Fundamental results about semiprojective...

Fundamental results about semiprojectiveC ∗-algebras

Jack Spielberg, Arizona State University

Masterclass on semiprojectivity, Copenhagen, August 2012

1 / 192

GOALS

1. The dimension drop algebras are semiprojective.

2. The UCT Kirchberg algebras having finitely generatedK -theory, and free K1, are semiprojective.

2 / 192

REFERENCES (for first part)

[ELP 98] S. Eilers, T. Loring, G. Pedersen, Stability ofanticommutation relations: an application of noncommutative CWcomplexes, J. reine angew. Math. 499 (1998), 101 - 143.[ELP 99] S. Eilers, T. Loring, G. Pedersen, Morphisms ofextensions of C ∗-algebras: pushing forward the Busby invariant,Adv. Math. 147 (1999), 74 - 109.[Loring 97] T. Loring, Lifting Solutions to Perturbing Problems inC ∗-algebras, Fields Inst. Mon. Ser. 8, Amer. Math. Soc.,Providence, R.I., 1997.[OP 89] C. Olsen and G. Pedersen, Corona C ∗-algebras and theirapplications to lifting problems, Math. Scand. 64 (1989), 63 - 86.

3 / 192

[Pedersen 79] G. Pedersen, C ∗-algebras and their automorphismgroups, Academic Press, London, 1979.[Pedersen 90] G. Pedersen, The corona construction, inProceedings of the 1988 GPOTS-Wabash Conference, LongmanSci. 6 Tech., Harlow, 1990.[Pedersen 97] G. Pedersen, A strict version of thenon-commutative Urysohn lemma, Proc. Amer. Math. Soc. 125(1997), 2657 - 2660.

4 / 192

1. Basic results

In this first section, we will prove some very basic lifting results.First, let’s recall the fundamental definitions.

Definition 1.1A C ∗-algebra A is projective if every map of A to a quotientC ∗-algebra lifts:

B/IA

B

For now, let’s see just one example of a projective C ∗-algebra —this is everyone’s first lifting result.

5 / 192

Lemma 1.2Let x ∈ B/I be a positive contraction. Then there is a positivecontraction y ∈ B such that π(y) = x (where π : B → B/I is thequotient map).

Proof. Let z1 ∈ B be any lift of x . Then z2 = (z1 + z∗1 )/2 is aself-adjoint lift of x . Let h be the function

1

1

Then h(z2) is a positive contraction. By the functional calculus,π(h(z2)) = h(π(z2)) = h(x) = x . �

The universal C ∗-algebra generated by a positive contraction isCC = C0(0, 1], so what we have shown is that CC is projective.

6 / 192

Lemma 1.2Let x ∈ B/I be a positive contraction. Then there is a positivecontraction y ∈ B such that π(y) = x (where π : B → B/I is thequotient map).

Proof. Let z1 ∈ B be any lift of x . Then z2 = (z1 + z∗1 )/2 is aself-adjoint lift of x . Let h be the function

1

1

Then h(z2) is a positive contraction. By the functional calculus,π(h(z2)) = h(π(z2)) = h(x) = x . �

The universal C ∗-algebra generated by a positive contraction isCC = C0(0, 1], so what we have shown is that CC is projective.

7 / 192

It is nearly as easy to observe that CC⊕ CC is projective: letφ : CC⊕ CC→ B/I . Put b1 = φ(t, 0) and b2 = φ(0, t). Then b1

and b2 are orthogonal positive contractions, so c = b1− b2 is a selfadjoint contraction with b1 = c+ and b2 = c−. There is z ∈ B, aself-adjoint contraction, with π(z) = c. Then z+ and z− areorthogonal positive contractions with π(z+) = b1 and π(z−) = b2.We may define ψ : CC⊕ CC→ B by ψ(f , g) = (f (z+), g(z−)).

This shows that orthogonal positive contractions lift to orthogonalpositive contractions. This applies to finite collections byinduction. (It is also true for countable collections.)

8 / 192

Definition 1.3An ideal I in a C ∗-algebra is an inductive limit ideal if there is anincreasing sequence (In) of subideals of I with union dense in I :I1 C I2 C · · ·C I , and I =

⋃∞1 In.

Definition 1.4A C ∗-algebra A is semiprojective if for every map φ : A→ B/I ,where I is an inductive limit ideal, there is n, and ψ : A→ B/In,such that φ = πn ◦ ψ.

B/IA

B/In

B

φ

πnψ

9 / 192

Lemma 1.5The Toeplitz algebra T is semiprojective.

Proof. Let x ∈ B/I with x∗x = 1. Let y ∈ B with π(y) = x . Then

limn‖y∗y − 1 + In‖ = ‖y∗y − 1 + I‖ = 0,

so there is n such that ‖y∗y − 1 + In‖ < 1. Let yn = y + In ∈ B/In.Then ‖y∗n yn − 1‖ < 1, so y∗n yn is invertible. Let z = yn(y∗n yn)−1/2.Then z is a partial isometry, π(z) = x (since π(y∗n yn) = 1), and

z∗z = (y∗n yn)−1/2y∗n yn(y∗n yn)−1/2 = 1. �

10 / 192

Lemma 1.6C (S1) is semiprojective.

Proof. Let x ∈ B/I with x∗x = xx∗ = 1. By Lemma 1.5, there aren and z ∈ B/In such that π(z) = x and z∗z = 1. Then zz∗ is aprojection, and zz∗ − 1 ∈ I . Therefore limk ‖zz∗ − 1 + Ik‖ = 0.Choose k ≥ n such that ‖zz∗ − 1 + Ik‖ < 1. Letw = z + Ik ∈ B/Ik . Then w∗w = 1 and ‖ww∗ − 1‖ < 1. Thenww∗ is an invertible projection, hence ww∗ = 1. �

11 / 192

Lemma 1.7C is semiprojective.

Proof. Let x ∈ B/I with x = x∗ = x2. By Lemma 1.2 there isy ∈ B, a positive contraction, with π(y) = x . Choose n such that‖y − y2 + In‖ < 1/4. Let yn = y + In ∈ B/In. Then‖yn − y2

n‖ < 1/4, so 1/2 6∈ sp(yn). Let z = χ(1/2,∞)(yn). Thenz = z∗ = z2, and

π(z) = χ(1/2,∞)(π(yn)) = χ(1/2,∞)(x) = x . �

12 / 192

Lemma 1.8A is semiprojective if and only if A is semiprojective.

Proof. First suppose that A is semiprojective. Let e be theadjoined unit in A. Let φ : A→ B/I , where I is an inductive limitideal. Since φ(e) is a projection in B/I , Lemma 1.7 implies thatthere are n, and a projection p ∈ B/In, such that πn(p) = φ(e).Replacing B by B/In, and I by I/In, we may assume that p ∈ B.Now replace B by pBp and I by pIp. Thus we may assume that Bis unital, with unit p, and that π(p) = φ(e). Since A issemiprojective, there are m, and ψ0 : A→ B/Im such thatπm ◦ ψ0 = φ|A. Define ψ : A→ B/Im by ψ(a + λe) = ψ0(a) + λp.

13 / 192

Now suppose that A is semiprojective. Let φ : A→ B/I , where I isan inductive limit ideal. We have φ : A→ B/I . Since A issemiprojective, there are n and ψ : A→ B/In such thatπn ◦ ψ = φ. Since πn ◦ ψ(A) = φ(A) = φ(A) ⊆ B/I , we know thatψ(A) ⊆ B/In. We may let ψ = ψ|A. �

We note that one consequence of Lemmas 1.8 and 1.6 is thatSC = C0(0, 1) is semiprojective.

14 / 192

Now suppose that A is semiprojective. Let φ : A→ B/I , where I isan inductive limit ideal. We have φ : A→ B/I . Since A issemiprojective, there are n and ψ : A→ B/In such thatπn ◦ ψ = φ. Since πn ◦ ψ(A) = φ(A) = φ(A) ⊆ B/I , we know thatψ(A) ⊆ B/In. We may let ψ = ψ|A. �

We note that one consequence of Lemmas 1.8 and 1.6 is thatSC = C0(0, 1) is semiprojective.

15 / 192

Lemma 1.9The universal C ∗-algebra generated by a partial isometry issemiprojective.

Proof. Let x ∈ B/I with xx∗x = x , where I is an inductive limitideal. Let y ∈ B with π(y) = x . Since x∗x is a projection, we have‖y∗y − (y∗y)2 + I‖ = 0. Choose n so that‖yy∗ − (y∗y)2 + In‖ < 1/4, and put yn = y + In. As in the proof ofLemma 1.7, we know that 1/2 6∈ sp(y∗n yn). Thenf (t) = t−1/2χ(1/2,∞)(t) is continuous on sp(y∗n yn). Leth = f (y∗n yn) ∈ B/In. then π(h) = x∗x . Put z = ynh. Thenπ(z) = x(x∗x) = x , and since tf (t)2 = χ(1/2,∞)(t), we have

zz∗z = z · hy∗n ynh = z · y∗n ynf (y∗n yn)2

= ynh · χ(1/2,∞)(y∗n yn) = ynh = z . �

16 / 192

2. Projectivity for CMn

We showed in Lemma 1.2 that CC is projective. It is afundamental result (proved by Loring) that

CMn = C0((0, 1],Mn) ∼= Mn ⊗ C0(0, 1]

is also projective. In order to give the proof, we need to presentCMn by generators and relations. An obvious set of generators is{te21, . . . , ten1}. If we call them u2, . . ., un, they satisfy someobvious relations:

I ‖uj‖ ≤ 1.

I u∗i uj = δiju∗2u2, all i , j . (p’wise ⊥ ranges; common init space)

I uiuj = 0, all i , j . (ranges are ⊥ to init space)

17 / 192

Lemma 2.1Let A be the universal C ∗-algebra given by these generators andrelations. Then A is isomorphic to CMn.

Proof. It is clear that the assignment uj 7→ tej1 extends to asurjective ∗-homomorphism σ : A→ CMn. We show that this mapis injective. Let π be an irreducible representation of A. Noticethat the positive contraction k = (u∗2u2 +

∑n2 uju

∗j )1/2 is central in

A (exercise). Therefore there is t0 ∈ (0, 1] such that π(k) = t0 · 1.Let ξ1 be a unit vector in the range of π(u∗2u2). (π(u∗2u2) 6= 0 sinceotherwise π(uj) = 0 for all j .) Then π(u∗j )ξ1 = 0 for all j . We have

‖π(uj)ξ1‖2 = 〈π(u∗j uj)ξ1, ξ1〉 = 〈π(k2)ξ1, ξ1〉 = t20 〈ξ1, ξ1〉 = t20 .

Let ξj = t−10 π(uj)ξ1. Then {ξ1, . . . , ξn} is an orthonormal set. Letp ∼ span {ξ1, . . . , ξn}. Then it is easy to check that p ∈ π(A)′, sop = 1. Therefore π = evt0 ◦ σ. It follows that σ is injective. �

18 / 192

In order to proceed, we need a technical lemma from [Loring 97](there traced back to GP). (We will need the same lemma againlater on.)

Lemma 2.2Let x, a ∈ A/I with x∗x ≤ a. Choose y, b ∈ A such that

b ≥ 0, π(b) = a, π(y) = x

(where π : A→ A/I is the quotient map). Then there is z ∈ Asuch that

π(z) = x , z∗z ≤ b, zz∗ ≤ yy∗.

The point is that we can start with any lift of a. Then any lift of xcan be modified so that the original inequality is preserved, and the“reverse quantity” zz∗ is controlled by the corresponding quantityfor the first lift of x .

19 / 192

In order to proceed, we need a technical lemma from [Loring 97](there traced back to GP). (We will need the same lemma againlater on.)

Lemma 2.2Let x, a ∈ A/I with x∗x ≤ a. Choose y, b ∈ A such that

b ≥ 0, π(b) = a, π(y) = x

(where π : A→ A/I is the quotient map). Then there is z ∈ Asuch that

π(z) = x , z∗z ≤ b, zz∗ ≤ yy∗.

The point is that we can start with any lift of a. Then any lift of xcan be modified so that the original inequality is preserved, and the“reverse quantity” zz∗ is controlled by the corresponding quantityfor the first lift of x .

20 / 192

Proof. Let c = (y∗y − b)+ + b. The following is clear:

(1) c ≥ b.

Since y∗y − b ≤ (y∗y − b)+, we have

(2) c ≥ y∗y .

Since x∗x − a ≤ 0, we have

(3) π(c) = a.

For t > 0 let yt = y(t + c)−1/2b1/2. Then

(4) π(yt)→ x as t → 0+,

since π((t + c)−1/2b1/2) =

(a

t + a

)1/2

is a right approximate unit

for x , as t → 0+.

21 / 192

Next,

y∗t yt = b1/2(t + c)−1/2y∗y(t + c)−1/2b1/2

≤ b1/2 c

t + cb1/2, by (2),

≤ b.(5)

Similarly (by (1)),

(6) yty∗t ≤ yy∗.

Finally,

(7) (yt) is Cauchy as t → 0+.

22 / 192

To prove this, we have

ys − yt = y [(s + c)−1/2 − (t + c)−1/2]b1/2

= yθb1/2.

‖ys − yt‖2 = ‖b1/2θy∗yθb1/2‖≤ ‖b1/2θcθb1/2‖, by (2)

= ‖c1/2θbθc1/2‖≤ ‖c1/2θcθc1/2‖, by (1)

= ‖cθ‖2.

23 / 192

θ =1

(s + c)1/2− 1

(t + c)1/2=

(t + c)1/2 − (s + c)1/2

(t + c)1/2(s + c)1/2

=t − s

(s + c)1/2(t + c)1/2((t + c)1/2 + (s + c)1/2

) .

cθ = (t1/2 − s1/2)

(c

s + c

)1/2( c

t + c

)1/2 t1/2 + s1/2

(t + c)1/2 + (s + c)1/2.

‖cθ‖ ≤ |t1/2 − s1/2| → 0, as s, t → 0+.

Then z = limt→0+ yt does the job. �

24 / 192

Corollary 2.3

Let a1, a2, x ∈ A/I , with

ai ≥ 0, x∗x ≤ a1, xx∗ ≤ a2.

Given bi ∈ A with bi ≥ 0 and π(bi ) = ai , there is w ∈ A with

w∗w ≤ b1, ww∗ ≤ b2, π(w) = x .

Proof. First apply the lemma to a1, x , and b1 to get z ∈ A withπ(z) = x and z∗z ≤ b1. Then apply it again to a2, x∗, b2, and z∗

to get z ′. Then we can take w = (z ′)∗. �

25 / 192

Theorem 2.4CMn is projective.

Proof. Let u2, . . ., un ∈ A/I satisfy the relations for CMn. Leta1 = u∗2u2, and aj = uju

∗j for 2 ≤ j ≤ n. Then a1, . . ., an are

pairwise orthogonal positive contractions. Choose b1, . . ., bn ∈ Apairwise orthogonal positive contractions with π(bj) = aj . By thecorollary, we can find v2, . . ., vn ∈ A with π(vj) = uj such that

v∗n vn ≤ b1 vnv∗n ≤ bn

v∗n−1vn−1 ≤ v∗n vn vn−1v∗n−1 ≤ bn−1

· · · · · ·v∗2 v2 ≤ v∗3 v3 v2v∗2 ≤ b2.

Thus v∗2 v2 ≤ v∗j vj for all j .

26 / 192

Finally, we let wj ,t = vj(t + v∗j vj)−1/2(v∗2 v2)1/2 for t > 0. As in the

proof of Lemma 2.2, wj ,t → wj as t → 0+, where

I π(wj) = limt→0+ π(wj ,t) = π(vj) = uj

I w∗j wj ≤ v∗2 v2 ≤ b1

I wjw∗j ≤ vjv

∗j ≤ bj .

Moreover, w∗j ,twj ,t = |v2|(|vj |2

t + |vj |2

)|v2| → |v2|2, since

|v2|2 ≤ |vj |2. Thus w∗j wj = w∗2 w2 for all j . It follows that w2, . . .,wn satisfy the relations defining CMn. �

27 / 192

Definition 2.5Here is a definition (from [ELP 98]): a map β : B1 → B2 isconditionally projective if every map from B2 to a quotient, whosecomposition with β factors through the quotient map, lifts.Pictorially,

B1

B2

D

D/I

β

(In other words, if you can lift part of B2 — namely, β(B1) — thenyou can finish lifting all of B2.)

To be precise, we should say “if the square commutes, then thereis a dotted map so that the whole diagram commutes.”

28 / 192

Exercise 2.6

1. If A is projective, then the unital map C→ A is conditionallyprojective.

2. Define β : C→ CMn ⊕ CMn by β(1) = (0, χ[0,1]). Then β isconditionally projective.

29 / 192

3. Multiplier algebra basics

If AC B, A is essential in B if A⊥ ∩ B = 0. The multiplier algebraof A, M(A), is the maximal C ∗-algebra in which A sits as anessential ideal. This characterization implies that whenever AC X ,there is a map X → M(A), with kernel A⊥ ∩ X .

For an intrinsic definition, define a left- (respectively right-)centralizer of A to be a linear map L (respectively R): A→ A suchthat L(a1a2) = L(a1)a2 (respectively R(a1a2) = a1R(a2)). Define adouble centralizer of A to be a pair (L,R), where L (respectively,R) is a left (respectively, right) centralizer of A, and are such thata1L(a2) = R(a1)a2. Then the set of double centralizers is themaximal idealizer of A: an element a ∈ A is identified with the pair(La,Ra) (left and right multiplication by a). If A is an ideal in aC ∗-algebra X , then each x ∈ X defines a double centralizer(Lx ,Rx). This defines the map X → M(A).

30 / 192

3. Multiplier algebra basics

If AC B, A is essential in B if A⊥ ∩ B = 0. The multiplier algebraof A, M(A), is the maximal C ∗-algebra in which A sits as anessential ideal. This characterization implies that whenever AC X ,there is a map X → M(A), with kernel A⊥ ∩ X .

For an intrinsic definition, define a left- (respectively right-)centralizer of A to be a linear map L (respectively R): A→ A suchthat L(a1a2) = L(a1)a2 (respectively R(a1a2) = a1R(a2)). Define adouble centralizer of A to be a pair (L,R), where L (respectively,R) is a left (respectively, right) centralizer of A, and are such thata1L(a2) = R(a1)a2. Then the set of double centralizers is themaximal idealizer of A: an element a ∈ A is identified with the pair(La,Ra) (left and right multiplication by a). If A is an ideal in aC ∗-algebra X , then each x ∈ X defines a double centralizer(Lx ,Rx). This defines the map X → M(A).

31 / 192

A ∗-homomorphism π : A→ B is proper if it satisfies one of thefollowing equivalent conditions:

1. π(A) contains an approximate unit for B.

2. For every approximate unit (ei ) of A, the net (π(ei )) is anapproximate unit for B.

3. π(A) generates B as a hereditary subalgebra.

Because of 2, it is tempting to use the term approximately unitalin place of proper.

The strict topology on M(A) is the weak topology defined by theseminorms x 7→ ‖xa‖ and x 7→ ‖ax‖, for a ∈ A. M(A) is the strictcompletion of A, and if ei is an approximate unit for A, thenei → 1 strictly. If π is proper, we can define π : M(A)→ M(B) byπ(x) · b = limi π(xei )b (for any choice of approximate unit for A).Then π is strictly continuous. Thus, proper maps are preciselythose that extend to strictly continuous unital maps betweenmultiplier algebras.

32 / 192

A ∗-homomorphism π : A→ B is proper if it satisfies one of thefollowing equivalent conditions:

1. π(A) contains an approximate unit for B.

2. For every approximate unit (ei ) of A, the net (π(ei )) is anapproximate unit for B.

3. π(A) generates B as a hereditary subalgebra.

Because of 2, it is tempting to use the term approximately unitalin place of proper.The strict topology on M(A) is the weak topology defined by theseminorms x 7→ ‖xa‖ and x 7→ ‖ax‖, for a ∈ A. M(A) is the strictcompletion of A, and if ei is an approximate unit for A, thenei → 1 strictly. If π is proper, we can define π : M(A)→ M(B) byπ(x) · b = limi π(xei )b (for any choice of approximate unit for A).Then π is strictly continuous. Thus, proper maps are preciselythose that extend to strictly continuous unital maps betweenmultiplier algebras.

33 / 192

Algebras having a countable approximate unit are special; we callthem σ-unital. Unital algebras and separable algebras are examples.More generally, an algebra is σ-unital if and only if it contains astrictly positive element. A positive element a ∈ A is called strictlypositive if it satisfies one of the following equivalent conditions:

I f (a) 6= 0 for all (pure) states f of A.

I π(a) 6= 0 for all (irreducible) representations π of A.

I aAa is dense in A.

If a is strictly positive, then (a1/n) is a countable approximate unit.Conversely, if (en) is a countable approximate unit, thena =

∑n 2−nen is strictly positive. If a is strictly positive, and h is

the function from the proof of Lemma 1.2, then h(a) is also strictlypositive. Thus a σ-unital algebra always contains a strictly positivecontraction.

34 / 192

4. Direct sums and matrices

We observed in section 1 that CC⊕ CC is projective. This factcan be used to show that projectivity is preserved by direct sums(for σ-unital algebras).

Proposition 4.1

If A1 and A2 are projective and σ-unital, then A1⊕A2 is projective.

Proof. Let hi be a strictly positive contraction in Ai . We have aproper map µ : CC⊕ CC→ A1 ⊕ A2 byµ(f1, f2) = (f1(h1), f2(h2)). We will writeC = CC⊕ CC = C1 ⊕ C2. Now let φ : A1 ⊕ A2 → B/I be given.Since C is projective, there is a map ν : C → B such that πν = φµ(where π : B → B/I is the quotient map).

35 / 192

4. Direct sums and matrices

We observed in section 1 that CC⊕ CC is projective. This factcan be used to show that projectivity is preserved by direct sums(for σ-unital algebras).

Proposition 4.1

If A1 and A2 are projective and σ-unital, then A1⊕A2 is projective.

Proof. Let hi be a strictly positive contraction in Ai . We have aproper map µ : CC⊕ CC→ A1 ⊕ A2 byµ(f1, f2) = (f1(h1), f2(h2)). We will writeC = CC⊕ CC = C1 ⊕ C2. Now let φ : A1 ⊕ A2 → B/I be given.Since C is projective, there is a map ν : C → B such that πν = φµ(where π : B → B/I is the quotient map).

36 / 192

Let Di = ν(Ci )Bν(Ci ), and let Ji = Di ∩ I = Di IDi C Di . Weclaim that φ(Ai ) ⊆ π(Di ):

φ(Ai ) = φ(µ(Ci )Aiµ(Ci ))

⊆ φµ(Ci )π(B)φµ(Ci )

= π(ν(Ci )Bν(Ci ))

= π(Di ).

Since Ai is projective, there is a map ψi : Ai → Di such thatπψi = φ|Ai

. Then ψ = ψ1 ⊕ ψ2 does the job. �

37 / 192

Next, we will show that tensoring with Mn preserves projectivity,for σ-unital algebras.

Lemma 4.2Let A be σ-unital. There is a proper map µ : CMn → Mn ⊗ A suchthat µ(eij ⊗ 1) = eij ⊗ 1.

Proof. Let h ∈ A be a strictly positive contraction, and defineµ(teij) = eij ⊗ h. �

Theorem 4.3Let A be σ-unital and projective. Then Mn ⊗ A is projective.

Proof. Let φ : Mn ⊗ A→ B/I . Let h be a strictly positivecontraction in A, and let µ be as in Lemma 4.2. Since CMn isprojective, there is a map ν : CMn → B such that πν = φµ:

B

B/IMn ⊗ ACMn

πν

µ φ

38 / 192

Next, we will show that tensoring with Mn preserves projectivity,for σ-unital algebras.

Lemma 4.2Let A be σ-unital. There is a proper map µ : CMn → Mn ⊗ A suchthat µ(eij ⊗ 1) = eij ⊗ 1.

Proof. Let h ∈ A be a strictly positive contraction, and defineµ(teij) = eij ⊗ h. �

Theorem 4.3Let A be σ-unital and projective. Then Mn ⊗ A is projective.

Proof. Let φ : Mn ⊗ A→ B/I . Let h be a strictly positivecontraction in A, and let µ be as in Lemma 4.2. Since CMn isprojective, there is a map ν : CMn → B such that πν = φµ:

B

B/IMn ⊗ ACMn

πν

µ φ

39 / 192

Let D = ν(CMn)Bν(CMn), and let J = D ∩ I = DID C D. Just asin the proof of projectivity for direct sums, we have thatφ(Mn ⊗ A) ⊆ π(D). Now we have

B

B/I

D

D/JMn ⊗ ACMn

ππν

µ φ

We have ν : M(CMn)→ M(D) such that πν = φµ.

40 / 192

Since A is projective, there is a map α0 : e11 ⊗ A→ ν(e11)D ν(e11)such that πα0 = φ|e11⊗A. We may define a map α : Mn ⊗ A→ Dby α(eij ⊗ a) = ν(ei1)α0(e11 ⊗ a)ν(e1j). We check that πα = φ.We know that πα|e11⊗A = πα0 = φ|e11⊗A. Then we have

πα(a⊗ eij) = π(ν(ei1)α0(e11 ⊗ a)ν(e1j)

)= πν(ei1)φ(e11 ⊗ a)πν(e1j)

= φµ(ei1)φ(e11 ⊗ a)φµ(e1j)

= φ(µ(ei1)φ(e11 ⊗ a)µ(e1j)

)= φ(eij ⊗ a). �

41 / 192

We note that the above proofs also show that semiprojectivity ispreserved by direct sums and tensoring with Mn, for σ-unitalalgebras. When lifting the algebras Ai in the case of direct sums,or A in the case of a matrix algebra, it is necessary in general topass to a partial quotient in the usual way.

42 / 192

5. PullbacksPullbacks are an elementary construction for C ∗-algebras. Given∗-homomorphisms βi : Bi → C , i = 1, 2, the pullback is aC ∗-algebra A with maps αi : A→ Bi such that

1. β1α1 = β2α2.

2. If D is a C ∗-algebra, and δi : D → Bi are ∗-homomorphismswith β1δ1 = β2δ2, then there is a unique ∗-homomorphismε : D → A such that δi = αiε, i = 1, 2.

The picture is

D

B1

A C

B2

β1

β2

α1

α2

δ1

δ2

∃!ε

43 / 192

The pullback is easily constructed asB1 ⊕

(β1,β2)B2 := {(b1, b2) ∈ B1 ⊕ B2 : β1(b1) = β2(b2)}. The maps

to B1 and B2 are the projections. It is easy to check that this hasthe desired universal property. The usual abstract nonsense verifiesthe uniqueness of the pullback.

As an application, we consider extensions. Given

0→ Ai→ X

π→ B → 0, the canonical map θ : X → M(A) inducesa map η : B → C (A) (the Busby invariant of the extension), givingthe diagram

0 −−−−→ A −−−−→ M(A)q−−−−→ C (A) −−−−→ 0∥∥∥ θ

x xη0 −−−−→ A

i−−−−→ Xπ−−−−→ B −−−−→ 0

44 / 192

0 −−−−→ A −−−−→ M(A)q−−−−→ C (A) −−−−→ 0∥∥∥ θ

x xη0 −−−−→ A

i−−−−→ Xπ−−−−→ B −−−−→ 0

Because the diagram commutes, the universal property of thepullback gives a map X → M(A) ⊕

(q,η)B; of course this map is

x 7→ (θ(x), π(x)). It is easy to check that it is bijective, so that Xreally is the pullback of (q, η). The point for us is that in order togive a map into X , we have to give maps into M(A) and B withthe appropriate commutation property.

45 / 192

We will use this idea to study certain maps of extensions. Let

0→ Ajij→ Xj

πj→ Bj → 0 be given, for j = 1, 2. Then we havecorresponding diagrams like the above for the two extensions.:

0 A2 X2 B2 0

0 A1 X1 B1 0

0 A2 M(A2) C (A2) 0

0 A1 M(A1) C (A1) 0

i2 π2

i1 π1

q2

q1

θ2

θ1

η2

η1

46 / 192

We will use this idea to study certain maps of extensions. Let

0→ Ajij→ Xj

πj→ Bj → 0 be given, for j = 1, 2. Then we havecorresponding diagrams like the above for the two extensions.:

0 A2 X2 B2 0

0 A1 X1 B1 0

0 A2 M(A2) C (A2) 0

0 A1 M(A1) C (A1) 0

i2 π2

i1 π1

q2

q1

θ2

θ1

η2

η1

47 / 192

Now suppose that we have a proper map α : A1 → A2. Thisinduces (unital) maps α : M(A1)→ M(A2) andα : C (A1)→ C (A2). We may picture all of this in the followingcommuting diagram:

0 A2 X2 B2 0

0 A1 X1 B1 0

0 A2 M(A2) C (A2) 0

0 A1 M(A1) C (A1) 0

i2 π2

i1 π1

q2

q1

θ2

θ1

η2

η1

α

α α α

48 / 192

Now suppose that we have a proper map α : A1 → A2. Thisinduces (unital) maps α : M(A1)→ M(A2) andα : C (A1)→ C (A2). We may picture all of this in the followingcommuting diagram:

0 A2 X2 B2 0

0 A1 X1 B1 0

0 A2 M(A2) C (A2) 0

0 A1 M(A1) C (A1) 0

i2 π2

i1 π1

q2

q1

θ2

θ1

η2

η1

α

α α α

49 / 192

We are missing maps X1 → X2 and B1 → B2:

0 A2 X2 B2 0

0 A1 X1 B1 0

0 A2 M(A2) C (A2) 0

0 A1 M(A1) C (A1) 0

i2 π2

i1 π1

q2

q1

θ2

θ1

η2

η1

α

α α α

ξ β

50 / 192

Notice that the map ξ creates two new squares (front left, andcenter), while the map β creates one (right end). It is easy to seethat if ξ makes the front left square commute, then the centersquare automatically commutes as well. For suppose thatξi1 = i2α. For x1 ∈ X1, to see that αθ1(x1) = θ2ξ(x1), it sufficesto check them as multipliers on a dense subset of A2. Since α(A1)contains an approximate unit for A2, we can use elements of theform α(a1)a2 (or a2α(a1) if multiplying on the right). Then

(θ2ξ)(x1) · α(a1)a2 = θ2(ξ(x1)i2(α(a1))a2 = θ2(ξ(x1)(ξi1)(a1))a2

= (θ2ξ)(x1i1(a1))a2 = (θ2ξ)(i1(θ1(x1)a1))a2 = (θ2i2α)(θ1(x1)a1)a2

= (θ2i2)((αθ1)(x1)α(a1))a2 = (αθ1)(x1) · α(a1)a2.

51 / 192

0 A2 X2 B2 0

0 A1 X1 B1 0

0 A2 M(A2) C (A2) 0

0 A1 M(A1) C (A1) 0

i2 π2

i1 π1

q2

q1

θ2

θ1

η2

η1

α

α α α

ξ β

(θ2ξ)(x1) · α(a1)a2 = θ2(ξ(x1)i2(α(a1))a2 = θ2(ξ(x1)(ξi1)(a1))a2

= (θ2ξ)(x1i1(a1))a2 = (θ2ξ)(i1(θ1(x1)a1))a2 = (θ2i2α)(θ1(x1)a1)a2

= (θ2i2)((αθ1)(x1)α(a1))a2 = (αθ1)(x1) · α(a1)a2.

52 / 192

Now if ξ is given such that ξi1 = i2α, then there is a unique map βsuch that the right front square commutes: βπ1 = π2ξ. (To defineβ, lift an element of B1 to X1, then apply π2ξ. Different lifts leadto the same result, by commutativity of the left front square.)Moreover, with this map β, the right end square commutes. (Tosee that η2β = αη1, notice that it is enough to check thatη2βπ1 = αη1π1, since π1 is onto. This is a simple diagram chase.)It is more interesting to see that we can recover ξ from β.

53 / 192

Theorem 5.1Let α : A1 → A2 be a proper map. If β : B1 → B2 satisfiesη2β = αη1, then there exists a unique map ξ : X1 → X2 such thatξi1 = i2α and π2ξ = βπ1.

Proof. Since X2 equals the pullback M(A2) ⊕(β,η2)

B2, we can

construct the map ξ by giving appropriate maps of X1 into M(A2)and B2. These are αθ1 and βπ1. We check thatq2 ◦ αθ1 = η2 ◦ βπ1 — this is a simple diagram chase. Thus theredoes exist a unique map ξ : X1 → X2 such that αθ1 = θ2ξ andβπ1 = π2ξ.Finally, we show that ξi1 = i2α. First, π2ξi1 = βπ1i1 = 0, so thatξi1(A1) ⊆ i2(A2). For a1 ∈ A1, we have

θ2(ξi1(a1)) = αθ1i1(a1) = α(a1) = α(a1) = θ2(i2α(a1)).

Since θ2 is one-to-one on i2(A2), we have ξi1 = i2α. �

54 / 192

Definition 5.2Call a C ∗-algebra corona semiprojective if the semiprojectivityproperty holds in the special case where the quotient (into whichthe algebra is mapped) is isomorphic to a corona algebra.

Exercise 5.3Prove that corona semiprojectivity implies semiprojectivity.

(Hints:a quotient B/I maps into C (I ). If I =

⋃n In, show that the square

B/In −−−−→ M(I )/Iny yB/I −−−−→ C (I )

is a pullback.)

55 / 192

Definition 5.2Call a C ∗-algebra corona semiprojective if the semiprojectivityproperty holds in the special case where the quotient (into whichthe algebra is mapped) is isomorphic to a corona algebra.

Exercise 5.3Prove that corona semiprojectivity implies semiprojectivity. (Hints:a quotient B/I maps into C (I ). If I =

⋃n In, show that the square

B/In −−−−→ M(I )/Iny yB/I −−−−→ C (I )

is a pullback.)

56 / 192

6. Corona extendibility

Exercise 5.3 is a crucial result. The reason is that corona algebrasare big, rambly C ∗-algebras, with lots of nooks and crannies whereyou can hide, or find, something that you need. So it ought to beeasier to lift a map into a corona algebra than a map into anordinary C ∗-algebra. When we use this exercise later on, we willalso need to enlarge the algebra from which we are lifting. This ismotivation, however paltry, for the next definition.

Definition 6.1A map ξ : X1 → X2 is corona extendible if every map of X1, intothe corona algebra of a σ-unital C ∗-algebra, factors through ξ:

X1

X2

C (D)

ξ ∃

57 / 192

In order to establish semiprojectivity for (certain) extensions by C,we need to know that some of the maps involved are coronaextendible. In this section, we will prove a theorem from [ELP 98]that does this. It is just a very beautiful (?) and clever diagramchase, with one deep fact about corona algebras inserted at theright place. In fact, this is the place where “corona magic” showsup in our treatment.

Recall the notion of conditional projectivity from Exercise 2.5: amap β : B1 → B2 is conditionally projective if

B1

B2

D

D/I

β

58 / 192

In order to establish semiprojectivity for (certain) extensions by C,we need to know that some of the maps involved are coronaextendible. In this section, we will prove a theorem from [ELP 98]that does this. It is just a very beautiful (?) and clever diagramchase, with one deep fact about corona algebras inserted at theright place. In fact, this is the place where “corona magic” showsup in our treatment.

Recall the notion of conditional projectivity from Exercise 2.5: amap β : B1 → B2 is conditionally projective if

B1

B2

D

D/I

β

59 / 192

Theorem 6.2For j = 1, 2, let 0→ A

ij→ Xjπj→ Bj → 0, with Busby invariant ηj ,

and suppose that A is σ-unital. Let ξ : X1 → X2 and β : B1 → B2

be maps such that the diagram

0 −−−−→ Ai1−−−−→ X1

π1−−−−→ B1 −−−−→ 0∥∥∥ ξ

y β

y0 −−−−→ A

i2−−−−→ X2π2−−−−→ B2 −−−−→ 0

commutes. If β is conditionally projective, then ξ is coronaextendible.

60 / 192

Here is an example of how the theorem can be used, phrased as anexercise.

Exercise 6.3

Define ξ : CC→ CC by ξ(f )(t) =

{f (2t), if 0 < t ≤ 1/2

f (1) if 1/2 ≤ t ≤ 1..

Show that ξ is corona extendible.

(Hint: Fit ξ into a commutative diagram

0 −−−−→ SC i1−−−−→ CC π1−−−−→ C −−−−→ 0∥∥∥ ξ

y β

y0 −−−−→ SC i2−−−−→ CC π2−−−−→ C [1/2, 1] −−−−→ 0.)

We also give the following as an exercise.

Exercise 6.4The inclusion SC→ CC is not corona extendible.

61 / 192


Exercise 6.3


{f (2t), if 0 < t ≤ 1/2

f (1) if 1/2 ≤ t ≤ 1..

Show that ξ is corona extendible.(Hint: Fit ξ into a commutative diagram

0 −−−−→ SC i1−−−−→ CC π1−−−−→ C −−−−→ 0∥∥∥ ξ

y β

y0 −−−−→ SC i2−−−−→ CC π2−−−−→ C [1/2, 1] −−−−→ 0.)



62 / 192


Exercise 6.3


{f (2t), if 0 < t ≤ 1/2

f (1) if 1/2 ≤ t ≤ 1..

Show that ξ is corona extendible.(Hint: Fit ξ into a commutative diagram

0 −−−−→ SC i1−−−−→ CC π1−−−−→ C −−−−→ 0∥∥∥ ξ

y β

y0 −−−−→ SC i2−−−−→ CC π2−−−−→ C [1/2, 1] −−−−→ 0.)



63 / 192

The fact about corona algebras alluded to above is the following.We will prove it in the next section.

Theorem 6.5Let A and D be σ-unital C ∗-algebras, with A ⊆ C (D). Let N bethe idealizer of A in C (D): N = {x ∈ C (D) : xA,Ax ⊆ A}. (So wehave AC N.) Then the canonical map θ : N → M(A) is surjective.

64 / 192

Proof. (of Theorem 6.2.) Let ψ : X1 → C (D), where D isσ-unital. Let φ = ψ|A, and let I (φ) be the idealizer of φ(A) inC (D).

A X1

C (D)

A X1

φ(A) I (φ) C (D)

φ ψ φ ψψ

Let E (φ) = I (φ)/φ(A). Since i1(A)C X1 we know thatψ(X1) ⊆ I (φ). Then also we may define ψ : B1 → E (φ) in theobvious way, so that ψπ1 = πψ.

0 −−−−→ φ(A) −−−−→ I (φ)π−−−−→ E (φ) −−−−→ 0

φ

x ψ

x ψ

x0 −−−−→ A

i1−−−−→ X1π1−−−−→ B1 −−−−→ 0

65 / 192

Proof. (of Theorem 6.2.) Let ψ : X1 → C (D), where D isσ-unital. Let φ = ψ|A, and let I (φ) be the idealizer of φ(A) inC (D).

A X1

C (D)

A X1

φ(A) I (φ) C (D)

φ ψ φ ψψ

Let E (φ) = I (φ)/φ(A). Since i1(A)C X1 we know thatψ(X1) ⊆ I (φ). Then also we may define ψ : B1 → E (φ) in theobvious way, so that ψπ1 = πψ.

0 −−−−→ φ(A) −−−−→ I (φ)π−−−−→ E (φ) −−−−→ 0

φ

x ψ

x ψ

x0 −−−−→ A

i1−−−−→ X1π1−−−−→ B1 −−−−→ 0

66 / 192

Then φ(A)C I (φ), so we have the canonical mapθ : I (φ)→ M(φ(A)), and of course θ : E (φ)→ C (φ(A)). ByTheorem 6.5 we know that θ is surjective, and hence θ is alsosurjective.

0 0

0

0

0

0

A X1 B1

φ(A) M(φ(A)) C (φ(A))

φ(A) I (φ) E (φ)

i1 π1

q

i π

θ θ

φ ψ ψ

We will use the following big diagram:

67 / 192

0

0

0

0

0

0

0

0

0

0

A

A

A

M(A)

X2

X1

C (A)

B2

B1

φ(A) M(φ(A)) C (φ(A))

φ(A) I (φ) E (φ)

q

i2 π2

i1 π1

q

i π

ξ

θ2

β

η2

θ θ

φ

φ

φ

ψ

φ

ψ

68 / 192

It is clear that the front, back, top, and bottom of this diagramcommute. Let’s check that the center and right squares alsocommute. First we show that θψ = φθ2ξ. It is enough to checkthis when multiplying elements of φ(A). So let x1 ∈ X1, anda ∈ A. We have

(θψ)(x1)φ(a) = θ(ψ(x1)φ(a))

= θ(ψ(x1a))

= φ(x1a)

(φθ2ξ)(x1)φ(a) = φ(θ2ξ(x1))φ(a)

= φ(θ2(ξ(x1))a)

= φ(ξ(x1)a)

= φ(x1a),

and similarly when multiplying on the other side.

69 / 192

Next we show that θ ψ = φη2β. Since π1 is onto, it is enough toshow that θ ψπ1 = φη2βπ1. But

θ ψπ1 = θπψ = qθψ = qφθ2ξ = φqθ2ξ = φη2βπ1.

Thus the big diagram above commutes.

Now, since β is conditionally projective, and because θ is onto,there is a map ρ : B2 → E (φ) such that ψ = ρβ and θρ = φη2:

70 / 192

Next we show that θ ψ = φη2β. Since π1 is onto, it is enough toshow that θ ψπ1 = φη2βπ1. But

θ ψπ1 = θπψ = qθψ = qφθ2ξ = φqθ2ξ = φη2βπ1.

Thus the big diagram above commutes.

Now, since β is conditionally projective, and because θ is onto,there is a map ρ : B2 → E (φ) such that ψ = ρβ and θρ = φη2:

71 / 192

0

0

0

0

0

0

0

0

0

0

A

A

A

M(A)

X2

X1

C (A)

B2

B1

φ(A) M(φ(A)) C (φ(A))

φ(A) I (φ) E (φ)

q

i2 π2

i1 π1

q

i π

ξ

θ2

β

η2

θ θ

φ

φ

φ

ψ

φ

ψ

ρ

72 / 192

Next, we notice that the map ρ, satisfying θρ = φη2, is exactlywhat is needed to use Theorem 5.1:

0

0

0

0

0

0

0

0

0

0

A

A

A

M(A)

X2

X1

C (A)

B2

B1

φ(A) M(φ(A)) C (φ(A))

φ(A) I (φ) E (φ)

q

i2 π2

i1 π1

q

i π

ξ

θ2

β

η2

θ θ

φ

φ

φ

ψ

φ

ψ

φ ψ′ ρ

73 / 192

Next, we notice that the map ρ, satisfying θρ = φη2, is exactlywhat is needed to use Theorem 5.1:

0

0

0

0

0

0

0

0

0

0

A

A

A

M(A)

X2

X1

C (A)

B2

B1

φ(A) M(φ(A)) C (φ(A))

φ(A) I (φ) E (φ)

q

i2 π2

i1 π1

q

i π

ξ

θ2

β

η2

θ θ

φ

φ

φ

ψ

φ

ψ

φ ψ′ ρ

74 / 192

Thus there exists a (unique) map ψ′ : X2 → I (φ) such thatψ′i2 = iφ. We also have φθ2 = θψ′ and πψ′ = ρπ2. Finally, noticethat

(ψ′ξ)i1 = ψ′i2 = iφ,

π(ψ′ξ) = ρπ2ξ = ρβπ1 = ψπ1.

Since such a map ψ′ξ : X1 → I (φ) is uniquely determined by ψ, wemust have ψ′ξ = ψ. Thus, ψ′ is the desired extension of ψ. �

75 / 192

7. Corona magic

In this section we will prove Theorem 6.5:

Let A and D be σ-unital C ∗-algebras, with A ⊆ C (D). Let N bethe idealizer of A in C (D): N = {x ∈ C (D) : xA,Ax ⊆ A}. (So wehave AC N.) Then the canonical map θ : N → M(A) is surjective.

76 / 192

We begin with a corollary of the technical results in section 2.

Lemma 7.1Let a1, a2, a3 ∈ A/I with a1 ≤ a2 ≤ a3. Let b1, b3 ∈ A withb1 ≤ b3 and π(bi ) = ai for i = 1, 3. Then there is b2 ∈ A withb1 ≤ b2 ≤ b3 and π(b2) = a2. Moreover, such lifts b1 and b3 exist.

Proof. We may prove the existence of b1 and b3 by first choosingb1 ∈ Asa with π(b1) = a1, then choosing w ∈ A+ withπ(w) = a3 − a1, and setting b3 = b1 + w .Now, for any choice of b1 and b3, we have 0 ≤ a2 − a1 ≤ a3 − a1,and b3 − b1 ∈ A+ with π(b3 − b1) = a3 − a1. Letx = (a2 − a1)1/2. Then x∗x ≤ a3 − a1. By (part of) Lemma 2.2,there is z ∈ A with π(z) = x and z∗z ≤ b3 − b1. Thenπ(z∗z) = a2 − a1. Set b2 = b1 + z∗z . �

77 / 192



Proof. We may prove the existence of b1 and b3 by first choosingb1 ∈ Asa with π(b1) = a1, then choosing w ∈ A+ withπ(w) = a3 − a1, and setting b3 = b1 + w .

Now, for any choice of b1 and b3, we have 0 ≤ a2 − a1 ≤ a3 − a1,and b3 − b1 ∈ A+ with π(b3 − b1) = a3 − a1. Letx = (a2 − a1)1/2. Then x∗x ≤ a3 − a1. By (part of) Lemma 2.2,there is z ∈ A with π(z) = x and z∗z ≤ b3 − b1. Thenπ(z∗z) = a2 − a1. Set b2 = b1 + z∗z . �

78 / 192



Proof. We may prove the existence of b1 and b3 by first choosingb1 ∈ Asa with π(b1) = a1, then choosing w ∈ A+ withπ(w) = a3 − a1, and setting b3 = b1 + w .Now, for any choice of b1 and b3, we have 0 ≤ a2 − a1 ≤ a3 − a1,and b3 − b1 ∈ A+ with π(b3 − b1) = a3 − a1. Letx = (a2 − a1)1/2. Then x∗x ≤ a3 − a1. By (part of) Lemma 2.2,there is z ∈ A with π(z) = x and z∗z ≤ b3 − b1. Thenπ(z∗z) = a2 − a1. Set b2 = b1 + z∗z . �

79 / 192

We use this lemma to prove a result of Olsen and Pedersen(though we have taken the proof from [Loring 97]). A C ∗-algebrais said to have the countable Riesz separation property (CRISP) ifwhenever a1 ≤ a2 ≤ · · · ≤ b2 ≤ b1, there exists z such thatan ≤ z ≤ bn for all n.

Theorem 7.2([OP 89]) Let A be σ-unital. Then C (A) has the CRISP.

80 / 192

Proof. Let an, bn ∈ C (A) with a1 ≤ a2 ≤ · · · ≤ b2 ≤ b1. Leth ∈ A be a strictly positive contraction. We will lift the inequalitiesto M(A), find an interpolating element there, and show that itdrops to the right thing.

Suppose inductively that we have found

x1 ≤ · · · ≤ xn−1 ≤ yn−1 ≤ · · · ≤ y1

in M(A), with q(xi ) = ai and q(yi ) = bi , and such that‖h(yi − xi )h‖ < 1/i , for i < n. By Lemma 7.1, there is xn ∈ M(A)with q(xn) = an and xn−1 ≤ xn ≤ yn−1. A second application ofthe lemma gives y ∈ M(A) with q(y) = bn and xn ≤ y ≤ yn−1.

81 / 192

Proof. Let an, bn ∈ C (A) with a1 ≤ a2 ≤ · · · ≤ b2 ≤ b1. Leth ∈ A be a strictly positive contraction. We will lift the inequalitiesto M(A), find an interpolating element there, and show that itdrops to the right thing.

Suppose inductively that we have found

x1 ≤ · · · ≤ xn−1 ≤ yn−1 ≤ · · · ≤ y1

in M(A), with q(xi ) = ai and q(yi ) = bi , and such that‖h(yi − xi )h‖ < 1/i , for i < n. By Lemma 7.1, there is xn ∈ M(A)with q(xn) = an and xn−1 ≤ xn ≤ yn−1. A second application ofthe lemma gives y ∈ M(A) with q(y) = bn and xn ≤ y ≤ yn−1.

82 / 192

For k ∈ N define uk ∈ M(A) by

uk = (y − xn)1/2(1− h1/k)(y − xn)1/2.

Then uk ≥ 0, q(uk) = bn − an, and uk ≤ y − xn. Since (h1/k) isan approximate unit for A, we have that ‖hukh‖ → 0 as k →∞.So choose k so that ‖hukh‖ < 1/n, and let yn = xn + uk . Thenq(yn) = bn, xn ≤ yn ≤ y ≤ yn−1, and ‖h(yn − xn)h‖ < 1/n.

83 / 192

We now claim that there is w ∈ M(A) such that xn → w andyn → w strictly. For this, let a ∈ A, with 0 ≤ a ≤ 1 forconvenience. Let ε > 0 be given, and choose b ∈ A such that‖a− hb‖ < ε.

To see that such an element b exists, choose k so that‖a− h1/ka‖ < ε. Then choose a function g ∈ C0(0, 1] such that|tg(t)− t1/k | < ε for 0 < t ≤ 1.

(For example, let g(t) =

{ε1−k , if 0 < t ≤ εk

t(1/k)−1 if εk ≤ t ≤ 1.)

Let b = g(h)a. Then

‖a− hb‖ = ‖a− hg(h)a‖ ≤ ‖a− h1/ka‖+ ‖h1/k − hg(h)‖ < 2ε.

84 / 192

We now claim that there is w ∈ M(A) such that xn → w andyn → w strictly. For this, let a ∈ A, with 0 ≤ a ≤ 1 forconvenience. Let ε > 0 be given, and choose b ∈ A such that‖a− hb‖ < ε.

To see that such an element b exists, choose k so that‖a− h1/ka‖ < ε. Then choose a function g ∈ C0(0, 1] such that|tg(t)− t1/k | < ε for 0 < t ≤ 1.

(For example, let g(t) =

{ε1−k , if 0 < t ≤ εk

t(1/k)−1 if εk ≤ t ≤ 1.)

Let b = g(h)a. Then

‖a− hb‖ = ‖a− hg(h)a‖ ≤ ‖a− h1/ka‖+ ‖h1/k − hg(h)‖ < 2ε.

85 / 192

Now we have for m < n,

‖(xn − xm)a‖ ≤ ‖xn − xm‖ε+ ‖(xn − xm)hb‖≤ ‖y1 − x1‖ε+ ‖b‖ · ‖(xn − xm)1/2‖ · ‖(xn − xm)1/2h‖≤ ‖y1 − x1‖ε+ ‖b‖ · ‖(y1 − x1)1/2‖ · ‖h(xn − xm)h‖1/2

≤ ‖y1 − x1‖ε+ ‖b‖ · ‖(y1 − x1)1/2‖ · ‖h(ym − xm)h‖1/2,

and the last part tends to zero as m→∞. Therefore (xna) isCauchy in A.

Define L : A→ A by L(a) = limn xna. Similarly, define R : A→ Aby R(a) = limn axn. Then (L,R) is a double centralizer of A, sothere exists w ∈ M(A) inducing (L,R). Moreover, we see thatxn → w strictly.

86 / 192

Now we have for m < n,

‖(xn − xm)a‖ ≤ ‖xn − xm‖ε+ ‖(xn − xm)hb‖≤ ‖y1 − x1‖ε+ ‖b‖ · ‖(xn − xm)1/2‖ · ‖(xn − xm)1/2h‖≤ ‖y1 − x1‖ε+ ‖b‖ · ‖(y1 − x1)1/2‖ · ‖h(xn − xm)h‖1/2

≤ ‖y1 − x1‖ε+ ‖b‖ · ‖(y1 − x1)1/2‖ · ‖h(ym − xm)h‖1/2,

and the last part tends to zero as m→∞. Therefore (xna) isCauchy in A.

Define L : A→ A by L(a) = limn xna. Similarly, define R : A→ Aby R(a) = limn axn. Then (L,R) is a double centralizer of A, sothere exists w ∈ M(A) inducing (L,R). Moreover, we see thatxn → w strictly.

87 / 192

Since ‖h(yn − xn)h‖ → 0, we have

‖yna− xna‖ ≤ ‖yn − xn‖ε+ ‖(yn − xn)hb‖≤ ‖y1 − x1‖ε+ ‖(y1 − x1)1/2‖ · ‖b‖ · ‖h(yn − xn)h‖,

and again the last part tends to zero as n→∞. Thus yn → wstrictly as well. Now w − xm = s- limn(xn − xm) ≥ 0, so xn ≤ w forall n. Similarly, w ≤ yn for all n. We set z = q(y). �

88 / 192

Now we prove Theorem 6.5.Proof. During the proof we will use independently the inclusionsof A into M(A) and C (D). Let m ∈ M(A) with 0 ≤ m ≤ 1. (Itsuffices to show that such m are in the image of θ.) Let h ∈ A bea strictly positive contraction. Put

xn = m1/2h1/nm1/2 and yn = 1− (1−m)1/2h1/n(1−m)1/2.

Clearly we have that

0 ≤ x1 ≤ x2 ≤ · · · ≤ m ≤ · · · ≤ y2 ≤ y1 ≤ 1,

and that xn → m and yn → m strictly. Moreover,

xn, yn ∈ A + 1 ⊆ N ⊆ C (D).

Since C (D) has the CRISP, there is z ∈ C (D) such thatxn ≤ z ≤ yn for all n.

89 / 192

We claim that z ∈ N, and that θ(z) = m. Let a ∈ A. Then‖(yn − xn)a‖ → 0.

Hence

‖(z − xn)a‖ ≤ ‖(z − xn)1/2‖ · ‖(z − xn)1/2a‖≤ ‖a(z − xn)a‖1/2

≤ ‖a(yn − xn)a‖1/2

≤ ‖a‖ · ‖(yn − xn)a‖1/2

→ 0, as n→∞.

Therefore za = limn xna ∈ A. Similarly, az ∈ A, so we have z ∈ N.Now

θ(z)a = za = limn

xna = ma,

and similarly, aθ(z) = am. Therefore θ(z) = m. �

90 / 192

We claim that z ∈ N, and that θ(z) = m. Let a ∈ A. Then‖(yn − xn)a‖ → 0. Hence


≤ ‖a(yn − xn)a‖1/2

≤ ‖a‖ · ‖(yn − xn)a‖1/2

→ 0, as n→∞.


θ(z)a = za = limn

xna = ma,


91 / 192

We claim that z ∈ N, and that θ(z) = m. Let a ∈ A. Then‖(yn − xn)a‖ → 0. Hence


≤ ‖a(yn − xn)a‖1/2

≤ ‖a‖ · ‖(yn − xn)a‖1/2

→ 0, as n→∞.


θ(z)a = za = limn

xna = ma,


92 / 192

Exercise 7.3If A and C are σ-unital, and if π : C → A is onto, thenπ : M(C )→ M(A) is onto.

(Hints: Let 0 ≤ m ≤ 1 in M(A), and let h be a strictly positivecontraction in A. Imitate the proof of Theorem 6.5 to sandwich mbetween an ∈ A and bn ∈ 1 + A, with h(yn − xn)h→ 0. Imitatethe proof of Theorem 7.2 to lift an to C and bn to 1 + C . Stillimitating, show that there is an element between these lifts inM(C ), and that it is mapped to m.)

93 / 192

8. Pushouts

Pushouts are a little bit trickier than pullbacks. Terry’s book[Loring 97] treats them nicely, but we won’t need even all of that.Dually to the situation with pullbacks, we need to use pushouts toconstruct maps out of the central algebra of an extension. We willonly consider extensions in which the quotient is isomorphic to C:0→ A→ X → C→ 0. This situation is elegantly dissected by anecessary theorem of Gert Pedersen [Pedersen 97].

Theorem 8.1Let 0→ A→ X

p→ C→ 0 be an extension, and assume that A isσ-unital. There is a positive contraction h ∈ X such that h − h2 isa strictly positive element of A. Moreover, h is strictly positive inX .

94 / 192

8. Pushouts

Pushouts are a little bit trickier than pullbacks. Terry’s book[Loring 97] treats them nicely, but we won’t need even all of that.Dually to the situation with pullbacks, we need to use pushouts toconstruct maps out of the central algebra of an extension. We willonly consider extensions in which the quotient is isomorphic to C:0→ A→ X → C→ 0. This situation is elegantly dissected by anecessary theorem of Gert Pedersen [Pedersen 97].

Theorem 8.1Let 0→ A→ X

p→ C→ 0 be an extension, and assume that A isσ-unital. There is a positive contraction h ∈ X such that h − h2 isa strictly positive element of A. Moreover, h is strictly positive inX .

95 / 192

Proof. (This proof is taken from [Loring 97].) Let a ∈ A be astrictly positive element. By scaling a, we may assume that thereis λ ∈ (0, 1) such that 0 ≤ a ≤ λ. Let x ∈ X be a positivecontraction with p(x) = 1.

We define h ∈ X by

h = x1/2(1− a)x1/2 + (1− x)1/2a(1− x)1/2.

Then h ∈ x + A, so p(h) = 1. It is clear that h ≥ 0. We also have

h = x − x1/2ax1/2 + (1− x)1/2a(1− x)1/2,

so

1− h = (1− x)− (1− x)1/2a(1− x)1/2 + x1/2ax1/2

= (1− x)1/2(1− a)(1− x)1/2 + x1/2ax1/2

≥ 0.

96 / 192

Proof. (This proof is taken from [Loring 97].) Let a ∈ A be astrictly positive element. By scaling a, we may assume that thereis λ ∈ (0, 1) such that 0 ≤ a ≤ λ. Let x ∈ X be a positivecontraction with p(x) = 1. We define h ∈ X by

h = x1/2(1− a)x1/2 + (1− x)1/2a(1− x)1/2.

Then h ∈ x + A, so p(h) = 1. It is clear that h ≥ 0.

We also have

h = x − x1/2ax1/2 + (1− x)1/2a(1− x)1/2,

so

1− h = (1− x)− (1− x)1/2a(1− x)1/2 + x1/2ax1/2

= (1− x)1/2(1− a)(1− x)1/2 + x1/2ax1/2

≥ 0.

97 / 192


h = x1/2(1− a)x1/2 + (1− x)1/2a(1− x)1/2.


h = x − x1/2ax1/2 + (1− x)1/2a(1− x)1/2

,

so

1− h = (1− x)− (1− x)1/2a(1− x)1/2 + x1/2ax1/2

= (1− x)1/2(1− a)(1− x)1/2 + x1/2ax1/2

≥ 0.

98 / 192


h = x1/2(1− a)x1/2 + (1− x)1/2a(1− x)1/2.


h = x − x1/2ax1/2 + (1− x)1/2a(1− x)1/2,

so

1− h = (1− x)− (1− x)1/2a(1− x)1/2 + x1/2ax1/2

= (1− x)1/2(1− a)(1− x)1/2 + x1/2ax1/2

≥ 0.

99 / 192

Thus 0 ≤ h ≤ 1. Since p(h) = 1, we have thath − h2 ∈ ker (p) = A.

Before proving that h − h2 is strictly positive in A, we will showthat h is strictly positive in X . For this, let π be a nondegeneraterepresentation of X such that π(h) = 0. Applying π to thedefinition of h above, we get

(a) π(x)1/2(1− π(a))π(x)1/2 = 0

(b) (1− π(x))1/2π(a)(1− π(x))1/2 = 0.

Since 0 ≤ a ≤ λ, we have 0 < 1− λ ≤ 1− a. Then 1− a isinvertible, so it follows from (a) that π(x) = 0. Then it followsfrom (b) that π(a) = 0. Since a is strictly positive in A, we havethat π|A = 0. Since X is generated by A and x , we have π = 0, acontradiction.

100 / 192



(a) π(x)1/2(1− π(a))π(x)1/2 = 0

(b) (1− π(x))1/2π(a)(1− π(x))1/2 = 0.


101 / 192



(a) π(x)1/2(1− π(a))π(x)1/2 = 0

(b) (1− π(x))1/2π(a)(1− π(x))1/2 = 0.


102 / 192

Now let σ be a nondegenerate representation of A such thatσ(h − h2) = 0. There is a nondegenerate extension σ of σ to arepresentation of X on the same Hilbert space. Then σ(h) is aprojection. Since h is strictly positive in X , the kernel of thisprojection is trivial; i.e. σ(h) = 1. But then 1− σ(h) = 0.

Applying σ to the formula derived above for 1− h,

1− h = (1− x)1/2(1− a)(1− x)1/2 + x1/2ax1/2,

we find that

(c) (1− σ(x))1/2(1− σ(a))(1− σ(x))1/2 = 0

(d) σ(x)1/2σ(a)σ(x)1/2 = 0.

Again, since 1− a is invertible, we conclude from (c) that1− σ(x) = 0, and hence that σ(x) = 1. But then (d) implies thatσ(a) = 0. Since σ|A = σ, and a is strictly positive in A, we thenhave that σ = 0, a contradiction. �

103 / 192

Now let σ be a nondegenerate representation of A such thatσ(h − h2) = 0. There is a nondegenerate extension σ of σ to arepresentation of X on the same Hilbert space. Then σ(h) is aprojection. Since h is strictly positive in X , the kernel of thisprojection is trivial; i.e. σ(h) = 1. But then 1− σ(h) = 0.Applying σ to the formula derived above for 1− h,

1− h = (1− x)1/2(1− a)(1− x)1/2 + x1/2ax1/2,

we find that

(c) (1− σ(x))1/2(1− σ(a))(1− σ(x))1/2 = 0

(d) σ(x)1/2σ(a)σ(x)1/2 = 0.


104 / 192

Now let σ be a nondegenerate representation of A such thatσ(h − h2) = 0. There is a nondegenerate extension σ of σ to arepresentation of X on the same Hilbert space. Then σ(h) is aprojection. Since h is strictly positive in X , the kernel of thisprojection is trivial; i.e. σ(h) = 1. But then 1− σ(h) = 0.Applying σ to the formula derived above for 1− h,

1− h = (1− x)1/2(1− a)(1− x)1/2 + x1/2ax1/2,

we find that

(c) (1− σ(x))1/2(1− σ(a))(1− σ(x))1/2 = 0

(d) σ(x)1/2σ(a)σ(x)1/2 = 0.


105 / 192

We use the theorem as follows. There is a map k0 : CC→ Xdefined by k0(t) = h, and a map i1 : SC→ A defined byi1(e2πit − 1) = e2πih − 1 (hence also i1(t − t2) = h− h2). We thushave a commutative diagram

SC i0−−−−→ CC

i1

y yk0

A −−−−→k1

X .

The pushout of the diagram consisting of the maps i0 and i1 is aC ∗-algebra that is universal for the role that X plays above.Namely, it is a C ∗-algebra A ∗

SCCC, with maps j0 : CC→ A ∗

SCCC

and j1 : A→ A ∗SC

CC such that

106 / 192

We use the theorem as follows. There is a map k0 : CC→ Xdefined by k0(t) = h, and a map i1 : SC→ A defined byi1(e2πit − 1) = e2πih − 1 (hence also i1(t − t2) = h− h2). We thushave a commutative diagram

SC i0−−−−→ CC

i1

y yk0

A −−−−→k1

X .

The pushout of the diagram consisting of the maps i0 and i1 is aC ∗-algebra that is universal for the role that X plays above.Namely, it is a C ∗-algebra A ∗

SCCC, with maps j0 : CC→ A ∗

SCCC

and j1 : A→ A ∗SC

CC such that

107 / 192

1. j0i0 = j1i1.

2. If k0 : CC→ X and k1 : A→ X are maps such thatk0i0 = k1i1, then there exists a unique map γ : A ∗

SCCC→ X

such that kµ = γjµ for µ = 1, 2.

The picture is like this:

SC

CC

A

A ∗SC

CC X

i0

i1

j0

j1

k1

k0

∃!γ

108 / 192

It is easy enough to prove the existence of the pushout by givinggenerators and relations. Namely, we let A ∗

SCCC be generated by

homomorphic images j0(CC) and j1(A) with relationsj0(f ) = j1(f (h)), for f ∈ SC.

We will now show that the pushoutis isomorphic to X (and that in fact, γ is an isomorphism). Weformulate this as a proposition

Proposition 8.2

In the situation discussed above, we have

1. The map j1 is injective.

2. j1(A) is an ideal in the pushout.

3. The map γ is an isomorphism.

109 / 192

It is easy enough to prove the existence of the pushout by givinggenerators and relations. Namely, we let A ∗

SCCC be generated by

homomorphic images j0(CC) and j1(A) with relationsj0(f ) = j1(f (h)), for f ∈ SC. We will now show that the pushoutis isomorphic to X (and that in fact, γ is an isomorphism). Weformulate this as a proposition

Proposition 8.2

In the situation discussed above, we have

1. The map j1 is injective.

2. j1(A) is an ideal in the pushout.

3. The map γ is an isomorphism.

110 / 192

Proof. We first prove (1). Let π1 : A→ B(H) be a faithfulnondegenerate representation. Since i1(SC) contains anapproximate unit for A, the representation π1 ◦ i1 of SC is alsonondegenerate. Since SC is an ideal in CC, there is arepresentation π0 of CC on H such that π0 ◦ i0 = π1 ◦ i1. By theuniversal property of the pushout, there is a representation π ofA ∗

SCCC on H such that π ◦ jµ = πµ, for µ = 0, 1. In particular,

since π1 is faithful, then so is j1.

111 / 192

Now we prove (2). The finite products of elements of j1(A) andj0(CC) form a total set in the pushout. In fact, we may replace Awith a dense subset of A. For this, we use (h− h2)A(h− h2), sinceh − h2 is strictly positive in A. But

j1((h − h2)a(h − h2)) = j0(t − t2)j1(a)j0(t − t2).

Then for any f ∈ CC,

j0(f )j1((h − h2)a(h − h2)) = j1(f (h)(h − h2)a(h − h2)) ∈ j1(A),

(and similarly when multiplying in the opposite order).

112 / 192

Finally we prove (3). For this, we consider also the map p : X → Cfrom the original extension. We will define a mapδ : X → A ∗

SCCC inverse to γ. Let x ∈ X . Write

x =(x − p(x)h

)+ p(x)h, where x − p(x)h ∈ A. Of course

(8) xy =[(

x − p(x)h)

+ p(x)h][(

y − p(y)h)

+ p(y)h].

In A ∗SC

CC, j0(t) plays the role of h, and j1(A) plays the role of A.

Therefore δ(x) = j1 ◦ k−11

(x − p(x)h

)+ p(x)j0(t) is multiplicative

by the “same algebra” that works in (8). Since γj1 = k1, andγj0 = k0 : t 7→ h, γ ◦ δ = idX . Checking that δ ◦ γ is the identityon A and t finishes the proof that γ is invertible. �

113 / 192

Finally we prove (3). For this, we consider also the map p : X → Cfrom the original extension. We will define a mapδ : X → A ∗

SCCC inverse to γ. Let x ∈ X . Write

x =(x − p(x)h

)+ p(x)h, where x − p(x)h ∈ A. Of course

(8) xy =[(

x − p(x)h)

+ p(x)h][(

y − p(y)h)

+ p(y)h].

In A ∗SC

CC, j0(t) plays the role of h, and j1(A) plays the role of A.

Therefore δ(x) = j1 ◦ k−11

(x − p(x)h

)+ p(x)j0(t) is multiplicative

by the “same algebra” that works in (8). Since γj1 = k1, andγj0 = k0 : t 7→ h, γ ◦ δ = idX . Checking that δ ◦ γ is the identityon A and t finishes the proof that γ is invertible. �

114 / 192

Thus, given an extension 0→ A→ X → C→ 0, we may writeX = A ∗

SCCC, where the element h of Theorem 8.1 is understood.

Exercise 8.3(Enders) Let 0→ A→ X → C, where A is σ-unital and Xsemiprojective. Then A is semiprojective.

115 / 192

9. Semiprojectivity for extensions

In this section we will prove an amazing theorem, which will giveeasy demonstrations of semiprojectivity for the dimension-dropalgebras and for the algebra of continuous functions on a finitegraph. It is a special case of a more general theorem from[ELP 98].The goal is to prove that in certain circumstances, if A issemiprojective and σ-unital, and 0→ A→ X → C→ 0, then X issemiprojective.

The idea is this. Let φ : X → B/I , with I aninductive limit ideal. Since X ∼= A ∗

SCCC, finding a lift

ψ : X → B/In is equivalent to finding compatible lifts of A andCC. We get a lift ψ0 : A→ B/In, since A is assumed to besemiprojective. Then there is a lift µ : CC→ B/In since CC isprojective.

116 / 192

9. Semiprojectivity for extensions

In this section we will prove an amazing theorem, which will giveeasy demonstrations of semiprojectivity for the dimension-dropalgebras and for the algebra of continuous functions on a finitegraph. It is a special case of a more general theorem from[ELP 98].The goal is to prove that in certain circumstances, if A issemiprojective and σ-unital, and 0→ A→ X → C→ 0, then X issemiprojective. The idea is this. Let φ : X → B/I , with I aninductive limit ideal. Since X ∼= A ∗

SCCC, finding a lift

ψ : X → B/In is equivalent to finding compatible lifts of A andCC. We get a lift ψ0 : A→ B/In, since A is assumed to besemiprojective. Then there is a lift µ : CC→ B/In since CC isprojective.

117 / 192

Thus

CC

SC A

X

B/In

B/I

ψ0

qn

φ

µ

We know that qnψ0 = φ|A, and qnµ = φ|CC. But in order to get amap : X → B/In, we need the compatibility conditionµ|SC = ψ0|SC. We have no way of arranging this. The “trick” is tobuild in some room. We begin with an elementary example.

118 / 192

Consider the function η pictured below:

η1

113

23

Define ξ : CC→ CC by ξ(f ) = f ◦ η, and α : SC→ SC byα(f ) = f ◦ η. We have

0 −−−−→ SC −−−−→ CC −−−−→ C −−−−→ 0yα yξ ∥∥∥0 −−−−→ SC −−−−→ CC −−−−→ C −−−−→ 0

119 / 192

An algebra like In that is built over the unit interval admits asimilar sort of construction. The main thing is that the idealA1 C X1 sits inside another ideal A2 C X2 in such a way that A2

contains an identity for A1. In order to make use of this “extraroom” to prove that X1 is semiprojective, we need to be able toextend a map of X1 to a quotient to a map of X2 to the samequotient; i.e. we need that the map X1 → X2 be corona extendible.Finally, we only need to assume semiprojectivity for A2 (though inapplications we often have A1

∼= A2).

120 / 192

Theorem 9.1([ELP 98]) Let the following diagram be given, in which A1 isσ-unital, A2 is semiprojective, A2 contains an identity, m, forα(A1), and ξ is corona extendible:

0 −−−−→ A1 −−−−→ X1 −−−−→ C −−−−→ 0

α

y ξ

y ∥∥∥0 −−−−→ A2 −−−−→ X2 −−−−→ C −−−−→ 0.

Then X1 is semiprojective.

121 / 192

Proof. Let h ∈ X1 be as in Theorem 8.1. Let φ1 : X1 → B/I begiven, where I is an inductive limit ideal. We may assume byExercise 5.3 that B/I is a corona algebra. Since ξ is coronaextendible, there is a map φ2 : X2 → B/I such that φ2ξ = φ1.Since A2 is semiprojective, the map φ2|A2 lifts to B/In, for some n.In order to simplify the notation, we replace B by B/In. Weconsider the following diagram, where j2 = ξj1, and in which thesolid arrows commute. The dashed arrow µ0 exists because of theprojectivity of CC, and we have qµ0 = φ2j2. However, asmentioned above, we do not have commutativity of the uppertriangle.

122 / 192

SC

CC

A1

X1

A2

X2

B

B/I

i q

i1

i2

j1

j2

α

ξ

ψ2

φ2

µ0

123 / 192

Now we begin the process of adjusting µ0 so as to achieve theneeded commutativity. Define µ1 : CC→ B by

µ1(f ) =(1−ψ2(m)

)µ0(f )

(1−ψ2(m)

)−ψ2

((1−m)j2(f )(1−m)−j2(f )

).

(Note that the second term makes sense, since A2 C X2 and hence(1−m)j2(f )(1−m)− j2(f ) = −mj2(f )− j2(f )m + mj2(f )m is anelement of A2.) We now have for f ∈ CC,

qµ1(f ) =(1− φ2(m)

)φ2j2(f )

(1− φ2(m)

)− φ2

((1−m)j2(f )(1−m)− j2(f )

)= φ2j2(f ) + φ2

(−mj2(f )− j2(f )m + mj2(f )m

)− φ2

(−mj2(f )− j2(f )m + mj2(f )m

)= φ2j2(f ).

124 / 192

We still don’t have commutativity of the upper triangle, andmoreover, µ1 is not multiplicative. However for g ∈ SC (andf ∈ CC), we have µ1(f )ψ2i2(g) = ψ2i2(fg).

To see this, first note that (1−m)α(A) = 0. Henceµ1(f )ψ2(α(a)) = ψ2(j2(f )α(a)).(Recall

µ1(f ) =(1−ψ2(m)

)µ0(f )

(1−ψ2(m)

)−ψ2

((1−m)j2(f )(1−m)−j2(f )

).)

Therefore

µ1(f )ψ2i2(g) = µ1(f )ψ2

(αi1(g)

)= ψ2

(j2(f )α(i1(g))

)= ψ2i2(fg).

125 / 192

Let ζ(t) = t, so that ζ is the standard generator of CC. PutT = µ1(ζ). Then T is a self-adjoint element of B. For g ∈ SC wehave T nψ2i2(g) = ψ2i2(ζng), so that p(T )ψ2i2(g) = ψ2i2(pg) forall polynomials p without constant term. Taking limits, we findthat f (T )ψ2i2(g) = ψ2i2(fg), for all f continuous on σ(T ) andvanishing at 0.

126 / 192

Letting h be the function

1

1

we have that h(T ) is a positive contraction, and that

q(h(T )) = h(q(T )) = h(qµ1(ζ)) = h(φ2j2(ζ)) = φ2j2(ζ),

since 0 ≤ ζ ≤ 1.

127 / 192

We define µ2 : CC→ B by µ2(f ) = f (h(T )). Then µ2 is a∗-homomorphism, qµ2 = φ2j2, and

(9) µ2(f )ψ2i2(g) = (f ◦h)(T )ψ2i2(g) = ψ2i2((f ◦h)g) = ψ2i2(fg),

since h(t) = t for t ∈ [0, 1].

128 / 192

Next, consider the pullback

CC ⊕(π,π)

CC = {(f1, f2) ∈ CC⊕ CC : f1(1) = f2(1)}.

We define ν : CC ⊕(π,π)

CC→ B by

ν(f1, f2) = ψ2i2(f1 − f2) + µ2(f2).

This is defined because f1 − f2 ∈ SC. Moreover, by (9), ν ismultiplicative (the cross-terms move inside ψ2i2). We have

qν(f1, f2) = q(ψ2i2(f1 − f2) + µ2(f2))

= φ2j2i(f1 − f2) + φ2j2(f2)

= φ2j2(f1)

= φ2j2p1(f1, f2),

where p1 : CC ⊕(π,π)

CC→ CC is the projection onto the first

summand.129 / 192

CC

CC ⊕(π,π)

CC

B

B/I

φ2j2

ν

p1 q

We now need to use the room we get from squashing SC and CCto the middle of the interval.

130 / 192

CC

CC ⊕(π,π)

CC

B

B/I

φ2j2

ν

p1 q

We now need to use the room we get from squashing SC and CCto the middle of the interval.

131 / 192

Let I = C0(1/3, 2/3) ⊆ SC. Then I C SC, and we haveSC/I = C0((0, 1/3] ∪ [2/3, 1)) and CC/I = C0((0, 1/3] ∪ [2/3, 1]).The quotient map π : CC/I → C is split by the mapσ : C→ CC/I defined by σ(1) = χ[2/3,1]. Thus we have thediagram

0

0

0

0SC CC C

SC/I CC/I C

i π

i π

ω0 ω1

σ

132 / 192

Putting the last two pictures together (at least partly), we have

CC

SC

CC

CC ⊕(π,π)

CC/I

CC ⊕(π,π)

CC

B/I

B/In

Bi ⊕ 0 ν

φ2j2

qn

qn,∞

id⊕ ω1

p1τ

νn

133 / 192

Here, the map τ is defined by τ = id⊕ σπ; that is,τ(f ) = (f , f (1)χ[2/3,1]). Let e ∈ SC be a positive contraction suchthat e|[1/3,2/3] = 1. Then e is an identity on I . Sinceφ2j2 ◦ p1 ◦ (id⊕ ω1)(0, e) = 0, then also q ◦ ν(0, e) = 0. Choose nso large that ‖qn ◦ ν(0, e)‖ < 1. It follows that qn ◦ ν(0⊕ I ) = 0.Then there is a map νn : CC ⊕

(π,π)CC/I → B/In such that

qn ◦ ν = νn ◦ (id⊕ ω1). Define µ3 = νn ◦ τ : CC→ B/In. Wecheck that µ3 makes the upper triangle commute. For g ∈ SC,

µ3i(g) = νnτ i(g)

= νn(g , 0)

= qnν(g , 0)

= qnψ2i2(g). �

134 / 192

10. Applications

Theorem 10.1(Loring) The dimension drop algebras are semiprojective.

Proof. Recall that In = {f ∈ CMn : f (1) ∈ C · 1}. Thus we havethe exact sequence 0→ SMn → In → C→ 0. We will squash SMn

and In to the middle third of the interval by means of the functionη pictured below:

η1

113

23

135 / 192

Define ξ : In → In by ξ(f ) = f ◦ η, and α : SMn → SMn byα(f ) = f ◦ η. Since SMn is σ-unital, semiprojective, and containsan identity element for α(SMn), Theorem 9.1 implies that if ξ iscorona extendible, then In is semiprojective.

(The picture is

0 −−−−→ SMn −−−−→ In −−−−→ C −−−−→ 0yα yξ ∥∥∥0 −−−−→ SMn −−−−→ In −−−−→ C −−−−→ 0,

where the maps of SMn to In are the inclusion.)

136 / 192

To show that ξ is corona extendibile, we use Theorem 6.2. Leti1 : SMn → In be the inclusion, and let i2 : SMn → In be given byi2(f ) = f ◦ η. We let

B = In/i2(SMn) = C0((0, 1/3],Mn)⊕C0([2/3, 1),Mn)˜ ∼= CMn⊕CMn.

and let β : C→ B be the map β(1) = χ[2/3,1]. We have thediagram

0 −−−−→ SMni1−−−−→ In −−−−→ C −−−−→ 0∥∥∥ ξ

y β

y0 −−−−→ SMn

i2−−−−→ In −−−−→ B −−−−→ 0.

By exercise 2.5, we know that β is conditionally projective. ByTheorem 6.2, we have that ξ is corona extendible. �

137 / 192

Theorem 10.2(Loring) Let X be a finite graph. Then C (X ) (the algebra ofcontinuous functions on X ) is semiprojective.

Proof. We let V be the set of vertices of X , and E the set ofedges. We direct the edges arbitrarily, so that each e ∈ E has asource s(e) and a range s(e) in V . Further we will identify eachedge with the interval [0, 1], so that s(e) is identified with 0, andr(e) is identified with 1. The proof is by induction on thecardinality of V . If |V | = 1, let V = {v}. ThenC (X ) = C

((X \ {v})

), so it is enough to show that C0(X \ {v})

is semiprojective. But C0(X \ {v}) =⊕

e∈E C0(0, 1) issemiprojective.

138 / 192

Now let n > 1, and suppose inductively that the result is true forgraphs with fewer than n vertices. Let X have n vertices, and letv0 and v1 be distinct vertices. Set X1 = X \ {v1}, and

X2 = X \ {v1, v2}. Then X = X1, so C (X ) = C0(X1). Therefore it

suffices to prove that C0(X1) is semiprojective. But X2 is a graph

with n − 1 vertices, so C (X2) = C0(X2) is semiprojective, by theinductive hypothesis. Hence C0(X2) is semiprojective. Let η be asin the proof of Theorem 10.1, and define α : C0(X2)→ C0(X2) andξ : C0(X1)→ C0(X1) by composition with η on each edge. (Thusξ(f ) is constant on the balls of radius 1/3 centered at the verticesof X .)

139 / 192

We have the diagram

0 −−−−→ C0(X2) −−−−→ C0(X1) −−−−→ C −−−−→ 0yα yξ ∥∥∥0 −−−−→ C0(X2) −−−−→ C0(X1) −−−−→ C −−−−→ 0.

Since C0(X2) is σ-unital, semiprojective, and contains an identityfor α(C0(X2)), Theorem 9.1 implies that if ξ is corona extendible,then C0(X1) is semiprojective.

140 / 192

To prove the corona extendibility, we consider the diagram

0 −−−−→⊕E

SC i1−−−−→ C0(X1) −−−−→⊕

V \{v0}C −−−−→ 0∥∥∥ yξ yβ

0 −−−−→⊕E

SC i2−−−−→ C0(X1) −−−−→ B −−−−→ 0.

Here, the map i1 is just the inclusion, and i2 is the composition ofthe inclusion with α.

141 / 192

The algebra B can be identified as

⊕v∈V \v0

((⊕

s(e)=vC0(0, 1/3]

)⊕(⊕

r(e)=vC0[2/3, 1)

))˜⊕((

⊕s(e)=v0

C0(0, 1/3])⊕(⊕

r(e)=v0C0[2/3, 1)

)).

The map β carries the v th summand to the scalar multiples of thev th adjoined unit. Since each summand (in large parentheses) ofB is projective, conditional projectivity of β follows fromconsiderations like those in exercise 2.5. Then Theorem 6.2 impliesthat ξ is corona extendible. �

142 / 192

Exercise 10.3

1. {f ∈ CMn : f (1) is diagonal} is semiprojective. (Hint: letAj = {f ∈ CMn : f (1) ∈ span{e11, . . . , ejj}}. ThenAj C Aj+1.) (When n = 2, this algebra is denoted qC.)

2. Let X be a finite graph, and n > 0. For v ∈ V , let Av be anabelian subalgebra of Mn, and letA = {f ∈ C (X ,Mn) : f (v) ∈ Av}. Prove that A issemiprojective.

143 / 192

11. Semiprojectivity for O∞

As a warm-up to the general result, we will start with the proofthat O∞ is semiprojective. This was proved by Blackadar in[Blackadar 04]. We will give a somewhat more involvedconstruction that, one hopes, will make the general proofintelligible. In fact, the proof for O∞ is -relatively- easy becausethere are no Cuntz-Krieger relations in its presentation to worryabout.Recall that O∞ has the presentation C ∗〈S1,S2, . . . | S∗i Sj = δij〉.Within O∞ we have the Toeplitz versions of On:T On = C ∗〈S1,S2, . . . ,Sn | S∗i Sj = δij〉, and O∞ =

⋃∞n=1 T On.

144 / 192

It is very easy to see that each T Ok is semiprojective. Supposethat φ : T Ok → B/I , where I is an inductive limit ideal. φ(S1) isan isometry, so by Lemma 1.9 there are n1, and an isometryz1 ∈ B/In1 , such that z1 + I = φ(S1).

Now consider any element y2 ∈ B/In1 lifting φ(S2). Theny ′2 = (1− z1z∗1 )y2 is also a lift of φ(S2). The proof of Lemma 1.9shows that there is n2 ≥ n1 such that

‖(y ′2)∗y ′2 − 1 + In2‖ < 1,

and hence that the polar part of y ′2 + In2 is an isometry. Call thispolar part z2. By the definition of y ′2 we know that z1 + In2 and z2have orthogonal ranges.

Iterating this process k times, we end up with a lift of φ into B/Ink .

145 / 192



‖(y ′2)∗y ′2 − 1 + In2‖ < 1,



146 / 192



‖(y ′2)∗y ′2 − 1 + In2‖ < 1,



147 / 192

It is clear that the above proof will not work for O∞! As we willsee later, the key to the proof lies in the details for the speciallifting associated to the telescope for the algebra. The telescope isdefined by

B ={

f ∈ C([0,∞],O∞

): f ([0, n]) ⊆ T On, for n = 1, 2, . . .

}.

B contains ideals In = {f ∈ B : f |[n,∞] = 0}, so thatB/In ∼= B|[n,∞]. Then In ⊆ In+1, and

I =⋃∞

n=1 In = {f ∈ B : f (∞) = 0}. Thus B/I ∼= O∞.

148 / 192

Theorem 11.1The map : O∞

∼=→ B/I lifts to B/I3.

Proof. Let π : B → B/I . We will construct a ∗-homomorphismφ : O∞ → B/I3 such that π ◦ φ = id. We do this inductively, byconstructing a sequence of ∗-homomorphisms φn : O∞ → B/I3satisfying

π ◦ φn|T On = id|T On(10)

φn|T On−1 = φn−1|T On−1 .(11)

Then φ = limn φn is the desired lift.

149 / 192

To accomplish the inductive step, we will have to blend theprevious homomorphism φn−1 with an auxiliary one in order to fixthe nth isometry. The basic trick to do this is the following.Let A be a C ∗-algebra, and let f , g ∈ C ([0, 1],A) be isometrieswith orthogonal ranges: f ∗f = g∗g = 1, f ∗g = 0. Letp(t) = (cos π2 t)f (t) + (sin π

2 t)g(t). Then p is an isometry.Moreover, p(1) = g(1), so that if g is an ‘appropriate lift’ (ofevaluation at 1), then so is p.

150 / 192

We adapt this trick by defining versions of it on the intervals[n − 1, n]:

an(t) =

1, if t ≤ n − 1

cos π2 t, if n − 1 < t < n

0, if t ≥ n

bn(t) =

0, if t ≤ n − 1

sin π2 t, if n − 1 < t < n

1, if t ≥ n.

151 / 192

Pictorially,

n − 1 n

1an bn

Thus an ∈ In, a2n + b2n = 1, bnIn−1 = 0, and bnSn ∈ B.

152 / 192

The auxiliary homomorphisms that we need must ‘hide’ the laterisometries somewhere out of the way. We will arrange this in thefollowing way (where ‘out of the way’ means ‘under the range ofSn+2’). For each n ≥ 1, and for j > n, let zn(j) ∈ T On+2 such that

I zn(j)zn(j)∗ ≤ Sn+2S∗n+2

I zn(j)∗zn(j ′) = δjj ′ .

For example, we could let zn(j) = S j−nn+2S1.

Now we define ∗-homomorphisms ψn : O∞ → T On+2 by

ψn|T On = id

ψn(Sj) = zn(j), for j > n.

Notice that bn+2ψn(x) ∈ B for all x ∈ O∞.

153 / 192

The auxiliary homomorphisms that we need must ‘hide’ the laterisometries somewhere out of the way. We will arrange this in thefollowing way (where ‘out of the way’ means ‘under the range ofSn+2’). For each n ≥ 1, and for j > n, let zn(j) ∈ T On+2 such that

I zn(j)zn(j)∗ ≤ Sn+2S∗n+2

I zn(j)∗zn(j ′) = δjj ′ .

For example, we could let zn(j) = S j−nn+2S1.

Now we define ∗-homomorphisms ψn : O∞ → T On+2 by

ψn|T On = id

ψn(Sj) = zn(j), for j > n.

Notice that bn+2ψn(x) ∈ B for all x ∈ O∞.

154 / 192

Now we will construct the φn. We begin by setting φ1 = ψ1. Sinceψ1(O∞) ⊆ T O3, we do have that φ1 : O∞ → B/I3. Moreoverequation (10) above is clearly true, and equation (11) holdsvacuously. It will turn out that we need one more condition on thehomomorphisms φn, along with (10) and (11):

(12) bn+3 φn(Sj) = bn+3 ψn(Sj) for j ≥ n.

Equation (12) holds trivially for φ1.

155 / 192

Now suppose inductively that we have defined φ1, . . ., φn−1satisfying (10) – (12). We define φn by

φn(Sj) =

{φn−1(Sj), if j ≤ n − 1

an+2φn−1(Sj) + bn+2ψn(Sj), if j ≥ n.

Notice that since ψn(O∞) ⊆ T On+2, bn+2 ψn(Sj) ∈ B. Moreover,ψn(Sj) is an isometry on the set {bn+2 6= 0}, so the basic trickapplies — if we have that φn−1(Sj) and ψn(Sj) have orthogonalranges. Let’s check that first.

In fact, it is enough to check thatbn+2ψn(Sj)

∗φn−1(Sj) = 0. Recalling that j ≥ n, we have

bn+2ψn(Sj)∗φn−1(Sj) = bn+2ψn(Sj)

∗ψn−1(Sj), by (12),

= bn+2ψn(Sj)∗ · Sn+1S∗n+1ψn−1(Sj)

=

{bn+2S∗n · Sn+1S∗n+1ψn−1(Sj), if j = n

bn+2ψn(Sj)∗Sn+2S∗n+2 · Sn+1S∗n+1ψn−1(Sj), if j > n

= 0.

156 / 192

Now suppose inductively that we have defined φ1, . . ., φn−1satisfying (10) – (12). We define φn by

φn(Sj) =

{φn−1(Sj), if j ≤ n − 1

an+2φn−1(Sj) + bn+2ψn(Sj), if j ≥ n.

Notice that since ψn(O∞) ⊆ T On+2, bn+2 ψn(Sj) ∈ B. Moreover,ψn(Sj) is an isometry on the set {bn+2 6= 0}, so the basic trickapplies — if we have that φn−1(Sj) and ψn(Sj) have orthogonalranges. Let’s check that first. In fact, it is enough to check thatbn+2ψn(Sj)

∗φn−1(Sj) = 0. Recalling that j ≥ n, we have

bn+2ψn(Sj)∗φn−1(Sj) = bn+2ψn(Sj)

∗ψn−1(Sj), by (12),

= bn+2ψn(Sj)∗ · Sn+1S∗n+1ψn−1(Sj)

=

{bn+2S∗n · Sn+1S∗n+1ψn−1(Sj), if j = n

bn+2ψn(Sj)∗Sn+2S∗n+2 · Sn+1S∗n+1ψn−1(Sj), if j > n

= 0.

157 / 192

Therefore we know that φn(Sj) is an isometry, for all j . Equation(11) holds by construction, and equation (10) holds sinceπ(an+2) = 0, π(bn+2) = 1, and ψn(Sn) = Sn. Moreover, sincebn+3an+2 = 0, equation (12) holds. All that is left is to verify thatφn is indeed a representation of O∞, i.e. that φn(Sj ′)

∗φn(Sj) = 0if j 6= j ′. If j , j ′ < n, then this is true because φn−1 is arepresentation.

158 / 192

Consider the case j < n and j ′ ≥ n. We have

φn(Sj ′)∗φn(Sj) =

(an+2φn−1(S∗j ′) + bn+2ψn(S∗j ′)

)φn−1(Sj)

= an+2φn−1(S∗j ′Sj) + bn+2ψn(S∗j ′)ψn−1(Sj), by (12),

= bn+2ψn(S∗j ′) · Sj

=

{bn+2S∗n · Sj , if j ′ = n

bn+2ψn(S∗j ′)Sn+2S∗n+2 · Sj , if j ′ > n

= 0.

159 / 192

Lastly, consider the case where j , j ′ ≥ n (and j 6= j ′). We have

φn(Sj ′)∗φn(Sj) =

(an+2φn−1(S∗j ′) + bn+2ψn(S∗j ′)

)·(an+2φn−1(Sj) + bn+2ψn(Sj)

)= a2n+2φn−1(S∗j ′Sj) + b2

n+2ψn(S∗j ′Sj)

+ an+2bn+2

(ψn−1(S∗j ′)ψn(Sj) + ψn(S∗j ′)ψn−1(Sj)

)= 0,

since, e.g.,

ψn−1(S∗j ′)ψn(Sj) = ψn−1(S∗j ′)Sn+1S∗n+1 · ψn(Sj)

=

{ψn−1(S∗j ′)Sn+1S∗n+1 · Sn, if j = n

ψn−1(S∗j ′)Sn+1S∗n+1 · Sn+2S∗n+2ψn(Sj), if j > n

= 0. �

160 / 192

12. From telescopes to the general case

Now, how does this proof for the telescope help us provesemiprojectivity for O∞? Let’s make a few observations about thetelescope.

1. The ideals In are such that we can lift T On modulo In. In thegeneral case, we will have to choose a subsequence of the ideals sothat we can lift T On modulo Ikn . (In fact, at the stage n we needto lift T On+2, so as to have a place to hide stuff.)

2. The element an is a unit on In−1, and is central in B. This istoo much to hope for in general, so we need to modify the basictrick.

161 / 192

Let’s address the second problem first. The key tool that saves usis the existence of quasi-central approximate units. The definitionis:

Definition 12.1Let A be a C ∗-algebra, and let I be an ideal in A. A quasi-centralapproximate unit for I (with respect to A) is an approximate unit(eλ) for I such that ‖aeλ − eλa‖ → 0 for all elements a ∈ A.

Quasi-central approximate units always exist. Here is a simpleconsequence. Let (eλ) be a quasi-central approximate unit for I ,and put fλ = (1− e2λ)1/2. Then (fλ) is also quasi-central, and fλx ,xfλ → 0 for all x ∈ I .The basic trick still works, asymptotically, if we use a quasi-centralapproximate unit. We formulate it as a lemma.

162 / 192

Lemma 12.2Let A be a C ∗-algebra, and let I be an ideal in A. Let x, y ∈ Asuch that

I x is an isometry.

I π(y) is an isometry (in A/I ).

I y∗x ∈ I .

Let (eλ) be a quasi-central approximate unit for I , and putfλ = (1− e2λ)1/2. Let

wλ = eλx + fλy .

Then ‖w∗λwλ − 1‖ → 0, and π(wλ) = π(y).

163 / 192

The point is that we will often have an isometry that satisfies acertain relation, but doesn’t lift the thing it ought to. We will beable to find something that is an appropriate lift, but isn’t anisometry in A; it becomes an isometry only modulo I . If the twoelements have orthogonal ranges (modulo I ), then the elements wλblend the two so as to give an appropriate lift, and areapproximately isometries. With a good enough approximation (i.e.large enough λ), the polar part of wλ is just what we need.

We should note that in the case of the telescope, this is preciselywhat we did (but with no worries about approximation). (Theconstant function) Sn is not in B/I3. In fact, we implicitly used anarbitrary lift y of Sn to B/I3 (i.e. an extension of the function to[3,∞]). Multiplication with bn+2 killed off the initial part of theextension, so that the arbitrariness didn’t matter.

164 / 192

The point is that we will often have an isometry that satisfies acertain relation, but doesn’t lift the thing it ought to. We will beable to find something that is an appropriate lift, but isn’t anisometry in A; it becomes an isometry only modulo I . If the twoelements have orthogonal ranges (modulo I ), then the elements wλblend the two so as to give an appropriate lift, and areapproximately isometries. With a good enough approximation (i.e.large enough λ), the polar part of wλ is just what we need.

We should note that in the case of the telescope, this is preciselywhat we did (but with no worries about approximation). (Theconstant function) Sn is not in B/I3. In fact, we implicitly used anarbitrary lift y of Sn to B/I3 (i.e. an extension of the function to[3,∞]). Multiplication with bn+2 killed off the initial part of theextension, so that the arbitrariness didn’t matter.

165 / 192

The proof of the lemma is completely straightforward: thecomputations that verified the basic trick before workasymptotically, by the quasi-centrality. Now we sketch the proofthat O∞ is semiprojective.

Theorem 12.3O∞ is semiprojective.

Proof. Let O∞ ⊆ B/I , where I is an inductive limit ideal. (SinceO∞ is simple, we don’t have to name the map into the quotient).We return to the first observation above. Choose1 ≤ k1 ≤ k2 ≤ · · · , and maps θn : T On+2 → B/Ikn such that

I π ◦ θn = id|T On+2 .

I θn|T On+1 = πkn ◦ θn−1.

166 / 192

It isn’t hard to see why this is possible. Having chosen θn−1,choose any lift y of Sn+2 in B/Ikn−1 . Thenz =

(1−

∑i<n θn−1(SiS

∗i ))y is also a lift. Choose kn ≥ kn−1 so

that ‖z∗z − 1 + Ikn‖ < 1, and let θn(Sn+2) be the polar part ofz + Ikn .It is important to remember that the θn map to deeper and deeperquotients of B. However the first quotient, B/Ik1 , is where we willconstruct our lift to establish semiprojectivity.

We choose maps ψn : O∞ → T On+2 just as before, so that

I ψn|T On = id|T On .

I ψn(SjS∗j ) ≤ Sn+2S∗n+2, for j > n.

167 / 192

It isn’t hard to see why this is possible. Having chosen θn−1,choose any lift y of Sn+2 in B/Ikn−1 . Thenz =

(1−

∑i<n θn−1(SiS

∗i ))y is also a lift. Choose kn ≥ kn−1 so

that ‖z∗z − 1 + Ikn‖ < 1, and let θn(Sn+2) be the polar part ofz + Ikn .It is important to remember that the θn map to deeper and deeperquotients of B. However the first quotient, B/Ik1 , is where we willconstruct our lift to establish semiprojectivity.We choose maps ψn : O∞ → T On+2 just as before, so that

I ψn|T On = id|T On .

I ψn(SjS∗j ) ≤ Sn+2S∗n+2, for j > n.

168 / 192

Now, let φ1 = θ1 ◦ ψ1 : O∞ → B/Ik1 . Then

I π ◦ φ1(S1) = S1.

I φ1(SjS∗j ) ≤ θ1(S3S∗3 ).

We will construct a sequence φn : O∞ → B/Ik1 satisfying

1. π ◦ φn|T On = id|T On .

2. φn|T On−1 = φn−1|T On−1 .

3. φn(Sj)− θn ◦ ψn(Sj) ∈ Ikn for j ≥ n.

(Compare with the three inductive hypotheses in the case of thetelescope.)

169 / 192

Suppose inductively that we have constructed φ1, . . ., φn−1. Wedefine φn as follows.

I φn(Sj) = φn−1(Sj) for j < n.

Let (aλ) be a quasi-central approximate unit for Ikn , and letbλ = (1− a2λ)1/2. For j ≥ n, let θn(Sj) ∈ B/Ik1 be an arbitrary lift

of θn(Sj). For j ≥ n, we check that φn−1(Sj)∗ · θn ◦ ψn(Sj) ∈ Ikn .

In fact, this uses conditions (1) - (3) in exactly the same way as forthe analogous step in the proof for the telescope (so we omit thedetails). Thus, the lemma implies that

wλ = aλφn−1(Sj) + bλθn ◦ ψn(Sj)

is asymptotically an isometry. (We omit a decoration on wλ toidentify j .)

170 / 192

Suppose inductively that we have constructed φ1, . . ., φn−1. Wedefine φn as follows.

I φn(Sj) = φn−1(Sj) for j < n.

Let (aλ) be a quasi-central approximate unit for Ikn , and letbλ = (1− a2λ)1/2. For j ≥ n, let θn(Sj) ∈ B/Ik1 be an arbitrary lift

of θn(Sj). For j ≥ n, we check that φn−1(Sj)∗ · θn ◦ ψn(Sj) ∈ Ikn .

In fact, this uses conditions (1) - (3) in exactly the same way as forthe analogous step in the proof for the telescope (so we omit thedetails). Thus, the lemma implies that

wλ = aλφn−1(Sj) + bλθn ◦ ψn(Sj)

is asymptotically an isometry. (We omit a decoration on wλ toidentify j .)

171 / 192

We have to make one additional move at this point. We want this(asymptotic) isometry to have range orthogonal to the ranges ofthe isometries already constructed: φn(S1), . . ., φn(Sj−1). To dothis, we let

zλ =(1−

∑i<j

φn(Si )φn(Si )∗)wλ.

We need to check that for i < j we have S∗i wλ ∈ Ikn . This is thesame kind of verification as the one we already omitted.

Now, choose λ large enough so that ‖z∗λzλ − 1‖ < 1, and set

φn(Sj) = polar(zλ).

It remains to verify (1) - (3) for φn. Again, the details areanalogous to the case of the telescope. Finally, we defineφ = limn φn. �

172 / 192

We have to make one additional move at this point. We want this(asymptotic) isometry to have range orthogonal to the ranges ofthe isometries already constructed: φn(S1), . . ., φn(Sj−1). To dothis, we let

zλ =(1−

∑i<j

φn(Si )φn(Si )∗)wλ.

We need to check that for i < j we have S∗i wλ ∈ Ikn . This is thesame kind of verification as the one we already omitted.Now, choose λ large enough so that ‖z∗λzλ − 1‖ < 1, and set

φn(Sj) = polar(zλ).

It remains to verify (1) - (3) for φn. Again, the details areanalogous to the case of the telescope. Finally, we defineφ = limn φn. �

173 / 192

13. Basics of graph algebras

A directed graph is a collection of dots, and arrows between thedots. E 0 is the set of dots (vertices), E 1 is the set of arrows(edges). The source and range maps are functions s, r : E 1 → E 0

so:

s(e) r(e)e

A vertex is called singular if s−1(v) is either empty or infinite. LetSE = {v ∈ E 0 : v is not singular}.

The idea of the graph C ∗-algebra is motivated by symbolicdynamics. For example, suppose E is finite and transitive: for anyvertices v1, v2 ∈ E 0, there is a path from v1 to v2. Let X be thespace of infinite paths, a Cantor set. Each edge defines a partialhomeomorphism of X : φf (e1, e2, e3, . . .) = (f , e1, e2, . . .), ifs(f ) = r(e1). Model this by partial isometries.

174 / 192

13. Basics of graph algebras

A directed graph is a collection of dots, and arrows between thedots. E 0 is the set of dots (vertices), E 1 is the set of arrows(edges). The source and range maps are functions s, r : E 1 → E 0

so:

s(e) r(e)e

A vertex is called singular if s−1(v) is either empty or infinite. LetSE = {v ∈ E 0 : v is not singular}.The idea of the graph C ∗-algebra is motivated by symbolicdynamics. For example, suppose E is finite and transitive: for anyvertices v1, v2 ∈ E 0, there is a path from v1 to v2. Let X be thespace of infinite paths, a Cantor set. Each edge defines a partialhomeomorphism of X : φf (e1, e2, e3, . . .) = (f , e1, e2, . . .), ifs(f ) = r(e1). Model this by partial isometries.

175 / 192

Definition 13.1Let E be a directed graph. The C ∗-algebra of E is the universalC ∗-algebra generated by E 0 ∪ E 1 with relations

1. E 0 are pairwise orthogonal projections.

2. e∗f = δe,f s(e), for e, f ∈ E 1. (In particular, E 1 are partialisometries.)

3. ee∗ ≤ r(e), for e ∈ E 1.

4. v =∑{e:s(e)=v} ee∗, for v ∈ SE .

Condition (4) is the Cuntz-Krieger relation (at v).

For example, if E has one vertex and n edges, the edges areisometries with pairwise orthogonal final projections. TheCuntz-Krieger relation forces the range projections to sum to 1;C ∗(E ) = On.

176 / 192

Definition 13.1Let E be a directed graph. The C ∗-algebra of E is the universalC ∗-algebra generated by E 0 ∪ E 1 with relations

1. E 0 are pairwise orthogonal projections.

2. e∗f = δe,f s(e), for e, f ∈ E 1. (In particular, E 1 are partialisometries.)

3. ee∗ ≤ r(e), for e ∈ E 1.

4. v =∑{e:s(e)=v} ee∗, for v ∈ SE .

Condition (4) is the Cuntz-Krieger relation (at v).For example, if E has one vertex and n edges, the edges areisometries with pairwise orthogonal final projections. TheCuntz-Krieger relation forces the range projections to sum to 1;C ∗(E ) = On.

177 / 192

Some basic properties are:

I C ∗(E ) is nuclear, in the UCT class.

I If E is transitive, and is not a cycle, then C ∗(E ) is aKirchberg algebra (simple and purely infinite).

I C ∗(E ) is unital if and only if E 0 is finite.

Definition 13.2Let E be a directed graph, and let S ⊆ SE . The relative Toeplitzgraph algebra T C ∗(E , S) is defined exactly as C ∗(E ), except thatrelation (4) holds only at vertices in S .

For example, if E has one vertex and n edges, we may let S = ∅(the only choice). The relative Toeplitz graph algebra has thesame relations as On except that the range projections need notsum to 1. Thus we obtain T On.

178 / 192

Now suppose that we have an inclusion of graphs F ⊆ E . We candescribe the relation between the graph algebras of E and F . Let

SF ,E = {v ∈ SF : s−1(v) ∩ F 1 = s−1(v) ∩ E 1}.

Thus, we should not impose the Cuntz-Krieger relation at a vertexof F if some edges of E originating there are missing in F . This isall that needs to be considered.

Theorem 13.3

1. Let F ⊆ E . The inclusion F 1 ⊆ E 1 defines a faithful mapT C ∗(F ,SF ,E )→ C ∗(E ).

2. If F1 ⊆ F2 ⊆ · · · , and E =⋃

n Fn, then

C ∗(E ) =⋃

n T C ∗(Fn,SFn,E ).

This generalizes the example O∞ =⋃

n T On.

179 / 192

Theorem 13.4Let E be a finite graph, and let S ⊆ SE . Then T C ∗(E ,S) issemiprojective.

Proof. Use Lemma 1.9 instead of Lemma 1.5, and imitate theproof that T On is semiprojective. �

180 / 192

14. K -theory for graph algebras

Many people have published papers on the K -theory of graphalgebras. Here is the version from [ES 11].

Theorem 14.1Let E be a directed graph.

K0C ∗(E ) = Cc(E 0,Z)/〈δv −∑

e∈s−1(v)

δr(e) | v ∈ SE 〉

K1C ∗(E ) ={

f ∈ Cc(SE ,Z) : f (v) =∑

e∈r−1(v)

f (s(e))}.

181 / 192

As an example, show that the C ∗-algebra of the following graphhas K∗ = (0,Z). (Moreover, if n + 1 “strands” are attached at thecentral vertex u, then K∗ = (0,Zn).) Since this graph is transitive,and is not a cycle, its C ∗-algebra is a Kirchberg algebra. Sincethere are infinitely many vertices, it is the stable version ofBlackadar’s algebra P∞.

u

v01v02v03· · · v11 v12 v13 · · ·e03 e02 e13e12

g03 g02 g01 g13g12g11

e01e11f01f02f03 f11 f12 f13

Corollary 14.2

If E is a directed graph, then K1C ∗(E ) is a free abelian group.

182 / 192

15. Semiprojectivity for P∞

In this section we will sketch the proof that P∞ is semiprojective.It turns out that it is a mere hop, skip, and a jump from there to

Theorem 15.1The UCT Kirchberg algebras with finitely generated K -theory andtorsion-free K1 are semiprojective.

Blackadar showed ([Blackadar 04]) that finitely generated K -theoryis necessary. [Blackadar 04, Neubuser 00, Spielberg 09] all giveproofs that semiprojectivity of Kirchberg algebras is invariant underMorita equivalence. As Søren mentioned, the Kirchberg-Phillipstheorem implies that UCT Kirchberg algebras with the sameK -theory are Morita equivalent. This is why semiprojectivity for(some) Kirchberg algebras can be proven using graph algebras.

183 / 192

We work with stable P∞ via the graph in section 14. As describedin Theorem 13.3, each finite subgraph determines a subalgebra,which can be defined intrinsically to the subgraph as a relativeToeplitz algebra. If we let E be the whole graph, we can let En bethe subgraph consisting of all vertices and edges with (second)index less than or equal to n. Notice that the only vertex in En

that emits more edges in E than it does in En is the singular vertexu. Since it is precisely such vertices where the Cuntz-Kriegerrelation is omitted from the presentation of the relative Toeplitzalgebra, we find that the relations for the generators of T C ∗(En)are exactly the same as the relations for these generators in thelarger algebra C ∗(E ).

184 / 192

In the context of graph algebras, the basic trick must begeneralized in two ways:

I partial isometries instead of isometries

I initial projections also need not be orthogonal.

The formulas for the initial and final projections of the combinedpartial isometry will also be important. There are four versionsneeded. The context for all is a C ∗-algebra A with an ideal I , andquotient map π. Let (aλ) be a quasi-central approximate unit forI , and put bλ = (1− a2λ)1/2.

185 / 192

Lemma 15.2Let e ′, e ′′ ∈ A be such that

I e ′ and π(e ′′) are partial isometries.

I e ′∗e ′ − e ′′∗e ′′ ∈ I .

I e ′∗e ′′ ∈ I .

Then eλ = aλe ′ + bλe ′′ is asymptotically a partial isometry,π(eλ) = π(e ′′), and

I e∗λeλ ≈ e ′∗e ′

I eλe∗λ ≈ a2λe ′e ′∗ + b2λe ′′e ′′∗ + 2aλbλ(e ′e ′′∗ + e ′′e ′∗).

186 / 192

Lemma 15.3Let g ′, g ′′, r0 ∈ A be such that

I g ′, π(g ′′), and π(r0) are partial isometries.

I g ′g ′′∗, g ′∗g ′′ ∈ I .

I r∗0 r0 − g ′∗g ′, r0r∗0 − g ′′∗g ′′ ∈ I .

Then gλ = a2λg ′ + b2λg ′′ + aλbλ(g ′r∗0 + g ′′r0) is asymptotically a

partial isometry, π(gλ) = π(g ′′), and

I g∗λgλ ≈ a2λg ′∗g ′ + b2λg ′′∗g ′′ + 2aλbλ(r0 + r∗0 )

I gλg∗λ ≈ a2λg ′g ′∗ + b2λg ′′g ′′∗ + 2aλbλ(g ′r∗0 g ′′∗ + g ′′r0g ′∗).

187 / 192

The choice of names for the elements in these lemmas comes fromthe edges in the graph. Namely, we are thinking of representativesof the relations for the piece of the graph:

••

g

e

It turns out that the results of the two lemmas above don’tnecessarily satisfy the Cuntz-Krieger relation at the left vertex.The next lemma is the trick to fix this problem.

188 / 192

Lemma 15.4Let e ′, e ′′ be as in Lemma 15.2, and let g ′, g ′′, r0 be as in Lemma15.3. Suppose that e ′ and g ′ satisfy the relations dictated by thebit of graph pictured above, and similarly for π(e ′′) and π(g ′′).(Thus, e ′e ′∗ + g ′g ′∗ = g ′∗g ′, and similarly for the ′′ case.) Let

r = e ′′e ′∗ + g ′′r0g ′∗.

Then r has the same properties as r0. Let

gλ = a2λg ′ + b2λg ′′r0r∗ + aλbλ(g ′r∗ + g ′′r0).

Then gλ is asymptotically a partial isometry (apply Lemma 15.3 tog ′, g ′′r0r∗, and r), π(gλ) = π(g ′′), and if eλ is defined as inLemma 15.2, then eλ and gλ satisfy the Cuntz-Krieger relationasymptotically.

The fourth version of lemma is like the third, but treats the casewhere e ′ and e ′′ satisfy the hypotheses of Lemma 15.3.

189 / 192

References I

B. Blackadar, Shape theory for C ∗-algebras, Math. Scand. 56(1985), 249-275.

B. Blackadar, Semiprojectivity in simple C ∗-algebras, Operatoralgebras and applications, 1-17, Adv. Stud. Pure Math., 38,Math. Soc. Japan, Tokyo, 2004.

E. Effros and J. Kaminker, Homotopy continuity and shapetheory for C ∗-algebras, in Geometric Methods in OperatorAlgebras, eds. Araki and Effros, Pitman Res. Notes Math. 123,Longman, Harlow, 1986.

S. Eilers, T. Loring, G. Pedersen, Stability of anticommutationrelations: an application of noncommutative CW complexes, J.reine angew. Math. 499 (1998), 101 - 143.

190 / 192

References II

S. Eilers, T. Loring, G. Pedersen, Morphisms of extensions ofC ∗-algebras: pushing forward the Busby invariant, Adv. Math.147 (1999), 74 - 109.

M. Ephrem and J. Spielberg, K -theory of C ∗-algebras ofdirected graphs, Houston J. Math. 37 (2011), 435 - 447.

T. Loring, Lifting Solutions to Perturbing Problems inC ∗-algebras, Fields Inst. Mon. Ser. 8, Amer. Math. Soc.,Providence, R.I., 1997.

B. Neubuser, Semiprojektivitat und realisierunen von reinunendlichen C ∗-algebren, preprint, Munster, 2000.

C. Olsen and G. Pedersen, Corona C ∗-algebras and theirapplications to lifting problems, Math. Scand. 64 (1989), 63 -86.

191 / 192

References III

G. Pedersen, C ∗-algebras and their automorphism groups,Academic Press, London, 1979.

G. Pedersen, The corona construction, in Proceedings of the1988 GPOTS-Wabash Conference, Longman Sci. 6 Tech.,Harlow, 1990.

G. Pedersen, A strict version of the non-commutative Urysohnlemma, Proc. Amer. Math. Soc. 125 (1997), 2657 - 2660.

J. Spielberg, A functorial approach to the C ∗-algebras of agraph, International J. Math. 13 (2002), 245-277.

J. Spielberg, Semiprojectivity for certain purely infiniteC ∗-algebras,Trans. Amer. Math. Soc. 361 (2009), 2805 - 2830.

W. Szymanski, The range of K -invariants for C ∗-algebras ofinfinite graphs, Indiana Univ. Math. J. 51 (2002), 239-249.

192 / 192

Fundamental results about semiprojective...

Documents

Transcript of Fundamental results about semiprojective...