New York Journal of Mathematics New York J. Math.

New York Journal of Mathematics

New York J. Math. 17(2011) 1–20.

C

-algebra of the Z

-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.

ISSN 1076-9803/2011

1

MENASSIE EPHREM

(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 to y 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:

Ee⁰=E^∗

Ee¹={(p, q)∈E^∗×E^∗ :q=pe for some e∈E¹} o(p, q) =p, t(p, q) =q.

The following lemma, due to Spielberg [16], is straightforward.

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 groupoid G(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

XX

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 5⁰, etc. we can writepas 35⁰12⁰4 which is an element ofZ⁺∗Z⁺ (the free product of two copies ofZ⁺). In other words, the set of vertices of Ee 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, that is, (k1, k2, . . . , kn) <(m1, m2, . . . , mn) if either k1 < m1, or k1 = m1,

MENASSIE EPHREM

. . .,kd−1 =md−1, and k_d< m_d. We set

∂Λ⁺={(k₁, k₂, . . . , kn−1,∞) :k_i∈N∪ {∞}, k_i =∞ ⇒k_i+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. . . a_d :a_i ∈G_k ⇒a_i+1 ∈Gk±1 for 1 ≤i < d} S {0}. We note that X is a Λ-tree, as studied in [4].

Let ∂X = {a₁a₂. . . a_d : a_i ∈ G_k ⇒ a_i+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 a_k, b₁ ∈G_i∪∂G_i (i.e., they belong to sets with the same index), thenpq:=a₁a₂. . . ak−1(a_k+b₁)b₂. . . b_m. Observe that sincea_k ∈Λ, the sum a_k+b1 is defined and is in the same set asb1.

(ii) If a_k 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) :=Pk

i=1ai.

And define l_i : X∪∂X −→ N∪ {∞} to be the i^th component ofl, i.e., l_i(p) is the i^th component of l(p). It is easy to see that both l and l_i 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) ifpq.

Hence

V(p)\V(q) =

V(p) ifpq and qp

∅ ifq p.

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, that s≤m. We consider two cases:

Case I. There exists k < s such that a_k 6= b_k (or they belong to different Gi’s).

Then x∈V(a1a2. . . a_k), y∈V(b1b2. . . b_k) and V(a₁a₂. . . a_k)∩V(b₁b₂. . . b_k) =∅.

Case II. a_i=b_i 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 a_k 6=b_k 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 ∈

∂G_i, witha_s6=b_s. Assuming, without loss of generality, thata_s < b_s, 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 thatk_i < r_i; letjbe the largest such. Hencea_s+e_j ≤b_s, whereej is then-tuple with 1 at thej^thspot and 0 elsewhere. Letting c=a_s+e_j, we see that x∈A=V(a₁a₂. . . as−1;c), y ∈B =V(c), and A∩B =∅.

(b) s < m. Theny=a₁a₂. . . as−1b_sb_s+1. . . b_m (m≥s+ 1).

Sincebs+1 ∈(Gi∪∂Gi)\{0}fori= 1,2, choosec=en∈Gi(same index asb_s+1 is in). Thenx∈A=V(a₁a₂. . . as−1;a₁a₂. . . as−1b_sc), y∈B =V(a₁a₂. . . as−1b_sc), 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, and x=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.

Suppose now thatxp2. Thenx=ra, for somer∈X anda∈∂X. By extendingr if necessary, we may assume thata∈∂Λ⁺. Then we may write p2 =rby for someb∈Λ⁺, andy∈∂X witha < b. Letb⁰ =b−(0, . . . ,0,1), and s1 = rb⁰. Notice that x s1 p2 and s1 6= p2. If x q2 then we

MENASSIE EPHREM

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 l_i(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 choosingx_s, since V(px₁. . . x_s) =V(px₁. . . x_sa)∪V(px₁. . . x_sb), either V(px1. . . xsa) or V(px1. . . xsb) is not finitely covered. And we let x_s+1 = a or b accordingly. Now let A_j = V(px₁. . . x_j) for j ≥ 1 and let x = px1x2. . . ∈ ∂X. Notice that A0 ⊇ A1 ⊇ A2. . ., and x ∈ T∞

j=0Aj. Choose A⁰ ∈ U, q, r ∈ X, such that x ∈ V(q;r) ⊆ A⁰. Clearly q x and rx. Once again, we distinguish two subcases:

(a) x r. Then, for a large enough k we get q px₁x₂. . . x_k and px1x2. . . xk r. Therefore Ak = V(px1x2. . . xk) ⊆ A⁰, which con- tradicts to thatA_k 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, . . . , B_s∈ U. For large enough k,q px₁x₂. . . x_k. Therefore A_k=V(px1x2. . . x_k)⊆V(q) =V(q;r)∪V(r)⊆A⁰∪Sn

j=1Bj, which is a finite union. This is a contradiction.

Case II. Letp₁∈X such thatl_i(p₁)≥1 andV(pp₁) is not finitely covered, for somei∈ {1, . . . , n−1}.

Having chosen p₁, . . . , p_s let p_s+1 with l_i(p_s+1) ≥ 1 and V(pp₁. . . p_s+1) not finitely covered, for somei∈ {1, . . . , n−1}. If no such ps+1 exists then we are back in to CaseIwithV(pp₁p₂. . . p_s) 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=0A_j. Choose A⁰ ∈ U such that x ∈ V(q;r) ⊆ A⁰. 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 enoughk,q pp1. . . p_k r, implying A_k⊆A⁰, a contradiction.

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 x ∈ A, y ∈ B. Clearly k = m. Furthermore, px = ry and qx = sy.

Suppose that l(p) ≤ l(r). Then r = pµ for some µ ∈ X, hence px= pµy, implying x = µy. Hence qx = qµy = sy, implying qµ = s. Therefore t= (px, k, qx) = (pµy, k, qµy), that is,t= (px, k, qx) for some x∈A∩µB.

MENASSIE EPHREM

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) =S

c∈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{χ_S :S ∈ D}.

ForA=V(p)∈ E,

[a, b]A= [a, b]_V(p) ={(ax, l(a)−l(b), bx) :x∈V(p)}

={(ax, l(a)−l(b), bx) :x=pt, t∈∂X}

={(apt, l(a)−l(b), bpt) :t∈∂X}

= [ap, bp]_∂X.

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)=χ_U(a,b)−χ_U(c,d), wheneverac, bd. This gives us:

C^∗(Λ) = span{χ_U(a,b):a, b∈X}.

4. Generators and relations

Forp∈X, let sp=χ_U(p,0), where 0 is the empty word. Then:

s^∗_p(x, k, y) =χ_U(p,0)((x, k, y)⁻¹)

=χ_U(p,0)(y,−k, x)

=χ_U(0,p)(x, k, y).

Hence s^∗_p =χ_U(0,p). And forp, q∈X,

spsq(x, k, y) = X

y∼mz

χ_U(p,0)((x, k, y)(y, m, z))χ_U(q,0)((y, m, z)⁻¹)

= X

y∼_mz

χ_U(p,0)(x, k+m, z) χ_U(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) =χ_U(pq,0)(x, k, y); that is,spsq=χ_U(pq,0)=spq. Moreover,

sps^∗_q(x, k, y) = X

y∼mz

χ_U(p,0)((x, k, y)(y, m, z))χ_U(0,q)((y, m, z)⁻¹)

= X

y∼mz

χ_U(p,0)(x, k+m, z) χ_U(0,q)(z,−m, y)

= X

y∼_mz

χ_U(p,0)(x, k+m, z) χ_U(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 s_ps^∗_q(x, k, y) =χ_U(p,q)(x, k, y); that is, s_ps^∗_q=χ_U(p,q).

Notice also that s^∗_ps_q(x, k, y) = X

y∼_mz

χ_U(0,p)((x, k, y)(y, m, z))χ_U(q,0)((y, m, z)⁻¹)

MENASSIE EPHREM

= X

y∼mz

χ_U(0,p)(x, k+m, z) χ_U(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^∗_ps_q =s_r. And if q p then there exists r ∈X such that p=qr. Then (s^∗_ps_q)^∗ =s^∗_qs_p =s_r. Hence s^∗_psq =s^∗_r. In short,

s^∗_psq =







s_r ifq=pr s^∗_r ifp=qr 0 otherwise.

We have established that

(4.1) C^∗(Λ) = span{s_ps^∗_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{χ_U(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 ∈C_c(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(µe⁰_n)∪V(µe⁰⁰_n), where e⁰_n=en= (0, . . . ,0,1)∈G₁ and e⁰⁰_n=e_n∈G₂.

Take a basic open set A=V(µ)\ Sm1

k=1V(νk)

. It is possible to rewrite A as V(p)\ Sm2

k=1V(r_k)

with µ6=p. Here is a relatively simple example (pointed out to the author by Spielberg): V(µ)\V(µe⁰_n) =V(µe⁰⁰_n), where e⁰_n=e_n∈G₁ and e⁰⁰_n=e_n∈G₂.

Lemma 4.1. Suppose A=V(µ)\(Ss

k=1V(µν_k))6=∅. Then we can write A as A=V(p)\ Sm1

k=1V(prk)

where l(p) is the largest possible, that is, if A=V(q)\ Sm2

j=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)\ Sm2

j=1V(qs_j)

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)\(Ss

k=1V(prk)), p qy. But l(p) < l(q) ⇒ p q. Let q =pr, since p 6=q, r 6= 0. Let r =a1a2. . . a_d.

Eithera₁ ∈G₁\{0}ora₁∈G₂\{0}. Suppose, for definiteness,a₁ ∈G₁\{0}.

Take t = (0, . . . ,0,∞) ∈ ∂G2. Since l(rk) > l(t) for each k = 1, . . . , s, we get pr_k ptfor each k= 1, . . . , s, moreover pt∈V(p). Hence pt∈A. But pr pt ⇒ q pt ⇒ pt /∈ V(q) ⇒ pt /∈ V(q)\ Sm2

j=1V(qs_j)

which is a contradiction to A=V(q)\ Sm2

j=1V(qsj)

. Thereforel(p) = l(q). In fact, p=q.

Case II. There existsk∈ {1, . . . , s}withli(ν_k) = 0, for eachi= 1, . . . , n−1.

After rearranging, suppose thatl_i(ν_k) = 0 for eachk= 1, . . . , αand each i = 1, . . . , n−1; and that for each k = α + 1, . . . , s, li(ν_k) ≥ 1 for some i≤n−1. We can also assume thatl(ν₁) is the largest ofl(ν_k)’s for k≤α.

Then

A=V(µ)\

s

[

k=1

V(µν_k)

!

=

"

V(µ)\

α

[

k=1

V(µνk)

!#

\

"

V(µ)\

s

[

k=α+1

V(µνk)

!#

.

Letme_n=l(ν₁) which is non zero. We will prove that if we can rewriteAas V(q)\ Sm2

k=1V(qsk)

withl(µ)≤l(q) then q=µr with 0≤l(r)≤m(en).

Clearly if µq, thenA∩V(q) =∅. So, ifA∩V(q)\ Sm2

k=1V(qs_k) 6=∅ then µ q. Now let q = µr, and let ν₁ = a₁a₂. . . a_d. Observe that since for each j, aj ∈ Λ⁺ and that lk(ν1) = 0 for each k ≤ n−1, we have l_k(a_j) = 0 for all k ≤ n−1. Also, by assumption, l(ν₁) > 0, therefore either ad ∈ G1 \ {0} or ad ∈ G2 \ {0}. Suppose, for definiteness, that a_d∈G1\ {0}. Leta⁰_d=a_d−e_nand letν⁰=a1a2. . . a⁰_d(or justa1a2. . . ad−1, if a⁰_d = 0). If V(µν⁰) ∩A = ∅ then we can replace ν₁ by ν⁰ in the ex- pression of A and and (after rearranging the ν_i⁰s) choose a new ν1. Since A6=∅ this process of replacement must stop with V(µν⁰)∩A 6=∅. Letting e⁰_n = en ∈ G1 and e⁰⁰_n = en ∈ G2, then V(µν⁰) = V(µν⁰e⁰_n)∪V(µν⁰e⁰⁰_n) = V(µν₁)∪V(µν⁰e⁰⁰_n). Since V(µν₁)∩A =∅,A∩V(µν⁰e⁰⁰_n)6=∅ hence ν⁰e⁰⁰_n∈/ {ν₁, . . . , να}. Take t⁰ = (0, . . . ,0,∞) ∈ ∂G1 and t⁰⁰ = (0, . . . ,0,∞) ∈ ∂G2. Then µν⁰e⁰⁰_nt⁰, µν⁰e⁰⁰_nt⁰⁰ ∈ V(µ) \(Sα

k=1V(µν_k)). Moreover, for each k = α+ 1, . . . , s, we havel(ν⁰e⁰⁰_nt⁰), l(ν⁰e⁰⁰_nt⁰⁰)< l(ν_k), implyingµν⁰e⁰⁰_nt⁰, µν⁰e⁰⁰_nt⁰⁰∈ V(µ)\ Ss

k=α+1V(µν_k)

. Hence µν⁰e⁰⁰_nt⁰, µν⁰e⁰⁰_nt⁰⁰ ∈ V(q)\ Sm2

k=1V(qs_k) . Thereforeqµν⁰e⁰⁰_n⇒µr µν⁰e⁰⁰_n⇒0≤l(r)≤l(ν⁰e⁰⁰_n) =l(ν⁰)+e_n=me_n. Therefore there is only a finite possibler’s we can choose form. [In fact, since rν⁰e⁰⁰_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 {χ_U(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.

Theorem 4.2. C^∗(G₀) is an AF algebra.

MENASSIE EPHREM

Proof. Suppose thatU ={χ_U(p1,q1), χ_U(p2,q2), . . . χ_U(ps,qs)} is a (finite) sub- set of the generating 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 A_F:= \

A∈F

A\ [

A /∈F

A.

Define

C:={A_F: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,T

A∈FA=V(p), for somep∈X, if it is not empty. Hence,

A_F=V(p)\

k

[

i=1

V(pr_i)

for some p ∈ X and some r_i ∈ X. Let p_F ∈X be such that A_F = V(p)\ Sk

i=1V(pri) and l(pF) is maximum (as in Lemma 4.1). Then A_F =p_F ∂X\

k

[

i=1

V(r_i)

!!

=p_FC_F, whereC_F=∂X\

Sk

i=1V(r_i)

. NowV(p_α) =p_F1C_F1∪p_F2C_F2∪. . .∪p_FkC_Fk where {F₁, F2, . . . , Fk} = {F ⊆ S : V(pα) ∈ F}. Notice that pFiCFi ⊆ V(p_α) for eachi, hencep_αp_Fi. Hencep_FiC_Fi =p_αt_iC_Fi, for somet_i∈X.

Therefore V(p_α) = p_αU₁ ∪p_αU₂ ∪. . .∪p_αU_k where U_i = t_iC_Fi. 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

[p⁰, q⁰]_C⁰∩D⁰is non-empty. Clearlyp(C∩D) T

p⁰(C⁰∩ D⁰)6=∅, andq(C∩D) T

q⁰(C⁰∩D⁰)6=∅. Therefore, among other things, pC∩p⁰C⁰6=∅and qD∩q⁰D⁰ 6=∅, but by construction,{pC, qD, p⁰C⁰, q⁰D⁰} is pairwise disjoint. HencepC=p⁰C⁰ andqD=q⁰D⁰. Suppose, without loss of generality, thatl(p)≤l(p⁰). Then p⁰ =pr and q⁰ =qsfor somer, s∈X, hence [p⁰, q⁰]_C⁰∩D⁰ = [pr, qs]_C⁰∩D⁰. Let (px,0, qx) ∈ [p, q]C∩DT

[pr, qs]_C⁰∩D⁰. Then px = prt and qx = qst, for some t ∈ C⁰ ∩D⁰, hence x = rt = st.

Therefore r = s (since l(r) = l(p⁰) −l(p) = l(q⁰)−l(q) = l(s)). Hence pC =p⁰C⁰ =prC⁰, and qD=q⁰D⁰ =qrD⁰, implyingC =rC⁰ and D=rD⁰. This gives us C ∩D = rC⁰ ∩rD⁰ = r(C⁰ ∩D⁰). Hence [p⁰, q⁰]_C⁰∩D⁰ =

[pr, qr]_C⁰∩D⁰ = [p, q]_r(C⁰_∩D⁰₎ = [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)\S_k

i=1V(µν_i), we can rewrite C ∩D as µW, where W = ∂X\Sk

i=1V(ν_i) 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\Sk

i=1V(ν_i).

Consider the collection D:={χ_[p,q]W : [p, q]W ∈ B}. We will show that, for each 1≤α≤s,χ_U(pα,qα) is a sum of elements ofD and thatD is a self- adjoint system of matrix units. For the first, letV(p_α) =p_αU₁∪p_αU₂∪. . .∪ pαU_k 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,

χ_U(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,

χ_[p,q]W ·χ_[r,s]V(x₁,0, x₂) =χ_[p,q]W(x₁,0, y₂)·χ_[r,s]V(y₂,0, x₂)

=χ_[p,q]W(x₁,0, y₂)·χ_[q,s]W(y₂,0, x₂)

=χ_[p,s]W(x₁,0, x₂).

MENASSIE EPHREM

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 α_t : C_c(G) −→ C_c(G) by α_t(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

Λˆ

α_t(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), tis_ps^∗_q(x, l(p)−l(q), y).

Hence if l(p) =l(q) then α_t(s_ps^∗_q) =s_ps^∗_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

Λˆ

α_t(s_ps^∗_q)dt= Z

Λˆ

hl(p)−l(q), tis_ps^∗_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 C_c( 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

Λˆ

f(t)α_t(g(t⁻¹s))dt

and

f^∗(s) =α(f(s⁻¹)^∗).

The functions of the form f(t) = hλ, tis_ps^∗_q from ˆΛ into A form a gen- erating set for A×_α Λ. Moreover the fixed-point algebraˆ C^∗(G₀) can be imbedded into A×_αΛ as follows: for eachˆ b ∈C^∗(G₀), define the function b: ˆΛ−→Aasb(t) =b(the constant function). ThusC^∗(G₀) is a subalgebra of A×_αΛ.ˆ

Proposition 5.2. The C^∗-algebra B :=C^∗(G₀) is a hereditary C^∗-subalge- bra of A×_αΛ.ˆ

Proof. To prove the theorem, we prove that B · A×_αΛˆ · B ⊆B. Since A×_α Λ is generated by functions of the formˆ f(t) = hλ, tis_ps^∗_q, it suffices to show that b₁ ·f ·b₂ ∈ B whenever b₁, b₂ ∈ B and f(t) = hλ, tis_ps^∗_q for λ∈Λ, p, q∈X.

(b₁·f·b₂)(z) = Z

Λˆ

b₁(t)α_t((f ·b₂)(t⁻¹z))dt

= Z

Λˆ

b₁α_t Z

Λˆ

f(w)α_w(b₂(w⁻¹t⁻¹z))dw

dt

= Z

Λˆ

Z

Λˆ

b₁α_t(f(w)α_w(b₂))dw dt

= Z

Λˆ

Z

Λˆ

b1αt(hλ, wis_ps^∗_q)b2dw dt, sinceαw(b2) =αt(b2) =b2

= Z

Λˆ

Z

Λˆ

b₁hλ, wiα_t(s_ps^∗_q)b₂dw dt

= Z

Λˆ

Z

Λˆ

b1hλ, wihl(p)−l(q), tis_ps^∗_qb2dw dt

= Z

Λˆ

hλ, widw Z

Λˆ

hl(p)−l(q), tidt b₁s_ps^∗_qb₂

= 0 unless λ= 0 andl(p)−l(q) = 0.

And in that case (in the case whenλ= 0 andl(p)−l(q) = 0) we get (b1·f· b₂)(z) =b₁s_ps^∗_qb₂ ∈B (sincel(p) =l(q)). Therefore B is hereditary.

LetI_B denote the ideal inA×_αΛ generated byˆ B. The following corollary follows from Theorem 4.2and Proposition 5.2.

Corollary 5.3. I_B 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 ˆα_λ(f) ∈ C( ˆΛ, A) by: ˆα_λ(f)(t) = hλ, tif(t).

Extend ˆα_λ continuously.

As before we use ·to represent multiplication inA×_αΛ.ˆ

MENASSIE EPHREM

Lemma 5.4. αˆ_λ(I_B)⊆I_B for each λ≥0.

Proof. Since the functions of the form f(t) =hλ, tis_ps^∗_q make a generating set for A×_αΛ, it suffices to show that ifˆ λ >0 then ˆα_λ(f ·b·g) ∈ I_B for f(t) =hλ₁, tis_p1s^∗_q1,g(t) =hλ₂, tis_p2s^∗_q2, and b=sp0s^∗_q0, withl(p0) =l(q0).

First

(f·b·g)(z)

= Z

Λˆ

f(t)αt((b·g)(t⁻¹z))dt

= Z

Λˆ

f(t)αt

Z

Λˆ

b(w)αw(g(w⁻¹t⁻¹z))dw

dt

= Z

Λˆ

f(t) Z

Λˆ

bα_tw(g(w⁻¹t⁻¹z)dw)

dt

= Z

Λˆ

f(t) Z

Λˆ

bαw(g(w⁻¹z)dw)

dt

= Z

Λˆ

Z

Λˆ

f(t)bα_w(g(w⁻¹z))dw dt

= Z

Λˆ

Z

Λˆ

hλ₁, tis_p1s^∗_q1sp0s^∗_q0hλ₂, w⁻¹ziα_w(sp2s^∗_q2)dw dt

= Z

Λˆ

Z

Λˆ

hλ₁, tis_p1s^∗_q1s_p0s^∗_q0hλ₂, w⁻¹zihl(p₂)−l(q₂), wis_p2s^∗_q2dw dt.

Hence ˆ

α_λ(f ·b·g)(z)

=hλ, zi Z

Λˆ

Z

Λˆ

hλ₁, tis_p1s^∗_q1s_p0s^∗_q0hλ₂, w⁻¹zihl(p₂)−l(q₂), wis_p2s^∗_q2dw dt

= Z

Λˆ

Z

Λˆ

hλ₁, tis_p1s^∗_q1sp0s^∗_q0hλ, w⁻¹zihλ, wihλ₂, w⁻¹zi

hl(p₂)−l(q₂), wis_p2s^∗_q2dw dt

= Z

Λˆ

Z

Λˆ

hλ₁, tis_p1s^∗_q1s_p0s^∗_q0hλ+λ₂, w⁻¹zihλ+l(p₂)−l(q₂), wis_p2s^∗_q2dw dt;

letting λ⁰ =λ∈G₁, then this last integral gives us

= Z

Λˆ

Z

Λˆ

hλ₁, tis_p1s^∗_q1s^∗_λ⁰s_λ⁰sp0s^∗_q0s^∗_λ⁰s_λ⁰hλ+λ2, w⁻¹zi

hλ+l(p₂)−l(q₂), wis_p2s^∗_q2dw dt

= Z

Λˆ

Z

Λˆ

hλ₁, tis_p1s^∗_λ⁰_q1sλ⁰p0s^∗_λ⁰_q0hλ+λ2, w⁻¹zi

hλ+l(p₂)−l(q₂), wis_λ⁰_p2s^∗_q2dw dt

= (f⁰·b⁰·g⁰)(z),