2.3 Cube complexes with isometric group actions
3.1.2 Using N to construct a product of trees
Given a groupGwith a proper, cocompact action on a CAT(0) cube complex X such that stabG(h) is separable for every hyperplane h, we will construct
an embedding ofX in a product of trees, by first constructing for each hi a
tree Ti by considering Nhi. We will use the method described in the proof
of lemma 1.38. We begin by defining a halfspace system.
Definition. Given any hyperplane hi ∈ {h1, . . . ,hk} consider the set of hy-
perplanes Nhi. For each n ∈ N, nhi separates X into two connected com-
ponents which we denote byXnhi and X
∗
nhi. We denote the set of halfspaces
obtained in this way by Hi = {Xnhi, X
∗
(Hi,≤,∗) where ≤ is the order given by inclusion of halfspaces and ∗ is the
order reversing involution given by interchanging the two halfspaces defined by any hyperplane.
We define the boundary map ∂ :Hi →Hi/∼ to be the map which takes
halfspaces to their boundaries, i.e. ∂(Xnh
i) = {Xnhi, X
∗
nhi} = ∂(X
∗
nhi). For
simplicity of notation, we equate the equivalence class{Xnh
i, X
∗
nhi}with the
hyperplane nhi.
Lemma 3.7. (Hi,≤,∗) is a halfspace system.
Proof. By lemma 1.36 the set of halfspaces of a CAT(0) cube complex form
a halfspce system. Hi is a subset of the halfspacesH of X, with the partial
order ≤ and the involution ∗ on Hi agreeing with the partial order and
involution on H. Suppose X1 and X2 are any two elements of Hi. Since
Hi ⊂ H the set {X3 ∈ Hi|X1 ≤ X3 ≤ X2} is contained in the set {X3 ∈
H|X1 ≤ X3 ≤ X2} and hence is finite. Similarly, since the partial order
on Hi agrees with the partial order on H, at most one of the inequalities
X1 ≤X2, X1 ≤X2∗, X1∗ ≤X2, X1∗ ≤X2∗ holds. These two observations on
the properties of triple (H,≤,∗) show that it is a halfspace system.
Lemma 3.8. For each i ∈ {1, . . . , k} let (Hi,≤,∗) be the halfspace system
defined above. Then the components of the cube complex corresponding to
(Hi,≤,∗) (as defined in lemma 1.37) are trees. There is an injective map
ξ1 from the set of components of X\Nhi into the vertex set of one of these
trees.
Proof. For any i∈ {1, . . . , k} we construct a CAT(0) cube complexCi using
(Hi,≤,∗) as follows: Take the set of vertices to be all sections for∂ such that
v(n1h)v(n2h)∗ for any n1h, n2h∈Hi.
Leth=hi. We define a mapξifrom the set of components ofX\Nhin the
set of vertices by mapping each componentDof X\Nhto the section ξi(D)
defined by setting ξi(D)(nh) to be the halfspace Xnh or Xn∗h containing D. SinceDis non-empty, the sectionξi(D) will satisfyξi(D)(n1h)ξi(D)(n2h)∗
for all n1h, n2h ∈N and hence is a vertex of Ci
We join two vertices u and v by an edge if and only if the values of the sectionsuand v differ on exactly one hyperplane. If two componentsD and
D0 are adjacent inX\Nhthen they are separated by exactly one hyperplane h. Hence the values of the sections ξi(D) and ξi(D0) differ on exactly one
boundary {Xh, Xh∗}, and so by definition the vertices ξi(D) and ξi(D0) are
adjacent inC0.
Choose any component of X \Nh and denote it by E. Let Ti be the
component ofCicontainingξi(E), the vertex corresponding to the component
E. X is connected, and any two vertices are separated by at most finitely many hyperplanes. Hence for any componentD ofX\Nh, the vertexξi(D)
also lies inTiand so the canonical map fromX\NhtoCi gives an embedding
in a single component ofCi.
To see that Ti is a tree, suppose that Ti contains a cycle. Then there is
a finite set of hyperplanesH0 ={n1h, . . . , nkh}such that for each nih, there
is a pair of verticesv, v0 in the cycle such that the values ofv and v0 on nih
differ.
Consider the finite set of halfspaces {Xnih, X
∗
nih|nih ∈ H
0}. Since ≤ is
an order on the set of all halfspaces from this set, we can choose a minimal halfspace in this set, i.e a halfspace Xnh with nh ∈/ H0 such that for all nih ∈ H0 Xnih Xnh and X
∗
nih Xnh. Without loss of generality, let this
halfspace beXn1h. By the definition of the set H
0, there exists a vertex v in
the cycle such thatv(n1h) =Xn1h. Suppose that the vertexuis adjacent tov
and lies in the cycle. The values ofuandv differ on precisely one hyperplane, call itnh. Then u(nh) = v(nh)∗.
Suppose thatnh 6=n1h. Asnhlies in the setH0 we know thatXnh Xn1h
andXn∗h Xn1h. Suppose thatv(nh) = Xnh. By the definition of a vertex we
haveXnh =v(nh)v(n1h)∗ =Xn∗1h andX
∗
nh =v(nh)
∗ =u(nh)
u(n1h)∗ =
Xn∗1h. Similarly, taking v(nh) = Xn∗h yields the same relations. However, at least one of the relationsXnh ≤Xn1h, Xnh ≤X
∗ n1h, X ∗ nh ≤Xn1h, X ∗ nh ≤X ∗ n1h
must hold, and hence we must have nh =n1h. This means that the vertex
defined by u(n1h) = v(n1h)∗, u(nih) = v(nih)∀nih ∈ H0 \ {n1h} is the only
vertex adjacent to u which lies in the cycle. This contradicts the existence of such a cycle, hence Ti is a tree.
Nhi to hyperplanes inNhi. Hence the action ofN on the set{Xnhi, X
∗
nhi|n∈
N}maps halfspaces to halfspaces in such a way that, for anym∈N,m(Xnhi)
is either X(mn)hi or X
∗
(mn)hi. We define an action of N on Ti by taking the
value of m(v) on the hyperplane nhi to be m(v(nhi)) for eachm, n∈N and
each vertexv in Ti.
We check that the resulting section m(v) is a vertex as follows:
v(nih)v(n2h)∗ ⇒ v(n1h)∩v(n2h)6=∅
⇒ m(v(n1h)∩v(n2h))6=∅
⇒ m(v(n1h))∩m(v(n2h))6=∅
= m(v(n1h))m(v(n2))∗
Hence m(v) is a vertex of Ti. Similarly, we can show that if ξi(D) is the
image inTi of the component D of X\Nhi then m(ξi(D)) = ξi(m(D)). Let
Y denote the product of trees T1×. . .×Tn. Then there is a natural action
of N on Y resulting from the action ofN on Ti.
Lemma 3.9. Suppose N is a normal subgroup of a finitely generated group
G with index n and that N acts on the left on a product of trees Y. Then G
acts on the product of n copies of Y.
Proof. In order to show this, we use Serre’s construction as described in
“Groups acting on graphs”, [13]. We have N normal in G with finite index n. Let Y1, . . . , Yn be copies of Y. We have an action of N on each Yi. Let
x1, . . . , xn be left coset representatives for N in G. G acts on the left on the
set of cosets {x1N, . . . , . . . , xnN}. Define an action of G on the index set {1, . . . .n} bygi=j if and only if g(xiN) =xjN.
Consider any g ∈G. For eachi∈ {1, . . . , n}there is a unique expression gxi = xgihi where gi ∈ {1, . . . , n} and hi ∈ N. Let (w1, . . . , wn), wi ∈ Yi
represent a general point ofY1 ×. . .×Yn. We define
g(w1, . . . , wn) = (hg−11wg−11, . . . , hg−1nwg−1n)
remains to check that f(g(w1, . . . , wn)) = (f g)(w1, . . . , wn).
For each i ∈ {1, . . . , n} there is a unique expression gxi = xgihi and a
unique expression f xi = xf iki where gi, f i ∈ {1, . . . , n} and hi, ki ∈ N.
Then (f g)xi = f(gxi) = f(xgihi) = (f xgi)hi = xf gikgihi = xf giji since
multiplication in the group is associative, and we have the expression ji =
kgihi. In order to check that we have an action, we will need the following
expression forj(f g)−1i; j(f g)−1i =jg−1f−1i =kg(g−1f−1i)hg−1f−1i =k(gg−1)f−1ihg−1f−1i =kf−1ihg−1f−1i. Then f(g(w1, . . . , wn)) = f(hg−11wg−11, . . . , hg−1nwg−1n) = f(y1, . . . , yn) whereyi =hg−1iwg−1i = (kf−11yf−11, . . . , kf−1nyf−1n) = (kf−11hg−1f−11wg−1f−11, . . . , kf−1nhg−1f−1nwg−1f−1n) = (jg−1f−11wg−1f−11, . . . , jg−1f−1nwg−1f−1n) (f g)(w1, . . . , wn) = (j(f g)−11w(f g)−11, . . . , j(f g)−1nw(f g)−1n)
Hence G acts on the finite productY ×. . .×Y as required.
Lemma 3.10. Let G be a group which acts properly and cocompactly by
isometries on a finite dimensional, locally finite CAT(0) cube complex X
such that stabG(h) is separable for any hyperplane h. Then for some k ∈N
there is an isometric map from X to a product of trees T1×. . .×Tk
Proof. By lemma 3.5 and remark 3.6 there exists a finite index subgroupN
and a finite set of hyperplanesh1, . . . ,hk such that the action ofN on the set
of hyperplanes generates all hyperplanes of X and such that, for all m ∈N and any hyperplaneh∈X, mh does not cross h.
Each vertexxinX is uniquely defined by the set of half spaces containing it. Since each hyperplane in X is the image under the action of N of some unique hyperplane in the set {h1, . . . ,hk}, each pair of halfspaces Xh, Xh∗ is contained in the set Hi for some unique i. Hence for each vertex x the set
of halfspaces {Xh(∗)|x ∈ Xh(∗)} can be decomposed into the disjoint union of the sets {Xh(∗) ∈ Hi|x ∈ X
(∗)
h }, i = 1, . . . , k. Hence each vertex in X is uniquely defined by the list of components D1, . . . , Dk containing it, where
Di ∈X\Nhi.
Define the map from the vertex set of the cube complexXto the vertex set of the product of treesT1×. . .×Tk defined as follows: for each vertexxinX
and eachi∈ {1, . . . , k}letDi(x) be the component ofX\Nhi containingx.
As defined in lemma 3.8 for eachithere is an injective mapξi :X\Nhi 7→Tiξ.
We define ξ:X(0)7→T
1×. . .×Tk byξ(x) = ξ1(D1(x)), . . . , ξk(Dk(x)).
Ifu and u0 are adjacent vertices inX then they are separated by exactly one hyperplane. Hence they map to vertices inT1×. . .×Tkwhich differ in only
one co-ordinate and are adjacent in that co-ordinate tree. Extending this to a shortest edge path between any two vertices inX, we see that the distance between two vertices is precisely the number of hyperplanes separating them. Similarly, we have one edge in the shortest edge path between corresponding vertices in the product of trees for each of these hyperplanes. Hence on the level of edge metrics distance is preserved and soX(1)embedsd
1-isometrically
in the one skeleton of T1×. . .×Tk.
Since both the cube complexX and the product of treesT1×. . .×Tk are
CAT(0), the mapξ extends to a map fromn-cubes ton-cubes for allnwhich isd2-isometric. To see this, note the map ξ preserves incidence of edges and
consider the image underξ of the 1-skeleton of a n-cube in X where n≥ 2. If n = 2 then, since T1 ×. . .×Tk is CAT(0) and hence simply connected,
the image of the 1-skeleton of every 2-cube in X must be the 1-skeleton of a 2-cube in T1 ×. . .×Tk. If n > 2, then since T1×. . .×Tk is non-positively
curved the image of the 1-skeleton of every n-cube in X is the 1-skeleton of an n-cube in T1 ×. . .×Tk. Similarly, we can show that for any pair of
verticesξ(u) andξ(v) in the image ofX inT1×. . .×Tk, if n is the minimal
value for which ξ(u) and ξ(v) are vertices in the boundary of an n-cube in T1×. . .×Tk, then u and v lie in the boundary of an n-cube in X.
Corollary 3.11. LetGbe a group which acts freely, properly and cocompactly
such that stabG(h) is separable for each hyperplane h of X. Then G embeds
quasi-isometrically in a finite product of trees.
Proof. By lemma 1.42, for any such G and X there is a quasi-isometric
embedding of G inX. By lemma 3.10 there is an isometric map from X to a product of trees T1×. . .×Tk. The composition of these two maps gives a
quasi-isometric map from the groupGto the product of treesT1×. . .×Tk.