Brownlowe, N., Carlsen, T. M., and Whittaker, M. (2015) Graph algebras and orbit equivalence. Ergodic Theory and Dynamical Systems

Transcription

1 n Brownlowe, N., Carlsen, T. M., and Whittaker, M. (015) Graph algebras and orbit equivalence. rgodic Theory and Dynamical Systems There may be differences between this version and the published version. You are advised to consult the publisher s version if you wish to cite from it. Deposited on: December 015 nlighten Research publications by members of the University of Glasgow

2 GRAPH ALGBRAS AND ORBIT QUIVALNC NATHAN BROWNLOW, TOK MIR CARLSN, AND MICHAL. WHITTAKR Abstract. We introduce the notion of orbit equivalence of directed graphs, following Matsumoto s notion of continuous orbit equivalence for topological Markov shifts. We show that two graphs in which every cycle has an exit are orbit equivalent if and only if there is a diagonal-preserving isomorphism between their C -algebras. We show that it is necessary to assume that every cycle has an exit for the forward implication, but that the reverse implication holds for arbitrary graphs. As part of our analysis of arbitrary graphs we construct a groupoid G (C (),D()) from the graph algebra C () and its diagonal subalgebra D() which generalises Renault s Weyl groupoid construction applied to (C (), D()). We show that G (C (),D()) recovers the graph groupoid G without the assumption that every cycle in has an exit, which is required to apply Renault s results to (C (), D()). We finish with applications of our results to out-splittings of graphs and to amplified graphs. 1. Introduction The relationship between orbit equivalence and isomorphism of C -algebras has been studied extensively in the last 0 years. The first result of this type was the celebrated theorem of Giordano, Putnam and Skau [5, Theorem.4], in which they showed that orbit equivalence for minimal dynamical systems on the Cantor set is equivalent to isomorphism of their corresponding crossed product C -algebras. The importance of Giordano, Putnam and Skau s result cannot be overstated. In general there is no direct method of checking whether two Cantor minimal systems are orbit equivalent. However, because the crossed product C -algebras are classifiable, Giordano, Putnam and Skau s result means that orbit equivalence can be determined using K-theory. The work in [5] has been generalised in many directions, including Tomiyama s results on topologically free dynamical systems on compact Hausdorff spaces [4], and Giordano, Matui, Putnam and Skau s extension of [5, Theorem.4] to minimal Z d -actions on the Cantor set [6]. More recently, Matsumoto and Matui have shown in [15] that two irreducible onesided topological Markov shifts (X A, σ A ) and (X B, σ B ) are continuously orbit equivalent if and only if the corresponding Cuntz-Krieger algebras O A and O B are isomorphic and det(i A) = det(i B). The proof of Matsumoto and Matui s theorem relies on two key results. The first of these is [1, Theorem 1.1], in which Matsumoto proves that the following statements are equivalent: (1) (X A, σ A ) and (X B, σ B ) are continuously orbit equivalent, () there exists a -isomorphism φ : O A O B which maps the maximal abelian subalgebra D A onto D B, and 010 Mathematics Subject Classification. Primary: 46L55; Secondary: 46L35, 37B10. Key words and phrases. C -algebra, directed graph, orbit equivalence, groupoid. This research was supported by the Australian Research Council. 1

3 NATHAN BROWNLOW, TOK MIR CARLSN, AND MICHAL. WHITTAKR (3) the topological full group of (X A, σ A ) and the topological full group of (X B, σ B ) are spatially isomorphic. (In [14, Theorem 1.1], Matsumoto showed that this is equivalent to the topological full groups being abstractly isomorphic.) The second key result is [, Proposition 4.13], which, as noticed by Matui (see [16, Theorem 5.1]), implies that there exists a -isomorphism φ : O A O B that maps the maximal abelian subalgebra, or diagonal, D A onto D B if and only if the corresponding groupoids G A and G B are isomorphic. In this paper we initiate the study of orbit equivalence of directed graphs, and we prove the analogous result to [, Proposition 4.13] for graph algebras. In particular, as part of our main result we prove that if and are two graphs in which every cycle has an exit, then the following are equivalent: (1) There is an isomorphism from C () to C ( ) which maps the diagonal subalgebra D() onto D( ). () The graph groupoids G and G are isomorphic as topological groupoids. (3) The pseudogroups of and are isomorphic. (4) The graphs and are orbit equivalent. It is natural to ask whether every cycle having an exit is necessary for our results. In our main result we in fact prove that (1) (), (3) (4) and () = (3) all hold for arbitrary directed graphs. It is only the implication (3) = () that requires that every cycle has an exit (and we provide examples that show that (3) = () does not hold in general without the assumption that every cycle has an exit). Our analysis of these implications for arbitrary graphs provides our most technical innovation, which is the introduction of a groupoid G (C (),D()) associated to (C (), D()) that we call the extended Weyl groupoid. Our construction generalises Renault s Weyl groupoid construction from [, Definition 4.11] applied to (C (), D()). We show that G (C (),D()) and G are isomorphic as topological groupoids for an arbitrary graph, which can be deduced from Renault s results in [] only when every cycle in has an exit. We conclude our paper with two applications of our main theorem. Our first application shows that if two general graphs and are conjugate then there is an isomorphism from C () to C ( ) which maps D() onto D( ). As a corollary, we strengthen a result of Bates and Pask [3, Theorem 3.] on out-splitting of graphs. Our second application adds three additional equivalences to ilers, Ruiz, and Sørensen s complete invariant for amplified graphs [4, Theorem 1.1]. In addition, we expect that our results will have applications in the study of Leavitt path algebras and further applications to graph algebras. The paper is organized as follows. Section provides background on graphs, their groupoids and their C -algebras. In Section 3 we define orbit equivalence of graphs and associate with each graph a pseudogroup which is the analogue of the topological full group Matsumoto has associated with each irreducible one-sided topological Markov shift, and we show that two graphs are orbit equivalent if and only if their pseudogroups are isomorphic. In Section 4 we construct the extended Weyl groupoid G (C (),D()) from (C (), D()), and we show that G (C (),D()) and G are isomorphic as topological groupoids. We use this result to show that if there is a diagonal-preserving isomorphism from C () to C ( ), then G and G are isomorphic as topological groupoids. In

4 GRAPH ALGBRAS AND ORBIT QUIVALNC 3 Section 5 we finish the proof of our main theorem and provide examples. inally, in Section 6 we give the two applications of our main theorem. Remark 1.1. We have learned that Xin Li has also considered orbit equivalence for directed graphs, and has independently proved that two graphs in which every cycle has an exit are orbit equivalent if and only if there is a diagonal-preserving isomorphism between their C*-algebras. Acknowledgement. We would like to thank the referee for several suggestions which have improved the paper.. Background on the groupoids and C -algebras of directed graphs We begin with some background on graphs and their C -algebras. In this section we recall the definitions of the boundary path space of a directed graph, graph C -algebras and graph groupoids..1. Graphs and their C -algebras. We refer the reader to [19] for a more detailed treatment on graphs and their C -algebras. However, we note that the directions of arrows defining a graph are reversed in this paper. We used this convention so that our results can easily be compared with the work of Matsumoto and Matui s work on shift spaces. A directed graph (also called a quiver) = ( 0, 1, r, s) consists of countable sets 0 and 1, and range and source maps r, s : 1 0. The elements of 0 are called vertices, and the elements of 1 are called edges. A path µ of length n in is a sequence of edges µ = µ 1... µ n such that r(µ i ) = s(µ i+1 ) for all 1 i n 1. The set of paths of length n is denoted n. We denote by µ the length of µ. The range and source maps extend naturally to paths: s(µ) := s(µ 1 ) and r(µ) := r(µ n ). We regard the elements of 0 as path of length 0, and for v 0 we set s(v) := r(v) := v. or v 0 and n N we denote by v n the set of paths of length n with source v, and by n v the paths of length n with range v. We define := n N n to be the collection of all paths with finite length. or v, w 0 let v w := {µ : s(µ) = v and r(µ) = w}. We define 0 reg := {v 0 : v 1 is finite and nonempty} and 0 sing := 0 \ 0 reg. If µ = µ 1 µ µ m, ν = ν 1 ν ν n and r(µ) = s(ν), then we let µν denote the path µ 1 µ µ m ν 1 ν ν n. A cycle in is a path µ such that µ 1 and s(µ) = r(µ). If µ is a cycle and k is a positive integer, then µ k denotes the cycle µµ µ where µ is repeated k-times. We say that the cycle µ is simple if µ is not equal to ν k for any cycle ν and any integer k. Notice than any cycle µ is equal to ν k for some simple cycle ν and some positive integer k. An edge e is an exit to the cycle µ if there exists i such that s(e) = s(µ i ) and e µ i. A graph is said to satisfy condition (L) if every cycle has an exit. A Cuntz-Krieger -family {P, S} consists of a set of mutually orthogonal projections {P v : v 0 } and partial isometries {S e : e 1 } satisfying (CK1) S e S e = P r(e) for all e 1 ; (CK) S e S e P s(e) for all e 1 ; (CK3) P v = e v 1 S e S e for all v 0 reg.

5 4 NATHAN BROWNLOW, TOK MIR CARLSN, AND MICHAL. WHITTAKR The graph C -algebra C () is the universal C -algebra generated by a Cuntz-Krieger -family. We denote by {p, s} the Cuntz-Krieger -family generating C (). There is a strongly continuous action γ : C () T, called the gauge action, satisfying γ z (p v ) = p v and γ z (s e ) = zs e, for all z T, v 0, e 1. If {Q, T } is a Cuntz-Krieger -family in a C -algebra B, then we denote by π Q,T the homomorphism C () B such that π Q,T (p v ) = Q v for all v 0, and π Q,T (s e ) = T e for all e 1. an Huef and Raeburn s gauge invariant uniqueness theorem [7] says that π Q,T is injective if and only if there is an action β of T on the C -algebra generated by {Q, T } satisfying β z (Q v ) = Q v and β z (T e ) = zt e, for all z T, v 0, e 1, and Q v 0 for all v 0. If µ = µ 1 µ n n and n, then we let s µ := s µ1 s µn. Likewise, we let s v := p v if v 0. Then C () = span{s µ s ν : µ, ν, r(µ) = r(ν)}. The C -subalgebra D() := span{s µ s µ : µ } of C () is a maximal abelian subalgebra if and only if every cycle in has an exit (see [17, xample 3.3])... The boundary path space of a graph. An infinite path in is an infinite sequence x 1 x... of edges in such that r(e i ) = s(e i+1 ) for all i. We let be the set of all infinite paths in. The source map extends to in the obvious way. We let x = for x. The boundary path space of is the space := {µ : r(µ) 0 sing}. If µ = µ 1 µ µ m, x = x 1 x and r(µ) = s(x), then we let µx denote the infinite path µ 1 µ µ m x 1 x. or µ, the cylinder set of µ is the set Z(µ) := {µx : x r(µ) }, where r(µ) := {x : r(µ) = s(x)}. Given µ and a finite subset r(µ) 1 we define ( ) Z(µ \ ) := Z(µ) \ Z(µe). The boundary path space is a locally compact Hausdorff space with the topology given by the basis {Z(µ \ ) : µ, is a finite subset of r(µ) 1 }, and each such Z(µ \ ) is compact and open (see [5, Theorem.1 and Theorem.]). Moreover, [5, Theorem 3.7] shows that there is a unique homeomorphism h from to the spectrum of D() given by { (.1) h (x)(s µ s 1 if x Z(µ), µ) = 0 if x / Z(µ). Our next lemma gives a description of the topology on the boundary path space, which we will need in the proof of Proposition 3.3. Lemma.1. very nonempty open subset of is the disjoint union of sets that are both compact and open. Proof. Let U be a nonempty open subset of. or each x U let B x := {(µ, ) : µ, is a finite subset of r(µ) 1, x Z(µ \ ) U}. e

6 GRAPH ALGBRAS AND ORBIT QUIVALNC 5 If (µ, ) B x, then x Z(µ) and x / Z(µe) for each e. Let µ x be the shortest µ such that (µ, ) B x for some finite subset of r(µ) 1, and let x := { : (µ x, ) B x }. Then (µ x, x ) B x and Z(µ \ ) Z(µ x \ x ) for all (µ, ) B x. It follows that if x, y U, then either Z(µ x \ x ) = Z(µ y \ y ) or Z(µ x \ x ) Z(µ y \ y ) =. Since U = x U Z(µ x \ x ) and each Z(µ x \ x ) is open and compact, this shows that U is the disjoint union of sets that are both compact and open. or n N, let n := {x : x n}. Then n = µ nz(µ) is an open subset of. We define the shift map on to be the map σ : 1 given by σ (x 1 x x 3 ) = x x 3 for x 1 x x 3 and σ (e) = r(e) for e 1. or n 1, we let σ n be the n-fold composition of σ with itself. We let σ 0 denote the identity map on. Then σ n is a local homeomorphism for all n N. When we write σ n (x), we implicitly assume that x n. We say that x is eventually periodic if there are m, n N, m n such that σ m(x) = σn (x). Notice that x is eventually periodic if and only if x = µννν for some path µ and some cycle ν with s(ν) = r(µ). By replacing ν by a subcycle if necessary, we can assume that ν is a simple cycle..3. Graph groupoids. In [11], Kumjian, Pask, Raeburn, and Renault defined groupoid C -algebras associated to a locally-finite directed graph with no sources. Their construction has been generalized to compactly aligned topological k-graphs in [6]. We will now explain this construction in the case that is an arbitrary graph. The resulting groupoid is isomorphic to the one constructed by Paterson in [18]. Let G := {(x, m n, y) : x, y, m, n N, and σ m (x) = σ n (y)}, with product (x, k, y)(w, l, z) := (x, k + l, z) if y = w and undefined otherwise, and inverse given by (x, k, y) 1 := (y, k, x). With these operations G is a groupoid (cf. [11, Lemma.4]). The unit space G 0 of G is {(x, 0, x) : x } which we will freely identify with via the map (x, 0, x) x throughout the paper. We then have that the range and source maps r, s : G are given by r(x, k, y) = x and s(x, k, y) = y. We now define a topology on G. Suppose m, n N and U is an open subset of m such that the restriction of σ m to U is injective, V is an open subset of n such that the restriction of σ n to V is injective, and that σm (U) = σn (V ), then we define (.) Z(U, m, n, V ) := {(x, k, y) G : x U, k = m n, y V, σ m (x) = σ n (y)}. Then G is a locally compact, Hausdorff, étale topological groupoid with the topology generated by the basis consisting of sets Z(U, m, n, V ) described in (.), see [11, Proposition.6] for an analogous situation. or µ, ν with r(µ) = r(ν), let Z(µ, ν) := Z(Z(µ), µ, ν, Z(ν)). It follows that each Z(µ, ν) is compact and open, and that the topology inherits when we consider it as a subset of G by identifying it with {(x, 0, x) : x } agrees with the topology described in the previous section. Notice that for all µ, ν, U a compact open subset of Z(µ), and V a compact open subset of Z(ν), the collection {Z(U, µ, ν, V ) : σ µ (U) = σ ν (V )} is a basis for the topology of G. According to [6, Proposition 6.], G is topologically amenable in the sense of [1, Definition..8]. It follows from [1, Proposition 3.3.5] and [1, Proposition 6.1.8] that the reduced and universal C -algebras of G are equal, and we denote this C -algebra by C (G ).

7 6 NATHAN BROWNLOW, TOK MIR CARLSN, AND MICHAL. WHITTAKR Proposition. (Cf. [11, Proposition 4.1]). Suppose is a graph. Then there is a unique isomorphism π : C () C (G ) such that π(p v ) = 1 Z(v,v) for all v 0 and π(s e ) = 1 Z(e,r(e)) for all e 1, and such that π(d()) = C 0 (G 0 ). Proof. Using calculations along the lines of those used in the proof of [11, Proposition 4.1], it is straight forward to check that {Q, T } := {Q v := 1 Z(v,v) and T e := 1 Z(e,r(e)) : v 0, e 1 } is a Cuntz-Krieger -family. The universal property of {p, s} implies that there is a -homomorphism π := π Q,T : C () C (G ) satisfying π(p v ) = Q v and π(s e ) = T e. An argument similar to the one used in the proof of [11, Proposition 4.1] shows that C (G ) is generated by {Q, T }, so π is surjective. The cocycle (x, k, y) k induces an action β of T on C (G ) satisfying β z (Q v ) = Q v and β z (T e ) = zt e, for all z T, v 0, e 1 (see [1, Proposition II.5.1]), and since Q v = 1 Z(v,v) 0 for all v 0, the gauge invariant uniqueness theorem of C (G ) ([, Theorem.1]) implies that π is injective. Since D() is generated by {s µ s µ : µ } and π(s µ s µ) = 1 Z(µ,µ), we have that π maps D() into C 0 (G 0 ). An application of the Stone-Weierstrass theorem implies that C 0 (G 0 ) is generated by {1 Z(µ,µ) : µ }. Hence π(d()) = C 0 (G 0 ). Suppose G is a groupoid, the isotropy group of x G 0 is the group Iso(x) := {γ G : s(γ) = r(γ) = x}. In [], an étale groupoid is said to be topologically principal if the set of points of G 0 with trivial isotropy group is dense. We will now characterize when G is topologically principal. Proposition.3. Let be a graph. Then the graph groupoid G is topologically principal if and only if every cycle in has an exit. Proof. Let x. We claim that (x, 0, x) has nontrivial isotropy group if and only if x is eventually periodic. Indeed, suppose (x, m n, x) Iso(x) with m n, then σ m (x) = σ n (x) and x is eventually periodic. On the other hand, suppose x = µλ, then (x, ( µ + λ ) µ, x) Iso(x), proving the claim. Now observe that if v is a vertex such that there are two different simple cycles α and β with s(α) = s(β) = v, then any cylinder set Z(δ) for which r(δ) v contains a y such that (y, 0, y) has trivial isotropy. To see this, pick λ r(δ) v, then y = δλαβα βα 3 β has trivial isotropy since it is not eventually periodic. Assume that every cycle in has an exit and suppose for contradiction that U is an nonempty open subset of such that (x, 0, x) has nontrivial isotropy group for every x U. Note that U since y with y < implies that the isotropy group of (y, 0, y) is trivial. Let x U. Since x has nontrivial isotropy group, there exist ζ 1 and a cycle η such that ζ 1 η Z(ζ 1 η k ) U for some k N. Since η has an exit and (x, 0, x) has nontrivial isotropy group for every x U, it follows that there is a ζ r(ζ 1 ) such that Z(ζ 1 ζ ) U and such that r(ζ ) r(ζ 1 ) =, for otherwise there would be two distinct simple cycles based at r(ζ 1 ). By repeating this argument we get a sequences of paths ζ 1, ζ, ζ 3,... such that s(ζ n+1 ) = r(ζ n ), r(ζ n+1 ) r(ζ n ) = and Z(ζ 1 ζ... ζ n ) U for all n. The element y = ζ 1 ζ ζ 3... then belongs to U, but since it only visits each vertex a finite number of times, (y, 0, y) must have trivial isotropy, which contradicts the assumption that (x, 0, x) has nontrivial isotropy group for every x U. Thus, G is topologically principal if every cycle in has an exit.

8 GRAPH ALGBRAS AND ORBIT QUIVALNC 7 Conversely, if µ is a cycle without exit and x = µµµ..., then (x, 0, x) is an isolated point in G 0 with nontrivial isotropy group. Thus, G is not topologically principal if there is a cycle in without an exit. Since the reduced and universal C -algebras of G are equal, it follows from [1, Proposition II.4.(i)] that we can regard C (G ) as a subset of C 0 (G ). or f C (G ) and j Z, we let Φ j (f) denote the restriction of f to {(x, k, y) G : k = j}, and for m N we let Σ m (f) := m j j= m (1 )Φ m+1 j(f). Proposition.4. Let be a graph and let f C (G ). Then each Φ k (f) and each Σ m (f) belong to C (G ), and (Σ m (f)) m N converges to f in C (G ). Proof. Let j Z. The map (x, k, y) k is a continuous cocycle from G to Z. or each z T there is a unique automorphism γ z on C (G ) such that γ z (g)(x, k, y) = z k g(x, k, y) for g C (G ) and (x, k, y) G, and that the map z γ z is a strongly continuous action of T on C (G ) (see [1, Proposition II.5.1]). It follows that the integral γ T z(f)z j dz, where dz denotes the normalized Haar measure on T, is welldefined and belongs to C (G ) (see for example [0, Section C.]). Let (x, k, y) G. If k j, then γ z (f)z j dz(x, k, y) = z k j dz f(x, k, y) = 0, T and if k = j, then γ z (f)z j dz(x, k, y) = T T T z k j dz f(x, k, y) = f(x, k, y). Thus, Φ j (f) = T γ z(f)z j dz from which it follows that Φ j (f) C (G ). Since each Σ m (f) is a linear combination of functions of the form Φ j (f), each Σ m (f) belongs to C (G ). or m N, let σ m : T R be the ejér s kernel defined by σ m (z) = m (1 j m + 1 )z j. j= m Then σ m (z) 0 for all z T, T σ m(z) dz = 1, and Σ m (f) = = m j= m m j= m ( 1 j m + 1 ) ( 1 j m + 1 ) Φ j (f) γ z (f)z j dz = T T γ z (f)σ m (z) dz. Thus Σ m (f) γ z (f) σ m (z)dz = f. T

9 8 NATHAN BROWNLOW, TOK MIR CARLSN, AND MICHAL. WHITTAKR If g C c (G ), then there is an m 0 N such that Φ j (g) = 0 for j > m 0. sufficiently large, it follows that m ( g Σ m (g) = g 1 j ) Φ j (g) m + 1 j= m m m ( ) j g Φ j (g) + Φ j (g) m + 1 j= m j= m j= m m ( ) j Φ j (g) m + 1 m 0 ( ) j Φ j (g) for m m 0 m + 1 j= m 0 m 0 j= m 0 j m + 1 Φ j(g) 0 as m. or m Thus, for any ɛ > 0 there exists g C c (G ) and an M N such that f g < ɛ/3 and g Σ m (g) < ɛ/3 for any m M, and then f Σ m (f) f g + g Σ m (g) + Σ m (g f) < ɛ for any m M. This shows that (Σ m (f)) m N converges to f in C (G ). 3. Orbit equivalence and pseudogroups In this section we introduce the notion of orbit equivalence of two graphs, which is a natural generalisation of Matsumoto s continuous orbit equivalence for topological Markov shifts from [1]. We also define the pseudogroup of a graph using Renault s pseudogroups associated to groupoids [], and then show that two graphs are orbit equivalent if and only if their pseudogroups are isomorphic. Definition 3.1. Two graphs and are orbit equivalent if there exist a homeomorphism h : and continuous functions k 1, l 1 : 1 N and k 1, l 1 : 1 N such that (3.1) σ k 1(x) (h(σ (x))) = σ l 1(x) for all x 1, y 1. xample 3.. Consider the graphs (h(x)) and σ k 1 (y) (h 1 (σ (y))) = σ l 1 (y) (h 1 (y)), e f e 1 f Then = {e 1 e e..., e e... } and = {f 1 f f 1 f..., f f 1 f f 1... }. The map h : given by h(e 1 e e... ) = f 1 f f 1 f... and h(e e... ) = f f 1 f f 1...

10 GRAPH ALGBRAS AND ORBIT QUIVALNC 9 is a homeomorphism. Consider k 1, l 1 : 1 N given by k 1 (e 1 e e... ) = 1 and k 1 (e e... ) = 0, and l 1 (e 1 e e... ) = 0 = l 1 (e e... ). Then k 1 and l 1 are continuous, and we have σ k 1(x) (h(σ (x))) = σ l 1(x) (h(x)) for all x 1. Similarly the functions k 1, l 1 : 1 N given by k 1(f 1 f f 1 f... ) = 0 and k 1(f f 1 f f 1... ) = 1, and l 1(f 1 f f 1 f... ) = 1 and l 1(f f 1 f f 1... ) = 0, are continuous and satisfy σ k 1 (y) (h 1 (σ (y))) = σ l 1 (y) (h 1 (y)) for all y 1. Hence and are orbit equivalent. Sections 5 and 6 contain further examples of orbit equivalent graphs. In Section 3 of [], Renault constructs for each étale groupoid G a pseudogroup in the following way: Define a bisection to be a subset A of G such that the restriction of the source map of G to A and the restriction of the range map of G to A both are injective. The set of all open bisections of G forms an inverse semigroup S with product defined by AB = {γγ : (γ, γ ) (A B) G () } (where G () denote the set of composable pairs of G), and the inverse of A is defined to be the image of A under the inverse map of G. ach A S defines a unique homeomorphism α A : s(a) r(a) such that α(s(γ)) = r(γ) for γ A. The set {α A : A S} of partial homeomorphisms on G 0 is the pseudogroup of G. When is a graph, then we call the pseudogroup of the étale groupoid G the pseudogroup of and denote it by P. We will now give two alternative characterizations of the partial homeomorphisms of that belong to P. Proposition 3.3. Let be a graph, let U and V be open subsets of, and let α : V U be a homeomorphism. Then the following are equivalent: (1) α P. () or all x V, there exist m, n N and an open subset V such that x V V, and such that σ m(x ) = σ n (α(x )) for all x V. (3) There exist continuous functions m, n : V N such that σ m(x) (x) = σ n(x) (α(x)) for all x V. Proof. (1) (): Suppose α P. Let A S be such that α = α A. Let x V. Then there is a unique γ A such that s(γ) = x, and then r(γ) = α(x). Since A is an open subset of G, there are m, n N, an open subset U of m such that the restriction of σ m to U is injective, and an open subset V of n such that the restriction of σ n to V is injective and σ m(u ) = σ n (V ), and such that γ Z(U, m, n, V ) A. Then x V V and σ m(x ) = σ n (α(x )) for all x V. () = (3): Assume that for all x V, there exist m, n N and an open subset V such that x V V, and such that σ m(x ) = σ n (α(x )) for all x V. According to Lemma.1, V is the disjoint union of sets that are both compact and open. Since is locally compact, it follows that there exists a family {V i : i I} of mutually disjoint compact and open sets and a family {(m i, n i ) : i I} of pairs of nonnegative integers such that V = i I V i and σ m i (x) = σn i (α(x)) for x V i. Define m, n : V N by setting m(x) = m i and n(x) = n i for x V i. Then m and n are continuous and (x) = σ n(x) (α(x)) for all x V. σ m(x)

11 10 NATHAN BROWNLOW, TOK MIR CARLSN, AND MICHAL. WHITTAKR (3) = (1): Assume that m, n : V N are continuous functions such that σ m(x) (x) = σ n(x) (α(x)) for all x V. Then there exist for each x V a compact and open subset V x such that x V x V, m(x ) = m(x) and n(x ) = n(x) for all x V x, the restriction of σ n(x) to V x is injective, and the restriction of σ m(x) of α(v x ) is injective. According to Lemma.1, V is the disjoint union of sets that are both compact and open. It follows that there exists a family {V i : i I} of mutually disjoint compact and open sets and a family {(m i, n i ) : i I} of pairs of nonnegative integers such that V = i I V i, m(x) = m i and n(x) = n i for all x V i, the restriction of σ n i to V i is injective, and the restriction of σ m i of α(v i) is injective. Let A := i I Z(α(V i), m i, n i V i ). Then A S and α = α A, so α P. Suppose that and are two graphs and that there exists a homeomorphism h :. Let U and V be open subsets of and let α : V U be a homeomorphism. We denote by h P h 1 := {h α h 1 : α P }. We say that the pseudogroups of and are isomorphic if there is a homeomorphism h : such that h P h 1 = P. We can now state the main result of this section. Proposition 3.4. Let and be two graphs. Then and are orbit equivalent if and only if the pseudogroups of and are isomorphic. To prove this proposition we will use the following result. Lemma 3.5. Suppose two graphs and are orbit equivalent, h : is a homeomorphism and k 1, l 1 : 1 N and k 1, l 1 : 1 N are continuous functions satisfying (3.1). Let n N. Then there exist continuous functions k n, l n : n N and k n, l n : n N such that (3.) σ kn(x) (h(σ(x))) n = σ ln(x) (h(x)) and σ k n (y) (h 1 (σ n (y))) = σ l n (y) (h 1 (y)), for all x n, y n. Proof. There is nothing to prove for n = 0 and n = 1. We will prove the general case by induction. Let m 1 and suppose that the lemma holds for n = m. Let x m+1. Then σ k 1(σ m(x)) (h(σ m+1 (x))) = σ l 1(σ m(x)) (h(σ m (x))) and Let (3.3) (3.4) Then σ km(x) (h(σ m (x))) = σ lm(x) (h(x)). k m+1 (x) := k 1 (σ m (x)) + max{l 1 (σ m (x)), k m (x)} l 1 (σ m (x)) l m+1 (x) := l m (x) + max{l 1 (σ m (x)), k m (x)} k m (x). σ k m+1(x) (h(σ m+1 (x))) = σ l m+1(x) (h(x)). Since k 1, l 1, k m, and l m are continuous, it follows that k m+1, l m+1 : m+1 N defined by (3.3) and (3.4) are also continuous. Similarly, if we define k m+1, l m+1 : m+1 N by letting k m+1(y) := k 1(σ m (y)) + max{l 1(σ m (y)), k m(y)} l 1(σ m (y)) l m+1(y) := l m (y) + max{l 1(σ m (y)), k m(y)} k m(y)

12 GRAPH ALGBRAS AND ORBIT QUIVALNC 11 for y m+1, then k m+1 and l m+1 are continuous, and σ k m+1 (y) (h 1 (σ m+1 (y))) = σ l m+1 (y) (h 1 (y)) for all y m+1. Thus, the lemma also holds for n = m + 1, and the general result holds by induction. Proof of Proposition 3.4. Suppose and are orbit equivalent. Then there exists a homeomorphism h : and, for each n N, there exists continuous functions k n, l n : n N satisfying the first equation of (3.). Let (α : V U) P, and let m, n : V N be continuous functions such that σ m(x) Let y h(v ). Then and σ l n(h 1 (y)) (α(h 1 (y))) So if we let σ k m(h 1 (y)) (h 1 (y)) (h(α(h 1 (y)))) = σ k n(h 1 (y)) (α(h 1 (y))) (h(σ m(h 1 (y)) = σ k n(h 1 (y)) (α(h 1 (y))) (x) = σ n(x) (α(x)) for all x V. (h(σ n(h 1 (y)) (α(h 1 (y))))) (h(σ m(h 1 (y)) (h 1 (y)))), (h 1 (y)))) = σ l m(h 1 (y)) (h 1 (y)) (y). (3.5) m (y) := l m(h 1 (y))(h 1 (y)) + max{k n(h 1 (y))(α(h 1 (y))), k m(h 1 (y))(h 1 (y))} and k m(h 1 (y))(h 1 (y)) (3.6) n (y) := l n(h 1 (y))(α(h 1 (y))) + max{k n(h 1 (y))(α(h 1 (y))), k m(h 1 (y))(h 1 (y))} k n(h 1 (y))(α(h 1 (y))), then σ m (y) (y) = σ n (y) (h(α(h 1 (y)))). Since h 1, m, n, and α are continuous, it follows that m, n : h(v ) N defined by (3.5) and (3.6) are also continuous. Thus, it follows from Proposition 3.3 that h α h 1 P. A similar argument proves that if α P, then h 1 α h P. Thus h P h 1 = P and the pseudogroups of and are isomorphic. Now suppose that h : is a homeomorphism such that h P h 1 = P. ix e 1 and let α e := σ Z(e). Then α e is a homeomorphism from Z(e) to α e (Z(e)) and since α e (x) = σ (x) for all x Z(e), it follows from Proposition 3.3 that α e P. Thus h α e h 1 P by assumption. It follows from Proposition 3.3 that there are continuous functions m e, n e : h(z(e)) N such that σ n e (y) (h(α e (h 1 (y)))) = σ m e (y) (y) for y h(z(e)). Define functions k 1, l 1 : 1 N by k 1 (x) = n x 1 (h(x)) and l 1 (x) = m x 1 (h(x)), which are continuous because the Z(e) are pairwise-disjoint compact open sets covering 1. Then for each x = x 1 x we have σ l 1(x) (h(x)) = σ m x 1 (h(x)) (h(x)) = σ n x 1 (h(x)) (h(α x1 (x))) = σ k 1(x) (h(σ (x))). Hence k 1 and l 1 satisfy the first equation from (3.1). A similar argument gets the second equation from (3.1). Thus and are orbit equivalent.

13 1 NATHAN BROWNLOW, TOK MIR CARLSN, AND MICHAL. WHITTAKR 4. The extended Weyl groupoid of (C (), D()) Proposition. says that the pair (C (), D()) is an invariant of G, in the sense that if and are two graphs such that G and G are isomorphic as topological groupoids, then there is an isomorphism from C () to C ( ) which maps D() onto D( ). In this section we show that G is an invariant of (C (), D()), in the sense that if there is an isomorphism from C () to C ( ) which maps D() onto D( ), then G and G are isomorphic as topological groupoids. To prove this result we build a groupoid from (C (), D()) that we call the extended Weyl groupoid, which generalises Renault s Weyl groupoid construction from [] applied to (C (), D()). Recall from [] that Weyl groupoids are associated to pairs (A, B) consisting of a C -algebra A and an abelian C -subalgebra B which contains an approximate unit of A. The Weyl groupoid construction has the property that if G is a topologically principal étale Hausdorff locally compact second countable groupoid and A = C red (G) and B = C 0(G 0 ), then the associated Weyl groupoid is isomorphic to G as a topological groupoid. We will modify Renault s construction for pairs (C (), D()) to obtain a groupoid G (C (),D()) such that G (C (),D()) and G are isomorphic as topological groupoids, even when G is not topologically principal. We will then show that if and are two graphs such that there is an isomorphism from C () to C ( ) which maps D() onto D( ), then G (C (),D()) and G (C ( ),D( )), and thus G and G are isomorphic as topological groupoids. As in [] (and originally defined in [9]), we define the normaliser of D() to be the set N(D()) := {n C () : ndn, n dn D() for all d D()}. According to [, Lemma 4.6], nn, n n D() for n N(D()). Recalling the definition of h given in (.1), for n N(D()), we let dom(n) := {x : h (x)(n n) > 0} and ran(n) := {x : h (x)(nn ) > 0}. It follows from [, Proposition 4.7] that, for n N(D()), there is a unique homeomorphism α n : dom(n) ran(n) such that, for all d D(), (4.1) h (x)(n dn) = h (α n (x))(d)h (x)(n n). rom [, Lemma 4.10] we also know that α n = αn 1 and α mn = α m α n for each m, n N(D()). The following lemma gives an insight into how the homeomorphisms α n work. We collect further properties of these homeomorphisms in Lemma 4.. Lemma 4.1. Let be a graph. or each µ, ν with r(µ) = r(ν) we have s µ s ν N(D()) with dom(s µ s ν) = Z(ν), ran(s µ s ν) = Z(µ) and α sµs (νz) = µz for all z r(ν). ν Proof. Let µ, ν with r(µ) = r(ν). or each λ we have s ν s ν if µ = λµ (4.) (s µ s ν) s λ s λ(s µ s ν) = s νλ s νλ if λ = µλ 0 otherwise. So (s µ s ν) s λ s λ (s µs ν) D(), and it follows that (s µ s ν) d(s µ s ν) D() for all d D(). A similar argument shows that (s µ s ν)d(s µ s ν) D() for all d D(), and

14 hence s µ s ν N(D()). We have GRAPH ALGBRAS AND ORBIT QUIVALNC 13 h (x)((s µ s ν) s µ s ν) = h (x)(s ν s ν) = { 1 if x Z(ν) 0 if x Z(ν), and hence dom(s µ s ν) = Z(ν). A similar calculation gives ran(s µ s ν) = Z(µ). Now suppose z r(ν). We use (4.1) and (4.) to get h (α sµs (νz))(s ν λs λ) = h (νz) ( (s µ s ν) s λ s λ(s µ s ν) ) h (νz) ( (s µ s ν) s µ s ν h (νz)(s ν s ν) if µ = λµ = h (νz)(s νλ s νλ ) if λ = µλ 0 otherwise { 1 if µz Z(λ) = 0 otherwise = h (µz)(s λ s λ). It follows that h (α sµs (νz)) = h ν (µz), and hence α sµs (νz) = µz. ν ) Denote by iso the set of isolated points in. Notice that x belongs to iso if and only if the characteristic function 1 {x} belongs to C 0 ( ). or x iso, we let p x denote the unique element of D() satisfying that h (y)(p x ) is 1 if y = x and zero otherwise. Lemma 4.. Let be a graph, n N(D()) and x iso dom(n). Then (a) np x n = h (x)(n n)p αn(x), (b) n p αn(x)n = h (x)(n n)p x, and (c) np x = p αn(x)n. Proof. We use quation (4.1) with d = nn to get which implies that h (x)(n n) = h (x)(n nn n) = h (α n (x))(nn )h (x)(n n), (4.3) h (α n (x))(nn ) = h (x)(n n). Note that this is a positive number because x dom(n). or (a) we again use (4.1) to get h (y) (( h (α n (x))(nn ) ) 1 npx n ) = ( h (α n (x))(nn ) ) 1 h (y)(np x n ) = ( h (α n (x))(nn ) ) 1 h (α n (y))(p x )h (y)(nn ) { 1 if y = α n (x) = 0 otherwise. By the defining property of p αn(x) we now have p αn(x) = ( h (α n (x))(nn ) ) 1 npx n. Using (4.3) gives np x n = h (x)(n n)p αn(x), which is (a). Identity (b) follows from (a) by replacing n with n and x with α n (x) and then use (4.3).

15 14 NATHAN BROWNLOW, TOK MIR CARLSN, AND MICHAL. WHITTAKR To prove (c) we first notice that h (y) ( (h (x)(n n)) 1 n np x ) = (h (x)(n n)) 1 h (y)(n n)h (y)(p x ) = Hence by the defining property of p x we have (4.4) n np x = h (x)(n n)p x. We now use (4.4) and (a) to get (c): { 1 if x = y 0 if x y. np x = n((h (x)(n n)) 1 n np x ) = (h (x)(n n)) 1 np x n n = p αn(x)n. Lemma 4.3. Suppose x iso. If x is not eventually periodic, then p x C ()p x = p x D()p x = Cp x. If x = µηηη for some µ and some simple cycle η with s(η) = r(µ), then p x C ()p x is isomorphic to C(T) by the isomorphism mapping p x s µ s η s µp x to the identity function on T, and p x D()p x = Cp x. Proof. Let (G ) x x denote the isotropy group {γ G : s(γ) = r(γ) = x} of (x, 0, x). Assume that x is not eventually periodic. Then (G ) x x = {(x, 0, x)}. Proposition. implies that there is an isomorphism from p x C ()p x to C ((G ) x x), and consequently p x C ()p x = p x D()p x = Cp x, completing the first assertion in the lemma. Assume then that x = µηηη for some µ and some simple cycle η with s(η) = r(µ). We then have that (G ) x x = {(x, k η, x) : k Z}. Now Proposition. implies that there is an isomorphism from p x C ()p x to C(T) which maps p x s µ s η s µp x to the identity function on T, and that p x D()p x = Cp x. The extended Weyl groupoid associated to (C (), D()) is built using an equivalence relation defined on pairs of normalisers and boundary paths. or isolated boundary paths x the equivalence relation is defined using a unitary in the corner of C () determined by p x. Lemma 4.4. Let be a graph. or x iso, n 1, n N(D()), x dom(n 1 ) dom(n ), and α n1 (x) = α n (x), we denote Then U (x,n1,n ) := (h (x)(n 1n 1 n n )) 1/ p x n 1n p x. (1) U (x,n1,n )U (x,n 1,n ) = U (x,n 1,n ) U (x,n 1,n ) = p x, and () U (x,n 1,n ) = U (x,n,n 1 ). Moreover, if n 3 N(D()), x dom(n 3 ), and α n3 (x) = α n1 (x) = α n (x), then (3) U (x,n1,n )U (x,n,n 3 ) = U (x,n1,n 3 ). Proof. Suppose that x iso, n 1, n N(D()), x dom(n 1 ) dom(n ), and α n1 (x) = α n (x). irst note that since x dom(n 1 ) dom(n ), we have h (n 1n 1 ), h (n n ) > 0, and the formula for U (x,n1,n ) makes sense. We now claim that (4.5) p x n 1n p x n n 1 p x = h (x)(n 1n 1 n n )p x.

16 GRAPH ALGBRAS AND ORBIT QUIVALNC 15 To see this, we apply identities (a) and (b) of Lemma 4. to get p x n 1n p x n n 1 p x = p x n 1(n p x n )n 1 p x ) = p x n h (x)(n n )p αn (x) n1 p x = h (x)(n n )p x n 1p αn1 (x)n 1 p x We now use (4.5) to get = h (x)(n 1n 1 )h (x)(n n )p x = h (x)(n 1n 1 n n )p x. U (x,n1,n )U (x,n 1,n ) = (h (n 1n 1 n n )) 1 p x n 1n p x n n 1 p x = p x Similarly, and using that p x C ()p x is commutative, we have U (x,n 1,n )U (x,n1,n ) = (h (n 1n 1 n n )) 1 p x n n 1 p x n 1n p x = (h (n 1n 1 n n )) 1 p x n 1n p x n n 1 p x = p x. So (1) holds. Identity () holds because n 1n 1, n n D(), and hence U (x,n 1,n ) := (h (x)(n 1n 1 n n )) 1/ p x n n 1 p x = (h (x)(n n n 1n 1 )) 1/ p x n n 1 p x = U (x,n,n 1 ). We use identities (a) and (c) of Lemma 4. to get U (x,n1,n )U (x,n,n 3 ) = (h (x)(n 1n 1 n n )h (x)(n n n 3n 3 )) 1/ p x n 1n p x n n 3 p x So (3) holds. = (h (x)(n n )) 1 (h (x)(n 1n 1 n 3n 3 )) 1/ p x n 1( h (x)(n n )p αn (x) ) n3 p x = (h (n 1n 1 n 3n 3 )) 1/ p x n 1p αn3 (x)n 3 p x = (h (n 1n 1 n 3n 3 )) 1/ p x n 1n 3 p x = U (x,n1,n 3 ). Notation 4.5. or x, n 1 and n as in Lemma 4.4 we denote λ (x,n1,n ) := h (x)(n 1n 1 n n ). So U (x,n1,n ) = λ 1/ (x,n 1,n ) p xn 1n p x. It follows from identity (1) of Lemma 4.4 that U (x,n1,n ) is a unitary element of p x C ()p x. We denote by [U (x,n1,n )] 1 the class of U (x,n1,n ) in K 1 (p x C ()p x ). Notice that since p x C ()p x is commutative (see Lemma 4.3), [U (x,n1,n )] 1 = 0 if and only if U (x,n1,n ) is homotopic to p x. Proposition 4.6. Let be a graph. or each x 1, x and n 1, n N(D()) such that x 1 dom(n 1 ) and x dom(n ) we write (n 1, x 1 ) (n, x ) if either (a) x 1 = x iso, α n1 (x 1 ) = α n (x ), and [U (x1,n 1,n )] 1 = 0; or (b) x 1 = x / iso and there is an open set V such that x 1 V dom(n 1 ) dom(n ) and α n1 (y) = α n (y) for all y V. Then is an equivalence relation on {(n, x) : n N(D()), x dom(n)}.

17 16 NATHAN BROWNLOW, TOK MIR CARLSN, AND MICHAL. WHITTAKR Proof. The only nontrivial parts to prove are that is symmetric and transitive when the boundary paths are isolated points. Suppose (n 1, x 1 ) (n, x ) with x := x 1 = x iso. We know from Lemma 4.4() that U (x,n,n 1 ) = U (x,n 1,n ). So [U (x,n1,n )] 1 = 0 = [U (x,n,n 1 )] 1 = [U (x,n 1,n )] 1 = 0, and hence (n, x ) (n 1, x 1 ). or transitivity, suppose (n 1, x 1 ) (n, x ) and (n, x ) (n 3, x 3 ) with x := x 1 = x = x 3 iso. We know from Lemma 4.4() that U (x,n1,n )U (x,n,n 3 ) = U (x,n1,n 3 ). So [U (x,n1,n )] 1 = 0 = [U (x,n,n 3 )] 1 = [U (x,n1,n 3 )] 1 = [U (x,n1,n )] 1 [U (x,n,n 3 )] 1 = 0, and hence (n 1, x 1 ) (n 3, x 3 ). Proposition 4.7. Let be a graph, and the equivalence relation on {(n, x) : n N(D()), x dom(n)} from Proposition 4.6. Denote the collection of equivalence classes by G (C (),D()). Define a partially-defined product on G (C (),D()) by [(n 1, x 1 )][(n, x )] := [(n 1 n, x )] if α n (x ) = x 1, and undefined otherwise. Define an inverse map by [(n, x)] 1 := [(n, α n (x))]. Then these operations make G (C (),D()) into a groupoid. Proof. We only check that composition and inversion are well-defined. That composition is associative and every element is composable with its inverse (in either direction) is left to the reader. To see that composition is well-defined, suppose [(n 1, x 1 )] = [(n 1, x 1)] and [(n, x )] = [(n, x )] with [(n 1, x 1 )] and [(n, x )] composable. We need to show that [(n 1, x 1)] and [(n, x )] are also composable with (4.6) [(n 1 n, x )] = [(n 1n, x )]. We immediately know that x 1 = x 1, x = x, x = α 1 n (x 1 ), α n1 (x 1 ) = α n 1 (x 1), and α n (x ) = α n (x ). This gives α 1 n (x 1) = α 1 n (x 1 ) = α 1 n (α n (x )) = α 1 (α n (x )) = x. So α n (x ) = x 1, and hence [(n 1, x 1)] and [(n, x )] are composable. To see that (4.6) holds we have two cases: Case 1 : Suppose x 1 / iso. Then x 1 = x 1 / iso, x = αn 1 (x 1 ) / iso, and x = α 1 n (x 1) / iso. We also know there exists an open set V 1 such that x 1 V 1 dom(n 1 ) dom(n 1) with α n1 V1 = α n 1 V1, and an open set V such that x V dom(n ) dom(n ) with α n V = α n V. Let V := V αn 1 (V 1 ), which is an open set containing x. We claim that n V dom(n 1 n ) dom(n 1n ). To see this, let x V. Then using (4.1) we have h (x)((n 1 n ) n 1 n ) = h (α n (x))(n 1n 1 )h (x)(n n ), which is positive because α n (x) dom(n 1 ) and x dom(n ). So V dom(n 1 n ). A similar argument gives V dom(n 1n ), and so the claim holds. or each x V we have α n (x) = α n (x) V 1, which means α n1 n (x) = α n1 (α n (x)) = α n 1 (α n (x)) = α n 1 n (x).

18 GRAPH ALGBRAS AND ORBIT QUIVALNC 17 So α n1 n V = α n 1 n V. Hence (n 1 n, x ) (n 1n, x ), and (4.6) holds in this case. Case : Suppose x 1 iso. Then x 1 = x 1 iso, x = αn 1 (x 1 ) iso, and x = α 1 n (x 1) iso. We also have α n1 (x 1 ) = α n 1 (x 1), α n (x ) = α n (x ), and hence α n1 n (x ) = α n1 (α n (x )) = α n1 (x 1) = α n 1 (x 1) = α n 1 (α n (x )) = α n 1 n (x ). To get (n 1 n, x ) (n 1n, x ) in this case it now suffices to show that U (x,n 1 n,n 1 n ) is homotopic to p x. We use that α n (x ) = x 1 and α n (x ) = x 1 = x 1 and apply Lemma 4.(c) twice to get p x n n 1n 1n p x = (n p x ) n 1n 1(n p x ) = (p αn (x )n ) n 1n 1p αn (x )n = n p x1 n 1n 1p x1 n. Now we can write U (x,n 1 n,n 1 n ) = λ 1/ (x,n 1 n,n 1 n )p x n n 1n 1n p x = λ 1/ (x,n 1 n,n 1 n )n p x1 n 1n 1p x1 n = λ 1/ (x,n 1 n,n 1 n )λ1/ (x 1,n 1,n 1 )n = λ 1/ (x,n 1 n,n 1 n )λ1/ (x 1,n 1,n 1 )n U (x1,n 1,n 1 ) n ( λ 1/ (x 1,n 1,n 1 )p x 1 n 1n 1p x1 ) n Since (n 1, x 1 ) (n 1, x 1) implies that U (x1,n 1,n 1 ) is homotopic to p x1, we see that U (x,n 1 n,n 1 n ) is homotopic to λ 1/ (x,n 1 n,n 1 n )λ1/ (x 1,n 1,n 1 )n p x1 n. We use Lemma 4.(c) to get Hence U (x,n 1 n,n 1 n ) is homotopic to n p x1 n n p αn (x )p αn (x ) n = p x n n p x. λ 1/ (x,n 1 n,n 1 n )λ1/ (x 1,n 1,n 1 )n p x1 n = λ 1/ (x,n 1 n,n 1 n )λ1/ (x 1,n 1,n )p x n n p x 1 = λ 1/ (x,n 1 n,n 1 n )λ1/ (x 1,n 1,n 1 )λ1/ (x,n,n )U (x,n,n ). But (n, x ) (n, x ) implies that U (x,n,n ) is homotopic to p x, and hence U (x,n 1 n,n 1 n ) is homotopic to p x. This complete the proof that composition is well-defined. To see that inversion is well-defined, suppose [(n 1, x 1 )] = [(n 1, x 1)]. We need to show that [(n 1, α n1 (x 1 ))] = [((n 1), α n 1 (x 1)]. We again have two cases. Case 1: Suppose that x 1 = x 1 / iso. We know that there is open V such that x 1 V dom(n 1 ) dom(n 1) and α n1 V = α n 1 V. A straightforward argument shows that the open set V := α n1 (V ) satisfies α n1 (x 1 ) V dom(n 1) dom((n 1) ) and α n 1 V = α (n 1 ) V. So [(n 1, α n1 (x 1 ))] = [((n 1), α n 1 (x 1)] in this case. Case : Suppose that x 1 = x 1 iso. We have to show that U (αn1 (x 1 ),n 1,(n 1 ) ) is homotopic to p αn1 (x 1 ). We use Lemma 4.(c) to get U (αn1 (x 1 ),n 1,(n 1 ) ) = λ 1/ (α n1 (x 1 ),n 1,(n 1 ) ) p α n1 (x 1 )n 1 (n 1) p αn1 (x 1 ) = λ 1/ (α n1 (x 1 ),n 1,(n 1 ) ) n 1p x1 (n 1).

19 18 NATHAN BROWNLOW, TOK MIR CARLSN, AND MICHAL. WHITTAKR Since (n 1, x 1 ) (n 1, x 1), we have that U (x1,n 1,n 1 ) is homotopic to p x1. Hence U (αn1 (x 1 ),n 1,(n 1 ) ) is homotopic to λ 1/ (α n1 (x 1 ),n 1,(n 1 ) ) n 1U (x1,n 1,n 1 ) (n 1) = λ 1/ (α n1 (x 1 ),n 1,(n 1 ) ) λ 1/ (x 1,n 1,n 1 )n 1p x1 n 1n 1p x1 (n 1). Now, using (4.3) we have λ 1/ (α n1 (x 1 ),n 1,(n 1 ) ) λ 1/ (x 1,n 1,n 1 ) = h (x 1 )(n 1n 1 ) 1 h (x 1 )((n 1) n 1) 1. So U (αn1 (x 1 ),n 1,(n 1 ) ) is homotopic to h (x 1 )(n 1n 1 ) 1 h (x 1 )((n 1) n 1) 1 n 1 p x1 n 1n 1p x1 (n 1) = ( )( h (x 1 )(n 1n 1 ) 1 n 1 p x1 n 1 h (x 1 )((n 1) n 1) 1 n 1p x1 (n 1) ) = p αn1 (x 1 ), where the last equality follows from Lemma 4.(a). We equip G (C (),D()) with the topology generated by {{[(n, x)] : x dom(n)} : n N(D())}. It can be proven directly that G (C (),D()) is a topological groupoid with this topology, however, it also follows from our next result. Proposition 4.8. Let be a graph. Then G (C (),D()) is a topological groupoid, and G (C (),D()) and G are isomorphic as topological groupoids. Remark 4.9. If G is topologically principal, which we know from Proposition.3 is equivalent to satisfying condition (L), then G (C (),D()) is isomorphic to the Weyl groupoid G C0 (G 0 ) of (C (G ), C 0 (G 0 )) as in []. In this case the isomorphism of G and G (C (),D()) proved below follows from [, Proposition 4.14]. However, for a general graph, the proof does not follow from [, Proposition 4.14] and becomes much more technical. To prove Proposition 4.8 we need the following result. The proof can be deduced from the proof of [, Proposition 4.8], but we include a proof for completeness. As in [], we let supp (f) := {y G : f(γ) 0} for f C (G ). Lemma Let be a graph and π : C () C (G ) the isomorphism from Proposition.. Let n N(D()), and f := π(n). Then supp (f) satisfies (i) s(supp (f)) = dom(n); (ii) (x, k, y) supp (f) = α n (y) = x; and (iii) y dom(n) = (α n (y), k, y) supp (f) for some k Z. Proof. Identity (i) follows because h (y)(n n) = π(n n)(y, 0, y) = f f(y, 0, y) = γ G s(γ)=(y) f(γ). or (ii) we first consider the function f gf where g is any element of π(d()) = C 0 (G (0) ). Using the convolution product we have (4.7) f gf(y, 0, y) = f(γ) g(r(γ)) for all y. γ G s(γ)=(y)

20 GRAPH ALGBRAS AND ORBIT QUIVALNC 19 Alternatively, we can also apply (4.1) to, say, g = π(d) to get (4.8) f gf(y, 0, y) = π(n dn)(y, 0, y) = h (y)(n dn) = h (α n (y))(d)h (y)(n n) = g(α n (y), 0, α n (y)) f(y, 0, y). Now suppose for contradiction that (x, k, y) supp (f) but α n (y) x. Choose g C 0 (G (0) ) a positive function with g(x, 0, x) = 1 and g(α n(y), 0, α n (y)) = 0. Then (4.7) gives f gf(y, 0, y) f(x, k, y) g(x, 0, x) > 0, whereas (4.8) gives f gf(y, 0, y) = g(α n (y), 0, α n (y)) f(y, 0, y) = 0. So (ii) holds. Implication (iii) follows immediately from (i) and (ii). Proof of Proposition 4.8. Let (x, k, y) G. Then there are µ, ν and z such that x = µz, y = νz, and k = µ ν. We know from Lemma 4.1 that s µ s ν N(D()), y dom(s µ s ν), and that α sµs ν (y) = x. Define φ : G G (C (),D()) by φ((x, k, y)) = [(s µ s ν, y)]. It is routine to check that φ is well-defined, in the sense that if µ, ν, µ, ν, z, z, µz = µ z, νz = ν z, and µ ν = µ ν, then [(s µ s ν, νz)] = [(s µ s ν, ν z )]. It is also routine to check that φ is a groupoid homomorphism. We now have to show that φ is a homeomorphism. To show that φ is injective, assume that φ((x, k, y)) = φ((x, k, y )). Then x = x and y = y. Suppose for contradiction that k k. Then y must be eventually periodic, because otherwise we would have α sκs (y) α λ s κ s (y) for κ λ = k and κ λ = k. λ Thus x = µηηη and y = νηηη for some µ, ν and a simple cycle η such that s(η) = r(µ) = r(ν). It follows that φ((x, k, y)) = [(s µ(η) ms ν(η), y)] and n φ((x, k, y)) = [(s µ(η) m s ν(η) n, y)] where m, n, m, n are nonnegative integers such that µ(η) m ν(η) n = k and µ(η) m ν(η) n = k. Suppose that η has an exit. Then y / iso, and there is a ζ such that s(ζ) = s(η), ζ η, and ζ η 1 η η ζ (where η = η 1 η η η ). Then for any open set U with y U dom(s µ(η) ms ν(η) n ) dom(s µ(η) m s ν(η) n ), there is a positive integer l such that = Z(ν(η)l ζ) U, and that α sµ(η) ms ν(η) n(z) α s µ(η) m s ν(η) n (z) for any z Z(µ(η)l ζ). This contradicts the assumption that φ((x, k, y)) = φ((x, k, y)). If η does not have an exit, then y iso. Without loss of generality assume k > k, then we can use Lemma 4.(c) to compute [p y (s µ(η) ms ν(η) n) s µ(η) m s ν(η) n p y] 1 = [p y s ν s η k k s νp y ] 1 = [(p y s ν s η s νp y ) k k ] 1, and the second assertion in Lemma 4.3 implies that [(p y s ν s η s νp y ) k k ] 1 0. Thus, [U (y,sµ(η) ms ν(η) n,s µ(η) m s ν(η) n ) ] 1 0 and hence (s µ(η) ms ν(η) n, y) (s µ(η) m s ν(η) n, y). But this means φ((x, k, y)) φ((x, k, y)), which is a contradiction. So we must have k = k, and hence φ is injective. To show that φ is surjective, let [(n, x)] be an arbitrary element of G (C (),D()). Let f := π(n), where π : C () C (G ) is the isomorphism from Proposition.. We

21 0 NATHAN BROWNLOW, TOK MIR CARLSN, AND MICHAL. WHITTAKR know from (iii) of Lemma 4.10 that (α n (x), k, x) supp (f) for some k Z. Suppose first that x / iso. Choose µ, ν, a clopen neighborhood U of α n (x), and a clopen neighborhood V of x such that U Z(µ), V Z(ν), σ µ (U) = σ ν (V ), k = µ ν, and Z(U, µ, ν, V ) supp (f). Then α sµs (y) = α ν n(y) for all y V, and hence φ(α n (x), µ ν, x) = [(s µ s ν, x)] = [(n, x)]. Now suppose that x iso is not eventually periodic. Choose µ, ν and z such that x = νz, α n (x) = µz, and k = µ ν. It follows from Lemma 4.3 that [U (x,n,sµs ν ) ] 1 = 0 (because K 1 (C) = 0), and thus that φ((α n (x), k, x)) = [(s µ s ν, x)] = [(n, x)]. Assume then that x is eventually periodic. Then there are µ, ν and a simple cycle η such that s(η) = r(µ) = r(ν), x = νηηη, α n (x) = µηηη, and k = µ ν. By Lemma 4.3 there are positive integers l and m such that [U (x,n,sµs ν)] 1 = l m such that [p x s µ s η ls η ms νp x ] 1 = [(p x s µ s η s νp x ) l (p x s µ s ηs νp x ) m ] 1 = l m, and hence [U (x,n,sµs ν ) ] 1 = [p x s ν s η ls η ms νp x ] 1. Since h (s νη ms νη m) 1/ U (x,n,sµs ν ) (p x s ν s η ls η ms νp x ) = U (x,n,sµη ms νη l), We have [U (x,n,sµη ms νη l) ] 1 = [U (x,n,sµs ν ) ] 1 [p x s ν s η ls η ms νp x ] 1 = 0. Hence φ((α n (x), µ(η) m ν(η) l, x)) = [(s µ(η) ms ν(η) l, x)] = [(n, x)], which shows that φ is surjective. To see that φ is open, let µ, ν and let U and V be clopen subsets of such that U Z(µ), V Z(ν), and σ µ (U) = σ ν (V ). Then there is a p V D() such that h (x)(p V ) = 1 if x V, and h (x)(p V ) = 0 if x \ V ; and then φ(z(u, µ, ν, V )) = {[s µ s νp V, x] : x dom(s µ s νp V )}. This shows that φ is open. To prove that φ is continuous we will show that φ 1 ({[(n, y)] : y dom(n)}) is open for each n N(D()). ix n N(D()) and z dom(n). We claim that there is an open subset V (n,z) in G such that φ 1 ([(n, z)]) V (n,z) φ 1 ({[(n, y)] : y dom(n)}). Let π : C () C (G ) be the isomorphism from Proposition., and f := π(n). We know from (iii) of Lemma 4.10 that (α n (z), k, z) supp (f) for some k Z. Suppose that there are two different integers k 1 and k such that both (α n (z), k 1, z) and (α n (z), k, z) belong to supp (f). Then there are µ 1, ν 1, µ, ν such that (α n (z), k 1, z) Z(µ 1, ν 1 ), (α n (z), k, z) Z(µ, ν ) and Z(µ 1, ν 1 ), Z(µ, ν ) supp (f). Without loss of generality we can assume that µ 1 = µ, and then we have µ 1 = µ. We also have ν 1 = ν ξ or ν = ν 1 ξ for some ξ \ 0 ; we assume that µ = µ 1 ξ and denote µ := µ 1 = µ. We claim that z is an isolated point, and that s(z(µ, ν 1 )) s(z(µ, ν 1 ξ)) = {(z)}. To see this, suppose (x) s(z(µ, ν 1 )) s(z(µ, ν 1 ξ)). Then x = ν 1 y for some y such that α n (y) = µy, and x = ν 1 ξy for some y such that α n (y) = µy. It follows that y = y = ξy, and hence y = ν 1 ξξξ.... So s(z(µ, ν 1 )) s(z(µ, ν 1 ξ)) = {(ν 1 ξ... )} = {(z)},