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2.4.1 Preparation for composition

Let Z, Ω be spaces, let π : Z → Ω be a surjection and λ a family of measures along π. Let XX :=X×π,πX and let π1,π2 be the projection maps from ZZ to the first and second copy of Z, respectively. The family of measures λinduces families of measures λ2 and λ1 along π1 and π2, respectively, as in Lemma 1.3.17. For xZ the measure λ1x on π1−1(x) is given by δx×λπ(x). And λ1x is defined similarly. This data gives Figure 2.2

Observation 2.4.1. The composite families of measures λλ2 and λλ1 on ZZ are the same family of measures along ππ1 =ππ2:ZZ →Ω. We denote this family of measures by λ×λ, where {(λ◦λ)u =λu×λu}u∈Ω, that is, for u∈Ω,fCc(X∗X),

Z

fd(λ×λ)u = Z

f(x, y) dλu(x) dλu(y).

Observation 2.4.2. Let f1, f2 ∈ B+(Z) and let m be a measure on Z. By an abuse of notation we write f1f2 for the restriction of f1f2 to ZZ. Then mλ1(f1f2) =mλ2(f1f2) means m(λ(f1)f2) = m(f1λ(f2)). In this situation, we say that λ1 and λ2 are symmetric with respect to m.

ZZ Z

Z

λ2

π2

π1

λ1 π λ

λ π

Figure 2.2

Notice that we are abit loose with the notation in Observation 2.4.2 because m(λ(f1)f2) = m(f1λ(f2)) means m((λ(f1)◦π)f2) =m(f1(λ(f2)◦π)).

Proposition 2.4.3. LetZ,be spaces,π :Z →Ωa surjection andλaπ-family of measures onZ.

Let π12 be the projection maps fromZZ onto the first and the second copy of Z.

i) Letµbe a measure onΩ. Thenλ1 andλ2 are symmetric with respect tom=µλin the sense of Observation 2.4.2.

ii) Let m be a measure on Z. If λ1 and λ2 are symmetric with respect to m and there is a non-negative Borel function e on Z with λ(e) = 1, then there is a measure µ onwith µλ=m.

iii) The measureµin(ii)withµλ=m is unique.

Recall from the discussion that followed Definition 1.3.1 that we work with proper families of measures only. Thus we always have a function e as in (ii) above.

Proof. (i): If µλ = m then λ1 and λ2 are symmetric with respect to m because mλ2 = (µ◦λ)λ2=µ◦(λ◦λ2) =µ◦(λ◦λ1) = (µ◦λ)λ1=mλ1. The equality λλ2 =λλ1 follows

from Observation 2.4.1.

(ii): For g∈ B+(Ω) defineµ(g) :=m((gπ)·e). In Observation 2.4.2 let f1=f, λ(e) = 1 and take g=λ(f) in the definition ofµ in the previous sentence. Then

m(f) =m(f·λ(e)) =m((λ(f)◦π)·e) =µ(λ(f)) =µλ(f).

(iii): Let µ0 be another measure on Ω which satisfies the condition µ0λ= m. Since λ is a proper family of measures, the integration map Λ :Cc(Z)→Cc(Ω) is surjective. Soµλ=µ0λ implies µ=µ0.

Forπ:Z →Ωthe fibre product ZZ is the groupoid of the equivalence relation defined by xy if and only if π(x) =π(y). For an equivalence groupoid relation(x, y)−1 = (y, x),sX∗X =π2

and rX∗X =π1. Now we study the case when the measures λ1 and λ2 are not symmetric, but weakly symmetric. The measures λ1 and λ2 are called weakly symmetric if there is a continuous homomorphism ∆ : ZZ → R+ with mλ2 = ∆·(m◦λ1). In Section 1.4, we saw that a homomorphism from a groupoid Gto an abelian group R is also called an R-valued 1-cocycle.

It is a well-known fact that ZZ is a proper groupoid (see Lemma 2.4.9 for the proof). Assume that the measures λ1 and λ2 are weakly symmetric. Let ∆ : ZZ → R+ be the R+-valued 1-cocycle that implements the weak equivalence. Thenlog◦∆ :ZZ →Ris anR-valued 1-cocycle.

Proposition 1.4.10 says that log◦∆ =bsbr for some continuous function b:Z →R. Thus

∆ = eb◦s eb◦r. Write b= eb, then b >0 and

∆ = eb◦s

eb◦r = ebs

ebr = bs

br = bπ2

bπ1. Now we have mλ2 =b◦πb◦π2

1

mλ1, which is equivalent to(b◦π1)(m◦λ2) = (b◦π2)(m◦λ1).

An easy calculation shows that(b◦π1)(m◦λ2) = (bm)◦λ2 and (b◦π2)(m◦λ1) = (bm)◦λ1. Thus we get

Proposition 2.4.4. LetZ,Ω,πand λbe as in Proposition2.4.3and letmbe a measure onZ with respect to whichλ1andλ2are weakly symmetric. Letbe theR+-valued1-cocycle that implements the weak equivalence. Then there is a functionb:Z →R with

i) b(y)b(x) = ∆(x, y)for all (x, y)∈ZZ;

ii) λ1 andλ2 are symmetric with respect to the measurebm, that is,bmλ1 =bmλ2. 2.4.2 Composition of topological correspondences

Let (X, α,∆1) and (Y, β,∆2) be correspondences1 from (G1, λ1) to (G2, λ2) and from (G2, λ2) to (G3, λ3), respectively. This is pictured in Figure 2.3

X Y

(G1, λ1) (G2, λ2) (G3, λ3)

1

α

2

β

Figure 2.3

We need to create a G1-G3-bispace Ω equipped with aG3-invariant andG1-quasi-invariant family of measuresµ={µu}

u∈H3(0). The C(G1)-C(G3)-Hilbert moduleH(Ω)should be isomorphic to the Hilbert module H(X)⊗C(G2)H(Y).

Let Z :=XY be the fibre product over G(0)2 for the maps sX and rY. ThenZ carries the diagonal action of G2. Since the action ofG2 on X is proper, its action on Z is proper. Define the space Ω =Z/G2.

1See the paragraph ‘Important Conventions 2.1’ on page 41.

Observation 2.4.5. The spaceZ is aG1-G3-bispace. The momentum maps are rZ(x, y) =rX(y) and sZ(x, y) =sY(z). For (γ1,(x, y))∈G1Z and ((x, y), γ3)∈ZG3, the actions areγ1·(x, y) = (γ1x, y) and(x, y)·γ3= (x, yγ3), respectively. These actions make Ωinto a G1-G3-bispace.

Lemma 2.4.6. The obvious right action ofG3 onis proper.

Proof. See [42, Proposition 7.6].

The quotient map π:Z →Ω carries the family of measures λ2Z as in Proposition 1.3.21. We write λ={λω}ω∈Ω instead of λ2Z ={λ2ωZ}ω∈Ω. Recall that for fCc(Z),

Z

fω=[x,y]:=

Z

GrY2 (y)

f(xγ, γ−1y) dλr2Y(y)(γ).

Proposition 1.3.21 shows thatλis a continuous family of measures with full support.

For a fixed uG(0)3 we define a measure mu on the spaceZ as follows: for fCc(Z), Z

Z

f dmu= Z

Y

Z

X

f(x, y) dαrY(y)(x) dβu(y). (2.4.7) Lemma 2.4.8. The family of measures{mu}

u∈G(0)3 is aG3-invariant continuous family of measures on Z.

Proof. The G3-invariance of the family of measuresβ makes {mu}

u∈G(0)3 G3-invariant.

LetfCc(X) and gCc(X), then Z

fgdmu=B((A(f)◦rY)g)(u).

Using a density argument as in the proof of Lemma 1.3.20 we conclude that {mu}

u∈G(0)3 is a continuous family of measures.

We wish to prove that up to equivalence {mu}u∈G(0) can be pushed down from Z to Ω to a G3-invariant family of measures {µu}. Before we proceed we prove a small lemma. Denote XG2Y :={(x, γ2, y)X×G2×Y :sX(x) =rG2) =rY(y)}. And letXG2 =X×sX,rG

2 G2 and G2Y =G2×rG

2,rY Y.

Lemma 2.4.9. Let(X, α,∆1)and(Y, β,∆2)be correspondences from(G1, λ1) to(G2, λ2)and from (G2, λ2)to (G3, λ3), respectively. LetZ,Ω,λ,mu, λi fori= 1,2be as discussed above. For each uG(0)3 there is a function bu on Z such that λ1 andλ2 are symmetric with respect to bu·mu. Furthermore,bsatisfiesb(x, y)b(xγ, γ−1y)−1= ∆((x, y),(xγ, γ−1y)) = ∆2(γ, γ−1y).

We shall writeb instead of bu. We work with a single µu at a time, so we prefer to drop the suffix u.

Proof. The proof follows in the steps below:

i) λ1 andλ2 are weakly symmetric with respect tomu for each uG(0)3 .

ii) ZZ is a proper groupoid.

iii) Appeal to Proposition 2.4.4 and get the result.

(i): Now we show that λ1 andλ2 are weakly symmetric families of measures. Figure 2.4 shows all maps and the families of measures along the maps:

ZZ Z

Z

λ2

π1

λ1 π2 π λ λ π

Figure 2.4 Let fCc(Z∗Z), then

(muλ2)(f) = Z Z

f((x, y),(xγ, γ−1y)) dλr2Y(y)(γ) dmu(x, y)

= Z Z Z

f((x, y),(xγ, γ−1y)) dλr(y)2 (γ) dαrY(y)(x) dβu(y).

Change variables (xγ, γ−1y)7→(x, y). Recall that the family α is G2-invariant and β is G2 -quasi-invariant. Now calculating further:

R. H. S.= Z Z Z

f((xγ−1, γy),(x, y)) ∆2−1, γy) dλr(y)2 (γ) dαrY(y)(x) dβu(y)

= (muλ1)(f ·∆2◦invG2nY),

where invG2nY is the inverse function on the groupoid G2nY.

(ii): Observe that rZ∗Z×sZ∗Z: ZZZ ×Z is the inclusion map. Hence to show that ZZ is proper, it suffices to prove that ZZZ ×Z is closed. To see this, we observe that Ω is Hausdorff, hence dia(Ω) := {(ω, ω) : ω ∈Ω} is closed in Ω×Ω. Since π is continuous, (π×π)−1(dia(Ω))⊂Z×Z is closed where π×π:Z×Z →Ω×Ωis the canonical map.

(iii): Due to (i) and (ii), we may apply Proposition 2.4.4 which gives a function b:Z →R such that λ1 and λ2 are symmetric with respect to bmu.

Remark 2.4.10. The cocycle ∆ :ZZ →R,∆((x, y),(xγ, γ−1y)) = ∆2(γ, γ−1y), implements the weak symmetry between λ1 and λ2. We observe:

i) since∆ does not depend onx,∆is G1-invariant;

ii) ∆2 is defined on G2∗(Y /G3)(see Remark 2.1.5). Hence ∆((z, z03) = ∆(zγ3, z0γ3) = ∆(z, z0) with sZ(z) =sZ(z0) =rG33),γ3G3. Thus ∆depends only on γ and[y].

The function b appearing in Lemma 2.4.9 can be computed explicitly. Let p ={pz}z∈Z be a family of probability measures on ZZ as in Lemma 1.3.29. Then Corollary 2.4.4 gives

b(z0) = exp(b)(z0) = exp Z

log◦∆((z, z0)) dpz0(z)

. (2.4.11)

This implies that bis continuous on Z.

Remark 2.4.12. i) TheG1-invariance of∆from Remark 2.4.10 and Equation 2.4.11 clearly implies that b is G1-invariant.

ii) The G3-invariance of∆(Remark 2.4.10 and Equation 2.4.11) implies b is G3-invariant. Indeed, for γ3G3

Due to Proposition 2.4.3 the measure µu is independent of the choice of the function e. Sometimes we abuse notation and writef instead of fπ. We think off as a function onZ itself.

Recall that Ωis a G1-G3-bispace (see Observation 2.4.5).

Proposition 2.4.14. The family of measuresu}

u∈G(0)3 is a G3-invariant continuous family of measures onalong the momentum maps.

Proof. We check the invariance first and then check the continuity. Let fCc(Ω) and γG3, and

Thus {µu}u∈G

3(0) is G3-invariant.

Now we check that µis a continuous family of measures. LetM, µand Λdenote the integration maps which the families of measures m, µ and λ induce between the corresponding spaces of continuous compactly supported functions. By the construction itself,M :Cc(Z)→Cc(G(0)3 ) is the composite ofCc(Z)−→Λ Cc(Ω)−→µ Cc(G(0)3 ). Due the definition ofµthe following diagram commutes:

Cc(Z) Cc(Ω)

Cc(G(0)3 ).

Λ

M µ

Lemma 2.4.8 shows that M is continuous, Proposition 1.3.21 shows that Λ is continuous and surjective. Hence µ is continuous.

The family of measures µ on Ω is the required family of measures for the composite corre-spondence. We still need to show that it is G1-quasi-invariant. Let fCc(G1nΩ) and uG(0)3 , then

Z

f−1,[x, y]) dλr1([x,y])(η) dµu[x, y]

= Z Z Z

f−1,[x, y])e(xγ, γ−1y)b(x, y) dλr1X(x)(η) dλs2X(x)(γ) dαrY(y)(x) dβu(y).

We apply Fubini’s Theorem to the last step to get

r1X(x)(η) dλs2X(x)(γ) dαrY(y)(x)7→dλr1X(x)(η) dαrY(y)(x)dλs2X(x)(γ).

Now we change (η−1,[x, y])7→(η,[η−1x, y]). Then

r1X(x)(η) dαrY(y)(x)7→∆1(η, η−1x) dλr1X(x)(η) dαrY(y)(x).

We incorporate this change and apply Fubini’s theorem again to get the same sequence of the integrals and compute further:

Z Z Z

f(η,[η−1x, y])e(η−1xγ, γ−1y)b(η−1x, y) ∆1(η, η−1x) dλr1X(x)(η) dλs2X(x)(γ) dαrY(y)(x) dβu(y)

= Z Z Z

f(η,[η−1x, y])b(η−1x, y)

b(x, y)1(η, η−1x)e(η−1x, y)b(x, y) dλr1X(x)(η) dαrY(y)(x)λs2X(x)(γ) dβu(y).

But e(η−1xγ, γ−1y) dλr1X(x)(η) = 1 and b(x, y) dαrY(y)(x) dβu(y) = dµu[x, y]. the last term equals Z Z

f(η,[η−1x, y])b(η−1x, y)

b(x, y)1(η, η−1x) dλr1([x,y](η) dµu[x, y].

Thus if ∆1,2 :G1nΩ→R+ is defined as

1,2−1,[x, y]) =b(η−1x, y)−11−1, x)b(x, y), (2.4.15)

then the above computation gives Z

f−1,[x, y]) dλ1(η) dµu[x, y] = Z Z

f(η,[η−1x, y]) ∆1,2(η, η−1[x, y]) dλ1(η) dµu[x, y], for alluG(0)3 . One must check that the function∆1,2 makes sense. We prove the following lemma for this purpose.

Lemma 2.4.16. The function1,2defined in Equation(2.4.15)is a well-definedR+-valued continuous 1-cocycle on the groupoidG1nΩ.

Proof. Let (xγ, γ−1y)∈[x, y], then

1,2−1,[xγ, γ−1y]) =b(η−1xγ, γ−1y)−11−1, xγ)b(xγ, γ−1y)

=b(η−1x, y)−11−1, x)b(x, y) b(η−1x, y) b(η−1xγ, γ−1y)

b(xγ, γ−1y) b(x, y)

!

= ∆1,2−1,[x, y])2−1, γ−1y)∆2−1, γ−1y)−1

= ∆1,2−1,[x, y]).

In the above computations, to get the third equality, we used the last claim in Lemma 2.4.9. Due to the continuity of b and ∆1,∆1,2 is continuous. Checking that∆1,2 is a groupoid homomorphism is a routine computation.

Proposition 2.4.17. The family of measuresu}

u∈G(0)3 isG1-quasi-invariant. The adjoining function for the quasi-invariance is given by Equation(2.4.15).

Proof. Clear from the discussion above.

Definition 2.4.18 (Composition). For correspondences

(X, α,∆1) : (G1, λ1)→(G2, λ2) and (Y, β,∆2) : (G2, λ2)→(G3, λ3),

their composite correspondence (Ω, µ,∆1,2) : (G1, λ1)→(G3, λ3) is defined by:

i) a spaceΩ := (X∗Y)/G2,

ii) a family of measures µ = {µu}u∈G(0) 3

that lifts to {bα×βu}u∈G(0) on Z for a cochain b ∈ C0G

3(Z∗Z,R+) satisfying d0(b) = ∆.

The∆ above is the one in Remark 2.4.10. C0G

3 is the zeroth cochain group of the G3-invariant R+-valued continuous cochain complex of XX (see Definition 1.4.7). For a composite correspon-dence the adjoining function ∆1,2 is the one given by Equation (2.4.15).

Theorem 2.4.19. Let(X, α) : (G1, λ1) → (G2, λ2) and (Y, β) : (G2, λ2) → (G3, λ3) be topological correspondences. Assume that the topologies are locally compact, Hausdorff and second count-able. Let (Ω, µ) : (G1, λ1) → (G3, λ3) be a composite of the correspondences. Then H(Ω) and H(X) ˆ⊗C(G2)H(Y)are isomorphic correspondences fromC(G1, λ1) toC(G3, λ3).

We make a remark before commencing the proof of the theorem.

Remark2.4.20.The function∆in Remark 2.4.10, is a cocycle inC1G

3(Z∗Z;R+), andb∈C0(Z∗Z;R+) is a cochain. Remark 2.4.12 says thatb∈C0G

3(Z∗Z;R+), and Corollary 2.4.4 gives that ∆ =d0(b).

Let H(Ω, b) denote the C(G3, λ3)-Hilbert module obtained using {µu=e·b·α×βu}

u∈G(0)3 . Letb0 be another G3-equivariant 0-cochain with∆ =d0(b0) and let H(Ω, b0) be the C(G3, λ3)-Hilbert module obtained by using {µ0u = e·b0 ·α×βu}

u∈G(0)3 as family of measures. Corollary 2.5.18 gives an isomorphism from the C-correspondence H(Ω, b) toH(Ω, b0). Hence in the statement of Theorem 2.4.19 we need not refer to a certain fixed 0-cochain. In the proof of the theorem, we work with a fixed cochain b∈C0G

3(Z∗Z;R+).

Proof of Theorem 2.4.19. We need to prove that H(Ω) and H(X) ˆ⊗C(G2)H(Y) are isomorphic C(G3, λ3)-Hilbert modules and the representations of C(G1, λ1)on H(Ω)and H(X) ˆ⊗C(G2)H(Y) are isomorphic. We divide the proof into two parts: the first dealing with the isomorphism of Hilbert modules and the other dealing with the isomorphism of representations.

Due to the Stone-Weierstraß Theorem, the set A := {f ⊗g :fCc(X) and gCc(Y)} is linearly dense inCc(Z)in the inductive limit topology, where (f⊗g)(x, y) =f(x)g(y). We observe the following two facts:

i) The Hilbert moduleH(X) ˆ⊗C(G2)H(Y) is the completion of ACc(Z) with respect to the norm given by the inner product hf ⊗g, f×giCc(G3):=hg,hf, fiH(X)giH(Y).

ii) As λ is a (proper) continuous family of measure along π : Z → Ω, we have a surjection Λ0 :Cc(Z)→Cc(Ω) given by

Λ0(F)[x, y] = Λ(F b−1/2)[x, y] = Z

G2

F(xγ, γ−1y)b−1/2(xγ, γ−1y) dλs2X(x)(γ) for FCc(Z).

For b as in Proposition 2.4.4, the multiplication by b−1/2 is an isomorphism from Cc(Z) to itself. Then Λ is a surjection from Cc(Z) toCc(Ω), since {λu}

u∈G(0)3 is a continuous family of measures. Thus the composite Λ0:Cc(Z) b

−1/2

−−−→Cc(Z)−→Λ Cc(Ω) is a continuous surjection.

Letf, f0Cc(X),g, g0Cc(Y)andψCc(G3). ThenΛ0(f⊗g+f0⊗g0) = Λ0(f⊗g)+Λ0(f0⊗g0).

Furthermore,

We show thatΛ0 is an isomorphism of pre-Hilbert modules, hence it extends to an isomorphism of Hilbert modules. Later we show that Λ0 also intertwines the representations.

The isomorphism of Hilbert modules: Now we compute the norm off⊗g∈ H(X) ˆ⊗C(G2)H(Y).

In the calculation below, the inner product on the left is taken in H(X) ˆ⊗C(G2)H(Y), and

subscripts to other inner products tell in what space the inner product is defined. For γG3,

After plugging in the definitions, the last term of the above equation becomes Z Z

The last equality is due to Lemma 2.4.9, which says that dλr2Y(y)) dµrG

Using Remark 2.4.12 we add a factor of γ inb(x, y). The previous term equals Z Z Z

f(x)g(y)f(xγ)g(γ−1yγ) b(x, yγ) b(xγ, γ−1yγ)

!1/2

rY(y)(x) dλr2Y(y)(γ) dβrG

3(γ)(y). (2.4.22) By Lemma 2.4.4 we relate the factors of b to see that last equation is equal to

Z Z Z

f(x)g(y)f(xγ)g(γ−1yγ)

21/2(γ, γ−1yγ) dαrY(y)(x) dλr2Y(y)(γ) dβrG

3(γ)(y). (2.4.23) Finally, we apply Fubini’s Theorem to λr2Y(y) and αrY(y) to get

0(f ⊗g·b−1/2),Λ0(f×gb−1/2)i(γ)

= Z Z Z

f(x)g(y)f(xγ)g(γ−1yγ)21/2(γ, γ−1yγ) dλr2Y(y)(γ) dαrY(y)(x) dβrG

3(γ)(y).

Comparing the values of both inner products, we conclude that

hf⊗g, fgiCc(G3) =hΛ0(f ⊗g·b−1/2), Λ0(f⊗gb−1/2)iCc(G3). (2.4.24) The isomorphism of representations: We denote the actions of C(G1, λ1)onH(X) ˆ⊗C(G2)H(Y) and H(Ω) by ρ1 and ρ2, respectively, that is, ρ1: C(G1, λ1) → B(H(X) ˆ⊗C(G

2)H(Y)) and ρ2: C(G1, λ1) → B(H(Ω)) are the *-homomorphisms that give the C-correspondences from C(G1, λ1) to C(G3, λ3). We are going to show that Λ0 intertwinesρ1 andρ2.

Let φCc(G1), f, gCc(X), then (ρ2(φ)Λ0)(f⊗g)[x, y]

= (φ∗Λ0(f ⊗g))[x, y]

= Z

G1

φ(η)Λ0(f⊗g))[η−1x, y] ∆1/21,2(η,[η−1x, y]) dλr1X(x)(η)

= Z Z

φ(η)f−1xγ)g(γ−1y)b−1/2−1xγ, γ−1y) ∆1/21,2(η,[η−1x, y]) dλr1X(x)(η) dλs2X(x)(γ).

Equation (2.4.15) gives ∆1,2(η,[η−1x, y]) = ∆1,2(η,[η−1xγ, γ−1y]) = ∆1(η, η−1xγ)b(ηb(xγ,γ−1xγ,γ−1−1y)y). Thus

R. H. S.= Z Z

φ(η)f−1xγ) ∆1/21 (η, η−1xγ) dλr1X(x)(η)

g(γ−1y) b−1/2(xγ, γ−1y) dλs2X(x)(γ)

= Z

(φ∗f)(xγ)g(γ−1y)b−1/2(xγ, γ−1y) dλs2X(x)(γ)

= Λ0((φ∗f)⊗g)[x, y]

= Λ02(φ)(f⊗g))[x, y].