Abstract
A Morita equivalence similar to that found by Green for crossed products by groups will be established for crossed products by inverse semigroups. More precisely, let S be an inverse semigroup, T a finite subinverse semigroup of S and A an Salgebra or a Talgebra. Then the crossed product \(A \rtimes T\) is Morita equivalent to a certain crossed product \(B \rtimes S\).
Introduction
In a classical paper [8], Green showed that for a closed subgroup H of a locally compact group G, and a Galgebra A there exists a Morita equivalence between \(A \rtimes H\) and \(C_0(G/H,A) \rtimes G\) via an imprimitivity bimodule over these algebras ([8, Prop. 3]). This useful result was discussed and generalized in many directions, for example, in [1, 7, 17, 21].
In this note we shall establish an analogous imprimitivity theorem for an inverse semigroup S and a finite subinverse semigroup \(T' \subseteq S\) for crossed products in Sieben’s sense [20]. As a corollary of this, we show this holds true also for a given \(T'\)algebra A, and thus this may be usefully combined with induction like in Kasparov [9, 10]. Actually, this note was motivated by the fact that the Baum–Connes map [2] for groups G is a kind of extrapolation of Green–Julg isomorphisms for crossed products by G of induced algebras by compact subgroups \(H \subseteq G\), as noted by Meyer and Nest in [14]. In establishing that, Kasparov’s induction plays a fundamental role. To potentially carry this result over from groups to inverse semigroups, we need induction for compact (and thus finite) subinverse semigroups \(T' \subseteq S\), and this is now provided in this note. Actually, in the meanwhile we have made considerable progress in this direction and were able to establish a Baum–Connes map for fibered Salgebras [4, 5] founding on this note.
We are going to give a brief summary of this article. At first we rewrite the inverse semigroup crossed product \(A \rtimes T'\) as a groupoid crossed product \(A \rtimes T\) to have a grouplike construction. Then we adapt and follow Green’s proof [8, p. 199204] in a natural way. The action on a certain quotient space \(S_T/T\) (G/H in Green [8]) is similar to the regular representation action by Khoshkam and Skandalis [11]. After establishing Green’s imprimitivity Theorem 3.7, we apply it to the induced algebra (in the sense of Kasparov [9, 10]) A of a \(T'\)algebra D, and restrict to ideals to get the second Green imprimitivity theorem, Corollary 3.9.
We believe that the results are correct and the proofs go verbatim through also for all subinverse semigroups \(T'\) such that their associated groupoid T is discrete. (In other words, when the elements of \(T^{(0)}\) can be expressed purely algebraically as in Definition 2.5.)
Preparing definitions and crossed products
We begin by recalling crossed products in the sense of Khoshkam and Skandalis [11] and Sieben [20], but use several notions from [6]. Let S denote an inverse semigroup.
Definition 2.1
An SalgebraA is a \(C^*\)algebra A endowed with an Saction in the following sense: there exists a semigroup homomorphism \(\alpha : S \rightarrow \text{ End }(A)\), written as \(s(a):= \alpha _s(a)\), such that
for all \(a,b \in A\) and \(s \in S\).
Such a Galgebra (whose definition is equivalent to [6, Def. 3.1]) is a special case of Galgebras in the sense of [20] and [11].
Definition 2.2
Let \({{\mathbb {F}}}(S,A)\), or \({{\mathbb {F}}}\) for brevity, be the universal \(*\)algebra over \({\mathbb {C}}\) generated by disjoint copies of A and S subject to the relations that the \(*\)algebraic relations of A are respected, the multiplication and involution of S are respected, and the relations
hold true for all \(a \in A\) and \(s \in S\). We shall identify S and A as subsets of \({{\mathbb {F}}}\). The algebraic crossed product\(A \rtimes _{alg} S \subseteq {{\mathbb {F}}}\) denotes the linear span of all elements of the form as (\(a \in A, s \in S\)), which are usually denoted by \(a \rtimes s\), and is a \(*\)subalgebra of \({{\mathbb {F}}}\).
Definition 2.3
We denote by \(E(S) \subseteq S\) the idempotent elements of S, and by \(E({{\mathbb {B}}}(S)) \subseteq {{\mathbb {F}}}\) the set of all projections of the form \(e_0 (1e_1) \ldots (1e_n) \in {{\mathbb {F}}}\) with \(e_0,\ldots , e_n \in E(S)\) and \(n \ge 0\). The subset \({{\mathbb {B}}}(S):= \{s p \in {{\mathbb {F}}}\,\, s \in S , p \in E({{\mathbb {B}}}(S))\} \subseteq {{\mathbb {F}}}\) is an inverse semigroup in \({{\mathbb {F}}}\) (under multiplication and taking adjoint). We even shall write \(a \rtimes s:=a s\) when \(a \in A\) and \(s \in {{\mathbb {B}}}(S)\).
As remarked by the referee, note that \({{\mathbb {B}}}(S)\) is what is known as the universal Booleanization of S, see [13] and the references therein. Note that then the identities
hold in \({{\mathbb {F}}}\) for all \(a \in A\) and \(s p \in {{\mathbb {B}}}(S)\) (\(s \in S, p \in E({{\mathbb {B}}}(S))\)).
Definition 2.4
Hence, it is natural to call an expression \(a \rtimes s p \in {{\mathbb {F}}}\) with \(s \in S,p \in E({{\mathbb {B}}}(S))\) and \(a \in A_{s s^*} := s s^*(A)\) to be standard.
The reader should be cautioned that the first relation of (2) is not true in general for \(s \in {{\mathbb {B}}}(S)\) (consider for example \((1e)(a)=0\) for the trivial Saction on A), however the second relation of (2) and identity (1) remain true for \(s \in {{\mathbb {B}}}(S)\) (see Lemma 2.8 below).
The full crossed product\(A \rtimes S\) is the closure of the image of \(A \rtimes _{alg} S\) under the universal \(*\)representation \(\pi\) of \({{\mathbb {F}}}\) on Hilbert space ([11, Def. 5.4] or [6, 5.16, 6.2, 8.4]). It is easy to see with the reduced representations [11, p. 271] that \(\pi\) is injective on \(A \rtimes _{alg} S\), and so the latter is a pre\(C^*\)algebra with a \(C^*\)norm. Sieben’s crossed product \(A {\widehat{\rtimes }} S\) is defined to be the image of \(A \rtimes _{alg} S\) under the universal \(*\)representation \(\tau\) of \({{\mathbb {F}}}\) on Hilbert space satisfying \(\tau (s(a)  s a s^*) = 0\) (see [20]). We write \(a {\widehat{\rtimes }} s\) for \(\tau (a s)\). Note, in particular, that \({\widehat{\rtimes }}\) is compatible:
Notice that this identity is not true for \(\rtimes\), and this compatibility is actually the essential difference between the full crossed product and Sieben’s crossed product.
Definition 2.5
Let us now be given a finite subinverse semigroup \(T' \subseteq S\) of S. Denote by T the (finite) groupoid associated to \(T'\) (cf. [15]). More precisely, let \(T^{(0)} \subseteq {{\mathbb {F}}}\) be the set of all nonzero minimal projections of \(E({{\mathbb {B}}}(T'))\) and
The multiplication within T is that inherited from \({{\mathbb {F}}}\).
It is easy to see that \(T^{(0)}\) consists exactly of the projections of the form \(e_0 \prod _{e < e_0, e \in E(T')} (1e)\) with \(e_0 \in E(T')\), and so \(T^{(0)} = E(T')\), but this fact will not be used in the following.
Definition 2.6
Define
We endow \(S_T\) with an equivalence relation: \(s \equiv u\) if and only if there exists \(t \in T\) such that \(s t = u\) (\(s,u \in S_T\)). We denote by \(S_T/T\) the settheoretical quotient of \(S_T\) by \(\equiv\).
We shall exclusively work with representatives in this quotient; writing \(s \in S_T/T\) means implicitly that \(s \in S_T\) and we use no class brackets; if then \(s \in S_T\) is meant or the class \(s \in S_T/T\) becomes apparent from the context. For an assertion \({\mathcal A}\) we let \([{\mathcal A}]\) be the real number 0 if \({\mathcal A}\) is false, and 1 if \({\mathcal A}\) is true.
Definition 2.7
Let \(C_0(S_T/T)\) denote the commutative \(C^*\)algebra of (continuous) complexvalued functions vanishing at infinity of the (discrete) set \(S_T/T\) with the pointwise operations. The delta function \(\delta _s\) in \(C_0(S_T/T)\) is denoted by s (\(s \in S_T/T\)). The algebra \(C_0(S_T/T)\) is endowed with the Saction \(s(u):= [s u \in S_T] \,\, s u\), where \(s \in S\) and \(u \in S_T/T\) (of course, \(s u \in S_T\) is equivalent to \(s^* s \ge u u^*\)). We let \(A \otimes C_0(S_T/T)\) be the \(C^*\)algebraic tensor product endowed with the diagonal action by S.
Lemma 2.8

(i)
If\(s_1, \ldots ,s_n \in S_T\)are mutually different then\(\sum _{i=1}^n a_i \rtimes s_i = 0\)(sum of standard elements) implies\(a_1 = \cdots = a_n = 0\).

(ii)
TheSaction on anSalgebraAextends naturally to an inversesemigroup\({{\mathbb {B}}}(S)\)action onA (i.e. one sets\((1e)(a):= a  e(a)\)for all\(a \in A\)and\(e \in E(S)\)).

(iii)
The formulas\((a \rtimes s) (b \rtimes u) = a s(b) \rtimes s u\)and\((b \rtimes u)^* = u^*(b^*) \rtimes u^*\)hold in\({{\mathbb {F}}}\)for all\(s, { u} \in {{\mathbb {B}}}(S)\), \(a \in A_{s s^*} := s s^*(A)\)and\(b \in A\).
Proof

(i)
We may assume that all \(s_i\) have the same source projection in \(T^{(0)}\) (otherwise multiply \(\sum a_i \rtimes s_i\) from the right with each single minimal mutually orthogonal projection of \(T^{(0)}\)). Hence we may fix \(e_{1},\ldots ,e_m \in E(S)\) such that for all \(1 \le i \le n\) we have \(s_i= u_i (1 e_1) \ldots (1 e_m)\) for certain mutually different \(u_i \in S\). Expanding, we get \(0=\sum _{i=1}^n a_i s_i = \sum _{i=1}^n a_i u_i  \sum _{i=1}^n a_i u_i e_{1} \pm \ldots\) in \({{\mathbb {F}}}\). Now note that by the reduced representation of the algebraic crossed product \(A \rtimes _{alg} S\) in [11, p. 271] the elements \(u_i, u_j e_1\) et cetera in the last sum are linearly independent, as far as they are all different. Certainly, however, we may conclude that \(a_i u_i = 0\) for all \(1 \le i \le n\), because assuming that \(u_i = u_j e_{k_1} \ldots e_{k_\nu }\) would yield \(s_i = u_j e_{k_1} \ldots e_{k_\nu } (1e_1) \ldots (1e_m) =0\) (however \(0 \notin S_T\)). This yields the claim.

(ii)
By the linear independence of the elements of E(S), it is easy to see that we have already a welldefined semigroup homomorphism \(\alpha : E({{\mathbb {B}}}(S)) \rightarrow \text{ End } (A)\) defined by \(\alpha _e(a)= e(a)\) and \(\alpha _{1e}(a) = a  e(a)\) for all \(a \in A\) and \(e \in E(S)\). To extend it to \({{\mathbb {B}}}(S)\), we consider an ambiguous representation \(0 \ne s p = u q \in {{\mathbb {B}}}(S)\) for some \(u,s \in S\) and \(p,q \in E({{\mathbb {B}}}(S))\). Then \(s p q = u p q\) and so \(s = u\) by a similar argument as in (i). Thus \(s^* s p = s^* s q\) in \(E({{\mathbb {B}}}(S))\). Hence, the definition \(\alpha _{s p}:= \alpha _s \alpha _p = \alpha _s \alpha _{s^* s p} = \alpha _{u} \alpha _q\) is welldefined (\(\alpha _s\) denotes the given Saction).

(iii)
Let \(s,u \in S\), \(p,q \in E({{\mathbb {B}}}(S))\), \(a \in A_{s p s^*}\) and \(b \in A\). We have, for example, by (1) for the extended action of (ii), (2) and because \(a = s p s^*(a) \in A_{s p s^*}\) that
$$\begin{aligned} a s p \cdot b u q = s p s^*(a) \cdot s p \cdot b \cdot u q = s p s^*(a) \cdot s(b) \cdot s p\cdot u q = a \cdot s p(b) \cdot s p u q \end{aligned}$$in \({{\mathbb {F}}}\).
\(\square\)
Lemma 2.9
IfAis anSalgebra and\(I \subseteq A\)anSinvariant ideal inAthen\(I \rtimes S \subseteq A \rtimes S\)and\(I {\widehat{\rtimes }} S \subseteq A {\widehat{\rtimes }} S\)canonically.
Proof
A representation of \({{\mathbb {F}}}(S,I)\) is given by a covariant triple \((\sigma ,U,H)\) for some Hilbert space H, a \(*\)homomorphism \(\sigma : I \rightarrow B(H)\) and an inverse semigroup homomorphism \(U: S \rightarrow B(H)\) satisfying the analogous defining relations as in (2). Proceed as in [3, Lemma A.4] to show that the representation of \({{\mathbb {F}}}(S,I)\) extends to \({{\mathbb {F}}}(S,A)\). Hence \(\overline{{{\mathbb {F}}}(S,I)} \supseteq I \rtimes S \rightarrow A \rtimes S \subseteq \overline{{{\mathbb {F}}}(S,A)}\) is isometric. \(\square\)
The imprimitivity theorem
We shall now introduce an imprimitivity bimodule in the sense of Rieffel [19]. Recall that \(T' \subseteq S\) denotes a finite subinverse semigroup, T its associated finite groupoid of Definition 2.5 and A an Salgebra.
Definition 3.1
We introduce the spaces
The spaces \(B_0 \subseteq A \rtimes S\) and \(E_0\) are regarded as pre\(C^*\)algebras. We make \(X_0\) to a right preHilbert module over \(B_0\) (cf. [19, Def. 2.8]) by the following operations
for \(a,b \in A\), \(c \in A_{t t^*}\), \(s,u \in S_T\) and \(t \in T\), and to a left preHilbert module over \(E_0\) by
for \(a,b \in A\), \(r \in S_T/T\), \(s \in S\) and \(j, g, h \in S_T\).
Note that \(B_0\) is actually a normclosed \(C^*\)algebra and in fact the ordinary groupoid crossed product \(B_0= A \rtimes T\). For a discrete groupoid \({\mathcal G}\), \(A \rtimes _{alg} {\mathcal G}\) is defined to be the function space \(\{f:{\mathcal G}\rightarrow A \, f(g) \in A_{g g^{1}} =A_{r(g)}, \, \text{ carrier }(f) \text{ is } \text{ finite }\}\) (written \(f \equiv \sum _{g \in {\mathcal G}} f(g) \rtimes g\)) with the natural involution \((f(g) \rtimes g)^* = g^{1}(f(g)^*) \rtimes g^{1}\) and convolution \((f_1(g) \rtimes g)(f_2(h) \rtimes h) = f_1(g) \, g(f_2(h)) \rtimes gh\) (if g, h are composable, otherwise the product is zero). For a finite groupoid, \(A \rtimes _{alg} {\mathcal G}\) is already normclosed in the \(C^*\)norm of the the left regular representation, and thus in any \(C^*\)norm, and so is \(A \rtimes {\mathcal G}\).
Because of identity (3) we may write every element in an algebraic crossed product as the sum of standard elements. Because of the linear independence statement of Lemma 2.8.(i) we may then extend the formulas of Definition 3.1 for standard expressions by linearity. We need however remark, that
Lemma 3.2
The formulas of Definition 3.1remain however valid also for nonstandard expressions as stated.
Proof
For example, by considering the formula of the map \(E_0 \times X_0 \rightarrow X_0\), given elementary elements \(a \otimes r \rtimes s s^*s \in E_0\) and \(b \rtimes j p \in X_0\) with \(a \in A, r \in S_T/T, s,j \in S\) and \(s^* s, p \in E({{\mathbb {B}}}(S))\), we go over to their standard form \([s s^* r \in S_T] s s^*(a) \otimes s s^* r \rtimes s \in E_0\) and \(j j^*(b) \rtimes j p \in X_0\) according to (3). Then their module product in \(X_0\) is
by (1), (3) and because \([s s^* r \in S_T]\) cancels because \(s s^* r \equiv s j p\) implies \(s s^* r = s j p t\) for some \(t \in T^{(0)}\), implies \(s s^* r = r\) (since the source projection of \(s s^* r = s j p t \in S_T\) is in \(T^{(0)}\) and thus cannot become smaller than the one of \(r \in S_T\)), implies \(s s^* r = r \in S_T\). This is the same element as taking the module product formula in \(X_0\) for the given nonstandard elements \(a \otimes r \rtimes s \in E_0\) and \(b \rtimes j p \in X_0\). The above formulas for nonstandard expressions can then be extended also linearly. \(\square\)
Proposition 3.3
Straightforward, but again rather timeconsuming, extensive and tedious computations show that we have
for all\(x,y,z \in X_0\), \(b \in B_0\)and\(f \in E_0\) (cf. [19, Def. 6.10]).
Proof
For convenience of the reader we demonstrate the first identity of line (5), which is the most difficult of all of these, and the others should not present further new difficulties (for the \(*\)algebraic operations in \(B_0\) use the formulas of Lemma 2.8.(iii)). We have
for \(a,b,c \in A\), \(r \in S_T/T\), \(s \in S\) and \(g,h \in S_T\), and
We only need to show that the scalar coefficients of the expressions (6) and (7) coincide, because observe that the vector parts coincides by a single application of identity (1). Note that the scalar \([g^* s^* h \in T]\) appears both in (6) and (7). Suppose that (6) is nonzero. Then \(sg \in S_T\), and thus \(s^* s \ge g g^*\) because \(g \in S_T\) and \(s g \in S_T\) (the source projection of g is in \(T^{(0)}\) and cannot be made smaller in sg). On the other hand, \(r \equiv sg\) and so there exists some \(t \in T\) such that \(r = sg t\). Hence, \(s^* r = s^* s g t = g t \in S_T\). Thus \([s^* r \in S_T]\) appearing in (7) is nonzero. Since \(g^* s^* h \in T\), both its source and range projection are in \(T^{(0)}\). Hence, since \(h \in S_T\), \(s^* h\) must be in \(S_T\) too in order not to lose information on the source projection of \(s^* h\). Since \(g \in S_T\), the source projection of \(g^*\) and the range projection of \(s^* h\) must perfectly fit together such that \(g^* s^* h \in T\). But this implies that \(g \cdot (g^* s^* h) = s^* h\) and hence \(g \equiv s^* h\). This implies \(s^* r \equiv g \equiv s^* h\). We have obtained that \([s^* r \equiv s^* h]\,[s^* h \in S_T]\) appearing in (7) is nonzero. Hence (7) is nonzero. In the other direction suppose that (7) is nonzero. Since \(g^* s^* h \in T\) and \(g,h \in S_T\), we can completely analogously deduce as before that \(s g \equiv h\). This already implies that \([s g \in S_T]\) appearing in (6) is nonzero. Since \(s^* h, s^* r \in S_T\) and \(h,r \in S_T\), we get \(s s^* \ge h h^*, r r^*\). Hence \(s^* r \equiv s^* h\) implies \(r \equiv h \equiv s g\). Thus (6) is nonzero. \(\square\)
Lemma 3.4
The inner products of \(X_0\) are positive.
Proof
Let \((a_\alpha )_\alpha\) be an approximate identity of A. Let \(x=\sum _{s=1}^m b_k \rtimes h_k\) in \(X_0\) and choose for every different equivalence class \(h_k T\) in \(S_T/T\) exactly one representative \(g_i := h_k \in S_T\), where \(1 \le i \le n\) with \(n \le m\). (We have a new index i because for \(h_{k_1} T= h_{k_2} T\) we would choose a common \(g_i\).) Set \(x_{i,\alpha } = a_\alpha \rtimes g_i \in X_0\). Set \(x_\alpha = \sum _{i=1}^n \langle x_{i,\alpha },x_{i,\alpha } \rangle _{E_0} x \in X_0\). Then
where identity (8) follows from the fact that \(g_i g_i^* h_k \in S_T\) and \(h_k \in S_T\) implies \(g_i g_i^* \ge h_k\), and so \(g_i \equiv g_i g_i^* h_k\) implies \([g_i \equiv h_k] \ne 0\), which, on the other hand, implies \(g_i g_i^* = h_k h_k^*\) and thus \([g_i g_i^* h_k \in S_T] \, [g_i \equiv g_i g_i^* h_k] \ne 0\). Identity (9) follows because we have chosen for each \(h_k\) one but at most one equivalent \(g_i\). We used also (1) and Lemma 2.8.(ii) there. Also (1) is used to easily compute that \(\langle x, xx_\alpha \rangle _{B_0} \rightarrow 0\). Consequently,
as in Green [8], page 202, with the last identity of (5). The argument for the positivity of \(\langle x, x\rangle _{E_0}\) is similar. Here we choose, for example, \(x_\alpha = x \sum _{e \in T^{(0)}} \langle a_\alpha \rtimes e, a_\alpha \rtimes e \rangle _{B_0}\). \(\square\)
Proposition 3.5
We have the inequalities
for all\(x \in X_0\), \(f \in E_0\)and\(b \in B_0\) (cf. [19, Def. 6.10]).
Proof
For the proof of the first inequality regard \(X_0\) as an preHilbert \(B_0\)module. Let \({\mathcal M}(A)\) denote the multiplier algebra of A. For a nonzero standard element \(f=a \otimes r \rtimes s \in E_0\) we compute \(f^* f\) with the formulas of Lemma 2.8.(iii), and for another \(x \in X_0\) we obtain, with Proposition 3.3,
where \(z := \big (\Vert f \Vert _{E_0}^2  s^*(a^* a ) \big )^{1/2} \otimes s^* r \rtimes s^* s\) and \(p := \Vert f \Vert _{E_0}^2 \otimes s^* r \rtimes s^* s\) are elements in \(\big ({\mathcal M}(A) \otimes C_0(S_T/T) \big ) \rtimes S\). Of course, we had here temporarily to replace our coefficient algebra A by \({\mathcal M}(A)\) in order to include \(\Vert f\Vert _{E_0}\) and have therefore some slightly larger new \(E_0\). (Note that for general \(f \in E_0\), \((\Vert f \Vert _{E_0}^2  f^* f)^{1/2}\) need not be in \(E_0\) and that is why we need to consider elementary elements f.)
By applying the norm \(\Vert \cdot \Vert _{B_0}\) in \(B_0\) to this inequality, we obtain \(\Vert f x \Vert \le \Vert f\Vert _{E_0} \, \Vert x\Vert\) (where \(\Vert x\Vert := \Vert \langle x,x\rangle _{B_0}\Vert ^{1/2}\)) for such elementary elements \(f \in E_0\), and by taking sums of such elements we readily obtain \(\Vert f x \Vert \le \Vert f\Vert _{\ell ^1 (S,A \otimes C_0(S_T/T) )} \, \Vert x\Vert\) for all \(f \in E_0\). Hence, the \(E_0\)module multiplication on \(X_0\) is a \(*\)homomorphism \(E_0 \rightarrow {\mathcal L}(X_0)\) which is an \(\ell ^1\)contractive representation into a pre\(C^*\)algebra. Since by definition the \(C^*\)norm closure of \(E_0\) is the enveloping \(C^*\)algebra of \(\ell ^1(S,A \otimes C_0(S_T/T) )\) (cf. [11]) and so induced by the sum over all \(\ell ^1\)contractive representations, we must get \(\Vert f\Vert _{{\mathcal L}(X_0)} \le \Vert f\Vert _{E_0}\). It is well known from the theory of Hilbertmodules that one has \(\langle f x, f x \rangle _{B_0} \le \Vert f\Vert ^2_{{\mathcal L}(X_0)} \langle x, x \rangle _{B_0}\) for adjointable operators f (see for instance Lance [12], Prop. 1.2), and hence the first inequality of (10). The second inequality of (10) is proved similarly (but is easier as \(B_0\) is even normclosed). \(\square\)
Definition 3.6
Denote by \(E_X \subseteq \overline{E_0}\) the norm closure of \(\langle X_0,X_0\rangle _{E_0}\) under the \(C^*\)norm \(\Vert \cdot \Vert _{E_0}\), and by \(B_X \subseteq \overline{B_0}\) the norm closure of \(\langle X_0,X_0\rangle _{B_0}\) under the \(C^*\)norm \(\Vert \cdot \Vert _{B_0}\). We now apply the argument following [18, Prop. 3.1] to see that \(X_0\) may be completed in seminorm \(\Vert x\Vert = \Vert \langle x,x\rangle _{B_0}\Vert ^{1/2}\) (after factoring out the elements of norm 0) to obtain an \(E_XB_X\) imprimitivity bimodule X.
Theorem 3.7
Let\(T'\)be a finite subinverse semigroup of an inverse semigroupSand denote byTits associated finite groupoid (Definition2.5). LetAbe anSalgebra. Then we have a\(C^*\)algebraic Morita equivalence
via the\(E_XB_X\)imprimitivity bimoduleXand isomorphisms\(E_X \cong C_0(S_T/T,A) {\widehat{\rtimes }} S\)and\(B_X \cong A {\widehat{\rtimes }} T'\).
Proof
The \(C^*\)algebra \(B_X = B_0\) is canonically isomorphic to the groupoid crossed product \(A \rtimes T\), which is canonically isomorphic to the inverse semigroup crossed product \(A {\widehat{\rtimes }} T'\) by [16, Thm. 7.2]. To meet exactly the assumptions in [16], switch to the carrier algebra\({\tilde{A}} = p(A)\) for \(p = \sum _{e \in T^{(0)}} e\) of A, which does not change the crossed product, that is, \({\tilde{A}} {\widehat{\rtimes }} T' = A {\widehat{\rtimes }} T'\).
Denote by \(C_0(S_T/T,A) \subseteq A \otimes C_0(S_T/T)\) the norm closure of the linear span of
Note that \(C_0(S_T/T, A)\) is an Sinvariant ideal in \(A \otimes C_0(S_T/T)\) and so
embeds by Lemma 2.9. Using (11), let
be the canonical map (\(a \in A, r \in S_T/T\) and \(s \in S\)). (Note that there is always a canonical map \(A \rtimes S \rightarrow A {\widehat{\rtimes }} S\).) It is surjective, because given a nonzero elementary element \(a a^* \otimes r {\widehat{\rtimes }} s\) in \(C_0(S_T/T,A) {\widehat{\rtimes }} S\) with \(a \in A_{r r^*}, r \in S_T/T\) and \(s \in S\) we note that
by the permeability (compatibility) of \({\widehat{\rtimes }}\) for projections in \(E({{\mathbb {B}}}(S))\), see (4), so that we may assume that the given element \(a a^* \otimes r {\widehat{\rtimes }} s\) satisfies \(a \in A_{r r^*}, r \in S_T/T\) and \(s \in {{\mathbb {B}}}(S)\) with \(r r^* = s s^*\). Hence we get
If \(\sigma\) were not injective, then its kernel \(J \subseteq E_X\) were nonzero, and so would correspond to a nonzero ideal I in \(B_X\) via the \(E_XB_X\) imprimitivity module X (see [18, Cor. 3.1]), which then would contain a nonzero element of the form \(a \rtimes e \in I\) with \(e \in T^{(0)}\). A nonzero element of the form \(f=\langle a \rtimes e,a \rtimes e \rangle _{E_0}\) would be in J, however \(\sigma\) is nonzero on f. We have obtained our result. \(\square\)
Definition 3.8
Now assume that D is a \(T'\)algebra. Define, similarly as in [10, §5 Def. 2],
It is a \(C^*\)algebra under the pointwise operations and the supremum’s norm and becomes an Salgebra under the Saction \((s f)(u) := [s^* u \in S_T ] \,\,f ( s^* u)\) for \(s \in S\), \(u \in S_T\) and \(f \in \text{ Ind }_{T'}^S(D)\).
Corollary 3.9
Let\(T' \subseteq S\)be a finite subinverse semigroup of an inverse semigroupS. LetDbe a\(T'\)algebra. Then we have a\(C^*\)algebraic Morita equivalence
Proof
Let A denote \(\text{ Ind }_{T'}^S(D)\). Consider the \(T'\)invariant ideal \(A_0\) of A consisting of all functions which vanish outside T. Let us again view \(B_X\) as \(B_X = A \rtimes T\) as in the last proof before. Then \(A_0 \rtimes T\) embeds canonically as an ideal J in \(A \rtimes T=B_X\), and by [18, Cor. 3.1], associated to J is the submodule in X generated by
and the ideal I in \(E_X\) generated by (and actually norm closure of)
Note that we are identifying \(E_X \cong C_0(S_T/T,A) {\widehat{\rtimes }} S\) by the isomorphism \(\sigma\) stated in (12). Using Lemma 2.9, the ideal I is canonically isomorphic to \(K {\widehat{\rtimes }} S \subseteq E_X\), where K denotes the Sinvariant ideal in \(C_0(S_T/T,A)\) which is the norm closure of the linear span of
To see that the identical embedding \(I \rightarrow K {\widehat{\rtimes }} S\) is surjective, write a given nonzero element \(g(a) \otimes g {\widehat{\rtimes }} s \in K {\widehat{\rtimes }} S\) (\(a \in A_0, g \in S_T/T\) and \(s \in S\)) as
with \(s s^*g, s^* g \in S_T\) by the compatibility of \({\widehat{\rtimes }}\), see (4). We have an Sequivariant isomorphism \(\psi :A \rightarrow K\) defined by
where \(f \in A = \text{ Ind }_{T'}^S(D)\) and \(f_{g}\in A\) denotes the function \(f_{g}(k)= [k \equiv g]f(k)\) for all \(k \in S_T\).
There is a \(T'\)equivariant epimorphism \(\Phi : D \rightarrow A_0\) given by \(\Phi (d )(t) = t^* (d)\) for \(t \in T\) and \(d \in D\). It is an isomorphism on the carrier algebra of D, so that \(D {\widehat{\rtimes }} T' \cong A_0 {\widehat{\rtimes }} T' \cong A_0 \rtimes T \cong J\). Consequently we have obtained, by restricting to the ideals I and J in Theorem 3.7 and applying [18, Cor. 3.1], our result. \(\square\)
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Acknowledgements
We thank the Universidade Federal de Santa Catarina in Florianópolis for the support we received when developing the content of this paper in 2014. This note presents a slightly modified version (more detailed proofs) of a preprint in arXiv from 2014.
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Burgstaller, B. An elementary Green imprimitivity theorem for inverse semigroups. Semigroup Forum 100, 141–152 (2020). https://doi.org/10.1007/s00233020100848
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Keywords
 Imprimitivity theorem
 Inverse semigroup
 Induction
 Crossed product
 Morita equivalence