( M ) such that T C is nonempty and disjoint from C .Proof. Suppose that C is gate-convex. If C ⊆ H C ( M ) is a chain and x ∈ T C , then the gate-projection of x to C also lies in T C . Thus either T C = ∅ or C ∩ T C = ∅ .Conversely, suppose that C is convex and that there does not exist a chain C ⊆ H C ( M ) suchthat T C is nonempty and disjoint from C . We define a map π : M → C as follows.Let σ C ⊆ H ( M ) be the set of halfspaces containing C . For every x ∈ M , the halfspaces in H x = σ C ∪ ( σ x \ H ( C | x )) airwise intersect. Any chain C ⊆ H x contains a cofinal subset C ′ that is contained in either σ C or σ x \ H ( C | x ) . If C ′ ⊆ σ C , then C ⊆ T C ′ . If C ′ ⊆ σ x \ H ( C | x ) , then x ∈ T C ′ and, by ourassumption on C , we have T C ′ ∩ C = ∅ . Thus, since T C = T C ′ = ∅ , Remark 2.10 implies that T C is a halfspace, and we have T C ∈ H x .Now, we can apply Lemma 2.9 and conclude that T H x = ∅ . Observing that H x is an ultrafilter,we deduce that this intersection consists of a single point π ( x ) . This defines the map π : M → C .Since σ C ⊆ H x , we have π ( x ) ∈ C . It follows that W ( x | C ) = W ( x | π ( x )) for all x ∈ M . Hence π ( x ) ∈ I ( x, y ) for every x ∈ M and y ∈ C , showing that C is gate-convex. (cid:3) Products.
Given two median algebras M and M , we denote by M × M their product . Thisis the only median algebra with underlying set M × M such that the coordinate projections to M and M are median morphisms.We will need the following result. When M is the vertex set of a CAT(0) cube complex, thisis [CS11, Lemma 2.5]. When M is the underlying median algebra of a complete median space,compare [Fio18, Proposition 2.10] . In full generality, we will rely on Lemma 2.9. Lemma 2.12.
For a finite-rank median algebra M , the following are equivalent:(1) M splits as a nontrivial product of median algebras M × M ;(2) there exists a nontrivial partition W ( M ) = W ⊔ W such that W and W are transverse.When this happens, each set W i is naturally identified with W ( M i ) .Proof. First, suppose that M = M × M . Let π i denote the factor projections. For all x, y ∈ M ,we have I ( x, y ) = I ( π ( x ) , π ( y )) × I ( π ( x ) , π ( y )) . Thus, every convex set C ⊆ M is necessarily ofthe form C × C , where each C i ⊆ M i is convex. It follows that every halfspace of M is either ofthe form h × M with h ∈ H ( M ) , or of the form M × h with h ∈ H ( M ) . This proves theimplication (1) ⇒ (2) and the final statement of the lemma.We are left to prove that (2) ⇒ (1). The partition W ( M ) = W ⊔ W determines a nontrivial,transverse partition H ( M ) = H ⊔ H , where H i is the set of halfspaces associated to W i . Given x ∈ M , recall that σ x denotes the set of halfspaces containing x . Setting σ ix := σ x ∩ H i , we obtaina transverse partition σ x = σ x ⊔ σ x .Let A x , B x ⊆ M denote the intersection of all halfspaces in σ x and σ x , respectively. These areconvex subsets with A x ∩ B x = { x } . Given y ∈ A x and z ∈ B x , the intersection of all halfspaces in σ y ⊔ σ z is nonempty, since this set satisfies the hypotheses of Lemma 2.9. Moreover, the intersectionmust consist of a single point, since σ y ⊔ σ z contains a side of every wall of M . We denote this pointby φ ( y, z ) . This defines a map φ : A x × B x → M .If ( y, z ) , ( y ′ , z ′ ) ∈ A x × B x are distinct, then either y and y ′ are separated by a wall in W , or z and z ′ are separated by a wall in W . The same wall will separate φ ( y, z ) and φ ( y ′ , z ′ ) , hence φ is injective. Given w ∈ M , the argument of the previous paragraph also shows that there existsa point y w lying in the intersection of the halfspaces in σ w ⊔ σ x ; note that, in particular, we have y w ∈ A x . Similarly, there exists z w ∈ B x lying in every halfspace of the set σ x ⊔ σ w . Observing that w = φ ( y w , z w ) , we conclude that φ is also surjective. Thus, φ is a bijection.Finally, we show that φ is a morphism of median algebras. If this were not the case, there wouldexist y , y , y ∈ A x and z , z , z ∈ B x such that the points m ( φ ( y , z ) , φ ( y , z ) , φ ( y , z )) and φ ( m ( y , y , y ) , m ( z , z , z )) are distinct. Hence there would exist a halfspace h ∈ H ( M ) containing the latter point, but notthe former. Without loss of generality h ∈ H . Then m ( y , y , y ) ∈ h , while at least two of the In particular, the equivalence of parts (1) and (2) in [Fio18, Proposition 2.10] does not require a metric, ormeasurability of the partition. By contrast, we stress that the equivalence with part (3) of [Fio18, Proposition 2.10]does require a complete metric. Without it, a counterexample is provided by [0 , \ { (0 , } with the ℓ metric. hree points φ ( y , z ) , φ ( y , z ) , φ ( y , z ) must lie in h ∗ . Since h ∈ H , the latter implies that atleast two among y , y , y lie in h ∗ , contradicting the fact that m ( y , y , y ) ∈ h . (cid:3) We say that M is irreducible if it is not isomorphic to a nontrivial product M × M . It is standardto deduce the following from Lemma 2.12 (cf. [CS11, Proposition 2.6] or [Fio18, Proposition 2.12]). Corollary 2.13.
Let M be a median algebra of rank r < + ∞ . Then, there exist irreducible medianalgebras M , . . . , M k (unique up to permutation) such that M ≃ M × . . . × M k and k ≤ r . Thegroup Aut M preserves this splitting, possibly permuting the factors. Automorphisms of finite-rank median algebras.
This section constitutes the heart of the paper. We will define convex cores in Subsection 3.3,where we also prove Theorem C. We will then study semisimplicity of automorphisms in Sub-section 3.4, proving Corollary D, which in turn implies Corollary A. We conclude the section bystudying non-transverse automorphisms in Subsection 3.5.Throughout Section 3, we fix a median algebra M of rank r < + ∞ , a group G , and an action G y M by median automorphisms.3.1. Dynamics of halfspaces.Definition 3.1.
We say that two halfspaces h , k ∈ H ( M ) are facing if h ∗ ∩ k ∗ = ∅ and h = k ∗ .The following G –invariant subsets of H ( M ) will play a fundamental role throughout Section 3. H ( G ) := { h ∈ H ( M ) | ∃ g ∈ G such that g h ( h } , H ( G ) := { h ∈ H ( M ) | the orbit G · h is finite } , H / ( G ) := { h ∈ H ( M ) \ H ( G ) | ∀ g ∈ G, g h and h are either facing, transverse, or equal } , H ( G ) := { h ∈ H ( M ) | ∀ g ∈ G either g h ∈ { h , h ∗ } or g h and h are transverse } , H / ( G ) := { h ∈ H ( M ) \ H ( G ) | ∃ g ∈ G such that g h and h are facing } . We chose this notation by analogy with the terminology of Caprace and Sageev from [CS11, Sub-section 3.3]; we will explain the connection in detail in Remark 4.10. When it is necessary to specifythe median algebra acted upon by G , we will also write H • ( G, M ) and H • ( G, M ) .Note that H ( G ) ∗ = H ( G ) , H ( G ) ∗ = H ( G ) and H ( G ) ∗ = H ( G ) . Remark 3.2.
Since rk M = r , the G –orbit of any halfspace h ∈ H ( G ) contains at most r halfspaces. In particular, we have H ( G ) ⊆ H ( G ) and it follows that H / ( G ) ⊆ H / ( G ) . Remark 3.3.
The halfspaces in H / ( G ) pairwise intersect. Indeed, suppose for the sake of con-tradiction that there exist h , k ∈ H / ( G ) with h ∩ k = ∅ . Choose g , g ∈ G such that g h and h arefacing, and g k and k are facing. Then g h ⊆ g k ∗ ( k ⊆ h ∗ ( g h . Hence g − g h ( h , contradictingthe assumption that h
6∈ H ( G ) . Remark 3.4.
Given two halfspaces h , k ∈ H ( M ) , exactly seven possibilities may occur as to theirrelative position. We might have h = k , h = k ∗ , h ( k , k ( h , h and k might be facing, h ∗ and k ∗ might be facing, or, finally, h and k might be transverse.Since H / ( G ) is G –invariant, the following is immediate from Remarks 3.3 and 3.4: Remark 3.5. If h ∈ H / ( G ) , then there does not exist g ∈ G such that g h = h ∗ , nor does thereexist g ∈ G such that g h ∗ and h ∗ are facing. In particular, we have H / ( G ) ⊆ H ( G ) ⊔ H / ( G ) . emma 3.6. We have G –invariant partitions: H ( M ) = H ( G ) ⊔ H ( G ) ⊔ H / ( G ) ⊔ H / ( G ) ∗ , H ( M ) = H ( G ) ⊔ H ( G ) ⊔ H / ( G ) ⊔ H / ( G ) ∗ , where H / ( G ) ⊆ H / ( G ) ⊆ H ( G ) ⊔ H / ( G ) and H ( G ) ⊆ H ( G ) .Proof. The second partition is immediate from Remark 3.4. Regarding the first, note that the sets H / ( G ) and H / ( G ) ∗ are disjoint since, as shown in Remark 3.2, we have H / ( G ) ⊆ H / ( G ) and H / ( G ) ∗ ⊆ H / ( G ) ∗ . Thus it is also clear that the sets H ( G ) , H ( G ) , H / ( G ) and H / ( G ) ∗ are pairwise disjoint. The fact that these four sets cover H ( M ) , as well as the rest of the lemma,follow from Remarks 3.2 and 3.5. (cid:3) Remark 3.7. If H ≤ G is a finite-index subgroup, then it is clear that H ( H ) = H ( G ) and H ( H ) = H ( G ) . It follows that we have H / ( H ) = H / ( G ) . The sets H ( H ) and H / ( H ) ,however, will differ from H ( G ) and H / ( G ) in general.When G = h g i , we simply write H • ( g, M ) or H • ( g ) , rather than H • ( h g i , M ) . In this case, we cangive a simpler characterisation of the sets H • ( g ) as follows: • h ∈ H ( g ) if and only if there exists n ∈ Z \ { } such that g n h = h ; • h ∈ H ( g ) if and only if there exists n ∈ Z such that g n h ( h ; • h ∈ H / ( g ) if and only if, for every n ∈ Z \ { } , the halfspaces g n h and h are either facingor transverse.We conclude this subsection with two results on the trivial case where the G –orbit of everyhalfspace of M is finite. They will prove useful in Subsection 3.3. Lemma 3.8. If H ( M ) = H ( G ) , then G has a finite orbit in M .Proof. Suppose for the sake of contradiction that no finite-index subgroup of G fixes a point of M .Consider two distinct points x, y ∈ M . Any two minimal elements of H ( x | y ) are transverse, andso are any two maximal elements. Thus, there are at most r among minimal and maximal elementsof H ( x | y ) , and a finite-index subgroup G ≤ G leaves each of them invariant. By Remark 2.10, everyhalfspace of H ( x | y ) contains a minimal element of H ( x | y ) and is contained in a maximal element.Thus, G preserves the set H ( x | y ) . Hence, for every g ∈ G , we have W ( x | y ) = W ( x, gx | y, gy ) and W ( x | gx ) = W ( x, y | gx, gy ) . Hence W ( x | y ) and W ( x | gx ) are transverse for all g ∈ G .By our assumption, G does not fix x , hence there exists g ∈ G such that W ( x | g x ) = ∅ .Repeating the argument above for the points x and g x (rather than x and y ), we find a finite-indexsubgroup G ≤ G and an element g ∈ G such that W ( x | g x ) is nonempty and transverse to W ( x | g x ) . Since g ∈ G , the set W ( x | g x ) is also transverse to W ( x | y ) .Iterating, we find g , . . . , g r ∈ G such that the sets W ( x | y ) , W ( x | g x ) , . . . , W ( x | g r x ) are nonemptyand pairwise transverse. This violates the assumption that rk M = r . (cid:3) Remark 3.9. If H ( M ) = H ( G ) and G is finitely generated, then G y M factors through theaction of a finite group.Indeed, by Remark 3.2, the G –stabiliser of any halfspace of M has index ≤ r in G . Since G is finitely generated, it contains only finitely many subgroups of index ≤ r . Their intersection isa finite-index characteristic subgroup H ⊳ G that preserves every halfspace of M . Hence H fixesevery point of M , and the G –action factors through the finite group G/H .3.2.
Some technical results.
We gather here a few technical results that we do not deem of muchinterest on their own. They will be needed in the proofs of some results in later sections, but weprove them here as they only require the terminology of Subsection 3.1. The reader can safely skipthis subsection and return when they encounter a reference to something proved here. he following is needed to prove Lemma 3.11, which is used in the proof of Proposition 3.24. Remark 3.10.
Consider a chain C ⊆ H ( G ) such that T C ∈ H ( M ) . Then T C
6∈ H ( G ) .Indeed, set k = T C and suppose for the sake of contradiction that there exists g ∈ G such that k ( g k . Pick a point x ∈ g k \ k . Since x k , there exists h ∈ C such that x h . Since h ∈ H ( G ) ,there exists m ≥ such that g m h = h . Then x ∈ g k ⊆ g m k ⊆ g m h = h , a contradiction. Lemma 3.11.
Suppose that H / ( G ) = ∅ . Then, for every h ∈ H ( G ) there exist x ∈ h and y ∈ h ∗ such that H ( x | y ) ⊆ H ( G ) .Proof. Since h ∈ H ( G ) , there exists g ∈ G such that g h ( h . Pick a point x ∈ h \ g h . Recall that σ x ⊆ H ( M ) is the set of halfspaces that contain x , and define σ := σ x ∩ H ( G ) . We will showthat the intersection of all elements of the set σ ⊔ { h ∗ } is nonempty. Picking any point y in thisintersection will then yield H ( x | y ) ∩ H ( G ) = ∅ , hence H ( x | y ) ⊆ H ( G ) .Let us show that H = σ ⊔ { h ∗ } satisfies the hypotheses of Lemma 2.9. First, for any chain C ⊆ H ,either h ∗ is a lower bound for C , or the subset C ∩ σ is cofinal. In the latter case, the intersection ofall elements of C ∩ σ is nonempty, since it contains x , so it must be a halfspace k by Remark 2.10.By Remark 3.10, we have k
6∈ H ( G ) . Therefore k ∈ H ( G ) , hence k ∈ σ ⊆ H . This shows thatevery chain in H has a lower bound in H .To conclude the proof, we need to show that any two elements of H intersect. Since all elementsof σ contain x , this boils down to showing that h ∗ cannot be disjoint from a halfspace j ∈ σ . If thiswere the case, a finite-index subgroup H ≤ G would preserve j . For every h ∈ H , we would have hx ∈ h j = j ⊆ h . However, since g h ( h and x ∈ h \ g h , we have g − n x ∈ h ∗ for all n ≥ . For n sufficiently large, g − n will lie in H , so we have obtained a contradiction. (cid:3) The next lemma will only be used in the proof of Proposition 3.36 in Subsection 3.5.
Lemma 3.12.
Let C ⊆ M be a G –invariant convex subset. If C ⊆ H C ( M ) is a chain such that T C ∈ H ( M ) \ H C ( M ) , then T C
6∈ H ( G, M ) .Proof. Set k := T C . Since k ∗ contains the complement of any element of C ∈ H C ( M ) , we have k ∗ ∩ C = ∅ . Thus, the fact that k H C ( M ) implies that k ∩ C = ∅ .Suppose for the sake of contradiction that there exists g ∈ G such that g k ) k . Replacing C witha cofinal subset, we can assume that some point x ∈ g k \ k lies in h ∗ for all h ∈ C . Claim: for every h ∈ C and n ≥ , there exists a halfspace j n ( h ) ∈ C such that, for every j ∈ C with j ⊆ j n ( h ) , the halfspaces h and g n j are transverse.Proof of Claim. Since we can always replace g with a proper power, it suffices to consider n = 1 .To begin with, for every j ∈ C , we have h ∩ g j ⊇ k ∩ g k = ∅ and x ∈ h ∗ ∩ g j . Suppose for the sakeof contradiction that the claim fails. Then there exists a cofinal subset C ′ ⊆ C such that either h ∩ g j ∗ = ∅ for all j ∈ C ′ , or h ∗ ∩ g j ∗ = ∅ for all j ∈ C ′ (equivalently, either h ⊆ g j or h ∗ ⊆ g j ). Itfollows that g k = T g C = T g C ′ contains either h or h ∗ , which both lie in H C ( M ) . This contradictsthe fact that g k ∩ C = g ( k ∩ C ) = ∅ . (cid:3) Pick any halfspace h ∈ C . Inductively choose h i +1 ∈ C with h i +1 ⊆ j i +1 ( h ) ∩ j i ( h ) ∩ · · · ∩ j ( h i ) .By the Claim, the infinitely many halfspaces g i h i are pairwise transverse. This contradicts the factthat M has finite rank, proving the lemma. (cid:3) The following two remarks are needed in the proof of Lemma 4.5 in Subsection 4.
Remark 3.13.
Consider g ∈ Aut M and h ∈ H ( g ) such that T n ∈ Z g n h = ∅ and T n ∈ Z g n h ∗ = ∅ .Then h is transverse to every element of H ( g ) .Indeed, suppose for the sake of contradiction that k ∈ H ( g ) is not transverse to h . Possiblyreplacing k and h with their complements, we can assume that k ⊆ h . Replacing g with a nontrivial ower, we can also assume that g k = k and g h ( h . But then we have k = g n k ⊆ g n h for all n ∈ Z ,contradicting the fact that T n ∈ Z g n h = ∅ .The next remark is also needed in the proof of Proposition 3.40 in Subsection 3.5. Remark 3.14.
Let N ⊆ M be a G –invariant subalgebra. Consider h ∈ H ( N ) . By Remark 2.3,there exists k ∈ H ( M ) with h = k ∩ N . If h ∈ H ( G, N ) , every such k lies in H ( G, M ) .Indeed, there exists g ∈ G such that g h ( h . It is clear that k
6∈ H ( g, M ) , or a power of g wouldhave to preserve k and h . For every n ∈ Z , the sets g n h ∩ h and g n h ∗ ∩ h ∗ are nonempty. Hence g n k ∩ k and g n k ∗ ∩ k ∗ are nonempty, and we have k
6∈ H / ( g, M ) ⊔ H / ( g, M ) ∗ . We conclude that k ∈ H ( g, M ) ⊆ H ( G, M ) .3.3. Convex cores.
It is often possible to reduce to the case of actions satisfying H / ( G ) = ∅ or H / ( G ) = ∅ by restricting to the following canonical subspaces. Definition 3.15. (1) The (convex) core C ( G ) is the intersection of all halfspaces in H / ( G ) .(2) The reduced (convex) core C ( G ) is the intersection of all halfspaces in H / ( G ) .We will write C ( G, M ) or C ( G, M ) when it is necessary to specify the median algebra.The core and the reduced core are G –invariant convex subsets of M , although they can of coursebe empty in general. Note that we always have C ( G ) ⊆ C ( G ) . Remark 3.16. If H ≤ G is a commensurated subgroup, then the core C ( H ) is G –invariant. Indeed,for every g ∈ G , Remark 3.7 implies that: H • ( H ) = H • ( gHg − ∩ H ) = H • ( gHg − ) = g H • ( H ) . Thus, the sets H ( H ) , H / ( H ) and H ( H ) are G –invariant, and so is C ( H ) .If H is normal, then also the reduced core C ( H ) and the sets H ( H ) , H / ( H ) are G –invariant.The following is the main result of this subsection. Theorem 3.17.
Let G be finitely generated. Then:(1) the core C ( G ) is nonempty;(2) if G y M has no wall inversions, the reduced core C ( G ) is nonempty. Example 3.18. If G y M has wall inversions, the reduced core C ( G ) can be empty. For instance,consider M = [ − , \ { } and the group G generated by the reflection in .We now obtain a sequence of lemmas leading up to Lemma 3.21. Theorem 3.17 will then imme-diately follow by applying Lemma 2.9. Lemma 3.19.
For some d ≥ , let B d be the (closed) ball of radius d in a Cayley graph of G . Let H ≤ G be a subgroup with the property that B d ⊆ g H ∪ · · · ∪ g d H for some g , . . . , g d ∈ G . Then H has index ≤ d in G .Proof. Let S ⊆ G be the generating set corresponding to the chosen Cayley graph. We prove thelemma by induction on d . In the base case d = 1 , a coset of H contains { } ∪ S , hence H = G .Suppose that there exist g , . . . , g d ∈ G such that B d ⊆ g H ∪ · · · ∪ g d H . Replacing each g i withits inverse, it is more convenient to assume that B d ⊆ Hg ∪ · · · ∪ Hg d . If B d − is contained in theunion of d − of these cosets, then the inductive hypothesis implies that H has index ≤ d − < d .Thus, we can assume that B d − intersects each of the cosets Hg , . . . , Hg d .Hence, for every ≤ i ≤ d and every s ∈ S , there exists ≤ j ( i, s ) ≤ d such that Hg i s = Hg j ( i,s ) .Since S generates G , this implies that, for every i and g ∈ G , there exists ≤ j ( i, g ) ≤ d such that Hg i g = Hg j ( i,g ) . Hence G = Hg ∪ · · · ∪ Hg d as required. (cid:3) emma 3.20. If G is finitely generated, there exists a finite subset F ⊆ G with the followingproperty. For every h ∈ H / ( G ) , there exists f ∈ F such that f h and h are facing.Proof. Choose any locally finite Cayley graph of G and let B ⊆ G be the (closed) ball of radius r in it. Take F := B − B . Consider h ∈ H / ( G ) .Let G ( h ) ≤ G denote the stabiliser of h . Choose elements g , . . . , g k ∈ G such that the cosets g i G ( h ) are pairwise distinct. We take k to be the index of G ( h ) in G if this is ≤ r , and k = r + 1 otherwise. Lemma 3.19 ensures that we can choose g , . . . , g k within B .If i = j , Remark 3.5 guarantees that the halfspaces g i h and g j h are either transverse or facing.If k ≤ r , then every coset of G ( h ) is of the form g i G ( h ) , and there exists f ∈ { g , . . . , g k } ⊆ B suchthat f h and h are facing. Otherwise, we have k > r = rk M and there exist i = j such that g i h and g j h are facing. Setting f = g − i g j , we have f ∈ B − B = F , and h and f h are facing. (cid:3) Lemma 3.21.
Let G be finitely generated.(1) If C ⊆ H / ( G ) is a chain, then T C is a halfspace of M lying in H / ( G ) .(2) If G y M has no wall inversions and C ⊆ H / ( G ) is a chain, then T C ∈ H / ( G ) .Proof. Recall that H / ( G ) ⊆ H / ( G ) . The first part of the proof will deal with a general chain C ⊆ H / ( G ) and a general action G y M . We will introduce the specific hypotheses of the twoparts of the lemma only towards the end.Set C := T C . Given g ∈ G , we denote by C ( g ) ⊆ C the subset of those halfspaces h such that h and g h are facing. We denote by S ( g ) ⊆ C the subset of halfspaces h such that g h = h .By Lemma 3.20, there exists a finite subset F ⊆ G such that, for every h ∈ H / ( G ) , there exists f ∈ F such that h and f h are facing. Thus, C is the union of finitely many subsets C ( f ) with f ∈ F , and there exists f ∈ F such that C ( f ) is cofinal in C .Fix some halfspace k ∈ C ( f ) . The set of those halfspaces h ∈ C ( f ) with h ⊆ k is also cofinalin C . For every such h , we have k ∗ ⊆ h ∗ ⊆ f h , hence f − k ∗ ⊆ h . This shows that f − k ∗ ⊆ C . Inparticular, C is nonempty and Remark 2.10 implies that C ∈ H ( M ) .We now show that C
6∈ H ( G ) . Suppose for the sake of contradiction that gC ( C for some g ∈ G . For every h ∈ H / ( G ) , the halfspaces h , g h , . . . , g r h cannot be pairwise transverse, henceRemark 3.5 guarantees the existence of ≤ i ≤ r such that h and g i h are either facing or equal.Thus, C is the union of the finitely many subsets C ( g i ) ∪ S ( g i ) , and there exists ≤ i ≤ r suchthat either C ( g i ) or S ( g i ) is cofinal in C . Hence either g i C ∗ ∩ C ∗ = ∅ or g i C = C . However, gC ( C implies that g i C ( C and g i C ∗ ∩ C ∗ = C ∗ = ∅ , a contradiction.We conclude the proof separately for the two parts of the lemma. Part (2): recalling that C ( f ) is cofinal in C , we see that f C ∗ ∩ C ∗ = ∅ . Since G y M has no wall inversions, this implies that f C and C are facing. We have already shown that C ∈ H ( M ) \ H ( G ) , so this means that C lies in H / ( G ) , as required. Part (1): as above, we have f C ∗ ∩ C ∗ = ∅ , hence C
6∈ H / ( G ) ∗ . We are left to show that C
6∈ H ( G ) . If this were not the case, there would exist a finite-index subgroup H ≤ G such that hC = C for all h ∈ H . By Remark 3.7, we would have H / ( H ) = H / ( G ) ⊇ C . Thus, applyingLemma 3.20 to H and proceeding as above, we would obtain h ∈ H such that C ( h ) is cofinal in C . This would yield h C ∗ ∩ C ∗ = ∅ , contradicting the fact that h C = C . (cid:3) Proof of Theorem 3.17.
Under the respective hypotheses, Remark 3.3 and Lemma 3.21 show thatthe sets H / ( G ) and H / ( G ) satisfy the hypotheses of Lemma 2.9. (cid:3) We can now characterise the halfspaces of our two cores. Recall that, for every convex subset C ⊆ M , Remark 2.3 shows that the subset H C ( M ) ⊆ H ( M ) is naturally identified with H ( C ) (and isomorphic to it as a pocset). emma 3.22. (1) We have H C ( G ) ( M ) ⊆ H ( G ) ⊔ H ( G ) and H C ( G ) ( M ) ⊆ H ( G ) ⊔ H ( G ) . Taking intersec-tions with the cores, this induces G –invariant partitions: H ( C ( G )) = H ( G, C ( G )) ⊔ H ( G, C ( G )) , H ( C ( G )) = H ( G, C ( G )) ⊔ H ( G, C ( G )) . (2) If G is finitely generated, then H ( G ) ⊆ H C ( G ) ( M ) . If moreover G y M has no wallinversions, then H ( G ) ⊆ H C ( G ) ( M ) .Proof. We begin with part (1). The first sentence is clear. By Remark 2.3, every k ∈ H ( C ( G )) isof the form k = h ∩ C ( G ) for a unique h ∈ H C ( G ) ( M ) (and similarly for the reduced core).If h ∈ H ( G, M ) , then h is preserved by a finite-index subgroup of G , and so is h ∩ C ( G ) . In thiscase, we have k ∈ H ( G, C ( G )) . If h ∈ H ( G, M ) , it follows from Remark 2.3 that h ∩ C ( G ) lies in H ( G, C ( M )) .If h ∈ H ( G, M ) , then there exists g ∈ G with g h ( h . In this case, g k ⊆ k and, since k is theintersection with C ( G ) of a unique halfspace of M , we must have g k ( k . Hence k ∈ H ( G, C ( G )) and, similarly, h ∩ C ( G ) ∈ H ( G, C ( G )) .Now, let us prove part (2). Consider a halfspace h ∈ H ( M ) . Applying Lemma 3.21 andLemma 2.9 to the sets H / ( G ) ∪ { h } and H / ( G ) ∪ { h ∗ } , we see that h H C ( G ) ( M ) if and onlyif there exists k ∈ H / ( G ) such that k ∩ h = ∅ or k ∩ h ∗ = ∅ . Similarly, if G y M has no wallinversions, h H C ( G ) ( M ) if and only if there exists k ∈ H / ( G ) such that k ∩ h = ∅ or k ∩ h ∗ = ∅ .Halfspaces k ∈ H / ( G ) and h ∈ H ( G ) can never be disjoint. Indeed, there exists a finite-indexsubgroup H ≤ G such that g h = h for all g ∈ H . By Remark 3.7, we have H / ( G ) = H / ( H ) , sothere exists g ∈ H such that g k and k are facing. If k and h were disjoint, then g h ∗ and h ∗ would befacing, contradicting the fact that g h = h .Finally, k ∈ H / ( G ) and h ∈ H ( G ) can never be disjoint either. Indeed, there exists g ∈ G suchthat k and g k are facing. If k and h were disjoint, then g h ∗ and h ∗ would be facing, violating thefact that, for every g ∈ G , either g h lies in { h , h ∗ } or it is transverse to h . (cid:3) As a consequence of Theorem 3.17, we obtain the following generalisation of a classical result ofSageev on
CAT(0) cube complexes (see e.g. [CFI16, Proposition B.8] and [Sag95, Theorem 5.1]).
Proposition 3.23.
Suppose that G is finitely generated and H ( G ) = ∅ . Then:(1) there exists a G –invariant median subalgebra C ⊆ M isomorphic to { , } k with ≤ k ≤ r ;(2) if G acts with no wall inversions, then G fixes a point of M .Proof. By part (1) of Theorem 3.17 and part (1) of Lemma 3.22, we can pass to the core and assumethat H ( M ) = H ( G ) . Part (1) then immediately follows from Lemmas 3.8 and 2.5.Now, if we could deduce that the G –action on C ≃ { , } k has no wall inversions from the factthat G y M has no wall inversions, then part (2) would also follow immediately. Unfortunately,this is not true in general and part (2) will require more work. Our objective will be to construct a specific subalgebra C ≃ { , } k such that G y C has no wall inversions.We begin by passing to the reduced core and assuming that H ( M ) = H ( G ) . This is allowedbecause of part (2) of Theorem 3.17, part (1) of Lemma 3.22, and Remark 2.4. By Remark 3.9, wecan moreover assume that G is a finite group.Recall that every set of pairwise-transverse walls of M has cardinality ≤ r = rk M . Thus,there exists a nonempty subset U ⊆ W ( M ) that is maximal among G –invariant sets of pairwise-transverse walls. Since H ( M ) = H ( G ) and G y M has no wall inversions, every G –orbit in W ( M ) consists of pairwise-transverse walls. This implies that U is also maximal among (not necessarily G –invariant) sets of pairwise-transverse walls. Indeed, if there existed w ∈ W ( M ) with U ⊔ { w } pairwise-transverse, then U ⊔ G · w would be pairwise-transverse and G –invariant. n conclusion, there exists a maximal set of pairwise-transverse walls U ⊆ W ( M ) that is also G –invariant. Let h +1 , . . . , h + k be a choice of halfspace for every element of U , and let h − , . . . , h − k betheir complements. Choose a finite subset F ⊆ M that intersects each of the k sectors of the form h ± ∩ · · · ∩ h ± k . Since G is finite, we can take F to be G –invariant.Let N ⊆ M be the median subalgebra generated by F . Note that N is G –invariant and, by[Bow13, Lemma 4.2], it is finite. It was observed in [Bow16, Section 3] that every finite medianalgebra is naturally isomorphic to the –skeleton of a finite CAT(0) cube complex. Let X be thecube complex with –skeleton isomorphic to N . Since the induced action G y X (0) preserves themedian operator of X , it extends to an action by cubical automorphisms on the entire X .By our choice of F , the intersections k i = h + i ∩ N are pairwise-transverse halfspaces of N . Bypart (1) of Remark 2.3, this is a maximal collection of pairwise-transverse halfspaces of N . Thecorresponding walls of N determine a maximal set of pairwise-transverse hyperplanes of X , whichis also G –invariant. Thus, they correspond to a maximal cube C ⊆ X that is preserved by G .We conclude the proof by showing that G acts on C without wall inversions (which implies that G fixes a vertex of C ). Suppose for the sake of contradiction that there exists an element g ∈ G such that g k i = k ∗ i for some i . This implies that g h + i ∩ h + i ∩ N = ∅ . However, since U is G –invariant,there exists an index j with g h + i = h ± j . Thus, g h + i ∩ h + i ∩ N contains one of the two nonempty sets h ± j ∩ h + i ∩ F . This is the required contradiction. (cid:3) Recall from Definition 1.2 that the action G y M is said to be essential if H ( M ) = H ( G ) .One can often reduce to studying essential actions by relying on the following result. Proposition 3.24. If G is finitely generated and G y M has no wall inversions, then there existsa nonempty, G –invariant, convex subset C ⊆ M such that H C ( M ) ⊆ H ( G, M ) .Proof. By Theorem 3.17, the core C ( G ) ⊆ M is nonempty. By Remark 2.4, the action G y C ( G ) hasno wall inversions. Thus, it is not restrictive to assume that M = C ( G ) , i.e. that H / ( G, M ) = ∅ .Let U ⊆ W ( M ) be the set of walls that bound halfspaces in H ( G, M ) . Let π U : M → M ( U ) bethe corresponding restriction quotient, as defined in Subsection 2.2. The image of the induced map π ∗U : H ( M ( U )) → H ( M ) coincides with the set of ∼ U –saturated halfspaces of M . By Lemma 3.11,no element of H ( G, M ) is ∼ U –saturated. Hence the image of π ∗U is exactly H ( G, M ) .We conclude that H ( G, M ( U )) = H ( M ( U )) . It is clear that G y M ( U ) has no wall inversions,so part (2) of Proposition 3.23 yields a G –fixed point x ∈ M ( U ) . The fibre C := π − U ( x ) isnonempty, G –invariant and convex. Since C consists of a single ∼ U –equivalence class, H C ( M ) and H ( G, M ) are disjoint. Hence H C ( M ) ⊆ H ( G, M ) . (cid:3) Remark 3.25.
Even if G y M does have wall inversions, the conclusion of Proposition 3.24always holds for a subgroup H ≤ G of index at most r . This can be proved exactly as above,simply appealing to part (1) of Proposition 3.23 rather than part (2).Recall from Definition 1.2 that the action G y M is minimal if M does not contain any proper, G –invariant, convex subsets (a convex subset is proper if it is nonempty and not the entire M ). Corollary 3.26. (1) If G y M is essential, then it is minimal.(2) Suppose that G is finitely generated and G y M has no wall inversions. If G y M isminimal, then it is essential.Proof. Part (2) follows from Proposition 3.24. Let us prove part (1).Suppose for the sake of contradiction that G y M is essential, but there exists a proper, G –invariant, convex subset C ⊆ M . Let σ C ⊆ H ( M ) be the set of halfspaces containing C . Notethat σ C is G –invariant and, since C = M , we have σ C = ∅ . very chain C ⊆ σ C has a lower bound in C . Indeed, T C is nonempty, so it must be a halfspaceof M containing C . Zorn’s lemma guarantees the existence of a minimal element h ∈ σ C . However,since G y M is essential, there exists g ∈ G such that g h ( h . Since σ C is G –invariant, we have g h ∈ σ C , obtaining the required contradiction. (cid:3) Remark 3.27.
When wall inversions are allowed, one can still conclude that G y M is essentialif the restriction H y M is minimal for every subgroup H ≤ G of index ≤ r . This follows fromRemark 3.25.Proposition 3.24 and Corollary 3.26 prove Theorem C from the introduction.3.4. Semisimple automorphisms.
This subsection is devoted to deducing from Proposition 3.24that automorphisms of finite-rank median algebras are (almost) semisimple.The key result here is Lemma 3.29, which makes significant use of the notion of zero-completion of a median algebra. We briefly recall all necessary information on this object, which was originallyintroduced by Bandelt and Meletiou in [BM93] and which we further developed in [Fio20b, Subsec-tion 4.1]. The reader should keep in mind the situation where M is the vertex set of a CAT(0) cubecomplex, in which case the zero-completion is just the Roller compactification.In general, the zero-completion of M is the median algebra M defined as follows. Consider theset of all intervals I := { I ( x, y ) | x, y ∈ M } . Points of M are functions x : I → M satisfying: • x ( I ) ∈ I for all I ∈ I ; • π I ∩ J ( x ( I )) = π I ∩ J ( x ( J )) for all I, J ∈ I with I ∩ J = ∅ , where π I ∩ J : M → I ∩ J denotesthe gate-projection.The median operator of M is given by m ( x, y, z )( I ) = m ( x ( I ) , y ( I ) , z ( I )) . We can define a map ι : M → M by setting ι ( x )( I ) = π I ( x ) , where π I : M → I is the gate-projection.In [Fio20b, Subsection 4.1], we showed the following: Theorem 3.28.
The zero-completion M is a median algebra with rk M = rk M . The map ι : M → M is an (Aut M ) –equivariant, injective median morphism with convex image. Thus, we will not distinguish between M and ι ( M ) ⊆ M . Moreover, we will identify the sets H M ( M ) and H ( M ) , which is allowed by part (2) of Remark 2.3. We write ∂M := M \ M .Note that a point ( x : I → M ) ∈ M lies in a halfspace h ∈ H ( M ) ≃ H M ( M ) if and only if x ( I ) ∈ h for every I such that h ∈ H I ( M ) (see [Fio20b, pp. 1356–1357]).We are now ready to state the main result of this subsection. Lemma 3.29.
Consider g ∈ Aut M . Suppose that H ( g ) = H ( M ) . Denote by H + (resp. H − ) theset of halfspaces h ∈ H ( M ) such that there exists n ≥ with g n h ( h (resp. h ( g n h ) .(1) For every h ∈ H ( M ) , we have T n ∈ Z g n h = ∅ and S n ∈ Z g n h = M .(2) We have a h g i –invariant partition H ( M ) = H − ⊔ H + . Any two halfspaces in H + intersect,and so do any two halfspaces in H − . Moreover, H − = ( H + ) ∗ .(3) There exists a unique point ξ + ∈ ∂M (resp. ξ − ∈ ∂M ) that lies in every halfspace in H + (resp. H − ). We have gξ + = ξ + , gξ − = ξ − and M ⊆ I ( ξ − , ξ + ) .(4) There exists x ∈ M such that the sets W ( g n x | g n +1 x ) are pairwise disjoint for n ∈ Z .(5) If no h ∈ H ( M ) is transverse to g h , then part (4) holds for all x ∈ M .Proof. We begin with part (1). For every h ∈ H ( M ) = H ( g ) , there exists k ∈ Z such that g k h ( h .Note that T n ∈ Z g nk h and its complement S n ∈ Z g nk h ∗ are convex and h g k i –invariant. Thus, if T n ∈ Z g nk h were nonempty, it would be a h g k i –invariant halfspace, contradicting the assumption that H ( g ) = ∅ . We conclude that T n ∈ Z g n h ⊆ T n ∈ Z g nk h = ∅ . Since this holds for every h ∈ H ( M ) ,we also have T n ∈ Z g n h ∗ = ∅ , hence S n ∈ Z g n h = M . et us prove part (2). The only statement requiring a proof is that any two halfspaces h , k ∈ H + must intersect. The fact that H − has the same property can be proved similarly.Suppose for the sake of contradiction that h ∩ k = ∅ and that there exist k, m ≥ with g k h ( h and g m k ( k . Replacing k and m by km , we can assume that k = m . Then, for every n ≥ , wehave g kn k ( k ⊆ h ∗ , hence k ⊆ g − kn h ∗ ( h ∗ . Thus, ∅ 6 = k ⊆ T n ∈ Z g kn h ∗ , which contradicts part (1).We now prove part (3). Let I ⊆ M be an interval and let σ I ⊆ H ( M ) be the set of halfspacesthat contain it. By Lemma 2.9, the intersection of all halfspaces in the set σ I ⊔ ( H + ∩ H I ( M )) isnonempty. Since H + contains a side of every wall of M , this intersection consists of a single point,which we denote by z + I . Note that z + I ∈ I . Similarly, there exists a unique point z − I ∈ I that lies inevery halfspace of the set σ I ⊔ ( H − ∩ H I ( M )) . It is clear that every wall of I separates the points z ± I , hence I = I ( z − I , z + I ) . It is straightforward to check that the functions I z + I and I z − I define the required points ξ + , ξ − ∈ ∂M .Let us address part (4). We begin with the following: Claim: for every x ∈ M , the set W ( x, ξ + | gx, g x, . . . , g r x ) is empty.Proof of Claim. Consider a halfspace h ∈ H ( x, ξ + | gx, g x, . . . , g r x ) . Recall from part (3) that M ⊆ I ( ξ − , ξ + ) . Thus, since ξ + ∈ h ∗ , we must have ξ − ∈ h .For each ≤ i ≤ r , either g i h ( h or the halfspaces g i h and h are transverse. Indeed, observingthat h ∈ H ( ξ + | g i x, ξ − ) and g i h ∈ H ( g i x, ξ + | ξ − ) , we see that the three intersections h ∩ g i h , h ∩ g i h ∗ and h ∗ ∩ g i h ∗ are nonempty, as they contain the points ξ − , g i x and ξ + , respectively.Since h , g h , . . . , g r h cannot be pairwise transverse, there exists ≤ k ≤ r such that g k h ( h .However, this implies that h ∈ H + , contradicting the fact that ξ + ∈ h ∗ . (cid:3) Now, pick any point x ∈ M and define iteratively x i +1 = m ( x i , gx i , ξ + ) for i ≥ . Observe that: W ( gx i | x i +1 ) = W ( gx i | ξ + , x i )= W (cid:0) m ( gx i − , g x i − , ξ + ) | ξ + , m ( x i − , gx i − , ξ + ) (cid:1) = W (cid:0) gx i − , g x i − | ξ + , x i − (cid:1) = W (cid:0) m ( gx i − , g x i − , ξ + ) , m ( g x i − , g x i − , ξ + ) | ξ + , m ( x i − , gx i − , ξ + ) (cid:1) = W (cid:0) gx i − , g x i − , g x i − | ξ + , x i − (cid:1) = · · · = W ( gx , . . . , g i +1 x | ξ + , x ) . For i ≥ r − , the Claim implies that W ( gx i | x i +1 ) = ∅ , hence gx i = x i +1 ∈ I ( x i , ξ + ) .It follows that the sets W ( g n x r − | g n +1 x r − ) = W ( x n + r − | x n + r ) are pairwise disjoint for n ≥ . Applying g − k to these sets, with k ≥ , we deduce that the sets W ( g n x r − | g n +1 x r − ) areactually pairwise disjoint for all n ∈ Z . This proves part (4).Finally, observe that, under the hypothesis of part (5), we actually have W ( gx | x ) = ∅ forevery x ∈ M . In order to see this, suppose for the sake of contradiction that there exists a halfspace h ∈ H ( x | gx ) = H ( x , ξ + | gx ) . As in the proof of the Claim, this implies that either g h ( h or g h and h are transverse. The former is ruled out by the fact that ξ + ∈ h ∗ , whereas the latterwould contradict the hypothesis of part (5).As in part (4), the fact that W ( gx | x ) is empty implies that the sets W ( g n x | g n +1 x ) arepairwise disjoint for n ∈ Z . This concludes the proof of part (5) and the entire lemma. (cid:3) Recall from Definition 1.1 that g ∈ Aut M is semisimple if there exists x ∈ M such that the sets W ( g n x | g n +1 x ) are pairwise disjoint for n ∈ Z . We say that g acts stably without wall inversions ifthe action h g i y M has no wall inversions. Corollary 3.30. (1) If g ∈ Aut M acts stably without wall inversions, then g is semisimple.(2) For every g ∈ Aut M , there exists ≤ i ≤ r such that g i is semisimple. roof. This follows from part (4) of Lemma 3.29, along with Proposition 3.24 for part (1) andRemark 3.25 for part (2). (cid:3)
Corollary 3.30 implies Corollary D from the introduction. This essentially also yields Corollary A,although we will only provide a complete proof towards the end of Section 4.3.5.
Non-transverse automorphisms.
In this subsection, we consider automorphisms of M sat-isfying the following property. Definition 3.31.
An element g ∈ Aut M acts non-transversely if there does not exist w ∈ W ( M ) such that w and g w are transverse. An action G y M is non-transverse if every g ∈ G actsnon-transversely .The following is the main motivating example for Definition 3.31. Example 3.32.
Consider actions by median automorphisms G y T i , where each T i is a medianalgebra of rank . For instance, these could correspond to isometric G –actions on R –trees. Considerthe diagonal G –action G y T × . . . × T k . If there exists a G –equivariant, injective median morphism M ֒ → T × . . . × T k , then the action G y M is non-transverse by part (1) of Remark 2.3.Let us make two simple observations that will be useful later in this subsection. Remark 3.33. If g acts non-transversely, then, for every h ∈ H / ( g ) , the halfspaces h and g h are facing, and so are h and g − h . If g acts non-transversely and stably without inversions, then h ∈ H ( g ) if and only if g h = h . Remark 3.34.
Let g ∈ Aut M act non-transversely and stably without inversions. Consider ahalfspace h ∈ H ( g ) ∩ H C ( g ) ( M ) . Then either g h ( h or g h ) h .Indeed, since h ∈ H ( g ) , there exists m ∈ Z such that g m h ( h . Helly’s lemma guarantees theexistence of a point x ∈ h ∩ g m h ∗ ∩C ( g ) . Let C x ⊆ C ( g ) be the convex hull of h g i· x . By Remark 3.33, g preserves every halfspace in H ( g ) , which implies that H C x ( M ) ∩ H ( g ) = H h g i· x ( M ) ∩ H ( g ) = ∅ .Since C x ⊆ C ( g ) , it follows that H C x ( M ) ⊆ H ( g ) , hence H ( C x ) = H ( g, C x ) .Now, Lemma 3.29 shows that, for every k ∈ H ( C x ) , the intersections g k ∩ k and g k ∗ ∩ k ∗ are bothnonempty. Since x ∈ h ∩ g m h ∗ , we have h ∩ C x ∈ H ( C x ) , hence g h ∩ h and g h ∗ ∩ h ∗ are nonempty.It follows that either g h ( h or g h ) h or g h and h are transverse. The latter is ruled out by thefact that g acts non-transversely.Recall from Definition 1.1 that the minimal set Min( g ) is the set of points x ∈ M such thatthe sets W ( g n x | g n +1 x ) are pairwise disjoint for n ∈ Z . When it is necessary to specify the medianalgebra under consideration, we will also write Min( g, M ) .For non-transverse automorphisms, Min( g ) turns out to be a convex subset of M : Proposition 3.35.
Let g ∈ Aut M act non-transversely and stably without inversions. Then Min( g ) = C ( g ) . Moreover, for every x ∈ M , we have I ( x, gx ) ∩ C ( g ) = ∅ .Proof. If x
6∈ C ( g ) , then there exists h ∈ H / ( g ) with x ∈ h ∗ . By Remark 3.33, we have gx ∈ g h ∗ ⊆ h and g − x ∈ g − h ∗ ⊆ h . In particular, W ( g − x | x ) ∩ W ( x | gx ) = ∅ , hence x Min( g ) . This showsthe inclusion Min( g ) ⊆ C ( g ) .Let us prove that I ( x, gx ) ∩ C ( g ) = ∅ for all x ∈ M . Given x ∈ M , consider the set of halfspaces K = H / ( g ) ∪ ( σ x ∩ σ gx ) . The argument in the above paragraph shows H ( x, gx |C ( g )) ∩H / ( g ) = ∅ .This implies that the halfspaces in K intersect pairwise. Any chain in K contains a cofinal subsetcontained in either H / ( g ) or σ x ∩ σ gx . By Lemma 3.21, any chain in H / ( g ) admits a lower bound A priori, requiring g to act non-transversely is weaker than asking that the action h g i y M be non-transverse. n H / ( g ) . It is clear that the set σ x ∩ σ gx has the same property. Thus, K satisfies the hypothesesof Lemma 2.9, which shows that I ( x, gx ) ∩ C ( g ) = ∅ .We are only left to prove that C ( g ) ⊆ Min( g ) . Recall that, by Lemma 3.22, we have a partition: H ( C ( g )) = H ( g, C ( g )) ⊔ H ( g, C ( g )) . By Remarks 2.3 and 2.4, g also acts on C ( g ) non-transversely and stably without inversions. ByRemark 3.33, every element of H ( g, C ( g )) is preserved by g . If C is the convex hull of any h g i –orbitin C ( g ) , it follows that H C ( M ) ⊆ H ( g, M ) , hence H ( C ) = H ( g, C ) .Thus, part (5) of Lemma 3.29 shows that Min( g, C ) = C . We deduce that Min( g, C ( g )) = C ( g ) and C ( g ) ⊆ Min( g, M ) , completing the proof. (cid:3) Proposition 3.36. If g ∈ Aut M acts non-transversely and stably without inversions, then C ( g ) isgate-convex.Proof. Consider a chain C ⊆ H C ( g ) ( M ) such that T C = ∅ . Then k := T C ∈ H ( M ) . We onlyneed to show that k must intersect C ( g ) , since then we can invoke Lemma 2.11.By part (1) of Lemma 3.22, we can replace C with a cofinal subset and assume that either C ⊆H ( g, M ) or C ⊆ H ( g, M ) . In the latter case, g preserves every halfspace in C , by Remark 3.33.Thus, g k = k , which implies that k ∈ H ( g ) . By part (2) of Lemma 3.22, k intersects C ( g ) .Let us suppose instead that C ⊆ H ( g, M ) . By Remark 3.34, we can pass to a cofinal subsetof C and assume that g h ( h for all h ∈ C (possibly also replacing g with g − ). Thus, g k ⊆ k . If g k = k , we conclude that k intersects C ( g ) as above. Otherwise g k ( k and Lemma 3.12 implies that k ∈ H C ( g ) ( M ) . (cid:3) The next three examples show, respectively, that Proposition 3.36 does not extend to actions ofgeneral finitely generated groups, does not extend to the core C ( g ) , and can fail if g does not actnon-transversely. Example 3.37.
There are isometric actions of finitely generated free groups F n on complete R –trees T for which the reduced core C ( F n , T ) is not closed (hence not gate-convex).This is because, for every isometric G –action on an R –tree, the reduced core coincides with theminimal G –invariant sub-tree. As in [GL95, Example II.6], the latter needs not be complete, henceit can be non-closed. We recall the argument for the reader’s convenience.Let G y T be an action on a complete R –tree with dense orbits (and no global fixed point).For instance, several examples with G = F were constructed in [Lev93, Theorem 5]. By [CM87,Proposition 3.1], the minimal invariant sub-tree T ⊆ T is a union of axes of elements of G . Since allorbits are dense, branch points accumulate on every point of T . Thus, every geodesic is nowhere-dense in T . It follows that T is a countable union of nowhere-dense subsets. This would violateBaire’s theorem if T were complete.Note that isometric G –actions on R –trees never have wall inversions, see e.g. Remark 4.3 below. Example 3.38.
Let T be the rooted simplicial tree with root v of degree and every other vertexof degree . Orient every edge of T away from v . Every vertex w ∈ T is contained in two orientededges moving away from w ; we label them by and , respectively. This gives a coding of thevertices of T in terms of finite sequences of s and s, with v corresponding to the empty sequence.Let S n ⊆ T be the set of n vertices represented by sequences of length n ≥ . If an edge e ⊆ T intersects S n − and S n , we assign e a length of − n . The metric completion b T of the resultingmetric on T is a complete R –tree. Note that b T coincides with the closed ball of radius around v ,and T ⊆ b T is the corresponding open ball.Points of b T \ T are represented by infinite sequences of s and s. Thus, it is natural to identifythe set b T \ T with the dyadic integers Z , i.e. with the set of expansions of the form P i ≥ a i i with i ∈ { , } . We then have an identification between S n and Z / n Z , i.e. with the set of expansionsof the form P ≤ i Consider the median space X = [0 , \ { (0 , } (endowed with the restriction ofthe ℓ metric on R ). Let g ∈ Isom X be the reflection in the diagonal through (0 , and (1 , .Then C ( g ) = [0 , \ ( { } × [0 , ∪ [0 , × { } ) , which is not gate-convex in X .We conclude this subsection with the following result. Although this will not be used any furtherin this paper, we think it is likely to prove useful in the future (especially since its proof seems tobe surprisingly nontrivial). Proposition 3.40. Let N ⊆ M be a G –invariant median subalgebra.(1) We have C ( G, N ) ⊆ N ∩ C ( G, M ) .(2) If G y N has no wall inversions, we have C ( G, N ) ⊆ N ∩ C ( G, M ) .(3) If G y M is non-transverse, then C ( G, N ) = N ∩ C ( G, M ) and C ( G, N ) ⊇ N ∩ C ( G, M ) .Proof. We can assume that N is nonempty, since the proposition is trivial otherwise.We begin by proving parts (1) and (2). They are immediate from Claim 1: Claim 1: If k ∈ H / ( G, M ) (resp. if G y N has no wall inversions and k ∈ H / ( G, M ) ), theneither N ⊆ k or k ∩ N ∈ H / ( G, N ) (resp. k ∩ N ∈ H / ( G, N ) ).Proof of Claim 1. Recall that H / ( G, M ) ⊆ H / ( G, M ) . If k ∈ H / ( G, M ) , then there exists g ∈ G such that g k ∗ ∩ k ∗ = ∅ , so k ∗ cannot contain the nonempty G –invariant set N . Thus, either N ⊆ k , or h := k ∩ N is a halfspace of N . By Remark 3.14, h does not lie in H ( G, N ) . If G y N has no wall inversions, the fact that g h ∗ ∩ h ∗ = ∅ implies that h 6∈ H ( G, N ) . Hence h ∈ H / ( G, N ) .If k ∈ H / ( G, M ) , the halfspace h cannot lie in H ( G, N ) . Otherwise, a finite-index subgroup H ≤ G would preserve h . By Remark 3.7, we would have H / ( G, M ) = H / ( H, M ) , hence therewould exist h ∈ H with h k ∗ ∩ k ∗ = ∅ , contradicting that h h = h . Thus, h ∈ H / ( G, N ) . (cid:3) In order to conclude the proof of the proposition, we need to obtain the reverse inclusions C ( G, N ) ⊇ N ∩ C ( G, M ) and C ( G, N ) ⊇ N ∩ C ( G, M ) when G y M is non-transverse. Thesewill follow once we show that, for every halfspace h ∈ H / ( G, N ) (resp. h ∈ H / ( G, N ) ), thereexists k ∈ H / ( G, M ) (resp. k ∈ H / ( G, M ) ) with h = k ∩ N . First, we need two more claims. Claim 2: Consider k ∈ H N ( M ) with k ∩ N ∈ H / ( G, N ) . If k , k , k ∈ G · k ∪ G · k ∗ are pairwisedisjoint, then k , k , k ∈ G · k ∗ .Proof of Claim 2. If the claim fails, there exist k ∈ G · k ∗ and disjoint halfspaces k , k ∈ G · k ∪ G · k ∗ contained in k . Since k ∩ N ∈ H / ( G, N ) , no two elements of G · k are disjoint (by Remark 3.5).Thus, k and k must both lie in G · k ∗ . Since k ∈ H N ( M ) , there exist points x, y, z ∈ N lying,respectively, in k ∗ , k , k . The median m ( x, y, z ) ∈ N lies in k ∩ k ∗ . Hence k ∩ N ( k ∩ N , whichcontradicts the fact that k ∩ N 6∈ H ( G, N ) . (cid:3) Claim 3: Consider k ∈ H N ( M ) with k ∩ N ∈ H / ( G, N ) . Then the orbit G · k can be partitionedinto chains C i such that any two halfspaces in different chains are facing.Proof of Claim 3. By Zorn’s lemma, there exists a maximal subset F ⊆ G · k of pairwise-facinghalfspaces. Since k ∩ N ∈ H / ( G, N ) , there exists g ∈ G such that g k ∩ N and k ∩ N are facing.Since G y M is non-transverse, no two halfspaces in G · k are transverse, and we conclude that g k and k are facing. This shows that F ≥ . ince k ∩ N ∈ H / ( G, N ) , Remark 3.5 shows that ( k ∩ N ) ∗ does not lie in the G –orbit of k ∩ N .In particular, k ∗ G · k . By maximality of F , for every j ∈ G · k \ F , there exists f ∈ F such that j and f are not facing. Since f = j ∗ , this is equivalent to saying that j does not contain f ∗ . Let usshow that the halfspace f is uniquely determined by j .Suppose for the sake of contradiction that there exist f , f ∈ F and a halfspace j ∈ G · k \ F thatcontains neither f ∗ nor f ∗ . Since f and f are facing, f ∗ and f ∗ are disjoint. By Claim 2, j cannotbe disjoint from both f ∗ and f ∗ . Since G y M is non-transverse, j is transverse to neither f ∗ nor f ∗ ,hence j must be contained in either f ∗ or f ∗ . However, since ( k ∩ N ) ∗ does not lie in the G –orbitof k ∩ N , we have f ∩ f ∩ N = ∅ . So this implies that either j ∩ N ( f ∩ N or j ∩ N ( f ∩ N ,contradicting the fact that k ∩ N 6∈ H ( G, N ) .Now, for each f ∈ F , let C ( f ) ⊆ G · k be the subset of halfspaces that do not face f (including f itself). The above discussion shows that the sets C ( f ) are pairwise disjoint and that their union isthe entire G · k . Let us show that C ( f ) is a chain.Every halfspace in C ( f ) faces every halfspace in F \ { f } . Since F ≥ , this implies that thehalfspaces in C ( f ) pairwise intersect. No two elements of C ( f ) are transverse. Thus, if C ( f ) werenot a chain, it would contain halfspaces k , k with k ∗ ∩ k ∗ = ∅ . Since f does not face any element of C ( f ) , the halfspaces k and k would both contain f , contradicting Claim 2.Finally, let us show that, if f , f ∈ F are distinct, then each k ∈ C ( f ) faces each k ∈ C ( f ) .Since f ∗ is contained in k , but not in k , we cannot have k ⊆ k . Similarly, we do not have k ⊆ k .Thus, if k and k were not facing, they would have to be disjoint. Then k ⊆ k ∗ ⊆ f and, since ( k ∩ N ) ∗ does not lie in the G –orbit of k ∩ N , we have k ∩ N ( k ∗ ∩ N ( f ∩ N . Again, thiscontradicts the fact that k ∩ N 6∈ H ( G, N ) . (cid:3) Now, consider a halfspace h ∈ H / ( G, N ) . By Remark 2.3, there exists k ∈ H N ( M ) with h = k ∩ N . Note that the chains C i provided by Claim 3 are the only maximal totally-orderedsubsets of G · k . Thus, the partition provided by the claim is unique and G permutes the chains C i .Let g i be the union of all halfspaces in C i . Since h = k ∩ N 6∈ H ( G, N ) , all halfspaces in C i havethe same intersection with N . Thus, g i ∩ N = k i ∩ N for some k i ∈ G · k . In particular, g i is notthe entire M , hence g i ∈ H ( M ) . Since G permutes the halfspaces g i , which are pairwise facing, wehave g i ∈ H / ( G, M ) . We conclude that there exists g ∈ H / ( G, M ) with g ∩ N = k ∩ N = h .Finally, if h ∈ H / ( G, N ) , we must have g ∈ H / ( G, M ) . Otherwise, g would lie in g ∈H ( G, M ) . This would clearly imply that h = g ∩ N lies in H ( G, N ) , a contradiction. (cid:3) Example 3.41. The following show that the various hypotheses in Proposition 3.40 are necessary.(1) If G y N has wall inversions, we can have C ( G, N ) N ∩ C ( G, M ) . Simply consider thesituation in Example 3.18 with M = [ − , and N = [ − , \ { } . Then C ( G, M ) = { } and C ( G, N ) = ∅ .(2) If G y M is not non-transverse, we can have C ( G, N ) N ∩ C ( G, M ) . Take M = [0 , , N = { (0 , , (0 , , (1 , } , and let G be the group generated by the reflection in the diagonalthrough (0 , and (1 , . Then C ( G, M ) = M and C ( G, N ) = { (0 , } .(3) If G y M is not non-transverse, we can also have C ( G, N ) N ∩ C ( G, M ) . Let ϕ be ahomeomorphism of [ − , that fixes − , , , and no other point. Consider M = [ − , and N = [ − , \ [ − , . Let G ≤ Aut M be the group generated by the transformation ( x, y ) ( ϕ ( y ) , ϕ ( x )) . Then C ( G, M ) = M and C ( G, N ) = [0 , . Remark 3.42. In part (3) of Proposition 3.40, the equality N ∩ C ( G, M ) = C ( G, N ) also holdsif, instead of requiring the action G y M to be non-transverse, we ask that M be the underlyingmedian algebra of a median space X equipped with an isometric G –action.Indeed, it then follows from part (1) of Lemma 4.5 below that, if k ∈ H ( G, M ) , then k ∩ N liesin H ( G, N ) . It is thus immediate that every element of H / ( G, N ) is the intersection with N ofan element of H / ( G, M ) (which is what occupied most of the proof of Proposition 3.40). he following is a consequence of Proposition 3.17 and the first two parts of Proposition 3.40. Corollary 3.43. If G is finitely generated, then C ( G ) intersects every nonempty, G –invariant sub-algebra of M . Moreover, C ( G ) intersects every nonempty, G –invariant subalgebra N ⊆ M such that G y N has no wall inversions. Compatible metrics. Let M be a finite-rank median algebra. Let G y M be an action by median automorphisms.In this section, we consider G –invariant pseudo-metrics on M that are compatible with the medianoperator. The main results are Corollary 4.6 and Proposition 4.9. We also prove Corollary A. Definition 4.1. A pseudo-metric η : M × M → [0 , + ∞ ) is compatible if, for all x, y, z ∈ M , wehave η ( x, y ) = η ( x, m ( x, y, z )) + η ( m ( x, y, z ) , y ) .Let PD ( M ) and D ( M ) be the sets, respectively, of all compatible pseudo-metrics on M and of allcompatible metrics. To avoid confusion, we will denote metrics by the letter δ and pseudo-metricsby η . We write PD ( M ) G and D ( M ) G for the sets of G –invariant pseudo-metrics and metrics.If δ ∈ D ( M ) , then the pair ( M, δ ) is a median space (not necessarily complete or connected). Infact, one can view every median space as the data of an underlying median algebra and a compatiblemetric on it. We have D ( M ) G = ∅ exactly when the action G y M arises from an isometric G –action on a median space. Remark 4.2. Consider δ ∈ D ( M ) and let ( X, δ ) be the metric completion of ( M, δ ) . Then X is a complete median space with rk X = rk M . This follows from [CDH10, Proposition 2.21] and[Fio20b, Lemma 2.5]. Note that the subset M ⊆ X is a dense median subalgebra. Remark 4.3. If there exists δ ∈ D ( M ) G such that ( M, δ ) is connected, then G y M has no wallinversions. Indeed, by Remark 4.2, the metric completion X of ( M, δ ) is a complete, finite-rankmedian space. By [Fio20b, Proposition B], every halfspace of X is either open or closed. By part (1)of Remark 2.3, every halfspace of M is then either open or closed in M . Since M is connected, aproper subset of M and its complement can never be both closed or both open. Hence G y M hasno wall inversions. Remark 4.4. Consider η ∈ PD ( M ) and let q : ( M, η ) → ( X, η ) be the quotient metric space. Then ( X, η ) is a median space and q : M → X is a surjective median morphism. By Remark 2.2, we have rk X ≤ rk M . Lemma 4.5. If D ( M ) G = ∅ , then:(1) for every g ∈ G and h ∈ H ( g ) , we have T n ∈ Z g n h = ∅ ;(2) the sets H ( G ) and H ( G ) are transverse.Proof. Part (2) immediately follows from part (1) and Remark 3.13, so we only prove part (1).Consider g ∈ G and h ∈ H ( g, M ) . Pick δ ∈ D ( M ) G and let X be the metric completion of ( M, δ ) . By Remark 4.2, this is a complete, finite-rank median space with an isometric G –action.By Remark 3.14, there exists k ∈ H ( g, X ) such that h = k ∩ M . Let n ∈ Z be such that g n k ( k .[Fio20b, Proposition B] ensures that δ ( g nk k , k ∗ ) > for some k ≥ . Hence δ ( g nk h , h ∗ ) ≥ δ ( g nk k , k ∗ ) diverges for k → + ∞ , proving part (1). (cid:3) Lemmas 4.5 and 2.12 immediately imply: Corollary 4.6. If D ( M ) G = ∅ , then the partitions in Lemma 3.22 give rise to G –invariant splittings C ( G ) = C ( G ) × C ( G ) and C ( G ) = C ( G ) × C ( G ) . emma 4.7. Consider g ∈ Aut M with D ( M ) h g i = ∅ . Let W ( g ) be the set of walls determined by H ( g ) . Then, for all x ∈ Min( g ) and y ∈ M , we have: G n ∈ Z W ( g n x | g n +1 x ) = W ( g ) ⊆ [ n ∈ Z W ( g n y | g n +1 y ) . Proof. Part (1) of Lemma 4.5 shows that every h ∈ H ( g ) satisfies T n ∈ Z g n h = ∅ . Thus, for every y ∈ M , the orbit h g i · y intersects both h and h ∗ . If w is the wall associated to h , it follows thatthere exists n ∈ Z such that w ∈ W ( g n y | g n +1 y ) . This shows that W ( g ) ⊆ S n ∈ Z W ( g n y | g n +1 y ) forevery y ∈ M .We are left to show that, for every x ∈ Min( g ) , we have W ( x | gx ) ⊆ W ( g ) . Observe that,if k ∈ H ( x | gx ) , then we have g n x ∈ k for all n ≥ , and g n x ∈ k ∗ for all n ≤ . Hence,for every k ∈ Z , the two intersections g k k ∩ k and g k k ∗ ∩ k ∗ are both nonempty. In particular, k 6∈ H / ( g ) ⊔ H / ( g ) ∗ . We also have k 6∈ H ( g ) , since a proper power of g cannot stabilise k . Weconclude that H ( x | gx ) ⊆ H ( g ) , which completes the proof. (cid:3) Example 4.8. Lemma 4.7 can fail if D ( M ) h g i = ∅ . It suffices to consider M = R and let g be anorientation-preserving homeomorphism with a single fixed point.Given g ∈ Aut M and η ∈ PD h g i ( M ) , we define the translation length : ℓ ( g, η ) := inf x ∈ M η ( x, gx ) ∈ [0 , + ∞ ) . The following is the main result of this section. Proposition 4.9. Let g ∈ Aut M act stably without inversions.(1) If η ∈ PD ( M ) h g i , then η ( x, gx ) = ℓ ( g, η ) for all x ∈ Min( g ) . Moreover, ℓ ( g n , η ) = | n |· ℓ ( g, η ) .(2) If δ ∈ D ( M ) h g i and y ∈ M , then y ∈ Min( g ) if and only if δ ( y, gy ) = ℓ ( g, δ ) .(3) If g acts non-transversely, then, for all η ∈ PD ( M ) h g i and all y ∈ M , we have: η ( y, gy ) = ℓ ( g, η ) + 2 η ( y, C ( g )) . Proof. We begin with part (1). We can assume that η is a genuine metric. Indeed, if q : M → X isthe quotient median space, it follows from Remarks 4.4 and 2.2 that q (Min( g, M )) ⊆ Min( g, X ) .Thus, it suffices to consider δ ∈ D ( M ) h g i . By [CDH10, Theorem 5.1], there exists a h g i –invariantmeasure µ on W ( M ) such that δ ( u, w ) = µ ( W ( u | w )) for all u, w ∈ M . Recall that, by Corollary 3.30,the set Min( g ) is nonempty. Consider points x ∈ Min( g ) and y ∈ M .Let Ω = S n ∈ Z W ( g n y | g n +1 y ) ⊆ W ( M ) . By Lemma 4.7, Ω contains W ( y | gy ) and W ( x | gx ) .Moreover, every h g i –orbit in Ω intersects W ( y | gy ) in at least one element, and W ( x | gx ) in at mostone element. Since µ is h g i –invariant, it follows that: δ ( y, gy ) = µ ( W ( y | gy )) ≥ µ ( W ( x | gx )) = δ ( x, gx ) . This inequality holds for every x ∈ Min( g ) and y ∈ M , so δ ( x, gx ) = ℓ ( g, δ ) for all x ∈ Min( g ) .For every n ∈ Z \ { } , we have Min( g ) ⊆ Min( g n ) . Thus, if η ∈ PD ( M ) h g i and x ∈ Min( g ) , then: ℓ ( g n , η ) = η ( x, g n x ) = | n | · η ( x, gx ) = | n | · ℓ ( g, η ) . This proves part (1).We now address part (2). Consider δ ∈ D ( M ) h g i and y ∈ M with δ ( y, gy ) = ℓ ( g, δ ) . The triangleinequality gives δ ( y, g n y ) ≤ n · ℓ ( g, δ ) for all n ≥ . Part (1) implies that δ ( y, g n y ) = n · ℓ ( g, δ ) .Thus, for all k ≤ m ≤ n in Z , we have δ ( g k y, g n y ) = δ ( g k y, g m y ) + δ ( g m y, g n y ) . This impliesthat m ( g k y, g m y, g n y ) = g m y . Hence the sets W ( g i y | g i +1 y ) are pairwise disjoint, and y must lie in Min( g ) . This proves part (2). inally, let us prove part (3). By Proposition 3.36, the set C ( g ) is gate-convex. Let π : M → C ( g ) be the gate-projection. Since C ( g ) is h g i –invariant, we have π ( gy ) = gπ ( y ) for every y ∈ M . Hence: W ( y | gy ) ⊆ W ( y | π ( y )) ∪ W ( π ( y ) | gπ ( y )) ∪ W ( π ( gy ) | gy ) . Since π is the gate-projection, we have: W ( y | gπ ( y )) = W ( y | π ( y )) ⊔ W ( π ( y ) | gπ ( y )) , W ( π ( y ) | gy ) = W ( π ( y ) | gπ ( y )) ⊔ W ( π ( gy ) | gy ) . By Proposition 3.35, the set W ( y | π ( y )) ∩ W ( gy | π ( gy )) = W ( y, gy |C ( g )) is empty, hence: W ( y | gy ) = W ( y | π ( y )) ⊔ W ( π ( y ) | gπ ( y )) ⊔ W ( π ( gy ) | gy ) . Part (1) and Proposition 3.35 yield η ( x, gx ) = ℓ ( g, η ) for all η ∈ PD ( M ) h g i and x ∈ Min( g ) . Thus: η ( y, gy ) = η ( y, π ( y )) + η ( π ( y ) , gπ ( y )) + η ( π ( gy ) , gy ) = ℓ ( g, η ) + 2 η ( y, C ( g )) . (cid:3) We can now deduce Corollary A from Corollary 3.30. Proof of Corollary A. Since X is connected, Remark 4.3 shows that g ∈ Isom X acts stably withoutinversions. By Corollary 3.30, there exists a point x ∈ Min( g ) . By part (1) of Proposition 4.9, thepoint x realises the translation length of g .If X is geodesic and g does not fix x , let α be a geodesic segment joining x and gx . Since x ∈ Min( g ) , the segments g n α glue to form the required axis of g . (cid:3) The following motivates our notation from Subsection 3.1 by comparing it to [CS11]. Remark 4.10. Suppose that there exists δ ∈ D ( M ) G . Fix a basepoint x ∈ M . By analogy with[CS11, Subsection 3.3], one could call a halfspace h ∈ H ( M ) : • fully inessential if the points of G · x have uniformly bounded distance from h and h ∗ ; • fully essential if G · x contains points arbitrarily far from h ∗ and arbitrarily far from h ; • half-essential if G · x contains points arbitrarily far from h ∗ , but G · x has uniformlybounded distance from h .These three classes of halfspaces are closely related to our sets H ( G ) , H ( G ) and H / ( G ) , re-spectively, and they motivate our notation. The sets H • ( G ) have the advantage of being evidentlyindependent of the choice of a metric δ .It is clear that the elements of H ( G ) are fully inessential. As in the proof of part (1) of Lemma 4.5,one sees that the elements of H ( G ) are fully essential. If G is finitely generated, it follows frompart (1) of Theorem 3.17 that no fully essential halfspace can lie in H / ( G ) .In conclusion, if G is finitely generated, then H ( G ) always coincides with the set of fully essentialhalfspaces and H ( G ) is always contained in the set of fully inessential halfspaces. However, H / ( G ) can contain both half-essential and fully inessential halfspaces (the latter e.g. when M is a locallyinfinite tree T and G fixes an infinite-valence vertex, while acting freely on its neighbours).5. Actions of polycyclic groups. This section is devoted to the proof of Corollary E and its immediate consequence Corollary B. .1. Lineal median algebras. Let M be a finite-rank median algebra. Recall from Subsection 3.4,that we denote by M the zero-completion of M . Definition 5.1. Given ξ ± ∈ M , we say that M is lineal with endpoints ξ ± if M ⊆ I ( ξ − , ξ + ) . Lemma 5.2. M is lineal if and only if H ( M ) can be partitioned into two ultrafilters.Proof. If M is lineal with endpoints ξ ± , then the intersections between H ( M ) ≃ H M ( M ) and thesubsets H ( ξ − | ξ + ) , H ( ξ + | ξ − ) ⊆ H ( M ) provide the required partition into two ultrafilters.Conversely, suppose that H ( M ) = σ ⊔ σ ∗ is a partition into two ultrafilters. For every interval I = I ( x, y ) ⊆ M , we have a partition H I ( M ) = ( σ ∩ H I ( M )) ⊔ ( σ ∗ ∩ H I ( M )) . By Remark 2.3, thiscorresponds to a partition H ( I ) = σ I ⊔ σ ∗ I , where σ I and σ ∗ I are ultrafilters.The sets σ I and σ ∗ I satisfy the hypotheses of Lemma 2.9. In order to see this, consider a chain C ⊆ σ I . Then either C ∩ H ( x | y ) or C ∩ H ( y | x ) is cofinal in C . Hence T C is nonempty and it isa halfspace of I . Note that, given any h ∈ C , the halfspace h ∗ lies in σ ∗ I and is disjoint from T C .Hence T C σ ∗ I , and T C ∈ σ I as required.This shows that, for every interval I ⊆ M , there exist two points z ± I ∈ I such that I = I ( z − I , z + I ) and z + I (resp. z − I ) lies in the intersection of all halfspaces in σ ∩ H I ( M ) (resp. in σ ∗ ∩ H I ( M ) ). Itis straightforward to check that the functions I z + I and I z − I define the required points ξ ± inthe zero-completion of M . (cid:3) Remark 5.3. If H ( M ) = σ ⊔ σ ∗ = τ ⊔ τ ∗ are distinct partitions into ultrafilters, we can write: H ( M ) = [( σ ∩ τ ) ⊔ ( σ ∩ τ ) ∗ ] ⊔ [( σ ∗ ∩ τ ) ⊔ ( σ ∗ ∩ τ ) ∗ ] . Note that every halfspace in σ ∩ τ is transverse to every halfspace in σ ∗ ∩ τ . Thus, Lemma 2.12 givesa nontrivial product splitting of M . It follows that there are only finitely many distinct partitionsof H ( M ) into two ultrafilters. Remark 5.4. Suppose M is lineal and G y M is an action by median automorphisms. Then thereexist a finite-index subgroup H ≤ G and H –fixed points ξ ± ∈ M such that M ⊆ I ( ξ − , ξ + ) . Thisfollows from Remark 5.3 and Lemma 5.2.5.2. Lineal median spaces. We now restrict to the setting of (not necessarily complete or con-nected) finite-rank median spaces.The proof of the next lemma requires a certain familiarity with our previous work in [Fio20b].Still, this is a very predictable result and we hope that the reader will not find this troubling. Lemma 5.5. (1) If X is a lineal, finite-rank median space, then its metric completion b X is also lineal.(2) Every lineal median space of rank r can be isometrically embedded in ( R r , d ℓ ) .Proof. Part (2) follows from part (1) and [Fio20b, Proposition 2.19]. Thus, we only prove part (1).By [CDH10, Theorem 5.1] and [Fio20b, Section 3], the set H ( b X ) can be equipped with a measure µ such that d ( x, y ) = µ ( H ( x | y )) for all x, y ∈ X . By Lemma 5.2 and Remark 2.3, the subset H X ( b X ) is partitioned into two ultrafilters. In the terminology of [Fio20b, p. 1349], the elements of H ( b X ) \ H X ( b X ) are non-thick , hence H ( b X ) \ H X ( b X ) has measure zero by [Fio20b, Corollary 3.7].The fact that b X is lineal can now be deduced by retracing the proof of Lemma 5.2 above andobserving that, invoking [Fio20b, Corollary 3.11(1)], it yields the following stronger result: Themedian space X is lineal if and only if there is a partition H ( M ) = σ ⊔ σ ∗ ⊔ K , where σ and σ ∗ areultrafilters and µ ( K ) = 0 . (cid:3) Lemma 5.6. If X is a lineal, finite-rank median space, then Isom X is virtually (locally finite)–by–abelian. roof. If b X is the metric completion of X , we have an embedding Isom X ֒ → Isom b X and b X is againlineal by part (1) of Lemma 5.5. Thus, it is not restrictive to assume that X is complete.We begin with the following special case. Set r = rk X . Claim: if H ≤ Isom X fixes points x ∈ X and ξ ∈ X , then H is locally finite.Proof of Claim. Denote by H ( t ) ⊆ H ( x | ξ ) ∩ H ( X ) the set of halfspaces at distance exactly t > from x . By [Fio20b, Proposition B], any chain in H ( t ) has length ≤ r , while any anti-chain haslength ≤ r by definition of rank. Dilworth’s lemma then yields H ( t ) ≤ r .The action H y H ( X ) preserves the set H ( t ) for each t > . This yields a homomorphism ι : H → Q t> Sym( H ( t )) . Each element in ker ι preserves every halfspace of X at positive distancefrom x , thus fixing every point of X . We conclude that ι is injective.The claim now follows from the (standard) observation that direct products of uniformly finitegroups are locally finite. (cid:3) By Corollary 2.13, there is a maximal product splitting X = X × . . . × X k . The product ofthe groups Isom X i sits inside Isom X as a finite-index subgroup. By Lemma 5.2, each X i is againlineal. Thus, it is not restrictive to assume that X is irreducible.Now, the points ξ, η ∈ X such that X ⊆ I ( ξ, η ) are unique by Remark 5.3. An index– subgroupof Isom X fixes ξ and η . By [Fio18, Theorem F], there exist a finite-index subgroup G ≤ Isom X and a homomorphism χ : G → R r such that every finitely generated subgroup of ker χ fixes a pointof X , as well as ξ and η .If H ≤ ker χ is finitely generated and x ∈ X is a point fixed by it, we obtain a restrictionhomomorphism ρ : H → Isom I ( x, ξ ) × Isom I ( x, η ) . By the Claim, the image of ρ is finite. On theother hand, the kernel of ρ preserves every halfspace of I ( x, ξ ) ∪ I ( x, η ) , hence every halfspace of X . It follows that ρ is injective and H is finite. In conclusion, ker χ is locally finite. (cid:3) Actions on lineal median spaces.Definition 5.7. Let G be the class of groups G with the following property. If X is a finite-rankmedian space and G y X is an essential isometric action without wall inversions, then X is lineal. Remark 5.8. It follows from part (3) of Lemma 3.29 that Z ∈ G . Remark 5.9. If H ≤ G has finite index and H ∈ G , then G ∈ G . Indeed, if an action G y X isessential and without wall inversions, then so is the restriction H y X . Proposition 5.10. Suppose that there exists a subgroup H ⊳ G such that H and G/H lie in G .Suppose moreover that H is finitely generated. Then G ∈ G .Proof. Let X be a finite-rank median space, and let G y X be an essential isometric action withoutwall inversions. Since H is finitely generated, the reduced core C ( H ) is nonempty by Theorem 3.17.Since H is normal, Remark 3.16 shows that C ( H ) is G –invariant. Part (1) of Corollary 3.26 thenguarantees that C ( H ) = M . By Corollary 4.6, we have a product splitting M = C ( H ) × C ( H ) . Note that Remark 3.16 also guarantees that the sets H ( H ) and H ( H ) are G –invariant. So, thefactors in the above splitting of M are preserved by G .Since H ∈ G , the space C ( H ) is lineal (possibly a single point). By Lemma 5.2, it suffices toshow that C ( H ) is lineal. By Remark 3.9, a finite-index characteristic subgroup H ⊳ H fixes C ( H ) pointwise. Thus, the essential action G y C ( H ) factors through an essential action of G/H . Since G/H ∈ G has finite index in G/H , Remark 5.9 shows that G/H ∈ G . Hence C ( H ) is lineal. (cid:3) Recall that a group is polycyclic if it is solvable and all its subgroups are finitely generated.Proposition 5.10 and Remarks 5.8 and 5.9 imply the following. orollary 5.11. All virtually polycyclic groups lie in G . Remark 5.12. Let X be a lineal, finite-rank median space. If G is virtually polycyclic, then everyisometric action G y X factors through a virtually abelian group. This follows from Lemma 5.6and the fact that polycyclic groups are virtually torsion-free [Hir46, Theorem 3.21].We are now ready to prove Corollary E from the introduction. Proof of Corollary E. By Theorem 3.17, the core C ( H ) is nonempty. By Corollary 4.6, we have asplitting C ( H ) = C ( H ) ×C ( H ) . By Remark 3.16, the G –action leaves C ( H ) invariant and preservesthe product splitting. Since G y X is minimal, we have C ( H ) = X and we can set P = C ( H ) and F = C ( H ) . By part (2) of Proposition 3.23, the action H y P fixes a point. The action H y F is essential, so F is lineal by Corollary 5.11. By Remark 5.12, the action H y F factors through avirtually abelian group. By part (2) of Lemma 5.5, the space F isometrically embeds in R r . (cid:3) Proof of Corollary B. By Remark 4.3, the action A y X has no wall inversions. 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