k-theory for crossed products and rotationallan/pimsner-voiculescu.pdf · k-theory for crossed...

K-THEORY FOR CROSSED PRODUCTS AND ROTATION

ALGEBRAS

ALLAN YASHINSKI

Abstract. In this paper, we define and discuss basic notions related to dis-

crete crossed products C∗-algebras. Then we give a proof of the Pimsner-

Voiculescu six-term exact sequence. This is the main tool used for computa-tion of the K-theory groups of a crossed product C∗-algebraba by the group

Z. Next, the rotation algebras Aθ are introduced and their basic properties

are discussed. It is shown that Ki(Aθ) ∼= Z⊕ Z for i = 0, 1, and we concludewith an explicit calculation of the generators of these groups.

1. Discrete Crossed Products

Suppose A is a C∗-algebra and G is a discrete group acting on A by automor-phisms. That is, we have a group homomorphism α : G→ Aut(A). Let αg denotethe automorphism α(g) of A. We shall construct a new C∗ algebra built out of Aand the action of G.

To do this, let AG be the vector space of finite formal sums of the form∑g∈G agg

where ag ∈ G. Addition and scalar multiplication are defined by∑g∈G

agg +∑g∈G

bgg =∑g∈G

(ag + bg)g

and

c(∑g∈G

agg) =∑g∈G

cagg, c ∈ C.

Multiplication is determined by the requirement that gag−1 = αg(a). So for finitesums, we have

(∑g∈G

agg)(∑h∈G

bhh) =∑g∈G

∑h∈G

aggbhh

=∑g∈G

∑h∈G

aggbhg−1gh

=∑g∈G

∑h∈G

agαg(bh)gh

=∑h∈G

(∑g∈G

agαg(bg−1h))h

We give AG the structure of a ∗-algebra with the adjoint defined by

(ag)∗ = αg−1(a∗)g−1

1

2 ALLAN YASHINSKI

and extended by linearity. Then ∗ is a conjugate-linear involution and moreover,

((ag)(bh))∗ = (aαg(b)(gh))∗

= α(gh)−1((aαg(b))∗)(gh)−1

= αh−1g−1(αg(b∗)a∗)h−1g−1

= αh−1(b∗)αh−1g−1(a∗)h−1g−1

= αh−1(b∗)h−1αg−1(a∗)g−1

= (bh)∗(ag)∗

for all a, b ∈ A and g, h ∈ G.We define a C∗-norm on AG by setting

‖f‖ = supπ‖π(f)‖

where the supremum ranges over all ∗-representations of AG. To show the supre-mum is bounded, first note that a ∗-representation π of AG gives rise to a repre-sentation π of A defined by π(a) = π(ae) where e ∈ G is the identity. We havethat

‖π(ae)‖ = ‖π(a)‖ ≤ ‖a‖ .Moreover, by the C∗-identity,

‖π(ag)‖ = ‖π(ag)∗π(ag)‖12

=∥∥π(αg−1(a∗)g−1ag)

∥∥ 12

=∥∥π(αg−1(a∗a)e)

∥∥ 12

≤∥∥αg−1(a∗a)

∥∥ 12

= ‖a∗a‖12

= ‖a‖

Thus, we see that for any representation π of AG,∥∥∥∥∥∥π(∑g∈G

agg)

∥∥∥∥∥∥ ≤∑g∈G‖π(agg)‖ ≤

∑g∈G‖ag‖

and consequently, ∥∥∥∥∥∥∑g∈G

agg

∥∥∥∥∥∥ ≤∑g∈G‖ag‖ <∞.

To show that the supremum in the norm is well defined, we will show that thereactually are ∗-representations of AG. Let π be a representation of A on a Hilbertspace H, which exists by the GNS construction. We will construct a representationof AG on the Hilbert space K = `2(G,H) of square summable functions from G toH. We define a representation π : AG→ B(K) by

(1) (π(ag)f)(h) = π(αh−1(a))(f(g−1h))

K-THEORY FOR CROSSED PRODUCTS AND ROTATION ALGEBRAS 3

where f ∈ K, and extend by linearity. We verify that π is ∗-preserving:

〈π((ag)∗)f, f ′〉K =⟨π(αg−1(a∗)g−1)f, f ′

⟩K

=∑h∈G

⟨(π(αg−1(a∗)g−1)f)(h), f ′(h)

⟩H

=∑h∈G

⟨π(αh−1g−1(a∗))(f(gh)), f ′(h)

⟩H

=∑h∈G

⟨f(gh), π(α(gh)−1(a))(f ′(h))

⟩H

=∑h∈G

⟨f(h), π(αh−1(a))(f ′(g−1h))

⟩H

=∑h∈G

〈f(h), (π(ag)f ′)(h)〉H

= 〈f, π(ag)f ′〉K

and so π((ag)∗) = π(ag)∗.Moreover, π preserves multiplication, as

(π(ag)π(bh)f)(k) = π(αk−1(a))((π(bh)f)(g−1k))

= π(αk−1(a))π(αk−1g(b))(f(h−1g−1k))

= π(αk−1(aαg(b)))(f((gh)−1k))

= (π(aαg(b)(gh))f)(k)

= (π(agbh)f)(k).

The C∗-algebra completion of AG in this norm is the crossed product A ×α G.As a consequence of the definition of the norm, every ∗-representation π of AGextends uniquely by continuity to a representation π of A×α G.

Remark. The crossed product A×αG that we constructed is called the full crossedproduct. When defining the norm in AG, if one only uses representations of the form(1), then upon completing to a C∗-algebra, one obtains what is called the reducedcross product. In general, the full and reduced crossed products may differ, but itis a theorem that if G is an amenable group, then they are equal.

1.1. Crossed Products by Z. Consider the case where A is unital with unit 1and G = Z. Let e ∈ Z denote the identity and let g denote the generator. Theaction of Z on A is determined by the automorphism α(g), which we will denoteby α. There is an isometric embedding i : A → A ×α Z given by i(a) = ae. Letu ∈ A ×α Z be the unitary defined by u = 1g. Then since every element of AZis a finite sum of the form

∑n ang

n, it is easily seen that the image i(A) and theunitary u generate AZ. So as a C∗-algebra, A×αZ is generated by a copy of A anda unitary u such that uau∗ = α(a) for all a ∈ A. In fact, more than that is true:

Proposition 1.1. A×αZ is the universal C∗-algebra generated by A and a unitaryu subject to the relation uau∗ = α(a) for all a ∈ A. In other words, given aC∗-algebra B, a unitary v ∈ B and a ∗-homomorphism φ : A → B such that

4 ALLAN YASHINSKI

vφ(a)v∗ = φ(α(a)) for all a ∈ A, there is a unique surjective ∗-homomorphismψ : A×α Z→ C∗(φ(A), v) such that φ = ψ ◦ i and ψ(u) = v.

Proof. Let π : B → B(H) be a faithful representation, which exists by the Gelfand-Naimark Theorem. We define a representation σ : AZ → B(H) by σ(agn) =π(φ(a)vn) and extend by linearity. We verify that σ is multiplicative:

σ(agnbgm) = σ(aαn(b)gn+m)

= π(φ(aαn(b))vn+m)

= π(φ(a)φ(αn(b))vnvm)

= π(φ(a)vnφ(b)vm)

= π(φ(a)vn)π(φ(b)vm)

= σ(agn)σ(bgm)

A representation σ of AG extends uniquely by continuity to a representation σ ofA×α Z. We define ψ : A×α Z→ B by ψ = π−1 ◦ σ. Then

(ψ ◦ i)(a) = ψ(ae) = π−1(σ(ae)) = π−1(π(φ(a))) = φ(a)

and alsoψ(u) = π−1(σ(1g)) = π−1(π(v)) = v.

The uniqueness of ψ follows from the requirements that ψ(ae) = φ(a) and ψ(u) =v combined with the fact that Ae and u generate AZ, which is dense in A×αZ. �

2. The Pimsner-Voiculescu Six-Term Exact Sequence

In this section, we give a proof of the Pimsner-Voiculescu six-term exact se-quence. It is a particular exact sequence in K-theory that allows computation ofthe K-theory groups of crossed products by Z.

2.1. The Extension. Suppose A is a unital C∗-algebra and let α be an automor-phism of A. By Proposition 1.1, A×αZ is the universal C∗-algebra generated by Aand a unitary u such that uau∗ = α(a) for all a ∈ A. Let T denote the Toeplitz al-gebra, which is the universal C∗-algebra generated by an isometry v. Let K denotethe C∗-algebra of compact operators on an infinite dimensional separable Hilbertspace. The first step in proving the Pimsner-Voiculescu Theorem is to construct ashort exact sequence of C∗-algebras:

0 −−−−→ K⊗A −−−−→ Tα −−−−→ A×α Z −−−−→ 0

where Tα is the C∗-subalgebra of (A×α Z)⊗ T generated by A⊗ 1 and u⊗ v.

2.1.1. The ∗-homomorphisms. In Tα, we will write a to denote the element a ⊗ 1for a ∈ A. Let V = u⊗ v and P = 1−V V ∗ = 1⊗ (1− vv∗). We have the followingbasic algebraic results in Tα.

Lemma 2.1. The following hold in Tα:

(1) V is an isometry.(2) P is a projection.(3) V a = α(a)V for any a ∈ A.(4) V ∗a = α−1(a)V ∗ for any a ∈ A.(5) aP = Pa for any a ∈ A.


Proof. For (1), V ∗V = (u∗⊗v∗)(u⊗v) = u∗u⊗v∗v = 1⊗1. Since V is an isometry,V V ∗ is a projection and hence P = 1− V V ∗ is a projection. For (3), we have

V a = (u⊗ v)(a⊗ 1)

= ua⊗ v= uau∗u⊗ v= α(a)u⊗ v= (α(a)⊗ 1)(u⊗ v)

= α(a)V.

For (4), let b = α−1(a∗). Then by taking the adjoint equation of (3), we haveV ∗a = V ∗α(b)∗ = b∗V ∗ = α−1(a)V ∗. Finally, for (5), we have aP = (a ⊗ 1)(1 ⊗1− vv∗) = a⊗ (1− vv∗) = (1⊗ 1− vv∗)(a⊗ 1) = Pa. �

In K, let ei,j denote the usual matrix units for i, j ≥ 0, that is, ei,jek,l = δi,jei,land e∗i,j = ej,i. In Tα, we define “matrix units” by ei,j(a) = αi(a)V iPV ∗j fori, j ≥ 0.

Lemma 2.2. There is an injective ∗-homomorphism ϕn : Mn(C) ⊗ A → Tα suchthat ϕ(ei,j ⊗ a) = ei,j(a).

Proof. Define ϕn : Mn(C) → Tα by ϕ(ei,j ⊗ a) = ei,j(a) and extend by linearity.Then ϕ is multiplicative because

ei,j(a)ej,k(b) = αi(a)V iPV ∗jαj(b)V jPV ∗k

= αi(a)V iPbV ∗jV jPV ∗k

= αi(a)V iPbPV ∗k

= αi(a)V ibPPV ∗k

= αi(a)αi(b)V iPV ∗k

= αi(ab)V iPV ∗k

= ei,k(ab).

When j 6= k,

ei,j(a)ek,l(b) = αi(a)V iPV ∗jαk(b)V kPV ∗l

= αi(a)V iPbV ∗jV kPV ∗l

= αi(a)V ibPV ∗jV kPV ∗l

= 0

because V ∗P = PV = 0, and one of those two appears in that expression whenj 6= k. We also have

ei,j(a)∗ = (αi(a)V iPV ∗j)∗

= V jPV ∗iαi(a∗)

= V jPa∗V ∗i

= V ja∗PV ∗i

= αj(a∗)V jPV ∗i

= ej,i(a∗)

6 ALLAN YASHINSKI

Moreover, ϕn is injective because if ϕn(x) = 0, where x =∑i,j

ei,j ⊗ ai,j , then

0 = ϕn(x)

= ϕn(∑i,j

ei,j ⊗ ai,j)

=∑i,j

αi(ai,j)ViPV ∗j

=∑i,j

αi(ai,j)ui−j ⊗ vi(1− vv∗)v∗j

Since the elements vi(1− vv∗)v∗j are linearly independent in T , we conclude thatαi(ai,j)u

i−j = 0 for each i, j. Since α is an automorphism and u is a unitary, thenai,j = 0 for all i, j. Therefore, x = 0 and ϕn is injective as claimed.

�

Lemma 2.3. There is a unique injective ∗-homomorphism ϕ : K ⊗ A → Tα suchthat ϕ(ei,j ⊗ a) = ei,j(a).

Proof. The C∗-algebra K⊗A is the direct limit of the sequence Mn(C)⊗A underthe inclusion maps

M1(C)⊗A ↪→M2(C)⊗A ↪→M3(C)⊗A ↪→ ... ↪→ K⊗ALet jn : Mn(C)⊗ A→ Mn+1(C)⊗ A and kn : Mn(C)⊗ A→ K⊗ A denote the

inclusion maps. Notice that ϕn = ϕn+1 ◦jn. By the universal property of the directlimit, there is a unique ∗-homomorphism ϕ : K ⊗A→ Tα such that ϕn = ϕ ◦ kn.

We will show that ϕ is isometric. Notice that since ϕn is an injective ∗-homomorphism,it is isometric. Let x ∈ K ⊗ A and let x = limxk where xk ∈ Mnk(C) ⊗ A. Then‖ϕ(x)‖ = ‖limϕ(xk)‖ = lim ‖ϕ(xk)‖ = lim ‖ϕnk(xk)‖ = lim ‖xk‖ = ‖x‖. Since ϕ isisometric, it is injective. �

Lemma 2.4. There is a unique ∗-homomorphism ψ : Tα → A ×α Z such thatψ(a) = a for all a ∈ A and ψ(V ) = u.

Proof. Let f : T → A×α Z be the unique ∗-homomorphism given by f(v) = 1. Bythe universal property of the tensor product, there is a unique ∗-homomorphismid⊗f : (A×αZ)⊗T → A×αZ such that id⊗f : x⊗y 7→ xf(y) for x ∈ A×αZ, y ∈ T .

Let ψ be the restriction of id⊗f to Tα. Then ψ(a) = ψ(a ⊗ 1) = af(1) = aand ψ(V ) = ψ(u ⊗ v) = uf(v) = u. Since A ⊗ 1 and u ⊗ v generate Tα, anyother ∗-homomorphism that agrees with ψ on this generating set must agree withψ everywhere. Thus, ψ is unique. �

2.1.2. The Short Exact Sequence. We will show that

0 −−−−→ K⊗A ϕ−−−−→ Tαψ−−−−→ A×α Z −−−−→ 0

is a short exact sequence. By Lemmas 2.3 and 2.4, the sequence is exact at K ⊗Aand A×α Z. It remains to be shown that the sequence is exact at Tα.

Let Sα denote that dense ∗-subalgebra of Tα generated by all elements of A andthe element V .

Lemma 2.5. Every element of Sα can be written uniquely as a finite sum∑i,j

ai,jViV ∗j.


Proof. Every element of Sα is a finite sum of words constructed from V, V ∗, andelements of A, so it suffices to prove the result for each of these words. If x is sucha word, then by Lemma 2.1 parts (3) and (4), x = ay, where a ∈ A and y is a word

in V and V ∗. Since V ∗V = 1, any such y can be written as V iV ∗j for some i andj.

To show uniqueness, suppose x ∈ Sα, and x =∑i,j

ai,jViV ∗j =

∑i,j

bi,jViV ∗j .

Then

0 =∑i,j

ai,jViV ∗j −

∑i,j

bi,jViV ∗j

=∑i,j

(ai,j − bi,j)V iV ∗j

=∑i,j

(ai,j − bi,j)ui−j ⊗ viv∗j

Since the elements viv∗j are linearly independent in T , (ai,j − bi,j)ui−j = 0 for alli, j. Because u is a unitary, we conclude that ai,j = bi,j for all i, j. �

Let (P )alg denote the two-sided ideal generated by P in Sα, and let (P ) be theclosed two-sided ideal generated by P in Tα. Then the closure of (P )alg in Tα is(P ). This is because any finite sum of the form

∑i,j xiPyj where xi, yj ∈ Tα can

be approximated by an element in (P )alg by choosing suitable approximations inSα for each xi and yj . Hence the ideal generated by P in Tα contains the closureof (P )alg. Therefore the closure of (P )alg is (P ).

Lemma 2.6. In Tα, the ideal (P ) is the image ϕ(K ⊗A).

Proof. For any elementary tensor of the form x = ei,j ⊗ a ∈ K ⊗ A, ϕ(x) =

αi(a)V iPV ∗j ∈ (P ). Since K ⊗ A is the closure of the span of elements of thisform, ϕ(K ⊗A) ⊆ (P ), as (P ) is a closed ideal.

Conversely, suppose z ∈ (P )alg. Then z is a finite sum of the form z =∑n xnPyn. By Lemma 2.5, for each n we can write xn =

∑i,j ai,j,nV

iV ∗j and

yn =∑k,l bk,l,nV

kV ∗l. So

z =∑n

(∑i,j

ai,j,nViV ∗j)P (

∑k,l

bk,l,nVkV ∗l)

=∑

n,i,j,k,l

ai,j,nViV ∗jPbk,l,nV

kV ∗l.

So to show z ∈ ϕ(K⊗A), it suffices to show that z0 = aV iV ∗jPbV kV ∗l ∈ ϕ(K⊗A)

for a, b ∈ A. Using Lemma 2.1, z = aV iV ∗jPbV kV ∗l = aαi−j(b)V iV ∗jPV kV ∗l.

Unless j = k = 0, then V jPV ∗k = 0 and z0 = 0. So we may assume j = k = 0, andz0 = aαi−j(b)V iPV ∗l = ϕ(ei,l ⊗ α−i(a)α−j(b)). This shows (P )alg ⊆ ϕ(K ⊗ A).Since the closure of (P )alg is (P ), and the image of a ∗-homomorphism is closed,(P ) ⊆ ϕ(K ⊗A). �

Proposition 2.7. The sequence

0 −−−−→ K⊗A ϕ−−−−→ Tαψ−−−−→ A×α Z −−−−→ 0

8 ALLAN YASHINSKI

is exact.

Proof. It is clear that imϕ ⊆ kerψ, because ϕ(K ⊗ A) = (P ) and ψ(P ) = ψ(1 −V V ∗) = 1− uu∗ = 0.

To show the converse, we will first construct a ∗-homomorphism ρ : A ×α Z →Tα/(P ). Note that in Tα,

V ∗V − V V ∗ = 1− V V ∗ = P,

and

α(a)− V aV ∗ = α(a)− α(a)V V ∗ = α(a)(1− V V ∗) = α(a)P.

So V is unitary mod (P ), and V aV ∗ = α(a) mod (P ). Thus, there is a unique ∗-homomorphism ρ : A×αZ→ Tα/(P ) such that ρ(a) = a+(P ) and ρ(u) = V +(P ).Consider the following commutative diagram:

0 −−−−→ K⊗A ϕ−−−−→ Tαψ−−−−→ A×α Z −−−−→ 0

ϕ

y∼= ∥∥∥ ρ

y0 −−−−→ (P )

i−−−−→ Tαp−−−−→ Tα/(P ) −−−−→ 0

If x ∈ kerψ, then p(x) = ρ(ψ(x)) = 0, and so x ∈ (P ) = imϕ. We conclude thatkerψ = imϕ.

�

2.2. The Isomorphism.

2.2.1. Quasihomomorphisms. Let A and B be two C∗-algebras. For our purposes,a quasihomomorphism from A to B is a pair of ∗-homomorphisms (f0, f1) from Ato E, where E is a C∗-algebra containing K ⊗ B as an ideal, such that f0(a) −f1(a) ∈ K ⊗ B for all a ∈ A. We will write (f0, f1) : A → E . K ⊗ B for such aquasihomomorphism.

Given a quasihomomorphism (f0, f1) : A → E . K ⊗ B, one can construct aK-theory map (f0, f1)∗ : Ki(A)→ Ki(B) as follows. Let E be the C∗-subalgebra ofA⊕E given by E = {(a, e)|f0(a) = e mod K⊗B}. Define the maps ι : K⊗B → Eand π : E → A by ι(x) = (0, x) and π(a, e) = a. Then we have the following splitexact sequence:

0 −−−−→ K⊗B ι−−−−→ Eπ−−−−→ A −−−−→ 0

with two splittings f0, f1 : A→ E given by f0(a) = (a, f0(a)) and f1(a) = (a, f1(a)).So there is a split exact sequence

0 −−−−→ Ki(K ⊗B)ι∗−−−−→ Ki(E)

π∗−−−−→ Ki(A) −−−−→ 0.

Since π◦f0 = id = π◦f1, we have that π∗◦f0∗ = π∗◦f1∗ and so π∗(f0∗−f1∗) = 0.Thus for any element x ∈ Ki(A), f0∗(x) − f1∗(x) ∈ kerπ∗. So f0∗(x) − f1∗(x) =ι∗(y) for some unique y ∈ Ki(K ⊗ B). We define (f0, f1)∗(x) = κB

−1∗ (y) where

κB : B → K⊗B is the canonical inclusion.


2.2.2. The inclusion. Our aim is to show that the inclusion j : A→ Tα induces anisomorphism j∗ : Ki(A)→ Ki(Tα).

Lemma 2.8. The pair of ∗-homomorphisms (id,Ad(1 ⊗ v)) : Tα → Tα . K ⊗ A isa quasihomomorphism from Tα to A.

Proof. For the element x = aV iV ∗j = aui−j ⊗ viv∗j ∈ Tα, we have

id(x)−Ad(1⊗ v)(x) = aui−j ⊗ viv∗j − aui−j ⊗ vi+1v∗j+1

= aui−j ⊗ (viv∗j − vi+1v∗j+1)

= aui−j ⊗ vi(1− vv∗)v∗j

= aV iPV ∗j ∈ K ⊗A

Since id−Ad(1⊗ v) is linear and continuous, then for any x ∈ Tα, id(x)−Ad(1⊗v)(x) ∈ K⊗A as Tα is the closure of the span of elements of the form aV iV ∗j . �

Let T denote the C∗-subalgebra of T ⊗ T generated by K⊗T and T ⊗ 1. In T ,let v = v ⊗ 1, p = 1− vv∗, and v′ = (1− vv∗)⊗ v. The following lemma is due toCuntz.

Lemma 2.9. (Cuntz) There is a continuous path φt, t ∈ [−1, 1] of ∗-homomorphisms

from T to T such thatφ−1(v) = v(1− p) + v′

φ0(v) = vφ1(v) = v(1− p) + p

Let Tα be the C∗-subalgebra of A×αZ⊗T generated by A⊗1 and the elements

V := u⊗ v and V ′ := u⊗ v′. Let P = 1− V V ∗. We will construct an embedding

of K ⊗ Tα into Tα.

Lemma 2.10. There is an injective ∗-homomorphism ϕ : K ⊗ Tα → Tα.

Proof. We first consider the ∗-homomorphism k′ : Tα → Tα given by k′(a) = aPfor a ∈ A and k′(V ) = V ′. It is evident that this is a ∗-homomorphism when the

elements of Tα and Tα are viewed as tensors. Then for the generators of Tα, wehave

k′ : a⊗ 1 7→ a⊗ p⊗ 1, u⊗ v 7→ u⊗ p⊗ v

where p = 1− vv∗ is the rank one projection in T .

Now define the “matrix units” ei,j(x) = V ik′(x)V ∗j for an element x ∈ Tα. Since

k′(x)V = V ∗k′(x) = 0, we have that ei,j(x)ek,l(y) = δj,kei,l(xy) and ei,j(x)∗

=

ej,i(x∗). Then ϕ : K ⊗ Tα → Tα is given by ϕ(ei,j ⊗ x) = ei,j(x) and extended

by linearity and continuity. The proof that it is an injective ∗-homomorphism issimilar to the proof of Lemma 2.3. �

Let k : Tα → Tα be given by k = id⊗φ0 where φ0 is as in Proposition 2.9.

Lemma 2.11. The pair of ∗-homomorphisms (k, k ◦Ad(1⊗ v)) : Tα → Tα .K⊗Tαis a quasihomomorphism from Tα to Tα.

10 ALLAN YASHINSKI

Proof. This is due to Lemma 2.8 and the fact that the following diagram commutes:

K ⊗A ϕ−−−−→ Tα

id⊗jy k

yK ⊗ Tα

ϕ−−−−→ Tα

�

Theorem 2.12. The inclusion j : A → Tα induces an isomorphism j∗ : Ki(A) →Ki(Tα).

Proof. We will show that the inverse of j∗ is q := (id,Ad(1 ⊗ v))∗. To show thatq ◦ j∗ = idKi(A), we first claim that j∗ − (Ad(1 ⊗ v) ◦ j)∗ = ϕ∗ ◦ κA∗ whereκA : A → K ⊗ A is the canonical embedding. For a ∈ A, Ad(1 ⊗ v)(j(a)) +ϕ(κA(a)) = a ⊗ vv∗ + a ⊗ 1 − vv∗ = a ⊗ 1 = j(a), so j = Ad(1 ⊗ v) ◦ j + ϕ ◦ κA.Moreover, Ad(1 ⊗ v) ◦ j and ϕ ◦ κA are orthogonal ∗-homomorphisms as Ad(1 ⊗v)(j(a))ϕ(κA(a)) = (a⊗vv∗)(a⊗1−vv∗) = 0. Thus, j∗ = (Ad(1⊗v)◦j+ϕ◦κA)∗ =(Ad(1⊗ v) ◦ j)∗ + (ϕ ◦ κA)∗ and so j∗ − (Ad(1⊗ v) ◦ j)∗ = ϕ∗ ◦ κA∗. So for a classx ∈ Ki(A), since (id∗−Ad(1⊗v)∗)(j∗(x)) = j∗(x)−Ad(1⊗v)∗(j∗(x)) = ϕ(κA∗(x)),we have q(j∗(x)) = κA

−1∗ (κA∗(x)) = x. Thus q ◦ j∗ = idKi(A).

To show that j∗ ◦ q = idKi(Tα), we will first show that j∗ ◦ q = (k, k ◦Ad(1⊗v))∗.Let q = (k, k ◦ Ad(1 ⊗ v))∗. Let E = {(x, y) ∈ Tα ⊕ Tα|x = y mod K ⊗ A} and

F = {(x, y) ∈ Tα ⊕ Tα|k(x) = y mod K ⊗ Tα}. Define f0, f1 : Tα → E by f0(x) =(x, x), f1(x) = (x,Ad(1 ⊗ v)(x)) and define g0, g1 : Tα → F by g0(x) = (x, k(x)),g1(x) = (x, (k ◦Ad(1⊗v))(x)). Let k : E → F be given by k(x, y) = (x, k(y)). Thisis well-defined, for if x = y mod K ⊗ A, then k(x) = k(y) mod K ⊗ Tα becausethe following diagram commutes:

K ⊗A ϕ−−−−→ Tα

id⊗jy k

yK ⊗ Tα

ϕ−−−−→ TαFor n = 0, 1 we have the commutative diagrams

0 −−−−→ K⊗A 0⊕ϕ−−−−→ Eπ−−−−→ Tα −−−−→ 0

id⊗jy k

y ∥∥∥0 −−−−→ K⊗ Tα

0⊕ϕ−−−−→ Fπ−−−−→ Tα −−−−→ 0

Efn←−−−− Tα

k

y ∥∥∥F

gn←−−−− Tαand at the level of K-theory,


0 −−−−→ Ki(K ⊗A)0⊕ϕ∗−−−−→ Ki(E)

π∗−−−−→ Ki(Tα) −−−−→ 0

id⊗j∗

y k∗

y ∥∥∥0 −−−−→ Ki(K ⊗ Tα)

0⊕ϕ∗−−−−→ Ki(F )π∗−−−−→ Ki(Tα) −−−−→ 0

Ki(E)fn∗←−−−− Ki(Tα)

k∗

y ∥∥∥Ki(F )

gn∗←−−−− Ki(Tα)

For x ∈ Ki(Tα), there is a unique y ∈ Ki(K ⊗A) such that

f0∗(x)− f1∗(x) = (0⊕ ϕ)∗(y)

and so q(x) = κA−1∗ (y). Applying k∗, we get

g0∗(x)− g1∗(x) = k∗(f0∗(x)− f1∗(x))

= k∗((0⊕ ϕ)∗(y))

= (0⊕ ϕ)∗((id⊗j)∗(y))

and so q(x) = κTα−1∗ ((id⊗ j)∗(y)). Since the diagram

Ki(A)j∗−−−−→ Ki(Tα)

κA∗

y∼= κTα∗

y∼=Ki(K ⊗A)

(id⊗j)∗−−−−−→ Ki(K ⊗ Tα)

we have that

q(x) = κTα−1∗ ((id⊗ j)∗(y)) = j∗(κA

−1∗ (y)) = j∗(q(x)).

So q = j∗ ◦ q as claimed.To finish the proof, we will show that q = idKi(Tα) is an isomorphism. Let

k′′ : Tα → Tα be the restriction of id⊗φ−1 where φ−1 is as in Lemma 2.9. Recallingthe definition of k, we see that k is homotopic to k′′ by Lemma 2.9. We claim thatk ◦ Ad(1 ⊗ v) and k′ are orthogonal ∗−homomorphisms. This is a consequence ofthe fact that

(k ◦Ad(1⊗ v))(V ia1V∗j)(k′(V ka2V

∗l))

= k(uia1u∗j ⊗ vi+1v∗j+1)V ′

ka2P V

′∗l

= k((ui ⊗ vi)(a1 ⊗ vv∗)(u∗j ⊗ v∗j))V ′ka2P V′∗l

= k(V ia1(1− P )V ∗j)V ′ka2P V

′∗l

= V ia1(1− P )V ∗jV ′ka2P V

′∗l

= 0

12 ALLAN YASHINSKI

because V ∗V ′ = 0, V ∗P = 0, (1− P )V ′ = 0, and a2P = P a2. Moreover, we checkthat for a ∈ A,

k′′(a) = a

= a(1− P ) + aP

= (k ◦Ad(1⊗ v))(a) + k′(a)

and

k′′(V ) = k′′(u⊗ v)

= u⊗ (v(1− p) + v′)

= u⊗ v(vv∗) + u⊗ v′

= (1⊗ v)(u⊗ v)(1⊗ v∗) + u⊗ v′

= (k ◦Ad(1⊗ v))(V ) + k′(V )

So k ◦ Ad(1 ⊗ v) + k′ = k′′ and k∗ = k′′∗ = (k ◦ Ad(1 ⊗ v))∗ + k′∗. Note thatk′ = ϕ ◦ κTα and so we have k∗ − (k ◦ Ad(1⊗ v))∗ = k′∗ = ϕ∗ ◦ κTα∗. For any x ∈Ki(Tα), k∗(x)− (k ◦Ad(1⊗v))∗(x) = ϕ∗(κTα∗(x)). So q(x) = κTα

−1∗ (κTα∗(x)) = x.

Thus we have shown that j∗ ◦ q = q = id. �

2.3. The Pimsner-Voiculescu Six-Term Exact Sequence. We begin this sec-tion with a simple lemma.

Lemma 2.13. Consider the following commutative diagram of abelian groups,

Aϕ′−−−−→ B

ψ′−−−−→ C

f

y g

y h

yD

ϕ−−−−→ Eψ−−−−→ F

where f , g, and h are isomorphisms. If the bottom row is exact at E, then the toprow is exact at B.

Proof. We have ψ′ ◦ ϕ′ = h−1 ◦ ψ ◦ g ◦ g−1 ◦ ϕ ◦ f = h−1 ◦ ψ ◦ ϕ ◦ f = 0 becauseψ ◦ ϕ = 0. Thus the im ϕ′ ⊆ kerψ′.

Suppose that x ∈ kerψ′. Then

0 = h(0) = h(ψ′(x)) = ψ(g(x))

and so g(x) ∈ kerψ. Therefore, g(x) = ϕ(y) for some y ∈ D. Then f−1(y) ∈ A,and

ϕ′(f−1(y)) = (g−1 ◦ ϕ ◦ f)(f−1(y)) = (g−1 ◦ ϕ)(y) = g−1(g(x)) = x.

Therefore, kerψ′ ⊆ im ϕ′. �

From the short exact sequence

0 −−−−→ K⊗A ϕ−−−−→ Tαψ−−−−→ A×α Z −−−−→ 0


we obtain the six term exact sequence of K-theory

K0(K ⊗A)ϕ∗−−−−→ K0(Tα)

ψ∗−−−−→ K0(A×α Z)

δ1

x δ0

yK1(A×α Z)

ψ∗←−−−− K1(Tα)ϕ∗←−−−− K1(K ⊗A)

Let κ : A → K ⊗ A be given by κ(a) = e0,0 ⊗ a. By stability of K-theory,κ∗ : Ki(A) → Ki(K ⊗ A) is an isomorphism. We also have the isomorphismj∗ : Ki(A)→ Ki(Tα) induced by inclusion.

Proposition 2.14. There is a unique group homomorphism f : Ki(A) → Ki(A)such that the following diagram commutes.

Ki(A)f−−−−→ Ki(A)

κA∗

y∼= j∗

y∼=Ki(K ⊗A)

ϕ∗−−−−→ Ki(Tα)

Moreover, f = id−α−1∗ .

Proof. The uniqueness follows from the fact that if j∗ ◦ f = ϕ∗ ◦ κA∗, then f =j−1∗ ◦ ϕ∗ ◦ κA∗.

Let γ : A → Tα be the ∗-homomorphism given by γ = AdV ◦ j ◦ α−1. Thenγ(a) = AdV (j(α−1(a))) = uα−1(a)u∗ ⊗ vv∗ = a ⊗ vv∗ = a(1 − P ). Note that(ϕ ◦ κA)(a) = ϕ(e0,0 ⊗ a) = aP . So γ and ϕ ◦ κA are orthogonal and note thatγ+ϕ◦κA = j. So j∗ = (γ+ϕ◦κA)∗ = γ∗+(ϕ◦κA)∗ = AdV∗◦j∗◦α−1

∗ +ϕ∗◦κA∗ =j∗ ◦ α−1

∗ + ϕ∗ ◦ κA∗ because AdV∗ = idKi(Tα). Taking j−1∗ of both sides, we obtain

idKi(A) = α−1∗ + j−1

∗ ◦ ϕ∗ ◦ κA∗ = α−1∗ + f . So f = id−α−1

∗ . �

We are now ready to prove the main theorem.

Theorem 2.15. (Pimsner-Voiculescu Sequence) For any C∗-algebra A, there is asix-term exact sequence

K0(A)id−α−1

∗−−−−−→ K0(A)ι∗−−−−→ K0(A×α Z)x y

K1(A×α Z)ι∗←−−−− K1(A)

id−α−1∗←−−−−− K1(A)

where ι : A→ A×α Z is the inclusion.

14 ALLAN YASHINSKI

Proof. We first prove it for the case where A is unital. Note that ψ ◦ j = ι, andconsider the commutative diagram

K0(A)id−α−1

∗−−−−−→ K0(A)ι∗−−−−→ K0(A×α Z)

κA∗

y j∗

y ∥∥∥K0(K ⊗A)

ϕ∗−−−−→ K0(Tα)ψ∗−−−−→ K0(A×α Z)

δ1

x δ0

yK1(A×α Z)

ψ∗←−−−− K1(Tα)ϕ∗←−−−− K1(K ⊗A)∥∥∥ j∗

x κA∗

xK1(A×α Z)

ι∗←−−−− K1(A)id−α−1

∗←−−−−− K1(A)

Applying Lemma 2.13 repeatedly, we obtain the following exact sequence from thetop and bottom rows of the previous diagram

K0(A)id−α−1

∗−−−−−→ K0(A)ι∗−−−−→ K0(A×α Z)

κA−1∗ ◦δ1

x κA−1∗ ◦δ0

yK1(A×α Z)

ι∗←−−−− K1(A)id−α−1

∗←−−−−− K1(A)

When A is nonunital, we consider the C∗-algebra A = {a + λ1|a ∈ A, λ ∈ C}obtained by adjoining a unit. Then there is a split exact sequence

0 −−−−→ Af−−−−→ A

g−−−−→ C −−−−→ 0

where f(a) = a, g(a+λ1) = λ, and the splitting h : C→ A is defined by h(λ) = λ1.This split exact sequence induces a split exact sequence

0 −−−−→ Ki(A)f∗−−−−→ Ki(A)

g∗−−−−→ Ki(C) −−−−→ 0

of K-theory groups.The automorphism α of A extends to an automorphism α given by α(a+ λ1) =

α(a) + λ1. Let β : C → C denote the trivial automorphism. Then, using theuniversal property of crossed products along with the maps f, g, and h, we obtaina split exact sequence

0 −−−−→ A×α Z −−−−→ A×α Z −−−−→ C×β Z −−−−→ 0

which induces the split exact sequence

0 −−−−→ Ki(A×α Z) −−−−→ Ki(A×α Z) −−−−→ Ki(C×β Z) −−−−→ 0.


Next, we form the large commutative diagram whose horizontal rows consist ofthe Pimsner-Voiculescu sequence for A, A, and C respectively.

0 0 0y y y. . . −−−−→ K1(A) −−−−→ K1(A×α Z) −−−−→ K0(A) −−−−→ . . .y y y. . . −−−−→ K1(A) −−−−→ K1(A×α Z) −−−−→ K0(A) −−−−→ . . .y y y. . . −−−−→ K1(C) −−−−→ K1(C×β Z) −−−−→ K0(C) −−−−→ . . .y y y

0 0 0

In this diagram, the columns are exact and the rows corresponding to A and Care exact because we proved the exactness of the Pimsner-Voiculescu sequence forunital C∗-algebras. To show the exactness of the Pimsner-Voiculescu sequence forA, it suffices to complete the following diagram chase:

0 0 0 0y y y yA

a−−−−→ Bb−−−−→ C

c−−−−→ D

m

y n

y p

y q

yE

e−−−−→ Ff−−−−→ G

g−−−−→ H

r

y s

y t

y u

yI

i−−−−→ Jj−−−−→ K

k−−−−→ Ly y y y0 0 0 0

Suppose that in this commutative diagram, the columns and the second and thirdrows are exact. We will show that the first row is exact at C. We have thatq ◦ c ◦ b = g ◦ f ◦ n = 0. But since q is injective, c ◦ b = 0. This proves thatim b ⊆ ker c. Conversely, suppose x ∈ ker c. Then (g ◦ p)(x) = (q ◦ c)(x) = 0, and sothere exists y ∈ F such that f(y) = p(x). So (j ◦ s)(y) = (t ◦ f)(y) = (t ◦ p)(x) = 0.Since s(y) ∈ ker j, there exists z ∈ I such that i(z) = s(y). Since r is surjective,there is w ∈ E such that r(w) = z. We have (s ◦ e)(w) = (i ◦ r)(w) = i(z) = s(y).So y − e(w) ∈ ker s and hence there is v ∈ B such that n(v) = y − e(w). Thus(p ◦ b)(v) = (f ◦ n)(v) = f(y − e(w)) = f(y) = p(x). Since p is injective, b(v) = x.So ker c ⊆ im b. �

16 ALLAN YASHINSKI

3. Rotation Algebras

For a real number θ, the rotation algebra Aθ is defined to be the universalC∗-algebra generated by unitaries u and v such that

uv = e2πiθvu.

The rotation algebra Aθ has a well-known interpretation as a crossed product.Consider the homeomorphism of the circle T = R/Z given by t 7→ t − θ. Thisrotation induces an automorphism Rθ of C(T) given by (Rθf)(t) = f(t− θ).

Proposition 3.1. The rotation algebra Aθ is isomorphic to the crossed productC(T)×Rθ Z.

Proof. Let u, v ∈ Aθ denote the generating unitaries, and let w ∈ C(T) ×Rθ Zdenote the unitary such that wfw∗ = Rθ(f) for all f ∈ C(T). Let z ∈ C(T) denotethe function z(t) = e2πit. In C(T)×Rθ Z,

z(t)w = ww∗z(t)w = w(Rθ−1(z))(t) = we2πi(t+θ) = e2πiθwz(t).

By universality of Aθ, there is a ∗-homomorphism φ : Aθ → C(T)×Rθ Z such thatφ(u) = z and φ(v) = w.

The functional calculus provides a ∗-homomorphism ρ : C(T) → Aθ such thatρ(f) = f(u). Since vρ(z)v∗ = vuv∗ = e−2πiθu = ρ(Rθ(z)) and z generates C(T),we have that vρ(f)v∗ = ρ(Rθ(f)) for any f ∈ C(T). By Proposition 1.1, there is a∗-homomorphism ψ : C(T)×Rθ Z→ Aθ such that ψ(z) = u and ψ(w) = v. Clearly,φ and ψ are inverses. �

3.1. K-theory of Aθ. Using the interpretation of Aθ as a crossed product, we canuse the Pimsner-Voiculescu exact sequence to determine the K-theory groups ofAθ.

Proposition 3.2. For any real number θ, K0(Aθ) ∼= Z⊕ Z and K1(Aθ) ∼= Z⊕ Z.

Proof. The Pimsner-Voiculescu sequence becomes

K0(C(T))id−R−θ∗−−−−−−→ K0(C(T)) −−−−→ K0(C(T)×Rθ Z)x y

K1(C(T)×Rθ Z) ←−−−− K1(C(T))id−R−θ∗←−−−−−− K1(C(T))

Since a rotation of the circle is homotopic to the identity map, it follows that themap R−θ, which is induced by a rotation, is homotopic to the identity automor-phism. As a consequence, the map id−R−θ∗ = id− id = 0. From the six-termexact sequence, we can extract two short exact sequences:

0 −−−−→ K0(C(T)) −−−−→ K0(C(T)×Rθ Z) −−−−→ K1(C(T)) −−−−→ 0

0 −−−−→ K1(C(T)) −−−−→ K1(C(T)×Rθ Z) −−−−→ K0(C(T)) −−−−→ 0

Since K0(C(T)) ∼= K1(C(T)) ∼= Z, we have

0 −−−−→ Z −−−−→ K0(C(T)×Rθ Z) −−−−→ Z −−−−→ 0

0 −−−−→ Z −−−−→ K1(C(T)×Rθ Z) −−−−→ Z −−−−→ 0

from which it follows that Ki(C(T)×Rθ Z) ∼= Z⊕ Z for i = 0, 1. �


Despite the fact that the isomorphism class of the K-theory groups of Aθ isindependent of θ, the structure of Aθ depends greatly on θ. The most striking ofthese differences depend on the irrationality of θ.

3.2. The case θ = 0.

Proposition 3.3. In the case where θ = 0, we have A0∼= C(T2).

Proof. In this case, A0 is the universal C∗-algebra generated by two commutingunitaries. So in particular, A0 is a unital commutative C∗-algebra. By the Gelfand-Naimark Theorem, A0

∼= C(X) for some compact Hausdorff space X. Recall thatX in the space of multiplicative linear functionals on A0 with the weak-* topology.We just need to show that X is homeomorphic to T2.

By universality of A0, there is a map φ : A0 → C(T2) such that φ(u) = z1 andφ(v) = z2, where z1 and z2 are the coordinate functions on T2. This induces acontinuous map of Gelfand duals φ∗ : T2 → X. Explicitly, (φ∗(s1, s2))(u) = s1

and (φ∗(s1, s2))(v) = s2. There is also a continuous map f : X → T2 given byf(ρ) = (ρ(u), ρ(v)) for a multiplicative linear functional ρ on A0. It is clear thatφ∗ and f are inverses. �

As a consequence of this proposition, the algebras Aθ, have earned the nickname“noncommutative tori.”

We remark at some of the C∗-algebraic properties of C(T2). Notice first, thatit is far from being simple. Indeed, for any closed subset Y ⊆ T2, IY = {f |f(y) =0,∀y ∈ Y } is a closed ideal. Moreover, C(T2) has many traces, as integration withrespect to any reasonable measure is a trace. However, C(T2) is projectionless. Iff ∈ C(T2) is a projection, then f(w)2 = f(w) for all w ∈ T2. So the only values fcan take are 0 or 1. But since T2 is connected, then either f = 0 or f = 1. Thus,C(T2) has no nontrivial projections.

3.3. Irrational Rotation Algebras.

3.3.1. Basic Properties. It can be shown that for irrational θ, Aθ posseses a uniqueunital faithful trace τ : Aθ → C. This trace has the property that

τ(∑i,j

ci,juivj) = c0,0

for finite sums. Viewing Aθ as C(T)×Rθ Z,

τ(∑i

fi(t)vi) =

∫ 1

0

f0(t)dt

where again the sum is finite. Here v denotes the unitary in C(T)×Rθ Z for whichvf(t)v∗ = f(t − θ). Using this trace, one can also show that Aθ contains nonontrivial closed ideals, that is, Aθ is simple. Already, this is a stark contrast tothe structure of C(T2).

3.3.2. Projections in Aθ. It has been conjectured in the past that like C(T2), Aθis projectionless. However, this could not be farther from the truth, as projectionsin Aθ exist in abundance.

Theorem 3.4. For every η ∈ (Z + Zθ) ∩ [0, 1], there exists a projection p ∈ Aθsuch that τ(p) = η.

18 ALLAN YASHINSKI

Proof. We first consider the case η = θ, and we shall construct the so-called Powers-Rieffel projection. Thinking of Aθ as C(T)×RθZ, the idea is to look for a projectionof the form p = v∗g(t) + f(t) + g(t)v, where f(t) and g(t) are real valued functionsin C(T). Thus p = p∗ is satisfied. We must choose f(t) and g(t) so that p2 = p.Recalling that vh(t) = h(t− θ)v and v∗h(t) = h(t+ θ)v∗, we calculate

p2 = (v∗g(t) + f(t) + g(t)v)2

= v∗g(t)v∗g(t) + v∗g(t)f(t) + v∗g(t)2v + f(t)v∗g(t) + f(t)2

+f(t)g(t)v + g(t)vv∗g(t) + g(t)vf(t) + g(t)vg(t)v

= g(t+ θ)g(t+ 2θ)v∗2 + (f(t)g(t+ θ) + f(t+ θ)g(t+ θ))v∗

(f(t)2 + g(t)2 + g(t− θ)2) + (f(t)g(t) + f(t− θ)g(t))v

g(t)g(t− θ)v∗2

Thus, we will have p2 = p if f(t) and g(t) are chosen to satisfy

g(t)g(t− θ) = 0

f(t)g(t) + f(t− θ)g(t) = g(t)

f(t)2 + g(t)2 + g(t− θ)2 = f(t)

The last requirement we make is that τ(p) = θ. But τ(p) =∫ 1

0f(t)dt, so we demand

that∫ 1

0f(t)dt = θ. These requirements can all be satisfied with the following

functions:

f(t) =

εt, if 0 ≤ t ≤ ε1, if ε ≤ t ≤ θ

ε−1(θ + ε− t), if θ ≤ t ≤ θ + ε0, if θ + ε ≤ t ≤ 1

g(t) =

0, if 0 ≤ t ≤ θ√

f(t)− f(t)2, if θ ≤ t ≤ θ + ε0, if θ + ε ≤ t ≤ 1

For any other η = m+ kθ ∈ (Z+Zθ)∩ [0, 1], we note that ukv = e2πikθvuk, andso the C∗-subalgebra on Aθ generated by uk and v is isomorphic to Akθ. We canconstruct the Powers-Rieffel projection in Akθ ⊆ Aθ, which will have trace equalto the fractional part of kθ, which is η. �

3.3.3. Generators of the K-theory of Aθ. We have already calculated that Ki(Aθ) ∼=Z⊕Z for i = 0, 1. In this section we will calculate the classes that can be taken asthe generators.

Theorem 3.5. The generators of K0(Aθ) are [1] and [p].

Proof. Recall that we have the short exact sequence

0 −−−−→ K0(C(T))ι∗−−−−→ K0(C(T)×Rθ Z)

κ−1∗ ◦δ0−−−−−→ K1(C(T)) −−−−→ 0

where ι : C(T) → C(T) ×Rθ Z is the inclusion, κ : C(T) → K ⊗ C(T) is thecanonical embedding, and δ0 : K0(C(T)×Rθ Z)→ K1(K⊗C(T)) is the exponentialmap associated to the short exact sequence

0 −−−−→ K⊗ C(T)ϕ−−−−→ TRθ

ψ−−−−→ C(T)×Rθ Z −−−−→ 0


Since [1] is the generator of K0(C(T)), we will take ι∗[1] = [1] as one of the gen-erators of K0(C(T) ×Rθ Z). To show that the Powers-Rieffel projection p givesthe other generator, we need to show that κ−1

∗ ◦ δ0 maps [p] to the generator ofK1(C(T)). We shall use the following recipe for calculating the exponential map.

Proposition 3.6. (Rordam et al. Proposition 12.2.2) Let

0 −−−−→ Iϕ−−−−→ A

ψ−−−−→ B −−−−→ 0

be a short exact sequence of C∗-algebras where A and B are unital. Let ϕ : I → Abe given by ϕ(x + α1) = ϕ(x) + α1A. Suppose that g ∈ K0(B) and g = [p] forsome projection in Mn(B), and let a be a self adjoint element of Mn(A) such that

ψ(a) = p. Then ϕ(u) = exp(2πia) for precisely one unitary element u in Mn(I),and δ0(g) = −[u].

Let a = V ∗g + f + gV ∈ TRθ , which is a self-adjoint lift of p. We shall calculateexp(2πia). To do this, we first evaluate a2. Recall that in TRθ , V h(t) = h(t− θ)Vand V ∗h(t) = h(t+ θ)V ∗ for any h ∈ C(T). We have that

a2 = (V ∗g(t) + f(t) + g(t)V )2

= V ∗g(t)V ∗g(t) + V ∗g(t)f(t) + V ∗g(t)2V + f(t)V ∗g(t) + f(t)2

+f(t)g(t)V + g(t)V V ∗g(t) + g(t)V f(t) + g(t)V g(t)V

= g(t+ θ)g(t+ 2θ)V ∗2 + (f(t)g(t+ θ) + f(t+ θ)g(t+ θ))V ∗

(f(t)2 + g(t− θ)2) + (f(t)g(t) + f(t− θ)g(t))V + g(t)2V V ∗

+g(t)g(t− θ)V ∗2

= V ∗g(t) + (f(t)− g(t)2) + g(t)V + g(t)2V V ∗

= (V ∗g(t) + f(t) + g(t)V )− g(t)2(1− V V ∗)= a− g(t)2(1− V V ∗)

using the relations between f(t) and g(t). If we denote the projection E = 1−V V ∗,we have the simple relationship a2 = a− g(t)2E. We will prove by induction that

an = a− (

n−2∑k=0

f(t)k)g(t)2E.

As we have already established this for n = 2, assume it holds for some n ≥ 2.Note that

g(t)2Ea = g(t)2E(V ∗g(t) + f(t) + g(t)V )

= g(t)2g(t+ θ)EV ∗ + g(t)2Ef(t) + g(t)3EV

= g(t)2Ef(t)

20 ALLAN YASHINSKI

because g(t)g(t+ θ) = 0 and EV = 0. So we have that

an+1 = (a− (

n−2∑k=0

f(t)k)g(t)2E)a

= a2 − (

n−2∑k=0

f(t)k)g(t)2Ea

= a− g(t)2E − (

n−2∑k=0

f(t)k)g(t)2Ef(t)

= a− (

n−1∑k=0

f(t)k)g(t)2E

and the claim is proved.We calculate

exp(2πia) =

∞∑n=0

(2πi)n

n!an

= 1 + 2πia+

∞∑n=2

(2πi)n

n!an

= 1 + 2πia+

∞∑n=2

(2πi)n

n!(a− (

n−2∑k=0

f(t)k)g(t)2E)

= 1 + (

∞∑n=1

(2πi)n

n!)a−

∞∑n=2

(2πi)n

n!(

n−2∑k=0

f(t)k)g(t)2E)

= 1−∞∑n=2

(2πi)n

n!(

n−2∑k=0

f(t)k)g(t)2E)

= 1− h(t)E


where h(t) =

∞∑n=2

(2πi)n

n!(

n−2∑k=0

f(t)k)g(t)2. If t /∈ (θ, θ + ε), then g(t) = 0 and so

h(t) = 0. If t ∈ (θ, θ + ε), then by definition, g(t)2 = f(t)− f(t)2. So

h(t) =

∞∑n=2

(2πi)n

n!(

n−2∑k=0

f(t)k)(f(t)− f(t)2)

=

∞∑n=2

(2πi)n

n!

f(t)n−1 − 1

f(t)− 1f(t)(1− f(t))

= −∞∑n=2

(2πi)n

n!(f(t)n − f(t))

= −∞∑n=2

(2πi)n

n!f(t)n + (

∞∑n=2

(2πi)n

n!)f(t)

= −(exp(2πif(t))− 2πif(t)− 1) + (exp(2πi)− 2πi− 1)f(t)

= 1− exp(2πif(t))

Thus we can write exp(2πia) = h(t)E + (1− E) where

h(t) =

{exp(2πif(t)), if t ∈ [θ, θ + ε]

0, if t /∈ [θ, θ + ε]

We see that h(t) has winding number −1, and so exp(2πia) corresponds to a gener-ator of K1(K⊗C(T)). Thus (κ−1

∗ ◦δ0)[p] is a generator of K1(C(T)) as desired. �

Theorem 3.7. The generators of K1(Aθ) are [u] and [v].

Proof. We have the short exact sequence

0 −−−−→ K1(C(T))ι∗−−−−→ K1(C(T)×Rθ Z)

κ−1∗ ◦δ1−−−−−→ K0(C(T)) −−−−→ 0

The identity function in C(T) can be taken as the generator of K1(C(T)). It’simage under ι∗ is just the class [u]. So we can take [u] as one of the generators ofK1(Aθ). To finish, we must show that [v] maps to the generator of K0(C(T)) underthe map κ−1

∗ ◦ δ1. We shall use the following concrete description of the index map:

Proposition 3.8. (Rordam et al. Proposition 9.2.3) Let

0 −−−−→ Iϕ−−−−→ A

ψ−−−−→ B −−−−→ 0

be a short exact sequence of C∗-algebras where A and B are unital. Let ϕ : I → Abe given by ϕ(x + α1) = ϕ(x) + α1A. Let v be a unitary in B and V be a partialisometry in A such that ψ(V ) = v. Then 1− V ∗V = ϕ(p) and 1− V V ∗ = ϕ(q) for

some projections p, q ∈ I, and δ1([v]) = [p]− [q].

We have that ψ(V ) = v, where V is the isometry in TRθ . Also, 1−V ∗V = 0 and1 − V V ∗ = ϕ(e0,0 ⊗ 1). Thus (κ−1

∗ ◦ δ1)([v]) = κ−1∗ (−[e0,0 ⊗ 1]) = −[1]. As [1] is

the generator of K0(C(T)), this completes the proof.�

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