New York J. Math. 17(2011) 1–20.
C
∗-algebra of the
Z
n-tree
Menassie Ephrem
Abstract. Let Λ = Zn with lexicographic ordering. Λ is a totally ordered group. Let X = Λ+∗Λ+. ThenX is a Λ-tree. Analogous to the construction of graphC∗
-algebras, we form a groupoid whose unit space is the space of ends of the tree. TheC∗
-algebra of the Λ-tree is defined as the C∗
-algebra of this groupoid. We prove some properties of thisC∗
-algebra.
Contents
1. Introduction 1
2. The Zn-tree and its boundary 3
3. The groupoid andC∗-algebra of the Zn-tree 7
4. Generators and relations 9
5. Crossed product by the gauge action 14
6. Final results 18
References 19
1. Introduction
Since the introduction of C∗-algebras of groupoids, in the late 1970’s, several classes of C∗-algebras have been given groupoid models. One such
class is the class of graphC∗-algebras.
In their paper [10], Kumjian, Pask, Raeburn and Renault associated to each locally finite directed graph E a locally compact groupoid G, and showed that its groupoidC∗-algebraC∗(G) is the universal C∗-algebra
gen-erated by families of partial isometries satisfying the Cuntz–Krieger relations determined byE. In [16], Spielberg constructed a locally compact groupoid G associated to a general graph E and generalized the result to a general directed graph.
We refer to [13] for the detailed theory of topological groupoids and their C∗-algebras.
A directed graph E = (E0, E1, o, t) consists of a countable set E0 of vertices and E1 of edges, and maps o, t : E1 → E0 identifying the origin
Received January 15, 2008; revised January 21, 2011.
2000Mathematics Subject Classification. 46L05, 46L35, 46L55.
Key words and phrases. Directed graph, Cuntz–Krieger algebra, Graph C∗
-algebra.
(source) and the terminus (range) of each edge. For the purposes of this discussion it is sufficient to consider row-finite graphs with no sinks.
For the moment, letT be a bundle of of row-finite directed trees with no sinks, that is a disjoint union of trees that have no sinks or infinite emitters, i.e., no singular vertices. We denote the set of finite paths of T by T∗ and the set of infinite paths by∂T.
For eachp∈T∗, define
V(p) :={px:x∈∂T, t(p) =o(x)}. Forp, q∈T∗, we see that:
V(p)∩V(q) =
V(p) ifp=qr for somer∈T∗ V(q) ifq =pr for somer∈T∗ ∅ otherwise.
It is fairly easy to see that:
Lemma 1.1. The cylinder sets {V(p) : p ∈ T∗} form a base of compact
open sets for a locally compact, totally disconnected, Hausdorff topology of ∂T.
We want to define a groupoid that has ∂T as a unit space. For x = x1x2. . ., andy=y1y2. . .∈∂T, we sayxis shift equivalent toy with lagk∈ Zand writex∼ky, if there existsn∈Nsuch thatxi=yk+i for each i≥n.
It is not difficult to see that shift equivalence is an equivalence relation.
Definition 1.2. Let G:= {(x, k, y)∈∂T ×Z×∂T :x∼k y}. For pairs in
G2:={((x, k, y),(y, m, z)) : (x, k, y),(y, m, z)∈ G}, we define (1.1) (x, k, y)·(y, m, z) = (x, k+m, z). For arbitrary (x, k, y)∈ G, we define
(1.2) (x, k, y)−1= (y,−k, x).
With the operations (1.1) and (1.2), and source and range maps s, r : G −→∂T given by s(x, k, y) = y, r(x, k, y) =x, G is a groupoid with unit space ∂T.
For p, q ∈ T∗, with t(p) = t(q), define U(p, q) := {px, l(p)−l(q), qx) : x ∈ ∂T, t(p) = o(x)}, where l(p) denotes the length of the path p. The sets {U(p, q) : p, q ∈ T∗, t(p) = t(q)} make G a locally compact r-discrete groupoid with (topological) unit space equal to∂T.
Now letE be a directed graph. We form a graph whose vertices are the paths ofE and edges are (ordered) pairs of paths as follows:
Definition 1.3. LetEe denote the following graph: e
E0=E∗
e
E1={(p, q)∈E∗×E∗ :q=pe for some e∈E1} o(p, q) =p, t(p, q) =q.
Lemma 1.4. [16, Lemma 2.4] Ee is a bundle of trees.
Notice that ifE is a row-finite graph with no sinks, then Ee is a bundle of row-finite trees with no sinks.
If G(E) is the groupoid obtained as in Definition1.2, where Ee plays the role ofT then, in [16] Spielberg showed, in its full generality, that the graph C∗-algebra of E is equal to theC∗-algebra of the groupoidG(E). We refer to [16], for readers interested in the general construction and the proof.
We now examine theC∗-algebraO2, which is theC∗-algebra of the graph
E = 0
a
X
X
b
Denoting the vertex ofE by 0 and the edges of E by aand b, as shown in the graph, the vertices ofEe are 0, a, b, aa, ab, ba, bb,etc. And the graphEe is the binary tree.
Take a typical path p of E, say p =aaabbbbbabbaaaa. Writing aaa as 3 andbbbbbas 5′, etc. we can writepas 35′12′4 which is an element ofZ+∗Z+ (the free product of two copies ofZ+). In other words, the set of vertices of
e
E isG+1 ∗G+2, where G+1 =Z+=G+
2, and the vertex 0 is the empty word. The elements of∂Ee are the infinite sequence ofn’s andm’s, where n∈G+1 and m∈G+2.
Motivated by this construction, we wish to explore theC∗-algebra of the case when Λ is an ordered abelian group, and X is the free product of two copies of Λ+. In this paper we study the special case when Λ =Znendowed
with the lexicographic ordering, where n∈ {2, 3, . . .}.
The paper is organized as follows. In Section 2 we develop the topology of theZn-tree. In Section3 we build the C∗-algebra of theZn-tree by first
building the groupoid G in a fashion similar to that of the graph groupoid. In Section 4, by explicitly exploring the partial isometries generating the C∗-algebra, we give a detailed description of the C∗-algebra. In Section5, we look at the crossed product of the C∗-algebra by the gauge action and
study the fixed-point algebra. Finally in Section 6we provide classification of the C∗-algebra. We prove that theC∗-algebra is simple, purely infinite,
nuclear and classifiable.
I am deeply indebted to Jack Spielberg without whom none of this would have been possible. I also wish to thank Mark Tomforde for many helpful discussions and for providing material when I could not find them otherwise.
2. The Zn
-tree and its boundary
Let n ∈ {2,3, . . .} and let Λ = Zn together with lexicographic ordering,
. . .,kd−1 =md−1, and kd< md. We set
∂Λ+={(k1, k2, . . . , kn−1,∞) :ki∈N∪ {∞}, ki =∞ ⇒ki+1 =∞}.
Let Gi = Λ+ = {a ∈ Λ : a > 0} for i = 1, 2, and let ∂Gi = ∂Λ+
for i = 1, 2. That is, we take two copies of Λ+ and label them as G1 and G2, and two copies of ∂Λ+ and label them as ∂G1 and ∂G2. Now consider the set X = G1 ∗G2. We denote the empty word by 0. Thus, X =S∞d=1{a1a2. . . ad :ai ∈Gk ⇒ai+1 ∈Gk±1 for 1 ≤i < d} S {0}. We note that X is a Λ-tree, as studied in [4].
Let ∂X = {a1a2. . . ad : ai ∈ Gk ⇒ ai+1 ∈ Gk±1, for 1 ≤ i < d− 1 andad−1 ∈ Gk ⇒ ad ∈ ∂Gk±1} S {a1a2. . . : ai ∈ Gk ⇒ ai+1 ∈ Gk±1 for each i}. In words,∂X contains either a finite sequence of elements of Λ from sets with alternating indices, where the last element is from∂Λ+, or an infinite sequence of elements of Λ from sets with alternating indices. Fora∈Λ+ and b∈∂Λ+, definea+b∈∂Λ+ by componentwise addition.
Forp=a1a2. . . ak∈X and q=b1b2. . . bm ∈X∪∂X, i.e.,m∈N∪ {∞},
definepq as follows:
(i) If ak, b1 ∈Gi∪∂Gi (i.e., they belong to sets with the same index),
thenpq:=a1a2. . . ak−1(ak+b1)b2. . . bm. Observe that sinceak ∈Λ,
the sum ak+b1 is defined and is in the same set asb1. (ii) If ak and b1 belong to sets with different indices, then
pq:=a1a2. . . akb1b2. . . bm.
In other words, we concatenatep and q in the most natural way (using the group law in Λ∗Λ).
For p ∈ X and q ∈ X∪∂X, we write p q to mean q extends p, i.e., there existsr∈X∪∂X such thatq =pr.
For p ∈∂X and q ∈ X∪∂X, we write p q to mean q extends p, i.e., for each r∈X,rp implies thatrq.
We now define two length functions. Define l:X∪∂X −→ (N∪ {∞})n
by l(a1a2. . . ak) :=Pki=1ai.
And define li : X∪∂X −→ N∪ {∞} to be the ith component ofl, i.e.,
li(p) is the ith component of l(p). It is easy to see that both l and li are
additive.
Next, we define basic open sets of∂X. For p, q∈X, we define
V(p) :={px:x∈∂X} and V(p;q) :=V(p)\V(q).
Notice that
(2.1) V(p)∩V(q) =
∅ ifpq and qp V(p) ifqp
V(q) if pq.
Hence
V(p)\V(q) =
Therefore, we will assume that pq whenever we write V(p;q). LetE :={V(p) :p∈X} S{V(p;q) :p, q∈X}.
Lemma 2.1. E separates points of∂X, that is, ifx, y∈∂X andx6=ythen there exist two sets A, B∈ E such that x∈A, y∈B, and A∩B =∅.
Proof. Suppose x, y∈∂X and x 6=y. Let x=a1a2. . . as, y =b1b2. . . bm.
Assume, without loss of generality, thats≤m. We consider two cases:
Case I. There exists k < s such that ak 6= bk (or they belong to different
Gi’s).
Then x∈V(a1a2. . . ak), y∈V(b1b2. . . bk) and
V(a1a2. . . ak)∩V(b1b2. . . bk) =∅.
Case II. ai=bi for each i < s.
Notice that ifs=∞, that is, if both xand y are infinite sequences then there should be a k∈N such that ak 6=bk which was considered in Case I.
Hence s <∞. Again, we distinguish two subcases:
(a) s = m. Therefore x = a1a2. . . as and y = a1a2. . . bs, and as, bs ∈
∂Gi, withas6=bs. Assuming, without loss of generality, thatas < bs,
let as = (k1, k2, . . . , kn−1,∞), and bs = (r1, r2, . . . , rn−1,∞) where (k1, k2, . . . , kn−1) < (r1, r2, . . . , rn−1). Therefore there must be an indexisuch thatki < ri; letjbe the largest such. Henceas+ej ≤bs,
whereej is then-tuple with 1 at thejthspot and 0 elsewhere. Letting
c=as+ej, we see that x∈A=V(a1a2. . . as−1;c), y ∈B =V(c), and A∩B =∅.
(b) s < m. Theny=a1a2. . . as−1bsbs+1. . . bm (m≥s+ 1).
Sincebs+1 ∈(Gi∪∂Gi)\{0}fori= 1,2, choosec=en∈Gi(same
index asbs+1 is in). Thenx∈A=V(a1a2. . . as−1;a1a2. . . as−1bsc),
y∈B =V(a1a2. . . as−1bsc), andA∩B =∅.
This completes the proof.
Lemma 2.2. E forms a base of compact open sets for a locally compact Hausdorff topology on ∂X.
Proof. First we prove that E forms a base. Let A = V(p1;p2) and B = V(q1;q2). Notice that if p1 q1 and q1 p1 then A∩B = ∅. Suppose, without loss of generality, that p1 q1 and let x ∈ A∩B. Then by con-struction, p1 q1 x and p2 x and q2 x. Since p2 x and q2 x, we can chooser ∈X such thatq1 r,p2 r,q2r, andx=ra for some a ∈ ∂X. If x p2 and x q2 then r can be chosen so that r p2 and rq2, hence x∈V(r)⊆A∩B.
can choose r so that r q2. Therefore x ∈ V(r;s1) ⊆ A∩B. If x q2, constructs2 the way as s1 was constructed, whereq2 takes the place ofp2. Then eithers1s2 ors2s1. Set
s=
s1 ifs1 s2 s2 ifs2 s1.
Then x∈V(r;s)⊆A∩B. The cases when A orB is of the formV(p) are similar, in fact easier.
That the topology is Hausdorff follows from the fact that E separates points.
Next we prove local compactness. Given p, q∈X we need to prove that V(p;q) is compact. SinceV(p;q) =V(p)\V(q) is a (relatively) closed subset ofV(p), it suffices to show thatV(p) is compact. LetA0 =V(p) be covered by an open cover U and suppose that A0 does not admit a finite subcover. Choose p1 ∈ X such that li(p1) ≥ 1 and V(pp1) does not admit a finite subcover, for somei∈ {1, . . . , n−1}. We consider two cases:
Case I. Suppose no such p1 exists.
Let a = en ∈ G1, b = en ∈ G2. Then V(p) = V(pa)∪V(pb). Hence either V(pa) or V(pb) is not finitely covered, say V(pa), then let x1 = a. After choosingxs, since V(px1. . . xs) =V(px1. . . xsa)∪V(px1. . . xsb),
either V(px1. . . xsa) or V(px1. . . xsb) is not finitely covered. And we let
xs+1 = a or b accordingly. Now let Aj = V(px1. . . xj) for j ≥ 1 and let
x = px1x2. . . ∈ ∂X. Notice that A0 ⊇ A1 ⊇ A2. . ., and x ∈ T∞j=0Aj.
Choose A′ ∈ U, q, r ∈ X, such that x ∈ V(q;r) ⊆ A′. Clearly q x and rx. Once again, we distinguish two subcases:
(a) x r. Then, for a large enough k we get q px1x2. . . xk and
px1x2. . . xk r. Therefore Ak = V(px1x2. . . xk) ⊆ A′, which
con-tradicts to thatAk is not finitely covered.
(b) x r. Notice l1(x) = l1(p) and since x = px1x2. . . r, we have l1(x) = l1(p) < l1(r). Therefore V(r) is finitely covered, say by B1, B2, . . . , Bs∈ U. For large enough k,q px1x2. . . xk. Therefore
Ak=V(px1x2. . . xk)⊆V(q) =V(q;r)∪V(r)⊆A′∪Snj=1Bj, which
is a finite union. This is a contradiction.
Case II. Letp1∈X such thatli(p1)≥1 andV(pp1) is not finitely covered, for somei∈ {1, . . . , n−1}.
Having chosen p1, . . . , ps let ps+1 with li(ps+1) ≥ 1 and V(pp1. . . ps+1) not finitely covered, for somei∈ {1, . . . , n−1}. If no such ps+1 exists then we are back in to CaseIwithV(pp1p2. . . ps) playing the role ofV(p). Now
let x =pp1p2. . .∈ ∂X and let Aj =V(pp1. . . pj). We get A0 ⊇A1 ⊇. . ., and x = pp1p2. . . ∈ T∞j=0Aj. Choose A′ ∈ U such that x ∈ V(q;r) ⊆ A′.
Notice thatq x and n−1 is finite, hence there existsi0 ∈ {1, . . . , n−1} such thatli0(x) =∞. Since li0(r)<∞, we havexr. Therefore, for large
ThereforeV(p) is compact.
3. The groupoid and C∗-algebra of the Zn
-tree
We are now ready to form the groupoid which will eventually be used to construct theC∗-algebra of the Λ-tree.
For x, y ∈ ∂X and k ∈ Λ, we write x ∼k y if there exist p, q ∈ X and
z∈∂X such that k=l(p)−l(q) and x=pz, y=qz. Notice that:
(a) If x∼ky theny∼−kx.
(b) x∼0 x.
(c) If x∼ky and y∼mz thenx=µt, y=νt, y=ηs, z=βsfor some
µ, ν, η, β∈X t, s∈∂X andk=l(µ)−l(ν), m=l(η)−l(β).
If l(η) ≤ l(ν) then ν = ηδ for some δ ∈ X. Therefore y = ηδt, implying s = δt, hence z = βδt. Therefore x ∼r z, where r =
l(µ) −l(βδ) = l(µ)−l(β)−l(δ) = l(µ)−l(β)−(l(ν) −l(η)) = [l(µ)−l(ν)] + [l(η)−l(β)] =k+m.
Similarly, ifl(η)≥l(ν) we getx∼r z, wherer =k+m.
Definition 3.1. LetG:={(x, k, y)∈∂X ×Λ×∂X :x∼ky}.
For pairs inG2:={((x, k, y),(y, m, z)) : (x, k, y),(y, m, z)∈ G}, we define
(3.1) (x, k, y)·(y, m, z) = (x, k+m, z).
For arbitrary (x, k, y)∈ G, we define
(3.2) (x, k, y)−1= (y,−k, x).
With the operations (3.1) and (3.2), and source and range maps s, r : G −→∂X given by s(x, k, y) =y, r(x, k, y) =x, G is a groupoid with unit space ∂X.
We want to makeG a locally compactr-discrete groupoid with (topolog-ical) unit space ∂X.
Forp, q∈X and A∈ E, define [p, q]A={(px, l(p)−l(q), qx) :x∈A}.
Lemma 3.2. For p, q, r, s∈X and A, B∈ E,
[p, q]A∩[r, s]B
=
[p, q]A∩µB if there exists µ∈X such that r =pµ, s=qµ
[r, s](µA)∩B if there exists µ∈X such that p=rµ, q=sµ
∅ otherwise.
Proof. Let t∈[p, q]A∩[r, s]B. Then t= (px, k, qx) = (ry, m, sy) for some
The case whenl(r)≤l(p) follows by symmetry. The reverse containment is
clear.
Proposition 3.3. Let G have the relative topology inherited from ∂X × Λ×∂X. Then G is a locally compact Housdorff groupoid, with base D = {[a, b]A:a, b∈X, A∈ E} consisting of compact open subsets.
Proof. That D is a base follows from Lemma3.2. [a, b]A is a closed subset
ofaA× {l(a)−l(b)} ×bA, which is a compact open subset of∂X×Λ×∂X. Hence [a, b]Ais compact open in G.
To prove that inversion is continuous, let φ : G −→ G be the inversion function. Then φ−1([a, b]A) = [b, a]A. Therefore φis continuous. In fact φ
is a homeomorphism.
For the product function, let ψ : G2 −→ G be the product function. Then ψ−1([a, b]A) =Sc∈X(([a, c]A×[c, b]A)∩ G2) which is open (is a union
of open sets).
Remark 3.4. We remark the following points:
(a) Since the set D is countable, the topology is second countable. (b) We can identify the unit space, ∂X, of G with the subset{(x,0, x) :
x∈∂X}ofGviax7→(x,0, x). The topology on∂X agrees with the topology it inherits by viewing it as the subset {(x,0, x) : x ∈∂X} ofG.
Proposition 3.5. For each A ∈ E and each a, b∈X, [a, b]A is a G-set. G
is r-discrete.
Proof.
[a, b]A={(ax, l(a)−l(b), bx) :x∈A}
⇒([a, b]A)−1={(bx, l(b)−l(a), ax) :x∈A}.
Hence, ((ax, l(a)−l(b), bx)(by, l(b)−l(a), ay))∈[a, b]A×([a, b]A)−1 T G2if
and only ifx=y. And in that case, (ax, l(a)−l(b), bx)·(bx, l(b)−l(a), ax) = (ax,0, ax)∈∂X, via the identification stated in Remark 3.4(b). This gives [a, b]A·([a, b]A)−1 ⊆∂X. Similarly, ([a, b]A)−1·[a, b]A ⊆∂X. ThereforeG
has a base of compact open G-sets, implyingG isr-discrete.
Define C∗(Λ) to be the C∗ algebra of the groupoid G. Thus C∗(Λ) =
span{χS :S ∈ D}.
ForA=V(p)∈ E,
And forA=V(p;q) =V(p)\V(q)∈ E,
[a, b]A={(ax, l(a)−l(b), bx) :x∈V(p)\V(q)}
={(ax, l(a)−l(b), bx) :x∈V(p)} \ {(ax, l(a)−l(b), bx) :x∈V(q)} = [ap, bp]∂X\[aq, bq]∂X.
Denoting [a, b]∂X byU(a, b) we get:
D={U(a, b) :a, b∈X} [ {U(a, b)\U(c, d) :a, b, c, d∈X, ac, bd}.
MoreoverχU(a,b)\U(c,d)=χU(a,b)−χU(c,d), wheneverac, bd. This gives us:
C∗(Λ) = span{χU(a,b):a, b∈X}.
4. Generators and relations
Forp∈X, let sp=χU(p,0), where 0 is the empty word. Then:
s∗p(x, k, y) =χU(p,0)((x, k, y)−1) =χU(p,0)(y,−k, x) =χU(0,p)(x, k, y).
Hence s∗p =χU(0,p). And forp, q∈X,
spsq(x, k, y) =
X
y∼mz
χU(p,0)((x, k, y)(y, m, z))χU(q,0)((y, m, z)−1)
= X
y∼mz
χU(p,0)(x, k+m, z) χU(q,0)(z,−m, y).
Each term in this sum is zero except when x=pz, with k+m =l(p), and z=qy, withl(q) =−m. Hence,k=l(p)−m=l(p)+l(q), andx=pz=pqy. Thereforespsq(x, k, y) =χU(pq,0)(x, k, y); that is,spsq=χU(pq,0)=spq.
Moreover,
sps∗q(x, k, y) =
X
y∼mz
χU(p,0)((x, k, y)(y, m, z))χU(0,q)((y, m, z)−1)
= X
y∼mz
χU(p,0)(x, k+m, z) χU(0,q)(z,−m, y)
= X
y∼mz
χU(p,0)(x, k+m, z) χU(q,0)(y, m, z).
Each term in this sum is zero except when x =pz, k+m =l(p), y =qz, and l(q) = m. That is, k = l(p)−l(q), and x = pz, y = qz. Therefore sps∗q(x, k, y) =χU(p,q)(x, k, y); that is, sps∗q=χU(p,q).
Notice also that
s∗psq(x, k, y) =
X
y∼mz
= X
y∼mz
χU(0,p)(x, k+m, z) χU(q,0)(z,−m, y).
is nonzero exactly whenz=px, l(p) =−(k+m), z=qy,and l(q) =−m, which implies thatpx=qy, l(p) =−k−m=−k+l(q). This implies that s∗psq is nonzero only if either pq orqp.
Ifpqthen there existsr∈Xsuch thatq=pr. But−k=l(p)−l(q)⇒ k=l(q)−l(p) =l(r). And qy=pry⇒x=ry. Therefores∗
psq =sr. And if
q p then there exists r ∈X such that p=qr. Then (s∗
psq)∗ =s∗qsp =sr.
Hence s∗psq =s∗r. In short,
s∗psq =
sr ifq=pr
s∗r ifp=qr 0 otherwise. We have established that
(4.1) C∗(Λ) = span{sps∗q:p, q∈X}.
LetG0 :={(x,0, y)∈ G:x, y∈∂X}. ThenG0, with the relative topology, has the basic open sets [a, b]A, whereA∈ E, a, b∈Xandl(a) =l(b). Clearly
G0 is a subgroupoid ofG. And
C∗(G0) = span{χU(p,q):p, q∈X, l(p) =l(q)}
⊆span{χ[p,q]A :p, q∈X, l(p) =l(q), A⊆∂X is compact open} ⊆C∗(G0).
The second inclusion is due to the fact that [p, q]Ais compact open whenever
A⊆∂X is, henceχ[a,b]A ∈Cc(G0)⊆C∗(G0).
We wish to prove that the C∗-algebra C∗(G0) is an AF algebra. But
first notice that for any µ∈X,V(µ) =V(µe′n)∪V(µe′′n), where e′n=en=
(0, . . . ,0,1)∈G1 and e′′n=en∈G2.
Take a basic open set A=V(µ)\ Sm1k=1V(νk)
. It is possible to rewrite A as V(p)\ Sm2k=1V(rk)
with µ6=p. Here is a relatively simple example (pointed out to the author by Spielberg): V(µ)\V(µe′n) =V(µe′′n), where e′
n=en∈G1 and e′′n=en∈G2.
Lemma 4.1. Suppose A=V(µ)\(Sks=1V(µνk))6=∅. Then we can write
A as A=V(p)\ Sm1k=1V(prk)
where l(p) is the largest possible, that is, if A=V(q)\ Sm2j=1V(qsj) then l(q)≤l(p).
Proof. We take two cases:
Case I. For eachk= 1, . . . , s, there existsi∈ {1, . . . , n−1}withli(νk)≥1.
Choose p =µ, rk = νk for each k (i.e., leave A the way it is). Suppose
now that A=V(q)\ Sm2j=1V(qsj)
with l(p) ≤ l(q). We will prove that l(p) = l(q). Assuming the contrary, suppose l(p) < l(q). Let x ∈ A ⇒ x = qy for some y ∈ ∂X. Since qy ∈ V(p)\(Ssk=1V(prk)), p qy. But
Eithera1 ∈G1\{0}ora1∈G2\{0}. Suppose, for definiteness,a1 ∈G1\{0}.
Therefore there is only a finite possibler’s we can choose form. [In fact, since rν′e′′n, there are at mostm of them to choose from.]
To prove thatC∗(G0) is an AF algebra, we start with a finite subsetU of the generating set {χU(p,q) :p, q ∈X, l(p) = l(q)} and show that there is a finite dimensional C∗-subalgebra ofC∗(G0) that contains the setU.
Proof. Suppose thatU ={χU(p1,q1), χU(p2,q2), . . . χU(ps,qs)} is a (finite) sub-set of the generating sub-set of C∗(G0). Let
S :={V(p1), V(q1), V(p2), V(q2), . . . , V(ps), V(qs)}.
We “disjointize” the set S as follows. For a subset Fof S, write
AF:= \
A∈F
A\ [
A /∈F A.
Define
C:={AF:F⊆ S}.
Clearly, the set C is a finite collection of pairwise disjoint sets. A routine computation reveals that for anyE ∈ S,E =S{C∈ C :C⊆E}. It follows from (2.1) that for anyF⊆ S,TA∈FA=V(p), for somep∈X, if it is not empty. Hence,
AF=V(p)\
k
[
i=1 V(pri)
for some p ∈ X and some ri ∈ X. Let pF ∈X be such that AF = V(p)\ Sk
i=1V(pri) and l(pF) is maximum (as in Lemma 4.1). Then
AF =pF ∂X\
k
[
i=1 V(ri)
!!
=pFCF,
whereCF=∂X\Ski=1V(ri)
. NowV(pα) =pF1CF1∪pF2CF2∪. . .∪pFkCFk where {F1, F2, . . . , Fk} = {F ⊆ S : V(pα) ∈ F}. Notice that pFiCFi ⊆ V(pα) for eachi, hencepαpFi. HencepFiCFi =pαtiCFi, for someti∈X. Therefore V(pα) = pαU1 ∪pαU2 ∪. . .∪pαUk where Ui = tiCFi. Similarly V(qα) =qαV1∪qαV2∪. . .∪qαVm, where each qαVi ∈ C is subset of V(qα).
Consider the set
B:={[p, q]C∩D :pC, qD∈ C and p=pα, q=qα,1≤α ≤s}.
Since C is a finite collection, this collection is finite too. We will prove that B is pairwise disjoint.
Suppose [p, q]C∩D T[p′, q′]C′∩D′is non-empty. Clearlyp(C∩D) T p′(C′∩
D′)6=∅, andq(C∩D) T q′(C′∩D′)6=∅. Therefore, among other things, pC∩p′C′6=∅and qD∩q′D′ 6=∅, but by construction,{pC, qD, p′C′, q′D′} is pairwise disjoint. HencepC=p′C′ andqD=q′D′. Suppose, without loss
of generality, thatl(p)≤l(p′). Then p′ =pr and q′ =qsfor somer, s∈X, hence [p′, q′]
C′∩D′ = [pr, qs]C′∩D′. Let (px,0, qx) ∈ [p, q]C∩DT[pr, qs]C′∩D′.
Then px = prt and qx = qst, for some t ∈ C′ ∩D′, hence x = rt = st. Therefore r = s (since l(r) = l(p′) −l(p) = l(q′)−l(q) = l(s)). Hence
[pr, qr]C′∩D′ = [p, q]
r(C′∩D′) = [p, q]C∩D. Therefore B is a pairwise disjoint
collection.
For each [p, q]C∩D ∈ B, since C∩D is of the form V(µ)\Ski=1V(µνi),
we can rewrite C ∩D as µW, where W = ∂X\Ski=1V(νi) and l(µ) is
maximal (by Lemma 4.1). Then [p, q]C∩D = [p, q]µW = [pµ, qµ]W. Hence
each [p, q]C∩D ∈ B can be written as [p, q]W where l(p) = l(q) is maximal
and W =∂X\Ski=1V(νi).
Consider the collection D:={χ[p,q]W : [p, q]W ∈ B}. We will show that,
for each 1≤α≤s,χU(pα,qα) is a sum of elements ofD and thatD is a self-adjoint system of matrix units. For the first, letV(pα) =pαU1∪pαU2∪. . .∪ pαUk andV(qα) =qαV1∪qαV2∪. . .∪qαVm. One more routine computation
gives us:
U(pα, qα) = [pα, qα]∂X = k,m[
i,j=1
[pα, pα]Ui·[pα, qα]∂X·[qα, qα]Vj
=
k,m[
i,j=1
[pα, qα]Ui∩Vj.
Since the union is disjoint,
χU(pα,qα)=
k,m
X
i,j=1
χ[pα,qα]Ui∩Vj.
And each χ[pα,qα]Ui∩Vj is in the collectionD. ThereforeU ⊆span(D).
To show thatDis a self-adjoint system of matrix units, letχ[p,q]W, χ[r,s]V ∈ D. Then
χ[p,q]W ·χ[r,s]V(x1,0, x2) = X
y1,y2
χ[p,q]W (x1,0, x2)(y1,0, y2)
·χ[r,s]V(y2,0, y1)
=X
y2
χ[p,q]W(x1,0, y2)·χ[r,s]V(y2,0, x2),
where the last sum is taken over all y2 such that x1 ∼0 y2 ∼0 x2. Clearly the above sum is zero if x1 ∈/ pW orx2 ∈/ sV. Also, recalling that qW and rV are either equal or disjoint, we see that the above sum is zero if they are disjoint. For the preselected x1, if x1 = pz then y2 = qz (is uniquely chosen). Therefore the above sum is just the single term χ[p,q]W(x1,0, y2)· χ[r,s]V(y2,0, x2). Suppose that qW = rV. We will show that l(q) = l(r), which implies that q=r and W =V.
Given this,
To show that l(q) = l(r), assuming the contrary, suppose l(q) < l(r) then r = qc for some non-zero c ∈ X, implying V = cW. Hence [r, s]V =
[r, s]cW = [rc, sc]W, which contradicts the maximality of l(r) = l(s). By
symmetry l(r) < l(q) is also impossible. Hence l(q) = l(r) and W = V.
This concludes the proof.
5. Crossed product by the gauge action
Let ˆΛ denote the dual of Λ, i.e., the abelian group of continuous homo-morphisms of Λ into the circle group T with pointwise multiplication: for t, s∈Λ,ˆ hλ, tsi =hλ, tihλ, si for each λ∈Λ, where hλ, ti denotes the value of t∈Λ atˆ λ∈Λ.
Define an action called the gauge action: α : ˆΛ −→ Aut(C∗(G)) as
follows. For t ∈ Λ, first defineˆ αt : Cc(G) −→ Cc(G) by αt(f)(x, λ, y) =
hλ, tif(x, λ, y) then extend αt:C∗(G)−→ C∗(G) continuously. Notice that
(A,Λ, α) is aˆ C∗- dynamical system.
Consider the linear map Φ ofC∗(G) onto the fixed-point algebra C∗(G)α
given by
Φ(a) = Z
ˆ Λ
αt(a)dt, fora∈C∗(G).
wheredt denotes a normalized Haar measure on Λ.b
Lemma 5.1. Let Φbe defined as above.
(a) The map Φ is a faithful conditional expectation; in the sense that Φ(a∗a) = 0 implies a= 0.
(b) C∗(G0) =C∗(G)α.
Proof. Since the action α is continuous, we see that Φ is a conditional expectation from C∗(G) onto C∗(G)α, and that the expectation is faithful. Forp, q∈X,αt(sps∗q)(x, l(p)−l(q), y) =hl(p)−l(q), tisps∗q(x, l(p)−l(q), y).
Hence if l(p) =l(q) then αt(sps∗q) =sps∗q for each t∈Λ. Thereforeˆ α fixes
C∗(G0). HenceC∗(G0)⊆C∗(G)α. By continuity of Φ it suffices to show that
Φ(sps∗q)∈C∗(G0) for all p, q∈X.
Z
ˆ Λ
αt(sps∗q)dt=
Z
ˆ Λ
hl(p)−l(q), tisps∗qdt= 0, when l(p)6=l(q).
It follows from (4.1) that C∗(G)α ⊆ C∗(G0). Therefore C∗(G)α = C∗(G0).
We study the crossed product C∗(G)×
αΛ. Recall thatb Cc(ˆΛ, A), which
is equal toC(ˆΛ, A), since ˆΛ is compact, is a dense *-subalgebra of A×αΛ.ˆ
Recall also that multiplication (convolution) and involution onC(ˆΛ, A) are, respectively, defined by:
(f ·g)(s) = Z
ˆ Λ
and
follows from Theorem 4.2and Proposition5.2.
Corollary 5.3. IB is an AF algebra.
We want to prove thatA×αΛ is an AF algebra, and to do this we considerˆ
the dual system. Define ˆα :Λ = Λˆˆ −→ Aut(A×αΛ) as follows: Forˆ λ∈Λ
and f ∈ C(ˆΛ, A), we define ˆαλ(f) ∈ C(ˆΛ, A) by: ˆαλ(f)(t) = hλ, tif(t).
Extend ˆαλ continuously.
Lemma 5.4. αˆλ(IB)⊆IB for each λ≥0.
1, then this last integral gives us
where f′(t) = hλ
1, tisp1s∗λ′q1, g′(t) = hλ+λ2, tisλ′p2s∗q2, and b′ =sλ′p0s∗λ′q0.
Therefore ˆαλ(f ·b·g)∈IB.
For each λ ∈ Λ define Iλ := ˆαλ(IB). Clearly each Iλ is an ideal of
A×αΛ and is an AF algebra. Letˆ λ1 < λ2 then λ2 −λ1 >0 ⇒ Iλ2−λ1 =
ˆ
αλ2−λ1(IB)⊆IB. Therefore Iλ2 = ˆαλ2(IB) = ˆαλ1(ˆαλ2−λ1(IB))⊆Iλ1. That
is,Iλ1 ⊇Iλ2 whenever λ1 < λ2. In particularIB =I0 ⊇Iλ for each λ≥0.
Furthermore, if f ∈ C(ˆΛ, A) is given by f(t) = hλ, tisps∗q, for λ ∈ Λ and
p, q∈X then ˆαβ(f)(t) =hβ, tif(t) =hβ, tihλ, tisps∗q =hβ+λ, tisps∗q.
For f ∈ C(G) given byf(t) = hλ, tisps∗q, let us compute f∗, f ·f∗, and
f∗·f so we can use them in the next lemma.
f∗(t) =αt(f(t−1)∗)
=αt
hλ, t−1isps∗q
∗
=hλ, t−1iα
t sqs∗p
=hλ, tihl(q)−l(p), tisqs∗p
=hλ+l(q)−l(p), tisqs∗p,
(f·f∗)(z) = Z
ˆ Λ
f(t)αt(f∗(t−1z))dt
= Z
ˆ Λ
hλ, tisps∗qαt hλ+l(q)−l(p), t−1zisqs∗p
dt
= Z
ˆ Λ
hλ, tisps∗qhλ+l(q)−l(p), t−1zihl(q)−l(p), tisqs∗pdt
= Z
ˆ Λ
hλ+l(q)−l(p), tisps∗qhλ+l(q)−l(p), t−1zisqs∗pdt
= Z
ˆ Λ
hλ+l(q)−l(p), zisps∗qsqs∗pdt
=hλ+l(q)−l(p), zisps∗qsqs∗p
=hλ+l(q)−l(p), zisps∗p,
and
(f∗·f)(z) =h λ+l(q)−l(p)+l(p)−l(q), zisqs∗q
=hλ, zisqs∗q.
Lemma 5.5. Let λ ∈ Λ, p, q ∈ X, f(t) = hλ, tisqs∗q, and let g(t) =
hλ, tisps∗q.
(a) If λ≥0 then f ∈IB.
Proof. To prove (a), sqs∗q ∈ C∗(G0) ⊆IB. Then f ∈ IB, since λ ≥0, by
Lemma 5.4. To prove (b), (g ·g∗)(z) = hλ+l(q)−l(p), zis
ps∗p. By (a),
g·g∗∈IB, implying g∈IB.
Theorem 5.6. A×αΛˆ is an AF algebra.
Proof. Letf(t) =hλ, tisps∗q. Chooseβ∈Λ large enough such thatβ+λ+
l(q)≥l(p). Then
ˆ
αβ(f)(z) =hβ, zihλ, zisps∗q
=hβ+λ, zisps∗q.
Applying Lemma5.5(b), ˆαβ(f) ∈IB. Thus ˆα−β(ˆαβ(f))∈I−β, implying
f ∈ I−β. Therefore A×αΛ =ˆ Sλ≤0Iλ. Since each Iλ is an AF algebra, so
isA×αΛ.ˆ
6. Final results
Let us recall that an r-discrete groupoid G is locally contractive if for every nonempty open subset U of the unit space there is an open G-set Z withs( ¯Z)⊆U andr( ¯Z)$s(Z). A subsetEof the unit space of a groupoid Gis said to be invariant if its saturation [E] =r(s−1(E)) is equal to E.
Anr-discrete groupoidGisessentially free if the set of allx’s in the unit space G0 withr−1(x)∩s−1(x) ={x} is dense in the unit space. When the only open invariant subsets of G0 are the empty set andG0 itself, then we say that Gis minimal.
Lemma 6.1. G is locally contractive, essentially free and minimal.
Proof. To prove thatGis locally contractive, letU ⊆ G0be nonempty open. Let V(p;q) ⊆U. Choose µ∈ X such that p µ, q µ and µq. Then V(µ) ⊆ V(p;q) ⊆ U. Let Z = [µ,0]V(µ). Then Z = ¯Z, s(Z) = V(µ) ⊆ U, r(Z) =µV(µ)$V(µ)⊆U. ThereforeG is locally contractive.
To prove thatG is essentially free, letx∈∂X. Thenr−1(x) ={(x, k, y) : x ∼k y} and s−1(x) = {(z, m, x) : z ∼m x}. Hence r−1(x) ∩s−1(x) =
{(x, k, x) :x∼k x}. Notice thatr−1(x)∩s−1(x) ={x}exactly whenx∼k x
To prove thatGis minimal, let E⊆ G0 be nonempty open and invariant, i.e., E =r(s−1(E)). We want to show thatE =G0. Since E is open, there exist p, q∈X such that V(p;q)⊆E. But
s−1(V(p;q)) ={(µx, l(µ)−l(pν), pνx) :q pµx}.
Let x ∈ G0. Choose ν ∈ X such that pν q and q pν. Then (x,−l(pν), pνx) ∈ s−1(V(p;q)) ⊆ s−1(E) and r(x,−l(pν), pνx) = x. That is,x∈r(s−1(V(p;q))), henceE =G0. ThereforeG is minimal.
Proposition 6.2 ([1, Proposition 2.4]). Let G be an r-discrete groupoid, essentially free and locally contractive. Then every non-zero hereditary C∗ -subalgebra of C∗
r(G) contains an infinite projection. Corollary 6.3. C∗
r(G) is simple and purely infinite.
Proof. This follows from Lemma 6.1and Proposition6.2.
Theorem 6.4. C∗(G) is simple, purely infinite, nuclear and classifiable.
Proof. It follows from the Takesaki–Takai Duality Theorem that C∗(G) is stably isomorphic to C∗(G) ×α Λˆ ׈α Λ. Since C∗(G) ×α Λ is an AFˆ
algebra and that Λ =Z2 is amenable,C∗(G) is nuclear and classifiable. We
prove that C∗(G) =Cr∗(G). From Theorem4.2 we get that the fixed-point algebra C∗(G0) is an AF algebra. The inclusion C
c(G0) ⊆ Cc(G) ⊆ C∗(G)
extends to an injective∗-homomorphismC∗(G0)⊆C∗(G) (injectivity follows since C∗(G0) is an AF algebra). Since C∗(G0) = Cr∗(G0), it follows that C∗(G0)⊆C∗
r(G). LetEbe the conditional expectation ofC∗(G) ontoC∗(G0)
and λbe the canonical map of C∗(G) onto Cr∗(G). IF Er is the conditional expectation of C∗
r(G) onto C∗(G0), then Er◦λ = E. It then follows that
λ is injective. Therefore C∗(G) = Cr∗(G). Simplicity and pure infiniteness
follow from Corollary6.3.
Remark 6.5. Kirchberg–Phillips classification theorem states that simple, unital, purely infinite, and nuclear C∗-algebras are classified by their K -theory [12]. In the continuation of this project, we wish to compute the K-theory of C∗(Zn).
Another interest is to generalize the study and/or the result to a more general ordered group or even a “larger” group, such as Rn
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Department of Mathematics and Statistics, Coastal Carolina University, Con-way, SC 29528-6054
