EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 1, 2024, 519-545 ISSN 1307-5543 – ejpam.com Published by New York Business Global On the construction of a groupoid from an ample Hausdorff groupoid with twisted Steinberg algebra not isomorphic to its non-twisted Steinberg algebra Rizalyn S. Bongcawel 1,∗, Lyster Rey B. Cabardo1, Gaudencio C. Petalcorin Jr.1, Jocelyn P. Vilela1 1 Department of Mathematics and Statistics, College of Science and Mathematics, Center of Mathematical and Theoretical Physical Sciences-PRISM, MSU-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. This study introduces an ample Hausdorff groupoid  ⋊R extracted from an ample Hausdorff groupoid G and a unital commutative ring R; a Hausdorff groupoid D which is the discrete twist over  ⋊ R. In the groupoid C*-algebra perspective, when R = C there is an isomorphism between the non-twisted groupoid C*-algebra (C∗(G)) and the twisted groupoid C*- algebra (C∗( ⋊ R;D)). However, in this paper, in a purely algebraic setting, the non-twisted Steinberg algebra (AR(G)) and the twisted Steinberg algebra (AR(D; Â⋊R)) are non-isomorphic. 2020 Mathematics Subject Classifications: 20L05, 22A22 Key Words and Phrases: Groupoids, Steinberg Algebra, Twisted Steinberg Algebra, Non- twisted Steinberg Algebra 1. Introduction The study of groupoids was initiated by Brandt in 1926 in [2]. Brandt utilizes the no- tion of groupoid in [4] and other researchers produced more studies related to groupoids. In [12], groupoid is defined as a small category in which every morphism is invertible. Groupoid was used in various areas like the fibre bundle theory, in differential theory, in foliation theory and in differential topology. In 1980s, Renault was motivated by the works of Feldman and Moore [5, 6] for von Neumann algebras and initiated the study of C*-algebras associated to groupoids in his PhD thesis [10]. This study proved itself useful as it caters many problem in C*-algebras. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v17i1.5051 Email addresses: rizalyn.bongcawel@g.msuiit.edu.ph (R. S. Bongcawel), lysterrey.cabardo@g.msuiit.edu.ph (L. B. Cabardo), gaudencio.petalcorin@g.msuiit.edu.ph (G. C. Petalcorin), jocelyn.vilela@g.msuiit.edu.ph (J. P. Vilela) https://www.ejpam.com 519 © 2024 EJPAM All rights reserved. R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 520 Significant works on characterization of Lei-type maps with C*-algebra are in [13] and [8]. Renault then introduced the twisted groupoid C*-algebras where the twist is done by incorporating a T-valued 2-cocycle to its multiplication and involution. This study yields extreme importance in the structures of large classes of C*-algebras as seen in the works of Renault[11], Tu[14] and Barlak and Li[4]. In [15], Williams, Renault and Muhly proved that the groupoid C*-algebra (C∗(G)) and the twisted groupoid C*-algebra (C∗(Â⋊R;D)) are isomorphic when R = C with the additional conditions for G to be second countable locally compact groupoid with a Haar system and abelian isotropy. Last 2010, thirty years after the introduction of twisted groupoid C*-algebras, Stein- berg algebra was introduced independently in [3, 9]. It is an algebraic analogue of groupoid C*-algebra. In 2021, Becky Armstrong, Lisa Clark, et al. introduced the twisted Steinberg algebra in [1]. It is a purely algebraic analogue of Renault’s twisted groupoid C*-algebra. This study is a generalisation of Steinberg algebra by twisting the convolution and invo- lution in two ways: a locally constant 2-cocyle σ and a discrete twist Σ over a Hausdorff étale groupoid G. In this paper, we consider a purely algebraic perspective, that is, in the notion of Steinberg algebra. Without the analysis requirements for our groupoid, our goal is to show that the non-twisted Steinberg algebra and the twisted Steinberg algebra are non- isomorphic. Our first task is to construct an ample Hausdorff groupoid  ⋊ R from an ample Hausdorff groupoid G and a unital commutative ring R with R× as the set of units of R. From the unit space of Â⋊R, we then construct a sequence Â× T i ↪→ D q ↪→ Â⋊R where D is a Hausdorff groupoid, T ≤ R× and (D, i, q) is our desired twist over  ⋊R. We then investigate properties of the Steinberg algebra of G over R or AR(G) and the twisted Steinberg algebra associated to the pair (Â⋊R, D) or AR(D; Â⋊R) and look at when isomorphism between the two fails to hold. 2. Preliminaries In this section, important concepts and notations on topological groupoids, Steinberg algebra and twisted Steinberg algebra arising from a discrete twist are presented. Definition 1. [7] Let G be a set and G(2) be a subset of G × G such that there is a (composition) map (γ, α) 7→ γα from G(2) to G. Suppose that there is an inverse map γ 7→ γ−1 on G such that (γ−1)−1 = γ. Then we say that G is a groupoid if the following are satisfied: (G1) if (γ, α), (α, β) ∈ G(2), then (γα, β), (γ, αβ) ∈ G(2) and the following equation is satisfied: (γα)β = γ(αβ); (G2) for all γ ∈ G, (γ−1, γ) ∈ G(2); (G3) if (γ, α) ∈ G(2), then (γ−1γ)α = α and γ(αα−1) = γ. We call G(2) as the set of all composable pairs. We write G(3) for the set of composable triples in G, that is, G(3) = {(α, β, γ) : (α, β), (β, γ) ∈ G(2)}. R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 521 Lemma 1. [12] Let G be a groupoid and γ, β ∈ G. We say that (γ, β) ∈ G(2) if and only if s(γ) = r(β). Definition 2. [12] Define the functions s and r from G to itself by s(α) = α−1α called the source of α ∈ G and r(α) = αα−1 called the range of α ∈ G, respectively. The common image of r and s is the unit space of G and is denoted by G(0), that is, G(0) := s(G) = r(G). Example 1. [12] (i) Let G be a group with identity e. Then G is a groupoid with G(2) = G×G; s(γ) = γ−1γ = e; r(γ) = γγ−1 = e; and G(0) = {e}. (ii) If {Gi|i ∈ I} is a family of groups with identities {ϵ1|i ∈ I}, then the disjoint union⋃ i∈I Gi has a groupoid structure with d(g) = c(g) = ϵi for every g ∈ Gi. The composition, defined only for pairs (g, h) ∈ ⋃ i∈I Gi ×Gi, is just the relevant group law. This is known as a group bundle. Lemma 2. [7] Let G be a groupoid. We have (i) (α, γ), (γ, β) ∈ G(2) and αγ = βγ imply α = β. Similarly, if (γ, α), (γ, β) ∈ G(2) and γα = γβ, then α = β. (ii) r(αβ) = r(β) and s(αβ) = s(β) for all α, β ∈ G(2). (iii) (αβ)−1 = β−1α−1 for all α, β ∈ G(2). (iv) r(x) = x = s(x) for all x ∈ G(0). If γ ∈ G, then (r(γ), γ) and (γ, s(γ)) belong to G(2), and r(γ)γ = γ = γs(γ). Definition 3. [12] Let xG = r−1(x), Gx = s−1(x), and xGy = xG ∩Gy. The isotropy of a groupoid G is the set Iso(G) := {γ ∈ G : r(γ) = s(γ)} = ⋃ x∈G(0) xGx. We say that G is principal if Iso(G) = G(0). Remark 1. [12] The isotropy of any groupoid is a group bundle. Lemma 3. [7] A groupoid G is principal if γ 7→ (r(γ), s(γ)) is injective. Definition 4. [12] A groupoid G is effective if the interior of the isotropy group of G is equal to its unit space. Definition 5. [12] Given groupoid G and H, we call a map ϕ : G → H a groupoid homomorphism if (ϕ× ϕ)(G(2)) ⊆ H(2) and ϕ(α)ϕ(β) = ϕ(αβ) for all (α, β) ∈ G(2). The following concepts is taken from [12]. A topological groupoid consists of a groupoid G and a topology compatible with the groupoid structure such that the composition and involution are continuous and G(2) has the induced topology from the product topology. Every groupoid is a topological groupoid R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 522 with the discrete topology. An open set B ⊆ G is an open bisection if r|B and s|B are homeomorphisms onto an open subset of G. A topological groupoid is étale if r (or equivalently s) is a local homeomorphism. An étale groupoid is ample if the topology of G has a basis of compact open bisections. Discrete group, discrete groupoids, and discrete space are some examples of an ample groupoid with the discrete topology. Definition 6. [12] Given a topological space X and a topological ring R, the open support of a function f : X → R is the set supp(f) := {x ∈ X : f(x) ̸= 0} = f−1(R\{0}). We say that f is compactly supported if supp(f) is contained in a compact set. We use the following notion for the characteristic function of a subset U of G: 1U : G → R defined by 1U (g) = { 1 if g ∈ U 0 if g /∈ U Let RG be the set of all functions f : G → R. Canonically RG has the structure of an R-module with operations defined pointwise. Definition 7. [12] Let AR(G) be the R-submodule of RG generated by the set {1U |U is a Hausdorff compact open subset of G}, that is, AR(G) = {f : G → R|f is continuous and supp(f) is compact}. The convolution of f, g ∈ AR(G) is defined as (f ∗ g)(x) := ∑ y∈G, s(y)=s(x) f(xy−1)g(y) = ∑ (z,y)∈G(2), zy=x f(z)g(y) for all x ∈ G. The R-module AR(G) with the convolution, is called the Steinberg algebra of G over R. The following example is a Steinberg algebra of R over Z. Example 2. [12] Consider the set of G = R\{0} which is a group under multiplication. By Example 1, G is a groupoid with the following structures: G(2) = G×G, s(γ) = γ−1γ = 1, r(γ) = γγ−1 = 1 and G(0) = {1}. We note that G is an ample Hausdorff groupoid with respect to the discrete topology. Its base is composed of singletons {r} where r ∈ G. We let G as our unital commutative ring and 1{r} : G → Z is the characteristic function of a subset {r} of G. Then AZ(G) := spanZ{1{r} : {r}is a compact open bisection} := {f : G → Z}. (1) Hence, AZ(G) together with the convolution (f ∗ g)(x) := ∑ y∈G, s(y)=s(x) f(xy−1)g(y) = ∑ (z,y)∈G(2), zy=x f(z)g(y) is a Steinberg algebra of Z over G. R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 523 The following discussions is taken from [1] where the Steinberg algebra is twisted via the discrete twist. Definition 8. Let G be a Hausdorff étale groupoid, and R be a commutative unital ring, and let T < R×. A discrete twist by T over G is a sequence G(0)×T i ↪→ Σ q ↪→ G, where the groupoid G(0)×T is regarded as trivial group bundle with fibres T , Σ is a Hausdorff étale groupoid with Σ(0) = i(G(0) ×{1}), and i and q are continuous groupoid homomorphisms that restricts to homeomorphism of unit spaces, such that the following condition holds: (1) The sequence is exact, in the sense that i({x} × T ) = q−1(x) for every x ∈ G(0), i is injective, and q is a quotient map (2) The groupoid Σ is a locally trivial G-bundle, in the sense that for each α ∈ G, there is an open bisection Bα of G containing α, and a continuous map Pα : Bα → Σ such that (i) q ◦ Pα = idBα (ii) the map (β, z) → i(r(β), z)Pα(β) is a homeomorphism from Bα×T to q−1(Bα). (3) The image of i is central in Σ, in the sense that i(r(ϵ), z)ϵ = ϵi(s(ϵ), z) for all ϵ ∈ Σ and z ∈ T . We will denote a discrete twist over G by (Σ, i, q). The following is an example of a discrete twist. Example 3. Consider the the set of integers Z as our groupoid and unital commutative ring. Then our groupoid Z will have the following structures: Z× = {−1, 1} , s(x) = x−1 + x = {0}, r(x) = x+ x−1 = {0}, Z(0) = {0} and Z(2) = {(x, y) ∈ Z×Z|s(x) = r(y)}. Note that Z is a Hausdorff étale groupoid having the discrete topology. Let the function σ : Z(2) → T ≤ R× be a continuous 2-cocycle. Choose T ≤ Z× = {1}. Then Z × T is a Hausdorff groupoid with respect to the product topology with multiplication given by (α, z)(β,w) := (αβ, σ(α, β)zw), and inversion given by (α, z)−1 := (α−1, σ(α, α−1)−1z−1) = (α−1, σ(α−1, α)−1z−1), for all (α, β) ∈ Z(2) and z, w ∈ T . Then, (Z× T, i, q) is a discrete twist by T over Z with the sequence Z(0) × T i ↪→ Z× T q ↪→ Z where i(x, z) = (x, z) and q(γ, z) = γ for all x, γ ∈ Z and z ∈ T . Definition 9. A continuous map Pα : Bα → Σ is called a continuous local section if it satisfies Definition 8(2i). If P (G(0)) = Σ(0) = i(G(0) × 1), then Pα is a continuous global section. R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 524 Definition 10. Let G be an ample Hausdorff groupoid and let (Σ, i, q) be a discrete twist by T ≤ R× over G. Denote C(Σ, R) as the collection of continuous functions from Σ to R. We say that f ∈ C(Σ, R) is T -equivariant if f(z · ϵ) = zf(ϵ) for all z ∈ T and ϵ ∈ Σ, and we define AR(G; Σ) := {f ∈ C(Σ, R) : f is T -equivariant and q(supp(f)) is compact}. Lemma 4. Let G be an ample Hausdorff groupoid, and let (Σ, i, q) be a discrete twist by T ≤ R× over G. Then AR(G; Σ) is an R-submodule of C(Σ, R). Definition 11. Let G be an ample Hausdorff groupoid, and let (Σ, i, q) be a discrete twist by T ≤ R× over G. Let P : G → Σ be any continuous global section. There is a multiplication called convolution on the R-module AR(G; Σ), given by (f ∗Σ g)(ϵ) := ∑ γ∈Gs(q(ϵ)) f(ϵP (γ))g(P (γ)−1), under which AR(G; Σ) is an R-algebra. We call AR(G; Σ) the twisted Steinberg algebra of G associated to the pair (G,Σ). The following is an example of a twisted Steinberg algebra of a discrete group Z over a commutative ring R called the twisted discrete group algebra. Example 4. Let R be a discrete commutative unital ring and consider an ample Hausdorff groupoid Z with the discrete topology. Let σ : Z 7→ R× be a continuous 2-cocycle which is locally constant. Then the set AR(Z, σ) = span {1{z} : Z 7→ R |{z} is compact open bisection of Z} with the twisted convolution (f ∗σ g)(z) := ∑ (x,y)∈Z(2) xy=z σ(x, y)f(x)g(y) is the twisted Steinberg algebra of Z over R associated to the pair (Z, σ) denoted as AR(Z, σ). 3. The groupoid Â⋊R and discrete twist (D, i, q) In this section, we will define what is Â⋊R from a groupoid G and a commutative unital ring R, investigate its properties and construct the discrete twist (D, i, q). Throughout, G is an ample Hausdorff groupoid. Let A = Iso(G) = {γ ∈ G : s(γ) = r(γ)} be the isotropy of G. For u ∈ G(0), we let Au = {γ ∈ A : s(γ) = u}. We define Âu = {χ : Au → R×|χ is a continuous group homomorphism} with Au and R× having the subspace and discrete topology, respectively. Define R = G/A = {γA : γ ∈ G}. Let γ̇ = γA ∈ R,  = {(χ, u) : u ∈ G(0), χ ∈ Âu} and Â⋊R = {(χ, u, γ̇) : (χ, u) ∈ Â, r(γ) = u}. Theorem 1. Let G be an ample Hausdorff groupoid and R be a commutative unital ring. Then R is an ample Hausdorff groupoid. Proof. Let m : R(2) → R be the composition map defined by m((α̇, β̇)) = α̇β = αβA where αβ ∈ G and (α, β) ∈ G(2) and i : R → R be defined by i(γ̇) = γ̇−1 = γ−1A. R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 525 Let (α̇, β̇), (γ̇, µ̇) ∈ R(2) such that (α̇, β̇) = (γ̇, µ̇). Then αA = γA and βA = µA. So that m((α̇, β̇)) = αβA = αAβA = γAµA = γµA = m((γ̇, µ̇)). Thus, m is well-defined. For α̇, γ̇ ∈ R with α̇ = γ̇, i(γ̇) = γ−1A = α−1A = i(α̇). Also, for γ̇, β̇ ∈ R with γ̇ = β̇, s(γ̇) = γ̇−1γ̇ = (γA)−1γA = γ−1AγA = β−1AβA = β̇−1β̇ = s(β̇); r(γ̇) = γ̇γ̇−1 = γAγ−1A = βAβ−1A = β̇β̇−1 = r(β̇). Thus, the inverse, source and range maps are also well-defined. Let (α̇, β̇), (β̇, γ̇) ∈ R(2). Then s(α̇) = β̇β̇−1 and s(β̇) = γ̇γ̇−1 since α and β are composable. Hence, s(α̇β) = (α̇β)−1(α̇β) = (αβA)−1(αβA) = β−1α−1AαβA = α−1αβ−1βA = s(αβ)A = s(β)A, = β−1βA. Also, r(γ̇) = γ̇γ̇−1 = β̇−1β̇ = β−1β A. Hence, s(α̇β) = r(γ̇). Thus, (α̇β̇, γ̇) ∈ R(2). We also have β̇γ̇ = β̇γ. Hence, r(β̇γ) = (β̇γ)(β̇γ)−1 = (βγA)(βγA = βγAγ−1β−1A = ββ−1γγ−1A = r(βγ)A = r(β)A = ββ−1A. Since s(α̇) = α̇−1α̇ = β̇β̇−1 = βAβ−1A = ββ−1A, then r(β̇γ̇) = s(α̇) and (α̇, β̇γ̇) ∈ R(2). Now, (α̇β̇)γ̇ = (αβA)γA = (αA(βγA)) = α̇(β̇γ) = α̇(β̇γ̇). Let γ̇ ∈ R. Now, (γ̇−1)−1 = (γ−1A)−1 = (γ1)−1A = γA = γ̇. For γ̇ ∈ R, r(γ̇−1) = γ̇−1(γ̇−1)−1 = γ̇−1γ̇. Hence, (γ̇, γ̇−1) ∈ R(2). Let (β̇, γ̇) ∈ R(2). Then , (β̇γ̇)γ̇−1 = β̇(γ̇γ̇−1) = β̇r(γ̇) = β̇s(β̇) = β̇ and γ̇−1(γ̇β̇) = (γ̇−1γ̇)β̇ = s(γ̇)β̇ = r(β̇)β̇ = β̇. Thus, R is a groupoid. Let R be a topological space with the quotient topology τR. Define the quotient map πR : G → R by πR(α) = αA = α̇. Define the topology for R × R and R(2) as follow: τR×R = {U × V : U, V ∈ τR} and τR(2) = {(U × V ) ∩ R(2) : U × V ∈ τR×R}. Let U R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 526 be an open set in R. Then V = π−1 R (U) is open in G. Let (γ̇, β̇) ∈ m−1(U) ⊆ R(2). Then π(γ)π(β) ∈ U , i.e., (γ̇, β̇) ∈ U × U and (γ̇, β̇) ∈ (U × U) ∩ R(2) ∈ τR(2) since U × U ∈ τR×R. Since (α̇, β̇) is chosen arbitrarily, every element in m−1(U) is con- tained in some open set in τR(2) and m is continuous. Also, let M be an open set in R. Then there exists V = π−1 R (M) = {γ ∈ G : πR(γ) ∈ M} ⊂ G. Note that i−1(M) = πR(V ) = {γ ∈ G : πR(γ −1) ∈ M} = {γ ∈ G : γ−1 ∈ M}. For any open set U ′ ∈ R, π−1 R (U ′) is open in G. In particular, V = π−1 R (M) is open in G. It follows that πR(V ) is open in R. Hence, i−1(M) is open in R and i is continuous. Thus, R is a topological groupoid. Suppose that γ̇ and β̇ are distinct points in R. Then for γ, β ∈ G, πR(γ) = γA and πR(β) = βA where γA ̸= βA. Let U̇ be an open set in G defined as U̇ = {α ∈ G : πR(α) ̸= βA}. Since γA ̸= βA, then γ ∈ U̇ . Hence, U̇ is an open neighborhood in G containing γ. Now, let V̇ be an open set in G defined as V̇ = {α ∈ G : πR(α) ̸= γA}. Since βA ̸= γA, then β ∈ V̇ . Hence, V̇ is an open neighborhood containing β. Let α ∈ U̇ ∩ V̇ . Then, πR(α) ̸= γA and πR(α) ̸= βA which is a contradiction since γA and βA are distinct. Thus, we have found open sets U̇ and V̇ in G such that U̇ ∩ V̇ is empty. It follows that π(V̇ ) and π(V̇ ) are open in R with π(V̇ )∩π(V̇ ) = ∅ and each contains distinct equivalence classes. Thus, R is Hausdorff. Since R is Hausdorff then R(0) is Hausdorff. Let B be a basis for a topology in G. Then πR(B) = {πR(B) : B ∈ B} is a basis for the quotient topology in R. Let V be an open subset of R and consider s|πR(B) : πR(B) → V . We denote s1 = s|πR(B). Let U be an open subset of V . Then π−1 R (U) is open in G. Since πR(B) is a ba- sis for the topology on R, then there exists basic element πR(B) in πR(B) containing s−1 1 (U). Now, let s−1 1 = (s|πR(B)) −1 : V → πR(B). Let B ⊆ πR(B) which is open in R. Then there exists W ⊆ G such that W is the inverse image of of B under πR. Then W = π−1 R (B) = {γ ∈ G : πR(γ) ∈ B}. Since πR is surjective, s−1 1 (B) = πR(W ). Since πR is a quotient map then it is an open map. Thus, for any open set U ′ in R, πR(U) is open in G. In particular, W = π−1 R (B) is open in G. It follows that πR(W ) is open in R. Hence, s−1 1 (W ) is open and the source map is a homeomorphism onto an open subset of R. Similarly, the range map is also homeomorphic onto an open subset of R. Therefore, R is an ample Hausdorff groupoid with respect to the quotient topology. From now on, we denote the elements of  ⋊ R by (χ, γ̇) with χ ∈ Âr(γ). A sub- set C of  is closed if and only if for all sequences where xn converges to x such that xn ∈ C, then x ∈ C. We will denote our topology for  as τ = {D : D = Cc, C is closed in Â} and Cc stands for the compliment of C. Also, τÂ×R = {U × V : U ∈ τ and V ∈ τR}, and τ(Â⋊R)×(Â⋊R) = {A × B : A,B ∈ τÂ⋊R} which gives us the relative topology for ( ⋊R)(2) as τ(Â⋊R)(2) = {(A × B) ∩ ( ⋊R)(2) : A × B ∈ τ(Â⋊R)×(Â⋊R)}. Let r((χ, γ̇)) = (χ, r(γ)) and s((χ, γ̇)) = (χ · γ, s(γ)), respectively. The set of composable pairs is ( ⋊R)(2) = {((χ, γ̇), (χ′, γ̇′))|χ′ = χ · γ}. Note that χ · γ(a) = χ(γaγ−1). Also, R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 527 for every u ∈ G, χ · u = χ since χ · u(a) = χ(uau−1) = χ(auu−1) = χ(ar(u)) = χ(au) = χ(as(a)) = χ(a). Also, i((χ, γ̇)) = (χ · γ, γ̇−1) and m((χ, γ̇), (χ′, γ̇′)) = (χ, γ̇γ̇′) are the inversion and com- position maps, respectively. Theorem 2. Let G be an ample Hausdorff groupoid and R be a commutative unital ring. Then Â⋊R = {(χ, u, γ̇) : (χ, u) ∈ Â, r(γ) = u} is a groupoid. Proof. Suppose that ((χ1, γ̇1), (χ ′ 1, γ̇ ′ 1)) , ((χ2, γ̇2), (χ ′ 2, γ̇ ′ 2)) ∈ (Â⋊R)(2) with ((χ1, γ̇1), (χ ′ 1, γ̇ ′ 1)) = ((χ2, γ̇2), (χ ′ 2, γ̇ ′ 2)). Then, (χ1, γ̇1) = (χ2, γ̇2) and (χ′ 1, γ̇ ′ 1) = (χ′ 2, γ̇ ′ 2). Now, m(((χ1, γ̇1), (χ ′ 1, γ̇ ′ 1))) = (χ1, γ̇1γ̇ ′ 1) = (χ2, γ̇2γ̇ ′ 2) = m(((χ2, γ̇2), (χ ′ 2, γ̇ ′ 2))). Let (χ, γ̇), (χ′, γ̇′) ∈ Â⋊R such that (χ, γ̇) = (χ′, γ̇′). Then i((χ, γ̇)) = (χ · γ̇, γ̇−1) = i((χ′, γ̇′)); r((χ, γ̇)) = (χ, r(γ)) = (χ′, r(γ′)) = r((χ′, γ̇′)); s((χ, γ̇)) = (χ · γ, s(γ)) = (χ′ · γ′, s(γ′)) = s((χ′, γ̇′)). Thus, the composition, inverse, range and source maps are well-defined. Let ((χ, γ̇), (χ′, γ̇′)) , ((χ′, γ̇′), (χ′′, γ̇′′)) ∈ (Â⋊R)(2). Then s((χ, γ̇)) = (χ · γ, s(γ)) = r((χ′, γ̇′)) = (χ′, r(γ′)); s((χ′, γ̇′)) = (χ′ · γ′, s(γ′)) = r((χ′′, γ̇′′)) = (χ′′, r(γ′′)). Now, s((χ, γ̇), (χ′, γ̇′)) = s((χ, γ̇γ̇′)) = (χ · γγ′, s(γγ′)) = (χ · γγ′, s(γ′)) = (χ · γγ′, r(γ′′)) = (χ′ · γ′, r(γ′′)) = (χ′′, r(γ′′)) = r(χ′′, γ̇′′); r[(χ′, γ̇′)(χ′′, γ̇′′)] = r((χ′, γ̇′γ̇′′)) = (χ′, r(γ′γ′′)) = (χ′, r(γ′)) = (χ · γ, s(γ)) = s((χ, γ̇)). Hence, ((χ, γ̇)(χ′, γ̇′), (χ′′, γ̇′′)) ∈ (Â⋊R)(2) and ((χ, γ̇), (χ′, γ̇′)(χ′′, γ̇′′)) ∈ (Â⋊R)(2). Now, [(χ, γ̇)(χ′, γ̇′)](χ′′, γ̇′′) = (χ, γ̇γ̇′)(χ′′, γ̇′′) = (χ, γ̇γ̇′γ̇′′) = (χ, γ̇)(χ′, γ̇′γ̇′′) = (χ, γ̇)[(χ′, γ̇′)(χ′′, γ̇′′)]. R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 528 Let (χ, γ̇) ∈ Â⋊R. Then ((χ, γ̇)−1)−1 = (χ · γ, γ̇−1) = (χ · γγ−1, (γ̇−1)−1) = (χ · r(γ), γ̇) = (χ · u, γ̇) = (χ, γ̇); r((χ, γ̇)−1) = r((χ · γ, γ̇−1)) = (χ · γ, r(γ−1) = (χ · γ, s(γ)) = s((χ, γ̇)). Hence, ((χ, γ̇), (χ, γ̇)−1) ∈ (Â⋊R)(2). Suppose that ((χ′, γ̇′), (χ, γ̇)) ∈ (Â⋊R)(2). Then, [(χ′, γ̇′)(χ, γ̇)](χ, γ̇)−1 = (χ, γ̇γ̇′)(χ′, γ̇′)−1 = (χ, γ̇γ̇′)(χ′ · γ′, (γ̇′)−1) = (χ, γ̇γ̇′(γ̇′)−1) = (χ, γ̇r((γ̇′)) = (χ, γ̇s(γ̇)) = (χ, γ̇) Also, (χ, γ̇)−1[(χ, γ̇)(χ′, γ̇′)] = (χ, γ̇)−1(χ, γ̇γ̇′) = (χ · γ, γ̇−1)(χ, γ̇γ̇′) = (χ · γ, γ̇−1γ̇γ̇′) = (χ · γ, s(γ̇)γ̇′) = (χ · γ, r(γ̇′)γ̇′) = (χ′, γ̇′) Therefore, Â⋊R is a groupoid. Lemma 5. Â⋊R is an ample Hausdorff groupoid. Proof. Let U be open in  ⋊ R. Then U = (A × B) ∩ ( ⋊ R) where A × B is open in Â × R. Let ((χ, γ̇), (χ′, γ̇′)) ∈ m−1(U). Then (χ, γ̇γ̇′) ∈ (A × B) and (χ, γ̇γ̇′) ∈  ⋊ R. Since A × B is open in Â × R, then there exists open set (A × B)′ containing (χ, γ̇γ̇′) such that (A × B)′ ⊆ A × B. Also there exists open set W in  ⋊R containing (χ, γ̇γ̇′). Since  ⋊ R ⊆ Â × R, then there exists open set W ′ in Â × R containing (χ, γ̇γ̇′) where W ′ = W ∩  ⋊R. Consider the set (A × B)′ ×W ′ ⊆ ( ×R) × ( ×R) where (A × B)′ ∈ τÂ×R and W ′ ∈ τÂ×R. It follows that (A × B)′ × W ′ is open in (Â × R) × (Â × R). Define M = ((A × B)′ × W ′) ∩ ( ⋊ R)(2) = {((χ, γ̇), (χ′, γ̇′)) ∈ R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 529 ( ⋊ R)(2) : m(((χ, γ̇), (χ′, γ̇′))) ∈ (A × B)′ ∩ W ′}. We claim that M ⊂ m−1(U). Let ((χ, γ̇), (χ′, γ̇′)) ∈ M . Then m(((χ, γ̇), (χ′, γ̇′))) ∈ (A× B)′ ∩W ′ ⊆ (A× B) ∩ Â⋊R and m(((χ, γ̇), (χ′, γ̇′))) ∈ (A×B)∩ Â⋊R, i.e., ((χ, γ̇), (χ′, γ̇′)) ∈ m−1(U) and M ⊂ m−1(U). Since (A×B)′×W ′ is open in (Â×R)× (Â×R), then M is open in (Â⋊R)(2). It follows that m−1(U) is open and m is continuous. Let U be an open set in Â⋊R. Then U = V ∩ Â⋊R where V is open in Â×R. Let (χ, γ̇) ∈ i−1(U). Then, i((χ, γ̇)) ∈ V ∩ Â⋊R. Since V is open in Â×R , then there exists open set Uv containing (χ, γ̇) where Uv ⊆ V . Also, there exists open set W containing (χ, γ̇)−1 where W ⊆  ⋊ R. Define U ′ v = {(χ′, γ̇′) ∈  ⋊ R : i((χ′, γ̇′)) ∈ Uv ∩ W}. Let (χ′, γ̇′, ) ∈ U ′ v. By definition, i((χ′, γ̇′)) ∈ Uv ∩ W ⊂ V ∩  ⋊ R which means that (χ′, γ̇′) ∈ i−1(U). Hence, U ′ v ⊆ i−1(U). Since Uv ⊆ V , then U ′ v ⊆ V ∩ W . Notice that V ∩ W is open in  ⋊ R. Hence, U ′ v is open in  ⋊ R. Thus, i−1(U) is open and our inverse map is continuous. Let (χ, γ̇) and (χ′, γ̇′) be distinct points in Â⋊R. We can choose open sets U and V in Â × R such that U ∩ V = ∅ with (χ, γ̇) ∈ U and (χ′, γ̇′) ∈ V . Let U ′ = U ∩  ⋊ R and V ′ = V ∩ Â⋊R. Since U and V are in τÂ⋊R, then U ′ and V ′ are open sets in Â⋊R containing (χ, γ̇) and (χ′, γ̇′), respectively. Note that U and V are disjoint, hence U ′ and V ′ are also disjoint and we have proved that Â⋊R is a Hausdorff groupoid. Since  ⋊R is Hausdorff, ( ⋊R)(0) is Hausdorff. Let B be a basis for the product topology on Â×R. Then, B′ = {B∩ Â⋊R : B ∈ B} is a basis for the relative topology on Â⋊R. Now, let r1 = r|B′ : B′ → U where U is an open subset of Â⋊R and let V be an open subset of U . Then V = (A×B)∩ Â⋊R where A×B is open in Â×R. Since B′ is a base for the topology on Â⋊R, then there exists a basic element B containing (χ, γ̇) such that r1(χ, γ̇) ∈ V . Then r−1 1 (U) is open in Â⋊R. Now, denote r−1 1 = (r|B′)−1 : U → B′. Let B be open subset of B′. Then B = A ∩ ( ⋊ R) where A ∈ B. Let (χ, γ̇) ∈ r1(B). Then r−1 1 (χ, γ̇) ∈ A ∩ ( ⋊ R). Since A is open in Â × R, then there exists open set A′ containing (χ, γ̇) where A′ ⊆ A. Also, there exists open set W ⊆  ⋊ R containing (χ, γ̇). Define A′′ = {(χ′, γ̇′) ∈ U : r−1 1 (χ′, γ̇′) ∈ A′ ∩W}. Let (χ′, γ̇′) ∈ A′′. By definition, r−1 1 (χ′, γ̇′) ∈ A′′ ∩W ⊂ A ∩ (Â⋊R). Hence, r−1 1 (χ′, γ̇′) ∈ A ∩ (Â⋊R) which means that (χ′, γ̇′) ∈ r1(B). Hence, A′′ ⊂ r1(B). Since A′′ ⊆ A′∪W and A′ ⊂ A, A′′ ⊆ A∩W . Notice that A ∩W is open in Â⋊R. Hence, A′′ is open in Â⋊R. Thus, r1(B) is open and r is a homeomorphism onto an open subset of  ⋊R. Similarly, s is homeomorphic onto an open subset of Â⋊R. Proposition 1. Â⋊R is a principal groupoid with unit space Â. Proof. Let (χ, u) ∈  where u ∈ G(0). Then, s(γ) = u for γ ∈ Au and s(γ) = r(γ) for γ ∈ A. Hence, (χ, u) = (χ, s(γ)) = (χ, r(γ)) = (χ · γ, s(γ)). Thus, (χ, u) ∈ (Â⋊R)(0) and  ⊆ (Â⋊R)(0). Let (χ · γ, s(γ)) ∈ (Â⋊R)(0) where (χ, γ̇) ∈ Â⋊R. Then (χ · γ, s(γ)) = (χ, r(γ)) = (χ · γ, u) since r(γ) = u. Thus, ( ⋊ R)(0) ⊆  and  is the unit space of Â⋊R. Let θ :  ⋊ R → Â × Â be defined by θ((χ, γ̇)) = (r((χ, γ̇)), s((χ, γ̇))) and let R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 530 (χ1, γ̇1), (χ2, γ̇2) ∈ Â⋊R such that θ((χ1, γ̇1)) = θ((χ2, γ̇2)). Then (r((χ1, γ̇1)), s((χ1, γ̇1))) = (r((χ2, γ̇2)), s((χ2, γ̇2))). Also, r((χ1, γ̇1)) = (χ1, r(γ1)) = r((χ2, γ̇2)) = (χ2, r(γ2)) and s((χ1, γ̇1)) = (χ1·γ1, s(γ1)) = s((χ2, γ̇2)) = (χ2 · γ2, s(γ2)). Hence, χ1 = χ2 and γ1 = γ2. Thus, (χ1, γ̇1) = (χ2, γ̇2) and θ is injective. Therefore, Â⋊R is a principal groupoid. We now introduce a sequence of groupoids and investigate whether it is our desired discrete twist over  ⋊R. Define  ∗ G × T = {(χ, z, γ) : χ ∈ Âr(γ), z ∈ T, and γ ∈ G}. Let r((χ, z, γ)) = (χ, r(γ)) and s((χ, z, γ)) = (χ · γ, s(γ)) be the range and source maps, respectively. The composition map and inverse map is (χ, z, γ)(χ′, z′, γ′) = (χ, zz′, γγ′) and (χ, z, γ)−1 = (χ · γ, z−1, γ−1), respectively. We note that (χ, z, γ) and (χ′, z′, γ′) are composable pairs if we have χ′ = χ · γ and χ · γ is defined by χ · γ(a) = χ(γaγ−1) where χ · u = χ. Lemma 6.  ∗ G × T is a Hausdorff groupoid. Proof. Let (χ, z, γ), (χ′, z′, γ′) ∈  ∗ G × T with (χ, z, γ) = (χ′, z′, γ′). Now, (χ, z, γ)−1 = (χ · γ, z−1, γ−1) = (χ′ · γ′, (z−1)′, γ′−1) = (χ′, z′, γ′)−1. Also, r((χ, z, γ)) = (χ, r(γ)) = (χ, s(γ′)) = r((χ′, z′, γ′)); s((χ, z, γ)) = (χ · γ, s(γ)) = (χ′ · γ′, s(γ′)) = s((χ′, z′, γ′)). Hence, the inverse range and source maps are well-defined. Composition is well-defined since for ((χ1, z1, γ1), (χ ′ 1, z ′ 1, γ ′ 1)), ((χ2, z2, γ2), (χ ′ 2, z ′ 2, γ ′ 2)) ∈ Â∗G×T (2) with ((χ1, z1, γ1), (χ ′ 1, z ′ 1, γ ′ 1)) = ((χ2, z2, γ2), (χ ′ 2, z ′ 2, γ ′ 2)),m(((χ1, z1, γ1)(χ ′ 1, z ′ 1, γ ′ 1))) = (χ1, z1z ′ 1, γ1γ ′ 1) = (χ2, z2z ′ 2, γ2γ ′ 2) = m(((χ2, z2, γ2), (χ ′ 2, z ′ 2, γ ′ 2))). Now, let ((χ1, z1, γ1), (χ2, z2, γ2)), ((χ2, z2, γ2), (χ3, z3, γ3)) ∈  ∗ G × T (2). Then s((χ1, z1, γ1)(χ2, z2, γ2)) = s((χ1, z1z2, γ1γ2)) = (χ1 · γ1γ2, s(γ1γ2)) = (χ2 · γ2, s(γ2)) = (χ3, r(γ3)) = ((χ3, z3, γ3)). Also, r((χ2, z2, γ2)(χ3, z3, γ3)) = r((χ2, z2z3, γ2γ3)) = (χ2, r(γ2γ3)) = (χ2, r(γ2)) = (χ1 · γ1, s(γ1)) R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 531 = s((χ1, z1, γ1)). Thus, ((χ1, z1, γ1)(χ2, z2, γ2), (χ3, z3, γ3)),((χ1, z1, γ1), (χ2, z2, γ2)(χ3, z3, γ3)) ∈ ( ∗ G × T )(2). Composition in  ∗ G × T is associative since ((χ1, z1, γ1)(χ2, z2, γ2))(χ3, z3, γ3) = (χ1, z1z2, γ1γ2)(χ3, z3, γ3) = (χ1, z1z2z3, γ1γ2γ3) = (χ1, z1, γ1)(χ2, z2z3, γ2γ3) = (χ1, z1, γ1)((χ2, z2, γ2)(χ3, z3, γ3)). For (χ, z, γ) ∈  ∗ G × T , ((χ, z, γ)−1)−1 = (χ · γ, z−1, γ−1)−1 = (χ · γ · γ−1, (z−1)−1, (γ−1)−1) = (χ · r(γ), z, γ) = (χ · u, z, γ) = (χ, z, γ). Also, r((χ, z, γ)−1) = r((χ · γ, z−1, γ−1)) = (χ · γ, r(γ−1)) = (χ · γ, s(γ)) = s((χ, z, γ)). Hence, ((χ, z, γ), (χ, z, γ)−1) ∈  ∗ G × T (2). Notice that ((χ1, z1, γ1)(χ2, z2, γ2))(χ2, z2, γ2) −1 = (χ1, z1z2, γ1γ2)(χ2, z2, γ2) −1 = (χ1, z1z2, γ1γ2)(χ2 · γ2, z−1 2 , γ−1 2 ) = (χ1, z1z2z3, γ1γ2γ −1 3 ) = (χ1, z1, γ1r(γ2)) = (χ1, z1, γ1s(γ1)) = (χ1, z1, γ1). Also, (χ1, z1, γ1) −1((χ1, z1, γ1)(χ2, z2, γ2)) = (χ1, z1, γ1) −1(χ1, z1z2, γ1γ2) = (χ1 · γ1, z−1 1 , γ−1 1 )(χ1, z1z2, γ1γ2) = (χ1 · γ1, z−1 1 z1z2, γ −1 1 γ1γ2) = (χ2, z2, s(γ1)γ2) = (χ2, z2, r(γ2)γ2) = (χ2, z2, γ2). Hence,  ∗ G × T is a groupoid. Endowed  ∗ G × T with the product topology define as τÂ∗×G = {(A ∗ B × C) ∈  ∗ G × T : A ∈ τÂ, B ∈ τG , C ∈ τT }. Since G, T and  ⋊ R are Hausdorff,  is R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 532 also Hausdorff. Let (χ, z, γ) and (χ′, z′, γ′) be distinct points in  ∗ G × T . Since  is Hausdorff, then there exists open neighborhoods A1 and A2 in  containing χ and χ′, respectively such that A1 ∩A2 = ∅. Also, there exists open neighborhoods G1 and G2 in G containing γ and γ′, respectively such that G1 ∩ G2 = ∅. For the Hausdorff space T , there exists open neighborhoods T1 and T2 in T containing z and z′, respectively wherein T1 ∩ T2 = ∅. Then by definition of τÂ∗G×T , U = A1 ∗G1 × T1 and V = A2 ∗G2 × T2 are open neighboorhoods in  ∗ G × T containing (χ, z, γ) and (χ′, z′, γ′), respectively such that U ∩ V = ∅. Therefore,  ∗ G × T is Hausdorff. Lemma 7. Let Â × T = {(χ, z, u) : (χ, u) ∈  and z ∈ T}. Then Â × T is the isotropy group of  ∗ G × T . Proof. Note that Iso( ∗ G × T ) = {(χ, z, γ) ∈  ∗ G × T : s(χ, z, γ) = r(χ, z, γ)}. Let (χ, z, γ) ∈ Iso(Â∗G×T ). Then s(χ, z, γ) = r(χ, z, γ), that is, (χ·γ, s(γ)) = (χ, r(γ)). Note that χ · γ = χ if and only if γ = u where u ∈ G(0). Also, s(γ) = r(γ) means γ = u ∈ G(0). Thus, (χ, z, γ) = (χ, z, u) and Iso( ∗ G × T ) = {(χ, z, u) ∈  ∗ G × T : (χ, u) ∈ Â, z ∈ T} = Â× T. Therefore, Â× T is the isotropy group for  ∗ G × T . Lemma 8. Define ∼ on  ∗G ×T by (χ, z, γ) ∼ (χ′, z′, γ′) if and only if χ = χ′ and there exists a ∈ Au such that χ(a)z = z′ and γ = a · γ′. Then ∼ is an equivalence relation on  ∗ G × T . Proof. Let (χ, z, γ) ∈  ∗ G × T . Choose a ∈ Au such that χ(a) = 1 ∈ R×. Then χ(a)z = 1(z) = z and s(a) = u. Since γ ∈ Au, s(γ) = u, s(γ) = u = s(a) = r(a) and a and γ are composable pairs in G. Then, (aa−1)γ = γ where aa−1 ∈ Au. Hence (χ, z, γ) ∼ (χ, z, γ). Let (χ, z, γ), (χ′, z′, γ′) ∈ Â∗G×T such that (χ, z, γ) ∼ (χ′, z′, γ′). Then χ(a)χ(a)−1z = χ(a)−1z′ and we have z = χ(a)−1z′. Also since γ′ ∈ Au then s(γ′) = u = r(a), that is, (γ′, a) ∈ G(2) and aa−1γ′ = a−1γ which is γ′ = a−1γ, a−1 ∈ Au. Thus, (χ′, z′, γ′) ∼ (χ, z, γ). Let (χ1, z1, γ1) ∼ (χ2, z2, γ2) and (χ2, z2, γ2) ∼ (χ3, z3, γ3). Then χ1 = χ3 and z3 = χ2(a)χ1(a)z1 = χ1(a)χ1(a)z1 = χ1(a)z1. Also, γ1 = a · γ2 = a · a · γ3 = a · γ3. Thus (χ1, z2, γ1) ∼ (χ3, z3, γ3). Therefore, ∼ is an equivalence relation on  ∗ G × T . Denote the set of equivalence classes of  ∗ G × T with respect to the equivalence relation ∼ on Lemma 8 by D =  ∗ G × T/ ∼= {[χ, z, γ] : (χ, z, γ) ∈  ∗ G × T}. Define the following structure for D and verify whether it is a groupoid. The range and source maps will be r([χ, z, γ]) = [χ, r(γ)] and s([χ, z, γ]) = [χ · γ, s(γ)], respectively. The composition and inverse map is [χ, z, γ][χ′, z′, γ′] = [χ, zz′, γγ′] and [χ, z, γ]−1 = [χ · γ, z−1, γ−1], respectively. We note that [χ, z, γ] and [χ′, z′, γ′] are composable pairs if χ′ = χ · γ where χ · γ is defined by χ · γ(a) = χ(γaγ−1) and χ · u = χ. Theorem 3. D is a Hausdorff étale groupoid with respect to the quotient topology with D(0) = i(Â× {1}). R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 533 Proof. Let [χ, z, γ], [χ′, z′, γ′] ∈ D with [χ, z, γ] = [χ′, z′, γ′]. Now, [χ, z, γ]−1 = [χ · γ, z−1, γ−1] = [χ′ · γ′, (z−1)′, γ′−1] = [χ′, z′, γ′]−1. Also, r([χ, z, γ]) = [χ, r(γ)] = [χ, s(γ′)] = r([χ′, z′, γ′]); s([χ, z, γ]) = [χ · γ, s(γ)] = [χ′ · γ′, s(γ′)] = s([χ′, z′, γ′]). Hence, the inverse, range and source maps are well-defined. Let ([χ1, z1, γ1], [χ ′ 1, z ′ 1, γ ′ 1]), ([χ2, z2, γ2], [χ ′ 2, z ′ 2, γ ′ 2]) ∈ D(2) with ([χ1, z1, γ1], [χ ′ 1, z ′ 1, γ ′ 1]) = ([χ2, z2, γ2], [χ ′ 2, z ′ 2, γ ′ 2]). Composition is well-defined since m(([χ1, z1, γ1], [χ ′ 1, z ′ 1, γ ′ 1])) = [χ1, z1z ′ 1, γ1γ ′ 1] = [χ2, z2z ′ 2, γ2γ ′ 2] = m(([χ2, z2, γ2], [χ ′ 2, z ′ 2, γ ′ 2])). Now, let ([χ1, z1, γ1], [χ2, z2, γ2]), ([χ2, z2, γ2], [χ3, z3, γ3]) ∈ D(2). Then s([χ1, z1, γ1][χ2, z2, γ2]) = s([χ1, z1z2, γ1γ2]) = [χ1 · γ1γ2, s(γ1γ2)] = [χ2 · γ2, s(γ2)] = [χ3, r(γ3)] = r([χ3, z3, γ3]). Also, r([χ2, z2, γ2][χ3, z3, γ3]) = r([χ2, z2z3, γ2γ3]) = [χ2, r(γ2γ3)] = [χ2, r(γ2)] = [χ1 · γ1, s(γ1)] = s([χ1, z1, γ1]). Thus, ([χ1, z1, γ1][χ2, z2, γ2], [χ3, z3, γ3]) and ([χ1, z1, γ1], [χ2, z2, γ2][χ3, z3, γ3]) are compos- able pairs. To show that composition in D is associative, ([χ1, z1, γ1][χ2, z2, γ2])[χ3, z3, γ3] = [χ1, z1z2, γ1γ2][χ3, z3, γ3] = [χ1, z1z2z3, γ1γ2γ3] = [χ1, z1, γ1][χ2, z2z3, γ2γ3] = [χ1, z1, γ1]([χ2, z2, γ2][χ3, z3, γ3]). R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 534 For [χ, z, γ] ∈ D we have, ([χ, z, γ]−1)−1 = [χ · γ, z−1, γ−1]−1 = [χ · γ · γ−1, (z−1)−1, (γ−1)−1] = [χ · r(γ), z, γ] = [χ · u, z, γ] = [χ, z, γ]. Also, r([χ, z, γ]−1) = r([χ ·γ, z−1, γ−1]) = [χ ·γ, r(γ−1)] = [χ ·γ, s(γ)] = s([χ, z, γ]). Hence, ([χ, z, γ], [χ, z, γ]−1) ∈ D(2). Notice that ([χ1, z1, γ1][χ2, z2, γ2])[χ2, z2, γ2] −1 = [χ1, z1z2, γ1γ2][χ2, z2, γ2] −1 = [χ1, z1z2, γ1γ2][χ2 · γ2, z−1 2 , γ−1 2 ] = [χ1, z1z2z3, γ1γ2γ −1 3 ] = [χ1, z1, γ1r(γ2)] = [χ1, z1, γ1s(γ1)] = [χ1, z1, γ1]. Also, [χ1, z1, γ1] −1([χ1, z1, γ1][χ2, z2, γ2]) = [χ1, z1, γ1] −1[χ1, z1z2, γ1γ2] = [χ1 · γ1, z−1 1 , γ−1 1 ][χ1, z1z2, γ1γ2] = [χ1 · γ1, z−1 1 z1z2, γ −1 1 γ1γ2] = [χ2, z2, s(γ1)γ2] = [χ2, z2, r(γ2)γ2] = [χ2, z2, γ2]. Hence, D is a groupoid. Let D be a topological space with the quotient topology τD. We define the quotient map πD : Â∗G×T → D by πD((χ, z, γ)) = [χ, z, γ] for (χ, z, γ) ∈ Â∗G×T and [χ, z, γ] ∈ D and τÂ∗G×T = {U × V : U ∈ τÂ, V ∈ τG×T } where τG×T is the product topology with the topology in G and T having the discrete topology. Let [χ, z, γ] and [χ′, z′, γ′] be distinct elements of D. Then there exists (χ, z, γ) ≁ (χ′, z′, γ′) in  ∗G ×T , that is, (χ, z, γ) ̸= (χ′, z′, γ′) such that πD((χ, z, γ)) = [χ, z, γ] and πD((χ ′, z′, γ′)) = [χ′, z′, γ′]. Since  ∗ G ×T is Hausdorff, there exists open neighborhoods U and V in Â∗G×T containing (χ, z, γ) and (χ′, z′, γ′), respectively such that U ∩V = ∅. Then, πD(U) and πD(V ) are open neighborhoods in D containing [χ, z, γ] and [χ′, z′, γ′], respectively such that πD(U) ∩ πD(V ) = ∅. Let [χ, z, γ] ∈ D and U and V be open subsets of D where [χ, z, γ] ∈ U . We need to show that r : U → V is a homeomorphism. Let V1 be an open subset of V such that r−1(V1) ⊆ U . Then, M = π−1 D (V1) is open in Â∗G×T so that r−1(V1) = πD(M) is open D. R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 535 Hence, r is continuous. Similarly, r−1 is continuous. Thus, r is a local homeomorphism. Therefore, D is a Hausdorff étale groupoid. Let [χ, z, γ] ∈ D such that s([χ, z, γ]) = r([χ, z, γ]). Note that s([χ, z, γ]) = [χ, z, γ]−1[χ, z, γ] = [χ · γ, z−1, γ−1][χ, z, γ] = [χ · γ, z−1z, γ−1γ] = [χ · γ, 1, s(γ)] r([χ, z, γ]) = [χ, z, γ][χ, z, γ]−1 = [χ, z, γ][χ · γ, z−1, γ−1] = [χ, zz−1, γγ−1] = [χ, 1, r(γ)]. Hence, [χ · γ, 1, s(γ)] = [χ, 1, r(γ)], that is, χ · γ = χ and r(γ) = s(γ). Then γ = u ∈ G(0). Hence, the elements in D(0) will look like [χ, 1, u]. Now, i(Â× {1}) = i(χ, 1, u) = [χ, 1, u]. Therefore, D(0) = i(Â× {1}). Note that Â × T is the isotropy group of  ∗ G × T by Lemma 7. Hence, Â × T is a group bundle by Remark 1. Then, define the sequence Â × T i ↪→ D q ↪→  ⋊ R where D is a Hausdorff étale groupoid by Proposition 3, and the maps i and q are defined by i((χ, z, u)) = [χ, z, u] and q([χ, z, γ]) = (χ, γ̇), respectively. Lemma 9. The maps i and q are continuous groupoid homomorphism that restricts to homeomorphism of unit spaces. Proof. Let (χ, z, u), (χ′, z′, u′) ∈ Â×T such that (χ, z, u) = (χ′, z′, u′). Then [χ, z, u] = [χ′, z′, u′]. Thus, i((χ, z, u)) = i((χ′, z′, u′)) and i is well-defined. Let [χ1, z1, γ1] and [χ2, z2, γ2] be elements in D such that q[χ1, z1, γ1] ̸= q[χ2, z2, γ2]. Then (χ1, γ̇1) ̸= (χ2, γ̇2). If χ1 ̸= χ2, then (χ1, z1, γ1) ≁ (χ2, z2, γ,). If γ̇1 ̸= γ̇2, then γ1A ̸= γ2A. Since Au ⊂ A, then we cannot find a ∈ Au such that γ1 = a · γ2. Hence, (χ1, z1, γ1) ≁ (χ2, z2, γ2). In both cases (χ1, z1, γ1) ≁ (χ2, z2, γ2) which means that [χ1, z1, γ1] ̸= [χ2, z2, γ2]. Thus, q is well-defined. We need Â × T ⊂  ∗ G × T to show that i is continuous. Let (χ, z, u) ∈ Â × T where (χ, u) ∈ Â, and u ∈ G(0). Since G(0) ⊂ G, then (χ, z, u) ∈  ∗ G × T . Since πD :  ∗ G × T → D is continuous, i = πD|Â×T : Â× T → D is also continuous. Let q ◦ πD :  ∗ G × T →  ⋊ R be defined by (q ◦ πD)(χ, z, γ) = q(πD(χ, z, γ)) and let U be an open subset of  ⋊ R. Then U = (A × B) ∩  ⋊ R where A × B is open in Â × R. Let (χ, z, γ) ∈ (q ◦ πD) −1(U). Then q ◦ πD(χ, z, γ) ∈ U , that is, q(πD(χ, z, γ)) = q([χ, z, γ]) = (χ, γ̇) ∈ A×B∩ Â⋊R. Then (χ, γ̇) ∈ A×B, that is, χ ∈ A and γ̇ ∈ B. Since πR is continuous, π−1 R (B) is open in G containing γ. Let M = A ∗B2 × {z} = {(χ, z, γ) ∈  ∗ G × T : πD(χ, z, γ) ∈ (q ◦ πD)−1(U)}. R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 536 Then M ⊆ (q ◦ πD)−1(U) and M is open in  ∗ G × T . Thus, (q ◦ πD)−1(U) is open and q ◦ πD is continuous. Since πD is continuous , q is continuous. Consider i× i : (Â× T )(2) → D(2) defined by (i× i)(((χ, z, u), (χ′, z′, u′))) = (([χ, z, u], [χ′, z′, u′])) ∈ D(2). Also, q × q : D(2) → (Â×R)(2) defined by (q × q)(([χ, z, γ], [χ′, z′, γ′])) = ((χ, γ̇), (χ′, γ̇′)) ∈ (Â⋊R)(2). Now, i((χ, z, u)(χ′, z′, u′)) = i((χ, zz′, uu′)) = [χ, zz′, uu′]; i((χ, z, u))i((χ′, z′, u′)) = [χ, z, u][χ′, z′, u′] = [χ, zz′, uu′]. Also, q([χ, z, γ][χ′, z′, γ′]) = q([χ, zz′, γγ′]) = (χ, γ̇γ̇′); q([χ, z, γ])q([χ′, z′, γ′]) = (χ, γ̇)(χ′, γ̇′) = (χ, γ̇γ̇′). Thus, i and q are continuous groupoid homomorphism. Let i|(Â×T )(0) : (Â × T )(0) → D(0). Since i is continuous by Lemma 9, then i|(Â×T )(0) is continuous. Let V be open in (Â× T )(0). Then V = A× B where A is open in  and B is open in T . Let M = A ∗ G(0) × B = {(χ, z, u) ∈  ∗ G × T : πD(χ, z, u) ∈ i(V )}. Then M = π−1 D (i(V )) and M is open in  ∗ G × T since A is open in Â, G(0) is open in G and B is open in T . Hence, i(V ) is open in D(0) and i−1 is continuous. Thus, i|Â×T is a homeomorphism of unit spaces. Now, let q|D(0) : D(0) → (Â⋊R)(0). Since q is continuous by Lemma 9, then q|D(0) is continuous. Let Y be an open subset of D(0). Then π−1 D (Y ) is open in  ∗ G × T , that is, π−1 D (Y ) = A ∗ B × C where A is open in Â, B is open in G and C is open in T . Let M = A × πR(B) ∩ ( ⋊R)(0) = {(χ, γ̇) ∈ ( ⋊R)(0) : q−1(χ, γ̇) ∈ Y }. Then M = q(Y ) and M is open in (Â⋊R)(0) since A is open in  and πR(B) is open in R. Then q(Y ) is open in (Â⋊R)(0) and q−1 is continuous. Therefore, q|D(0) is a homeomorphism. In order to show that (D, i, q) is a discrete twist over  ⋊ R. We prove first the following results. Theorem 4. The sequence Â× T i ↪→ D q ↪→ Â⋊R is exact, that is, (i) i({(χ, u)× T}) = q−1((χ, u)) for (χ, u) ∈ (Â⋊R)(0), (ii) i is injective, and (iii) q is a quotient map. Proof. (i) Let (χ, u) ∈ Â. Then i({(χ, u)} × T}) = [χ, z, u] for some z ∈ T and q−1((χ, u)) = [χ, z, u] for some z ∈ T . Hence, i({(χ, u)×T}) = q−1((χ, u)) for (χ, u) ∈ (Â⋊R)(0). R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 537 (ii) Let (χ1, z1, u1), (χ2, z2, u2) ∈  ⋊ T with q((χ1, z1, u1)) = q((χ2, z2, u2)). Then, [χ1, z1, γ1] = [χ2, z2, γ2], that is, (χ1, z1, γ1) ∼ (χ2, z2, γ2). Then χ1 = χ2 and we can choose a ∈ AU such that χ(a) = 1 ∈ R× such that z1 = χ(a)z2 = (1)z2 = z2. Also, choose a′ ∈ AU such that a;∈ G(0). Then γ1 = a · γ2 = γ2. Thus, (χ1, z1, γ1) = (χ2, z2, γ2) and i is injective. (iii) By Lemma 9, q is continuous. Let (χ, γ̇) ∈ Â⋊R. Then [χ, z, γ] where r(γ) = u is the pre-image of (χ, γ̇) in D. Thus, q is surjective and a quotient map. Theorem 5. D is a locally trivial G-bundle in the sense that for each (χ, γ̇) ∈ Â⋊R, there is an open bisection Bα of Â⋊R containing (χ, γ̇), and a continuous map Pα : Bα → D such that (i) q ◦ Pα = idBα (ii) the map (β, z) → i(r(β), z)Pα(β) is a homeomorphism from Bα × T to q−1(Bα). Proof. (i) Let (χ, γ̇) ∈ Â⋊R and Bα be an open bisection of Â⋊R containing (χ, γ̇) and let Pα : Bα → D defined by Pα((χ, γ̇)) = [χ, z, γ]. Let U be an open subset of D. Then π−1 D (U) is open in  ∗ G × T , that is, π−1 D (U) = A ∗B ×C where A is open in Â, B is open in G and C is open in T . Let (χ, z, γ) ∈ π−1 D (U). Then (χ, z, γ) ∈ A ∗B×C, that is, χ ∈ A and γ ∈ B. Since B is open in G, then πR(B) is open in D containing γ̇. Let M = A × πR(B) ∩  ⋊ R. Then (χ, γ̇) ∈ M and M is open in  ⋊ R. Since (χ, γ̇) is chosen arbitrarily, then for every element in P−1 α (U) there exists an open neighborhoodM containing (χ, γ̇). Thus, P−1 α (U) is open and Pα is continuous. Now, q◦Pα : Bα → Â⋊R. Then, q◦Pα((χ, γ̇)) = q(pα((χ, γ̇))) = q([χ, z, γ]) = (χ, γ̇). Hence, the image of Bα in Pα is just itself and we have q ◦ Pα = idBα . (ii) Let θ : βα × T → q−1(βα) where βα × T ⊆  ⋊ R × T and q−1(βα) ⊆ D. Let U be an open subset of q−1(βα). Then there exists U ′ ∈  ∗ G × T such that U = U ′/ ∼ ∈ D where U ′ is open in  ∗ G × T . Here, U ′ is the pre-image of U under πD. Since U ′ is open in  ∗ G × T , then U ′ = (A × B) × C where A is open in Â, B is open in G and C is open in the discrete topology for T . Note that θ−1(U) = {(χ, z, γ̇) ∈  ⋊ R × T : θ(χ, z, γ̇) ∈ U)}. Since U = U ′/ ∼, then there exists an element (χ, z, γ) in U ′ whose equivalence class in D is in U . Since (χ, z, γ) ∈ U ′, then χ ∈ Â, z ∈ C which is open in T , and γ ∈ B which is open in G. Since we have B to be an open set in G containing γ, then there exists open set M in R containing γ̇. Hence, (A×M)×C is an open set in Â⋊R× T . Since (χ, z, γ̇) is arbitrary, we have shown that every element in θ−1(U) is contained in some open set in Â⋊R× T . Thus, θ is continuous. Note that θ−1 : q−1(βα) → βα × T . Let U be an open subset of βα × T . Then, U = V ×T ′ where V is open in Â⋊R and T ′ is open in T . Let [χ, z, γ] ∈ θ(U). Then R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 538 θ−1([χ, z, γ]) ∈ V ×T ′, that is, (χ, z, γ) ∈ V ×T ′. Hence, z ∈ T ′ and (χ, γ̇) ∈ V . Since V is open in Â⋊R, then V = A×B ∩ Â⋊R where χ ∈ A, A open in  and γ̇ ∈ B, B open in R. Thus, π−1 R (B) is open in G. Let π−1 D (θ(U)) = M = A ∗ π−1 R (B) × T ′ which is open in  ∗ G × T . Hence, θ(U) is open q−1(Bα) and θ−1 is continuous. Therefore, θ is a homeomorphism. Theorem 6. The image i(Â×T ) is central in D in the sense that i(r([χ, z, γ]), z)[χ, z, γ] = [χ, z, γ]i(s([χ, z, γ]), z) for all [χ, z, γ] ∈ D and z ∈ T . Proof. Let [χ, z, γ] ∈ D and z′ ∈ T . Now, i(r([χ, z, γ]), z′)[χ, z, γ] = i((χ, r(γ)), z′)[χ, z, γ] = [χ, z′, r(γ)][χ, z, γ] = [χ, z′z, r(γ)γ] = [χ, z′z, γ]. Also, [χ, z, γ]i(s([χ, z, γ]), z′) = [χ, z, γ]i((χ · γ, s(γ), z′)) = [χ, z, γ][χ · γ, z′, s(γ))] = [χ, zz′, γs(γ)] = [χ, zz′, γ]. Hence, i(r([χ, z, γ]), z′)[χ, z, γ] = [χ, z, γ]i(s([χ, z, γ]), z′) and so the image of i is central in D. The following corollary follows from Theorems 4, 5 and Lemma 6. Corollary 1. (D, i, q) is a discrete twist over Â⋊R. 4. Non-isomorphic property of AZ(Z) and AZ(D; Â⋊R) In this section we present a case in which the non-twisted Steinberg algebra (AR(G)) and twisted Steinberg algebra (AR(D; Â⋊R)) is not isomorphic when G = Z and R = Z. Let G = Z and R = Z. The set of multiplicative units of Z is Z× = {−1, 1} = T and the unit space of Z is Z(0) = {x ∈ Z : x = s(y) = r(y), y ∈ Z} = {0}. The source and range maps are s(x) = (−x) + x = {0} and r(x) = x + (−x) = {0}, respec- tively. The isotropy group for Z is A = Z. Also, R = ZZ = {x + Z : x ∈ Z} = {0̇} where {0̇} = 0 + Z. For u ∈ Z(0), we have A0 = Z. Also, Â0 = {χ1, χ2|χi : Z → {1,−1} is a continuous group homomorphism}, i = 1, 2 where χ1 : Z → Z× defined by χ1(a) = 1 and χ2 : Z → Z× defined by χ2(a) = { 1 if a ∈ 2Z −1 if a ∈ 2Z+ 1. R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 539 Note that  ∗ Z× T = {(χ, z, x) : χ ∈ Â0, z ∈ T, x ∈ Z} = {(χ1, 1, x), (χ1,−1, x), (χ2, 1, x), (χ2,−1, x)}. So, D = ( ∗ Z× T/ ∼) = {[χ1, 1, x], [χ1,−1, x], [χ2, 1, x], [χ2,−1, x]|x ∈ Z}. Claim 1: D = {[χ1, 1, 0], [χ1,−1, 0], [χ2, 1, 0], [χ2,−1, 0]}. For i = 1, 2, [χi, 1, x] = {(χ′, z′, x′)|χi = χ′,∃ a ∈ Zwhereχi(a)(1) = z′ andx = a · x′} = {(χi, z ′, x′)|∃ a ∈ Zwhere 1 = χi(a)(1) = z′ andx = a · x′} = {(χi, 1, x ′)|∃ a ∈ Z, x = a · x′} = {(χi, 1, x/a)|a ∈ Z, x ∈ Z} = {(χi, 1, x)|a = 1, x ∈ Z} = {· · · (χi, 1,−1), (χi, 1, 0), (χi, 1, 1) · · · } [χi,−1, x] = {(χ′, z′, x′)|χi = χ′,∃ a ∈ Zwhereχi(a)(−1) = z′ andx = a · x′} = {(χi, z ′, x′)|∃ a ∈ Zwhere − 1 = χi(a)(−1) = z′ andx = a · x′} = {(χi,−1, x′)|∃ a ∈ Z, x = a · x′} = {(χi,−1, x/a)|a ∈ Z, x ∈ Z} = {(χi,−1, x)|a = 1, x ∈ Z} = {· · · (χi,−1,−1), (χi,−1, 0), (χi,−1, 1) · · · } Hence, (χi, 1, 0) ∈ [χi, 1, x] and (χi,−1, 0) ∈ [χi,−1, x] imply that [χi, 1, x] = [χi, 1, 0] and [χi,−1, x] = [χi,−1, 0]. Therefore, D = {[χ1, 1, 0], [χ1,−1, 0], [χ2, 1, 0], [χ2,−1, 0]}. and claim 1 is proved. Now, the source of [χ1,−1, 0] in D is, s([χ1,−1, 0]) = [χ1,−1, 0]−1[χ1,−1, 0] = [χ1 · 0, (−1)−1, 0][χ1,−1, 0] = [χ1, (−1)−1(−1), 0(0)] = [χ1, 1, 0]. Also, the range of [χ1,−1, 0] in D is, r([χ1,−1, 0]) = [χ1,−1, 0][χ1,−1, 0]−1 = [χ1,−1, 0][χ1 · 0, (−1)−1, 0] R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 540 = [χ1, (−1)(−1)−1, (0)0] = [χ1, 1, 0] For [χ2,−1, 0] in D, s([χ2,−1, 0]) = [χ2,−1, 0]−1[χ2,−1, 0] = [χ2 · 0, (−1)−1, 0][χ2,−1, 0] = [χ2, (−1)−1(−1), 0(0)] = [χ2, 1, 0]. and r([χ2,−1, 0]) = [χ2,−1, 0][χ2,−1, 0]−1 = [χ2,−1, 0][χ2 · 0, (−1)−1, 0] = [χ2, (−1)(−1)−1, (0)0] = [χ2, 1, 0] Our groupoid D is best understood with this illustration: [x1, 1, 0] [x1,−1, 0] [x2,−1, 0] [x2, 1, 0] Figure 1: Morphisms in D Hence, the unit space for D is D(0) = {[χ1, 1, 0], [χ2, 1, 0]}. When Z is endowed with the discrete topology, its base will be composed of singletons {z}, for all z ∈ Z. Let 1{z} denotes the characteristic function of {z} from Z to Z. The Steinberg algebra associated to Z is AZ(Z) := span{1{z} : Z → Z|{z} is a compact open bisection of Z} equipped with pointwise addition, a11{z1} + a21{z2} = (a1 + a2)1{z1+z2} and multiplication as follows; a11{z1} · a21{z2} = a1a21{z1+z2}. Claim 2: Â⋊R = (Â⋊R)(0) The source and range of (χ1, 0, 0̇) ∈ Â⋊R are: s((χ1, 0, 0̇)) = (χ1, 0, 0̇) −1(χ1, 0, 0̇) = (χ1 · 0, 0, 0̇)(χ1, 0, 0̇) = (χ1, 0, 0̇) R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 541 r((χ1, 0, 0̇)) = (χ1, 0, 0̇)(χ1, 0, 0̇) −1 = (χ1, 0, 0̇)(χ1 · 0, 0, 0̇) = (χ1, 0, 0̇). Also, the source and range for (χ2, 0, 0̇) ∈ Â⋊R are: s((χ2, 0, 0̇)) = (χ2, 0, 0̇) −1(χ2, 0, 0̇) = (χ2 · 0, 0, 0̇)(χ2, 0, 0̇) = (χ2, 0, 0̇) r((χ2, 0, 0̇)) = (χ2, 0, 0̇)(χ2, 0, 0̇) −1 = (χ2, 0, 0̇)(χ2 · 0, 0, 0̇) = (χ2, 0, 0̇). Hence, (Â⋊R)(0) = {(χ1, 0, 0̇), (χ2, 0, 0̇)} = Â⋊R. Theorem 7. If Â⋊R = (Â⋊R)(0), then AZ(Â⋊R) ∼= AZ(D; Â⋊R). Proof. Let F : AZ(Â⋊R) → AZ(D; Â⋊R) be defined by F (f)([χi, z, γ]) = z · f(r((χi, γ̇))) where z ∈ T, (χi, γ̇) ∈ Â⋊R. Linearity holds since for all f, g ∈ AZ(Â⋊R) and n ∈ Z, F (f + g)([χi, z, γ]) = z(f + g)(r((χi, γ̇))) = zf(r((χi, γ̇))) + zg(r((χi, γ̇))) = (F (f) + F (g))([χi, z, γ]) and F (nf) = znf(r((χi, γ̇))) = nzf(r(χi, γ̇))) = nF (f). Now, observe that F (fg)([χi, z, γ]) = z(fg)(r((χi, γ̇))) = zf(r((χi, γ̇)))g(r((χi, γ̇))). Also, (F (f)F (g))([χi, z, γ])(χi, γ̇) = ∑ ((χ′ i,γ̇ ′),(χ′′ i ,γ̇ ′′)∈(Â⋊R)(2), (χ′ i,γ̇ ′))(χ′′ i ,γ̇ ′′)=(χi,γ̇) F (f)(χ′ i, γ̇ ′)F (g)(χ′′ i , γ̇ ′′)−1 Since Â⋊R = (Â⋊R)(0), then for all (χi, γ̇) ∈ Â⋊R the only composable pairs in Â⋊R is of the form ((χi, γ̇), (χi, γ̇)) where s(χi, γ̇) = (χi, γ̇) = r(χi, γ̇). But (χ1, 0̇)(χ2, 0̇) ̸= (χ1, 0̇). Hence, (F (f)F (g))([χi, z, γ])(χ1, 0̇) = ∑ ((χ1,0̇),(χ1,0̇)∈(Â⋊R)(2), (χ1,0̇))(χ1,0̇)=(χ1,0̇) F (f)(χ1, 0̇)F (g)(χ1, 0̇) −1 = F (f)(χ1, 0̇)F (g)(χ1, 0̇) −1 = F (f)(χ1, 0̇)F (g)(χ1, 0̇) = zf(r((χ1, 0̇)))g(r((χ1, 0̇))). R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 542 Also, since (χ2, 0̇)(χ1, 0̇) ̸= (χ2, 0̇), we get (F (f)F (g))([χi, z, γ])(χ2, 0̇) = ∑ ((χ2,0̇),(χ2,0̇)∈(Â⋊R)(2), (χ2,0̇))(χ2,0̇)=(χ2,0̇) F (f)(χ2, 0̇)F (g)(χ2, 0̇) −1 = F (f)(χ2, 0̇)F (g)(χ2, 0̇) −1 = F (f)(χ2, 0̇)F (g)(χ2, 0̇) = zf(r((χ2, 0̇)))g(r((χ2, 0̇))). Thus, for all (χi, γ̇) ∈ Â⋊R, F (f)F (g)([χi, z, γ])(χi, γ̇) = zf(r((χi, γ̇)))zg(r((χi, γ̇))) = F (fg)([χi, z, γ]) and F is a Z-module homomorphism. Suppose that F (f)([χi, z, γ]) = 0 for [χi, z, γ] ∈ D. Then, zf(r((χi, γ̇))) = 0 for all z ∈ T and γ ∈ Z which means that f(r((χi, γ̇))) = 0. Since  ⋊ R = ( ⋊ R)(0), then for all (χi, γ̇) ∈  ⋊ R, r((χi, γ̇)) = (χi, γ̇). Hence for all (χi, γ̇) ∈  ⋊ R, f((χi, γ̇)) = f(r((χi, γ̇))) = 0 which implies that f = 0. Thus, F is injective. Now let h ∈ AZ(D; Â⋊R) and define fh :  ⋊ R → Z by fh((χi, 0̇)) = h([χi, 1, 0]). Notice that every element in AZ(Â⋊R) is a mapping from Â⋊R to Z. We are left to show that fh is continuous and supp(fh) is compact. Since  ⋊R and Z are both discrete, then fh is continuous. Since h ∈ AZ(D; Â⋊R), h = a11{[χ1,1,0]} + a21{[χ1,−1,0]} + a31{[χ2,1,0]} + a41{[χ2,−1,0]}. Thus, fh(χi, γ̇) = h([χi, z, γ]) = (a11{[χ1,1,0]} + a21{[χ1,−1,0]} + a31{[χ2,1,0]} + a41{[χ2,−1,0]})([χi, z, γ]) Hence, fh(χ1, 0̇) = h([χ1, 1, 0]) = (a11{[χ1,1,0]} + a21{[χ1,−1,0]} + a31{[χ2,1,0]} + a41{[χ2,−1,0]})([χ1, 1, 0]) = a1 fh(χ2, 0̇) = h([χ2, 1, 0]) = (a11{[χ1,1,0]} + a21{[χ1,−1,0]} + a31{[χ2,1,0]} + a41{[χ2,−1,0]})([χ2, 1, 0]) = a3 If a1 = a3 = 0, then supp(fh) = ∅ which is closed and bounded, that is, compact. If a1, a2 ∈ Z \ {0}, then supp(fh) = {(χi, γ̇) ∈  ⋊R : fh((χi, γ̇)) ̸= 0))} =  ⋊R which is compact. Thus, fh ∈ AZ( ⋊R). Now for [χi, z, γ] ∈ D and since [χi, z, 0] and [χi, z, γ] are the same equivalence classes in D, F (fh)([χi, z, γ]) = zfh(r((χi, γ̇))) R. S. Bongcawel et al. / Eur. J. Pure Appl. Math, 17 (1) (2024), 519-545 543 = zfh(χi, r(γ)) = zfh(χi, γ) = zh([χ,1, 0]) = h([χi, z, 0]) = h([χi, z, γ]) = h. Then F is surjective. Therefore, AZ(Â⋊R) ∼= AZ(D; Â⋊R). Since  ⋊ R = {(χ1, 0̇), (χ2, 0̇)} is a topological space with the discrete topology, {(χ1, 0̇)} and {(χ2, 0̇)} are the basic elements of its base which are compact and an open bisection since they are singletons. Hence, we introduce our characterictic functions that spans AZ(Â⋊R) as 1{(χ,γ̇)} : Â⋊R → Z defined by 1{(χ,γ̇)}(g) = { 1 if g ∈ {(χ, γ̇)} 0 if g /∈ {(χ, γ̇)}. Define the Steinberg algbera of  ⋊ R over Z as AZ( ⋊ R) = span {1{(χ1,0̇)}, 1{(χ2,0̇)}} equipped with the pointwise addition and multiplication as follows: a11{(χ1,0̇)} + a21{(χ1,0̇)} = (a1 + a2)1{(χ1,0̇)}+{(χ2,0̇)} a11{(χ1,0̇)} · a21{(χ2,0̇)} = (a1 · a2)1{(χ1,0̇)}{(χ2,0̇)}. By Theorem 7, AZ(Â⋊R) ∼= AZ(D; Â⋊R). Then the twisted Steinberg algebra of Â⋊R over the pair (D,Z) is defined as AZ(D; Â⋊R) ∼= span {1{(χ1,0̇)}, 1{(χ2,0̇)}}. Thus, dimension of AZ(D;  ⋊ R) is less than or equal to 2. Note that AZ(Z) is a Z-module. Since AZ(Z) is generated by the characteristic functions of the form 1{z} where z ∈ Z which is infinite, then AZ(Z) is infinite dimensional. Hence, if we map AZ(Z) to AZ(D; Â⋊R) it will never be injective. Thus, isomorphism fails to hold. Note that from Section 3, the isotropy group of Z is itself, that is, Iso(Z) = Z and the unit space of Z is Z(0) = {0}. Now, the interior of Iso(Z) = Z ̸= Z(0). By Definition 4, Z is not an effective groupoid. Conjecture: Let G be an effective ample Hausdorff groupoid and R be a unital commutative ring. Then AR(G) ∼= AR(D; Â⋊R). Conclusion: We have defined a groupoid  ⋊ R coming from the isotropy of an ample Hausdorff groupoid G. We have examined the properties of the groupoid  ⋊ R. We have successfully constructed a discrete twist on  ⋊ R thereby the presence of a twisted Steinberg algebra over  ⋊ R via the discrete twist (D, i, q). 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