Choosing N orbits of h which do not cross

Let a group Gact properly and cocompactly on a CAT(0) cube complex X. Then there exists a compact subset C of X such that GC =X. Since C is compact, there is a finite set of hyperplanes inXwhich intersect the subsetC. Denote these hyperplanes by h1,h2, . . . ,hn. Then the set of all hyperplanes

inX is given by {Gh1, Gh2, . . . , Ghn}, where Ghi ={ghi|g ∈G}.

In general, there may be some g ∈ G and hi ∈ {h1. . . ,hn} such that

ghi∩hi 6=hi and ghi∩hi 6=∅. In this case we say that ghi crosses hi.

For a given hyperplane h ∈ {h1, . . . ,hn}, let Lh denote the set {g ∈ G|gh crosses h} and Hh the group stabG(h) ={g ∈G|gh=h}.

Lemma 3.3. Let G be a group which acts properly and cocompactly on a CAT(0) cube complex X. Then for any hyperplane h ∈ {h1, . . . ,hn}, Lh = stabG(h)FhstabG(h) for some finite set Fh ⊂G.

Proof. Suppose G acts properly on X and let h be any hyperplane of X.

Any compact subset K of the hyperplaneh is a compact subset of X, hence {g ∈stabG(h)|gK∩K 6=∅} ⊂ {g ∈G|gK∩J K 6=∅} is a finite set, and the

action of stabG(h) on h is proper.

Suppose G acts cocompactly on X, and let h be any hyperplane in X. Let C be a compact subset of X such that GC = X. Any midplane in the hyperplane equivalence classhmust be the image of a midplane inC. LetM be the set of midplanes inC which are mapped by someg ∈Gto a midplane of h. For each Mi ∈M choose a gi ∈GH such that giMi ∈h. Then the set

C0 =S

Mi∈MgiMi is the union of a finite set of midplanes of h and hence is

a compact subset ofh. We claimHhC0 =h.

SupposeM0 is a midplane inh. ThenM0 =gMi for someg ∈Gand some

Mi ∈M. Then gi(g−1M) =giMi ∈ C0 and gig−1 ∈ Hh since any element of the group which maps a midplane inhto a midplane inh stabilisesh. Hence HhC0 =h and the action of Hh onh is cocompact.

Ifg ∈Lh then gh crossesh. Hence for someh1, h2 ∈Hh, gh1C0 and h2C0

intersect but are not equal. We will say that subsets which intersect but are not equal cross.

Sincegh1C0 andh2C0 cross, so doh−12 gh1C0andC0. SinceGacts properly

onXandC0is a compact subset ofX, the setFh ={f ∈G|f C0 crossesC0} ⊆ {f ∈ G|f C0 ∩ C0 6= ∅} is finite. We have shown that if g ∈ Lh, then g ∈HhFhHh, that isLh ⊆HhFhHh.

Suppose g ∈ HhFhHh. Then for some h1, h2 ∈ Hh and some f ∈ Fh, g =h1f h2. Then gh crosses h if and only if h1f h2h =h1fh crosses h. Since

h1 ∈stabG(h),h−11 ∈stabG(h) and h1fhcrossesh if and only if fh crossesh.

By the definition of the setFh, fh crosses h, hence gh crosses h and g ∈Lh. Hence HhFhHh ⊂ Lh and we have Lh = HhFhHh = stabG(h)FhstabG(h) as

required.

Lemma 3.4. Let G be a group which acts properly and cocompactly on a

CAT(0) cube complex such that stabG(h) is separable for each hyperplane h.

Then for each h ∈ {h1, . . . ,hn} there exists a finite index subgroup Kh of G

containing stabG(h) such that for all k ∈Kh, kh does not cross h.

Proof. By the hypothesis stabG(h) = Hh is separable for any h, so we have

Hh =THj where, for each j, Hj is a finite index subset of G. Hence for all

g ∈G\Hh , there exists a Hj with g /∈Hj.

As Fh = {f ∈ G|f C0 crossesC0} if f ∈ Fh then fh crosses h, and so if f ∈ Fh, f does not stabilise h. Hence Fh∩Hh = ∅ and for each f ∈ Fh we can choose a finite indexHj not containing f. We intersect these to form a

subgroup Kh which contains Hh = stabG(h) and contains no element of Fh. SinceLh =HhFhHh, it follows thatKh contains no element ofLh, and hence for all k ∈ Kh, kh does not cross h. Since Fh is finite and each Hj is finite

index inG, it follows that Kh is a finite index subgroup of G.

Lemma 3.5. Let G be a group which acts properly and cocompactly on a

CAT(0) cube complex X such that stabG(h) is separable for any hyperplane

h. Then we can find a finite index normal subgroup N of G such that mhi

Proof. Taking M = T

iKhi gives a subgroup of G such that hhi does not

cross hi for any h ∈M,hi ∈ {h1,h2, . . . ,hn}. M is an intersection of a finite

number of finite index subgroups of G, and hence is finite index in G. LetN =T

g∈GM

g_{. SinceM is finite index inG, there are a finite number}

of subgroups conjugate to M in G. Hence the intersection T

g∈GM

g _{is an}

intersection of a finite number of finite index subgroups, and so N is a finite index normal subgroup ofG. By definition, for allm ∈N, hi ∈ {h1, . . . ,hn},

mhi does not cross hi.

Remark 3.6. SinceN is finite index inG, we can choose a finite set of coset representatives forN inG,{γ1, . . . , γl}. Let{h1, . . . ,hk}denote the finite set

of hyperplanes{γihj|i∈1, . . . , l, j ∈1, . . . , n}. Then the set {Nh1, . . . , Nhk}

includes all the hyperplanes of the cube complexX.

Note that for any m ∈N and any hyperplane h inX mh does not cross h. This follows from the fact thatN is a normal subgroup: For anyh we can writeh =ghi for some g ∈Gand hi ∈ {h1, . . . ,hn} and so for any m∈N

mh∩h=mghi∩ghi =g(m0hi)∩ghi =g(m0hi∩hi)

for some m0 ∈N, which by lemma 3.5 is either the empty set or the hyper- planeghi.