Classifying C∗-algebras

Classifying C ∗ -algebras 

Klaus Thomsen 

Version pr. 18/3 2005 

1

Contents 

Chapter 1. Fundamentals 5 

1. Basics 5 

2. Inductive limits of C ∗ -algebras 38 

3. K0 and the classification of AF-algebras 61 

4. Choquet simplices and C ∗ -algebras 94 

5. Ideals and simple C ∗ -algebras 128 

Chapter 2. Simple AI-algebras 163 

Chapter 3. The range of the Elliott invariant for simple unital AI-algebras 165 

3

1.1. Banach algebras. 

CHAPTER 1 

Fundamentals 

1. Basics 

Definition 1.1. A Banach algebra A is an algebra over the complex numbers 

which is also a Banach space in a norm satisfying 

ab ≤ ab, a, b ∈ A. (1.1) 

A unit in a Banach algebra A is a non-zero element 1 such that 1a = a1 = a, a ∈ 

A. When A has a unit we say that A is unital. 

Example 1.2. Let X be a set and B(X) the set of bounded functions f : X → 

C. Then B(X) is an algebra over C in the obvious way (multiplication is defined 

pointwise : fg(x) = f(x)g(x), x ∈ X) and we introduce the norm 

f = sup {|f(x)| : x ∈ X} 

on B(X). In checking that B(X) is a Banach algebra the only non-trivial point is to 

show that B(X) is complete in this norm, i.e. is a Banach space. Let’s do that. So 

we consider a Cauchy sequence {fn} in B(X). For each x ∈ X the sequence {fn(x)} 

is Cauchy in C and therefore convergent by the completeness of C. Let f : X → C 

be the function given by f(x) = limn→∞ fn(x). Let ǫ > 0. Because the sequence 

{fn} is Cauchy with respect to the norm · we can find N ∈ N such that 

|fn(x) − fm(x)| ≤ fn − fm ≤ min {ǫ, 1} (1.2) 

for all n, m ≥ N and all x ∈ X. If we let m tend to ∞ in (1.2) we get 

|fn(x) − f(x)| ≤ min {ǫ, 1}, n ≥ N, x ∈ X. (1.3) 

From (1.3) it follows that |f(x)| ≤ |fN(x)| + 1 ≤ fN + 1 for all x ∈ X, showing 

that f ∈ B(X), and that fn − f ≤ ǫ, n ≥ N, showing that limn→∞ fn = f in 

B(X). - Thus B(X) is indeed a Banach space and hence a Banach algebra. Note 

that B(X) is unital; the unit is the function on X which takes the constant value 1. 

Example 1.3. Let V be a complex Banach space. The set B(V ) of bounded 

operators on V is a Banach algebra. The product is simply composition of operators 

and the norm is the operator norm 

A = sup {A(v) : v ≤ 1}. 

Example 1.4. If we take V = C n in Example 1.3 we see that the complex n by n 

matrices form a Banach algebra where the product is the usual product of matrices. 

This Banach algebra will be denoted by Mn. The unit of Mn is the matrix with all 

diagonal entries 1 and all other entries 0. 

5

6 1. FUNDAMENTALS 

Example 1.5. Let L1 (R) denote the space of complex functions on R which are 

absolutely integrable with respect to Lebesgue measure. More precisely L1 (R) is the 

space of measurable functions f : R → C for which 

 

|f(t)| dt < ∞ ; 

R 

but with two such functions being identified when they agree almost everywhere. 

L1 (R) is then a Banach algebra with norm 

 

|f(t)| dt 

and with the convolution product 

 

f ∗ g(x) = g(t)f(x − t) dt . 

R 

R 

Example 1.6. Let X be a set and B a Banach algebra. Let B(X, B) denote the 

vector space of functions f : X → B for which sup x∈X f(x) < ∞. Imitating the 

constructions from Example 1.2 one sees that B(X, B) is a Banach algebra. 

Closed subalgebras of Banach algebras are clearly Banach algebras. By combining 

this observation with the previous examples the reader can use his or her 

imagination to construct an unlimited reservoir of examples of Banach algebras. 

For example, the set of continuous functions f : [0, ∞) → Mn for which the limit 

limt→∞ f(t) exists in Mn and is an upper triangular matrix (meaning that all entries 

below the diagonal are zero), form a Banach algebra in a natural way. 

Definition 1.7. Let a be an element of a unital Banach algebra A. Then a is 

invertible if there is an element b ∈ A such that ab = ba = 1. The element b with 

this property is called the inverse of a and is denoted by a −1 . 

The spectrum σ(a) of a is the set 

σ(a) = {λ ∈ C : λ1 − a is not invertible } 

and ρ(a) = sup {|λ| : λ ∈ σ(a)} is called the spectral radius of a. 

Example 1.8. Consider the Banach algebra Mn. A matrix A ∈ Mn is invertible 

in the sense of Definition 1.7 if and only if it is invertible as a linear transformation 

C n → C n . This happens if and only if ker A = {0}. Thus, for any λ ∈ C, λ1 − A 

is invertible if and only if λ is not an eigenvalue of A. Hence the set σ(A) is the set 

of eigenvalues of A. 

Example 1.9. Let X be a compact Hausdorff space and consider the Banach 

algebra C(X) of continuous complex-valued functions on X. (This is a Banach 

subalgebra of the Banach algebra B(X) introduced in Example 1.2.) The constant 

function 1 is a unit in C(X) and a function f ∈ C(X) is invertible in C(X) if 

and only if f(x) = 0 for all x ∈ X; in which case the inverse f −1 is the function 

X ∋ x ↦→ 1 . Thus, for any λ ∈ C, λ1 − f is invertible if and only if f(x) = λ for 

f(x) 

all x ∈ X; i.e. if and only if λ is not in the image of f. Hence the spectrum σ(f) of 

f is the image set f(X). 

As the two last examples show, the notion of spectrum of a Banach algebra 

element is something which generalizes both the notion of eigenvalue for a matrix 

and the image set for a continuous function. Our first objective here is to prove that

1. BASICS 7 

the spectrum is always non-empty and say a little about its position in C. That the 

spectrum is non-empty is a non-trivial fact and we hope the reader will enjoy the 

following proof. 

The basic source of invertible elements is the following 

Lemma 1.10. Let A be a unital Banach algebra and x ∈ A. 

1) If x < 1, it follows that 1 − x is invertible with 

(1 − x) −1 = 

∞ 

x n . 

n=0 

2) If 1 − x < 1, it follows that x is invertible with 

x −1 = 

∞ 

(1 − x) n . 

Proof. We remark first that x 0 = 1 by definition. a) : The estimate 

shows that the partial sums of 

 

n=0 

m 

x j ≤ 

j=n 

∞ 

x n 

n=0 

m 

x j 

form a Cauchy sequence in A, so that the infinite sum converges by completeness. 

Since (1 − x) ∞ n=0 xn = 1 = ∞ n=0 xn (1 − x), this proves a). b) : Substitute x for 

1 − x in a). 

Lemma 1.11. Let A be a unital Banach algebra and let x ∈ A be an invertible 

element. Then every element y ∈ A with x − y < 1 

x−1 is also invertible and 

 

j=n 

x −1 − y −1 ≤ x−1 2 x − y 

1 − x −1 x − y . 

Proof. Note that 1 − x −1 y = x −1 (x − y) ≤ x −1 x − y < 1 so that 

x −1 y is invertible by Lemma 1.10. y must therefore also be invertible with y −1 = 

(x −1 y) −1 x −1 = ∞ 

n=0 (1 − x−1 y) n x −1 , so that x −1 − y −1 ≤ x −1 ∞ 

n=1 

x −1 y n = x −1 1−x−1 y 

1−1−x −1 y ≤ x−1 2 x−y 

1−x −1 x−y 

1 − 

. 

Note that the qualitative statements of Lemma 1.11 are that the invertible elements 

form an open subset of A and that the map x → x −1 is continuous on this 

subset. In particular we see that for any sequence {xn} in A we have the equivalence 

lim 

n→∞ xn = 0 ⇔ lim (1 − xn) 

n→∞ −1 = 1. (1.4) 

Lemma 1.12. Let {αn} be a sequence of real numbers such that αn+m ≤ αn +αm 

for all n, m ∈ N. Then the limit limn→∞ 1 

n αn exists and equals infn∈N 1 

n αn.


Proof. Fix ǫ > 0 and m ∈ N. Choose N ∈ N so large that 

2 m 1 

max {|α1|, 

N 2 |α2|, . . ., 1 

m |αm|} < ǫ. 

If n ≥ N we write n = lm + r where l ∈ N and r ∈ {1, 2, . . ., m}. Then 

Thus 

1 

n αn ≤ 1 

n αlm + 1 

n αr ≤ l 

n αm + 1 

n αr = 

lm 1 

n m αm + r 1 

n r αr = 1 

m αm + r 1 

n r αr − r 1 

n m αm ≤ 

1 

m αm + 2 m 1 

max {|α1|, 

N 2 |α2|, . . ., 1 

m |αm|} < 1 

m αm + ǫ. 

lim sup 

n→∞ 

1 

n αn ≤ sup 

n≥N 

1 

n αn ≤ 1 

m αm + ǫ. 

If we let ǫ → 0 and take the infimum over all m ∈ N we see that 

lim sup 

n→∞ 

1 

n αn ≤ inf 

n∈N 

1 

n αn. 

Since obviously infn∈N 1 

n αn ≤ lim infn→∞ 1 

n αn we conclude that limn→∞ 1 

n αn = infn∈N 1 

n αn. 

 

Lemma 1.13. Let A be a unital Banach algebra and a ∈ A. There is an element 

n. 

λ ∈ σ(a) such that |λ| ≥ limn→∞ a n 1 

n = infn∈N a n 1 

Proof. We prove first that limn→∞ an 1 

n = infn∈N an 1 

n. If ak = 0 for 

some k ∈ N, an = 0 ∀n ≥ k and the conclusion is trivial. So we assume that 

an > 0 ∀n ∈ N and set αn = log an. Note that 

By Lemma 1.12 limn→∞ αn 

n 

lim 

n→∞ an 1 

n = exp( lim 

n→∞ 

αn+k ≤ αn + αk, n, k ∈ N. 

αn = infn∈N . It follows that 

n 

1 

n αn) = exp(inf 

n∈N 

1 

n αn) = inf 

n∈N an 1 

n. 

Set r = limn→∞ an 1 

n. We will prove that σ(a) contains an element λ with 

|λ| ≥ r. If r = 0 it follows that 0 ∈ σ(a). Indeed, if a is invertible we have that 

1 = ana−n so that 1 ≤ ana−n for all n ∈ N and this implies that 

1 ≤ r lim a 

n→∞ −n 1 

n, 

which is impossible because r = 0. We can therefore assume that r > 0. 

Assume to get a contradiction that λ1 − a is invertible whenever |λ| ≥ r. Fix 

1 

2πi first n ∈ N and let ω = e n. Set Mk = {1, 2, . . ., n}\{k}, k = 1, 2, . . ., n. We will 

use the following identities : 

n 

and 

k=1 

n 

(1 − ω k z) = 1 − z n , z ∈ C, 

 

(1 − ω i z) = n, z ∈ C. 

k=1 i∈Mk

1. BASICS 9 

(If the reader find it difficult to prove these algebraic identities, we offer the following 

arguments : The identity n 

k=1 (1 − ωk z) = 1 − z n follows from the fact that the 

two polynomials have the same roots, namely {ω k : k = 1, 2, · · · , n}. The second 

identity follows by observing that the left hand side is a polynomial of degree ≤ n−1, 

which takes the same value at n distinct points, namely at z = ω k , k = 1, 2, · · · , n. 

It must therefore be the constant polynomial, and the constant value, n, is easily 

identified by checking for z = 0.) 

By assumption 1 − a is invertible when |λ| ≥ r and the above polynomial iden- 

λ 

tities show that 

n 

(1 − ω (1.5) 

and 

k=1 

n 

 

k=1 i∈Mk 

k a an 

) = 1 − 

λ λn Substituting (1.5) into (1.6) yields that 

n 1 

(1 − 

n 

an a 

λn)(1 − ωk 

λ )−1 = 1 

k=1 

or, alternatively, that (1 − an 

λ n) is invertible with 

(1 − an 

λ n)−1 = 1 

n 

(1 − ω ia 

) = n1. (1.6) 

λ 

n 

(1 − ω 

k=1 

k a 

λ )−1 

(1.7) 

when |λ| ≥ r. Set K = sup {(λ1 − a) −1 : |λ| ≥ r}. This is a finite number for the 

following reason. When |λ| > a+1 we have that (λ1−a) −1 = λ −1 (1− a 

λ )−1 = 

λ−1 ∞ a 

n=0 ( λ )n ≤ |λ| −1 ∞ a 

n=0 ( |λ| )n = 1 ≤ 1. Since M = {λ ∈ C : r ≤ |λ| ≤ 

|λ|−a 

a + 1} is a compact set and λ → (λ1 − a) −1 is continuous by Lemma 1.11, we 

see that 

K ≤ max {1, sup (λ1 − a) 

λ∈M 

−1 } < ∞. 

For each k ∈ N and |λ| ≥ r we have that 

(1 − ω ka 

r )−1 k a 

− (1 − ω 

λ )−1 = ω k a( 1 

r 

ω k a(λ − r)(r1 − ω k a) −1 (λ1 − ω k a) −1 

which gives the estimate 

1 

− )(1 − ωka 

λ r )−1 k a 

(1 − ω 

λ )−1 = 

k a 

λ )−1 ≤ |λ − r|aK 2 

(1 − ω ka 

r )−1 − (1 − ω 

when |λ| ≥ r. By substituting this estimate into (1.7) we get that 

(1 − an 

rn)−1 − (1 − an 

λn)−1 ≤ |λ − r|aK 2 

when |λ| ≥ r. It is important that this estimate is independent of n. 

We prove next that 

(1.8) 

lim 

n→∞ 1 − (1 − an 

r n)−1 = 0 . (1.9)


Let ǫ > 0. Choose λ ∈ C such that |λ| > r and aK2 |λ − r| ≤ ǫ. 

Since |λ| > r, 

2 

it follows that limn→∞ an 

λn 

 

r 

= 0. To see this, let t ∈ , 1 . Since limn→∞ 

a 

|λ| n 

λn 

1 

n = 

|λ| −1 limn→∞ a n 1 

n = r 

|λ| < t, there is an N so large that an 1 

n < t n for all n ≥ N. 

Hence limn→∞ an 

λ n = 0 because limn→∞ t n = 0. 

Now an application of (1.4) gives N ∈ N such that 

Combined with (1.8) this shows that 

1 − (1 − an 

λn)−1 ≤ ǫ 

, n ≥ N. 

2 

1 − (1 − an 

r n)−1 ≤ 1 − (1 − an 

λ n)−1 + (1 − an 

λ n)−1 − (1 − an 

r n)−1 ≤ ǫ, n ≥ N, 

proving (1.9). Combined with (1.4), (1.9) implies that limn→∞ an 

rn = 0. Thus 

( an rn ) 1 

n = r−1 (an) 1 

n < 1 for all sufficiently large n. But this is impossible since 

r = infn an 1 

n. 

Theorem 1.14. Let A be a unital Banach algebra and a ∈ A. Then σ(a) is a 

non-empty compact subset of C and 

ρ(a) = inf 

n∈N an 1 

n = lim a 

n→∞ n 1 

n ≤ a. 

Proof. If |λ| n > a n for some n ∈ N, λ n 1 − a n = λ n (1 − an 

λ n) is invertible by 

Lemma 1.10. Since λ n 1 − a n = (λ1 − a) n−1 

k=0 λk a n−k−1 , it follows that λ1 − a is 

invertible, i.e. λ /∈ σ(a). - This shows that 

ρ(a) ≤ inf 

n∈N an 1 

n = lim a 

n→∞ n 1 

n. 

But Lemma 1.13 implies that limn→∞ a n 1 

n ≤ ρ(a), i.e. we have that 

ρ(a) = inf 

n∈N an 1 

n = lim a 

n→∞ n 1 

n. 

Since a n 1 

n ≤ a, this identity implies that ρ(a) ≤ a. In particular, ρ(a) is a 

bounded subset of C. The invertible elements of A form an open subset of A by 

Lemma 1.11 and hence 

C\σ(a) = {λ ∈ C : λ1 − a is invertible} 

is open in C; i.e. σ(a) is closed and therefore compact. σ(a) is non-empty by Lemma 

1.13. 

Corollary 1.15. Let A be a unital Banach algebra. If all non-zero a ∈ A are 

invertible we have that 

A = C1. 

Proof. Let a ∈ A. By Theorem 1.14 σ(a) = ∅, so there is a λ ∈ C such that 

λ1 −a is not invertible. By assumption this implies that λ1 −a = 0, i.e. a = λ1.

1. BASICS 11 

1.2. Abelian C ∗ -algebras. An involution in a Banach algebra A is a map 

A ∋ a → a ∗ ∈ A such that 

1) (a ∗ ) ∗ = a, a ∈ A, 

2) (ab) ∗ = b ∗ a ∗ , a, b ∈ A, 

3 (a + λb) ∗ = a ∗ + λb ∗ , a, b ∈ A, λ ∈ C. 

Definition 1.16. A Banach algebra A with involution ∗ is a C ∗ -algebra when 

for all a ∈ A. 

aa ∗ = a 2 

(1.10) 

Welcome to the world of C ∗ -algebras. It is my hope that you will get an enjoyable 

stay and maybe even return from time to time when the surrounding world becomes 

to hard or boring (or both). 

Example 1.17. Let X be a set and B(X) the Banach algebra of bounded complex 

functions on X, cf. Example 1.1.2. We define an involution ∗ in B(X) by 

f ∗ (x) = f(x), x ∈ X. 

It is easy to check that ∗ is an involution on B(X). This turns B(X) into a C ∗ - 

algebra. The C ∗ -identity (1.10) is easily established: 

ff ∗ = sup {|f(x)f(x)| : x ∈ X} = sup {|f(x)| 2 : x ∈ X} 

= sup {|f(x)| : x ∈ X} 2 = f 2 

Example 1.18. Let H be a Hilbert space and B(H) the algebra of bounded 

operators on H. The operation T → T ∗ which associates to T ∈ B(H) its adjoint 

operator is an involution, cf. [GKP, Theorem 3.2.3]. Thus B(H) is a Banach algebra 

with involution. It follows from [GKP] that B(H) is a C∗-algebra. In particular, 

this is true when H is the Hilbert space Cn for any n ∈ N. Now B(Cn ) is the same 

as the space of complex n by n matrices Mn, so we see that Mn is a C∗- algebra. To 

be explicit the C∗-norm on Mn is given by 

 

 

 

n 

A = sup { | 

i=1 

n 

j=1 

Aijxj| 2 : (x1, x2, . . ., xn) ∈ C n , 

n 

|xi| 2 ≤ 1} 

when A = (Aij) ∈ Mn. (I regret that there is no simpler formula - but this is in the 

nature of things : There exists no algebraic expression which gives A in terms of 

its entries.) 

The involution on Mn is reflection in the diagonal followed by conjugation of the 

entries : 

(A ∗ )ij = Aji. 

An alternative proof of the fact that Mn is a C ∗ -algebra will follow below; set 

A = C in Proposition 1.53. 

We begin now to develop the most elementary general properties of C ∗ -algebras. 

Note that a general C ∗ -algebra is not assumed to be unital. Certainly, not all 

C ∗ -algebras have a unit as the following example will show. 

i=1


Example 1.19. Let X be a locally compact Hausdorff space and denote by 

C0(X) the continuous complex valued functions f on X that ’vanish at infinity’, i.e. 

satisfies the following condition : For every ǫ > 0 there is a compact subset K ⊆ X 

such that |f(x)| < ǫ for all x ∈ X\K. C0(X) is clearly a subalgebra of B(X) and 

is globally invariant under the involution ∗. Furthermore, C0(X) is closed (with 

respect to the supremum norm · ). This is seen as follows. Let {fn} be a sequence 

from C0(X) and f ∈ B(X) such that limn→∞ fn − f = 0. We first show that f 

is continuous : Fix x0 ∈ X and ǫ > 0. Choose N ∈ N such that fN − f < ǫ 

3 . 

Since fN is continuous at x0 there is an open neighbourhood V of x0 such that 

|fN(x)−fN(x0)| < ǫ 

3 for all x ∈ V . But then |f(x)−f(x0)| ≤ 2ǫ 

3 +|fN(x)−fN(x0)| ≤ ǫ 

for all x ∈ V , proving that f is continuous (as should be well-known). To check that 

f ∈ C0(X), take K ⊆ X compact such that |fN(x)| < ǫ 

for all x ∈ X\K. It follows 

3 

that |f(x)| ≤ ǫ 

3 + |fN(x)| < ǫ for all x ∈ X\K. - Thus C0(X) is closed in B(X) as 

asserted and we conclude that C0(X) is a C∗-algebra with the operations inherited 

from B(X); i.e. C0(X) is C∗-subalgebra of B(X). 

For every x ∈ X there is a function f ∈ C0(X) such that f(x) = 1. This follows 

from an appropriate version of Urysohn’s lemma. So if g is a unit in C0(X) we must 

have that g(x) = 1 for all x ∈ X, i.e. g must be the constant function 1 (the unit 

of B(X)). But the constant function 1 can only ‘vanish at infinity’ if X is compact. 

Indeed, if K is a compact subset of X such that 1 < 1 for x ∈ X\K, then X\K 

2 

must be empty, and thus X = K is compact. Hence we see that C0(X) can only be 

unital when X is compact. Conversely, if X is compact every continuous function 

on X ‘vanish at infinity’ (for every ǫ > 0 we can use X itself as the required compact 

subset) so C0(X) = C(X) and C(X) contains the constant function 1 as a unit. In 

other words: C0(X) is unital if and only if X is compact. 

Example 1.20. Let X be a locally compact Hausdorff space and A a C ∗ -algebra 

(e.g. A = Mn for some n ∈ N). Denote by C0(X, A) the set of continuous functions 

f : X → A which ’vanish at infinity’; i.e. satisfies that for every ǫ > 0 there 

is a compact subset K ⊆ X such that f(x) ≤ ǫ ∀x ∈ X\K. A repetition of 

the arguments from Example 1.19 shows that C0(X, A) is a C ∗ -algebra with the 

involution f ∗ (x) = f(x) ∗ and norm 

f = sup {f(x) : x ∈ X}. 

When X is compact we denote this algebra by C(X, A). The enthusiastic reader 

may prove that C0(X, A) is unital if and only if X is compact and A is unital. 

Let us now remedy the fact that a C ∗ -algebra A may not have a unit. Denote 

by M(A) the set of maps T : A → A (a priori not even linear) for which there is 

another map T ∗ : A → A such that 

(T(a)) ∗ b = a ∗ T ∗ (b), a, b ∈ A. (1.11) 

Lemma 1.21. Every element T of M(A) is a bounded linear operator on A and 

T ∗ ∈ M(A) with (T ∗ ) ∗ = T. 

Proof. We first observe that the following implication holds for any element 

a ∈ A : 

ax = 0 ∀x ∈ A ⇒ a = 0. (1.12) 

Indeed, if ax = 0 for all x ∈ A, then in particular a 2 = aa ∗ = 0 which implies 

that a = 0.

To show that T is linear, let a, b ∈ A and λ ∈ C. Then 

1. BASICS 13 

(T(λa + b) − λT(a) − T(b)) ∗ x = (T(λa + b) ∗ − λT(a) ∗ − T(b) ∗ )x = 

(λa + b) ∗ T ∗ (x) − λa ∗ T ∗ (x) − b ∗ T ∗ (x) = 0 

for all x ∈ A, so by (1.12) we have that T(λa + b) = λT(a) + T(b); i.e. T is linear. 

That T is bounded follows from the principle of uniform boundedness in the 

following way. For each a ∈ A with a ≤ 1, define Sa : A → A by Sa(b) = T(a) ∗ b. 

Each Sa is a bounded linear operator (Sa ≤ T(a)) and for every fixed b ∈ A we 

have that Sa(b) = T(a) ∗ b = a ∗ T ∗ (b) ≤ T ∗ (b). Therefore [GKP, Theorem 

2.2.9] implies that there is an M < ∞ such that Sa ≤ M for all a. Hence 

T(a) ∗ b = Sa(b) ≤ M for all a, b ∈ A with a ≤ 1, b ≤ 1. In particular, 

whenever a ≤ 1 and T(a) = 0 we have that 

∗ T(a) 

T(a) = T(a) ≤ M. 

T(a) 

It follows that T ≤ M. Finally, by applying the involution to the identity T(a) ∗ b = 

a ∗ T ∗ (b) we get immediately that T ∗ ∈ M(A) and that in fact (T ∗ ) ∗ = T. 

Proposition 1.22. M(A) is a C ∗ -algebra with the involution T ↦→ T ∗ and the 

norm T = sup {T(a) : a ≤ 1}. The product in the algebra is the composition 

of operators on A and the identity operator is in M(A) (and is therefore a unit of 

M(A)). 

Proof. We leave it to the reader to check that M(A) is a subalgebra of the 

algebra of bounded linear operators on A. That is, you should check that λT + 

S, TS ∈ M(A) when T, S ∈ M(A) and λ ∈ C. It is also trivial that the identity 

operator is in M(A). 

Let us next check the C ∗ -identity. Since TS ≤ T S because we are dealing 

with the operator norm, we see immeditaly that T ∗ T ≤ T ∗ T . Let a, b ∈ 

A, a ≤ 1, b ≤ 1. Then 

T(a) ∗ b = a ∗ T ∗ (b) ≤ T ∗ (b) ≤ T ∗ . 

When we set b = T(a) 

T(a) in this inequality we get that T(a) ≤ T ∗ (when T(a) = 

0). The conclusion is that T ≤ T ∗ , T ∈ M(A). When we substitute T ∗ into this 

inequality we get the converse version, T ∗ ≤ T , so we have that T ∗ = T . 

Now, for any a ∈ A with a ≤ 1 we have the estimate 

T(a) 2 = T(a) ∗ T(a) = a ∗ T ∗ T(a) ≤ T ∗ T , 

so by taking the supremum over all such a we conclude that 

T 2 ≤ T ∗ T . 

All in all we have the following inequalities 

T 2 ≤ T ∗ T ≤ T ∗ T = T 2 

from which the C ∗ -identity (1.10) follows. 

To show that M(A) is complete with respect to the operator norm, let {Tn} be 

a Cauchy sequence in M(A). For each a ∈ A, n, m ∈ N, we have that 

Tna − Tma ≤ Tn − Tma .


It follows that {Tna} is a Cauchy sequence in A for all a ∈ A. Since T ∗ n − T ∗ m = 

(Tn − Tm) ∗ = Tn − Tm for all n, m, we see that also {T ∗ n } is a Cauchy sequence 

in M(A). Consequently, {T ∗ na} is a Cauchy sequence in A for all a ∈ A. We can 

therefore define maps T, S : A → A by Ta = limn→∞ Tna, Sa = limn→∞ T ∗ na, a ∈ A. 

Note that 

(Ta) ∗ b = lim 

n→∞ (Tna) ∗ b = lim 

n→∞ a ∗ T ∗ n b = a∗ Sb 

for all a, b ∈ A, proving that T ∈ M(A) (and that T ∗ = S). To show that 

limn→∞ Tn = T in M(A), let ǫ > 0. There is a N ∈ N such that 

Tna − Tma ≤ Tn − Tm ≤ ǫ (1.13) 

for all n, m ≥ N and all a ∈ A with a ≤ 1. If we let m tend to infinity we get 

from (1.13) that 

Tna − Ta ≤ ǫ 

for all n ≥ N and all a ∈ A with a ≤ 1. Hence Tn − T ≤ ǫ for all n ≥ N. 

The C ∗ -algebra M(A) is called the multiplier algebra of A. 

Definition 1.23. Let ϕ : A → B be a linear map between the C ∗ -algebras A 

and B. ϕ is a ∗-homomorphism if it is an algebra homomorphism (i.e. satisfies 

that ϕ(ab) = ϕ(a)ϕ(b), ∀a, b ∈ A) and respects the involutions in the sense that 

ϕ(a) ∗ = ϕ(a ∗ ), a ∈ A. In addition, if ϕ is a bijection, we call it a ∗-isomorphism. 

A ∗-homomorphism ϕ : A → B is called unital when A and B are both unital 

and ϕ(1) = 1. 

It follows from Exercise 1.91 that a ∗-homomorphism between C ∗ -algebras is 

always of operator norm ≤ 1, and is isometric if and only if it is injective. 

When there is a ∗-isomorphism between two C ∗ -algebras A and B we say that 

they are isomorphic and write 

A ≃ B. 

Example 1.24. a) If X and Y are sets and r : Y → X an arbitrary map, then r 

defines a ∗-homomorphism ϕ : B(X) → B(Y ) by ϕ(f)(y) = f(r(y)), f ∈ B(X), y ∈ 

Y . Observe that when X and Y are compact topological spaces and r continuous, 

then ϕ maps C(X) into C(Y ). ϕ is a ∗-isomorphism if and only if r is a bijection. 

b) Let U ∈ Mn be a unitary matrix (i.e. U ∗ U = 1). Then 

ϕ(A) = UAU ∗ , A ∈ Mn, 

defines a ∗-homomorphism ϕ : Mn → Mn. ϕ is a ∗-isomorphism and because it 

maps to and from the same C ∗ -algebra, it is called an automorphism (of Mn.) 

c) If X is compact and A unital we can define unital ∗-homomorphisms ϕ : 

A → C(X, A) and ψ : C(X, A) → A by ϕ(a)(x) = a, x ∈ X, and ψ(f) = f(x0), 

respectively, where x0 is some fixed point in X. Then ψ ◦ ϕ is the identity map on 

A, but ϕ is only a ∗-isomorphism in the rather extreme case where X = {x0}. 

For any C ∗ -algebra A there is a ∗-homomorphism ϕA : A → M(A) defined in 

the following way : For any a ∈ A define Ta : A → A by 

Ta(x) = ax, x ∈ A. 

Since (Ta(y)) ∗ x = (ay) ∗ x = y ∗ a ∗ x = y ∗ Ta ∗(x) for all x, y ∈ A we see that Ta ∈ M(A) 

and that Ta ∗ = Ta ∗. The map ϕA(a) = Ta is easily seen to be a ∗-homomorphism.

1. BASICS 15 

Since Ta(x) ≤ ax, x ∈ A, we see immediately that Ta ≤ a. But 

Ta(a ∗ ) = aa ∗ = a 2 , so we can conclude that Ta = a which means that 

ϕA is an isometry. Thus ϕA(A) is a C ∗ -subalgebra of M(A) which is ∗- isomorphic, 

via ϕA, to A. 

We will from now on suppress ϕA in the notation and instead simply consider 

A as a C ∗ -subalgebra of M(A) when convenient. In particular, when dealing with 

an arbitrary (possibly not unital) C ∗ -algebra A the symbol 1 will denote the unit of 

M(A). Note that the result of Exercise 1.65 is that A = M(A) if and only if A is 

unital. 

While M(A) has the unit which A may lack, it is too large for many purposes. 

Therefore we set 

Â = span{1, A} = C1 + A . 

It is obvious that Â is a ∗-subalgebra of M(A). We claim that it is in fact a C∗- subalgebra, i.e. that it is also closed. This is obvious when A is unital because in 

this case Â = A and there is nothing to prove. The following lemma handles the 

case when A is not unital. 

Lemma 1.25. Assume that A is not unital. Then Â is a unital C∗-subalgebra of 

M(A) and there is a unital ∗-homomorphism χ : Â → C such that ker χ = A. 

Proof. It is straightforward to check that because 1 /∈ A and A is closed in 

M(A), the association λ ↦→ infa∈A λ1 − a defines a norm on C. Thus 

inf λ1 − a = K|λ|, λ ∈ C, 

a∈A 

where K = infa∈A 1 − a > 0. So if {λn} and {an} are sequences in C and A, 

respectively, such that λn1+an → x in M(A), then K|λn −λm| = infa∈A λn −λm + 

a ≤ λn + an − λm − am for all n, m ∈ N, showing that {λn} is Cauchy and hence 

convergent in C, say to λ. But then an = λn1 + an − λn1 converges, say to a ∈ A. 

It follows that x = λ1 + a ∈ Â. Thus Â is closed in M(A) and is a C∗-subalgebra. Since 1 /∈ A, we can define χ : Â → C by χ(λ1 + a) = λ, λ ∈ C, a ∈ A. It is 

straightforward to check that χ is a ∗-homomorphism. 

Note that Lemma 1.25 shows that A is a two-sided ideal in Â, being the kernel 

of a homomorphism. (This follows also from Exercise 1.65.) 

If now A is an arbitrary C∗-algebra and a ∈ A, we set 

σ(a) = {λ ∈ C : λ1 − a is not invertible in Â}. 

When A is unital Â = A, this definition agrees with the one introduced in Section 

1.1 (considering A as a Banach algebra only). In particular all the results about σ(a) 

obtained in the last section apply. 

Definition 1.26. Let A be a C ∗ -algebra and a ∈ A. Then 

1) a is normal when a ∗ a = aa ∗ . 

2) a is selfadjoint when a ∗ = a. 

3) a is unitary when aa ∗ = a ∗ a = 1. 

Clearly, a selfadjoint or unitary element is also normal. We will sometimes write 

Asa for the set of selfadjoint elements in A.


Lemma 1.27. When a ∈ A is normal, 

Proof. We have that 

ρ(a) = a. 

a 2n 

2 = a 2n 

(a 2n 

) ∗ = (aa ∗ ) 2n 

= (aa ∗ ) 2n−1 

(aa ∗ ) 2n−1 

= 

(aa ∗ ) 2n−1 

2 = (aa ∗ ) 2n−2 

(aa ∗ ) 2n−2 

2 = (aa ∗ ) 2n−2 

4 = . . . 

= aa ∗ 2n 

= a 2n+1 

, 

which implies that a2n = a2n, for all n ∈ N. Thus 

ρ(a) = lim a 

n→∞ 2n 

2−n 

= a 

in this case, cf. Theorem 1.14. 

Lemma 1.28. When a ∈ A is unitary, 

σ(a) ⊆ T = {λ ∈ C : |λ| = 1}. 

Proof. Note that a is invertible with a −1 = a ∗ so that 0 /∈ σ(a). The formula 

λ −1 (λ1 − a)a −1 = −(λ −1 − a −1 ), λ = 0, 

shows that λ ∈ σ(a) ⇔ λ −1 ∈ σ(a −1 ). So if λ ∈ σ(a) we have that λ −1 ∈ σ(a ∗ ). Thus 

|λ| ≤ a and |λ| −1 = |λ −1 | ≤ a ∗ = a, using Lemma 1.27. Since the unitarity 

condition (a ∗ a = aa ∗ = 1) implies that a = a ∗ = 1, this is only possible if 

|λ| = 1. 

Lemma 1.29. When a ∈ A is selfadjoint, 

n=0 

σ(a) ⊆ R. 

Proof. For each x ∈ A the sum ∞ x 

n=0 

n 

converges in A. We claim that 

n! 

∞ (x + y) n ∞ x 

= 

n! 

n ∞ y 

n! 

n 

, (1.14) 

n! 

when xy = yx. To prove this let ǫ > 0. Choose N ∈ N so large that 

∞ x 

 

n ∞ y 

n! 

n 

n! − 

m xn m y 

n! 

n 

< ǫ, m ≥ N, 

n! 

and 

n=0 

 

n=0 

∞ (x + y) n 

− 

n! 

n=0 

Then m (x+y) 

n=0 

n 

n! = m n=0 

 

n 

j=0 

m (x + y) n 

− 

n! 

n=0 

m 

n=0 

n=0 

n=0 

n=0 

n=0 

m (x + y) n 

< ǫ, m ≥ N. 

n! 

n=0 

n! x 

j!(n−j)! 

jyn−j n! = m n=0 

x n 

n! 

m 

n=0 

n x 

j=0 

j 

j! 

yn x 

= 

n! 

k,l∈Mm 

k y 

k! 

l 

l! , 

y n−j 

(n−j)! 

and hence 

where Mm = {(k, l) ∈ N 2 : m < k + l, k, l ≤ m}. Set L = max{1, x, y}. Since 

xk 

k! 

y l 

l! 

L2m 

≤ [ m when k, l ≤ m and k + l ≥ m, this gives the estimate 

]! 2 

m (x + y) 

 

n m x 

− 

n! 

n m 

n! 

n=0 

n=0 

n=0 

y n 

n! ≤ m2 

2 

L 2m 

[ m 

2 ]!.

Since limm→∞ m2 

2 

It follows that 

L2m [ m 

2 

1. BASICS 17 

= 0 there is an m ≥ N such that 

]! 

m (x + y) 

 

n m x 

− 

n! 

n m y 

n! 

n 

≤ ǫ. 

n! 

 

n=0 

∞ (x + y) n 

− 

n! 

n=0 

n=0 

∞ 

n=0 

x n 

n! 

n=0 

∞ 

n=0 

yn ≤ 3ǫ, 

n! 

proving (1.14). 

Now consider the given selfadjoint element a ∈ A and set eia = ∞ (ia) 

n=0 

n 

. Then n! 

(e ia ) ∗ ∞ (−ia) 

= 

n 

. 

n! 

n=0 

Hence (1.14) shows that (eia ) ∗eia = eia (eia ) ∗ = 1 , i.e. eia is a unitary. 

Let λ ∈ σ(a). Set zn = n−1 k=0 (iλ)k (ia) n−k−1 . Then zn ≤ nLn where L = 

max{1, |λ|, a} so that 

n=1 

n=1 

∞ 

n=1 

converges in A to an element x satisfying that 

e iλ − e ia ∞ (iλ) 

= 

n1 − (ia) n ∞ i(λ1 − a)zn 

= 

= i(λ1 − a)x = ix(λ1 − a). 

n! 

n! 

Since λ1 − a not is invertible, this shows that e iλ − e ia is not invertible. Since e ia is 

unitary, this implies that e iλ ∈ T by Lemma 1.28. Hence λ must be real. 

The C ∗ -algebra C0(X) of Example 1.19 is clearly Abelian and our next objective 

is to prove that any Abelian C ∗ -algebra is of this form. 

Definition 1.30. Let A be a ∗-algebra. A non-zero ∗-homomorphism ω : A → C 

is called a character of A. 

Theorem 1.31. Let A be an Abelian C ∗ -algebra. Then the set X of characters 

of A is a locally compact Hausdorff space in the weak ∗ -topology (inherited from A ∗ ) 

and A is ∗-isomorphic to C0(X) via the isometric map Φ : A → C0(X) given by 

zn 

n! 

Φ(a)(γ) = γ(a), γ ∈ X. 

X is compact if and only if A is unital. 

Proof. We assert that 

a = sup {|γ(a)| : γ ∈ X}, a ∈ A . (1.15) 

Note that this equality says that Φ is isometric. To prove this, let γ ∈ X. First we 

show that γ admits an extension γ ′ to Â which is a character of Â such that γ′ (1) = 1. 

This is trivial when A is unital because in this case Â = A and γ(1)2 = γ(1 2 ) = γ(1) 

so that γ(1) ∈ {0, 1}. Note that γ(1) = 0 is impossible because γ = 0. So assume 

that A is not unital. We can then define γ ′ (λ1 + z) = λ + γ(z), λ ∈ C, z ∈ A, and 

it is trivial to check that γ ′ is in fact a character of Â. 

Now, if γ(a) /∈ σ(a), we would have an element b ∈ Â such that b(γ(a)1 −a) = 1. 

But then 1 = γ ′ (1) = γ ′ (b)γ ′ (γ(a)1 − a) = γ ′ (b)(γ ′ (a) − γ ′ (a)) = 0, a contradiction.


Hence γ(a) ∈ σ(a). Since γ ∈ X was arbitrary this shows that sup {|γ(a)| : γ ∈ 

X} ≤ ρ(a). We conclude from Theorem 1.14 that 

a ≥ sup {|γ(a)| : γ ∈ X}. 

To prove the reverse inequality, we may assume that a = 0. Take t ∈ σ(a) such 

that |t| = ρ(a) = a. This is possible because σ(a) is compact. The set (t1 − a) Â 

is a ideal in Â which does not contain 1. Let I be an ideal in Â which contains 

(t1 − a) Â but not 1, and is maximal with respect to these properties. Such an ideal 

exists by Zorn’s lemma. Then 

I is closed in Â. (1.16) 

Â/I = C(1 + I). (1.17) 

To prove (1.16) and (1.17), note that I is also an ideal in Â which contains 

(t1 − a) Â. If 1 ∈ I, there would be an element x ∈ I such that 1 − x < 1. But 

then x would be invertible in Â by Lemma 1.10 and hence 1 = x−1x ∈ I, which 

was not the case. Hence 1 /∈ I. By maximality of I we conclude that I = I. It 

follows in particular that Â/I is a Banach algebra in the quotient norm : x + I = 

inf{x − z : z ∈ I}. Let χ ∈ Â/I be a non-zero element. If χ was not invertible in 

Â/I, χÂ/I would define a proper ideal in Â/I and {x ∈ Â : x+I ∈ χÂ/I} would be 

an ideal properly containing I but not 1, contradicting the maximality of I. Thus 

χ is indeed invertible and we conclude from Corollary 1.15 that Â/I = C(1 + I). 

It follows from (1.17) that Â = C1 + I. Define γ(λ1 + z) = λ, λ ∈ C, z ∈ I. 

It is straightforward to check that γ is linear and multiplicative. As above we see 

that γ(x) ∈ σ(x) for all x ∈ Â. In particular, γ(x) ∈ R when x = x∗ by Lemma 

1.29. An arbitrary element x can be written x = x1 + ix2 where x1 = 1 

2 (x + x∗ ) and 

x2 = 1 

2i (x − x∗ ) are selfadjoint. Then 

γ(x) = γ(x1) + iγ(x2) = γ(x1) − iγ(x2) = γ(x1 − ix2) = γ(x ∗ ), 

proving that γ is a ∗-homomorphism. Since |γ(a)| = |γ(t1−(t1−a))| = |t| = a = 0 

we see first that γ|A = 0, and hence that γ|A ∈ X, and then that 

a ≤ sup{|γ(a)| : γ ∈ X}. 

Hence (1.15) is established. 

We prove now that X is locally compact. By (1.15) γ ≤ 1 for all γ ∈ X, 

i.e. X is a subset of the unit ball in A ∗ . Let ω0 ∈ X. To show that X is locally 

compact it suffices to find an open neighbourhood V of ω0 whose closure is compact. 

Since ω0 = 0 there is an element a ∈ A such that ω0(a) > 1. Set V = {ω ∈ X : 

|ω(a)| > 1}. Then V is open in X (in the weak ∗ -topology restricted to X) . Since 

V ⊆ K = {ω ∈ X : |ω(a)| ≥ 1}, it suffices to prove that K is compact. K is clearly 

closed in the weak ∗ -topology; indeed the function f(ω) = ω(a) is continuous on X 

by definition of the weak ∗ -topology, so that K = f −1 ({λ ∈ C : |λ| ≥ 1}) is closed. 

But K is a subset of the unit ball of A ∗ which is compact in the weak ∗ -topology. 

Hence K is compact. 

For each a ∈ A, define Φ(a) : X → C by 

Φ(a)(ω) = ω(a), ω ∈ X. 

Then Φ(a) is continuous on X by definition of the weak ∗ -topology. Furthermore, 

for arbitrary ǫ > 0 we can use the argument above to show that {ω ∈ X : |Φ(a)(ω)| ≥

1. BASICS 19 

ǫ} is compact in X. Thus Φ(a) ∈ C0(X). It is then straightforward to check that Φ 

defines a ∗-homomorphism Φ : A → C0(X). By (1.15) Φ is an isometry. 

The surjectivity of Φ follows from the Stone-Weierstrass theorem in the following 

way. 1 Since an arbitrary f ∈ C0(X) admits a decomposition f = g1 + ig2 where 

g1, g2 ∈ C0(X) are real-valued, it suffices to show that any real-valued element 

f ∈ C0(X) is in Φ(Asa). In fact, by writing f = f ∨ 0 − (−f) ∨ 0, we may even 

assume that f ≥ 0. Note that Φ(Asa) is a norm closed algebra of real-valued 

functions; norm closed because Φ is an isometry and algebra because A is Abelian. 

Let ǫ > 0. It suffices to find c ∈ Asa such that Φ(c) − f ≤ 2ǫ. To this end take a 

compact subset K ⊆ X such that f(x) < ǫ, x /∈ K. Then 

{h ∈ CR(X) : h = Φ(a)|K for some a ∈ Asa} 

is an algebra of continuous real functions on K which separates the points of K 

and does not vanish at any point of K. By the Stone-Weierstrass theorem there is 

therefore an element a ∈ Asa such that |f(x) − Φ(a)(x)| ≤ ǫ, x ∈ K. Choose a 

compact subset K1 ⊆ X, K ⊆ K1, such that |Φ(a)(x)| ≤ ǫ, x /∈ K1. By repeating 

the previous argument we find b ∈ Asa such that |Φ(b)(x)−f(x)| ≤ ǫ, x ∈ K1. Now, 

by Exercise 1.66 we know that 

g = (Φ(a) ∧ Φ(b)) ∨ 0 

is in Φ(Asa), i.e. g = Φ(c) for some c ∈ Asa. Since |g(x) − f(x)| ≤ 2ǫ, x ∈ X, we 

have found the desired c. 

By Example 1.19 X is compact iff A is unital. 

Theorem 1.31 shows that there is associated a locally compact Hausdorff space 

to any abelian C ∗ -algebra, in such a way that the algebra is ∗-isomorphic to the 

C ∗ -algebra of continuous functions on the space that vanish at infinity. One must 

then necessarily ask which locally compact Hausdorff spaces can occur this way. The 

answer is very simple: They all do ! 

Lemma 1.32. Let X and Y be topological spaces, Y locally compact and Hausdorff. 

Let f : X → Y be a function with the property that h ◦ f is continuous for all 

h ∈ C0(Y ). It follows that f is continuous. 

Proof. Let V ⊆ Y be an open set. By Urysohn’s lemma there is for each x ∈ V 

a function h ∈ C0(Y ) such that h(x) = 1, h(y) ≥ 0 for all y ∈ Y and the support of h 

is contained in V . It follows that there is a family hα, α ∈ A, of real-valued functions 

in C0(Y ) such that V = 

α∈A h−1 

α (]0, ∞[). By assumption hα ◦ f is continuous for 

all α ∈ A. It follows that 

f −1 (V ) = f −1 

 

α∈A 

h −1 

α (]0, ∞[) 

 

= 

(hα ◦ f) −1 (]0, ∞[) 

is open. 

1 The Stone-Weierstrass theorem has different forms in different books. Here I shall apply 

a real-valued version; specifically, Theorem 7.32 in ’Principles of Mathematical analysis’ by W. 

Rudin. If the reader is familiar with other versions, like [GKP, Analysis now, Cor. 4.3.5], the 

following arguments can be changed accordingly, and in most cases greatly simplified. 

α∈A


Theorem 1.33. Let X be a locally compact Hausdorff space. Then every character 

ω of the C ∗ -algebra C0(X) is of the form 

ω(f) = f(xω) 

for a unique xω ∈ X. The map ω ↦→ xω is a homeomorphism of the character space 

of C0(X) onto X. 

Proof. Set I = ker ω = {f ∈ C0(X) : ω(f) = 0}, and note that I is a two-sided 

ideal in C0(X). Set 

F = {x ∈ X : f(x) = 0 ∀f ∈ I}. 

Assume to get a contradiction, that F = ∅. Let f ∈ C0(X), and let ǫ > 0. There 

is then a compact subset K ⊆ X such that |f(x)| ≤ ǫ for all x ∈ X\K. Since no 

point of K is in F, there is for each x ∈ K an element hx ∈ I such that hx(x) = 1. 

Let Ux be an open neighbourhood of x with the property that |hx(y) − 1| ≤ ǫ for all 

y ∈ Ux. Let {Uxi : i = 1, 2, . . ., N} be a finite subcover of the cover {Ux : x ∈ K} of 

K, and let ϕi, i = 1, 2, . . ., N, be non-negative elements of C0(X) such that 

i) 0 ≤ N 

i=1 ϕi(x) ≤ 1, x ∈ X, 

ii) N i=1 ϕi(y) = 1, y ∈ K, 

iii) supp ϕi ⊆ Uxi , i = 1, 2, . . ., N. 

Then f ′ = N fϕihxi i=1 ∈ I and 

|f ′ (y) − f(y)| = | 

for all y ∈ X, and 

N 

i=1 

f(y)ϕi(y) (hxi (y) − 1) | ≤ 

|f ′ (y) − f(y)| ≤ ǫ + |f ′ (y)| ≤ ǫ + 

N 

|f(y)|ϕi(y)ǫ = |f(y)|ǫ 

i=1 

N 

|f(y)|ϕi(y)(1 + ǫ) ≤ ǫ + ǫ(1 + ǫ) 

i=1 

for all y ∈ X\K. It follows that f ′ − f ≤ ǫ + fǫ + ǫ(1 + ǫ). Since ǫ > 0 was 

arbitrary, and since I is closed, we conclude that f ∈ I. But f was arbitrary, so we 

see that I = C0(X), which is impossible because ω = 0. This contradiction shows 

that F = ∅. 

Assume next that F contains two different points x1, x2. Define Φ : C0(X) → C 2 

by Φ(f) = (f(x1), f(x2)). Then Φ is surjective by Urysohn’s lemma. Since I ⊆ ker Φ, 

Φ induces a surjection C0(X)/I → C 2 . However, this is impossible because ω 

induces a linear injection C0(X)/I → C, so that C0(X)/I is one-dimensional. This 

contradiction shows that F consists of exactly one point, xω. Let h0 ∈ C0(X) be a 

function such that ω(h0) = 1. Then h = |h0| 2 is a non-negative function with the 

same property. It follows that f − ω(f)h ∈ I for all f ∈ C0(X), and hence 

f(xω) = ω(f)h(xω) (1.18) 

for all f ∈ C0(X). Choose g ∈ C0(X) such that g (xω) = 1, and note that (1.18) 

implies that 1 = |g n (xω) | = |ω(g)| n h(xω) for all n ∈ N. This is only possible if 

h(xω) = 1 (and |ω(g)| = 1). It follows that f (xω) = ω(f) for all f ∈ C0(X). 

The map ω → xω from the character space of C0(X) to X is surjective: If 

x ∈ X, µ(f) = f(x) defines a character µ such that xµ = x. Thus if we for a 

second let XC0(X) denote the character space of C0(X), we have a bijection δ : 

XC0(X) → X given by δ(ω) = xω. If Ψ : C0(X) → C 

XC0(X) is the ∗-isomorphism

1. BASICS 21 

of Theorem 1.31 we see that Ψ(f) = f ◦ δ. It follows then from Lemma 1.32 that δ 

is a homeomorphism. 

 

Let A be a C ∗ -algebra. For any set F ⊆ A we let C ∗ (F) denote the least C ∗ - 

subalgebra of A containing F. Thus C ∗ (F) is the intersection of all C ∗ -subalgebras 

of A which contains F. Alternatively, 

C ∗ (F) = span{x1x2 . . .xn : xi ∈ F ∪ F ∗ , i = 1, 2, . . ., n, n ∈ N}, 

where F ∗ = {a ∗ : a ∈ F }. 

Lemma 1.34. Let a be an invertible element of the C ∗ -algebra A. Then a −1 ∈ 

C ∗ (1, a). 

Proof. We consider first the case where a = a∗ . Let b ∈ Â such that ab = ba = 

1. Note that b must be self-ajoint since a is. Then C ∗ (a, b) is an Abelian C ∗ -algebra 

with unit, so there is a ∗-isomorphism Φ : C ∗ (a, b) → C(X) where X is a compact 

Hausdorff space, cf. Theorem 1.31. Set f = Φ(a), g = Φ(b). Since f is invertible in 

C(X) (with 1 = g), we see that the image f(X) ⊆ R of X under f does not contain 

f 

0. Thus f(X) is a compact subset of R not containing 0. We can therefore find a 

uniformly on f(X). It follows 

sequence {Pn} of polynomials such that Pn(t) → 1 

t 

that Pn ◦ f → 1 

f = g in C(X). Since Pn ◦ f ∈ C∗ (1, f) = Φ(C∗ (1, a)), we conclude 

that b ∈ C∗ (1, a). Consider then the general case. Since a is invertible it follows that 

a∗a is also invertible (with (a∗a) −1 = a−1 (a−1 ) ∗ )). Since a∗a is selfadjoint we know 

that (a∗a) −1 ∈ C∗ (1, a∗a) ⊆ C∗ (1, a). It follows that a−1 = (a∗a) −1a∗ ∈ C∗ (1, a). 

Note that it follows from Lemma 1.34 that the spectrum σ(a) of a (considered 

as an element of A) is the same as when we consider a as an element of C ∗ (1, a). 

Theorem 1.35. Let a be a normal element of the C ∗ -algebra A. Then there is 

a ∗-isomorphism Φa : C(σ(a)) → C ∗ (1, a) such that 

Φa(idσ(a)) = a, 

where idσ(a) denotes the identity map on σ(a) (i.e. idσ(a)(λ) = λ, λ ∈ σ(a)). 

Proof. The crucial observation for the proof is that 

σ(a) = {ω(a) : ω ∈ X} (1.19) 

where X is the space of characters of C ∗ (1, a). To prove (1.19), assume first that 

λ ∈ σ(a); i.e. that λ1 − a is not invertible in Â. Since a is normal C∗ (1, a) is 

Abelian, so there is a ∗-isomorphism Φ : C ∗ (1, a) → C(X) by Theorem 1.31. Since 

Φ(λ1 − a) = λ1 − Φ(a) is not invertible in C(X), there is an x ∈ X such that 

λ = Φ(a)(x). Then C ∗ (1, a) ∋ b → Φ(b)(x) defines a character ω of C ∗ (1, a) such 

that ω(a) = λ. Conversely, if λ = ω(a) for some character ω of C ∗ (1, a), then 

λ1 − a is not invertible in Â, for if it was it would be invertible already in C∗ (1, a) 

by Lemma 1.34 and this is clearly not possible : If b = (λ1 − a) −1 ∈ C ∗ (1, a) we 

would have that 1 = ω(1) = ω(b(λ1 − a)) = ω(b)ω(λ1 − a) = ω(b)(λ − ω(a)) = 0, a 

contradiction. Thus λ ∈ σ(a). 

Now define ψ : X → σ(a) by ψ(ω) = ω(a). It is clear that ψ is continuous 

and that it is surjective by (1.19). If ω1, ω2 ∈ X are two characters such that 

ψ(ω1) = ψ(ω2), then ω1(a ∗ ) = ω1(a) = ω2(a) = ω2(a ∗ ). Since ω1 and ω2 also agree


on 1, we see that they agree on any element of the form a n (a ∗ ) m , n, m ∈ N. These 

elements span a dense subset of C ∗ (1, a), so by linearity and continuity ω1 = ω2, i.e. 

ψ is also injective. 

Let ϕ : σ(a) → X be the inverse of ψ. Define Ψ : C ∗ (1, a) → C(σ(a)) by 

Ψ(b)(λ) = Φ(b)(ϕ(λ)), λ ∈ σ(a), where Φ : C ∗ (1, a) → C(X) is the ∗-homomorphism 

of Theorem 1.31. It is clear that Ψ is a ∗-isomorphism. When λ = ω(a), ω ∈ X, we 

have that ϕ(λ) = ω and hence Ψ(a)(λ) = Φ(a)(ω) = ω(a) = λ. Set Φa = Ψ −1 . 

Theorem 1.35 defines a ‘function calculus’ on the normal elements of a C∗-algebra in the following way : When f : σ(a) → C is a continuous function on the spectrum 

σ(a) of the normal element a ∈ A then f(a) will denote Φa(f) ∈ C∗ (1, a) ⊆ Â. 

If f is the restriction to σ(a) of a polynomial λ0 + λ1x + λ2x2 + · · · + λnxn , then 

f(a) = λ01 + λ1a + λ2a2 + · · · + λnan . Note that in general, when A has no unit, 

f(a) may not lie in A, only in Â. But we have the following. 

Lemma 1.36. Let A be a non-unital C ∗ - algebra and a ∈ A a normal element. 

When f ∈ C(σ(a)) and f(0) = 0, it follows that f(a) ∈ C ∗ (a) ⊆ A. 

Proof. By the Stone-Weierstrass theorem the functions of the form 

z ↦→ 

λn,mz n z m 

0≤n,m≤N 

(i.e. polynomials in z and z) are dense in C(σ(a)). Let Qn be a sequence of functions 

of this type such that Qn − f → 0. Since f(0) = 0 we have that Qn(0) → 0. So 

if we set Pn = Qn − Qn(0) we get a sequence of function of the same form, but 

with no constant term (i.e. with λ0,0 = 0), so that Pn − f → 0. Because there 

is no constant term in Pn we see that Pn(a) ∈ C ∗ (a) for all n and hence that 

f(a) = limn→∞ Pn(a) ∈ C ∗ (a). 

Lemma 1.37. Let a ∈ A be a normal element and f : σ(a) → C a continuous 

function. Then 

σ(f(a)) = f(σ(a)). 

Proof. Since Φa is a ∗-isomorphism, it is clear that σ(f(a)) = σ(f), the 

spectrum of f in C(σ(a)). But λ1 − f is invertible in C(σ(a)) if and only if 

λ1 − f does not take the value 0 on σ(a), i.e. if and only if λ /∈ f(σ(a)). Hence 

σ(f(a)) = σ(f) = f(σ(a)). 

Example 1.38. Let G be a group. G could be a topological group (like R for 

instance), but here we treat G as if it did not have a topology, i.e. as a discrete 

group. Let l2 (G) denote the Hilbert space of square summable functions on G, i.e. 

l2 (G) consists of the functions f : G → C such that 

 

 

sup |f(g)| 2 

: F ⊆ G, F finite < ∞ . 

g∈F 

For each g ∈ G define an operator ug on l 2 (G) by 

(ugf)(h) = f(g −1 h) , h ∈ G, f ∈ l 2 (G) .

1. BASICS 23 

It is then straightforward to check that uguk = ugk and u∗ g = ug−1 for all g, k ∈ G, 

ı.e. u is a unitary represenation of G on l 2 (G). The set 

 

g∈F 

λgug : λg ∈ C, g ∈ F , F finite 

is a ∗-subalgebra of B(l 2 (G)) and the closure of this set (with respect to the operator 

norm on B(l 2 (G))) is then a C ∗ -algebra which we call the reduced group C ∗ -algebra 

of G and denote by C∗ r (G). 

It is a simple exercise to show that C∗ r (G) is Abelian if and only if G is. 

1.3. Positivity and Hilbert C ∗ -modules. 

Definition 1.39. An element a of a C ∗ -algebra A is called positive when a = a ∗ 

(i.e. a is selfadjoint) and σ(a) ⊆ R + = [0, ∞[. 

Example 1.40. If A = C0(X) where X is a locally compact Hausdorff space, 

then an element f ∈ C0(X) is positive if and only if f(x) ≥ 0, x ∈ X. This is an 

immediate consequence of the observation that σ(f) = f(X). 

If A = Mn, then an element a ∈ Mn is positive if and only if a = a ∗ and all 

eigenvalues of a are non-negative. Note that an element a ∈ Mn is not neccesarily 

positive just because σ(a) ⊆ R + ; it may not be selfadjoint. This is the case for 

example if a is nilpotent : a2 

0 1 

= 0, e.g. a = in M2. 

0 0 

The set of positive elements in the C ∗ -algebra A will be denoted A+. 

Lemma 1.41. Let a = a ∗ ∈ A, a ≤ 1. Then a ∈ A+ if and only if 1 −a ≤ 1. 

Proof. By Lemma 1.34 a ∈ A+ ⇔ a ∈ C ∗ (1, a)+. So by Theorem 1.35 we must 

show that a real-valued function f ∈ C(σ(a)) with f ≤ 1 is non-negative if and 

only if 1 − f ≤ 1. And that is trivial. 

Lemma 1.42. Let t ∈ R + , a, b ∈ A+. Then ta + b ∈ A+. 

Proof. Set c = ta. Then σ(c) = tσ(a) ⊆ [0, ∞[, e.g. by Lemma 1.37. So 

c ∈ A+. To prove that c + b ∈ A+ it suffices to show that (2M) −1 (c + b) ∈ A+ 

where M > c + b. Since c 

M 

1− c 

M 

≤ 1 and 1− b 

M 

≤ 1, b 

M 

≤ 1 and hence that 1− c+b 

2M 

 

≤ 1 we know from Lemma 1.41 that 

) ≤ 1. 

= 1 

2 

(1− c 

M 

)+ 1 

2 

(1− b 

M 

Thus c+b 

2M ∈ A+ by Lemma 1.41 and hence also c + b ∈ A+. 

In words Lemma 1.42 says that A+ is a convex cone in A. The positive elements 

A+ define a partial ordering of Asa in the natural way: 

Lemma 1.43. Let a, b ∈ A. Then 

a ≤ b ⇔ b − a ∈ A+. 

σ(ab) ∪ {0} = σ(ba) ∪ {0}. 

Proof. If λ /∈ σ(ba), (λ1−ab)(1+a(λ1−ba) −1 b) = (1+a(λ1−ba) −1 b)(λ1−ab) = 

λ1. Thus if λ = 0, λ /∈ σ(ab). This means that C\ (σ(ba) ∪ {0}) ⊆ C\ (σ(ab) ∪ {0}), 

or that σ(ab) ∪ {0} ⊆ σ(ba) ∪ {0}. By symmetry we must have equality. 

Proposition 1.44. Let a be an element of the C ∗ - algebra A. Then a is positive 

if and only if there is an element b ∈ A such that a = b ∗ b.


Proof. If a is positive, the function f(t) = t 1 

2 is continuous on σ(a) and b = f(a) 

is selfadjoint with b2 = a. (Note that b ∈ A by Lemma 1.36.) For the converse 

assume that a = b∗b. Then a is selfadjoint and we can write a = f(a) − g(a) where 

f(t) = t ∨ 0 and g(t) = (−t) ∨ 0. Set c = f(a), d = g(a) and note that c, d are 

positive by Lemma 1.37 and cd = 0. We will prove that d = 0. Note first that 

(bd) ∗bd = d(c − d)d = −d3 ∈ −A+. Next write bd = s + it with s, t ∈ A selfadjoint, 

i.e. s = 1 

2 (bd + (bd)∗ ) and t = 1 

2i (bd − (bd)∗ ). Then (bd)(bd) ∗ = −(bd) ∗ (bd) + 2(s2 + 

t2 ) ∈ A+ + A+ ⊆ A+ where we have used Lemma 1.42 and Lemma 1.37. But 

σ((bd) ∗ (bd)) ∪ {0} = σ((bd)(bd) ∗ ) ∪ {0} by Lemma 1.43, so we have (bd) ∗ (bd) ∈ A+, 

i.e. we have shown that both (bd) ∗ (bd) and −(bd) ∗ (bd) are positive. But this means 

that σ((bd) ∗ (bd)) = {0} which implies that (bd) ∗ (bd) = 0 by Lemma 1.27. Hence 

−d3 = 0 and d4 = d4 = 0, i.e. d = 0. Thus a = c ∈ A+. 

 

Lemma 1.45. Let A be a C ∗ -algebra and a, b ∈ Asa two self-adjoint elements. 

Then 

a) a ≤ a1 in Â. 

b) If a ≤ b, then c ∗ ac ≤ c ∗ bc for all c ∈ A. 

c) If 0 ≤ a ≤ b, then a ≤ b. 

Proof. a) a1 − a is self-adjoint and it follows e.g. from (1.18) that σ(a1 − 

a) = a − σ(a). Since σ(a) ⊆ [−a, a] by Theorem 1.14 and Lemma 1.29, we 

conclude that a1 − a ∈ A+, i.e. a ≤ a1. 

b) By assumption and Proposition 1.44, b − a = d ∗ d for some d ∈ A. Hence 

c ∗ bc − c ∗ ac = c ∗ (b − a)c = (dc) ∗ dc ∈ A+. 

c) Set a1 = 1 + a, b1 = 1 + b. Then 

a1 = ρ(a1) = max{1 + t : t ∈ σ(a)} = 1 + ρ(a) = 1 + a. 

Similarly, b1 = 1 + b, so it suffices to show that that a1 ≤ b1. We may 

therefore assume that 1 ≤ a ≤ b. Then the spectra of a and b are contained in 

[1, ∞). We can therefore make sense of a −1/2 , and we conclude from a) and b) that 

ba −1 ≥ a −1/2 ba −1/2 ≥ a −1/2 aa −1/2 = 1. 

a 

b 

But then σ (ba−1 ) ⊆ [1, ∞), and hence σ 

Lemma 1.37. It follows that 

a 

 

a 

= ρ ≤ 1, 

b b 

and hence that a ≤ b. 

 

= σ 

 

(ba −1 ) −1 

⊆]0, 1] by 

Definition 1.46. Let A be a C ∗ -algebra. An inner-product A-module is a 

linear space E which is a right A-module (compatible with scalar multiplication: 

λ(xa) = (λx)a = x(λa), λ ∈ C, x ∈ E, a ∈ A), together with a map 

E × E ∋ (x, y) ↦→ 〈x, y〉 ∈ A 

such that 

i) 〈x, αy + βz〉 = α 〈x, y〉 + β 〈x, z〉, 

ii) 〈x, ya〉 = 〈x, y〉a,

iii) 〈x, y〉 ∗ = 〈y, x〉, 

iv) 〈x, x〉 ≥ 0, 

v) 〈x, x〉 = 0 ⇒ x = 0. 

1. BASICS 25 

It follows from iii) and i) that the inner product is conjugate linear in the first 

variable. If E satisfies all the conditions for an inner-product A-module, except v), 

we call E a semi-inner-product A-module. 

Proposition 1.47. When E is a semi-inner-product A-module and x, y ∈ E, 

then 

〈y, x〉 〈x, y〉 ≤ 〈x, x〉 〈y, y〉 

in A. 

Proof. Let a ∈ A. Then 

0 ≤ 〈xa − y, xa − y〉 

= a ∗ 〈x, x〉 a − 〈y, x〉a − a ∗ 〈x, y〉 + 〈y, y〉 (by ii) and iii)) 

≤ 〈x, x〉 a ∗ a − 〈y, x〉a − a ∗ 〈x, y〉 + 〈y, y〉 (by Lemma 1.45) 

(1.20) 

Assume first that 〈x, x〉 = 0. Then (1.20) gives us the inequality 〈y, x〉a+a ∗ 〈y, x〉 ∗ ≤ 

〈y, y〉. With a = 〈y, x〉 ∗ , it follows from Lemma 1.45 that 2 〈y, x〉 〈y, x〉 ∗ ≤ 〈y, y〉 ≤ 

〈y, y〉 1 in this case. The same conclusion holds when x is replaced by tx for any 

t ∈ R, i.e. we conclude that 2t 2 〈y, x〉 〈y, x〉 ∗ ≤ 〈y, y〉 1 for all t ∈ R. It follows 

that 

〈y, y〉 − 2t 2 σ (〈y, x〉 〈y, x〉 ∗ ) ⊆ [0, ∞) 

for all t ∈ R. This is only possible when σ (〈y, x〉 〈y, x〉 ∗ ) ⊆] − ∞, 0], i.e. when 

〈y, x〉 〈y, x〉 ∗ ≤ 0. Since 〈y, x〉 〈y, x〉 ∗ ≥ 0, we conclude that 〈y, x〉 = 0 when 〈x, x〉 = 

0. Hence the lemma holds when 〈x, x〉 = 0, and we can therefore assume that 

〈x, x〉 > 0. Working with x replaced by x 

√ 〈x,x〉 , we may then assume that 

〈x, x〉 = 1. In this case (1.20) yields the inequality 

〈y, x〉a + a ∗ 〈x, y〉 ≤ a ∗ a + 〈y, y〉, 

valid for all a ∈ A. This gives the desired inequality when we take a = 〈x, y〉. 

Given a semi-inner-product A-module E, we set 

x = 〈x, x〉 . (1.21) 

Proposition 1.48. In any semi-inner-product module over A, we have the inequalities, 

〈x, y〉 ≤ xy, 

and 

x + y ≤ x + y. 

Proof. It follows from Proposition 1.47 and c) of Lemma 1.45 that 〈x, y〉 2 ≤ 

x 2 y 2 , which is the first inequality. The second is a consequence of the first: 

x + y 2 = 〈x + y, x + y〉 = 〈x, x〉 + 〈x, y〉 + 〈y, x〉 + 〈y, y〉 

≤ 〈x, x〉 + 〈x, y〉 + 〈y, x〉 + 〈y, y〉 

≤ x 2 + y 2 + 2xy = (x + y) 2 .


Definition 1.49. Let A be a C ∗ -algebras. An inner-product A-module E is 

called a Hilbert C ∗ -module over A when E is complete in the norm (1.21). 

Example 1.50. a) A is a Hilbert C ∗ -module over A in a natural way: The rightmodule 

structure is the given one (from the algebra structure of A), and the inner 

product is defined by 

〈x, y〉 = x ∗ y. 

Note that the norm (1.21) is the given C ∗ -algebra norm. 

b) When A = C, a Hilbert C ∗ -module over A is the same as a Hilbert space. 

c) The direct sum A n = A ⊕ A ⊕ · · · ⊕A of n copies of A is a Hilbert C ∗ -module 

over A. The right-module structure is given by 

and the inner product is given by 

Note that 

(x1, x2, . . .,xn)a = (x1a, x2a, . . .,xna), 

〈(x1, x2, . . .,xn), (y1, y2, . . .,yn)〉 = x ∗ 1 y1 + x ∗ 2 y2 + · · · + x ∗ n yn. 

max 

i xi 2 = max 

i x ∗ ixi ≤ x ∗ 1x1 + x ∗ 2x2 + · · · + x ∗ nxn ≤ n max 

i xi 2 , 

proving that 

max 

i xi ≤ (x1, x2, . . ., xn) ≤ √ n max 

i xi, 

for all (x1, x2, . . .,xn) ∈ A n . These inequalities shows that the completeness of A n 

in the norm coming from the inner product follows from the completeness of A. 

Given a Hilbert C ∗ -module E over A, let LA(E) denote the set of maps T : E → 

E with the property that there is another map T ∗ : E → E such that 

for all x, y ∈ E. 

〈Tx, y〉 = 〈x, T ∗ y〉 

Lemma 1.51. Every element T ∈ LA(E) is a linear bounded operator on E, 

T ∗ ∈ LA(E) and (T ∗ ) ∗ = T. 

Proof. Rewrite the proof of Lemma 1.21. 

Proposition 1.52. LA(E) is a C ∗ -algebra with involution T ↦→ T ∗ and the norm 

T = sup{Tx : x ∈ E, x ≤ 1}. The product is the composition of operators. 

Proof. Rewrite the proof of Proposition 1.22. 

We let Mn(A) denote the set of n by n matrices with entries from A. Thus an 

element a = 

aij is a matrix 

⎛ 

⎞ 

a11 a12 a13 . . . a1n 

⎜a21 

a22 a23 . . . a2n⎟ 

⎜ 

⎟ 

⎜a31 

a32 a33 . . . a3n⎟ 

⎜ 

⎝ 

. 

⎟ 

. . . .. . ⎠ 

an1 an2 an3 . . . ann 

such that aij ∈ A for all i, j. Mn(A) is a complex vector space in the obvious way 

and an algebra with the product

(ab)ij = 

and we define an involution in Mn(A) by 

1. BASICS 27 

n 

k=1 

aikbkj 

(1.22) 

(a ∗ )ij = a ∗ ji . (1.23) 

It is straightforward to check that these definitions turn Mn(A) into a ∗-algebra. 

Set 

 

n 

a = sup 

n 

x ∗ ixi 

≤ 1 (1.24) 

for a = (aij). 

i,j,k 

x ∗ ka∗ 1 

ikaijxj 2 : (x1, x2, . . .,xn) ∈ A n , 

Proposition 1.53. Mn(A) is a C ∗ -algebra in the norm (1.24). 

Proof. The proof is essentially just an interpretation of the horrible expression 

defining the norm which allows one to work with it. For a = (aij) ∈ Mn(A), define 

a linear operator Ta : A n → A n in the way dictated by linear algebra : 

Ta(x) = 

n 

i=1 

a1ixi, 

n 

a2ixi, . . ., 

i=1 

n 

i=1 

anixi 

 

i=1 

, x = (x1, . . .,xn) ∈ A n . 

It is straightforward to check that a ↦→ Ta gives us an injective ∗-homomorphism 

from Mn(A) into LA (A n ). It follows therefore from Proposition 1.52 that Mn(A) is 

a C ∗ -algebra, except for the completeness and non-degeneracy of the norm. These 

properties follow from the inequalities 

max 

ij 

aij ≤ a ≤ n 3 

2 max 

ij 

aij (1.25) 

for elements a = (aij) ∈ Mn(A). 

To prove the first inequality (from left), let x, y ∈ A, x, y ≤ 1, and set 

x ′ = (0, 0, . . ., 0, x, 0, . . ., 0), y ′ = (0, 0, . . ., 0, y ∗ , 0, . . ., 0) 

where the non-zero entry occurs on the l’th, respectively, k’th coordinate. Then 

yaklx = 〈y ′ , Tax ′ 〉 ≤ Ta = a. 

If we set x = a∗ kl akl 

akl 2, y = a∗ kl 

akl , we get the inequality akl ≤ a. Since k, l ∈ 

{1, 2, . . ., n} were arbitrary this gives the first inequality. To prove the second, let 

x = (x1, . . ., xn) ∈ A n with x ≤ 1. Using c) of Lemma 1.45 we find that xj = 

x ∗ j 

xj 1 

2 ≤ n 

i=1 x∗ i 

xi 1 

2 = x ≤ 1 for all j, and hence n 

i,j,k x∗ k a∗ ik 

n i,j,k x∗k a∗ 1 

2 3 

ikaijxj ≤ n maxij aij 

2 1 

2 = n 3 

aijxj 1 

2 ≤ 

2 maxij aij. It follows that 

a ≤ n 3 

2 maxij aij. 

The completeness of Mn(A) in the C ∗ -norm follows straightforwardly : If {a n } 

is a Cauchy sequence in Mn(A) , the first inequality show that {a n ij} is a Cauchy 

sequence in A so that aij = limn→∞ a n ij exists for all i, j, and the second inequality 

shows that limn→∞ a n = a in Mn(A) when a = (aij).


1.4. Projections, partial isometries and finite dimensional C ∗ - algebras. 

Definition 1.54. A projection in a C ∗ -algebra is a selfadjoint idempotent, i.e. 

an element p such that p = p ∗ = p 2 . A partial isometry is an element v such that 

vv ∗ is a projection. 

and 

Lemma 1.55. If v is a partial isometry. Then 

vv ∗ v = v, v ∗ vv ∗ = v ∗ , 

v ∗ v is a projection. 

Proof. The calculation 

(v − vv ∗ v)(v − vv ∗ v) ∗ = vv ∗ − vv ∗ vv ∗ − vv ∗ vv ∗ + vv ∗ vv ∗ vv ∗ 

= vv ∗ − 2(vv ∗ ) 2 + (vv ∗ ) 3 = 0 

shows that v = vv ∗ v. It follows that (v ∗ v) 2 = v ∗ vv ∗ v = v ∗ v, proving that v ∗ v is a 

projection, i.e. also that v ∗ is a partial isometry. It follows that v ∗ vv ∗ = v ∗ . 

The projection v ∗ v is called the support projection and vv ∗ the range projection 

of v. Thus the range projection of v is the support projection of v ∗ and the support 

projection of v is the range projection of v ∗ . 

Example 1.56. Let X be a compact Hausdorff space. In C(X) the projections 

are the functions f ∈ C(X) with f(X) ⊆ {0, 1} and the partial isometries are the 

f’s with f(X) ⊆ T ∪ {0}. 

Example 1.57. Let H be a Hilbert space and let B(H) be the C ∗ -algebra of all 

bounded operators on H. When p is a projection in B(H) the range space pH is a 

closed subspace of H and p is the orthogonal projection H → pH ⊆ H corresponding 

to the decomposition H = pH ⊕ (pH) ⊥ . Conversely, when V is a closed subspace 

of H the orthogonal projection H → V ⊆ H corresponding to the decomposition 

H = V ⊕V ⊥ is a projection in B(H). In this way we have a bijective correspondence 

between projections in B(H) and closed subspaces in H. 

When v is a partial isometry in B(H) we have decompositions H = v ∗ vH ⊕ 

(v ∗ vH) ⊥ and H = vv ∗ H ⊕ (vv ∗ H) ⊥ . The restriction of v to v ∗ vH maps this closed 

subspace isometrically onto vv ∗ H and v annihilates (v ∗ vH) ⊥ , i.e. v is ’partially an 

isometry’. The name partial isometry (in the general case) orginates in this example 

which also explains the names ’support projection’ and ’range projection’. 

Definition 1.58. A set of matrix units in a C ∗ - algebra A is a subset 

where N, n1, n2, . . .,nN ∈ N, such that 

and 

for all d, b, i, j, k, l. 

{e d ij : i, j = 1, 2, . . ., nd, d = 1, 2, . . ., N}, 

e d ije b kl = δd,bδj,ke d il 

e d ∗ d 

ij = eji (1.26) 

(1.27)

1. BASICS 29 

Example 1.59. Let N, n1, n2, . . .,nN ∈ N and consider A = Mn1 ⊕ Mn2 ⊕ · · · ⊕ 

be the matrix in Mnd all of whose entries are zero, except the j’th 

} is a set of matrix units in Mnd , 

MnN . Let fd ij 

entry of the i’th row where there is a 1. Then {fd ij 

the standard matrix units. To get a set of matrix units in A = Mn1 ⊕ · · · ⊕ MnN we 

just put the fd ij ’s into their respective coordinates; i.e. we set 

e d ij = (0, 0, . . ., 0, fd ij , 0, . . ., 0) 

with f d ij on the d’th coordinate. The resulting set {ed ij : i, j = 1, 2, . . ., nd, d = 

1, 2, . . ., N} constitute the standard matrix units in A. This set of matrix units is 

the universal such set (when n1, n2, . . .,nN are given), in the sense that every other 

set of matrix units with these indices are images of this set under some automorphism 

of A. 

We shall prove the following theorem which characterizes the C ∗ -algebras whose 

vector space dimension is finite. 

Theorem 1.60. A finite dimensional C ∗ -algebra A is ∗-isomorphic to a finite 

direct sum of full matrix algebras; i.e. there are numbers N, n1, n2, . . ., nN ∈ N such 

that A ≃ Mn1 ⊕ Mn2 ⊕ · · · ⊕ MnN . 

To present the proof it is appropriate to isolate the following lemmas. 

Lemma 1.61. A finite dimensional Abelian C ∗ -algebra A is isomorphic to C n , 

where n = dim A. 

Proof. By Theorem 1.31 A ≃ C0(X) for some locally compact Hausdorff space 

X. If there were n + 1 distinct elements, say x1, x2, . . .,xn+1, in X we could find, 

by Urysohn’s lemma, continuous functions fi, i = 1, 2, . . ., n+1, in C0(X) such that 

fi(xj) = δi,j for all i, j. Since the fi’s are linearly independent we would have to 

conclude that dimA = dim(C0(X)) ≥ n+1, a contradiction. Hence X contains ≤ n 

points. Call them y1, y2, . . .,yk. Define C0(X) ∋ f → (f(y1), f(y2), . . ., f(yk)) ∈ C k . 

It is straightforward to prove that this is a ∗-isomorphism. Hence A ≃ C k and thus 

k = n. 

The C ∗ -algebra C n contains a distinguished set of projections, namely the elements 

ei, i = 1, 2, . . ., n, forming the standard vector space basis for C n . If A is a 

C ∗ -algebra which is isomorphic to C n , then the image of this set of projections in A 

will be called the minimal projections in A. 

Lemma 1.62. Let A be a non-zero finite dimensional C ∗ - algebra. Then A is 

unital. 

Proof. Since every selfadjoint element generates an Abelian C ∗ -subalgebra, 

such subalgebras always exist in abundance in any C ∗ -algebra. Since A is finitedimensional 

the set 

{dim B : B is an abelian C ∗ -subalgebra of A} (1.28) 

is finite. Let D ⊆ A be an abelian C ∗ -subalgebra of A such that dim D is maximal in 

the set (1.28). By Lemma 1.61, D ≃ C d where d = dimD, and hence, in particular, 

D is unital and we denote the unit of D by p. Set B = {a ∈ A : pa = ap = 0}. 

Then B is in fact {0}. To see this, observe that B is a finite dimensional C ∗ - 

algebra in itself, and hence, if nonzero, would contain a non-zero projection q, by


Lemma 1.61. But then Cq + D would be an Abelian C ∗ -subalgebra of A properly 

containing D since q /∈ D (qp = 0). This contradicts the maximality of dimD, 

and B must be zero as asserted. Now for any a ∈ A it is easy to check that 

(a − ap) ∗ (a − ap) = (a ∗ − pa ∗ )(a − ap) ∈ B = {0}, i.e. a = ap. Since p = p ∗ we find 

by taking adjoints that pa = a for all a ∈ A. Thus p is a unit in A. 

We now give the 

Proof. Proof of Theorem 1.60 : If A = {0} we take N = 1 and n1 = 0 and 

we are done. So assume that A = {0}. Let Z = {x ∈ A : xa = ax for all a ∈ A}, 

i.e. Z is the center of A. Since A is unital by Lemma 1.62, Z = {0}. Being a finite 

dimensional Abelian C ∗ -algebra Z is therefore isomorphic to C N for some N ∈ N, 

cf. Lemma 1.61. Let pi, i = 1, 2, . . ., N, be the minimal projections in Z. Note that 

N 

i=1 pi = 1. For each i, piA is C ∗ -subalgebra of A and the map 

a ↦→ (ap1, ap2, . . .,apN) 

defines a ∗-isomorphism A ≃ p1A ⊕ p2A ⊕ · · · ⊕ pNA. So now it will suffice to find 

nm ∈ N such that pmA ≃ Mnm for each m. 

To simplify notation set B = pmA. As above we can conclude that the center of 

B is isomorphic to Cd for some d. But if d ≥ 2 we would have a central projection 

q in B which was neither 0 nor pm. q would automatically also be central in A 

and hence contradict the minimality of pm. Hence d = 1 and the center of B is 

C1 (where we use the notation 1 for the unit (= pm) in B = pmA since this is the 

algebra in which we now work.) Let D be a maximal Abelian C∗-subalgebra of B, 

so that D ≃ Cnm for some nm ∈ N, and let q1, q2, . . .,qnm be the minimal non-zero 

projections in D. Since 1 = nm i=1 qi we have that 

B = 

nm 

qiBqj. (1.29) 

i,j=1 

Consider the C ∗ -subalgebra qkBqk for some k ∈ {1, 2, . . ., nm}. We claim that 

qkBqk = Cqk. (1.30) 

To prove this it suffices to take an arbitrary non-zero selfadjoint a = a ∗ ∈ qkBqk 

and show that a ∈ Cqk. So consider the, necessarily finite dimensional, Abelian 

C ∗ -algebra C ∗ (a, qk) ⊆ qkBqk. Then C ∗ (a, qk) ≃ C k for some k ∈ N and we must 

show that k = 1. Assume not, i.e. assume that k ≥ 2. In that case there is a 

projection p ∈ C ∗ (a, qk) such that p /∈ {0, qk}. Then 

span {p, q1, q2, . . .,qnm} 

is an Abelian C ∗ -subalgebra of B which contains D and has dimension ≥ nm + 

1 because {p, q1, q2, . . .,qnm} is a linearly independent set. This contradicts the 

maximality of D and hence k = 1, proving (1.30). 

Next we claim that 

span BqkB = B . (1.31) 

To prove this note that span BqkB is a C ∗ - subalgebra of B and also a two-sided 

ideal in B. By Lemma 1.62 there is a unit p in span BqkB. For any a ∈ B, we have 

that ap ∈ spanBqkB and hence pap = ap. By taking adjoints we see that ap = pa

1. BASICS 31 

for all a ∈ B. So p is a non-zero central projection in B and hence p = 1 because 

the center of B is C1. But p = 1 implies that span BqkB = B as asserted. 

Let i, j ∈ {1, 2, . . .,nm}. Then 

there is a non-zero partial isometry wij ∈ B such that qiBqj = Cwij. (1.32) 

When wij is a partial isometry as in (1.32), wijw ∗ ij is a non-zero projection qiBqi, 

and hence wijw ∗ ij = qi by (1.30). Similarly, w ∗ ij wij = qj, i.e. wij is a partial isometry 

with qj as support projection and qi as range projection. 

To prove (1.32), let x be a non-zero element of qiBqj. Then xx ∗ is non-zero 

element of qiBqi and, according to (1.30), this means that xx∗ = λqi for some 

λ ∈ C. But xx∗ ≥ 0 and non-zero, so λ > 0. Set v = λ−1 2x. Then vv∗ = qi. In 

particular, v is a partial isometry and hence so is v ∗ by Lemma 1.55, i.e. v ∗ v is a 

projection. This element lies in qjBqj = Cqj, so v ∗ v = qj. Thus wij = v is a partial 

isometry in B with support projection qj and range projection qi. It remains to 

check that qiBqj = Cwij. Since wij ∈ qiBqj we must show that 

qiBqj ⊆ Cwij. (1.33) 

Let z ∈ qiBqj\{0}. As above we see that z = µv for some partial isometry µ > 0 

and some v ∈ B with vv ∗ = qi and v ∗ v = qj. Note that vw ∗ ij ∈ qiBqi. It follows then 

from (1.30) that vw ∗ ij = κqi for some κ ∈ C. By using Lemma 1.55 we find that 

v = vv ∗ v = vqj = vw ∗ ij wij = κqiwij = κwijw ∗ ij wij = κwij. Hence z = µv = κµwij ∈ 

Cwij. This proves (1.33) and hence (1.32). 

Let {wij : i, j = 1, 2, · · ·nm} be the partial isometries found in (1.32). We set 

fij = w ∗ 1iw1j, i, j = 1, 2, . . ., nm. 

We find that fijfkl = 0 when j = k because fij ∈ qiBqj and fkl ∈ qkBql. If j = k 

we have 

fijfkl = w ∗ 1iw1jw ∗ 1jw1l = w ∗ 1iq1w1l = w ∗ 1iw1l = fil. 

Since it is obvious that f ∗ ij = fji we have that {fij : i, j = 1, 2, . . ., nm} is a set of 

matrix units in B. It follows from (1.32) that fij = µijwij for some µij ∈ C with 

|µij| = 1. Thus 

qiBqj = Cfij. (1.34) 

Let {eij : i, j = 1, 2, . . ., nm} be the standard system of matrix units in Mnm. We 

can then define ϕ : Mnm → B as the linear map with the property that ϕ(eij) = fij 

for all i, j. It is easy to check that ϕ is a ∗- homomorphism. If (aij) ∈ Mni and 

ϕ((aij)) = 

i,j aijfij = 0, then 0 = qk 

 

i,j aijfijql = aklfkl, so that akl = 0 for 

all k, l; i.e. ϕ is injective. To prove that ϕ is also surjective we see from (1.29) 

that it suffices to show that qiBqj is contained in the image of ϕ for all i, j. But 

qiBqj = Cfij by (1.34) so this is trivial. 

Example 1.63. Let G be a finite group and consider the reduced group C ∗ - 

algebra, C ∗ r (G), introduced in Example 1.38. Since G is finite both l2 (G) and C ∗ r (G) 

must be finite dimensional. By Theorem 1.60 

C ∗ r(G) ≃ Mn1 ⊕ Mn2 ⊕ · · · ⊕ Mnm . 

But which numbers n1, n2, · · · , nm, occur here ? To answer this, let π : G → B(H) 

be an irreducible unitary representation of G on the Hilbert space H. We can then


define a ∗-homomorphism ϕπ : C∗ r (G) → B(H) by 

ϕπ( 

λgug) = 

λgπ(g) . 

g∈G 

By Exercise 1.67 below, ϕπ maps C ∗ r(G) onto B(H). Let ˆ G denote the set of unitary 

equivalence classes of irreducible unitary representations of G. For each χ ∈ ˆ G we 

pick an irreducible unitary representation πχ of G representing χ. The dimension of 

the Hilbert space Hπχ on which πχ(G) acts is independent of the particular choice 

of πχ and we denote this number by dim χ. Define a ∗-homomorphism 

by 

g∈G 

ϕ : C ∗ r(G) → ⊕ χ∈ ˆ G B(Hπχ) ≃ ⊕ χ∈ ˆ G Mdim χ , 

ϕ(x) = (ϕπχ(x)) χ∈ ˆ G . 

We claim that ϕ is a ∗-isomorphism. To see this, note that ker ϕπχ is a two-sided 

ideal in C ∗ r (G) and, in particular, a finite dimensional C∗ -algebra in itself. Let pχ be 

the unit in ker ϕπχ, cf. Lemma 1.62. Since ker ϕπχ is a two-sided ideal in C ∗ r (G), 

pχ is a central projection in C ∗ r(G). Thus (1 −pπχ)C ∗ r(G) is also a two-sided ideal in 

C ∗ r (G) and ϕπχ maps (1 − pπχ)C ∗ r (G) ∗-isomorphically onto B(Hπχ). We claim that 

(1 − pχ)(1 − pχ1) = 0 when χ = χ1 . (1.35) 

Indeed, if this central projection in C ∗ r (G) was non-zero, the image of it, ϕπχ((1 − 

pχ)(1 − pχ1)) ∈ B(πχ) would be a projection invariant under πχ(g) for all g ∈ G and 

hence ϕπχ((1 − pχ)(1 − pχ1)) = 1 by irredicibility of πχ. Thus (1 − pχ)(1 − pχ1) = 

(1 − pχ) since ϕπχ is injective on (1 − pχ)C ∗ r(G). By symmetry (1 − pχ)(1 − pχ1) = 

is a ∗-isomorphism 

B(Hπχ) → B(Hπχ ) taking πχ(g) to πχ1(g) for all g ∈ G. By Exercise 1.82 below, 

1 

(1 − pχ1) and hence (1 − pχ) = (1 − pχ1). Then ϕπχ 1 ◦ ϕ −1 

πχ 

there is a unitary w : Hπχ → Hπχ 1 such that ϕπχ 1 ◦ ϕ −1 

πχ (x) = wxw∗ , x ∈ B(Hπχ). 

But this implies that πχ and πχ1 are unitarily equivalent, contradicting that χ = χ1 

in ˆ G. This proves (1.35). We can now easily deduce that ϕ is surjective; indeed, if 

(xχ) ∈ ⊕ χ∈ ˆ G B(Hπχ), we can find yχ ∈ (1 − pχ)C ∗ r (G) such that ϕπχ(yχ) = xχ for 

each χ ∈ ˆ G. Then y = 

χ∈ ˆ G yχ ∈ C∗ r (G) and ϕ(y) = x. 

To see that ϕ is also injective, let x ∈ ker ϕ. Consider one of the ∗-homomorphisms 

ψi : C∗ r(G) → Mni = B(Cni ) obtained by combining the abstract ∗-isomorphism 

C∗ r (G) ≃ Mn1 ⊕Mn2 ⊕· · ·⊕Mnm with the projection Mn1 ⊕Mn2 ⊕· · ·⊕Mnm → Mni . 

Then G ∋ g ↦→ ψi(ug) is a representation G on Cni which is irreducible since ψi maps 

C∗ r(G) onto B(Cni ). This representation is unitarily equivalent to πχ for some χ ∈ ˆ G. 

Since πχ(x) = 0, because x ∈ ker χ, we conclude that ψi(x) = 0. This holds for each 

i ∈ {1, 2, · · · , m} and hence x = 0. Consequently ϕ is a ∗-isomorphism as asserted. 

In short we have proved that 

C ∗ r (G) ≃ ⊕ χ∈ ˆ G Mdim χ . 

Therefore the numbers, n1, n2, · · · , nm, are the dimensions of the irreducible representations 

of G and m is the number of unitary equivalence classes of irreducible 

representations of G, i.e. m = | ˆ G|.

and 

1.5. Exercises and additional material. 

Exercise 1.64. Let A be a C ∗ -algebra. Show that 

1. BASICS 33 

a ∗ = a 

a ∗ a = a 2 

for all a ∈ A. Show that if A is unital, then 1 = 1 ∗ and 1 = 1. 

Exercise 1.65. Show that the canonical ∗-homomorphism ϕA : A → M(A) is 

surjective if and only if A is unital. Show that ϕA(A) is a two-sided ideal in A. 

Exercise 1.66. Let A be a norm closed (in the supremum norm) algebra of realvalued 

continuous and bounded functions on a topological space X. Let f, g ∈ A. 

Show that f ∧ g = min{f, g} ∈ A and f ∨ g = max{f, g} ∈ A. (Hint : f ∨ g = 

1 

1 

(f +g+|f −g|) and f ∧g = (f +g −|f −g|). Approximate t ↦→ |t| by polynomials, 

2 2 

uniformly on the image of f − g.) 

Exercise 1.67. Let H be a Hilbert space and A ⊆ B(H) a C ∗ -subalgebra of 

the bounded operators of H. Assume that A is finite dimensional and that the only 

non-zero projection in B(H) which is invariant under A (i.e. satisfies that pap = ap 

for all a ∈ A) is 1. Prove that A = B(H). (Hints : Note first that the assumption 

implies that 1 is the only non-trivial central projection in A. Since A is finite 

dimensional this implies that A ≃ Mm for some m ∈ N. Let {eij : i, j = 1, 2, · · · , m} 

be a set of matrix units in A. If f ∈ B(H) is a non-zero projection such that 

f ≤ e11, then p = m 

i=1 ei1fe1i is a projection in B(H) such that pap = ap for all 

a ∈ A. Hence p = 1 by assumption, and thus f = e11. But this means that e11 is a 

minimal projection in B(H), and it must therefore be a one-dimensional projection. 

Similarly, eii is a one-dimensional projection for each i. Since 

i eii = 1 this implies 

that dim H = m. But then Mm = B(H) because these vector spaces have the same 

dimension, namely m 2 .) 

Exercise 1.68. Let A be a C∗-algebra. We denote by l2(A) the sequences 

(a1, a2, a3, . . .) in A with the property that the sequence 

n 

{ 

i=1 

a ∗ i ai} ∞ n=1 

is convergent in A. Show that l2(A) is a Hilbert C∗-module over A with an inner 

product given by 

∞ 

< (ai) ∞ i=1 , (bi) ∞ i=1 >= 

i=1 

a ∗ i bi. 

Exercise 1.69. Let A = C[0, 1], and set E = {f ∈ A : f(t) = 0, t ≥ 1}. 

Show 2 

that E is a Hilbert C∗-module over A with an inner product given by 

< f, g >= f ∗ g. 

Exercise 1.70. Let I be any set, and assume that we have a Hilbert C ∗ -module 

Ei over A for all i ∈ I. Is it true, that the direct sum 

⊕i∈IEi


is a Hilbert C∗-module with an inner product given by 

 

< (ai), (bi) >= lim < ai, bi >, 

F 

when we take the limit over the net of finite subsets F ⊆ I, ordered by inclusion ? 

I think so, but please check if you can make it work. 

Exercise 1.71. Consider the C ∗ -algebra Mn (C), and an element m ∈ Mn(C). 

a) Show that the spectrum σ(m) of m is the set of eigenvalues of m. 

b) Give an example of an element m ∈ M2(C) whose spectum σ(m) is nonnegative, 

i.e. σ(m) ⊆ [0, ∞), but which is not positive. 

Exercise 1.72. Let X be a locally compact Hausdorff space and consider the 

C ∗ -algebra C0(X). Let f ∈ C0(X). Show that 

i∈F 

σ(f) = f(X). 

Exercise 1.73. Let A be unital C ∗ -algebra, and a ∈ Asa a self-adjoint element. 

Then 

a = inf{t > 0 : −t1 ≤ a ≤ t1}. 

Exercise 1.74. Let p and q be projections in the C ∗ -algebra A. Show that 

p ≤ q ⇔ pq = p . 

Exercise 1.75. A non-zero projection p in a C ∗ -algebra B is minimal when the 

following holds: If q ∈ B is a projection such that q ≤ p, then either p = q or q = 0. 

Let A be finite dimensional Abelian C ∗ -algebra of dimension n. Let {ei : i = 

1, 2, . . ., n} and {fi : i = 1, 2, . . ., n} be sets of minimal projections in A. Show that 

there is a permutation σ of {1, 2, . . ., n} such that 

ei = fσ(i), i = 1, 2, . . ., n. 

Definition 1.76. A continuous linear functional ω ∈ A ∗ is positive when a ∈ 

A+ ⇒ ω(a) ≥ 0. 

Example 1.77. By the Riesz representation theorem every positive functional 

on C(X), where X is some compact Hausdorff space, is given by integration with 

respect to a regular Borel measure on X. 

On Mn the positive functionals are the ones of the form 

Mn ∋ a → Tr(ha) 

for some positive element h ∈ Mn. The proof of this statement is an exercise. 

Lemma 1.78. (Schwarz inequality) Let ω ∈ A ∗ be positive. Then 

|ω(a ∗ b)| 2 ≤ ω(a ∗ a)ω(b ∗ b), a, b ∈ A. 

Proof. Set < a, b >= ω(a ∗ b). Then < ·, · > defines a quasi-inner product on A 

and the desired inequality follows from the Cauchy-Schwarz inequality for this. For 

the convenience of the reader we repeat the proof. First we observe that 

ω(a ∗ ) = ω(a), a ∈ A. (1.36)

1. BASICS 35 

To see this observe first that when a = a ∗ we can write a = a+ − a− where a+, a− ∈ 

A+, see the proof of Proposition 1.44. Hence ω(a) = ω(a+) − ω(a−) ∈ R in this 

case. In general we write a = b + ic where b, c ∈ Asa. Then ω(a) = ω(b) − iω(c) = 

ω(b − ic) = ω(a ∗ ), proving (1.36). 

Next let a, b ∈ A, λ ∈ C, |λ| = 1, and set c = λa. Then 

0 ≤ ω((tc + b) ∗ (tc + b)) = ω(c ∗ c)t 2 + (ω(b ∗ c) + ω(c ∗ b))t + ω(b ∗ b) = 

ω(a ∗ a)t 2 + (ω(b ∗ c) + ω(b ∗ c))t + ω(b ∗ b) = 

ω(a ∗ a)t 2 + 2Re ω(b ∗ c)t + ω(b ∗ b), t ∈ R. 

This is only possible if (Re ω(b ∗ c)) 2 ≤ ω(a ∗ a)ω(b ∗ b), i.e. if (Re λω(b ∗ a)) 2 ≤ ω(a ∗ a)ω(b ∗ b). 

Since this holds also when λ is chosen such that λω(b ∗ a) = |ω(b ∗ a)|, we have the 

stated inequality. 

Lemma 1.79. Let A be a unital C ∗ - algebra and ω ∈ A ∗ . Then ω is positive if 

and only if ω = ω(1). 

Proof. Assume first that ω is positive. By Lemma 1.78 we have that |ω(a)| 2 ≤ 

ω(a ∗ a)ω(1) ≤ ωω(1)a 2 for all a ∈ A. Taking the supremum over all a with 

a ≤ 1 yields ω 2 ≤ ωω(1), i.e. ω ≤ ω(1). Since the reverse inequality is 

trivial we have that ω(1) = ω. Conversely, assume that ω = ω(1). The first 

(and main) problem is to show that ω(a) ∈ R when a = a ∗ . To this end write 

ω −1 ω(a) = α + iβ, α, β ∈ R. For all γ ∈ R, ω −1 ω(a + iγ1) = α + i(β + γ). 

But σ(a) ⊆ [−a, a] so Lemma 1.37 gives that σ(a + iγ1) ⊆ [−a, a] + iγ. 

Since a + iγ1 is normal we know that a + iγ1 = ρ(a + iγ1) ≤ a 2 + γ 2 . Since 

|ω −1 ω(a + iγ1)| ≥ |β + γ|, we get that |β + γ| ≤ a 2 + γ 2 or β 2 + 2γβ ≤ a 2 . 

Since this holds for all γ ∈ R we must have β = 0, i.e. ω(a) = α ∈ R. Finally, for 

any non-zero a ∈ A, 1 − a∗ a 

a 2 ≤ 1 by Lemma 1.41 and Proposition 1.44, so 

|ω −1 ω(1) − a −2 ω −1 ω(a ∗ a)| ≤ 1. 

Since ω −1 ω(1) = 1 and a −2 ω −1 ω(a ∗ a) ∈ R this is only possible if ω(a ∗ a) ≥ 0, 

i.e. ω is positive. 

Proposition 1.80. Let A be a C ∗ -algebra (= 0) and a ∈ A. There exists a 

positive functional ω ∈ A ∗ of norm ω = 1 such that ω(a ∗ a) = a 2 . 

Proof. By Theorem 1.35 there is a ∗-isomorphism Φa ∗ a : C(σ(a ∗ a)) → C ∗ (1, a ∗ a) 

such that Φa ∗ a(idσ(a ∗ a)) = a ∗ a. Set t = ρ(a ∗ a) ∈ σ(a ∗ a) and define ω0 : C ∗ (1, a ∗ a) → 

C by ω0(b) = Φa ∗ a −1 (b)(t). Then ω0 is a positive linear functional of norm 1 on 

C ∗ (1, a ∗ a) such that ω0(a ∗ a) = ρ(a ∗ a) = a 2 and ω0(1) = 1. By the Hahn- 

Banach theorem there is an extension ω ∈ Â∗ of ω0 with ω = ω0 = 1. Since 

ω(1) = ω0(1) = 1, Lemma 1.79 shows that ω is positive on Â. Hence the restriction 

to A is also positive and ω(a∗a) = ω0(a∗a) = a2 . This shows in particular that 

ω’s restriction to A also has norm 1. 

Exercise 1.81. A famous theorem from the childhood of C∗-algebras says that 

every C∗-algebra is (isometrically) ∗-isomorphic to a norm-closed ∗-invariant subalgebra 

of the bounded operators LC(H) of some Hilbert space H. Since A ⊆ Â, it 

suffices to show this when A is unital. We will therefore assume that A is unital in 

the following. The proof can then be organized to consist of the following steps.


1) Let ω ∈ A ∗ be a positive functional of norm 1. Set Iω = {a ∈ A : ω(a ∗ a) = 

0}. Show that Iω is a linear subspace of A which is also a left A-module. 

2) Show that the quotient space A/Iω is an inner product space with inner 

product 

< a + Iω, b + Iω >= ω(a ∗ b). 

3) Let Hω be the Hilbert space obtained by completing the inner product space 

from 2). Show that there is a ∗-homomorphism πω : A → LC (Hω) such that 

πω(a)(b + Iω) = ab + Iω. 

4) Show that πω(a) = 0 ⇒ ω(a ∗ a) = 0. 

5) Let H be the Hilbert space direct sum 

H = ⊕ωHω, 

where we take the sum over all positive norm-one functionals ω on A. Show 

that there is an injective and isometric ∗-homomorphism π : A → LC(H). 

(Hint: Use Proposition 1.50.) 

6) Show that A is ∗-isomorphic to a closed ∗-invariant sub-algebra of LC(H). 

Exercise 1.82. Let H1 and H2 be two finite dimensional Hilbert spaces and 

ψ : B(H1) → B(H2) a ∗-isomorphism. Show that there is a unitary w : H1 → H2 

such that ψ(x) = wxw ∗ for all x ∈ B(H1). (Hint : Let {eij : i, j = 1, 2, · · · , dim H1} 

be a set of matrix units for B(H1) which spans B(H1). Show that {ψ(eij)} is a set of 

matrix units for B(H2) which spans B(H2). For each i, let λi and µi be unit vectors 

in the range of eii and ψ(eii), respectively. Show that we can define an operator 

v : H1 → H2 such that vλ1 = µ1 and vλi = 0, i = 2, 3, · · · , dim H2 = dim H2, and 

that the unitary 

will do the job.) 

w = 

ψ(ei1)ve1i 

i 

Exercise 1.83. Let Ai, i = 1, 2, . . ., n be C ∗ - algebras. Then the direct sum 

⊕ n i=1 Ai is an algebra with coordinate-wise multiplication 

(a1, a2, . . .,an)(b1, b2, . . .,bn) = (a1b1, . . .,anbn). 

Show that ⊕ n i=1 Ai is a C ∗ -algebra with involution 

and norm 

(a1, a2, . . .,an) ∗ = (a ∗ 1 , a∗ 2 , . . .,a∗ n ) 

(a1, a2, . . .,an) = max 

i 

ai. 

Show that ⊕ n i=1Ai is unital if and only if each Ai is unital. 

Show that ⊕ n i=1Ai is Abelian if and only if each Ai is Abelian. 

Exercise 1.84. Let A1, A2, A3, . . . be a sequence of C ∗ -algebras. The infinite 

product ∞ i=1 Ai is a ∗-algebra with coordinate-wise operations. 

Show that 

1) A = {(ai) ∈ ∞ i=1 Ai : supi ai < ∞} 

2) C = {(ai) ∈ ∞ i=1 Ai : limn→∞ ai = 0 } 

are C∗-algebras in the norm 

(ai) = sup 

i 

ai.

1. BASICS 37 

Exercise 1.85. Let A be a ∗-algebra and · 1 and · 2 two norms on A such 

that A is a C ∗ -algebra with respect to both norms. 

Show that a1 = a2 for all a ∈ A. 

In contrast, A has uncountably many Banach algebra norms, e.g. · = k · 1 

for any k ≥ 1. 

Exercise 1.86. Let A be a C ∗ -algebra. Construct a ∗-homomorphism ϕ : A → 

A ⊕ A which is unital when A is. Construct a ∗-homomorphism ψ : A → M2(A) 

which is unital when A is. 

Let k, n ∈ N. Show that Mk(Mn(A)) is ∗-isomorphic to Mkn(A). 

Exercise 1.87. Let X be a locally compact Hausdorff space. Show that 

ϕ(A)(x) = (Aij(x)), x ∈ X, A ∈ Mn(C0(X)), 

defines a ∗-isomorphism ϕ : Mn(C0(X)) → C0(X, Mn). 

Exercise 1.88. Consider the two C ∗ -algebras A = Mn1 ⊕ Mn2 ⊕ · · · ⊕ MnN and 

B = Mm1 ⊕ Mm2 ⊕ · · · ⊕ MmM , where ni ≥ 1 and mj ≥ 1 for all i, j. Show that 

A ≃ B if and only if N = M and there is a permuation σ of {1, 2, . . ., N} such that 

nσ(i) = mi, i = 1, 2, . . ., N. 

Exercise 1.89. Let X and Y be compact Hausdorff spaces and ϕ : C(X) → 

C(Y ) a unital ∗-homomorphism. 

Show that there is a continuous map f : Y → X such that ϕ(g) = g ◦ f, g ∈ 

C(X). (Hint : For each y ∈ Y, g → ϕ(g)(y) defines a character of C(X). Use 

Exercise 1.83.) 

Exercise 1.90. Let X and Y be locally compact Hausdorff spaces. A continuous 

function f : Y → X is proper when the inverse image of every compact subset of X 

is compact in Y , i.e. when 

K ⊆ X compact ⇒ f −1 (K) ⊆ Y compact. 

a) Let f : Y → X be a continuous proper map. Show that there is a ∗homomorphism 

ψ : C0(X) → C0(Y ) such that ψ(h) = h ◦ f. 

b) Let ϕ : C0(X) → C0(Y ) be a ∗-homomorphism. Show that there is a proper 

continuous map f : Y → X such that ϕ(h) = h ◦ f for all h ∈ C0(X). 

c) Let ϕ and f be as in b). Show that ϕ is injective if and only if f is surjective. 

d) Let ϕ and f be as in b). Show that ϕ is isometric if and only if f is surjective. 

e) Show that C0(X) ≃ C0(Y ) if and only if there is a homeomorphism f : Y → X. 

Exercise 1.91. Let ϕ : A → B be a ∗-homomorphism between C ∗ -algebras A 

and B. 

a) Show that ϕ(a)) ≤ a when a ∈ A is selfadjoint. (Hint: One possibility is 

to reduce to the case when A and B are abelian, and then use Exercise ??.) 

b) Show that ϕ(a)) ≤ a for all a ∈ A. 

c) Show that ϕ is injective if and only if ϕ is isometric. 

Exercise 1.92. Let α be an automorphism of Mn (i.e. α is a ∗-isomorphism α : 

Mn → Mn). Show that there is a unitary U ∈ Mn such that α(A) = UAU ∗ , A ∈ Mn.


(Hint : Let {eij} be the standard set of matrix units in Mn. Show that α(e11)Mne11 

is spanned by a partial isometry v and consider 

n 

α(ej1)ve1j.) 

j=1 

Exercise 1.93. Let p, q ∈ Mn be projections. Show that there is a partial 

isometry v ∈ Mn such that vv ∗ = p and v ∗ v = q if and only if Tr(p) = Tr(q). 

Exercise 1.94. Let a be a selfadjoint element of a C ∗ -algebra A. Show that a 

is a projection if and only if σ(a) ⊆ {0, 1}. 

Exercise 1.95. Let X be a locally compact, non-compact Hausdorff space (e.g. 

X = R). Then A = C0(X) is an Abelian non-unital C∗-algebra and we can consider 

A ⊆ Â ⊆ M(A). 

1) Show that M(A) is ∗-isomorphic to the C∗-algebra Cb(X) of continuous 

bounded functions on X. 

2) Show that Â is ∗-isomorphic to the C∗-algebra of continuous functions f : 

X → C which ’has a limit at infinity’, i.e. satisfies that there is a λ ∈ C 

such that 

∀ǫ > 0 ∃K ⊆ X compact : |f(x) − λ| ≤ ǫ ∀x ∈ X\K . 

3) Show that Â ≃ C(Y ) where Y is a compact Hausdorff space with a point 

y0 ∈ Y such that X is homeomorphic to Y \{y0}. (Y is the one-point 

compactification of X.) 

4) Show that M(A) ≃ C(β(X)), where β(X) is the Stone-Cech-compactification 

of X. 

2. Inductive limits of C ∗ -algebras 

2.1. The definition and the universal property. By a sequence of C ∗ - 

algebras we mean a countable family A1, A2, A3, . . . of C ∗ -algebras together with ∗ 

-homomorphisms ϕn : An → An+1, n = 1, 2, 3, . . .. Such a sequence will sometimes 

be denoted 

A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . 

and sometimes more compressed as (An, ϕn). The ∗-homomorphisms ϕn will be 

called the connecting maps. If the An’s are unital C ∗ -algebras and the connecting 

maps all unital, we call the sequence a unital sequence of C ∗ - algebras. In any case, 

we will use the notation ϕm,n for the composition ϕm,n = ϕm−1 ◦ ϕm−2 ◦ · · · ◦ ϕn : 

An → Am when n < m. So we have that ϕn = ϕn+1,n. It is convenient to define 

ϕn,n to be the identity map on An. 

Assume now that we are given a sequence of C ∗ -algebras (An, ϕn). We will 

construct from this sequence a new C ∗ -algebra as follows. First we introduce an 

equivalence relation ≡ on the disjoint union 

∞ 

n=1 

by writing a ≡ b when a ∈ An, b ∈ Am and there is a k ≥ max{n, m} such that 

ϕk,n(a) = ϕk,m(b). For a ∈ ∞ 

n=1 An we let [a] denote the equivalence class containing 

An

2. INDUCTIVE LIMITS OF C ∗ -ALGEBRAS 39 

a. It is straightforward to check that we can make ∞ 

n=1 An/ ≡ into a ∗-algebra by 

defining 

[a] + [b] = [ϕk,n(a) + ϕk,m(b)], a ∈ An, b ∈ Am, k ≥ max{n, m}, 

[a][b] = [ϕk,n(a)ϕk,m(b)], a ∈ An, b ∈ Am, k ≥ max{n, m}, 

λ[a] = [λa], a ∈ An, λ ∈ C, 

and 

[a] ∗ = [a ∗ ], a ∈ An. 

To get a C∗-algebra out of ∞ n=1 An/ ≡ we first introduce a semi-norm by 

[a] = lim 

k→∞ ϕk,n(a), a ∈ An. 

This limit exists because a ∗-homomorphism between C∗- algebras never increases 

the norm, so that the sequence ϕk,n(a), k ≥ n, is decreasing. It is 

straightforward to check that · defines a semi-norm on ∞ n=1 An/ ≡ satisfying 

xy ≤ xy, x ∗ = x, x ∗ x = x 2 , (2.1) 

for all x, y ∈ ∞ n=1 An/ ≡. In particular, it follows that 

∞ 

I = {x ∈ An/ ≡ : x = 0} 

is a two-sided ∗-invariant ideal in ∞ n=1 An/ ≡. The quotient space 

∞ 

( An/ ≡)/I 

is therefore a ∗-algebra and 

n=1 

n=1 

x + I = x, x ∈ 

∞ 

An/ ≡, 

defines norm on ( ∞ n=1 An/ ≡)/I. Let q : ∞ n=1 An/ ≡ → ( ∞ n=1 An/ ≡)/I denote the 

quotient map. Note that (2.1) also holds in ( ∞ n=1 An/ ≡)/I so that this ∗-algebra is 

almost a C∗-algebra, the only missing property is the completeness of the norm. We 

define the inductive limit of the sequence (An, ϕn) to be the C ∗ -algebra we obtain 

by completing ( ∞ 

n=1 An/ ≡)/I in the norm · and we denote it by lim 

−→ (An, ϕn), 

or simply by A∞ when no ambiguity is possible. 

Let ι : ∞ 

n=1 An/ ≡ → A∞ be the ∗-homomorphism obtained by composing the 

quotient map q with the canonical imbedding of ( ∞ 

n=1 An/ ≡)/I into its completion, 

A∞. Define 

by 

n=1 

ϕ∞,n : An → A∞ 

ϕ∞,n(a) = ι([a]), a ∈ An. 

These ∗-homomorphisms will be referred to as the canonical maps. It is clear 

from the construction that 

ϕ∞,n(An) ⊆ ϕ∞,n+1(An+1), n ∈ N, (2.2)


and that 

∞ 

ϕ∞,n(An) = A∞. (2.3) 

n=1 

Furthermore, we shall frequently use the following properties. Let a ∈ An, b ∈ Am. 

Then 

and 

ϕ∞,n(a) = ϕ∞,m(b) ⇔ lim 

k→∞ ϕk,n(a) − ϕk,m(b) = 0 (2.4) 

ϕ∞,n(a) = lim 

k→∞ ϕk,n(a). (2.5) 

Before we give some examples to familiarize the reader with inductive limits, we 

will first emphasize the categorial properties of the construction. Assume that we 

are given a C ∗ -algebra B and ∗-homomorphisms ψn : An → B, n ∈ N, which are 

compatible with the connecting maps of the sequence (An, ϕn), in the sense that 

ψn+1 ◦ ϕn = ψn 

for all n ∈ N. We can then define a ∗-homomorphism 

such that 

ψ : A∞ → B 

ψ ◦ ϕ∞,n = ψn 

(2.6) 

(2.7) 

for all n ∈ N. Indeed, we first obtain a ∗-homomorphism ψ0 : ∞ 

n=1 An/ ≡ → B 

by setting ψ0([a]) = ψn(a), a ∈ An. This is welldefined because of (2.6). Since 

ψ0([a]) = ψk(ϕk,n(a)) ≤ ϕk,n(a) for all k ≥ n, we see that ψ0([a]) ≤ [a]. 

Thus ψ0 induces a ∗-homomorphism ψ00 : ( ∞ n=1 An/ ≡)/I → B such that ψ00 ◦ q = 

ψ0. Since ψ00(q(x)) = ψ0(x) ≤ x = q(x), x ∈ ∞ n=1 An/ ≡, we see that ψ00 

extends by continuity to a ∗-homomorphism ψ : A∞ → B which satisfies (2.7) by 

construction. Note that the condition (2.7) determines ψ uniquely because of (2.3). 

For later reference we state this important property of the construction as a 

lemma. 

Lemma 2.1. Let 

A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . 

be a sequence of C ∗ -algebras and A∞ = lim 

−→ (An, ϕn) the corresponding inductive limit 

C ∗ -algebra. If ψn : An → B is a sequence of ∗-homomorphisms into the C ∗ -algebra 

B such that ψn+1 ◦ϕn = ψn, n ∈ N, then there is one and only one ∗-homomorphism 

ψ : A∞ → B such that ψ ◦ ϕ∞,n = ψn, n ∈ N. 

The property described in Lemma 2.1 is sometimes referred to as the universal 

property of the inductive limit C ∗ - algebra, although it is really a property of the 

family ϕ∞,n : An → A∞, n = 1, 2, 3, . . ., rather than of the C ∗ -algebra A∞ in itself. 

It is this universal property which characterizes the construction, see Exercis 2.18. 

Proposition 2.2. Let (An, ϕn) and (Bn, ψn) be two sequences of C ∗ -algebras. 

Assume that there are sequences k1 < k2 < k3 < . . . and m1 < m2 < m3 < . . . in N


and ∗-homomorphisms µi : Aki → Bmi , λi : Bmi → Aki+1 such that ϕki+1,ki = λi ◦ µi 

and µi+1 ◦ λi = ψmi+1,mi for all i ∈ N, i.e. such that the infinite diagram 

ϕk2 ,k ϕk 1 3 ,k ϕk 2 4 ,k3 Ak1 

 

Ak2 

 

Ak3 

 

µ1 

 

 

λ1 

 

 

 

µ2 

 

 

λ2 

 

 

 

µ3 

 

· · · 

 

λ3 

 

 

 

 

· · · 

Bm1 

 

Bm2 

 

Bm3 

ψm2 ,m1 ψm3 ,m2 ψm4 ,m3 commutes. It follows that lim(An, 

ϕn) ≃ lim(Bn, 

ψn). In fact, there is a ∗-isomorphism 

−→ −→ 

µ : lim(An, 

ϕn) → lim(Bn, 

ψn) such that µ ◦ ϕ∞,kn = ψ∞,mn ◦ µn for all n ∈ N. 

−→ −→ 

Proof. To simplify notation, set A∞ = lim(An, 

ϕn) and B∞ = lim(Bn, 

ψn). Let 

−→ −→ 

i ∈ N and a ∈ Ai. Then the sequence 

where kj ≥ i, is constant, and we set 

ψ∞,mj ◦ µj ◦ ϕkj,i, 

µi = lim 

j→∞ ψ∞,mj ◦ µj ◦ ϕkj,i. 

This is a ∗-homomorphism µi : Ai → B∞ and µi+1 ◦ ϕi = limj→∞ ψ∞,mj ◦ µj ◦ 

ϕkj,i+1 ◦ ϕi+1 = limj→∞ ψ∞,mj ◦ µj ◦ ϕkj,i = µi for all i. It follows therefore from 

the universal property of the inductive limit construction, Lemma 2.1 that there is 

a ∗-homomorphism µ : A∞ → B∞ such that µ ◦ ϕ∞,i = µi for all i. In particular, 

µ ◦ ϕ∞,kn ◦ ϕ∞,kn = limj→∞ ψ∞,mj ◦ µj ◦ ϕkj,kn = ψ∞,mn ◦ µn for all n ∈ N. 

It remains to show that µ is a ∗-isomorphism. To this end, let iN and b ∈ Bi. 

Then the sequence 

ϕ∞,kj+1 ◦ λj ◦ ψmj,i, 

where mj ≥ i, is constant, and we set 

λi = lim 

j→∞ ϕ∞,kj+1 ◦ λj ◦ ψmj,i. 

The λi’s are ∗-homomorphisms and λi+1 ◦ψi = λi for all i. By the universal property 

of the inductive limit construction, Lemma 2.1 that there is a ∗-homomorphism 

λ : B∞ → A∞ such that λ ◦ ψ∞,i = λi for all i. Note that 

λ ◦ µ ◦ ϕ∞,i = λ ◦ µi = λ ◦ lim 

j→∞ ψ∞,mj ◦ µj ◦ ϕkj,i 

= lim λ ◦ ψ∞,mj 

j→∞ ◦ µj 

λmj ◦ ϕkj,i = lim 

j→∞ 

◦ µj ◦ ϕkj,i 

= lim lim ϕ∞,kl+1 

j→∞ l→∞ ◦ λl ◦ ψml,mj ◦ µj ◦ ϕkj,i 

= lim lim ϕ∞,kl+1 ◦ ϕkl+1,kj ◦ ϕkj,i 

j→∞ l→∞ 

= ϕ∞,i. 

Since this holds for all i we see that λ ◦µ is the identity on A∞. A similar argument 

shows that µ ◦ λ is the identity on B∞. 

 

Example 2.3. The purpose of this relatively long example is to describe the C ∗ - 

algebra one gets as the inductive limit of a unital sequence of Abelian C ∗ -algebras.


So consider a sequence 

A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . 

of unital Abelian C ∗ -algebras An with unital connecting maps. By Theorem 1.31 

there are compact Hausdorff spaces Xn and ∗-isomorphisms λn : An → C(Xn) for 

all n ∈ N. Partly for convenience we restrict the attention to the case where each 

Xn is metrizable. 

By Exercise 1.89 we know that there are continuous maps fn : Xn+1 → Xn such 

that the diagram 

λn 

An 

⏐ 

 

ϕn 

−−−→ An+1 

C(Xn) 

commutes when ψn : C(Xn) → C(Xn+1) is the ∗-homomorphism induced by fn, viz. 

λn+1 

⏐ 

 

ψn 

−−−→ C(Xn+1) 

ψn(g) = g ◦ fn, g ∈ C(Xn). 

By Proposition 2.2 we therefore have that lim(An, 

ϕn) ≃ lim(C(Xn), 

ψn). Now, 

−→ −→ 

it is fairly clear that the inductive limit of a unital sequence of Abelian C∗-algebras must be unital and Abelian itself. Thus there must be a compact Hausdorff space 

X such that 

lim 

−→ (C(Xn), ψn) ≃ C(X). 

Also it must depend only on the Xn’s and the fn’s. Here it is. Set 

∞ 

X = {(xi) ∈ Xi : xn = fn(xn+1) for all n ∈ N}. 

i=1 

X is clearly a closed subset of the infinite product ∞ 

n=1 Xn and is therefore compact 

by Tychonoffs theorem. X is the inverse limit space (sometimes called the projective 

limit ) of the sequence 

X1 

 

X2 

 

X3 

X4 

 

f1 f2 f3 

. . . 

(by definition) and is usually denoted by lim(Xn, 

fn). Let us prove that lim(C(Xn), 

ψn) ≃ 

←− −→ 

C(X). Let pn : X → Xn be the map which projects to the n’th coordinate, 

i.e. pn (xi) = xn, and set Φn(g) = g ◦ pn, g ∈ C(Xn). Then the Φn’s are 

∗-homomorphisms Φn : C(Xn) → C(X) such that Φn+1 ◦ ψn = Φn. Indeed, 

Φn+1 ◦ ψn(g) = Φn+1(g ◦ fn) = g ◦ fn ◦ pn+1 = g ◦ pn bacause fn ◦ pn+1 = pn 

on X. By the universal property of the inductive limit, Lemma 2.1, there is a 

∗-homomorphism Φ : lim(C(Xn), 

ψn) → C(X) such that Φ ◦ ψ∞,n = Φn for all n. 

−→ 

We first prove that Φ is an isometry. So let g ∈ C(Xn) for some n. It will suffice 

to show that 

Φ(ψ∞,n(g)) = Φn(g) = ψ∞,n(g). 

By (2.5) 

ψ∞,n(g) = lim 

k→∞ ψk,n(g). 

For each k > n choose xk ∈ Xk such that 

|g ◦ fn ◦ fn+1 ◦ · · · ◦ fk−1(xk)| = g ◦ fn ◦ fn+1 ◦ · · · ◦ fk−1 = ψk,n(g). 

f4


By compactness of Xn there is a sequence m n 1 < mn 2 < mn 3 

sequence 

{fn ◦ fn+1 ◦ · · · ◦ fm n k −1(xm n k )} 

converges in Xn, say to yn. Next there is a subsequence m n+1 

1 

of mn 1 < mn2 < mn3 < . . . such that 

{fn+1 ◦ fn+2 ◦ · · · ◦ f n+1 

mk −1(xm n+1 

k 

)} 

< . . . in N such that the 

< mn+1 

2 

< mn+1 

3 

< . . . 

converges in Xn+1, say to yn+1. (It is here we use the assumption that Xn is 

metrizable.) Continuing in this way we get elements yj ∈ Xj, j ≥ n, such that 

yj = limk→∞ fj ◦fj+1 ◦· · ·◦f j 

m < . . . in N 

k−1(xm j ) for some sequence m 

k 

j 

1 < mj 2 < mj3 

and some subsequence m j+1 

1 < m j+1 

2 < m j+1 

3 < . . . of m j 

1 < m j 

2 < m j 

3 < . . . . Then 

fj(yj+1) = fj( lim fj+1 ◦ fj+2 ◦ · · · ◦ f j+1 

k→∞ 

mk −1(xm j+1 

k 

lim 

k→∞ fj ◦ fj+1 ◦ · · · ◦ f j+1 

mk −1(xm j+1) 

= yj 

k 

)) = 

for all j ≥ n so if we set yi = fi ◦ fi+1 ◦ · · · ◦ fn−1(yn), i < n, then the sequence 

y = (yi) is in X. Note that 

Φn(g) = g ◦ pn ≥ |g ◦ pn(y)| = |g(yn)| = lim |g(fn ◦ fn+1 ◦ · · · ◦ fm 

k→∞ n k −1(xmn))| = k 

lim 

k→∞ ψmn k ,n(g) = ψ∞,n(g). 

Thus 

Φ(ψ∞,n(g)) ≥ ψ∞,n(g) 

and the reversed inequality is automatic because Φ is a ∗-homomorphism between 

C∗-algebras. We have shown that Φ is an isometric ∗-homomorphism. It is clear 

that Φ is unital (Φ(ψ∞,n(1)) = 1), so to conclude that Φ is surjective and hence 

is a ∗-isomorphism, it suffices by the Stone-Weierstrass theorem to check that 

Φ(lim(C(Xn), 

ψn)) separates the points of X. So consider (xi) = (zi) in X. There 

−→ 

is an n ∈ N such that xn = zn and hence there is an element g ∈ C(Xn) such that 

g(xn) = g(zn). Since 

and similarly 

Φ(ψ∞,n(g)) (xi) = Φn(g) (xi) = g(xn), 

Φ(ψ∞,n(g)) (zi) = Φn(g) (zi) = g(zn), 

this completes the proof. 

Speaking in the language of category theory, the conclusion that 

lim 

−→ (C(Xn), ψn) ≃ C(lim(Xn, 

fn)), 

←− 

tells us that the contravariant functor X → C(X) from compact Hausdorff spaces 

to Abelian unital C∗-algebras is continuous. 

2.2. AF-algebras. We now come to an exciting and surprisingly rich class of 

C ∗ -algebras whose structure is very well understood and for which there is a neat 

classification, up to isomorphism, by fairly simple invariants. 

Definition 2.4. A C ∗ -algebra A is called approximately finite dimensional (AFalgebra, 

for short) when there is an increasing sequence 

A1 ⊆ A2 ⊆ A3 ⊆ . . .


of finite dimensional C∗-subalgebras in A whose union is dense, i.e. satisfies that 

∞ 

An = A. 

n=1 

We know from Theorem 1.60 that a finite dimensional C ∗ - algebra is ∗-isomorphic 

to a finite direct sum of matrix algebras : 

Mn1 ⊕ Mn2 ⊕ Mn3 ⊕ · · · ⊕ MnN 

The following lemma can therefore hardly come as a surprise. 

Lemma 2.5. A C ∗ -algebra A is an AF-algebra if and only if there is a sequence 

A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . 

of finite direct sums of matrix algebras such that A ≃ lim 

−→ (An, ϕn). 

Proof. If F1 ⊆ F2 ⊆ F3 ⊆ . . . is a sequence of finite dimensional C ∗ -subalgebras 

in A such that A = ∞ 

n=1 Fn, we can use Theorem 1.60 to get finite direct sums of 

matrix algebras An and ∗- isomorphisms ψn : Fn → An, n ∈ N. Let in denote the 

inclusion Fn → Fn+1. Then ϕn = ψn+1 ◦ in ◦ ψ −1 

n : An → An+1 makes the diagram 

ψn 

Fn 

⏐ 

 

An 

in 

−−−→ Fn+1 

ψn+1 

⏐ 

 

ϕn 

−−−→ An+1 

commute and we can define ψ : ∞ 

n=1 Fn → A∞ by ψ|Fn = ϕ∞,n ◦ ψn. Note that 

ψ(a) = limk→∞ ϕk,n ◦ ψn(a) = a because each ϕk,n and ψn is injective and 

hence isometric by Exercise 1.91. Thus ψ extends to an injective (isometric) ∗- 

homomorphism ψ : A → A∞ = lim 

−→ (An, ϕn). Obviously ϕ∞,n(An) = ψ(Fn) is in the 

image of ψ for all n and hence ψ must be surjective. 

Conversely, if A ≃ A∞ = lim 

−→ (An, ϕn) for a sequence of finite direct sums of 

matrix algebras, let ψ : A∞ → A be a ∗-isomorphism. Then Fn = ψ(ϕ∞,n(An)) is a 

finite dimensional C ∗ - subalgebra of A, Fn ⊆ Fn+1 for all n and ∞ 

n=1 Fn = A. 

The AF-algebras are thus just the C ∗ -algebras which arise as an inductive limit 

of a sequence of finite direct sums of matrix algebras. The structure of matrix 

algebras, or a finite direct sum of them, is as uncomplicated as one could wish 

and hence the problem, which must be overcome in order to gain a more detailed 

insight into the nature of AF-algebras, is to determine how such finite dimensional 

C ∗ -algebras can be put together to form an inductive limit. Thus the first task is to 

obtain a satisfying description of the ∗- homomorphisms between finite direct sums 

of matrix algebras. 

Let A = Mn1 ⊕Mn2 ⊕· · ·⊕MnN and B = Mm1 ⊕Mm2 ⊕· · ·⊕MmM . Let S = (sij) 

be an M × N matrix with entries from N such that 

N 

sijnj ≤ mi, i = 1, 2, . . ., M. (2.8) 

j=1 

We can then define a ∗-homomorphism ϕS : A → B such that 

ϕS(a1, a2, . . .,aN) = (ϕ1(a1, a2, . . .,aN), ϕ2(a1, a2, . . .,aN), . . ....,ϕM(a1, a2, . . ., aN))


for all (a1, a2, . . .,aN) ∈ A, where ϕi : A → Mmi is the ∗-homomorphism which 

sends (a1, a2, . . .,aN) ∈ A to the block diagonal matrix 

diag 

a1, a1, . . .,a1 , a2, a2, . . .,a2 , . . ....,aN, aN, . . ., aN, 

0, 0, . . ., 0 

 

si1 times si2 times 

siN times r times 

where r = mi − N 

j=1 sijnj. Such a ∗-homomorphism will be called a standard 

homomorphism. 

For example, if A = M2 ⊕ M3 and B = M7, the 1 × 2 matrix S = (s11, s12) = 

(2, 1) determines the standard homomorphism ϕS : M2 ⊕ M3 → M7 which sends 

(a, b) = ((aij), (bij)) ∈ M2 ⊕ M3 to 

diag ⎛ 

a11 a12 0 0 0 0 

⎞ 

0 

⎜a21 

⎜ 0 

⎜ 

a, a, b) = ⎜ 0 

⎜ 0 

⎝ 

0 

a22 

0 

0 

0 

0 

0 

a11 

a21 

0 

0 

0 

a12 

a22 

0 

0 

0 

0 

0 

b11 

b21 

0 

0 

0 

b12 

b22 

0 ⎟ 

0 ⎟ 

0 ⎟. 

⎟ 

b13⎟ 

⎠ 

b23 

0 0 0 0 b31 b32 b33 

Note that the matrix S = (sij) is easy to recover from ϕS; if {e d ij : i, j = 

1, 2, . . ., nd, d = 1, 2, . . ., N} is the standard system of matrix units in A, then 

sij = Tri(ϕS(e j 

11)) 

where Tri : B → C is the composition of the projection B → Mmi onto the i’th 

component of B with the trace Tr of Mmi , i.e. 

Tri(b1, b2, . . .,bM) = Tr(bi), (b1, b2, . . .,bM) ∈ B. 

A very useful way to represent a standard homomorphism is by using a diagram. 

The diagram, called a Bratteli diagram, corresponding to the standard homomorphism 

given by the matrix S = (sij) is obtained as follows. First draw a horizontal 

row of N vertices, labelled n1, n2, . . ., nN and below that another horizontal row of 

M vertices labelled m1, m2, . . .,mM. Then draw sij edges from the vertex labelled nj 

to the vertex labelled mi for all i, j. For example the standard map M2 ⊕M3 → M7 

described above has Bratteli diagram 

2 

3 

 

 

 

 

 

 

7 

The importance of the standard homomorphisms comes from the following two 

results. 

Lemma 2.6. Let A = Mn1 ⊕ Mn2 ⊕ · · · ⊕ MnN , B = Mm1 ⊕ Mm2 ⊕ · · · ⊕ MmM 

and let ψ : A → B be a ∗-homomorphism. 

There is then a unitary u ∈ B and a standard homomorphism ϕ : A → B such 

that 

uψ(a)u ∗ = ϕ(a), a ∈ A.


Proof. Let {e d ij : i, j = 1, 2, . . ., nd, d = 1, 2, . . ., N} be the standard system of 

matrix units in A. Set sij = Tri(ψ(e j 

11)), j = 1, 2, . . ., N, i = 1, 2, . . ., M. Note that 

Tri(ψ(e j 

j 

kk )) = Tri(ψ(ek1 )ψ(ej 

j 

1k )) = Tri(ψ(e1k )ψ(ej 

j 

k1 )) = Tri(ψ(e11 )) 

for all k = 1, 2, . . ., nj. Thus sijnj = nj j 

k=1 Tri(ψ(ekk )) and 

N 

j=1 

sijnj = 

N 

nj 

j=1 k=1 

Tri(ψ(e j 

kk 

)) = Tri(ψ( 

j,k 

e j 

kk )) = Tri(ψ(1)) ≤ Tri(1) = mi. 

Thus (sij) satisfies (2.8) and hence defines a standard homomorphism ϕ : A → B. 

Let πi : B → Mmi be the projection and set ψi = πi ◦ ψ, ϕi = πi ◦ ϕ. It will suffice 

to construct, for each i, a unitary ui ∈ Mmi such that uiψi(a)u∗ i = ϕi(a), a ∈ A, 

because then u = (u1, u2, . . .,uM) ∈ B will do the job. So fix i ∈ {1, 2, . . ., M}. 

Since 

Tr(ψi(e j 

11)) = Tr(ϕi(e j 

11)), 

we know from Exercise 1.93, that there is a partial isometry vj ∈ Mmi such that 

vjv∗ j 

j = ϕi(e11) and v∗ jvj = ψi(e j 

11). Set v = N nj j j 

j=1 k=1 ϕi(ek1 )vjψi(e1k ) and note 

that 

vv ∗ = 

 

 

) ) = 

 

j,k 

j,k 

ϕi(e j j 

k1 )vjψi(e1k ϕi(e j j 

k1 )ϕi(e11)ϕi(e j 

1k 

s,t 

) = 

j,k 

ψi(e t s1 )v∗ t ϕi(e t 1s 

ϕi(e j 

kk ) = ϕi(1), 

where we have used that vjψi(e j 

11 )v∗ j = vjv ∗ j vjv ∗ j 

Since 

j,k 

ϕi(e j j 

k1 )vjψi(e11 )v∗ j 

jϕi(e1k ) = 

= ϕi(e j 

11 ). Similarly, v∗ v = ψi(1). 

Tr(1 − ψi(1)) = mi − Tr(ψi(1)) = mi − Tr(v ∗ v) = mi − Tr(vv ∗ ) = 

mi − Tr(ϕi(1)) = Tr(1 − ϕi(1)), 

we can use Exercise 1.93 again to get a partial isometry w ∈ Mmi such that ww∗ = 

1 −ϕi(1) and w∗w = 1 −ψi(1). Then ui = v +w is a unitary in Mmi (CHECK!) and 

uiψi(e d kl )u∗ i = vψi(e d kl )v∗ = 

b,c,s,t 

ϕi(e b c1 )vbψi(e b 1c )ψi(e d kl )ψi(e s t1 )v∗ sϕi(e s 1t ) = 

ϕi(e d k1 )vdψi(e d 11 )v∗ d ϕi(e d 1l ) = ϕi(e d k1 )ϕi(e d 11 )ϕi(e d 1l ) = ϕi(e d kl ). 

Since this holds for all k, l = 1, 2, . . ., nd, d = 1, 2, . . ., N, we see that uiψi(a)u ∗ i = 

ϕi(a), a ∈ A. 

 

Theorem 2.7. Let A be an AF-algebra. There is then a sequence 

A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . 

of finite direct sums of matrix algebras with connecting maps that are all standard 

homomorphisms such that A ≃ lim 

−→ (An, ϕn). 

Proof. By Lemma 2.5 there is a sequence (An, ψn) of finite direct sums of 

matrix algebras such that A ≃ lim(An, 

ψn). So the only problem we have is that 

−→ 

we must substitute the ψn’s by standard homomorphisms. By Proposition 2.2 it


is enough to find standard homomorphisms ϕn : An → An+1 and ∗-automorphisms 

αn : An → An such that the diagram 

α1 

ψ1 ψ2 

A1 −−−→ A2 −−−→ A3 

⏐ 

 

α2 

⏐ 

 

α3 

⏐ 

 

ψ3 

−−−→ . . . 

ϕ1 ϕ2 ϕ3 

A1 −−−→ A2 −−−→ A3 −−−→ . . . 

commutes. This is done by induction using Lemma 2.6. The induction is started by 

taking α1 to be the identity map on A1. So assume that we have constructed the 

diagram up to 

αn−1 

An−1 

⏐ 

 

An−1 

ψn−1 

−−−→ An 

αn 

⏐ 

 

ϕn−1 

−−−→ An 

Then ψn ◦ α −1 

n : An → An+1 is unitarily conjugate to a standard homomorphism ϕn 

by Lemma 2.6; i.e. there is a unitary v ∈ An+1 such that ϕn = Ad v ◦ ψn ◦ α −1 

n is a 

standard homomorphism. If we set αn+1 = Ad v, the diagram 

αn 

An 

⏐ 

 

An 

ψn 

−−−→ An+1 

αn+1 

⏐ 

 

ϕn 

−−−→ An+1 

commutes and hence we have completed the induction step. 

We now consider the unital case. Note that a standard homomorphism ϕ : 

Mn1 ⊕ · · · ⊕MnN → Mm1 ⊕ · · · ⊕MmM given by the matrix (sij) is unital if and only 

if 

N 

sijnj = mi, i = 1, 2, . . ., M. (2.9) 

j=1 

As one would expect, a unital AF-algebra can be realized as the inductive limit 

of a sequence of finite direct sums of matrix algebras with unital standard homomorphisms 

as connecting maps. The starting point for the proof of this is first 

to write the given unital AF-algebra A as the closure of an increasing sequence 

A1 ⊆ A2 ⊆ A3 ⊆ . . . of finite dimensional C∗-subalgebras, viz. 

A = 

∞ 

An. 

n=1 

By Lemma 1.61 each An contains a unit for it self, say pn ∈ An. Then limn→∞ pna = 

a for all a ∈ A. Indeed, if a ∈ A and ǫ > 0 are given, there is an N ∈ N and b ∈ AN 

. Then 

such that a − b < ǫ 

2 

pna − a ≤ pna − pnb + pnb − b + b − a = pn(a − b) + b − a < ǫ 

for all n ≥ N. In particular we see that 

lim 

n→∞ 1 − pn = lim pn1 − 1 = 0. 

n→∞ 

So for all large enough n, 1−pn < 1 

2 . But 1−pn is a projection, so σ(1−pn) ⊆ {0, 1} 

(this follows from Exercise 1.93), and hence 1 − pn < 1 ⇔ pn = 1. So we see that


pn = 1 for all large enough n, say for n ≥ N. Thus A = 

n≥N An and 1 ∈ An for all 

n ≥ N. Consequently we might as well assume to begin with that 1 ∈ An for all n. 

It is now straightforward to check that with this assumption as point of departure, 

all the maps constructed in and betweeen Lemma 2.5 and Theorem 2.7 become 

unital. So we arrive at the following conclusion. 

Theorem 2.8. Let A be a unital AF-algebra. There is then a sequence 

A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . 

of finite direct sums of matrix algebras with unital standard homomorphisms as connecting 

maps such that A ≃ lim 

−→ (An, ϕn). 

Thanks to Theorem 2.7 and Theorem 2.8, an AF-algebra A can be described by 

use of infinite Bratteli diagrams obtained as follows. Write 

A ≃ lim 

−→ (An, ϕn) 

for some sequence (An, ϕn) of finite direct sums of matrix algebras with standard 

homomorphisms as connecting maps. The Bratteli diagram for A is then the infinite 

graph obtained by drawing the Bratteli diagrams of the standard homomorphisms 

ϕn : An → An+1 below one another. For example 

2 2 

 

 

 

4 4 

 

 

 

8 8 

 

 

 

16 

16 

 

 

 

 

. . . 

is the Bratteli diagram for the AF-algebra which is the inductive limit of the sequence 

M2 ⊕ M2 

ϕ1 

M4 ⊕ M4 

ϕ2 

M8 ⊕ M8 

ϕ3 

. . . 

where ϕn : M2n ⊕ M2n → M2n+1 ⊕ M2n+1 is the standard homomorphism given by 

the matrix 

2 0 

1 1 

for all n ∈ N. 

Note that the AF-algebra we get is unital because the connecting maps are. This 

can be used to simplify the Bratteli diagram a little. Because, if it is stated that the 

connecting maps are unital, then the numbers occuring in row 2 and onwards are 

determined completely by the numbers in the first row and the diagram itself, by 

repeated use of (2.9). And we can always assume, in the unital case, that A1 = C so 

that the first row of vertices only contains one vertex which is labelled 1. Since this


is a standard convention, one doesn’t even write that first label in. So usually, in the 

litterature, the AF-algebra above is defined as the unital AF-algebra with diagram 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

. . . 

In the non-unital case, however, it is important to indicate the integer labels 

on the vertices, because otherwise the matrix sizes involved in the sequence of C∗- algebras can not be determined from the diagram. 

We will now investigate an important special class of unital AF-algebras, namely 

the socalled UHF-algebras. These are the AF-algebras which we obtain as inductive 

limits of sequences 

Mn1 

ϕ1 

Mn2 

ϕ2 

Mn3 

ϕ3 

. . . 

with unital connecting maps. It should be clear at this point that a unital C∗-algebra A is (isomorphic to) a UHF-algebra if and only if there is a sequence 1 ∈ A1 ⊆ A2 ⊆ 

A3 ⊆ . . . of C∗-subalgebras in A such that each An is isomorphic to a matrix algebra 

and ∞ n=1 An = A. 

The UHF-algebras can be classified (without use of K-theory) in the following 

way. Let A be a UHF-algebra. Let P denote the set of prime numbers, remembering 

that 1 is not a prime. For each prime number p ∈ P we set 

FA(p) = sup{n ∈ N ∪ {0} : there is a unital ∗-homomorphism Mpn → A}. 

Then FA : P → N ∪ {∞}. The classification of the UHF-algebras can then be stated 

as follows. 

Theorem 2.9. a) Two UHF-algebras A and B are isomorphic if and only if 

FA = FB. 

b) For every function F : P → N ∪ {∞} there is a UHF-algebra A such that 

F = FA. 

For the proof of this theorem we need some lemmas. For applications later on, 

they are formulated and proved in a setting more general than what is required for 

the present applications. 

Lemma 2.10. Let a = a∗ be a selfadjoint element of the C∗-algebra A. If a2 − 

, then there is a projection p ∈ A such that a − p ≤ 2δ. 

a ≤ δ < 1 

4


Proof. Since σ(a 2 − a) = {t 2 − t : t ∈ σ(a)} ⊆ [−δ, δ] ⊆ [− 1 

4 

1 , ] an elementary 

4 

investigation of the function t ↦→ t2 −t shows that σ(a) ⊆ [−2δ, 2δ] ∪[1 −2δ, 1+2δ]. 

Thus the characteristic function 1 1 

[ ,2] for the interval [1, 

2] is continuous on σ(a). 

2 2 

Hence p = 1 1 

[ 2 ,2](a) is a projection in A and since |1 [ 1 ,2](t) − t| ≤ 2δ, t ∈ σ(a), we 

2 

see that p − a ≤ 2δ. 

 

Lemma 2.11. When p, q are projections in the C∗-algebra A, x ∈ A, x ≤ 1, 

and 

xx ∗ − p ≤ δ < 1 

3 , x∗x − q ≤ δ < 1 

3 

then there is a partial isometry v ∈ A such that vv∗ = p, v∗v = q and x−v ≤ 4 √ δ. 

Proof. Set a = pxq and note that 

aa ∗ − p = pxqx ∗ p − p ≤ px(q − x ∗ x)x ∗ p + pxx ∗ xx ∗ p − p 

≤ δ + pxx ∗ xx ∗ p − p ≤ δ + p(xx ∗ − p)xx ∗ p + p 2 xx ∗ p − p 

≤ 2δ + pxx ∗ p − p = 2δ + p(xx ∗ − p)p ≤ 3δ < 1 . 

Similarly, a∗a − q ≤ 3δ < 1. Hence, by Lemma 1.10, aa∗ is invertible in the 

unital C∗-algebra pAp and a∗a is invertible in the unital C∗-algebra qAq. We 

set v = (aa∗ ) −1 2a where the functional calculus is done within pAp. Then vv∗ = 

(aa∗ ) −1 2aa∗ (aa∗ ) −1 

2 = p. In particular, v∗v is a projection by Lemma 1.55. Note 

that v ∗ v ∈ qAq so that qv ∗ v = v ∗ v. On the other hand v ∗ v = a ∗ (aa ∗ ) −1 a and 

v ∗ va ∗ a = a ∗ (aa ∗ ) −1 aa ∗ a = a ∗ pa = a ∗ a. In combination these identities show 

that (q − v ∗ v)a ∗ a = 0. Since a ∗ a is invertible in qAq, this implies that q = v ∗ v. 

To estimate x − v note first that 

v − a 2 = (aa ∗ ) −1 2a − a 2 = ((aa ∗ ) −1 

2 − p)a 2 

= [(aa ∗ ) −1 

2 − p]aa ∗ [(aa ∗ ) −1 

2 − p] = sup 

t∈σ(aa ∗ ) 

≤ sup 

t∈σ(aa ∗ ) 

= p − aa ∗ ≤ 3δ . 

Furthermore, 

|1 − 2 √ t + t| 

|1 − t| (since 0 ≤ 1 − 2 √ t + t ≤ 1 − t on σ(aa ∗ ) ⊆ [0, 1]) 

x−px 2 = (1−p)xx ∗ (1−p) ≤ (1−p)xx ∗ (1−p)−(1−p)p(1−p)+(1−p)p(1−p) ≤ δ , 

and similarly, x − xq 2 = (1 − q)x ∗ x(1 − q) ≤ δ. Hence 

x − a = x − pxq ≤ x − px + p(x − xq) ≤ 2 √ δ . 

In total, v − x ≤ v − a + a − x ≤ √ 3δ + 2 √ δ ≤ 4 √ δ. 


A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . (2.10) 



C ∗ -algebra. 

If p1, p2, . . .,pN are mutually orthogonal projections in A∞ and ǫ > 0, then there 

is an m ∈ N and orthogonal projections q1, q2, . . ., qN ∈ Am such that ϕ∞,m(qi) − 

pi < ǫ, i = 1, 2, . . ., N.


If the sequence (2.10) is unital and N 

i=1 pi = 1, then the qi’s can be chosen such 

that, in addition, N 

i=1 qi = 1. 

Proof. We use induction in N. The argument for the induction start, N = 1, 

is contained in the proof of the induction step and hence we only give the latter. So 

assume that the assertion holds for N − 1 projections and consider N orthogonal 

projections p1, p2, . . ., pN in A∞. Choose δ > 0 such that 22δ < 1 

4 

and 48δ < 

ǫ. There is then, by induction hypothesis, a k ∈ N and orthogonal projections 

q1, q2, . . .,qN−1 ∈ Ak such that 

ϕ∞,k(qi) − pi < δ 

, i = 1, 2, . . ., N − 1. 

N 

Choose n ≥ k and a ∈ An such that 

ϕ∞,n(a) − pN < δ 

N . 

By substituting 1 

2 (a + a∗ ) for a we may assume that a is selfadjoint. Note that 

lim 

k→∞ ϕk,n(a) < pN + δ ≤ 1 + δ 

so upon substituting a ∈ An by ϕl,n(a) ∈ Al for some l ≥ n, we can assume that 

a ≤ 1 + δ ≤ 2. 

Set Q = q1 + q2 + · · · + qN−1 ∈ Ak. Note that 

so that 

Furthermore, 

ϕ∞,k(Q)ϕ∞,n(a) = ϕ∞,n(a)ϕ∞,k(Q) < δ 

ϕ∞,k(Q)ϕ∞,n(a) + ϕ∞,n(a)ϕ∞,k(Q) − ϕ∞,k(Q)ϕ∞,n(a)ϕ∞,k(Q) < 3δ. 

ϕ∞,n(a) 2 − ϕ∞,n(a) ≤ 

ϕ∞,n(a) 2 − ϕ∞,n(a)pN + ϕ∞,n(a)pN − pN + ϕ∞,n(a) − pN < 4δ. 

Choose m ≥ n such that 

and 

Set 

ϕm,k(Q)ϕm,n(a) + ϕm,n(a)ϕm,k(Q) − ϕm,k(Q)ϕm,n(a)ϕm,k(Q) < 3δ. 

ϕm,n(a) 2 − ϕm,n(a) < 4δ. 

b = ϕm,n(a) − ϕm,k(Q)ϕm,n(a) − ϕm,n(a)ϕm,k(Q) + ϕm,k(Q)ϕm,n(a)ϕm,k(Q) 

and note that b ≤ 2 + 3δ ≤ 3. Furthermore, 

b 2 − b ≤ 18δ + ϕ∞,n(a) 2 − ϕ∞,n(a) < 22δ, 

ϕ∞,m(b) − pN < 4δ, 

(2.11) 

and b ∈ {x ∈ Am : ϕm,k(Q)x = xϕm,k(Q) = 0}. The latter set is a C ∗ -subalgebra 

of Am so we can use Lemma 2.10 to conclude that there is a projection rN ∈ Am 

such that b − rN ≤ 44δ and ϕm,k(Q)rN = rNϕm,k(Q) = 0. Set ri = ϕm,k(qi), i = 

1, 2, . . ., N −1. Then r1, r2, . . ., rN are orthogonal projections in Am and ϕ∞,m(ri)− 

pi < 48δ < ǫ for all i = 1, 2, . . ., N. This completes the proof of the induction step 

and hence the proof in the non-unital case.


When (2.10) is unital and N 

i=1 pi = 1 we first conclude from the non-unital 

case that there is an k ∈ N and orthogonal projections r1, r2, . . .,rN ∈ Ak such that 

}. Then 

pi − ϕ∞,k(ri) < min{ǫ, 1 

N+1 

so for some m ≥ k 

1 − ϕ∞,k( 

N 

i=1 

ϕm,k(1 − 

ri) < N 

N + 1 

N 

ri) < 1 

i=1 

which implies that 1 = N 

i=1 ϕm,k(ri). Then qi = ϕm,k(ri), i = 1, 2, . . ., N, will have 

the desired properties. 

Proposition 2.13. Let 

A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . (2.12) 



C ∗ -algebra. If {e d ij : i, j = 1, 2, . . ., nd, d = 1, 2, . . ., N} is a set of matrix units in A∞ 

and ǫ > 0, then there is an M ∈ N and a set {f d ij : i, j = 1, 2, . . ., nd, d = 1, 2, . . ., N} 

of matrix units in AM such that 

ϕ∞,M(f d ij ) − edij < ǫ 

for all i, j = 1, 2, . . ., nd, d = 1, 2, . . ., N. 

If (2.12) is a unital sequence of C∗-algebras and N can be chosen such that N nd d=1 i=1 fd ii = 1. 

d=1 

nd 

i=1 ed ii = 1 then {f d ij} 

Proof. We first choose δ > 0 so small that 5δ < 1 

3 and 24√ δ + 4δ < ǫ. 

By Lemma 2.12 there is an m ∈ N and orthogonal projections q d ii ∈ Am, i = 

1, 2, . . ., nd, d = 1, 2, . . ., N, such that 

ϕ∞,m(q d ii ) − edii < δ 

for all d, i (and 

i,d qd ii = 1 when (2.12) is a unital sequence). For each d, j, choose 

wd j ∈ Ak, k ≥ m, such that 

ϕ∞,k(w d j ) − ed1j < δ. 

Since e d 1j ≤ 1 and limr→∞ ϕr,k(w d j) < e d 1j + δ ≤ 1 + δ, there is an r ≥ k such 

that 

ϕr,k(w d j) ≤ 1 + δ 

for all d, j. Set v d j = (1 + δ)−1 ϕr,k(w d j ) ∈ Ar and note that 

for all d, j. It follows that 

and 

for all d, j. Thus 

ϕ∞,r(v d j ) − ed 1j < (1 − (1 + δ)−1 )(1 + δ) + δ = 2δ 

ϕ∞,r(v d j)ϕ∞,r(v d j) ∗ − ϕ∞,m(q d 11) < 4δ + δ 

ϕ∞,r(v d j )∗ϕ∞,r(v d j ) − ϕ∞,m(q d jj ) < 4δ + δ

and 


ϕM,r(v d j )ϕM,r(v d j )∗ − ϕM,m(q d 11 ) ≤ 5δ 

ϕM,r(v d j )∗ϕM,r(v d j ) − ϕM,m(q d jj ) ≤ 5δ 

for some M ≥ r and all d, j. By Lemma 2.11 there is therefore a partial isometry 

sd j ∈ AM such that sd jsd ∗ 

j = ϕM,m(qd 11), sd∗ d 

j sj = ϕM,m(qd jj) and sd j − ϕM,r(vd j ) ≤ 

4 √ 5δ ≤ 12 √ δ. It follows that 

f d ij = sd ∗ d 

i sj , i, j = 1, 2, . . ., nd, d = 1, 2, . . ., N, 

is a set of matrix units in AM (with N nd d=1 i=1 fd ii = 1 when (2.12) is a unital 

sequence) such that 

ϕ∞,M(f d ij ) − ed ij ≤ 24√ δ + 4δ < ǫ 

for all d, i, j. 

Corollary 2.14. Let 

Mn1 

ϕ1 

Mn2 

ϕ2 

Mn3 

ϕ3 

. . . 

be a unital sequence of matrix algebras and A∞ = lim 

−→ (Mni , ϕi) the corresponding 

UHF-algebra. For any k ∈ N, there is a unital ∗-homomorphism ψ : Mk → A∞ if 

and only if k|ni for some i ∈ N. 

Proof. If k|ni there is a unital ∗-homomorphism µ : Mk → Mni , e.g. the 

standard homomorphism given by the 1×1 matrix ( ni 

k ). Then ψ = ϕ∞,ni ◦µ : Mk → 

A∞ is a unital ∗-homomorphism. 

Conversely, if ψ : Mk → A∞ is a unital ∗-homomorphism, let {eij} be the 

standard system of matrix units in Mk. Then {ψ(eij)} is a system of matrix units 

in A∞ with k 

i=1 ψ(eii) = 1 and by Proposition 2.13 there is a set {fij : i, j = 

1, 2, . . ., k} of matrix units in Mni for some i, such that k 

i=1 fii = 1 in Mni 

. We 

therefore obtain a unital ∗-homomorphism µ : Mk → Mni such that µ(eij) = fij 

for all i, j. By Lemma 2.6 there is then also a unital standard homomorphism 

Mk → Mni . But this is only possible if k|ni, cf. condition (2.9). 

Proof. Proof of Theorem 2.9 a) : By assumption A = lim(Mni 

−→ , ϕi) and B = 

lim 

−→ (Mmi , ψi) are the inductive limits of the two unital sequences (Mni , ϕi) and 

(Mmi , ψi), respectively. We will construct by induction two sequences k1 < k2 < . . . 

and l1 < l2 < . . . in N and unital ∗-homomorphisms µi : Mnk i → Mml i and 

λi : Mml i → Mnk i+1 such that 

ϕk2 ,k ϕk 1 3 ,k ϕk 

Mnk 

2 4 ,k 

Mnk 

3 

Mnk 1 

2 

3 

µ1 

 

Mml 1 

 

λ1 

 

 

 

 

ψl 2 ,l 1 

µ2 

 

Mml 2 

 

λ2 

 

 

 

 

ψl 3 ,l 2 

µ3 

 

Mml 3 

 

· 

· · 

 

λ3 

 

 

 

 

· · · 

ψl 4 ,l 3 

commutes. So assume that we have constructed everything up to


ϕkr,kr−1 Mnk 

Mnkr 

r−1 

µr−2 

 

 

λr−1 

 

 

 

 

µr−1 

Mml 

 

Mmlr r−1 

ψlr,lr−1 . 

Write mlr as a product of prime powers, say as 

mlr = p x1 

1 px2 2 · · ·pxs s . 

Then FB(pi) ≥ xi, i = 1, 2, . . ., s, by Corollary 2.14. But FA(pi) = FB(pi) by 

assumption, so FA(pi) ≥ xi for all i. There are therefore unital ∗-homomorphisms 

Mpi xi → A for all i. Another application of Corollary 2.14, now to A, shows that 

there are numbers ai ∈ N such that p xi 

i |nai for all i = 1, 2, . . ., s. Since nj|nj+1 for 

all j there is a number kr+1 > kr such that p xi 

i |nkr+1 for all i. But then 

mlr|nkr+1 

and hence there is a unital ∗-homomorphism λ : Mmlr → Mnk r+1 , for example the 

standard homomorphism determined by the 1 × 1 matrix ( nkr+1 ). It follows from 

mlr 

Lemma 2.6 that there is a unitary v ∈ Mnk such that Ad v ◦ λ ◦ µr−1 = ϕkr+1,kr. 

r+1 

We set λr = Ad v ◦ λ. 

To construct lr+1 and µr+1 we repeat the previous argument with the roles of 

A and B exchanged : Because FA = FB we see that there is a lr+1 > lr such that 

nkr+1|mlr+1 and hence such that there is a unital ∗-homomorphism µ : Mnk → 

r+1 

Mml r+1 . Again we conclude from Lemma 2.6 that there is a unitary w ∈ Mml r+1 

such that Ad w ◦ µ ◦ λr = ψlr,lr+1 and we set µr+1 = Ad w ◦ µ. Thus we get the 

following commutative diagram 

ϕkr+1 ,kr 

Mnkr 

 

Mnkr+1 µr 

 

 

λr 

 

 

 

 

µr+1 

Mmlr 

 

Mml . 

ψl 

r+1 

r+1 ,lr 

This completes the induction step. 

Having the infinite commutative diagram above we conclude from Proposition 

2.2 that A ≃ B. 

b) Let p1 < p2 < p3 < . . . be a numbering of P. For each n ∈ N we choose 

a n 1 ≤ a n 2 ≤ . . . in N such that limj→∞ a n j = F(pn). Set 

ri = p a1 i 

1 p a2 i 

2 · · ·p ai i 

i . 

Then ri|ri+1 and hence there is a unital ∗-homomorphism ϕi : Mri → Mri+1 . Let A 

be the inductive limit of the sequence 

Mr1 

ϕ1 

Mr2 

ϕ2 

Mr3 

ϕ3 

. . . 

It is straightforward to check, by use of Corollary 2.14, that F = FA.


It follows from Theorem 2.9 that there are uncountably many mutually nonisomorphic 

UHF-algebras. The invariant FA is sometimes refered to as a supernatural 

number because FA can be represented by a formal infinite product of prime powers: 

p a1 

1 p a2 

2 p a3 

3 · · · 

where ai = FA(pi). In this terminology the UHF-algebra with Bratteli diagram 

is called the UHF-algebra of type 3 ∞ . 

· 

· 

· 

· 

. 

It is not always clear at first glance if a given C ∗ -algebra is an AF-algebra. It is 

therefore often useful to have a criterion which does not refer to inductive limits or 

increasing sequences of subalgebras. We will conclude this first introduction to AFalgebras 

by giving such a criterion. A C ∗ -algebra A is separable when it is separable 

as a metric space, i.e. if it contains a dense sequence. It is easy to see that a finitedimensional 

C ∗ -algebra is separable and that the inductive limit of a sequence of 

separable C ∗ -algebras is again separable. So an AF-algebra is certainly separable. 

This condition is not enough, however, as for example the case A = C[0, 1] shows. 

Lemma 2.15. Let n1, n2, . . .,nN ∈ N and ǫ > 0 be given. There is then a δ > 0 

with the following property : When {f d ij : i, j = 1, 2, . . ., nd, d = 1, 2, . . ., N} are 

matrix units in a C∗-algebra A and B is a C∗-subalgebra of A containing elements 

{xd ij } such that xdij − fd ij < δ for all i, j, d, then there are matrix units {edij : i, j = 

1, 2, . . ., nd, d = 1, 2, . . ., N} in B such that 

x d ij − edij < ǫ 

for all i, j, d and a unitary u ∈ Â such that 

and 

for all i, j, d. 

u − 1 < ǫ 

uf d ij u∗ = e d ij 

Proof. The following is only a sketch of the proof. In Exercise 2.30 below the 

reader is asked to provide the missing details. 

By substituting each xd ii by 1 

2 (xdii+x d ∗ d 

ii ), we may assume that each xii is selfadjoint. 

If δ is small enough, Lemma 2.10 gives us a projection E ∈ B close to 

Ex d ij E will be close to fd ij for all i, j, d. So we may assume that xd ij 

i,d fd ii. Then 

∈ EBE for all


i, j, d. If δ is small enough Lemma 2.10 gives us projections ed 11 ∈ EBE close to xd11 and hence also close to fd 11 . Then (E − ed11 )xd22 (E − ed11 ) is close to fd 22 and hence, if 

δ is small enough, Lemma 2.10 will give us a projection ed 22 ∈ (E − ed11 )A(E − ed 11 ) 

close to (E − ed 11)xd 22(E − ed 11) and hence close to fd 22. Note that ed 11 and ed 22 are 

orthogonal. Then (E − ed 11 − ed 11)xd 33(E − ed 11 − ed 11) is close to fd 33 and we get in the 

same way a projection ed 33 ∈ B, orthogonal to both ed11 and ed22 , such that ed33 is close 

to fd 33 . By proceding in this way we get projections edii ∈ B such that edii ed′ kk = 0 

unless k = i, d = d ′ , and such that ed ii is close to fd ii. Furthermore, provided δ is 

small enough, Lemma 2.11 gives us partial isometries vd i ∈ B close to xd1i such that 

vd i vd∗ d 

i = eii, vd∗ d 

i vi = ed 11 for all i. Set ed ij = vd i vd∗ d 

j ∈ B and note that {eij} is a set 

of matrix units in B such that each ed ij is close to fd ij , say ed ij − fd ij < ǫ, for all 

i, j, d, if just δ is small enough. In particular, if only fd 11 − ed11 is small enough for 

all d, Lemma 2.11 (applied with x = ed 11fd 11 ) gives us partial isometries wd ∈ B such 

that wd − fd 11 is small and wdwd ∗ = ed 11, wd ∗wd = fd 11 for all d. Furthermore, since 

1 − 

i,d fd ii − (1 − 

i,d ed ii) is small, a similar argument provides us with a partial 

isometry v ∈ Â such that vv∗ = 1− 

i,d ed ii , v∗ v = 1− 

i,d fd ii 

and v−(1− 

i,d ed ii ) 

is small. Set u = 

i,d ed i1 wdf d 1i + v. Then u is a unitary in Â such that ufd ij u∗ = e d ij 

for all i, j, d. If only δ was small enough to begin with, we get in addition that 

u − 1 < ǫ. 

Theorem 2.16. A separable C ∗ -algebra A is an AF-algebra if and only if the 

following holds: 

For every finite set F ⊆ A and every ǫ > 0, there is a finite dimensional C ∗ - 

subalgebra B in A such that 

dist(x, B) < ǫ 

for all x ∈ F. 

Proof. It is clear from the definition of an AF-algebra that the condition is 

necessary: Since A = 

n An, where each An is finite dimensional, there is for each 

x ∈ F an integer nx ∈ N such that dist(x, Anx) < ǫ. Set n = maxx nx. Then 

dist(x, An) < ǫ for all x ∈ F. 

The problem is to prove the sufficiency of the condition. To this end we choose 

a dense sequence {yn} in A. Set xn = (max{yn, 1}) −1 yn, n ∈ N. Then {xn} is 

a dense sequence in the unit ball of A. We will construct a sequence F1 ⊆ F2 ⊆ 

F3 ⊆ . . . of finite dimensional C∗-subalgebras of A such that dist(xj, Fn) < 1 

, j = 

n 

1, 2, . . ., n, n ∈ N. This will complete the proof because it will then be obvious that 

A = 

n Fn. 

F1 is found by using the assumption on A with F = {x1} and ǫ = 1. Assume 

that F1, F2, . . ., Fn have been constructed. Let {fd ij : i, j = 1, 2, . . ., nd, d = 

1, 2, . . ., N} be a set of matrix units which span Fn. Choose δ > 0 as in Lemma 

. By assumption there is a 

2.15, corresponding to {n1, n2, . . .,nN} and ǫ = 1 

5(n+1) 

finite dimensional C∗-subalgebra F of A such that dist(x, F) < min{δ, ǫ} for all 

x ∈ {x1, x2, . . .,xn+1} ∪ {fd ij : i, j = 1, 2, . . ., nd, d = 1, 2, . . ., N}. By Lemma 2.15 

there are matrix units {ed ij : i, j = 1, 2, . . ., nd, d = 1, 2, . . ., N} in F and a unitary 

1 

u ∈ Â such that u − 1 < ǫ = 5(n+1) such that ufd iju∗ = ed ij for all i, j, d. Set 

Fn+1 = u∗Fu. Then Fn+1 is a finite dimensional C∗-subalgebra of A such that 

Fn ⊆ Fn+1. Furthermore, for every j ∈ {1, 2, . . ., n + 1}, there is an element fj ∈ F


such that fj − xj < ǫ. Then yj = u∗fju ∈ Fn+1 and yj = fj ≤ xj + ǫ ≤ 2. 

In addition yj − xj ≤ u∗fju − fj + ǫ < 5ǫ = 1 for all j = 1, 2, . . ., n + 1. This 

n+1 

completes the induction step and hence the proof. 

Corollary 2.17. An inductive limit of a sequence of AF-algebras is itself an 

AF-algebra. 

Proof. Let 

A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

· · · 

be a sequence of AF algebras and let A∞ be the inductive limit of the sequence and 

ϕn,∞ : An → A∞ the canonical maps. If {x n i : i ∈ N} is a dense sequence in An, 

then {ϕ∞,n(x n i ) : i, n ∈ N} is a countable dense set in A∞. Hence A∞ is separable. 

Let {x1, x2, . . .,xm} be a finite set in A∞ and let ǫ > 0. Since 

n ϕ∞,n(An) is 

dense in A∞ there is an N ∈ N and elements yi ∈ AN such that ϕ∞,N(yi) − xi < 

ǫ 

2 , i = 1, 2, . . .,m. Since AN is AF, there is a finite dimensional C∗-subalgebra F 

of AN containing elements zi ∈ F such that zi − yi < ǫ, 

i = 1, 2, . . ., m. Then 

2 

ϕ∞,N(F) is a finite dimensional C∗-subalgebra of A∞ and xi − ϕ∞,N(zi) < ǫ, i = 

1, 2, . . ., m. Hence A∞ satisfies the criterion of Theorem 2.16 and is consequently 

an AF algebra. 

2.3. Exercises. 

Exercise 2.18. Let (An, ϕn) be a sequence of C ∗ -algebras. Assume that there 

is a C ∗ -algebra D and a sequence of ∗-homomorphisms λn : An → D such that 

λn+1 ◦ ϕn = λn for all n and with the universal property of Lemma 2.1, i.e. such 

that whenever ψn : An → B, n ∈ N, is a sequence of ∗-homomorphisms into a 

common C ∗ -algebra B, satisfying that ψn+1 ◦ ϕn = ψn for all n, then there is one 

and only one ∗-homomorphism ψ : D → B such that ψ ◦ λn = ψn for all n ∈ N. 

Prove that there is a ∗-isomorphism Φ : D → A∞ such that Φ ◦ λn = ϕ∞,n for 

all n ∈ N. 

Exercise 2.19. Let (An, ϕn) and (Bn, ψn) be two sequences of C ∗ -algebras. 

Assume that there are sequences k1 < k2 < k3 < . . . and m1 < m2 < m3 < . . . in 

N and ∗-homomorphisms µi : Ami → Bki such that µi+1 ◦ ϕmi+1,mi = ψki+1,ki ◦ µi for 

all i ∈ N, i.e. such that the infinite diagram 

ϕm2 ,m1 ϕm3 ,m 

Am1 

2 

Am2 

Am3 

 

µ1 

Bk1 ψk 2 ,k 1 

 

µ2 

 

µ3 

ϕm 4 ,m 3 

. . . 

 

Bk2 

 

Bk3 

ψk3 ,k ψk 2 4 ,k3 commutes. 

Show that there is a unique ∗-homomorphism µ : A∞ → B∞ such that µ◦ϕ∞,mi = 

ψ∞,ki ◦ µi for all i ∈ N. 

Exercise 2.20. Let (An, ϕn) be a unital sequence of C∗-algebras. Show that 

A∞ = lim(An, 

ϕn) is unital with unit 1 = ϕ∞,1(1). 

−→ 

Exercise 2.21. Set An = C[0, 1 

n ], n ∈ N, and define ϕn : C[0, 1 

n ] → C[0, 1 

n+1 ] 

by 

Determine lim 

−→ (An, ϕn). 

ϕn(f) = f| [0, 1 

n+1 ]. 

 

. . .


Exercise 2.22. Let p, q be projections in the unital C∗-algebra A such that 

p − q < 1. Show that there is a unitary u ∈ A such that upu∗ = q. (Hint : Set 

u = v(v∗v) −1 

2 where v = 1 − p − q + 2pq.) 

Exercise 2.23. Let X = {0} ∪ { 1 : n ∈ N, n ≥ 1}, considered as a compact 

n 

subset of [0, 1]. Show that C(X) is a unital AF-algebra with Bratteli diagram 

etc. 

· 

 

 

· · 

 

 

· · · 

 

 

· · · · 

 

Exercise 2.24. Let F1 = [0, 1], F2 = [0, 1 

. 

. 

. 

. 

3 ]∪[2 

. 

3 , 1], F3 = [0, 1 

9 ]∪[2 1 , 9 3 ]∪[2 7 , 3 9 ]∪[8 , 1], 9 

Then K = ∞ 

n=1 Fn is the (middle third) Cantor set. Show that C(K) is a unital 

AF-algebra with Bratteli diagram 

· 

 

 

· · 

 

 

 

 

 

 

 

 

 

 

 

. . . . 

Exercise 2.25. Let A be the unital AF-algebra with Bratteli diagram


· 

 

 

 

· · · 

 

 

· 

 

 

 

· · · 

 

 

· 

 

 

 

· · · 

 

 

· 

 

 

. . . 

Show that A is isomorphic to the UHF-algebra of type 3 ∞ . 

Exercise 2.26. Let A be the AF-algebra with Bratteli diagram 

1 

 

2 

 

3 

 

4 

 

5 

 

. 

Show that A ≃ K, the compact operators on an infinite dimensional separable 

Hilbert space


Exercise 2.27. Let A be the unital AF-algebra with Bratteli diagram 

· 

 

 

 

· 

· 

 

 

 

 

 

· 

· 

 

 

 

 

 

· 

· 

 

 

 

 

 

· · 

 

 

 

 

 

 

. . . 

Show that A is isomorphic to the UHF-algebra of type 2 ∞ . 

Exercise 2.28. Let A be the unital AF-algebra with diagram 

· 

 

 

 

· 

· 

 

 

 

 

 

· 

· 

 

 

 

 

 

· 

· 

 

 

 

 

 

· · 

 

 

 

 

 

 

. . . 

Show that A is not isomorphic to a UHF-algebra, for example by showing that A 

does not contain a full matrix algebra Mn for any n ≥ 2 as a unital C ∗ -subalgebra. 

Exercise 2.29. Show that the C ∗ -algebra C[0, 1] is separable but not an AFalgebra. 

Exercise 2.30. Fill in the details missing in the proof of Lemma 2.15. 

Exercise 2.31. Let A be a unital separable C ∗ -algebra with the following property 

: For any finite set F ⊆ A and for any ǫ > 0 there is a C ∗ -subalgebra B ⊆ A 

such that B is ∗-isomorphic to Mn for some n ∈ N and dist(x, B) < ǫ for all x ∈ F. 

Show that A is a UHF-algebra.

y 

3. K0 AND THE CLASSIFICATION OF AF-ALGEBRAS 61 

Exercise 2.32. Define a unital ∗-homomorphism ϕn : C([0, 1], M2 n) → C([0, 1], M 2 n+1) 

ϕn(f)(t) = 

 

t f( ) 0 

2 

0 f( t+1 

2 ) 

Show that A∞ = lim(C([0, 

1], M2 −→ n), ϕn) is an AF-algebra. (Hint : Let a ∈ C([0, 1], M2n+1). Show that for every ǫ > 0, there is a k ≥ n such that ϕk,n(a) ∈ C([0, 1], M2k+1) is 

within the distance ǫ from an element of M2k+1 ⊆ C([0, 1], M2k+1). Then use Theorem 

2.16.) Try to show that A∞ is a UHF-algebra of type 2∞ . 

 

, t ∈ [0, 1] . 

3. K0 and the classification of AF-algebras 

3.1. Definition of K0. Let A be a C ∗ -algebra. 

Definition 3.1. Two projections p, q ∈ A are equivalent when there is a partial 

isometry v ∈ A such that vv ∗ = p, v ∗ v = q. We write p ≈v q in this case, or just 

p ≈ q when it is not necessary to specify the partial isometry. 

Lemma 3.2. 1) ≈ is an equivalence relation on the set of projections in A. 

2) When p, q, p1, q1 are projections in A such that pp1 = qq1 = 0 and p ≈ q, p1 ≈ q1, 

then p + p1 ≈ q + q1. 

3) When p, q ∈ A are projections such that p ≈ q and 1 − p ≈ 1 − q (in Â), 

then there is a unitary u ∈ Â such that upu∗ = q. 

4) When p, q ∈ A are projections in A and u ∈ Â is a unitary such that 

upu∗ = q, then p ≈ q and 1 − p ≈ 1 − q (in Â). 

Proof. 1) : Since p ≈p p, ≈ is a reflexive relation. Since p ≈v q ⇒ q ≈v∗ p, the 

relation is also symmetric. Finally, if p ≈v q, q ≈w r, then p ≈vw r. Indeed, 

while 

Thus 

vw(vw) ∗ = vww ∗ v ∗ = vqv ∗ = vv ∗ vv ∗ = p 2 = p, 

(vw) ∗ vw = w ∗ v ∗ vw = w ∗ qw = w ∗ ww ∗ w = r 2 = r. 

2): If p ≈v q and p1 ≈w q1, note first that vw ∗ = 0 because 

vw ∗ (vw ∗ ) ∗ = vw ∗ wv ∗ = vq1v ∗ = vv ∗ vq1v ∗ = vqq1v ∗ = 0. 

(v + w)(v + w) ∗ = vv ∗ + vw ∗ + wv ∗ + ww ∗ = vv ∗ + ww ∗ = p + p1. 

Similarly, (v + w) ∗ (v + w) = q + q1 because v ∗ w = 0. 

3) : If p ≈v q and 1 − p ≈w 1 − q, then 1 ≈v+w 1 by 2); i.e. v + w is a unitary 

and 

(v + w)q(v + w) ∗ = vqv ∗ + vqw ∗ + wqv ∗ + wqw ∗ = 

vv ∗ vv ∗ + vq(1 − q)w ∗ + w(1 − q)qv ∗ + w(1 − q)qw ∗ = vv ∗ vv ∗ = p 2 = p. 

4) It is straightforward to check that q ≈up p. 

Let Pn(A) denote the set of projections in Mn(A) and let P(A) denote the disjoint 

union 

∞ 

P(A) = Pn(A). 

n=1


When e ∈ Pn(A), f ∈ Pk(A) we define e ⊕ f to be the projection 

 

e 0 

0 f 

in Pn+k(A). Also, we extend the equivalence relation ≈ to P(A) by setting e≈f 

when 

e 0 f 0 

≈ 

0 0 0 0 

in Mm(A) for some m ≥ k, n. We leave the reader to check that ≈ is an equivalence 

relation also on P(A) and that it is indeed an extension of the equivalence relation 

≈ in P1(A); more precisely, that two projections p, q ∈ A are equivalent in A if and 

only if they are equivalent in P(A). 


⎛ 

⎞ 

p11 p12 p13 . . . p1n 

⎜p21 

p22 p23 . . . p2n⎟ 

⎜ 

⎟ 

p = ⎜p31 

p32 p33 . . . p3n⎟ 

⎜ 

⎝ . 

⎟ ∈ Pn(A). 

. . . .. . ⎠ 

pn1 pn2 pn3 . . . pnn 

Then 

⎛ 

p11 

⎜p21 

⎜ 

⎜p31 

⎜ . 

⎝pn1 

p12 

p22 

p32 

. 

pn2 

p13 

p23 

p33 

. 

pn3 

. . . p1n 

. . . p2n 

. . . p3n 

. .. . 

. . . pnn 

⎞ ⎛ 

0 0 0 

0⎟ 

⎜0 

p11 ⎟ ⎜ 

0⎟ 

⎜0 

p21 

⎟ 

. 

⎟ ≈ ⎜ 

⎜0 

p31 

⎟ ⎜ 

0⎠ 

⎝. 

. 

0 

p12 

p22 

p32 

. 

0 

p13 

p23 

p33 

. 

⎞ 

. . . 0 

. . . p1n ⎟ 

. . . p2n ⎟ 

. . . p3n⎟ 

. 

⎟ 

.. . ⎠ 

0 0 0 . . . 0 0 0 pn1 pn2 pn3 . . . pnn 

in Mn+1(A). 

Proof. Set 

⎛ 

⎞ 

0 0 0 . . . 0 1 

⎜1 

0 0 . . . 0 0⎟ 

⎜ 

⎟ 

⎜0 

1 0 . . . 0 0⎟ 

W = ⎜ 

⎟ 

⎜0 

0 1 . . . 0 0⎟ 

∈ Mn+1( 

⎜ 

⎝ .. 

⎟ 

. . . . . . ⎠ 

0 0 0 . . . 1 0 

Â). 

Then W is a unitary in Mn+1( Â) and a direct calculation shows that 

⎛ 

⎞ 

p11 p12 p13 . . . p1n 0 

⎜p21 

p22 p23 . . . p2n 0⎟ 

⎜ 

⎟ 

⎜p31 

p32 p33 . . . p3n 0⎟ 

W ⎜ . 

⎟ 

⎜ . . . .. . . 

⎟W 

⎟ 

⎝pn1 

pn2 pn3 . . . pnn 0⎠ 

0 0 0 . . . 0 0 

∗ ⎛ 

⎞ 

0 0 0 0 . . . 0 

⎜0 

p11 p12 p13 . . . p1n⎟ 

⎜ 

⎟ 

⎜0 

p21 p22 p23 . . . p2n⎟ 

= ⎜ 

⎟ 

⎜0 

p31 p32 p33 . . . p3n⎟ 

. 

⎜ 

⎝ 

. 

⎟ 

. . . . .. . ⎠ 

0 pn1 pn2 pn3 . . . pnn 

Lemma 3.4. 1) If e1, e2, f1, f2 ∈ P(A) with e1 ≈ e2 and f1 ≈ f2, then 

e1 ⊕ f1 ≈ e2 ⊕ f2.


2) e ⊕ f ≈ f ⊕ e for all e, f ∈ P(A). 

3) e ⊕ (f ⊕ g) ≈ (e ⊕ f) ⊕ g for all e, f, g ∈ P(A). 

Proof. 1) By assumption there is a k such that 

 

e1 0 e2 0 

≈ 

0 0 0 0 

within Mk(A) and 

 

f1 

0 

 

0 f2 

≈ 

0 0 

 

0 

, 

0 

also within Mk(A). It follows from Lemma 3.2 2) that 

⎛ ⎞ 

e1 0 0 0 

⎜ 0 0 0 0 ⎟ 

⎝ 0 0 f1 0⎠ 

0 0 0 0 

≈ 

⎛ 

e2 0 0 

⎜ 0 0 0 

⎝ 0 0 f2 

⎞ 

0 

0 ⎟ 

0⎠ 

0 0 0 0 

within M2k(A). But repeated use of Lemma 3.3 shows that 

⎛ ⎞ 

e1 0 0 0 

⎜ 0 0 0 0 ⎟ 

⎝ 0 0 f1 0⎠ 

0 0 0 0 

≈ 

⎛ ⎞ 

e1 0 0 0 

⎜ 0 f1 0 0 ⎟ 

⎝ 0 0 0 0⎠ 

0 0 0 0 

≈ e1 ⊕ f1 

and ⎛ 

e2 

⎜ 0 

⎝ 0 

0 0 

0 0 

0 f2 

⎞ 

0 

0 ⎟ 

0⎠ 

0 0 0 0 

≈ 

⎛ 

e2 

⎜ 0 

⎝ 0 

0 

f2 

0 

⎞ 

0 0 

0 0 ⎟ 

0 0⎠ 

0 0 0 0 

≈ e2 ⊕ f2 

To prove 2) observe first that we may assume that e, f ∈ Pk(A) for the same k by 

1). Set 

 

0 1 

W = ∈ M2k( 

1 0 

Â). 

W is a unitary such that 

 

e 0 

W W 

0 f 

∗ 

f 0 

= . 

0 e 

3) is a triviality since in fact e ⊕ (f ⊕ g) = (e ⊕ f) ⊕ g. 

It follows from Lemma 3.4 that set of equivalence classes in P(A) form an Abelian 

semigroup when we set 

[e] + [f] = [e ⊕ f], e, f ∈ P(A). 

This semigroup, P(A)/≈, will be denoted by V (A). We shall now apply a standard 

algebraic trick - the Grothendieck construction - to produce a group, K00(A), from 

V (A). First we describe the Grothendieck construction. 

Let S be an Abelian semi-group, i.e. S is a set endowed with a composition + 

such that s + t = t + s and (s + t) + u = s + (t + u) for all s, t, u ∈ S. Define an 

equivalence relation ∼ on S × S by 

(x, u) ∼ (y, v) ⇔ there is an element r ∈ S such that x + v + r = y + u + r.


We leave the reader to check that this is indeed an equivalence relation on S ×S. 

The equivalence class containing (x, y) ∈ S × S will be denoted by [x, y]. The set of 

equivalence classes, S × S/ ∼, will be denoted by G(S). Define a composition + in 

G(S) by 

[x, y] + [u, v] = [x + u, y + v]. 

Then G(S) is an Abelian group with [x, x] (for any x ∈ S) as neutral element and 

with −[x, y] = [y, x]. (CHECK !) There is a canonical semi-group homomorphism 

ϕ : S → G(S) given by 

ϕ(x) = [x + r, r] 

for some fixed r ∈ S. The crucial functorial property of this construction is described 

in the following 

Proposition 3.5. If H is an Abelian group and µ : S → H a semi-group 

homomorphism, then there is a unique group homomorphism ˆµ : G(S) → H such 

that ˆµ ◦ ϕ = µ. 

Proof. Note first that [x, y] = [x + r, r] − [y + r, r] for all x, y ∈ S. Thus 

G(S) = ϕ(S) − ϕ(S) and we see that any homomorphism G(S) → H is determined 

by its restriction to ϕ(S). This gives the uniqueness of ˆµ. To construct ˆµ, set 

ˆµ([x, y]) = µ(x) − µ(y), x, y ∈ S. It is straightforward to check that ˆµ is a welldefined 

group homomorphism. Since ˆµ ◦ ϕ(x) = ˆµ([x + r, r]) = µ(x + r) − µ(r) = 

µ(x), x ∈ S, the proof is complete. 

Definition 3.6. We define K00(A) to be the group G(V (A)), i.e. the group 

obtained by applying the Grothendieck construction to the semi-group V (A) of 

equivalence classes of projections in P(A). 

For every e ∈ P(A) we denote by [e] the element of K00(A) which is the image 

of [e] ∈ V (A) under the canonical homomorphism ϕ : V (A) → G(V (A)) = K00(A). 

We summarize the most basic aspects of the construction of K00(A) as follows. 

Lemma 3.7. 1) Every element of K00(A) has the form [e] − [f] for some 

e, f ∈ P(A). 

2) If e1, e2, f1, f2 ∈ P(A), then [e1] − [f1] = [e2] − [f2] in K00(A) if and only if 

there is a q ∈ P(A) such that 

e1 ⊕ f2 ⊕ q ≈ e2 ⊕ f1 ⊕ q. 

3) ([e1] − [f1]) + ([e2] − [f2]) = [e1 ⊕ e2] − [f1 ⊕ f2] for all e1, e2, f1, f2 ∈ P(A). 

4) The zero element of K00(A) is [0]. 

We now develop the basic functorial properties of the K00-construction. So let B 

be another C∗-algebra and ψ : A → B a ∗-homomorphism. ψ extends in a canonical 

way to a ∗-homomorphism Mn(A) → Mn(B) which we also denote by ψ, namely 

⎛ 

⎞ 

a11 a12 . . . a1n 

⎜a21 

a22 . . . a2n⎟ 

ψ ⎜ 

⎝ . 

⎟ 

. . .. . 

⎠ 

an1 an2 . . . ann 

= 

⎛ 

⎞ 

ψ(a11) ψ(a12) . . . ψ(a1n) 

⎜ψ(a21) 

ψ(a22) . . . ψ(a2n) ⎟ 

⎜ 

⎝ 

. 

⎟ 

. . .. . 

⎠ 

ψ(an1) ψ(an2) . . . ψ(ann) 

. 

In this way ψ induces a map ψ : P(A) → P(B) which in turn defines a map 

V (A) → V (B) given by 

[e] ↦→ [ψ(e)], e ∈ P(A).


It is clear that this is a semi-group homomorphism, so if we compose it with the 

canonical map V (B) → K00(B) we can use Proposition 3.5 to conclude that there 

is a group homomorphism 

such that 

ψ∗ : K00(A) → K00(B) 

ψ∗([e] − [f]) = [ψ(e)] − [ψ(f)], e, f ∈ P(A). 

In this way the construction of the K00-group becomes a functor from C ∗ -algebras 

to Abelian groups. In particular, 

Lemma 3.8. Let ψ : A → B and ϕ : B → C be ∗-homomorphisms between 

C ∗ -algebras then 

(ϕ ◦ ψ)∗ = ϕ∗ ◦ ψ∗. 

Although this lemma is trivial, it is the core of the usefulness of the whole 

construction because it implies immediately that two isomorphic C ∗ -algebras must 

have isomorphic K00-groups. In the rest of this section we develop the K00-theory 

for the purpose of using it for the functor which interests us more, namely K0. 

Lemma 3.9. Let ϕ, ψ : A → B be ∗- homomorphisms between C ∗ -algebras such 

that 

ϕ(A)ψ(A) = {0}. 

Then ϕ + ψ : A → B is a ∗-homomorphism and (ϕ + ψ)∗ = ϕ∗ + ψ∗ : K00(A) → 

K00(B). 

Proof. It is straightforward to check that ϕ + ψ is a ∗-homomorphism. To 

obtain the formula for (ϕ + ψ)∗ note first that if e, f are orthogonal projections in 

Mn(B) then 

in M2n(B) with 

 

e + f 0 

0 0 

v = 

≈v 

 

e f 

0 0 

 

e 0 

0 f 

So if p is a projection in Mn(A) then ϕ(p) and ψ(p) are orthogonal projections in 

Mn(B) and hence 

 

ϕ(e) 0 ϕ(e) + ψ(e) 0 

[ϕ(e)] + [ψ(e)] = [ ] = [ 

] = (ϕ + ψ)∗([e]) 

0 ψ(e) 0 0 

Thus (ϕ∗ + ψ∗)[e] = (ϕ + ψ)∗([e]) and the desired conclusion follows because of 

Lemma 3.7 1). 

 

Lemma 3.10. Let A and B be C ∗ -algebras. Then 

K00(A ⊕ B) ≃ K00(A) ⊕ K00(B). 

More precisely, if π A : A ⊕ B → A and π B : A ⊕ B → B denote the two projections, 

then the map K00(A ⊕ B) ∋ x ↦→ (π A ∗ (x), πB ∗ (x)) ∈ K00(A) ⊕ K00(B) is an 

isomorphism.


Proof. Let iA : A → A ⊕ B and iB : B → A ⊕ B be the maps iA(a) = (a, 0) 

and iB(b) = (0, b), respectively. Then π A ◦ iA = idA and π B ◦ iB = idB. If γ denotes 

the map from the statement of the lemma and µ : K00(A) ⊕ K00(B) → K00(A ⊕ B) 

the map µ(x, y) = iA∗(x) + iB∗(y), then 

And 

µ ◦ γ(x) = (iA ◦ π A )∗(x) + (iB ◦ π B )∗(x) 

= (iA ◦ π A + iB ◦ π B )∗(x) (by Lemma 3.9) 

= idA⊕B∗ (x) = idK00(A⊕B)(x) = x ( because iA ◦ π A + iB ◦ π B = idA⊕B). 

γ ◦ µ(x, y) = γ(iA∗(x) + iB∗(y)) = (π A ◦ iA)∗(x), (π B ◦ iB)∗(y) = (x, y). 

For the purpose of classifying C ∗ -algebras it is important to keep track of the 

semigroup V (A) which gave us the group K00(A). Thus we set 

K00(A) + = {[e] : e ∈ P(A)}. (3.1) 

In other words, K00(A) + is the image of V (A) in K00(A) under the canonical map 

V (A) → K00(A) coming with the Grothendieck construction. In particular, K00(A) + 

is a semi-group in K00(A) and hence it gives rise to a notion of positivity within 

K00(A), namely we write x ≤ y when y − x ∈ K00(A) + . 

To study the notion of positivity we first consider the general notion of preordered 

Abelian groups. 

Definition 3.11. A pre-ordered Abelian group G is an Abelian group endowed 

with a reflexive, transitive and translation invariant relation ≤. In other words, 

1) g ≤ g, 

2) g ≤ h, h ≤ k ⇒ g ≤ k, 

3) g ≤ h ⇒ g + k ≤ h + k. 

for all g, h, k ∈ G. 

In a pre-ordered Abelian group G the set G + = {x ∈ G : x ≥ 0} is a semigroup, 

called the positive cone in G. Note that G + determines the relation ≤ because 

x ≤ y ⇔ y − x ∈ G + . Conversely, a subsemigroup G + of G with 0 ∈ G + defines a 

relation x ≤ y on G by x ≤ y ⇔ y − x ∈ G + which makes G a pre-ordered Abelian 

group. 

A pre-ordered Abelian group is called partially ordered when G + ∩(−G + ) = {0}, 

i.e. if x ≤ y, y ≤ x ⇒ x = y. 

It is now a triviality to check that 

Lemma 3.12. K00(A) is a pre-ordered Abelian group with K00(A) + = {[e] : e ∈ 

P(A)} as positive cone. 

A homomorphism ϕ : G → H between pre-ordered Abelian groups is positive 

when ϕ(G + ) ⊆ H + . Note that the group homomorphism ϕ∗ : K00(A) → K00(B) 

induced by a ∗-homomorphism ϕ : A → B is positive. 

We will now construct the inductive limits in the category of pre-ordered Abelian 

groups. Let 

G1 

p1 

G2 

p2 

G3 

p3 

. . .


be a sequence of pre-ordered Abelian groups with positive connecting homomorphisms. 

Two elements g ∈ Gn, h ∈ Gm are equivalent, written g ∼ h, in the disjoint 

union 

∞ 

i=1 

when pk,n(g) = pk,m(h) for some k ≥ n, m. The equivalence classes of this equivalence 

relation form an Abelian group G with composition 

[g] + [h] = [pk,n(g) + pk,m(h)], k ≥ n, m, g ∈ Gn, h ∈ Gm. 

The homomorphisms qn : Gn → G given by qn(g) = [g] will be called the canonical 

homomorphisms . We set 

G + ∞ 

= qn(G + n) 

n=1 

which is a semigroup in G and we define the relation ≤ on G by x ≤ y ⇔ y−x ∈ G + . 

Proposition 3.13. (a) G is a pre-ordered Abelian group and, if each Gn is 

partially ordered, then so is G. 

(b) G is the inductive limit of the sequence (Gn, pn) in the category of pre-ordered 

Abelian groups, i.e. if H is a pre-ordered Abelian group and ri : Gi → H, i ∈ N, 

are positive group homomorphisms such that rn+1 ◦ pn = rn for all n, then there is 

a unique positive group homomorphism r : G → H such that r ◦ qn = rn. 

Gi 

Proof. (a) It is straightforward to check that G is a pre-ordered Abelian group. 

We show that G is partially ordered if each Gn is. So assume that x ∈ G + ∩(−G + ). 

It follows that there is a k ∈ N and elements g, h ∈ G + 

k 

x = qk(g), −x = qk(h). 

such that 

But then for some n ≥ k, −pn,k(h) = pn,k(g) ∈ G + n ∩ (−G+ n ) = {0} because Gn is 

partially ordered. Hence x = qn(pn,k(g)) = 0. 

(b) Since rn+1 ◦ pn = rn, we can define a homomorphism r : G → H by 

r([g]) = rn(g), g ∈ Gn, n ∈ N. Since r(qn(G + n)) = rn(G + n) ⊆ H + , r is a positive 

homomorphism. Since G = ∞ 

n=1 qn(Gn), it is clear that r is uniquely determined 

by the condition that r ◦ qn = rn for all n. 

Note that the preceding construction gives the inductive limit of a sequence of 

Abelian groups (not necessarily pre-ordered) because we can always consider an 

Abelian group as being pre-ordered by the positive cone {0}. 

To describe the relation between inductive limits of C ∗ -algebras and inductive 

limits of pre-ordered Abelian groups we first need some lemmas. The first is an 

immediate corollary of Lemma 2.11. 

Lemma 3.14. Let p, q be projections in the C∗-algebra A. Assume that p−q ≤ 

δ < 1 

3 . It follows that there is a partial isometry v ∈ A such that vv∗ = p, v∗v = q 

and v − pq ≤ 4 √ δ. 

Proof. Set x = pq and note that xx ∗ − p = pqp − p = p(q − p)p ≤ 

q − p, x ∗ x − q = q(p − q)q ≤ q − p. Apply Lemma 2.11.



A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . 

be a sequence of C ∗ -algebras and A = lim 

−→ (An, ϕn) the inductive limit C ∗ -algebra. 

Let p be a projection in Md(A) for some d ∈ N. For any ǫ > 0 there is an m ∈ N 

and a projection q ∈ Md(Am) such that 

ϕ∞,m(q) − p < ǫ . 

Proof. For any δ > 0 we can find i ∈ N and y ∈ Md(Ai) such that 

ϕ∞,i(y) − p < δ. 

By using 1 

2 (y + y∗ ) instead of y we may assume that y = y∗ . If δ ≤ 1 we find 

that ϕ∞,i(y) ≤ 2 and ϕ∞,i(y2 − y) < 4δ. By using (1.25) this implies that 

ϕ∞,i((y2 − y)kl) < 4δ for all k, l ∈ {1, 2, . . ., d}. There is therefore an m > i such 

that ϕm,i((y2 − y)kl) < 4δ for all k, l. But then (1.25) implies that ϕm,i(y2 − 

y) < 4d 3 

2δ. Thus, if 4d 3 

2δ < 1 

4 , we have a projection q ∈ Md(Am) such that 

q − ϕm,i(y) < 8d 3 

2δ by Lemma 2.10. It follows that 

ϕ∞,m(q) − p < (8d 3 

2 + 1)δ. 

We have completed the proof if (8d 3 

2 + 1)δ < ǫ. 

Theorem 3.16. Let 

A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . 

be a sequence of C∗-algebras and A = lim(An, 

ϕn) the inductive limit C 

−→ ∗-algebra. Let 

K00(A1) ϕ1∗ 

K00(A2) ϕ2∗ 

K00(A3) ϕ3∗ 

. . . 

be the sequence of pre-ordered Abelian groups obtained by applying the K00-functor 

and let G = lim(K00(An), 

ϕn∗) be the inductive limit in the category of pre-ordered 

−→ 

Abelian groups and qn : K00(An) → G the canonical homomorphisms. 

There is then a unique positive homomorphism q : G → K00(A) such that q◦qn = 

ϕ∞,n∗ for all n ∈ N and q is an isomorphism in the category of pre-ordered Abelian 

groups (i.e. q is a group isomorphism such that both q and q−1 are positive.) 

Proof. Note that ϕ∞,n+1 ∗ ◦ ϕn∗ = ϕ∞,n ∗ for all n, so the existence of a unique 

positive homomorphism q : G → K00(A) with the property that q ◦ qn = ϕ∞,n ∗ for 

all n follows from Proposition 3.13 (b). 

Let x ∈ K00(A) + so that x = [e] for some projection e ∈ Mn(A). By Lemma 

3.14 and Lemma 3.15 there is an m ∈ N and a projection p ∈ Mn(Am) such that 

e ≈ ϕ∞,m(p). Hence 

x = [e] = [ϕ∞,m(p)] = ϕ∞,m ∗ ([p]) = q(qm([p])), 

proving that q(G + ) = K00(A) + . Since K00(A) = K00(A) + − K00(A) + by Lemma 3.7 

1), this shows also that q is surjective. To show that q is a positive isomorphism 

with positive inverse it is now sufficient to prove that q is injective. 

Consider x ∈ ker q. Since G = 

i qi(K00(Ai)), there are n, i ∈ N and projections 

p1, q1 ∈ Mn(Ai) such that x = qi([p1] − [q1]). Since 

0 = q(x) = ϕ∞,i ∗ ([p1] − [q1]) = [ϕ∞,i(p1)] − [ϕ∞,i(q1)]


we know from Lemma 3.7 2) that there is a projection f ∈ P(A) such that ϕ∞,i(p1)⊕ 

f ≈ ϕ∞,i(q1)⊕f. From the argument above we know that there is a j ≥ i, an m ∈ N 

and a projection h ∈ Mm(Aj) such that f ≈ ϕ∞,j(h) in Mm(A). Set k = n + m and 

p = ϕj,i(p1) ⊕ h, q = ϕj,i(q1) ⊕ h. Both p and q are projections in Mk(Aj) and we 

know that 

ϕ∞,j(p) ≈ ϕ∞,i(p1) ⊕ f ≈ ϕ∞,i(q1) ⊕ f ≈ ϕ∞,j(q) 

in Mk(A). There is therefore a partial isometry w ∈ Mk(A) such that 

ww ∗ = ϕ∞,j(p), w ∗ w = ϕ∞,j(q). 

Let δ ∈]0, 1[ and find a ∈ Mk(Al) for some l ≥ j such that ϕ∞,l(a) − w < δ. 

Then 

ϕ∞,l(aa ∗ − ϕl,j(p)) < 4δ, ϕ∞,l(a ∗ a − ϕl,j(q)) < 4δ. 

By using (1.25) as in the proof of Lemma 3.15 we conclude that there is a r ≥ l such 

that 

ϕr,l(aa ∗ − ϕl,j(p)) < 4k 3 

2δ, ϕr,l(a ∗ a − ϕl,j(q)) < 4k 3 

2δ . 

In particular, ϕr,l(a)2 ≤ ϕl,j(p))+4k 3 

2δ ≤ 1+4k 3 

2δ. If just δ > 0 is small enough 

we see that x = (1 + 4k 3 

2δ) −1 2ϕr,l(a) is an element of Mk(Ar) such that x ≤ 1 and 

xx ∗ − ϕr,j(p) < 1 

3 , x∗x − ϕr,j(q) < 1 

3 . 

It follows from Lemma 2.11 that ϕr,j(p) ≈ ϕr,j(q) in Mk(Ar), and hence that 

[ϕr,j(p)] = [ϕr,j(q)] in K00(Ar). We conclude that 

x = qi([p1] − [q1]) = qj([ϕj,i(p1)] − [ϕj,i(q1)]) = qj([ϕj,i(p1)] + [h] − ([ϕj,i(q1)] + [h])) = 

qj([p] − [q]) = qr([ϕr,j(p)] − [ϕr,j(q)]) = qr(0) = 0. 

Hence q is injective and the proof is complete. 

An alternative, shorter and maybe more instructive way to formulate the result 

in Theorem 3.16 is that 

K00(lim An) = lim K00(An) 

−→ −→ 

as pre-ordered Abelian groups. 

To introduce the genuine K0-group we need to adjoin units to a C∗-algebra in 

a way which allows us to deal with unital and non-unital algebras in the same way. 

The method used to add a unit which we used in Section 1 is not sufficiently wellbehaved 

from a functorial point of view. The trouble is that a ∗-homomorphism 

ϕ : A → B may not extend to a unital ∗-homomorphism Â → ˆ B. To repair this 

defect it is necessary to add a unit in a way which changes the algebra even when it 

is unital to begin with. We summarize the construction in the following 

Lemma 3.17. Let A be a C ∗ -algebra and endow the vector space direct sum A⊕C 

with the involution 

(a, λ) ∗ = (a ∗ , λ) 

and the product 

(a, λ)(b, µ) = (ab + λb + µa, λµ). 

The resulting ∗-algebra is a C ∗ -algebra A + with unit (0, 1) and there is a unital ∗homomorphism 

χA : A + → C given by χA(a, λ) = λ such that 

ker χA = A ⊆ A + . 

Furthermore,


1) When A is unital, then (a, λ) → (a+λ1, λ) defines an isomorphism between 

A + and the C ∗ - algebra direct sum A ⊕ C. 

2) When ϕ : A → B is a ∗-homomorphism between C ∗ -algebras, then ϕ + (a, λ) = 

(ϕ(a), λ) is a unital ∗-homomorphism ϕ + : A + → B + such that χB ◦ ϕ + = 

χA. 

Proof. The C ∗ -norm on A + can be obtained as follows. Define a ∗-homomorphism 

ψ : A + → M(A ⊕ C) by 

ψ(a, λ)(b, µ) = (ab + λb, λµ). 

It is straightforward to check that ψ is an injective ∗-homomorphism, so we obtain 

a norm on A + by setting x = ψ(x), x ∈ A + . We leave it to the reader to check 

the completeness of this norm on A + . The rest of the statements in the lemma are 

straightforward to verify. 

We are now ready for the definition of the genuine K0-group. 

Definition 3.18. Let A be a C ∗ -algebra. Set 

K0(A) = ker (χA∗ : K00(A + ) → K00(C)). 

For every projection e ∈ Mn(A) ⊆ Mn(A + ) we have that χA(e) = 0 so it follows 

that the inclusion ιA : A → A + defines us a homomorphism ιA∗ : K00(A) → K0(A). 

In particular, we can set 

K0(A) + = ιA∗(K00(A) + ) 

to make K0(A) into a pre-ordered Abelian group. Alternatively, 

K0(A) + = {[e] : e ∈ P(A)}. 

If ψ : A → B is a ∗-homomorphism we have a unital ∗-homomorphism ψ + : 

A + → B + such that χB ◦ψ + = χA. It follows that ψ + ∗ : K00(A + ) → K00(B + ) maps 

K0(A) = ker χA∗ into ker χB∗ = K0(B). We denote the restriction 

ψ + ∗|K0(A) : K0(A) → K0(B) 

by ψ∗ and hope that we will be able to avoid any confusion with the map ψ∗ : 

K00(A) → K00(B). It is straightforward to check that ψ∗ : K0(A) → K0(B) 

is a positive homomorphism, i.e. satisfies that ψ∗(K0(A) + ) ⊆ K0(B) + and that 

(ϕ ◦ ψ) ∗ = ϕ∗ ◦ ψ∗, when ϕ : B → C is a ∗-homomorphism. In particular it follows 

that two C ∗ -algebras A and B can only be isomorphic if the groups K0(A) and 

K0(B) are isomorphic as pre-ordered Abelian groups. However, at this point it is 

not clear why K0(A) is any better than K00(A) for the purpose of distinguishing 

C ∗ -algebras. This question becomes even more pertinent by the next results which 

shows that for a large class of C ∗ -algebras these two groups agree. We will return 

to this issue later. 

Lemma 3.19. Let A be a unital C ∗ -algebra. The map ιA∗ : K00(A) → K0(A) is 

an isomorphism of pre-ordered Abelian groups. 

Proof. Let α : A + → A ⊕ C be the isomorphism from Lemma 3.17 1) and let 

β : K00(A ⊕ C) → K00(A) ⊕ K00(C) be the group isomorphism from Lemma 3.10. 

Remember that β(·) = (p1∗(·), p2∗(·)) where p1 : A⊕C → A and p2 : A⊕C → C are 

the two projections. Then β ◦ α∗ : K00(A + ) → K00(A) ⊕ K00(C) is an isomorphism 

such that β ◦ α∗ ◦ ιA∗(x) = (p1∗ ◦ α∗ ◦ ιA∗(x), p2∗ ◦ α∗ ◦ ιA∗(x)) = (idA∗(x), 0) =


(x, 0), x ∈ K00(A). This relation shows that ιA∗ is injective. Let x ∈ K0(A). Since 

p2 ◦ α = χA and χA∗(x) = 0 we find that β ◦ α∗(x) = (y, 0) for some y ∈ K00(A). 

Then 

β ◦ α∗ ◦ ιA∗(y) = (idA∗(y), 0) = (y, 0) = β ◦ α∗(x), 

proving that x = ιA∗(y) because β ◦ α∗ is injective. Thus ιA∗ maps onto K0(A) 

and since K0(A) + = ιA∗(K00(A) + ) by definition, ιA∗ must be an isomorphism of 

pre-ordered Abelian groups. 


A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . 



C ∗ -algebra. Consider the unital sequence 

A + 1 

and let ψ∞,n : A + n → lim 

then a ∗-isomorphism η : lim 

ϕ + 

1 

A + 2 

ϕ + 

2 

A + 3 

ϕ + 

3 

. . . 

) be the corresponding canonical maps. There is 

for all n. 

−→ (A+ n , ϕ+ n 

(A 

−→ + n , ϕ+ n ) → A+ such that η ◦ ψ∞,n = ϕ + ∞,n 

Proof. The existence of the ∗-homomorphism η follows from the universal property 

of the inductive limit construction. To show that η is injective, let a ∈ 

ker η, and let ǫ > 0. There is an n ∈ N and an element b ∈ A + n such that 

ψ∞,n(b) − a ≤ ǫ. Then ϕ + ∞,n (b) = η(ψ∞,n(b) − a) ≤ ǫ. Write b = (c, λ), 

where c ∈ An, λ ∈ C. Then ϕ + ∞,n (b) = (ϕ∞,n(c), λ) and |λ| ≤ ϕ + ∞,n (b) ≤ ǫ. 

Hence ϕ∞,n(c) = ϕ + ∞,n (c, 0) ≤ ϕ+ ∞,n (b) + |λ| ≤ 2ǫ. It follows ϕm,n(c) < 3ǫ 

for some m ≥ n, and hence ϕ + m,n (b) = (ϕm,n(c), λ) ≤ ϕm,n(c) + |λ| ≤ 4ǫ. It 

follows that ψ∞,n(b) ≤ 4ǫ and hence that a ≤ 5ǫ. Since ǫ > 0 was arbitrary we 

conclude that a = 0, and that η is injective. Let now a ∈ A + . There is then an n ∈ N, 

an element b ∈ An and a complex number λ such that (ϕ∞,n(b), λ) − a ≤ ǫ. Since 

η(ψ∞,n(b, λ)) = (ϕ∞,n(b), λ), we see that η has dense range in A + and therefore 

must be surjective. 


A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . 



C ∗ -algebra. Let 

K0(A1) ϕ1∗ 

K0(A2) ϕ2∗ 

K0(A3) ϕ3∗ 

. . . 

be the sequence of pre-ordered Abelian groups obtained by applying the K0-functor 

and let G = lim 

−→ (K0(An), ϕn∗) be the inductive limit in the category of pre-ordered 

Abelian groups and q ′ n : K0(An) → G the canonical maps. There is then a unique 

group homomorphism q ′ : G → K0(A) such that q ′ ◦ q ′ n = ϕ∞,n ∗ for all n and q ′ is 

an isomorphism of pre-ordered Abelian groups. 

Proof. Let η : lim 

−→ (A + n , ϕ+ n ) → A+ be the ∗-isomorphism from Lemma 3.20. By 

Theorem 3.16, there is an isomorphism 

q : lim 

−→ (K00(A + n), (ϕ + n)∗) → K00(lim 

−→ (A + n,ϕ + n))


of pre-ordered groups such that q ◦ qn = ψ∞,n ∗ . Furthermore, since we have a 

commuting diagram 

K0(A1) 

⏐ 

∩ 

K00(A + 1 ) 

ϕ1∗ 

−−−→ K0(A2) 

⏐ 

∩ 

(ϕ + 

1 )∗ 

−−−→ K00(A + 2 ) 

ϕ2∗ 

−−−→ K0(A3) 

⏐ 

∩ 

(ϕ + 

2 )∗ 

−−−→ K00(A + 3 ) 

ϕ3∗ 

−−−→ . . . 

(ϕ + 

3 )∗ 

−−−→ . . ., 

there is an injective map J : lim(K0(An), 

ϕn∗) → lim(K00(A 

−→ −→ + n ), (ϕ+ n )∗) such that 

J ◦q ′ n = qn|K0(An). Thus η∗ ◦q ◦J : lim(K0(An), 

ϕn∗) → K00(A 

−→ + ) is injective. This is 

in fact the map q ′ of the theorem, but to prove this, and the other assertions about 

it, we must first introduce more maps. 

Note that, because χAn+1 ◦ ϕ + n = χAn, we have a unique ∗-homomorphism χ : 

lim 

−→ (A+ n , ϕ+ n ) → C such that χ ◦ψ∞,n = χAn. Since χA ◦η ◦ψ∞,n = χA ◦ϕ∞,n + = χAn 

for all n, we see that χA ◦ η = χ. Similarly, since χAn+1∗ ◦ (ϕ+ n)∗ = χAn ∗ for all 

n, there is a unique group homomorphism χ ′ : lim(K00(A 

−→ + n), (ϕ + n)∗) → K00(C) such 

, we see that 

that χ ′ ◦ qn = χAn ∗ for all n. Since χ∗ ◦ q ◦ qn = χ∗ ◦ ψ∞,n∗ = χAn ∗ 

χ∗ ◦ q = χ ′ . All in all we have that 

χA∗ ◦ η∗ ◦ q = χ ′ . 

Now, it is clear that 

J(lim(K0(An), 

ϕn∗)) = ker χ 

−→ ′ . 

(If this is not clear to the reader, he or she should provide the missing details.) So 

we conclude that 

η∗ ◦ q ◦ J(lim 

−→ (K0(An), ϕn∗)) ⊆ ker χA∗ = K0(A). 

On the other hand, if x ∈ K0(A) = 

ker χA∗, then x = η∗ ◦ q(y) for some y ∈ lim 

−→ (K00(A + n), ϕ + n ∗ ) since η∗ ◦ q is an isomorphism, 

and because χ ′ (y) = χA∗ ◦ η∗ ◦ q(y) = 0 we see that y ∈ 

ker χ ′ = J(lim(K0(An), 

ϕn∗)). Thus η∗◦q◦J is a group isomorphism, lim(K0(An), 

ϕn∗) ≃ 

−→ −→ 

K0(A). Note that η∗◦q◦J◦q ′ n = η∗◦q◦qn|K0(An) = η∗◦ψ∞,n ∗ |K0(An) = (ϕ + ∞,n)∗|K0(An) = 

ϕ∞,n ∗ , so that η∗ ◦ q ◦ J = q ′ . In particular, 

q ′ 

q ′ n (K0(An) + ) ⊆ K0(A) + . 

n 

If z ∈ K0(A) + , then z = η∗ ◦ q(y) for some y ∈ qn(K00(A + n )+ ) since η∗ ◦ q is an 

isomorphism of pre-ordered Abelian groups. Then y = qn([f]) for some f ∈ P(A + n ) 

and 

[χAn(f)] = χAn∗ ([f]) = χ′ ◦ qn([f]) = χ ′ (y) = χA∗ ◦ η∗ ◦ q(y) = χA∗(z) = 0 

in K00(C). By definition of K00 this means that there is a projection r ∈ Mm(C) 

for some m such that χAn(f) ⊕ r ≈ r in Mk(C) for some k ≥ m. It follows that the 

trace of χAn(f) is zero and hence χAn(f) must be the zero projection. This means 

that f ∈ P(An) and [f] ∈ K0(An) + . Thus 

K0(A) + = q ′ 

q ′ n(K0(An) + ) 

and q ′ is an isomorphism of pre-ordered Abelian groups. 

n


It follows from Exercise 3.49 that K00(A) = K0(A) for a large class of C ∗ - 

algebras, including for example the AF-algebras. However,as will be shown in the 

example below, 

K0(C0(R 2 )) ≃ Z 

while 

K00(C0(R 2 )) = 0. 

So in general the two functors K00 and K0 are different and it is not generally the 

case that K0(A) + − K0(A) + = K0(A), although this holds for all inductive limits of 

unital C ∗ -algebras by Exercise 3.49 and Lemma 3.8 1). The main reason for working 

with K0 instead of K00 is that the former has better functorial properties. 

Example 3.22. This example introduces an important projection, called the 

Bott projection. The point is to show that projections in Mn(A) in general can not 

be ’diagonalized’. More precisely, we give an example of a unital C ∗ -algebra A which 

only contains the projections 0 and 1, while M2(A) contains a projection which is 

not Murray-von Neumann equivalent to any projection of the form ( a b ) for some 

a, b ∈ A. 

Let D = {z ∈ C : |z| ≤ 1} which is a compact Hausdorff space in the topology 

inherited from C. Thus B = C(D) is an abelian C ∗ -algebra, and since D is connected 

there are only the trivial projections 0 and 1 in C(D). Let T be the unit circle in C 

which is the topological boundary of D in C. Set 

A = {f ∈ B : f is constant on T}. 

Then A is a unital C∗-subalgebra of B which just like B only contains the projections 

0 and 1. Set 

 

2 1 − |z| 

P = 

z 1 − |z| 2 

z 1 − |z| 2 |z| 2 

 

. 

Since the functions f11(z) = 1 − |z| 2 , f12(z) = z 1 − |z| 2 , f21(z) = z 1 − |z| 2 

and f22(z) = |z| 2 all are constant on T we see that P ∈ M2(A). A straightforward 

check shows that P is a projection. We claim that there are no partial isometry 

V ∈ M2(A) such that V ∗ V = P and 

V V ∗ ∈ {( a b ) : a, b ∈ A} . 

To see this, assume that such a V exists. Since V V ∗ is a projection and A only 

contains the trivial projections we see that 

Since that 

V V ∗ ∈ {( 1 0 ) , ( 0 1 ) , ( 1 1 ), ( 0 0 )} . 

 

2 1 − |z| z 

Tr 

1 − |z| 2 

z 1 − |z| 2 |z| 2 

 

= 1, 

we conclude that the only possibilities are V V ∗ = ( 1 0 ) or V V ∗ = ( 0 1 ). In the last 

case we see that 

W = ( 0 1 

1 0 ) V 

will then be an element of M2(A) such that WW ∗ = ( 1 0 ) and W ∗W = P. We may 

therefore assume, without loss of generality, that 

V V ∗ = ( 1 0 ).


Now set 

S = 

 

1 − |z| 2 z 

, 

0 0 

which is an element of M2(B), but not of M2(A). A straigtforward check shows that 

It follows therefore that 

and 

Consequently 

S ∗ S = P, SS ∗ = ( 1 0 ). 

V S ∗ ( 1 0 ) = V S∗ SS ∗ = V S ∗ 

( 1 0 ) V S∗ = V V ∗ V S ∗ = V S ∗ . 

V S ∗ = ( g 

0 ) 

for some function g : D → C. Since 

 

|g| 2 

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 

= V S (V S ) = V S SV = V PV = V V V V = V V = ( 1 

0 ) , 

0 

we conclude that g takes values in T. Note that 


V = V V ∗ V = V S ∗ S = ( g 

0)S = 

g(z) 1 − |z| 2 g(z)z 

0 0 

 

. 

g(z)z = v12(z) (3.2) 

for all z ∈ D, where v12 ∈ A is the 1, 2-entry in V . Since v12 ∈ A there is complex 

number λ ∈ C such that w12(z) = λ for all z ∈ T. It follows from (3.2) that λ ∈ T, 

and then that 

z = λg(z) (3.3) 

for all z ∈ T. Define h : D → T such that h(z) = λg(z). Then h is clearly a 

continuous function h : D → T and (3.3) tells us that h is the identity map when 

restricted to the circle T. This is impossible; there are no continuous function D → T 

extending the identity on T. If this well-known topological fact is not familiar to 

the reader we offer in the following a C ∗ -algebraic proof; the involved lemma will be 

useful when we start to study the unitary groups of C ∗ -algebras anyway. 

Lemma 3.23. Let D be a unital C ∗ -algebra and u ∈ D a unitary such that 

1 − u < 2. It follows that there is self-adjoint element a ∈ D such that a < π 

and u = e ia . 

Proof. Note that σ(u) can not contain −1 because this would imply that the 

spectral radius of 1 − u is at least 2. Since the spectral radius of 1 − u is 1 − u, 

this is impossible by assumption. On the other hand, the spectrum of u is contained 

in the circle T since u is unitary so we see that 

σ(u) ⊆ T\{−1}. 

Let ϕ : T\{−1} →] − π, π[ be the inverse of ] − π, π[∋ x ↦→ e ix , and set a = ϕ(u). 

Then σ(a) is a compact subset of ] − π, π[ so a is self-adjoint and a < π. Since 

e ia = u, the proof is complete.


Let us use this lemma to show that there is no continuous function h : D → T 

such that h(z) = z for all z ∈ T. Assume that h is such a function. We can then 

define, for each s ∈ [0, 1], a unitary hs ∈ C(T) such that 

hs(z) = h(sz). 

Observe that the map [0, 1] ∋ s → hs ∈ C(T) is continuous; this follows from the 

uniform continuity of h. There is therefore a partition 0 = x0 < x1 < x2 < · · · < 

xN = 1 such that 

 

hxi − hxi−1 

≤ 1 

for all i = 1, 2, . . ., N. It follows then from Lemma 3.23 that there are continuous 

functions gj : T → [−π, π] such that 

hxj−1 hxj = eigj (3.4) 

for all j = 1, 2, . . ., N. Note that h0 is a constant; say h0 = e ir for some r ∈ R. It 

follows then from (3.4) that 

h = hxN = eiH0 , 

where H0(z) = r + N 

j=1 gj(z). Since h(z) = z for all z ∈ T, we conclude that 

z = e iH0(z) for all z ∈ T or, alternatiely, that 

e 2πit = e 2πiH(t) 

for all t ∈ [0, 1], where H : [0, 1] → R is the continuous function 

H(t) = 1 

2π H0 

 

2πit 

e . 

It follows that e 2πi(t−H(t)) = 1 for all t ∈ [0, 1], and hence that t − H(t) ∈ Z for all 

t ∈ [0, 1]. Thus t−H(t) must be constant, which is clearly absurd since H(0) = H(1). 

This contradiction proves the claim. 

Note that A ≃ E + , where E = {f ∈ C(D) : f|T = 0}, and that 

[P] − [( 0 1 )] ∈ K0 (E) . 

To show that [P] − [( 1 0 )] is not zero in K0(E) we use the following lemma which is 

often usefull. 

Lemma 3.24. Let B be a unital C ∗ -algebra and p, q ∈ Mn(B) projections such 

that [p] −[q] = 0 in K0(B). There is then an m ∈ N such that p ⊕1m is Murray-von 

Neumann equivalent to q ⊕ 1m, where 1m = diag(1, 1, . . ., 1) ∈ Mm(B). 

Proof. Since [p] − [q] = 0 in K00(B) there is a projection r ∈ Mm(B), for some 

m, such that p ⊕ r is Murray-von Neumann equivalent to q ⊕ r. The lemma follows 

because r ⊕ (1m − p) is Murray-von Neumann equivalent to 1m. 

Assume to get a contraduction that [P] − [( 1 0 )] = 0 in K0(E). It follows from 

Lemma 3.24 that P ⊕1m is Murray-von Neumann equivalent to diag(1, 0, 1, 1, . . ., 1) 

in Mm+2(A) for some m. Thus there is a partial isometry W in Mm+2(A) such 

that WW ∗ = diag(1, 0, 1, 1, . . ., 1) and W ∗ W = diag(P, 1m). We compare now 

W with T = diag(S, 1m) which also satisfies that TT ∗ = diag(1, 0, 1, 1, . . ., 1) and 

T ∗ T = diag(P, 1m). Since 

diag(1, 0, 1, 1, . . ., 1)WT ∗ = WT ∗ diag(1, 0, 1, 1, . . ., 1) = WT ∗ ,


we see that the second row and second column in WT ∗ consists entirely of zeroes. 

Let 

⎛ 

⎜ 

X = ⎜ 

⎝ 

x11 

x21 

. 

x12 

x22 

. 

. . . 

. . . 

.. . 

⎞ 

x1m+1 

x2,m+1 ⎟ 

. 

⎠ 

xm+1,1 xm+1,2 . . . xm+1,m+1 

be the element of Mm+1(B) obtained by removing the second row and second column 

in WT ∗ . Then X is a unitary because WT ∗ TW ∗ = diag(1, 0, 1, 1, . . ., 1). Since 

W = WT ∗ T we find that the elememt of Mm+1(A) obtained by removing the first 

column and the second row of W is 

⎛ 

⎞ 

zx11 x12 . . . x1m+1 

⎜ zx21 x22 . . . x2,m+1 ⎟ 

⎜ 

⎝ 

. 

⎟ 

. . .. . 

⎠ . 

zxm+1,1 xm+1,2 . . . xm+1,m+1 

Thus by taking determinants we obtain an element v ∈ A such that 

v(z) = zg(z) 

for all z ∈ D, where g(z) is the determinant of X(z). Since X is unitary, |g(z)| = 1, 

and we can obtain a contradiction in the same way as before. 

Thus we see that K0 (E) = 0. It is easy to see that Mn(E) contains no non-zero 

projections for any any n, which implies that K00 (E) = 0. Hence E is a C ∗ -algebra 

for which K00 (E) = K0 (E). It is easy to see that E ≃ C0 (R 2 ). 

3.2. The classification of AF-algebras. In this section we use the K0-group 

to classify AF-algebras up to isomorphism. 

In the following A is a C ∗ -algebra. 

Definition 3.25. A has cancellation of projections when the following holds: 

When n ∈ N and e, f ∈ Mn(A) are projections such that [e] = [f] in K0(A), it 

follows that e ≈ f, i.e. that e and f are Murray-von Neumann equivalent in Mn(A). 


A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . 



C ∗ -algebra. Assume that each Ai has cancellation of projections. Then A has cancellation 

of projections. 

Proof. Let e, f ∈ Mn(A) be projections such that [e] = [f] in K0(A). By 

Lemma 3.15 there is then an m ∈ N and projections e ′ , f ′ ∈ Mn(Am) such that 

ϕ∞,m(e ′ ) − e ≤ 1 

4 and ϕ∞,m(f ′ ) − f ≤ 1 

. It follows from Lemma 3.14 that 

4 

e ≈ ϕ∞,m(e ′ ) and f ≈ ϕ∞,m(f ′ ). In particular, ϕ∞,m∗ [e ′ ] = [e] = [f] = ϕ∞,m∗ [f ′ ]. 

And then Theorem 3.21 shows that ϕk,m∗ [e ′ ] = ϕ∞,m∗ [f ′ ] in K0 (Ak) for some k > 

m. Since Ak has cancellation of projections it follows that ϕk,m(e ′ ) ≈ ϕk,m(f ′ ) in 

Mn (Ak). Consequently, ϕ∞,m(e ′ ) ≈ ϕ∞,m(f ′ ) in Mn (A). 

Lemma 3.27. Let A be a C∗-algebra such that Â has cancellation of projections. 

1) When e, f ∈ P(A) are two projections, we have that [e] = [f] in K0(A) if 

and only if e ≈ f.


2) When e and f are projections in A such that [e] = [f] in K0(A), there is a 

unitary u ∈ Â such that ueu∗ = f. 

3) K0(A) is a partially ordered group, i.e. K0(A) + ∩ (−K0(A) + ) = {0}. 

Proof. 1) Assume that [e] = [f] in K0(A). Then [e] = [f] in K0( Â) and hence 

e ≈ f since Â has cancellation of projections. The converse is trivial. 

2) : From 1) we get a partial isometry v ∈ A such that vv ∗ = f, v ∗ v = e. Since 

[1 −e]+[e] = [1] = [1 −f]+[f] in K0( Â) by Lemma 3.2 we see that [1 −e] = [1 −f] 

in K0( Â). Since Â has cancellation of projections we conclude that 1 − f ≈ 1 − e 

in Â, i.e. there is a partial isometry w ∈ Â such that ww∗ = 1 − f, w∗w = 1 − e. 

Then u = v + w is a unitary in Â such that ueu∗ = f. 

3) : Let x ∈ K0(A) + ∩ (−K0(A) + ). There are projections e, f ∈ P(A) such that 

x = [e] = −[f] in K0(A). But this means that [e ⊕ f] = 0 in K0(A). By 1) this 

implies that e ⊕ f ≈ 0, and hence that e ⊕ f = 0. Thus e = f = 0 and x = 0 in 

K0(A). 

 

Let A be a C ∗ -algebra. A continuous linear functional ω ∈ A ∗ is a trace when 

ω(xy) = ω(yx), x, y ∈ A. 

Lemma 3.28. Let ω ∈ A∗ be a trace. There is then a trace ωn ∈ Mn(A) ∗ such 

that 

 

ωn (aij) n 

n 

i,j=1 = ω (aii). 

Moreover, every trace of Mn(A) is of this form. Specifically if µ ∈ Mn(A) ∗ is a trace, 

⎛ ⎞ 

a 0 0 . . . 0 

⎜0 

0 0 . . . 0⎟ 

⎜ ⎟ 

ω(a) = µ ⎜0 

0 0 . . . 0⎟ 

⎜ 

⎝ . 

⎟ 

. . . .. . ⎠ 

0 0 0 . . . 0 

defines a trace ω ∈ A ∗ such that µ = ωn. 

Proof. Note that 

 

 

ωn 

≤ 

 

(aij) n 

i,j=1 

i=1 

n 

|ω (aii)| ≤ nω max 

i aii 

 

 

≤ nω 

i=1 

This shows that ωn ∈ Mn(A) ∗ . When A = (aij) n 

i,j=1 

ωn (AB) = 

= 

n 

 

n 

ω 

i=1 

k=1 

aikbki 

n 

ω (bikaki) = ωn(BA). 

i,k=1 

 

= 

n 

ω (aikbki) = 

i,k=1 

(aij) n 

i,j=1 

 

 

. 

n 

and B = (bij) 

i,j=1 , we find that 

n 

ω (bkiaik) 

We leave the proof of the remaining part of the statement to the reader. 

Definition 3.29. A C ∗ -algebra A is rich on traces when the following holds: 

When n ∈ N and e, f ∈ Mn(A) are projections such that ω(e) = ω(f) for all traces 

ω ∈ Mn(A) ∗ , it follows that e ≈ f. 

i,k=1


Lemma 3.30. A C ∗ -algebra A which is rich on traces has cancellation of projections. 

Proof. Let n ∈ N and assume that e, f ∈ Mn(A) are projections such that 

[e] = [f] in K0(A). There is then a projection p ∈ Mm(A + ) such that e ⊕ p ≈ f ⊕ p 

in Mn+m(A + ). Let µ ∈ Mn(A) ∗ be a trace. By Lemma 3.28 there is a trace ω ∈ A ∗ 

such that µ = ωn. Define ˜ω : A + → C such that ˜ω(a, λ) = ω(a) + λ and note that 

˜ω ∈ A ∗ is a trace. Set ˜µ = ˜ωn+m ∈ Mn+m(A + ) ∗ which is a trace on Mn+m(A + ). 

Since e ⊕ p ≈ f ⊕ p we find that 

˜µ(e ⊕ p) = ˜µ(f ⊕ p). 

Note that ˜µ(e ⊕ p) = ˜ωn(e) + ˜ωm(p) = ωn(e) + ˜ωm(p) = µ(e) + ˜ωm(p) and similarly 

˜µ(f ⊕ p) = µ(f) + ˜ωm(p). Hence µ(e) = µ(f). Since µ was an arbitrary trace in 

Mn(A) ∗ it follows that e ≈ f since A is rich on traces. 

Lemma 3.31. A finite dimensional C ∗ -algebra is rich on traces. 

Proof. Let F be a finite-dimensinal C∗-algebra and e, f ∈ F projections such 

that ω(e) = ω(f) for every trace ω ∈ A∗ . To show that e ≈ f we may assume 

that F = Mn1 ⊕ Mn2 ⊕ · · · ⊕ MnN , cf. Theorem 1.60. Then we can write e = 

(e1, e2, . . .,eN), f = (f1, f2, . . .,fN) where ei, fi ∈ Mni . For each i ∈ {1, 2, . . ., N} 

we can define a trace Tri : F → C by 

Tri(x1, x2, . . .,xN) = Tr(xi), 

(x1, . . .,xN) ∈ F. Since Tr(ei) = Tri(e) = Tri(f) = Tr(fi) by assumption, we know 

from linear algebra that there is a unitary ui ∈ Mni such that uieiui ∗ = fi, i = 

1, 2, . . ., N. Set u = (u1, u2, . . ., uN) and note that ueu ∗ = f. 

Since Mn(F) is a finite dimensional C ∗ -algebra for all n, this shows that F is 

rich on traces. 

Lemma 3.32. Let A be an AF-algebra. 

1) When e, f ∈ P(A) are two projections, we have that [e] = [f] in K0(A) if 

and only if e ≈ f. 

2) When e and f are projections in A such that [e] = [f] in K0(A), there is a 

unitary u ∈ Â such that ueu∗ = f. 

3) K0(A) is a partially ordered group, i.e. K0(A) + ∩ (−K0(A) + ) = {0}. 

Proof. It follows from Lemma 3.20 that Â is the inductive limit of a sequence 

of finite-dimensional C∗-algebras (with unital connecting maps). Therefore Â has 

cancellation of projections by Lemma 3.31. Hence Lemma 3.27 applies. 

Definition 3.33. For any C ∗ -algebra A we set 

Σ(A) = {[e] ∈ K0(A) : e ∈ P1(A)}. 

Thus Σ(A), called the scale of A, consists of the elements in K0(A) which are 

represented by projections in A. Note that Σ(A) ⊆ K0(A) + . 

Lemma 3.34. Let A be a finite dimensional C ∗ -algebra and B = lim 

−→ (Bn, ψn) the 

inductive limit of a sequence 

B1 

ψ1 

B2 

of finite direct sums of matrix algebras. 

ψ2 

B3 

ψ3 

. . .


1) If r : K0(A) → K0(B) is a positive homomorphism such that r(Σ(A)) ⊆ 

Σ(B), then there is a ∗-homomorphism ϕ : A → Bk for some k ∈ N such 

that ψ∞,k ∗ ◦ ϕ∗ = r. 

2) If µ, λ : A → B are two ∗-homomorphisms such that µ∗ = λ∗ on K0(A), 

then there is a unitary u ∈ ˆ B such that 

Ad u ◦ λ = µ. 

Proof. 1) : Before reading the proof the reader is advised to do Exercise 3.50 

} be a complete set of matrix units in A. Since r(Σ(A)) ⊆ Σ(B) there 

first. Let {ed i,j 

are projections qd i ∈ B such that 

for all d, i. Furthermore, since 

d,i 

there is a projection p ∈ B such that 

r([e d ii]) = [q d i ] 

[q d i ] = r([1]) ∈ Σ(B) 

[diag q 1 1 , . . .,q1 n1 , q2 1 , . . .,q2 n2 , q3 1 , . . ....,qN nN 

] = 

d,i 

[q d i 

] = [p] 

in K0(B). By Theorem 3.21 or Theorem 3.16 there is a k0 ∈ N and projections 

hd i ∈ Bk0 and p ′ ∈ Bk0 such that qd 

d 

i = ψ∞,k0∗ hi and [p] = ψ∞,k0∗ [p′ ] in K0(B) 

and 

 

] = 

[diag h 1 1 , . . .,h1 n1 , h2 1 , . . .,h2 n2 , h3 1 , . . ....,hN nN 

d,i 

[h d i ] = [p′ ] 

in K0(Bk0). It follows that there is a partial isometry V ∈ MPN d=1 

nd (Bk) such that 

⎛ 

⎞ 

h 1 1 0 0 . . . 0 

V ∗ ⎜ 0 

⎜ 

V = ⎜ 

⎝ 

h1 0 

2 0 

h 

. . . 0 

1 . . 

3 

. 

. . . 

. .. 

0 

. 

0 0 0 . . . hN ⎟ 

nN 

and 

⎠ = diag h 1 1 , . . .,h1 n1 , h2 1 , . . ., h2 n2 , h3 1 , . . ....,hN nN 

V V ∗ ⎛ 

p 

⎜ 

= ⎜ 

⎝ 

′ ⎞ 

0 0 . . . 0 

0 0 0 . . . 0 ⎟ 

0 0 . . . 0⎟ 

. 

⎟ 

. . . .. . ⎠ 

0 0 0 . . . 0 

. 

There are then projections s d i ∈ Bk such that 

⎛ 

s d i 0 0 . . . 0 

⎞ 

⎜ 0 0 0 . . . 0⎟ 

⎜ 

⎟ 

⎜ 0 0 . . . 0⎟ 

⎜ 

⎝ .. 

⎟ 

. . . . . ⎠ 

0 0 0 . . . 0 

= V diag 0, 0, . . ., 0, h d i , 0, . . .,0)V ∗ . 

Note that the sd i ’s are mutually orthogonal: sdi sd′ 

i ′ = 0 unless d = d′ and i = i ′ . Note 

d d d d d d that ψ∞,k0∗ si = ψ∞,k0 ∗ hi = qi = r eii = r e11 = ψ∞,k0∗ s1 for all d, i. 

By Theorem 3.21 (or Theorem 3.16) this implies that there is a k > k0 such that


 

d d si = ψk,k0 s1 in K0 

(Bk) 

 

for all d, i. There are therfore partial isometries 

d i s1 and vdvi ∗ 

d 

d = ψk,k0 si for all i, d. By using that 

’s are mutually orthogonal it is the easy to check that 

 

ψk,k0 

vi d ∈ Bk such that vi ∗ i 

d vd = ψk,k0 

the sd i 

f d ij = v i dv j ∗ 

d 

d = 1, 2, . . ., N, i, j = 1, 2, . . ., nd, is a set of matrix units in Bk. There is therefore 

d = fij for all i, j, d. Since 

a ∗-homomorphism ϕ : A → Bk such that ϕ ed ij 

 

d d d d d 

ψ∞,k∗ ◦ ϕ∗ e11 = ψ∞,k∗ f11 = ψ∞,k ψk,k0 ∗ si = ψ∞,k0∗ si = r e11 , 

for all d, it follows that ψ∞,k∗ ◦ ϕ∗ = r as desired. 

2) : Since [µ(ed 11 )] = µ∗([ed 11 ]) = λ∗([ed 11 ]) = [λ(ed11 )] in K0(B) we conclude from 

Lemma 3.27 1) that there is a partial isometry vd ∈ B such that vdvd ∗ = µ(ed 11 ) and 

vd ∗ vd = λ(e d 11 

). Set 

v = 

µ(e d i1)vdλ(e d 1i). 

d,i 

Then vv ∗ = µ(1), v ∗ v = λ(1) and vλ(e d ij)v ∗ = µ(e d ij) for all d, i, j. Since [λ(1)] = 

[µ(1)] in K0(B) there is, by Lemma 3.27 2) a unitary s ∈ ˆ B such that s(1−λ(1))s ∗ = 

1 − µ(1). Then w = s(1 − λ(1)) ∈ ˆ B is a partial isometry such that ww ∗ = 

1 −µ(1), w ∗ w = 1 −λ(1) and we set u = v+w. Then u is a unitary in B + such that 

Adu ◦ λ = µ. (The calculations are essentially the same as in the proof of Lemma 

2.6.) 

and 

We can now prove the classification theorem for AF-algebras. 

Theorem 3.35. Let A and B be AF-algebras. 

1) If f : K0(A) → K0(B) is a positive group homomorphism such that f(Σ(A)) ⊆ 

Σ(B), then there is a ∗- homomorphism ϕ : A → B such that ϕ∗ = f. 

2) If ϕ, ψ : A → B are two ∗-homomorphisms such that ψ∗ = ϕ∗ on K0(A), 

then there is a sequence un of unitaries in B + such that 

ϕ(a) = lim 

n→∞ unψ(a)u ∗ n, a ∈ A. 

3) When f : K0(A) → K0(B) is an isomorphism of partially ordered groups 

such that f(Σ(A)) = Σ(B), then there is an isomorphism ϕ : A → B such 

that ϕ∗ = f. 

Proof. 1) : Let A and B be the inductive limits of 

A1 

B1 

ϕ1 

A2 

ψ1 

B2 

ϕ2 

A3 

ψ2 

B3 

ϕ3 

. . . 

ψ3 

. . . 

respectively, where each An and Bn is a finite direct sum of matrix algebras. We will 

construct sequences l1 < l2 < . . . and k1 < k2 < . . . in N and ∗-homomorphisms 

λi : Ali → Bki such that ψ∞,ki∗ ◦ λi∗ = f ◦ ϕ∞,li∗ and the diagram 

λ1 

Al1 

⏐ 

 

Bk1 

ϕl2 ,l1 −−−→ Al2 

λ2 

⏐ 

 

ψk 2 ,k 1 

−−−→ Bk2 

ϕl3 ,l2 −−−→ Al3 

λ3 

⏐ 

 

ψk 3 ,k 2 

−−−→ Bk3 

ϕl 4 ,l 3 

−−−→ . . . 

ψk 4 ,k 3 

−−−→ . . .


commutes. Assume that we have constructed everything up to n ∈ N. We then 

construct ln+1, kn+1 and λn+1 such that 

λn 

Aln 

⏐ 

 

Bkn 

ϕln+1 ,ln 

−−−−−→ Aln+1 

λn+1 

⏐ 

 

ψk n+1 ,kn 

−−−−−→ Bkn+1 

commutes and ψ∞,kn+1∗ ◦ λn+1∗ = f ◦ ϕ∞,ln+1∗ . First we take an arbitrary ln+1 > ln. 

By Lemma 3.34 1) there is an m > kn and a ∗- homomorphism λ : Aln+1 → Bm 

such that ψ∞,m∗ ◦ λ∗ = f ◦ ϕ∞,ln+1∗ . In comparison we have that ψ∞,kn∗ ◦ λn∗ = 

f ◦ ϕ∞,ln∗ = f ◦ ϕ∞,ln+1∗ ◦ ϕln+1,ln ∗ . Hence ψ∞,m∗ ◦ λ∗ ◦ ϕln+1,ln∗ = ψ∞,kn∗ ◦ λn∗. 

Let {ed ij } be the standard matrix units in Aln. By Theorem 3.21 (or Theorem 3.16) 

there is a kn+1 > m such that 

for all d. But then 

ψkn+1,m ∗ ◦ λ∗ ◦ ϕln+1,ln ∗ ([ed 11]) = ψkn+1,kn∗ ◦ λn∗([e d 11]) 

= ψkn+1,kn∗ ◦ λn∗ 

ψkn+1,m ◦ λ∗ ◦ ϕln+1,ln ∗ ∗ 

on K0(Aln) by Exercise 3.50. By Lemma 3.34 2) there is therefore a unitary u ∈ B + 

kn+1 

such that 

Ad u ◦ ψkn+1,m ◦ λ ◦ ϕln+1,ln = ψkn+1,kn ◦ λn. 

We set λn+1 = Ad u ◦ ψkn+1,m ◦ λ and observe that the desired diagram commutes 

with this choice for λn+1. Since Ad u induces the identity map on K0(Bkn+1) we find 

that 

ψ∞,kn+1∗ ◦ λn+1∗ = ψ∞,kn+1∗ ◦ ψkn+1,m ∗ ◦ λ∗ = ψ∞,m ∗ ◦ λ∗ = f ◦ ϕ∞,ln+1∗ . 

The diagram gives immediately the existence of a ∗-homomorphism ϕ : A → B 

such that 

ϕ ◦ ϕ∞,ln = ψ∞,kn ◦ λn, n ∈ N. 

Since ϕ∗ ◦ ϕ∞,ln∗ = ψ∞,kn∗ ◦ λn∗ = f ◦ ϕ∞,ln∗ for all n, we see that ϕ∗ = f. 

2) : Since 

(ϕ ◦ ϕ∞,n)∗ = ϕ∗ ◦ ϕ∞,n ∗ = ψ∗ ◦ ϕ∞,n ∗ = (ψ ◦ ϕ∞,n)∗ 

on K0(An), Lemma 3.34 2) provides a unitary un ∈ B + such that Ad un ◦ϕ◦ϕ∞,n = 

ψ ◦ ϕ∞,n. This gives us a sequence of unitaries, un, and since 

i ϕ∞,i(Ai) = A, it is 

clear that limn→∞ unϕ(a)u∗ n = ψ(a), a ∈ A. 

3) : We will construct sequences l1 < l2 < l3 < · · · and k1 < k2 < · · · in N and ∗homomorphisms 

λn : Aln → Bkn, µn : Bkn → Aln+1 such that ψ∞,kn∗ ◦λn∗ = f ◦ϕ∞,ln∗ 

and ϕ∞,ln+1∗ ◦ µn∗ = f −1 ◦ ψ∞,kn∗ for all n and 

λ1 

Al1 

 

ϕl2 ,l ϕl 1 3 ,l ϕl 2 4 ,l3 

Al2 

 

Al3 

 

 

µ1 

 

 

 

λ2 

 

 

µ2 

 

 

 

λ3 

 

. 

. . 

 

µ3 

 

 

 

 

. . . 

Bk1 

 

Bk2 

 

Bk3 

ψk2 ,k ψk 1 3 ,k ψk 2 4 ,k3 commutes. Once this has been obtained it is easy to finish the proof : There are 

∗-homomorphisms ϕ : A → B and ψ : B → A such that 

ϕ ◦ ϕ∞,ln = ψ∞,kn ◦ λn


and 

ψ ◦ ψ∞,kn = ϕ∞,ln+1 ◦ µn 

for all n and hence ϕ and ψ must be the inverse of each other. That ϕ∗ = f follows 

as in the proof of 1). 

The desired diagram is constructed by induction. Let us assume that everything 

has been constructed for n ≤ N, i.e. that we have everything up to 

ϕlN ,lN−1 AlN−1 

AlN 

λN−1 

 

 

 

 

 

 

µN−1 

 

λN 

BkN−1 

 

BkN 

ψkN ,kN−1 . 

To construct µN we repeat the argument used in the proof of 1). By Lemma 3.34 1) 

there is an m > lN and a ∗-homomorphism µ : BkN → Am such that ϕ∞,m∗ ◦ µ∗ = 

f −1 ◦ ψ∞,kN ∗ . Then 

ϕ∞,m ∗ ◦ (µ ◦ λN) ∗ = f −1 ◦ ψ∞,kN ∗ ◦ λN ∗ = f −1 ◦ f ◦ ϕ∞,lN ∗ = ϕ∞,lN ∗ 

so we conclude that there is a lN+1 > m such that 

ϕlN+1,m ∗ ◦ (µ ◦ λN) ∗ = ϕlN+1,lN ∗ . 

By Lemma 3.34 2) there is then a unitary v ∈ A + 

such that 

lN+1 

Ad v ◦ ϕlN+1,m ◦ µ ◦ λN = ϕlN+1,lN . 

Set µN = Ad v ◦ ϕlN+1,m ◦ µ. Then ϕ∞,lN+1∗ ◦ µN ∗ = ϕ∞,m ∗ ◦ µ∗ = f −1 ◦ ψ∞,kN ∗ . 

The construction of kN+1 > kN and λN+1 : AlN+1 → BkN+1 is similar and should be 

routine now. 

The classification result for unital AF-algebras can be improved a little by use 

of the following lemma. 

Lemma 3.36. Let A be a unital AF-algebra. Then 

Σ(A) = {x ∈ K0(A) : 0 ≤ x ≤ [1]}. 

Proof. When p is a projection in A we have that 

[1] = [p] + [1 − p] 

in K0(A), cf. the proof of Lemma 3.9. Hence 0 ≤ [p] ≤ [1], proving that 

Σ(A) ⊆ {x ∈ K0(A) : 0 ≤ x ≤ [1]}. 

To prove the reverse inclusion, let x ∈ K0(A) such that 0 ≤ x ≤ [1]. Then x = [p] 

for some projection p ∈ Mn(A) and by Lemma 3.27 1) there is a projection r ∈ 

Mm(A) such that p ⊕ r ≈ 1. Let v ∈ Mn+m(A) be a partial isometry such that 

vv ∗ = 1, v ∗ v = p ⊕r. Set q = vpv ∗ and note that q is a projection : q 2 = vpv ∗ vpv ∗ = 

vp(p ⊕ r)pv ∗ = vpv ∗ = q and that q = vpv ∗ ≤ v1v ∗ = 1. Hence q is a projection in 

A (strictly speaking, q is a projection in Mn+m(A) whose only non-zero entry is in 

the upper left-hand corner). Hence x = [p] = [q] ∈ Σ(A). 

Theorem 3.37. Let A and B be unital AF-algebras. 

1) If f : K0(A) → K0(B) is a positive group homomorphism such that f([1]) = 

[1], then there is a unital ∗- homomorphism ϕ : A → B such that ϕ∗ = f.


2) If ϕ, ψ : A → B are two unital ∗-homomorphisms such that ψ∗ = ϕ∗ on 

K0(A), then there is a sequence un of unitaries in B such that 

ϕ(a) = lim 

n→∞ unψ(a)u ∗ n, a ∈ A. 

3) When f : K0(A) → K0(B) is an isomorphism of partially ordered groups 

such that f([1]) = [1], then there is an isomorphism ϕ : A → B such that 

ϕ∗ = f. 

Proof. 1): By combining Lemma 3.36 with Theorem 3.35 1) we get a ∗homomorphism 

ψ : A → B such that ψ∗ = f. Since [ψ(1)] = f([1]) = [1] in K0(B) 

we conclude from Lemma 3.27 that there is a unitary u ∈ B such that uψ(1)u ∗ = 1. 

Set ϕ = Ad u ◦ ψ. 

2) and 3) follow straightforwardly from Lemma 3.36 and Theorem 3.35. 

3.3. The range of the invariant : Dimension groups. The classification 

result, Theorem 3.35, says that the isomorphism class of an AF-algebra is completely 

determined by the K0-group, considered as a partially ordered Abelian group, plus 

the position of the scale in this group or the position of [1] in the unital case, cf. 

Theorem 3.37. In this section we give an abstract description of the partially ordered 

groups which occur in this way, i.e. as the K0-group of an AF-algebra. We will then, 

subsequently, determine the subsets of the group which can play the role of the scale 

of an AF-algebra. 

We emphasize that the K0-group of an arbitrary C ∗ -algebra is not partially 

ordered. For an AF-algebra A, however, K0(A) is always partially ordered by Lemma 

3.27 3). 

Lemma 3.38. Let G be a partially ordered Abelian group. Then the following 

conditions are equivalent : 

1) G is upward directed, i.e. for all g, h ∈ G there is a k ∈ G such that g ≤ k 

and h ≤ k. 

2) G is downward directed, i.e. for all g, h ∈ G there is a k ∈ G such that 

k ≤ g and k ≤ h. 

3) G is generated by G + . 

4) G = G + − G + . 

Proof. 1) : ⇔ 2) follows by using the map g ↦→ −g. 

1) ⇒ 4) : Given a ∈ G, upward directedness implies that there is some x ∈ G 

such that a ≤ x and 0 ≤ x. Then a = x − (x − a) and x, x − a ∈ G + . 

4) ⇒ 1) : Given a1, a2 ∈ G, write ai = xi − yi for some xi, yi ∈ G + , i = 1, 2. Set 

a = x1 + x2 and note that a1, a2 ≤ a. 

3) ⇒ 4) : Let a ∈ G. Since a is in the group generated by G + , there are integers 

zi ∈ Z and elements gi ∈ G + , i = 1, 2, . . ., n, such that 

a = z1g1 + z2g2 + · · · + zngn. 

For each i, write zi = si − ti, where si, ti ∈ N. Then x = n 

i=1 sigi ∈ G + , y = 

n 

i=1 tigi ∈ G + and a = x − y. 4) ⇒ 3) is trivial. 

A partially ordered Abelian group is directed when the equivalent conditions of 

Lemma 3.38 are satisfied. Note that Exercise 3.49 shows that the K0-group of an 

AF-algebra is directed.


Proposition 3.39. Let G be a partially ordered Abelian group. Then the following 

conditions are equivalent : 

1) If x1, x2, y1, y2 ∈ G and xi ≤ yj for all i, j, then there exists z ∈ G such that 

xi ≤ z ≤ yj for all i, j. 

2) If x, y1, y2 ∈ G + and x ≤ y1 + y2, then x = x1 + x2 for some x1, x2 ∈ G + 

such that xj ≤ yj, j = 1, 2. 

3) If x1, x2, y1, y2 ∈ G + and x1 + x2 = y1 + y2, then there exist elements zij ∈ 

G + , i, j = 1, 2, such that xi = zi1 +zi2, i = 1, 2 and yj = z1j +z2j, j = 1, 2. 

Proof. 1) ⇒ 2) : We are given 0 ≤ x and 0 ≤ y2. Since y1 ≥ 0, we also have 

x − y1 ≤ x and x − y1 ≤ y2. By 1), there is some x2 ∈ G such that 0 ≤ x2 ≤ x and 

x − y1 ≤ x2 ≤ y2. Set x1 = x − x2. Then x1, x2 ∈ G + with x = x1 + x2 and x2 ≤ y2. 

As x − y1 ≤ x2, we also have x1 = x − x2 ≤ y1. 

2) ⇒ 3) : Since x2 ≥ 0, we have x1 ≤ y1 + y2, so by 2), x1 = z11 + z12 for some 

z11, z12 ∈ G + such that z1j ≤ yj, j = 1, 2. Set z2j = yj − z1j, j = 1, 2, so that 

z2j ∈ G + and yj = z1j + z2j. Since 

x1 + x2 = y1 + y2 = z11 + z21 + z12 + z22 

we conclude that x2 = z21 + z22. 

3) ⇒ 1) : We have that yj − xi ∈ G + for all i, j and 

(y1 − x1) + (y2 − x2) = (y1 − x2) + (y2 − x1). 

By 3) there exist elements zij ∈ G + , i, j = 1, 2, such that yi −xi = zi1 +zi2, i = 1, 2 

while also y1 −x2 = z11 +z21 and y2 −x1 = z12 +z22. Set z = x1 +z12, and note that 

x1 ≤ z. As y2 −x1 = z12 +z22, we obtain z = y2 −z22 ≤ y2. On the other hand, from 

y1 − x1 = z11 + z12, we see that z = y1 − z11 ≤ y1. Finally, since y1 − x2 = z11 + z21, 

we have that x2 ≤ x2 + z21 = y1 − z11 = z. Thus xi ≤ z ≤ yj for all i, j. 

Property 1) of Proposition 3.39 is called the Riesz interpolation property, while 

properties 2),3) are called the Riesz decomposition property. A partially ordered 

Abelian group is an interpolation group when it has one and hence both of these 

properties. 

In the present context it is of crucial importance that the interpolation property 

is preserved under inductive limits. 

Proposition 3.40. Let 

G1 

µ1 

G2 

µ2 

G3 

µ3 

. . . 

be a sequence of interpolation groups with positive connecting maps. Then the inductive 

limit group G = lim 

−→ (Gn, µn) is an interpolation group. 

Proof. Let qn : Gn → G denote the canonical homomorphisms. Let xi, yj ∈ G 

such that xi ≤ yj, i, j = 1, 2. There is a k ∈ N such that 

xi, yj ∈ qk(Gk) 

for all i, j. Take ai, bj ∈ Gk such that qk(ai) = xi, qk(bj) = yj for all i, j. Since 

yj − xi ∈ G + , there is an m > k such that 

µm,k(bj − ai) ∈ G + m 

for all i, j. Since Gm is an interpolation group, there is an element c ∈ Gm such that 

µm,k(ai) ≤ c ≤ µm,k(bj) for all i, j. Set z = qm(c). Then xi ≤ z ≤ yj for all i, j.


A partially ordered Abelian group G is lattice ordered when the following holds: 

For each pair a, b ∈ G there is an element, denoted by a ∨ b, such that 

and 

a, b ≤ a ∨ b 

c ∈ G, a, b ≤ c ⇒ a ∨ b ≤ c. 

In other words, the supremum of a and b exists. 

In is clear that a lattice ordered group is also an interpolation group. However, 

the property of being lattice ordered is not preserved in inductive limits. A 

prominent example of a lattice ordered group is Z n , ordered in the natural way by 

Z n+ = {(z1, z2, . . .,zn) ∈ Z n : zi ≥ 0, i = 1, 2, . . ., n}. 

A partially ordered group which is isomorphic to Z n , ordered in this way (for some 

n) is called a simplicial group. The K0-group of a finite-dimensional C ∗ -algebra is a 

simplicial group by Exercise 3.50, so we see from Theorem 3.21 that the K0-group of 

an AF-algebra is the inductive limit of a sequence of simplicial groups. In particular, 

the K0-group of an AF-algebra is an interpolation group. 

Now we only need one more order theoretic property in order to specify which 

partially ordered groups occur as K0 of an AF-algebra. A partially ordered Abelian 

group G is unperforated when the implication n ∈ N, ng ≥ 0 ⇒ g ≥ 0 holds for all 

g ∈ G. Note that Z n , ordered by N n , is unperforated and that the inductive limit of 

a sequence of unperforated partially ordered Abelian groups is again unperforated. 

Hence the K0-group of an AF-algebra is always unperforated. 

All in all we see that the K0-group of an AF-algebra enjoys the following properties 

1) directedness, 

2) the Riesz interpolation property, 

3) unperforatedness. 

A partial ordered Abelian group with these three properties will be called a 

dimension group. In other words, a dimension group is an interpolation group which 

is directed and unperforated. An obvious example of a dimension group is R, the 

real numbers, with its natural ordering. However, R is not countable, and can 

therefore not be the K0-group of an AF-algebra. The K0-group of an AF-algebra is 

the inductive limit of a sequence of countable groups and must therefore also itself 

be countable. This size restriction is the last restriction, and it is the only restriction 

which is not directly related to the partial ordering. But before we can present a 

proof of this assertion we need some preparations. The main problem is to show 

that a countable dimension group is the inductive limit of a sequence of simplicial 

groups. 

We first need the following strong form of the Riesz decomposition properties. 

Lemma 3.41. Let G be a dimension group and let x1, x2, . . .,xn ∈ G + and 

a1, a2, . . .,an ∈ Z such that 

a1x1 + a2x2 + · · · + anxn = 0. 

Then there exist elements y1, y2, . . ., ym ∈ G + and nonnegative integers bij, i = 

1, 2, . . ., n, j = 1, 2, . . ., m, such that 

xi = bi1y1 + bi2y2 + · · · + bimym


for all i = 1, 2, . . ., n, and 

for all j = 1, 2, . . ., m. 

a1b1j + a2b2j + · · · + anbnj = 0 

Proof. Let N×N have the lexicographic ordering, i.e. (a, b) ≤ (a1, b1) iff a < a1 

or a = a1 and b ≤ b1. (We remark that 0 ∈ N.) This is a total ordering, i.e. for any 

pair z1, z2 ∈ N × N, either z ≤ z1 or z1 ≤ z. For any finite tuple 

a1, a2, . . .,an 

of elements from Z, let the degree of a1, a2, . . .,an be the element (a, l) ∈ N × N 

defined as follows: a = max {|a1|, |a2|, . . ., |an|} and l is the number of times a 

occurs in the list |a1|, |a2|, . . ., |an|. The proof consists in establishing the following 

assertion : 

A) When (a, l) ∈ N × N and the conclusion of the lemma holds for all finite 

tuples from Z whose degree is strictly less than (a, l), then the conclusion 

of the lemma holds for all finite tuples from Z whose degree is (a, l). 

If this assertion has been proved the lemma follows by a simple contradiction 

argument as follows : If there was a finite tuple a1, a2, . . .,an of elements from Z 

for which the lemma failed, then we could use A) repeatedly to conclude that there 

would be such a tuple, for which the lemma failed, with degree = (0, 1). But the 

lemma is obviously true for such tuples, and we have the desired contradiction. 

In order to prove A) we first establish the lemma in the special case where ai ≥ 0 

for all i. In this case we have that 

0 ≤ xi ≤ aixi ≤ a1x1 + · · · + anxn = 0 

when ai > 0, so we see that ai > 0 ⇒ xi = 0 in this case. If xi = 0 for all 

i we can set m = 1, y1 = 0 and bi1 = 0 for all i = 1, 2, . . ., n, i.e. the lemma 

holds in this case. We can therefore assume that ai = 0 for at least one i. After 

reindexing there is then an m ∈ {1, 2, . . ., n} such that a1 = a2 = · · · = am = 0 

and xj = 0, j = m + 1, m + 2, . . .,n. By choosing yj = xj, j = 1, 2, . . ., m, and 

bjj = 1, j = 1, 2, . . ., m, and bij = 0 for all other i, j, we get the conclusion of the 

lemma in this case also. 

The case where ai ≤ 0 for all i follows from the case where ai ≥ 0 for all i by 

multiplying the ai’s by −1. 

Now we proceed to the proof of A), i.e. we consider a general finite tuple 

a1, a2, . . ., an and assume that the conclusion of the lemma holds for all tuples whose 

degree is strictly less than that of the given one. If the degree is (0, l) for some l, 

we are in the case where ai = 0 for all i and this case has already been covered. 

So we may assume that the degree is (a, l) with a > 0. Furthermore, by what we 

have just shown, we can disgard the case where all the ai’s are either non-negative 

or non-positive, i.e. we may assume that ai > 0 for some i and aj < 0 for some j. 

Reindex the ai’s such that |a1| = a. As it is harmless to multiply by −1, we 

can assume that a1 = a. Next reindex a2, a3, . . .,an and x2, x3, . . .,xn so that 

a1, a2, . . ., ar ≥ 0 and ar+1, ar+2, . . .,an < 0 for some r ∈ {2, 3, . . ., n−1}. Note that 

a ≥ −ai ≥ 0 for all i = r + 1, r + 2, . . ., n. We have 

ax1 = a1x1 ≤ a1x1 + a2x2 + · · · + arxr = 

− ar+1xr+1 − ar+2xr+2 − · · · − anxn ≤ a(xr+1 + xr+2 + · · · + xn),

which implies that 


x1 ≤ xr+1 + xr+2 + · · · + xn 

because G is unperforated. By n−r applications of the Riesz decomposition property 

we can write 

x1 = zr+1 + zr+2 + · · · + zn, 

where 0 ≤ zi ≤ xi, i = r + 1, r + 2, . . .,n. Then 

(a1 + ar+1)zr+1 + · · · + (a1 + an)zn + a2x2 + · · · + arxr+ 

ar+1(xr+1 − zr+1) + · · · + an(xn − zn) = 

a1(zr+1 + · · · + zn) + a2x2 + · · · + anxn = 0. 

As 0 ≤ a1 + ai < a1 = a for all i = r + 1, r + 2, . . .,n, we see that a appears in 

the list 

|a1 + ar+1|, . . ., |a1 + an|, |a2|, . . .,|an| 

at most l − 1 times. Consequently the degree of the tuple 

a1 + ar+1, . . .,a1 + an, a2, a3, . . .,an 

is strictly less than (a, l). Hence, by assumption, we can find elements y1, y2, . . .,ym ∈ 

G + and nonnegative integers cij, i = r + 1, r + 2, . . .,n, j = 1, 2, . . ., m, and dij, i = 

2, 3, . . ., n, j = 1, 2, . . ., m, such that 

zi = ci1y1 + · · · + cimym, i = r + 1, . . ., n, 

while 

xi = di1y1 + · · · + dimym, i = 2, 3, . . ., r, 

xi − zi = di1y1 + · · · + dimym, i = r + 1, . . ., n. 

(a1 + ar+1)cr+1,j + · · · + (a1 + an)cnj + a2d2j + · · · + andnj = 0 

for all j = 1, 2, . . .,m. 

Define bij by 

i=1 

b1j = cr+1,j + · · · + cnj, j = 1, 2, . . ., m, 

bij = dij, i = 2, 3, . . ., r, j = 1, 2, . . ., m, 

bij = cij + dij, i = r + 1, . . ., n, j = 1, . . ., m. 

Then xi = bi1y1 + · · · + bimym for all i = 1, 2, . . ., n, and 

n 

n 

r n 

n 

aibij = a1 cij + aidij+ ai(cij+dij) = 

i=r+1 

i=2 

i=r+1 

i=r+1 

(a1+ai)cij+ 

n 

aidij = 0 

for all j = 1, 2, . . ., m. This completes the proof of A) and hence of the lemma. 

Lemma 3.42. Let G be a dimension group, H1 a simplicial group, and let h1 : 

H1 → G be a positive homomorphism. Then there is a simplicial group H2 and 

positive homomorphisms h : H1 → H2 and h2 : H2 → G such that h1 = h2 ◦ h and 

ker h1 = ker h. 

Proof. We will first prove the following assertion by induction in k : 

(A) Let H be a simplicial group and h : H → G a positive homomorphism. 

Let v1, v2, · · · , vk be elements of ker h. There is then a simplicial group H ′ and 

positive homomorphisms, h ′ : H → H ′ and h ′′ : H ′ → G such that h = h ′′ ◦ h ′ and 

{v1, v2, · · · , vk} ⊆ ker h ′ . 

i=2


We start the induction by considering the case k = 1 : If H = {0} we can just 

take H ′ = H, h ′′ = h ′ = 0. Now assume that H = {0}. Let e1, e2, . . .,en be elements 

of H such that 

H = Ze1 ⊕ Ze2 ⊕ · · · ⊕ Zen 

and 

H + = Ne1 + Ne2 + · · · + Nen. 

Such elements exist because H is a simplicial group. Set xi = h(ei) ∈ G + , i = 

1, 2, . . ., n. Write v1 = a11e1 + · · · + a1nen for some a1i ∈ Z. Then 

a11x1 + · · · + a1nxn = h(a11e1 + · · · + a1nen) = h(v1) = 0. 

By Lemma 3.41 there exist elements y1, y2, . . ., ym ∈ G + and non-negative integers 

bij, i = 1, 2, . . ., n, j = 1, 2, . . ., m, such that 

and 

xi = bi1y1 + · · · + bimym, i = 1, 2, . . ., n, 

a11b1j + · · · + a1nbnj = 0, j = 1, 2, . . .,m. 

Let H ′ be the simplicial group H ′ = Z m , and let fi, i = 1, 2, . . ., m, be the standard 

basis for Z m . We may define homomorphisms h ′ : H → H ′ and h ′′ : H ′ → G so that 

and 

h ′ (ei) = bi1f1 + · · · + bimfm, i = 1, 2, . . .,n, 

h ′′ (fj) = yj, j = 1, 2, . . ., m. 

As each bij ≥ 0 and each yj ≥ 0, we see that both h ′ and h ′′ are positive homomor- 

phisms. Since 

h ′′ ◦ h ′ (ei) = h ′′ ( 

m 

bijfj) = 

j=1 

for all i, we see that h ′′ ◦ h ′ = h. In addition, 

h ′ (v1) = 

n 

i=1 

a1ih ′ (ei) = 

n 

i=1 

m 

j=1 

bijyj = xi 

m 

a1ibijfj = 0, 

so that v1 ∈ ker h ′ as desired. Thus (A) holds when k = 1. 

Now let k > 1 and assume that there exists a simplicial group H ′ and positive 

homomorphisms h ′ : H → H ′ and h ′′ : H ′ → G such that h = h ′′ ◦ h ′ and 

{v1, v2, . . .,vk−1} ⊆ ker h ′ . Note that h ′′ ◦h ′ (vk) = h(vk) = 0 so that h ′ (vk) ∈ ker h ′′ . 

Applying the case k = 1, we obtain a simplicial group H ′′ and positive homomorphisms 

ψ : H ′ → H ′′ and l : H ′′ → G such that h ′′ = l ◦ψ and h ′ (vk) ∈ ker ψ. Then 

ψ ′ = ψ ◦ h ′ : H → H ′′ is a positive homomorphism such that l ◦ ψ ′ = h ′′ ◦ h ′ = h. 

Clearly, {v1, . . .,vk−1} ⊆ ker ψ ′ and ψ ′ (vk) = ψ ◦ h ′ (vk) = 0. This completes the 

induction step. 

Using assertion (A) we can easily prove the lemma. Since H1 is a finitely generated 

Abelian group, so is ker h1. Thus 

j=1 

ker h1 = Zv1 + Zv2 + · · · + Zvk 

for some elements v1, v2, . . .,vk ∈ ker h1. There is then a simplicial group H2 and 

positive homomorphisms h : H1 → H2 and h2 : H2 → G such that h1 = h2 ◦ h 

and {v1, v2, · · · , vk} ⊆ ker h. It is then automatic that ker h1 ⊆ ker h and the 

factorization h1 = h2 ◦ h shows that ker h ⊆ ker h1.


Theorem 3.43. Let G be a countable dimension group. There is then a sequence 

G1 

p1 

G2 

p2 

G3 

p3 

. . . 

of simplicial groups and positive homomorphisms such that G ≃ lim 

−→ (Gn, pn) as partially 

ordered Abelian groups. 

Proof. Let G + = {x1, x2, . . . } be a numeration of G + . We shall construct 

simplicial groups G1, G2, . . . and positive homomorphisms gn : Gn → G and pn : 

Gn → Gn+1 for all n such that 

1) xn ∈ gn(G + n) for all n, 

2) gn ◦ pn−1 = gn−1 for all n > 1, 

3) ker gn = ker pn for all n. 

To start, set G1 = Z and define g1 : G1 → G such that g1(1) = x1. 

Now assume that G1, G2, . . .,Gn, g1, g2, . . ., gn, p1, p2, . . .,pn−1 have been constructed. 

Set H = Gn ⊕ Z, ordered by G + n ⊕ N, so that H is a simplicial group. We 

may define a positive homomorphism h : H → G by 

h(x, z) = gn(x) + zxn+1. 

Applying Lemma 3.42, we obtain a simplicial group Gn+1 and positive homomorphisms 

h ′ : H → Gn+1 and gn+1 : Gn+1 → G such that h = gn+1 ◦ h ′ and 

ker h = ker h ′ . As (0, 1) ∈ H + , we have h ′ ((0, 1)) ∈ G + n+1 with gn+1 ◦ h ′ ((0, 1)) = 

h((0, 1)) = xn+1, so that xn+1 ∈ gn+1(G + n+1 ). Let q : Gn → H be the natural 

injection (to the first coordinate), and set pn = h ′ ◦ q. Then pn is a positive homomorphism 

Gn → Gn+1 such that gn+1 ◦ pn = gn+1 ◦ h ′ ◦ q = h ◦ q = gn. In 

particular, ker pn ⊆ ker gn. On the other hand, q(ker gn) ⊆ ker h = ker h ′ , and 

hence ker gn ⊆ ker h ′ ◦ q = ker pn. Thus ker gn = ker pn as desired, and we have 

completed the induction step. 

Now let K = lim(Gn, 

pn) be the partially ordered inductive limit group and let 

−→ 

kn : Gn → K be the natural map. Because of condition 2), there is a unique positive 

homomorphism g : K → G such that g ◦ kn = gn for all n. Condition 1) then shows 

that xn ∈ gn(G + n ) = g ◦ kn(G + n ) ⊆ g(K+ ) for all n, i.e. G + = g(K + ). In particular, 

it follows that g is surjective. To show that g is also injective, let x ∈ ker g. Write 

x = kn(y) for some n ∈ N and some y ∈ Gn. Then gn(y) = g ◦ kn(y) = 0. Condition 

3) now implies that y ∈ ker pn so that x = kn+1 ◦ pn(y) = 0. 

Let G be a partially ordered Abelian group. An element u ∈ G + is an order unit 

when the following holds : For all g ∈ G there is an n ∈ N such that g ≤ nu. Thus 

the element [1] in the K0-group of a unital C ∗ -algebra is always an order unit and 

it is through this example the notion of an order unit becomes important here. 

Proposition 3.44. Let G be a countable dimension group and u ∈ G an order 

unit. There is then a sequence 

G1 

p1 

G2 

p2 

G3 

p3 

. . . 

of simplicial groups with order units un ∈ Gn, positive homomorphisms pn such that 

pn(un) = un+1, and an isomorphism ϕ : lim(Gn, 

pn) → G of partially ordered Abelian 

−→ 

groups such that 

ϕ(qn(un)) = u, n ∈ N, 

when qn : Gn → lim(Gn, 

pn) is the canonical map. 

−→


Proof. We know from Theorem 3.43 that there is a sequence 

H1 

r1 

H2 

r2 

H3 

r3 

. . . 

of simplicial groups and positive homomorphisms such that G ≃ lim(Hn, 

rn) as par- 

−→ 

tially ordered groups. Let v ∈ lim(Hn, 

rn) be the image of u under this isomorphism. 

−→ 

Then v is an order-unit in lim(Hn, 

rn). In particular, v is positive, i.e. v ∈ q 

−→ ′ n(H + n ) 

for some n ∈ N. Here q ′ n : Hn → lim(Hn, 

rn) is the canonical map. Since we can 

−→ 

ignore Hi, i < n, we may assume that v = q ′ 1 (u1) for some u1 ∈ H + 1 . Define un ∈ Hn 

by 

Set 

un+1 = rn(un), n ∈ N. 

Gn = {x ∈ Hn : ∃k ∈ N : −kun ≤ x ≤ kun}, 

n ∈ N. It is straightforward to check that Gn is a simplicial group in the ordering 

inherited from Hn, i.e. when the positive semigroup is G + n = H + n ∩ Gn, and that un 

is an order unit for Gn. Set pn = rn|Gn. Then 

G1 

p1 

G2 

p2 

G3 

p3 

. . . 

is a sequence of simplicial groups with positive connecting maps such that pn(un) = 

un+1 for all n. Let qn : Gn → lim(Gn, 

pn) be the canonical maps and note that there 

−→ 

is a positive group homomorphism 

ψ : lim(Gn, 

pn) → lim(Hn, 

rn) 

−→ −→ 

such that ψ ◦ qn = q ′ n |Gn. It is clear that ψ is injective and that ψ(qn(un)) = v for 

all n. It now suffices to show that ψ lim(Gn, 

pn) 

−→ + = lim(Hn, 

rn) 

−→ + . So let x = q ′ n (y) 

for some y ∈ H + n 

. Since v is an order unit in lim 

−→ (Hn, rn), there is a k ∈ N such that 

0 ≤ x ≤ kv, i.e. 0 ≤ q ′ n (y) ≤ kq′ 1 (u1) = kq ′ n (un). Since q ′ n (kun − y) ≥ 0, there is 

an m ≥ n such that rm,n(kun − y) ≥ 0. Then −kum ≤ 0 ≤ rm,n(y) ≤ kum, so that 

rm,n(y) ∈ H + m ∩ Gm = G + m. Hence qm(rm,n(y)) ∈ lim 

−→ (Gn, pn) + and ψ(qm(rm,n(y))) = 

q ′ m(rm,n(y)) = q ′ n(y) = x. 

 

Theorem 3.45. Let G be a countable dimension group and u ∈ G an order unit. 

There is then a unital AF-algebra A and an isomorphism ϕ : K0(A) → G of partially 

ordered groups such that ϕ([1]) = u. 

In short, 

(G, u) = (K0(A), [1]). 

Proof. By Proposition 3.44 we may assume that G is the inductive limit of a 

sequence 

G1 

p1 

G2 

p2 

G3 

p3 

. . . 

of simplicial groups with order units un ∈ Gn such that pn(un) = un+1 and qn(un) = 

u for all n. In fact, we may assume that there is a sequence {Nn} in N such that 

) and each 

Gn = ZNn , with its simplicial ordering. For each n, un = (dn 1 , dn2 , . . .,dn Nn 

dn i > 0 because un is an order unit in ZNn . Each pn : ZNn → ZNn+1 is given by a


Nn+1 ×Nn matrix {sij} with entries from N. Note that the condition pn(un) = un+1 

yields that 

Nn 

Set 

j=1 

sijd n j 

= dn+1 

i , i = 1, 2, . . ., Nn+1. 

An = Md n 1 ⊕ Md n 2 ⊕ · · · ⊕ Md n Nn 

and note that because of (2.8) and (2.9) there is a unital standard homomorphism 

ϕn : An → An+1 for each n. Set A = lim 

(An, ϕn). By combining Exercise 3.50 with 

−→ 

Theorem 3.21 one sees that there is an isomorphism ϕ : K0(A) → G = lim(Gn, 

pn) 

−→ 

such that ϕ([1]) = u. 

We now turn to the question of which subsets Σ of a dimension group G can 

be the scale of an AF-algebra A with G ≃ K0(A). When we restrict the attention 

to unital AF-algebras, Lemma 3.44 and Theorem 3.45 tells us the full story; the 

possible subsets Σ are exactly those of form Σ = {g ∈ G : 0 ≤ g ≤ e} for some order 

unit e in G. But what about the non-unital case? 

Definition 3.46. Let G be a dimension group. A subset Σ ⊆ G + is called a 

scale when the following hold : 

1) {g1 + g2 + · · · + gn : n ∈ N, g1, g2, · · · , gn ∈ Σ} = G + , 

2) 0 ≤ h ≤ g ∈ Σ ⇒ h ∈ Σ, g, h ∈ G, 

3) For every pair g, h ∈ Σ there is an element k ∈ Σ such that g ≤ k and 

h ≤ k. 

Theorem 3.47. Let Σ be a subset of a countable dimension group G. There 

is then an AF-algebra A and an isomorphism f : K0(A) → G of partially ordered 

groups such that f(Σ(A)) = Σ if and only if Σ is a scale. 

Proof. To prove the ’only if’ part we must show that Σ(A) is a scale in K0(A) 

when A is an AF algebra. We check that conditions 1),2) and 3) of Definition 3.46 

hold. Let A be the inductive limit of the sequence 

A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

· · · 

of finite dimensional C ∗ -algebras An. Let x ∈ K0(A) + . By Theorem 3.21 there is an 

N ∈ N and a y ∈ K0(AN) + such that x = ϕ∞,N ∗ (y), where ϕ∞,N : AN → A is the 

canonical map. Since AN is finite dimensional, it follows from Exercise 3.50 that y 

is in the semi-group of K0(AN) generated by Σ(AN). Since ϕ∞,N ∗ (Σ(AN)) ⊆ Σ(A), 

we see that x is in the semi-group of K0(A) generated by Σ(A). In particular, 1) 

holds. Let next x, y ∈ K0(A) such that 0 ≤ y ≤ x ∈ Σ(A). To check that 2) 

holds, note first that x = [p] for some projection p ∈ A. By Lemma 3.15 there is 

. Thus p and 

an N ∈ N and a projection q ∈ AN such that µ∞,N(q) − p < 1 

3 

ϕ∞,N(q) are equivalent by Lemma 3.14, i.e. x = [p] = [ϕ∞,N(q)]. Since 0 ≤ y, there 

is, by Theorem 3.21, a z ∈ K0(AM) + for some M ≥ N such that y = ϕ∞,M ∗ (z). 

Since y ≤ x = [ϕ∞,N(q)], there is a L ≥ M such that 0 ≤ ϕL,M ∗ (z) ≤ ϕL,N ∗ ([q])] in 

K0(AL). Since [ϕL,N(q)] ∈ Σ(AL), we see from Exercise 3.50 that ϕL,M ∗ (z) ∈ Σ(AL). 

Hence y = ϕ∞,L∗ (ϕL,M ∗ (z)) ∈ Σ(A). Thus 2) holds. To check 3), let x, y ∈ Σ(A). 

As above we see from Lemma 3.14 and Lemma 3.15 that there is an N ∈ N and 

projections p, q ∈ AN such that x = ϕ∞,N ∗ ([p]), y = ϕ∞,N ∗ ([q]). Let e be the unit


of AN. Then p ≤ e and q ≤ e in AN and hence z = ϕ∞,N ∗ ([e]) ∈ Σ(A) satisfies that 

x ≤ z, y ≤ z. Thus 3) holds also. 

To prove the ’if’ part, assume that Σ is a scale. Let e1, e2, e3, . . . be a numbering 

of the elements in Σ. It follows from 3) of Definition 3.46 that there is a sequence 

f1 ≤ f2 ≤ f3 ≤ . . . of elements in Σ such that fn ≥ ej, j = 1, 2, . . ., n, for all n ∈ N. 

Set 

Hn = {g ∈ G : −kfn ≤ g ≤ kfn for some k ∈ N}. 

It is straightforward to check that Hn is a dimension group for which fn is an 

order unit. By Theorem 3.45 there is a unital AF-algebra An and an isomorphism 

ψn : K0(An) → Hn of partially ordered groups such that ψn([1]) = fn. Then 

ψ −1 

n+1 ◦ ψn : K0(An) → K0(An+1) is a positive group homomorphism taking Σ(An) 

into Σ(An+1). By Theorem 3.35 1) there is a ∗-homomorphism ϕn : An → An+1 

such that ϕn∗ = ψ −1 

n+1 ◦ ψn. Let A be the inductive limit C ∗ -algebra of the sequence 

A1 

ϕ1 

A2 

ϕ2 

A3 

. By Corollary 2.17 A is an AF-algebra. Since 

Hn 

 

⏐ 

⏐ψn 

K0(An) −−−→ 

ϕn∗ 

ϕ3 

· · · 

⊆ 

−−−→ Hn+1 

 

⏐ 

⏐ψn+1 

K0(An+1) 

commutes, we get a group homomorphism ψ : K0(A) = lim 

−→ (K0(An), ϕn∗) → G such 

that ψ ◦ ϕ∞,n ∗ = ψn for all n. Since each ψn is injective, so is ψ. From the first part 

of the proof we know that an element x ∈ Σ(A) is of the form x = ϕ∞,N ∗ (y) for 

some N ∈ N and some y ∈ Σ(AN). Since ψ(x) = ψN(y) ∈ {h ∈ G : 0 ≤ h ≤ fN}, 

it follows from condition 2) of Definition 33.46 that ψ(x) ∈ Σ. This shows that 

ψ(Σ(A)) ⊆ Σ. On the other hand, if g ∈ Σ then 0 ≤ g ≤ fn for some n and hence 

g ∈ Hn. Since ψn(Σ(An)) = {h ∈ Hn : 0 ≤ h ≤ fn} there is a y ∈ Σ(An) such that 

ψn(y) = g. Then x = ϕ∞,n ∗ (y) ∈ Σ(A) and ψ(x) = ψn(y) = g. Hence ψ(Σ(A)) = Σ. 

Since Σ generates G by condition 1) of Definition 3.46, this implies in particular 

that ψ is surjective. 

3.4. Exercises to Chapter 3. 

Exercise 3.48. Let Tr : Mn → C be the usual trace. 

1) Show that there is a linear map Trk : Mk(Mn) → C given by 

⎛ 

a11 

⎜a21 

Trk 

⎜ 

⎝ 

. 

a12 

a22 

. 

⎞ 

. . . a1k 

. . . a2k⎟ 

. 

⎟ 

.. . 

⎠ = 

k 

Tr(aii) 

ak1 ak2 . . . akk 

which satisfies that Trk(AB) = Trk(BA), A, B ∈ Mk(Mn). 

2) Show that there is a semi-group isomorphism T : V (Mn) → N such that 

T([e]) = Trk(e), e ∈ Pk(Mn). 

3) Show that K00(Mn) ≃ Z. 

4) Show that under this isomorphism K00(Mn) + = N. 

i=1


Exercise 3.49. Let A = lim 

−→ (An, ϕn) be the inductive limit of the sequence 

A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

. . . 

of unital C ∗ -algebras, but not necessarily with unital connecting maps. Show that 

ιA∗ : K00(A) → K0(A) is an isomorphism of pre-ordered Abelian groups. (Hint : 

Combine Theorem 3.21, Theorem 3.16 and Lemma 3.19) 

Exercise 3.50. Let A = Mn1⊕Mn2⊕· · ·⊕MnN and let {ed ij : i, j = 1, 2, . . ., nd, d = 

1, 2, . . ., N} be the standard matrix units in A. Show that the map Z N → K0(A) 

given by 

is an isomorphism. Show also that 

and 

Σ(A) = { 

(z1, z2, . . ., zN) ↦→ 

N 

i=1 

zi[e i 11 ] 

K0(A) + = N[e 1 11] + N[e 2 11] + · · · + N[e N 11] 

N 

i=1 

ti[e i 11 ] : ti ∈ {0, 1, 2 . . ., ni}, i = 1, 2, . . ., N}. 

Exercise 3.51. Let G be a partially ordered Abelian group. Assume that G 

has the Riesz interpolation property. Show that G also has the following property: 

When x1, x2, · · · , xn and y1, y2, · · · , ym are elements of G such that xi ≤ yj for all 

i, j, there is an element z ∈ G such that xi ≤ z ≤ yj for all i, j. 

Exercise 3.52. Determine the Grothendieck group G(S) for the following semigroups 

: 

(1) S = N with +. 

(2) S = {n ∈ N : n > 7} with +. 

(3) S = [−1, 1] with · (multiplication). 

(4) S = [−1, 1]\{0} with ·. 

(5) S = {z ∈ C : Rez ≥ 0, Imz ≤ 0} with +. 

Exercise 3.53. a) Any subgroup of R is a partially ordered Abelian group with 

the order inherited from R. 

b) In G = Z ⊕ Z the following subsets all make G into a pre-ordered group : 

1) G + = {(x, y) ∈ Z 2 : x ≥ 0, y ≥ 0}. 

2) G + = {(x, y) ∈ Z 2 : x ≥ 0}. 

3) G + = {(x, y) ∈ Z 2 : x > 0 or x = 0, y ≥ 0}. 

In the second case the pre-order is not a partial order. 

c) Construct 7 examples of pre-ordered groups yourself. 

Exercise 3.54. Let A be a separable C ∗ -algebra. Prove that K00(A) and K0(A) 

are countable groups. (Hint : Remember that projections that are close are equivalent.) 

Exercise 3.55. Let 

A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

· · ·



−→ (An, ϕn). For each d ∈ N we have a new 

sequence of C ∗ -algebras, 

Md(A1) ϕ1 

Md(A2) ϕ2 

Md(A3) ϕ3 

· · · 

. Prove that lim 

−→ (Md(An), ϕn) ≃ Md(A). 

Exercise 3.56. Let ϕ, ψ : A → B be ∗-homomorphisms between C ∗ -algebras. 

Assume that there is a sequence {un} of unitaries in B + such that ϕ(a) = limn→∞ unψ(a)u ∗ n 

for all a ∈ A. Prove that ϕ∗ = ψ∗ : K0(A) → K0(B). 

Exercise 3.57. Let P be the projection introduced in Example 3.22. Show that 

[P] − [( 0 1 )] is not zero K00(A). 

Exercise 3.58. Let A be a C∗-algebra and define s : A → Mn(A) by 

⎛ ⎞ 

a 0 . . . 0 

⎜0 

0 . . . 0⎟ 

s(a) = ⎜ 

⎝ . 

⎟ 

. . .. 0 

⎠ 

0 0 . . . 0 

. 

Show that s∗ : K00(A) → K00(Mn(A)) is an isomorphism. 

Exercise 3.59. Let A, B be C ∗ -algebras and let p1 : A⊕B → A and p2 : A⊕B → 

B be the two projections. Show that the map K0(A ⊕ B) ∋ x → (p1∗(x), p2∗(x)) ∈ 

K0(A) ⊕ K0(B) gives an isomorphism of pre-ordered groups when 

(K0(A) ⊕ K0(B)) + = K0(A) + ⊕ K0(B) + . 

Exercise 3.60. a) Let A and B be C ∗ -algebras and ϕt : A → B, t ∈ [0, 1], be a 

family of ∗-homomorphisms such that t ↦→ ϕt(a) is continuous for all a ∈ A. Prove 

that ϕ0∗ = ϕ1∗ : K00(A) → K00(B) and ϕ0∗ = ϕ1∗ : K0(A) → K0(B). 

b) Define k : C → C[0, 1] by k(λ)(s) = λ. Prove that k∗ : K0(C) → K0(C[0, 1]) 

is an isomorphism of partially ordered groups. (Hint : Define ψ : C[0, 1] → C 

by ψ(f) = f(0). Then ψ ◦ k = idC. Conclude that k∗ is injective. Define 

ϕt : C[0, 1] → C[0, 1] by ϕt(f)(s) = f(st), s ∈ [0, 1] and use this to show that k∗ 

is surjective.) 

c) Show that K0(C([0, 1], A)) ≃ K0(A) as pre-ordered Abelian groups. 

Exercise 3.61. Determine the ordered K0-group of a UHF-algebra. 

Exercise 3.62. Determine the ordered K0-group of C(X), where X is the Cantor 

set, cf. Exercise 2.24 

4. Choquet simplices and C ∗ -algebras 

4.1. Compact convex sets and functional analysis. A compact convex set 

K is a convex subseteq of a locally convex real vector space which is compact in the 

relative topology. In this section we will develop the basic connection between the 

theory of compact convex sets and functional analysis. 

If A is a C ∗ -algebra, it follows from the Banach-Alouglu theorem that the unit 

ball of A ∗ is compact in the weak ∗ topology. If A has a unit, it follows that 

S(A) = {ω ∈ A ∗ : ω = ω(1) = 1}

4. CHOQUET SIMPLICES AND C ∗ -ALGEBRAS 95 

is a compact convex subset of A ∗ . By Lemma 1.79 S(A) consists exactly of the 

states of A, i.e. of the positive linear functionals of norm 1 on A. Consequently, 

S(A) is called the state space of A. By the Krein-Milman theorem every compact 

convex set is the closed convex hull of its extreme points, so this is in particular 

the case for S(A). The extreme points of S(A) are called pure states of A. It is 

symptomatic for the close relationship between the theory of compact convex sets 

and the theory of C ∗ -algebras that the former commences by determining the pure 

states of C(X) for a compact Hausdorff space X. 

Theorem 4.1. Let X be a compact Hausdorff space. A pure state of C(X) is of 

the form C(X) ∋ f ↦→ f(x) for some point x in X. 

Proof. Let ω be a pure state of C(X). We first prove that ω is a character, 

i.e. that ω(fg) = ω(f)ω(g), f, g ∈ C(X). It clearly suffices to show this for 

non-negative f and g with f ≤ 1. Set µ(h) = ω(fh), h ∈ C(X). Since f ≥ 0, 

µ is a positive linear functional. If µ = 0, we have in particular that µ(g) = 

ω(fg) = 0 = ω(g)µ(1) = ω(g)ω(f). If µ = ω, we have that ω(f) = 1 and that 

ω(fg) = µ(g) = ω(g) = ω(f)ω(g). If µ = 0 and µ = ω, we have that 

ω = µ(1) µ 

ω − µ 

+ (ω(1) − µ(1)) 

µ(1) ω(1) − µ(1) . 

Note that µ 

µ(1) , 

ω−µ 

ω(1)−µ(1) 

∈ S(C(X)). Since µ(1), ω(1) −µ(1) ∈ [0, 1], µ(1)+(ω(1) − 

µ(1)) = 1, the assumption that ω is pure, i.e. is an extreme point in S(C(X)), implies 

that ω = µ 

µ(1) 

. In particular these states must agree on g, i.e. ω(g) = µ(g) 

µ(1) 

= ω(fg) 

ω(f) . 

Since ω is a character, ker ω is a twosided ideal Iω in C(X). Note that C(X)/Iω ≃ 

C. Set Y = {y ∈ X : f(y) = 0, f ∈ Iω}. We assert that Y is not empty. Indeed, 

if it was we could for any z ∈ X find a function gz ∈ Iω such that gz(z) = 0. Then 

|gz|(y) > 0 for all y in an open neighbourhood Vz of z and |gz| = (gzgz) 1 

2 ∈ Iω. By 

compactness of X this would give us a finite set gz1, gz2, . . .,gzn of such functions 

such that n 

i=1 |gzi |(y) > 0 for all y ∈ X. Then h = n 

i=1 |gzi | ∈ Iω and hh −1 = 1. 

Thus 0 = ω(h)ω(h −1 ) = ω(1) = 1, contradiction. Hence Y = ∅ as asserted. Let x 

be an arbitrary point in Y . Then, for any f ∈ C(X), f − f(x)1 ∈ Iω, so ω(f) = 

ω(f − f(x)1) + ω(f(x)1) = f(x). 

Exercise 4.2. Prove the converse of Theorem 4.1 : For every x ∈ X the functional 

f ↦→ f(x) is a pure state of C(X). 

We have now, in effect, proved that the category of compact Hausdorff spaces is 

equivalent to the category of unital abelian C ∗ -algebras : To a compact Hausdorff 

space X we can associate the unital abelian C ∗ -algebra C(X), cf. Example 1.20. 

On the other hand, we can to any unital abelian C ∗ -algebra associate the compact 

Hausdorff space of characters of the algebra. By Theorem 1.31 and Theorem 4.1, 

these two operations are inverses of each other. 

In the right perspective the connection between the theory of compact convex sets 

and functional analysis is a generalisation of the bijective correspondance of compact 

Hausdorff spaces with unital abelian C ∗ -algebras. When K is a compact convex set, 

Aff K will denote the continuous realvalued affine functions on K, i.e. Aff K consists 

of the continuous functions f : K → R which respect the convex structure of K in 

the sense that f(tx + (1 − t)y) = tf(x) + (1 − t)f(y), t ∈ [0, 1], x, y ∈ K. Aff K 

is clearly a closed subspace of the continuous realvalued functions CR(K), i.e. of


the selfadjoint part of the C ∗ -algebra C(K). The state space , S(Aff K), of this 

subspace, i.e. 

S(Aff K) = {ω ∈ Aff K ∗ : ω = ω(1) = 1} 

is a compact convex subset of Aff K ∗ in the weak ∗ topology by the Banach-Alouglu 

theorem. The following theorem shows that S(Aff K) is K again, so that the real 

Banach space Aff K contains all information about K. Note that every point x ∈ K 

defines an element ωx of S(Aff K) by 

ωx(f) = f(x), f ∈ Aff K. 

The set of extreme points in a compact convex set K will be denoted by ∂eK. 

Theorem 4.3. The map x ↦→ ωx is an affine homeomorphism from K onto 

S(Aff K). 

Proof. Set Φ(x) = ωx. If {xα} is a net in K converging to x ∈ K, then 

limα f(xα) = f(x) for all f ∈ CR(K), in particular for all f ∈ Aff K. This shows 

that Φ is continuous. Assume x, y ∈ K, x = y. Let E be the locally convex 

real vectorspace of which K is a compact convex subseteq. By the Hahn-Banach 

separation theorem there is a continuous linear functional ω on E such that ω(x − 

y) = 0. The restriction ω|K defines an element of Aff K which takes different values 

at x and y. Thus ωx = ωy, showing that Φ is injective. 

To prove that Φ is also surjective, let first ω be an extreme point in S(Aff K). 

By the Hahn-Banach extension theorem there is a continuous real functional ω1 on 

CR(K) with the same norm as ω such that ω1|Aff K = ω. Extend ω1 to a linear 

functional ω2 on C(K) by 

ω2(f) = ω1( 1 

1 

(f + f)) + iω1( (f − f)), f ∈ C(K). 

2 2i 

Clearly, ω2 ≤ 2ω1 = 2ω. If 0 ≤ f ≤ 1, then ω2(1 − f) ≤ |ω2(1 − f)| ≤ 

ω11 − f ≤ ω = 1, from which we conclude that ω2(f) ≥ 0. Hence ω2 is a 

positive continuous linear functional on C(K), so by Lemma 1.79 ω2 = ω2(1) = 1, 

i.e. ω2 ∈ S(C(K)). It follows that 

F = {µ ∈ S(C(K)) : µ|Aff K = ω} 

is not empty. Note that F is a closed convex subseteq of S(C(K)). By the Krein- 

Milman theorem there is an extreme point ν in F. If ν = tµ1 + (1 − t)µ2, where 

t ∈]0, 1[ and µ1, µ2 ∈ S(C(K)), then ω = ν|Aff K = tµ1|Aff K + (1 − t)µ2|Aff K. Since 

µ1|Aff K, µ2|Aff K ∈ S(Aff K), the fact that ω is an extreme point in S(Aff K) implies 

that µ1|Aff K = µ2|Aff K = ω. Hence µ1, µ2 ∈ F, and since ν is an extreme point 

in F, we conclude that ν = µ1 = µ2. This shows that ν is in fact an extreme 

point in S(C(K)), not only in F. By Theorem 4.1 there is a point x ∈ K such 

that ν(f) = f(x), f ∈ C(K). In particular, ω(g) = ν(g) = g(x), g ∈ Aff K. 

Hence ω = ωx = Φ(x), i.e. ω ∈ Φ(K). Thus all extreme points in S(Aff K) are 

in Φ(K). Since Φ(K) is a compact convex subseteq of S(Aff K), it follows from 

the Krein-Milman theorem that Φ(K) = S(Aff K), i.e. Φ is surjective. As is wellknown 

a continuous bijection between compact Hausdorff spaces is automatically a 

homeomorphism, so the proof is complete. 

By Theorem 4.3 the real Banach space Aff K contains enough structure to recapture 

the compact convex set K defining it. In this way we can study K through


the space Aff K. In order to use this viewpoint effectively, it is necessary to have an 

abstract characterization of the real Banach spaces which occur in this way. 

A convex cone in a real vector space E is a subset C ⊆ E such that 0 ∈ C and 

t1, t2 ∈ R + , x1, x2 ∈ C ⇒ t1x1 + t2x2 ∈ C. Alternatively, C is a convex cone if and 

only if C is a convex subset of E such that t ∈ R + , x ∈ C ⇒ tx ∈ C. 

Definition 4.4. An ordered Banach space is a real Banach space X containing 

a closed convex cone X+. 

Any ordered Banach space carries a natural pre-order; namely, x ≤ y ⇔ y − x ∈ 

X+. Then X+ = {x ∈ X : x ≥ 0} and X+ is therefore referred to as the positive 

cone in X. 

Definition 4.5. An order unit space is an ordered Banach space X with a 

distinguished element u = 0 such that 

{x ∈ X : x ≤ 1} = {x ∈ X : −u ≤ x ≤ u}. 

Example 4.6. i) Let A be a unital C ∗ -algebra. Then Asa = {a ∈ A : a ∗ = a} is 

an ordered Banach space with positive cone consisting of the positive elements, i.e. 

Asa+ = A+ = {aa ∗ : a ∈ A}. 

If a ∈ Asa, then by spectral theory (Theorem 1.35), a ≤ 1 if and only if −1 ≤ a ≤ 

1. Hence Asa is an order unit space. 

ii) Let X be an ordered Banach space with the property that there is a constant 

K > 0 such that 

a ≤ x ≤ b ⇒ x ≤ K max{a, b} . 

(An ordered Banach space with this property is called normal .) If u is an interior 

point in X+, there is a norm · u which is equivalent to the given one and turns X 

into an order unit space with X+ as positive cone and u as order unit. This norm is 

xu = inf{t > 0 : −tu ≤ x ≤ tu}. 

Let us prove this. First we show that {t > 0 : −tu ≤ x ≤ tu} = ∅ for all x ∈ X. 

Indeed, since u is an interior point of X+ there is an ǫ > 0 such that the closed ball 

u ≤ 

of radius ǫ centered at u is contained in X+. For any x ∈ X, we have that − x 

ǫ 

x ≤ x 

x 

u because u − (u ± ǫ ǫ x ) ≤ ǫ when x = 0. Hence xu is welldefined and 

xu ≤ 1 

x. If −su ≤ x ≤ su and −tu ≤ y ≤ tu, then −(t+s)u ≤ x+y ≤ (s+t)u, 

ǫ 

proving that x+yu ≤ xu + yu. So · u is indeed a seminorm on X such that 

xu ≤ 1 

ǫ x. The normality condition implies that x ≤ Kuxu for all x ∈ X, 

so · u and · are indeed equivalent norms. 

It follows that most ”reasonable” ordered Banach spaces can be re-normed to 

become an order unit space. 

iii) If K is a compact convex set, Aff K is a real Banach space and 

is a closed convex cone. Since 

Aff K+ = {f ∈ Aff K : f(x) ≥ 0, x ∈ K} 

f = sup |f(x)| = inf{t > 0 : −t1 ≤ f ≤ t1}, 

x∈K 

we see that Aff K is an order unit space with the constant function 1 as order unit.


We now show that Example 4.6 iii) gives the most general example of an order 

unit space, showing that this notion is the right abstraction of Aff K. 

Let X be an order unit space with order unit u. Set 

S(X) = {ω ∈ X ∗ : ω = ω(u) = 1}. 

Then S(X) is a convex set which is compact in the weak ∗ -topology by the Banach- 

Alaoglu theorem. Note that if X is the selfadjoint part of the unital C ∗ -algebra A, cf. 

Example 4.6 i), then S(X) ≃ S(A), i.e. there is a homeomorphism f : S(X) → S(A) 

such that f(tµ1 +(1 −t)µ2) = tf(µ1)+(1 −t)f(µ2), t ∈ [0, 1], µ1, µ2 ∈ S(X). (Such 

a map will be called an affine homeomorphism and we say that S(X) and S(A) are 

affinely homeomorphic.) 

Lemma 4.7. S(X) = {ω ∈ X ∗ : ω(x) ≥ 0, x ∈ X+, ω(u) = 1}. 

Proof. ⊆ : Let ω ∈ S(X). When x ≥ 0 and x = 0, 0 ≤ x ≤ u and hence 

x 

u − x 

x 

≤ 1 so that ω(u − ) ≤ 1. It follows that ω(x) ≥ 0. ⊇ : Assume that 

x x 

ω(x) ≥ 0, x ∈ X+, and ω(u) = 1. If x ≤ 1 we have that −u ≤ x ≤ u and hence 

−1 ≤ ω(x) ≤ 1, proving that ω ≤ 1, i.e. ω ∈ S(X). 

Lemma 4.8. For x ∈ X, 

x = sup{|ω(x)| : ω ∈ S(X)}. 

Proof. If x = 0 there is nothing to prove, so assume that x = 0. For any 

0 < s < x we have that either su − x or x + su is not in X+. If su − x /∈ X+, 

Hahn-Banach separation gives us an element ω ∈ X∗ such that ω(y) ≥ 0, y ∈ X+, 

and ω(su − x) < 0. Then µ = ω ∈ S(X) by Lemma 4.7. (ω(u) = 0 since ω is 

ω(u) 

non-zero and non-negative on X+.) Note that s < µ(x). If instead x + su /∈ X+ 

we get in the same way an element µ ∈ S(X) such that s < −µ(x). It follows 

that s < sup{|µ(x)| : µ ∈ S(X)}. Since 0 < s < x was arbitrary, we see that 

x ≤ sup{|µ(x)| : µ ∈ S(X)}. The reversed inequality is trivial. 

Lemma 4.9. Let X be an order unit space. Then 

{ω ∈ X ∗ : ω ≤ 1} = co(S(X) ∪ −S(X)). 

Proof. Recall that co(S(X) ∪ −S(X)) = {tω + (1 − t)µ : ω, µ ∈ S(X) ∪ 

−S(X), t ∈ [0, 1]} = {tω − (1 − t)µ : ω, µ ∈ S(X), t ∈ [0, 1]}. Define µ : [0, 1] × 

S(X)×S(X) → X ∗ such that µ(t, µ, ν) = tµ−(1−t)ν. Then µ ([0, 1] × S(X) × S(X)) = 

co(S(X)∪−S(X)), so we conclude that co(S(X)∪−S(X)) is compact in the weak ∗ - 

topology because S(X) is and µ is continuous. Assume to get a contradiction that 

there is an element q ∈ {ω ∈ X ∗ : ω ≤ 1}\co(S(X) ∪ −S(X)). By the Hahn- 

Banach separation theorem there is then a weak ∗ -continuous functional which separates 

q and co(S(X) ∪ −S(X)). But a weak ∗ -continuous functional is given by 

an element of X, so we see that there is an element x ∈ X and a α ∈ R such 

that q(x) > α and µ(x) ≤ α, µ ∈ co(S(X) ∪ −S(X)). Note that α ≥ 0 and that 

|µ(x)| ≤ α for all µ ∈ S(X). By Lemma 4.8 it follows that x ≤ α, which is 

impossible since q(x) > α. This proves the lemma. 

For x ∈ X, define fx ∈ Aff S(X) by 

fx(ω) = ω(x), ω ∈ S(X).


Theorem 4.10. The map x ↦→ fx is an isometric isomorphism of X onto 

Aff S(X) which takes X+ onto Aff S(X)+ and u to the constant function 1. 

Proof. Set Φ(x) = fx. Then Φ is a linear map which is an isometry by Lemma 

4.8. It follows from Lemma 4.7 that Φ(X+) ⊆ Aff S(X)+. It remains only to show 

that Φ(X+) = Aff S(X)+. Since Φ is an isometry and X+ is closed in X, Φ(X+) is 

a closed convex cone in Aff S(X) and if Φ(X+) is not all of Aff S(X)+ the Hahn- 

Banach separation theorem gives us elements g ∈ Aff S(X)+, ν ∈ (Aff S(X)) ∗ and 

β ∈ R such that ν(z) ≤ β for all z ∈ Φ(X+) and ν(g) > β. Note first that β ≥ 0 

since 0 ∈ Φ(X+) and that ν(z) ≤ 0 for all z ∈ Φ(X+) since Φ(X+) is a cone. Thus 

while 

ν(g) > 0 (4.1) 

ν ◦ Φ(z) ≤ 0 (4.2) 

for all z ∈ X+. It follows from Lemma 4.9 that ν = t+ν+ − t−ν− where t± ≥ 0 and 

ν± ∈ S (Aff S(X)). Note that ν± are given by evaluation at two states ω± ∈ S(X) 

by Theorem 4.3. Then (4.2) implies that t+ω+(z) = t+ν+ ◦ Φ(z) ≤ t−ν− ◦ Φ(z) = 

t−ω−(z) for all z ∈ X+. In combination with (4.1) this implies in particular that 

t− = 0. Thus there is a γ ∈ S(X) and some t ≥ 0 such that 

ω− = t+ 

t− 

ω+ + tγ. 

Since g is affine this implies that g (ω−) = t+ 

t− g (ω+) + tg(γ), and hence that 

t−ν−(g) = t+ν+(g) + t−tg(γ). 

Thus ν(g) = t+ν+(g) − t−ν−(g) = −t−tg(γ). Note that −t−tg(γ) ≤ 0 since 

g ∈ Aff S(X)+ so that we have contradicted (4.1). Hence Φ(X+) must be all of 

Aff S(X)+. 

Corollary 4.11. Let X be an order unit space with order unit u. For x ∈ X, 

x ∈ X+ ⇔ xu − x ≤ 1 . 

We can now pass from compact convex sets to order unit spaces and from order 

unit spaces to compact convex sets. The functors are K → Aff K and X → S(X), 

respectively. By Theorem 4.3 and Theorem 4.10 the functors are inverses of each 

other, i.e. S(Aff K) = K by Theorem 4.3 and AffS(X) = X by Theorem 4.10. We 

shall later see that this equivalence of functors may be considered as an extension of 

the equivalence between compact Hausdorff spaces and unital abelian C ∗ -algebras. 

Let X be a compact metric space. A Borel measure µ on X is a probability 

measure when µ(X) = 1. The set of Borel probability measures on X will be 

denoted by M1(X). Since we assume that X is metrizable a Borel probability 

measure µ ∈ M1(X) is regular, i.e. satisfies that 

µ(B) = sup{µ(C) : C ⊆ B, C closed} = inf{µ(U) : B ⊆ U, U open} 

for all Borel subsets B of X. If X is not metrizable these properties will have to 

be added to the definition. There is a one-to-one correspondance between the Borel 

probability measures on X and the states of C(X) given by integration with respect 

to measures (the direction ’from measures to states’) and the Riesz representation


theorem (the direction ’from states to measures’). Thus the state sµ on C(X) given 

by the measure µ ∈ M1(X) is 

 

sµ(f) = f dµ, f ∈ C(X), 

and the Borel probability measure µs corresponding to the state s ∈ S(C(X)) is 

given by 

µs(V ) = sup{s(f) : f ∈ C(X), 0 ≤ f ≤ 1, supp(f) ⊆ V } 

when V ⊆ X is open. 

By using the identification M1(X) = S(C(X)), the space M1(X) of Borel probability 

measures becomes a compact convex set in the weak ∗ -topology. Note that the 

convex structure is the natural one where (tµ1+(1−t)µ2)(B) = tµ1(B)+(1−t)µ2(B) 

for all Borel sets B. By Theorem 4.1 and Exercise 4.2 the extreme points of the 

compact convex set M1(X) are exactly the point measures, i.e. measures δx obtained 

by fixing an x ∈ X and setting 

 

0, x /∈ B 

δx(B) = 

1, x ∈ B, 

for every Borel set B ⊆ X. Note that f dδx = f(x), f ∈ C(X). 

Exercise 4.12. 1) Show that a compact Hausdorff space X is metrizable 

if and only if C(X) is separable. 

(Sketch of a possible solution : If d is a metric, show that there is a 

sequence {Bn}n∈N of balls in X such that dist(·, Bn), n ∈ N, separates the 

points of X. By the Stone-Weierstrass theorem the complex polynomials in 

these functions, all of whose coefficients have rational real and imaginary 

parts, will be dense in C(X). - Conversely, if {fn} is a dense sequence in 

the unit ball of C(X), consider d(x, y) = ∞ 

n=1 2−n |fn(x) − fn(y)|.) 

2) Show that a subseteq of a separable normed vector space is itself separable. 

3) Let K be a compact convex set. Show that K is metrizable if and only if 

Aff K is separable. 

4) Let X be a metrizable compact Hausdorff space. Show that M1(X) is 

metrizable. 

Consider now a compact convex set K. 

Lemma 4.13. 1) For every Borel probability measure µ ∈ M1(K) there is 

a unique point b(µ) ∈ K such that 

 

f(b(µ)) = f dµ, f ∈ Aff K. 

2) b : M1(K) → K is an affine continuous surjection. 

Proof. 1) follows immediately from Theorem 4.3. 2) : When µ1, µ2 ∈ M1(K), t ∈ 

[0, 1], then f d(tµ1+(1−t)µ2) = t f dµ1+(1−t) f dµ2 for all f ∈ C(K). In particular, 

it holds for all f ∈ Aff K and hence b(tµ1 +(1−t)µ2) = tb(µ1)+(1−t)b(µ2), 

i.e. b is affine. When {µα} is a net in M1(K) converging to µ ∈ M1(K), we 

have in particular that limα f dµα = f dµ for all f ∈ Aff K, and hence that 

limα f(b(µα)) = f(b(µ)) for all f ∈ Aff K. By Theorem 4.3 this implies that 

limα b(µα) = b(µ).


The point b(µ) ∈ K will be called the barycenter of µ. 

Since the point measures {δx : x ∈ K} are the extreme points of M1(K), we know 

from the Krein-Milman theorem that every meausure µ ∈ M1(K) is the limit of a 

sequence (or net if K is not metrizable) of convex combinations of point measures. 

We shall need the fact that the elements in this approximating sequence (or net) 

can be chosen such that all of them have the same barycenter as µ. 

We say that µ ∈ M1(K) is supported on the Borel set B ⊆ K when µ(B) = 1. 

More generally, we say that a positive Borel measure ν is supported on B when 

ν(K\B) = 0. Note that a measure in M1(K) is a (finite) convex combination of 

point measures if and only if it is supported on a finite subseteq of K. Hence the 

result which we shall need can be stated as follows. 

Lemma 4.14. Let µ ∈ M1(K). There is then a net {να} in M1(K) such that 

each να is supported on a finite set, b(να) = b(µ) for all α and limα να = µ. 

Proof. We will define a directed set as follows. The elements of the set consists 

of finite partitions U = {Ui : i = 1, 2, . . ., n} of K into Borel sets, i.e. each Ui is a 

Borel set, Ui∩Uj = ∅ when i = j and 

i Ui = K. We order the set of Borel partitions 

in the following way : When U = {Ui} and V = {Vi} are two such partitions we 

write U ≤ V when every Vi is a subseteq of some Uj, i.e. when V is a refinement of 

U. In this way the Borel partitions form a directed set. Next we associate a finitely 

supported measure µU to each Borel partition in the following way. First choose 

a fixed, but arbitrary meausure µ0 ∈ M1(K). When U = {Ui : i = 1, 2, . . ., n} is 

a Borel partition, we set µi = µ0 when µ(Ui) = 0, and when µ(Ui) = 0 we define 

µi ∈ M1(K) as the unique Borel probability measure such that 

 

f dµi = 1 

µ(Ui) 

 

f1Ui 

dµ, f ∈ C(K). 

The existence and uniqueness of such a µi follows from the Riesz representation 

theorem. Set xi = b(µi) for all i and define 

n 

µU = µ(Ui)δxi . 

i=1 

Then µU has finite support (it is supported on {x1, x2, . . ., xn}) and we have that 

 

n 

n 

 

n 

 

f dµU = µ(Ui)f(xi) = µ(Ui) f dµi = f1Ui dµ = f dµ 

i=1 

i=1 

for all f ∈ Aff K, from which we conclude that b(µU) = b(µ). We assert that the net 

{µU} converges to µ. To prove this we must show that limU f dµU = f dµ for all 

f ∈ C(K). So fix such an f ∈ C(K) and use the compactness of K to find an open 

convex cover Wi, i = 1, 2, . . ., m, of K such that x, y ∈ Wi ⇒ |f(x)−f(y)| < ǫ for all 

i. Set U1 = W1 and Ui = Wi\ 

j


when λj = 0. To prove this we first prove that 

xj is in the closed convex hull of Vj. (4.4) 

If namely, (4.4) was not true the Hahn-Banach separation theorem would give us a 

g ∈ Aff K and a α ∈ R such that g(xj) > α while g(x) ≤ α for all x ∈ Vj. This would 

imply that λ −1 

g1Vj j dµ ≤ αλ−1 1Vj j dµ = α in contradiction with the estimate 

λ −1 

g1Vj j dµ = g dµj = g(b(µj)) = g(xj) > α. Next (4.4) is used to establish 

(4.3) as follows. Since U ≤ V there is a k such that Vj ⊆ Uk ⊆ Wk. Since we chose 

the Wi’s to be convex, we conclude from (4.4) that xj ∈ Wk. When x ∈ Vj ⊆ Wk 

we see from the choice of Wk that |f(x) − f(xj)| ≤ ǫ, proving (4.3). By using (4.3) 

we can now estimate 

 

n 

n 

 

| f dµV − f dµ| = | λif(xi) − f1Vi dµ| 

≤ 

n 

 

λi|f(xi) − 

i=1 

i=1 

f dµi| = 

n 

i=1 

 

λi| 

i=1 

(f(xi) − f)1Vi dµi| ≤ 

n 

λiǫ = ǫ. 

Thus we have shown that limU µU = µ as asserted. 

The reader can check on any compact convex subseteq of R 2 he or she can draw, 

that any point of the set can be realized as the barycenter of a Borel probability 

measure which is supported on the extreme points. This holds in general for all 

compact convex sets, except for the annoying fact that the set ∂eK of extreme 

points of K need not be a Borel set when K is not metrizable. Our next goal is to 

prove this in the metrizable case. 

Set 

S(K) = {f ∈ CR(K) : f(tx + (1 − t)y) ≤ tf(x) + (1 − t)f(y), t ∈ [0, 1], x, y ∈ K}, 

i.e. S(K) consists of the continuous convex functions on K. For any f ∈ CR(K) we 

define the upper envelope f by 

f(x) = inf{g(x) : −g ∈ S(K), g ≥ f}. 

A function g ∈ CR(K) is concave when −g is convex, so the upper envelope of f 

is the infimum over all continuous concave functions dominating f. In particular, it 

follows straightforwardly that f is concave and upper semicontinuous. (Recall that 

a function h : K → R is upper semicontinuous when h −1 (] − ∞, α[) is open for all 

α ∈ R.) Therefore f is Borel measurable. 

Lemma 4.15. Let f ∈ CR(K). Then 

is the closed convex hull of 

T = {(ω, t) ∈ K × R : t ≤ f(ω)} 

S = {(ω, t) ∈ K × R : t ≤ f(ω)}. 

Proof. Note that T is convex since f is concave and that T is closed since f 

is upper semicontinuous. Since f ≤ f, it is clear that S ⊆ T so we have almost for 

free that co(S) ⊆ T. Assume that there is a point (ω0, t0) ∈ T \co(S). If so we may 

i=1


apply the Hahn-Banach separation theorem to get a continuous linear functional g 

on (Aff K × R) ∗ and a real number α such that 

g(ω0, t0) > α and g(ω, t) ≤ α when t ≤ f(ω) 

Now g is of the form g(ω, t) = h(ω) + tλ for some weak ∗ -continuous functional h 

on Aff K ∗ and some λ ∈ R. Thus h(ω0) + t0λ > α and h(ω) + f(ω)λ ≤ α for all 

ω ∈ K. Inserting ω = ω0 and subtracting we get that λ(f(ω0) − t0) < 0 which 

implies that λ > 0 because t0 > f(ω0). Since λ > 0 we can now conclude that 

k(ω) = λ −1 (α − h(ω)) defines an element of Aff K such that k ≥ f. In particular, 

k ≥ f and t0 > k(ω0) ≥ f(ω0), contradicting that (ω0, t0) was an element of T. 

Lemma 4.16. Assume that K is metrizable. For every f ∈ CR(K), 

∂eK ⊆ {ω ∈ K : f(ω) = f(ω)}. 

Proof. Assume that f(ω) > f(ω). Since (ω, f(ω)) is in the set T of Lemma 

4.15, there is a sequence {ηn} of finite convex combinations, ηn = Nn 

i=1 tn i (ω n i , s n i ), 

where s n i ≤ f(ωn i ) for all i, n, such that limn→∞ ηn = (ω, f(ω)). By passing to 

a subsequence we may assume that ν = limn→∞ 

) = limn 

follows that b(ν) = limn b( Nn 

i=1 tn i δω n i 

 

f dν = lim n 

 

Nn 

f d( 

i=1 

t n i δω n i ) = lim n 

Nn i=1 tni δωn i exists in M1(K). It 

Nn i=1 tni ωn i = ω. Furthermore, 

Nn 

t 

i=1 

n i f(ωn i ) ≥ lim n 

Nn 

t n i sni i=1 

= f(ω), 

from which we see that f dν > f(ω). Now, by Lemma 4.14, there is a sequence 

of finitely supported measures ρn ∈ M1(K) such that b(ρn) = b(ν) = ω for all n and 

limn ρn = ν. For some large enough N ∈ N we have that f dρN > f(ω). Since ρN 

is finitely supported there are points xi ∈ K and numbers ri ∈]0, 1], i = 1, 2, . . ., mN 

such that ρN = mN riδxi i=1 . Then 

mN 

ω = b(ρN) = rixi. (4.5) 

Since f(ω) < f dρN = mN 

i=1 rif(xi), we conclude that not all xi’s equal ω so that 

(4.5) is a non-trivial convex combination, and hence we see that ω /∈ ∂eK. This 

proves that ∂eK ⊆ {ω : f(ω) = f(ω)}. 

Proposition 4.17. Let K be a metrizable compact convex set. There is a convex 

function F ∈ S(K) such that 

i=1 

∂eK = {x ∈ K : F(x) = F(x)}. 

Proof. Since K is metrizable we can find a metric d on K defining the topology. 

In fact, since Aff K is separable by Exercise 4.12, we can define a metric for the 

topology by 

∞ 

d(x, y) = 2 −n |fn(x) − fn(y)| 

n=1 

where {fn} is a dense sequence in the unit ball of Aff K. Note that with this 

definition the function d(x, · ) is convex for all x ∈ K. Since K is metrizable and


compact it is also separable, so we can find a dense sequence {xn} in K. Since 

sup x,y∈K d(x, y) < ∞, the sum 

F(x) = 

∞ 

n=1 

2 −n d(xn, x) 2 

converges uniformly in x ∈ K. Since t ↦→ t 2 is convex on R + it follows that F ∈ 

S(K). Let ω /∈ ∂eK. We assert that F(ω) > F(ω). To see this note that ω = 

1 

2 (µ1 + µ2) for some µ1 = µ2 in K since ω /∈ ∂eK. There is an n ∈ N such that 

d(xn, µ1) < d(xn, µ2). Then 

d(xn, ω) 2 ≤ 1 

2 (d(xn, µ1) + d(xn, µ2) 2 

= 1 

2 (d(xn, µ1) 2 + d(xn, µ2) 2 ) − 1 

4 

< 1 

2 (d(xn, µ1) 2 + d(xn, µ2) 2 ), 

d(xn, µ1) − d(xn, µ2) 2 

from which it follows that F(ω) < 1 

2 (F(µ1)+F(µ2)). Thus, if g ∈ S(K) and −g ≥ F, 

then 

−g(ω) = −g( 1 

2 (µ1 + µ2)) ≥ 1 

2 (−g(µ1) − g(µ2)) ≥ 1 

2 (F(µ1) + F(µ2)) = F(ω) + δ, 

where δ = 1 

2 (F(µ1) + F(µ2)) − F(ω). By taking the infimum over all such g we 

see that F(ω) ≥ F(ω) + δ, so that F(ω) > F(ω). We can now conclude that 

{ω ∈ K : F(ω) = F(ω)} ⊆ ∂eK. The reversed inclusion follows from Lemma 4.16. 

 

It follows from Proposition 4.17 that the extreme boundary ∂eK of a metrizable 

convex set K is the intersection of a countable number of open subseteqs, and hence, 

in particular, is a Borel set. Indeed, (F − F) −1 (] − ∞, α[) = 

β∈R F −1 (] − ∞, α + 

β[) ∩ F −1 (]β, ∞[) is open for all α ∈ R and 

∂eK = 

(F − F) −1 (] − ∞, 1 

n [) 

by Proposition 4.17. 

n∈N 

Lemma 4.18. S(K) − S(K) is dense in CR(K), i.e. every continuous realvalued 

function on K can be approximated uniformly by the difference between two 

continuous convex functions. 

Proof. By Theorem 4.3 Aff K separates the points of K. By the real Stone- 

Weierstrass theorem it follows that polynomials in elements from Aff K form a dense 

subalgebra of CR(K), i.e. every f ∈ CR(K) can be approximated arbitrarily closely 

(in the uniform norm) by a function of the form K ∋ ω → P(g1(ω), g2(ω), . . ., gn(ω)) 

for some polynomial P of n real variables. Set L = g1(K)×g2(K)×· · ·×gn(K) ⊆ R n . 

It is wellknown that a smooth function h : [0, 1] → R is convex if and only if 

h ′′ (t) ≥ 0 for all t ∈ [0, 1]. A polynomial Q : L → R is convex on L if and only 

if t ↦→ Q(tx + (1 − t)y) is convex on [0, 1] for all x, y ∈ L, so we conclude that 

Q is convex if and only if 

i,j 

∂ 2 Q 

∂xi∂xj (tx + (1 − t)y)(xi − yi)(xj − yj) ≥ 0 for all


x, y ∈ L, t ∈ [0, 1]. In particular, it suffices that the selfadjoint n × n-matrix 

∂2Q (χ) 

∂xi∂xj 

 

is positive in Mn(C) for all χ ∈ L. Since χ ↦→ ∂2Q ∂xi∂xj (χ) is continuous, there is a 

constant K ∈ R + such that 

2K + ∂2Q (χ) 

∂xi∂xj 

 

is positive in Mn(C) for all χ ∈ L. Set P1(x1, x2, . . .,xn) = P(x1, x2, . . .,xn) + 

K n 

i=1 x2 i and P2(x1, x2, . . .,xn) = K n 

i=1 x2 i . Then P1 and P2 are both convex 

polynomials on L and ω ↦→ P1(g1(ω), . . ., gn(ω)) and ω ↦→ P2(g1(ω), . . ., gn(ω)) are 

convex functions on K whose difference is ω ↦→ P(g1(ω), . . ., gn(ω)). 

Theorem 4.19. Let K be a metrizable compact convex set. Then every point 

x ∈ K is the barycenter of a Borel probability measure supported on the extreme 

boundary, ∂eK, of K. 

Proof. We introduce a partial ordering < on M1(K) by µ < ν iff f dµ ≤ 

f dν, f ∈ S(K). This relation is obviously reflexive (µ < µ) and transitive 

(µ < ν, ν < δ ⇒ µ < δ). That it is also anti-symmetric follows from Lemma 

4.18 : If µ < ν and ν < µ, then clearly g dµ = g dν for all g ∈ S(K), and 

so f dµ = f dν for all f ∈ CR(K) by Lemma 4.18, i.e. µ = ν. Consider 

L = {µ ∈ M1(K) : b(µ) = x} with the partial ordering < . Note that δx ∈ L 

so that L is a non-empty partially ordered set. If G is a totally ordered subseteq 

of L we have immediately that limµ∈G g dµ exists in R for all g ∈ S(K). It 

follows from Lemma 4.18 that limµ∈G f dµ exists for all f ∈ CR(K), so there is a 

ν ∈ M1(K) such that limG µ = ν. By the continuity of the barycenter map, we have 

that b(ν) = limG b(µ) = x, i.e. ν ∈ L. Clearly, µ < ν for all µ ∈ G, so that ν is 

an upper bound for G. We can now apply Zorns lemma to conclude that there is a 

maximal element µ in L. We use µ to define a Minkowski functional p on CR(K) by 

 

p(f) = f dµ, f ∈ CR(K). 

Since tf = tf, f + g ≤ f + g for all f, g ∈ CR(K), t ≥ 0, the linearity of 

the integral yields that p is a Minkowski functional. Let F ∈ S(K) such that 

∂eK = {ω ∈ K : F(ω) = F(ω)}, cf. Proposition 4.17. By the Hahn-Banach theorem 

there is a linear functional l : CR(K) → R such that l(F) = p(F) and l(f) ≤ p(f) for 

all f ∈ CR(K). If g ∈ CR(K) and g ≤ 0, then g ≤ 0 and l(g) ≤ p(g) = g dµ ≤ 0, 

proving that l is a positive linear functional. By the Riesz representation theorem 

there is a positive Borel measure ν on K such that l(f) = f dν, f ∈ CR(K). 

When g ∈ S(K), −g is concave and hence 

 

 

−l(g) = −g dν ≤ p(−g) = 

 

−g dµ = − 

g dµ. (4.6) 

Since any affine function h ∈ Aff K is both convex and concave, (4.6) implies that 

h dν = h dµ for all h ∈ Aff K. In particular, ν(K) = 1 dν = 1 dµ = 1, 

showing that ν is a Borel probability measure. Also it follows that b(ν) = b(µ) = x, 

i.e. ν ∈ L. Now (4.6) shows that µ < ν. By the maximality of µ we must have 

that µ = ν and hence F dµ = p(F) = l(F) = F dν = F dµ. It follows


from this that µ must be supported on ∂eK, i.e. that µ(∂eK) = 1. Indeed, if 

Ωn = {ω ∈ K : F(ω) − F(ω) ≥ 1 } had positive measure for some n, we could 

n 

conclude that F − F dµ ≥ 

1 F − F dµ ≥ Ωn n µ(Ωn) > 0, which is impossible. 

Therefore K\∂eK = 

n Ωn has measure 0, i.e. µ(∂eK) = 1. 

4.2. Choquet simplices. Among the compact convex sets there is a special 

class with particularly nice properties - the Choquet simplices, or just simplices. 

Definition 4.20. A compact convex set K is a Choquet simplex when Aff K 

has the Riesz decomposition property in the order given by the cone Aff K+ = {f ∈ 

Aff K : f(x) ≥ 0, x ∈ K}. 

Note that by Proposition 3.39 we could equally well require Aff K to have the 

Riesz interpolation property. In Section 4.1 we saw that the compact convex sets 

are in one to one correspondance with the order unit spaces. By definition the state 

space of an order unit space is a Choquet simplex if and only if the order unit space 

has the Riesz interpolation (or decomposition) property. 

There are other ways to define a Choquet simplex (e.g. as the base of a lattice 

cone), but we prefer the given one because it emphasizes the property which makes 

the simplices so important in connection with dimension groups. As it stands, 

however, it is not so easy to identify a Choquet simplex among other compact 

convex sets, so we seek a more geometric condition for a compact convex set to be 

a Choquet simplex. We shall only be interested in compact convex sets which are 

metrisable, and since the theory presents additional difficulties in the non-metrisable 

case we shall simply confine ourself to the metrizable case whenever it is convenient. 

Example 4.21. Let X be a compact metric space. M1(X) is a metrizable Choquet 

simplex. To see this, define a map Φ : CR(X) → Aff M1(X) by 

 

Φ(f)(µ) = f dµ, µ ∈ M1(X). 

Φ is then clearly a linear map which is easily seen to be an isometry such that 

f ≥ 0 ⇔ Φ(f) ≥ 0. If h ∈ (Aff M1(X))+, the function 

X ∋ x ↦→ h(δx) 

is continuous so it defines an element f of CR(X) which is clearly non-negative. Since 

Φ(f)(δx) = f(x) = h(δx), x ∈ X, we see that Φ(f) and h agree on the extremepoints 

of M1(X). Since h and Φ(f) are both continuous and affine, it follows from the 

Krein-Milman theorem that h = Φ(f). Hence Aff M1(X) is isomorphic to CR(X) 

as partially ordered vector spaces. Since CR(X) is obviously a lattice , i.e. the 

supremum of two elements exists in CR(X), cf. Definition 4.22 below, it follows that 

Aff M1(X) is also a lattice. In particularly, Aff M1(X) has the Riesz interpolation 

property, so M1(X) is a Choquet simplex. The fact that M1(X) is metrizable is left 

as an exercise, cf. Exercise 4.12. 

This example comprises a large stock of examples of Choquet simplices, including 

the classical simplices 

∆n = {(t1, t2, . . .,tn) ∈ [0, 1] n n 

: ti = 1}, 

n ∈ N. Indeed, if F is a finite set consisting of n elements, then M1(F) is affinely 

homeomorphic to ∆n as the reader can easily verify. 

i=1


The next goal here will be to prove that ’the way a point is expressed as a convex 

combination of extreme points is unique if and only if the set is a simplex’. 

Definition 4.22. An ordered Banach space X is a lattice when any pair x1, x2 ∈ 

X have a supremum in X; i.e. if there is an element x1 ∨ x2 ∈ X such that 

xi ≤ x1 ∨ x2, i = 1, 2, and such that xi ≤ y, i = 1, 2 ⇒ x1 ∨ x2 ≤ y. 

Hence X is a lattice if and only if any finite set has a supremum as well as an 

infimum. It is wellknown that CR(Y ) is a lattice (in the natural ordering) for any 

compact Hausdorff space Y . What may be less obvious is that the dual space of 

CR(Y ) is also a lattice. To make this precise we return for a moment to general 

setup where X is an order unit space with order unit u and positive cone X+. We 

turn the dual space X ∗ into an ordered Banach space by setting 

X ∗ + = {l ∈ X ∗ : l(X+) ⊆ [0, ∞[}. 

It follows from Lemma 4.9 that X∗ + − X∗ + = X∗ , so X∗ becomes an ordered Banach 

space in this way. In fact, since X+ − X+ = X, X∗ + is a proper cone in the sense 

that X ∗ + ∩ (−X∗ + ) = {0}. When the dual space X∗ is considered as an ordered 

Banach space in this way, we refer to X ∗ as the ordered dual of X. 

Lemma 4.23. If X is an order unit space with the Riesz interpolation property, 

then the ordered dual X ∗ is a lattice. 

Proof. Let l1, l2 ∈ X ∗ . For x ∈ X+ we define l1 ∨ l2(x) ∈ R by 

l1 ∨ l2(x) = 

n 

sup{ max{l1(ai), l2(ai)} : x = a1 + a2 + · · · + an, ai ∈ X+, i = 1, 2, . . ., n}. 

i=1 

Note that by Lemma 4.9 we have that li = l + i − l− i where l+ i , l− i ∈ X∗ + and l + i 

li, l − 

i ≤ li, i = 1, 2. Thus 

n 

max{l1(ai), l2(ai)} ≤ 

n 

l + 1 (ai) + l − 1 (ai) + l + 2 (ai) + l − 2 (ai) 

i=1 

i=1 

= l + 1 (x) + l − 1 (x) + l + 2 (x) + l − 2 (x) ≤ 2x(l1 + l2)) 

whenever ai ∈ X+ and 

i ai = x. Therefore the supremum occuring in (4.23) exists. 

It is (almost) obvious that l1 ∨ l2 is superadditive on X+, i.e. that l1 ∨ l2(x + y) ≥ 

l1 ∨ l2(x) + l1 ∨ l2(y), x, y ∈ X+. To prove the reversed inequality, we shall use 

the Riesz decomposition property. Indeed, if x + y = a1 + a2 + · · · + an, where 

ai ∈ X+ for all i, then we may write ai = bi + ci where 0 ≤ bi, 0 ≤ ci for all i and 

 

i bi = x, 

i ci = y, we find that 

n 

max{l1(ai), l2(ai)} = 

i=1 

≤ 

n 

max{l1(bi) + l1(ci)), l2(bi) + l2(ci)} 

i=1 

n 

max{l1(bi), l2(bi)} + max{l1(ci), l2(ci)} ≤ l1 ∨ l2(x) + l1 ∨ l2(y). 

i=1 

Since the decomposition x + y = 

i ai into positive elements was arbitrary, we 

conclude that l1 ∨ l2(x + y) ≤ l1 ∨ l2(x) + l1 ∨ l2(y). Hence l1 ∨ l2 is additive 

on X+ and we can extend it to an additive function l1 ∨ l2 : X → R by setting 

≤


l1 ∨l2(x−y) = l1 ∨l2(x)−l1 ∨l2(y) when x, y ∈ X+. As observed above, we have that 

|l1 ∨l2(x)| ≤ 2x(l1+l2)) when x ∈ X+. In general, x = a−b, where a, b ∈ X+ 

and a ≤ x and b ≤ x, so in general we have that |l1 ∨ l2(x)| ≤ 4x(l1 + 

l2)), x ∈ X. Hence l1 ∨ l2 is continuous. Furthermore, l1 ∨ l2(nx) = nl1 ∨ l2(x) 

when n ∈ N, so it follows that l1 ∨ l2(qx) = ql1 ∨ l2(x) for all rational numbers q. 

The density of Q in R, combined with the continuity of l1 ∨ l2, implies that l1 ∨ l2 

is linear, i.e. l1 ∨ l2 ∈ X∗ . Clearly, li ≤ l1 ∨ l2, i = 1, 2, and if l ∈ X∗ satisfies 

that li ≤ l, i = 1, 2, then n i=1 max{l1(ai), l2(ai)} ≤ 

i l(ai) = l(x) whenever 

x, a1, a2, · · · , an ∈ X+ are such that x = a1 + a2 + · · · + an. Hence l1 ∨ l2(x) ≤ l(x), 

proving that l1 ∨ l2 is indeed the supremum of l1 and l2 in X∗ . 

Lemma 4.24. For every f ∈ CR(K) and ω ∈ K, 

n 

f(ω) = sup{ tif(ωi) : ti ∈ [0, 1], ωi ∈ K, 

ti = 1, 

tiωi = ω , n ∈ N}. 

i=1 

Proof. Since f is concave we have that 

n 

f(ω) ≥ tif(ωi) ≥ 

i=1 

i 

n 

tif(ωi) 

whenever ti ∈ [0, 1], ωi ∈ K, 

inequalities making up the identity in the lemma. To prove the other, we use Lemma 

i=1 

i ti = 1, and 

i tiωi = ω. This proves one of the 

4.15 to find λα i ∈ [0, 1], ωα i ∈ K, tα i ∈ R and nα ∈ N such that nα i=1 λαi = 1, tα i ≤ 

f(ωα i ) and 

nα 

lim λ 

α 

i=1 

α i ωα i = ω, lim nα 

λ 

α 

i=1 

α i tαi = f(ω). 

Set 

nα 

µα = λ 

i=1 

α i δωα i 

and pass to a weak ∗ convergent subseteq µα 

one has b(µ) = limα ′ b(µα ′) = ω and 

 

f dµ = lim 

α ′ 

f dµα ′ ≥ lim α ′ 

i 

′. If µ denotes the limit of this subnet 

n α ′ 

 

i=1 

λ α′ 

i tα′ i = f(ω). 

Let ǫ > 0. By Lemma 4.14 we can find a finitely supported measure ν ∈ M1(K) such 

that f dν > f dµ − ǫ and b(ν) = b(µ) = ω. Since ν is finitely supported there 

are ti ∈ [0, 1], ωi ∈ K, such that 

i ti = 1, and 

i tiδωi = ν. Since n i=1 tiωi = 

b(ν) = ω, and 

n 

 

tif(ωi) = f dν > f dµ − ǫ ≥ f(ω) − ǫ, 

i=1 

we get the second inequality. 

Lemma 4.25. Let g ∈ Aff K such that g(x) > 0 for all x ∈ K, and let p ∈ Aff K∗ . 

Then 

inf{q(g) : q ∈ (Aff K) ∗ + , q ≥ p} = sup{p(h) : 0 ≤ h ≤ g}.


Proof. When q ≥ p and 0 ≤ h ≤ g, then p(h) ≤ q(h) ≤ q(g), proving that 

sup{p(h) : 0 ≤ h ≤ g} ≤ inf{q(g) : q ∈ (Aff K) ∗ + , q ≥ p}. Note that β = sup{p(h) : 

0 ≤ h ≤ g} is finite since p can be written as the difference between two positive 

functionals. Let ǫ > 0 and choose m ∈ N such that mǫ ≥ β − p(g). Define linear 

subspaces of Aff K ⊕ Aff K by 

C = {(x, −x) : x ∈ Aff K}, D = R(mg, g) + C . 

Note that D is a closed subspace of Aff K⊕Aff K and hence a real Banach space. Set 

D+ = D ∩ (Aff K+ × Aff K+) and note that (mg, g) is in the interior of D+ relative 

to D. Since (D, D+) is a normal ordered Banach space, we know from Example 4.6 

2) that there is a norm · ′ , equivalent to the one inherited from Aff K ⊕ Aff K, 

such that D is an order unit space with order unit (mg, g). Since g = 0 we have 

that (mg, g) /∈ C. We can therefore define a linear functional f on D so that 

f(t(mg, g) + (x, −x)) = tm(β + ǫ) + p(x) 

for all t ∈ R and x ∈ Aff K. We claim that f is positive. Given w ∈ D+\{0}, 

write w = t(mg, g) + (x, −x) for some t ∈ R and x ∈ Aff K. Then tmg + x ≥ 0 

and tg − x ≥ 0 in Aff K. Adding these inequalities, we find that t(m + 1)g ≥ 0 

and hence that t > 0. In addition, 0 ≤ tg − x ≤ tg + tmg = t(m + 1)g, whence 

0 ≤ t −1 (m + 1) −1 (tg − x) ≤ g and so t −1 (m + 1) −1 p(tg − x) ≤ β. Then tp(g) − 

p(x) ≤ t(m + 1)β, and consequently −p(x) ≤ tmβ + t(β − p(g)) ≤ tmβ + tmǫ, 

whence f(w) = tm(β + ǫ) + p(x) ≥ 0. Thus f is positive as claimed. In particular 

f’s norm, relative to · ′ , is f(mg, g). Note that (mg, g) is an interior point of 

Aff K+ × Aff K+, so that the order unit norm · ′ on D is the restriction of the 

order unit norm on Aff K ⊕ Aff K defined by (mg, g), cf. Example 4.6 2). Extend 

f norm-preservingly to Aff K ⊕ Aff K by using the Hahn-Banach theorem, and 

conclude from Lemma 4.7 that the extension is positive on Aff K+ × Aff K+. We 

let ˜ f denote the extension, and define q : Aff K → R by q(x) = ˜ f(x, 0). Then q ≥ 0 

and q(x) = ˜ f(x, 0) ≥ ˜ f(x, 0) − ˜ f(0, x) = f(x, −x) = p(x), x ∈ Aff K+. Hence q ≥ p. 

Since mq(g) = ˜ f(mg, 0) ≤ ˜ f(mg, 0) + ˜ f(0, g) = f(mg, g) = m(β + ǫ), we see that 

q(g) ≤ β + ǫ, proving that 

inf{q(g) : q ∈ (Aff K) ∗ +, q ≥ p} ≤ β + ǫ. 

Since ǫ > 0 was arbitrary, the proof is complete. 

We can now derive the desired geometric description of the Choquet simplices. 

To get the must useful formulation we introduce an additional partial ordering


4) Every element of K is the barycenter of a unique Borel probability measure 

supported on the extreme boundary ∂eK. 

5) The upper envelope of every continuous convex function is affine. 

Conditions 1),2) and 3) are equivalent also when K is not metrizable. 

Proof. We prove that 1) ⇒ 3) ⇒ 5) ⇒ 4) ⇒ 3) ⇒ 2) ⇒ 1). The reader should 

check that the implications 1) ⇒ 3) ⇒ 2) ⇒ 1) make no use of the assumption that 

K is metrizable so that the 3 first conditions remain equivalent in the general case. 

The implication 1) ⇒ 3) follows from Lemma 4.23. 3) ⇒ 5) : Let f be a continuous 

convex function. We must prove that f is affine. To do this we identify K with 

S(Aff K), cf. Theorem 4.3. Extend f ∈ S(K) to (Aff K) ∗ + by setting 

f(tω) = tf(ω), t ≥ 0, ω ∈ S(Aff K). 

By using that f is convex on K it follows easily that f is subadditive on (Aff K) ∗ + , 

i.e. that f(x + y) ≤ f(x) + f(y), x, y ∈ (Aff K) ∗ + . When we extend f to all of 

(Aff K) ∗ + in the same way, the concavity of f on K shows that f is superadditive on 

(Aff K) ∗ + , i.e. f(x + y) ≥ f(x) + f(y), x, y ∈ (Aff K)∗ + . It follows from Lemma 4.24 

that 

f(x) = sup{ 

n 

f(yi) : yi ∈ (Aff K) ∗ + , 

n 

yi = x} 

i=1 

i=1 

for all x ∈ (Aff K) ∗ + . Since Aff K∗ is a lattice it has the Riesz decomposition property, 

so when 

x + y = z1 + z2 + · · · + zn, x, y, z1, z2, · · · , zn ∈ (Aff K) ∗ + 

we may write zi = ai + bi, where 0 ≤ ai ≤ x, 0 ≤ bi ≤ y in Aff K∗ for all i and 

 

i ai = x, 

i bi = y. Hence 

 

f(zi) = 

f(ai + bi) ≤ 

f(ai) + 

f(bi) ≤ f(x) + f(y). 

i 

i 

i 

By taking the supremum over all such zi’s we get that f(x + y) ≤ f(x) + f(y), 

proving that f is additive on (Aff K) ∗ + . It follows that f is affine on K. (Note that 

we don’t know if f is continuous; only that it is upper semi-continuous.) 

5) ⇒ 4) : By using that the upper envelope of every continuous convex function 

is affine, we can now show that 

 

f(ω) = f dµ, µ ∈ M1(K), b(µ) = ω, (4.7) 

when f ∈ S(K). Since CR(K) is separable (because K is metrisable), we conclude 

from Exercise 4.12 that {g ∈ CR(K) : g ≥ f, −g ∈ S(K)} contains a dense sequence. 

There is therefore a sequence {hn} of concave continuous functions such that hn ≥ f 

for all n. Set 

gn = h1 ∧ h2 ∧ · · · ∧ hn . 

Then each gn is concave and majorises f. Furthermore, gn+1 ≤ gn for all n and 

limn→∞ gn = f pointwise on K. Therefore, by the Lebesgue theorem on monotone 

convergence, we find that 

 

f dµ = lim 

n 

gn dµ. 

i


Let {µβ} be a net in M1(K) of finitely supported measures with barycenter ω and 

converging to µ. Such a net exists by Lemma 4.14. Since f is affine we have that 

 

f(ω) = f dµβ ≤ gn dµβ ≤ gn(ω) 

for all β and all n. By taking the limit over β we get that 

 

f(ω) ≤ gn dµ ≤ gn(ω) 

for all n. Now take the limit over n and conclude that 

 

f(ω) ≤ f dµ ≤ f(ω), 

proving (4.7). Now 4) follows easily : If µ1, µ2 ∈ M1(K) are both supported on ∂eK 

and have barycenter ω, it follows from Lemma 4.16 that fdµi = f dµi, f ∈ 

CR(K), i = 1, 2. So (4.7) yields the conclusion that 

 

 

fdµ1 = f dµ1 = f(ω) = f dµ2 = f dµ2, f ∈ S(K). 

It follows now from Lemma 4.18 that µ1 = µ2. 

4) ⇒ 3) : Let l, t ∈ Aff K ∗ . By Lemma 4.9 we can write l = l1 − l2, t = t1 − t2 

where l1, l2, t1, t2 ∈ (Aff K) ∗ + . Note that r ∈ Aff K∗ will be an infimum of l and 

t if and only if r + l2 + t2 is an infimum of l + l2 + t2 and t + l2 + t2. So to 

show that an infimum exists for every pair l, t, it suffices to consider the case where 

l, t ∈ (Aff K) ∗ + . By assumption there are unique positive Borel measures µ, ν on 

K such that l(f) = f dµ and t(f) = f dν for all f ∈ Aff K and µ(K\∂eK) = 

ν(K\∂eK) = 0. Consider ν and µ as elements of CR(K) ∗ and let δ be the infimum 

of ν and µ in CR(K) ∗ , cf. Lemma 4.23. By the Riesz representation theorem, δ is 

given by a positive Borel measure on K; we denote this measure by δ also. Since 

f dδ ≤ min{ f dµ, f dν} for all f ∈ Aff K+, we see that δ|Aff K is a lower bound 

of l and t. To show that δ|Aff K is the greatest lower bound for l and t, let r ∈ Aff K∗ such that r ≤ t and r ≤ l. Write r = r1 − r2 where r1, r2 ∈ Aff K∗ +. Then r2(f) = 

f dχ, f ∈ Aff K, for some Borel measure χ supported on ∂eK. Set l1 = l +r2 and 

t1 = t + r2. Then 0 ≤ r + r2 ≤ l1 and r + r2 ≤ t1. Let µ1 and ν1 be the positive 

Borel measures supported on ∂eK such that l1(f) = f dµ1 and t1(f) = f dν1 

for all f ∈ Aff K. Then µ1 = µ + χ and ν1 = ν + χ. Let δ1 denote the infimum in 

CR(K) ∗ of the functionals CR(K) ∋ f ↦→ f dµ1 and CR(K) ∋ f ↦→ fdν1. Then 

δ1(f) = δ(f) + f dχ for all f ∈ CR(K). Since 0 ≤ r + r2 there is a positive Borel 

measure ǫ supported on ∂eK such that r(f) + r2(f) = f dǫ, f ∈ Aff K. Since 

l1 ≥ r + r2 and t1 ≥ r + r2 in Aff K∗ , there are positive Borel measures ǫl and ǫt, 

both supported on ∂eK, such that 

 

f dµ1 = f dǫ + f dǫl 

and 

 

f dν1 = 

 

f dǫ + 

f dǫt 

for all f ∈ Aff K. By the uniqueness assumption it follows that µ1 = ǫ + ǫl and 

ν1 = ǫ + ǫt. Therefore fdǫ ≤ f dµ1 and fdǫ ≤ f dν1 for all f ∈ CR(K)+, so


because δ1 is the greatest lower bound for CR(K) ∋ f ↦→ f dν1 and CR(K) ∋ f ↦→ 

fdµ1, we conclude that 

f dǫ ≤ δ1(f), f ∈ CR(K)+. 

Restricted to f ∈ Aff K+ this shows that r + r2 ≤ δ1|Aff K = δ|Aff K + r2, i.e. 

r ≤ δ|Aff K. 

3) ⇒ 2) : We prove that the strict ordering has the Riesz decomposition property. 

Given f1, f2 ∈ Aff K+ such that fi >> 0, i = 1, 2, set 

Fi = {f ∈ Aff K : 0

and h < δǫ. Then set 


hi = (1 + 2δ) −1 (gi + h + δǫ), i = 1, 2, 

and note that h1 + h2 = f. Since h < δǫ, we have that 

and hence 0


Lemma 4.31. Let {xn} and {yn} be sequences in A such that 

 

= 

n 

xnx ∗ n 

m 

y ∗ m ym . 

It follows that there is a sequence {znm : n, m ∈ N} such that 

 

z ∗ nizni = x ∗ nxn, 

for all n, m. 

i 

i 

zimz ∗ im = ymy ∗ m 

Proof. Set x = 

n xnx∗ 1 

n . Since x is positive, k 

k ∈ N. For each m, n, k ∈ N, consider 

ym( 1 

+ x)−1 2xn . 

k 

+ x is invertible in Â for all 

Let ǫ > 0 and set ak,l = ( 1 + x)−12 

− ( k 1 + x)−1 2. For all k, l ∈ N we have that 

l 

ym( 1 

k 

+ x)−1 2xn − ym( 1 

l 

+ x)−12xn 

2 = ymak,lxn 2 

= x ∗ nak,ly ∗ mymak,lxn ≤ x ∗ nak,lxak,lxn 

since y ∗ mym ≤ x so that x ∗ nak,ly ∗ mymak,lxn ≤ x ∗ nak,lxak,lxn. But 

x ∗ nak,lxak,lxn = x 1 

2ak,lxnx ∗ 1 

nak,lx 2 ≤ x 1 

2ak,lxak,lx 1 

2 

for a similar reason and hence 

ym( 1 

+ x)−12xn 

− ym( 

k 1 

+ x)−12xn 

l 2 ≤ x 2 a 2 k,l since x and ak,l commute. Since limk→∞ t2 

1 

k 

spectral theory that limk→∞ x2 ( 1 

k + x)−1 = x in A. By taking square-roots we see 

that limk→∞ x( 1 

+ x)−1 2 = x k 1 

2 in A. In particular, this shows that 

x 2 a 2 k,l = x2 ( 1 

< ǫ 

k + x)−1 + ( 1 

l + x)−1 − 2( 1 

k 

for all sufficiently large k, l. We conclude that {ym( 1 

k 

all n, m and we set znm = limk→∞ ym( 1 

k 

Note that 

+t = t, uniformly on [0, ∞[, we get from 

+ x)−12xn. 

+ x)−1 2( 1 

l 

 

+ x)−1 2 

+ x)−1 2xn} is Cauchy in A for 

ym( 1 

k + x)−1xy ∗ m − ymy ∗ m = ym(1 − ( 1 

k + x)−1x)y ∗ m = 

(1 − ( 1 

k + x)−1x) 1 

2y ∗ mym(1 − ( 1 

k + x)−1x) 1 

2 ≤ 

x − ( 1 

k + x)−1x 2 

for all m and that limk→∞ x − ( 1 

k + x)−1 x 2 = 0. Fix now ǫ > 0 and m ∈ N. 

Choose N ∈ N so large that 0 ≤ x − N 

n=1 xnx ∗ n ≤ ǫ1, and take K ∈ N such that

ymy ∗ m 


1 − ym( k + x)−1xy∗ m ≤ ǫ1 for all k ≥ K. Then 

0 ≤ ymy ∗ m − 

N 

i=1 

ym( 1 

+ x)−12xix 

k ∗ i (1 + x)−1 2y 

k ∗ m 

≤ ymy ∗ m − ym( 1 

k + x)−1 xy ∗ m + ǫym( 1 

k + x)−1 y ∗ m ≤ ǫ1 + ǫym( 1 

k + x)−1 y ∗ m 

for all k ≥ K. Note that 

ym( 1 

k + x)−1y ∗ m = ( 1 

+ x)−12y 

k ∗ mym( 1 

+ x)−12 

 

k 

≤ ( 1 

+ x)−12x( 

k 1 

+ x)−12 

= 

k x 

so that ǫ1 + ǫym( 1 

k + x)−1 y ∗ m ≤ 2ǫ1. It follows that 

0 ≤ ymy ∗ m − 

N 

i=1 

 

= lim ymy 

k→∞ 

∗ m − 

zimz ∗ im 

N 

i=1 

1 

k 

+ x ≤ 1 , 

ym( 1 

+ x)−12xix 

k ∗ i( 1 

+ x)−1 2y 

k ∗ 

m ≤ 2ǫ1 , 

proving that ymy∗ m − N i zimz∗ im. By 

exchanging m with n, ym with x∗ n and xn with ym in the previous argument we get 

that x∗ nxn = 

i z∗ nizni. 

i=1 zimz ∗ im ≤ 2ǫ. This shows that ymy ∗ m = 

When a, b are positive elements in A we write a ≈ b when there is a sequence 

{xn} in A such that a = ∞ n=1 x∗nxn and b = ∞ n=1 xnx∗ n. ≈ is an equivalence relation 

on A+ thanks to Lemma 4.31. Indeed, x = x 1 

2x 1 

2 = x for all x ∈ A+, showing that 

x ≈ x. If {xn} is a sequence establishing a ≈ b then {x∗ n } gives that b ≈ a. To 

prove transitivity we use Lemma 4.31. If namely a = 

n x∗nxn, b = 

n xnx∗ 

n, b = 

 

i z∗ ni zni = x ∗ n xn and 

i zimz ∗ im = ymy ∗ m 

m y∗ mym and c = 

m ymy ∗ m, then Lemma 4.31 gives us a sequence {znm} such that 

a while 

n,m znmz ∗ nm 

= 

m ymy ∗ m 

for all n. Then 

n,m z∗ nm znm = 

n x∗ n xn = 

= c, proving that a ≈ c. We next extend this 

equivalence relation to the selfadjoint part Asa of A by setting a ∼ b when there is 

a positive element p ∈ A+ such that a + p ≈ b + p. Again we argue that ∼ is an 

equivalence relation on Asa. If a = a∗ we know that a = a+ − a−, a+, a− ∈ A+, so 

a + a− = (a+) 1 

2(a+) 1 

2 = a + a−, showing that a ∼ a. That ∼ is symmetric on Asa is 

obvious. If a ∼ b, b ∼ c in Asa, then there are positive elements p, q ∈ A+ such that 

a + p ≈ b + p and b + q ≈ c + q in A+. Hence a + p + q ≈ b + p + q ≈ c + p + q in 

A+, showing that a ∼ c in Asa. For a ∈ Asa we denote the equivalence class under 

∼ containing a by [a]. 

We observe that a ∼ b if and only if there are positive elements a1, a2, b1, b2 ∈ A+ 

such that a = a1 − a2, b = b1 − b2 and a1 + b2 ≈ b1 + a2. Indeed, when a ∼ b and 

p ∈ A+ such that a + p ≈ b + p, then we write a = a+ − a−, b = b+ − b−, where 

a+, a−, b+, b− ∈ A+ and note that a++2p+b− = a+2p+b−+a− ≈ b+2p+b−+a− = 

b+ + 2p + a−, a = a+ + p − (a− + p), and b = b+ + p − (b− + p). Conversely, when 

a = a1−a2, b = b1−b2, a1, a2, b1, b2 ∈ A+ and a1+b2 ≈ b1+a2, then p = a2+b2 ∈ A+ 

and a + p = a1 + b2 ≈ b1 + a2 = b + p.


Lemma 4.32. Asa/ ∼ is a real vector space with the operations 

t[a] + [b] = [ta + b], t ∈ R, a, b ∈ Asa. 

Proof. It suffices to check that the addition and scalar multiplication is welldefined 

on Asa/ ∼. We must show that when a, b, c ∈ Asa, t ∈ R and a ∼ c, then 

ta ∼ tc and a+b ∼ c+b. If t ≥ 0 we write a = a1−a2, b = b1−b2, c = c1−c2 such that 

a1+c2 ≈ c1+a2. Then ta1+tc2 ≈ tc1+ta2, showing that ta ∼ tc. If instead t ≤ 0, we 

observe that −ta1 − tc2 ≈ −tc1 − ta2 while ta = −ta2 − (−t)a1, tc = −tc2 − (−t)c1, 

which shows that ta ∼ tc. We leave the reader to show that a + b ∼ c + b. 

Set A = Asa/ ∼. We go on to show that A is a Banach space. 

Lemma 4.33. Let x, y, a, b ∈ A+ such that a ≈ b and x + a ≈ y + b. Then 

b for all n ∈ N. 

x + 1 1 a ≈ y + n n 

Proof. Note that x + 1 1 a ≈ y + b ⇔ nx + a ≈ ny + b. We prove the last 

n n 

equivalence by induction in n. By assumption the equivalence holds in the case 

n = 1. Assume that it has been established for n. Then 

(n + 1)x + a ≈ x + ny + b ≈ x + ny + a ≈ ny + y + b = (n + 1)y + b. 

Lemma 4.34. Set A0 = {x−y : x, y ∈ A+, x ≈ y}. Then A0 is a closed subspace 

of Asa. 

Proof. It is straightforward to check that A0 is a subspace of Asa. Let z ∈ A0 

and ǫ > 0. We show that 

z = x − y for some x, y ∈ A+ such that x ≈ y and x + y ≤ z + ǫ . (4.8) 

We have that z = a − b with a, b ∈ A+, a ≈ b. Since z is selfadjoint we can write 

z = z+ − z− where z+, z− ∈ A+, z+z− = 0 and max{z+, z−} = z. Then 

z+ +b = z− +a, whence z+ + 1 

nb ≈ z− + 1a 

for all n ∈ N by Lemma 4.33. Let n ∈ N 

n 

be so large that 2 

na + b < ǫ and set x = z+ + 1 

n (a + b), y = z− + 1(a 

+ b). Then 

n 

1 b) + b = y. Furthermore, 

z = x − y and x = (z+ + 1 

n 

na ≈ (z− + 1 1 a) + n n 

x+y = z+ +z− + 2 

n (a+b) ≤ z+ +z−+ 2 

n (a+b) ≤ z++z−+ǫ = z+ǫ . 

We then use (4.8) to show that A0 is closed. Let {zn} be a sequence in A0 such that 

limn→∞ zn = z in Asa. To show that z ∈ A0 we pass to a subsequence {zni } such 

that zni − zni+1 < 2−i for all i. By applying (4.8) to zni+1 − zni we get elements 

xi, yi ∈ A+ such that zni+1 − zni = xi − yi, xi ≈ yi and xi + yi ≤ 2 −i . Then 

x = ∞ i=1 xi and y = ∞ i=1 yi exist in A+. Since xi ≈ yi for all i, we see that x ≈ y. 

Furthermore 

∞ 

∞ 

z − zn1 = − zni ) = (xi − yi) = x − y , 

i=1 

(zni+1 

proving that z = x − y + zn1 ∈ A0. 

Since x ∼ y in Asa if and only if x − y ∈ A0, we have that A = Asa/A0. (Note 

that this description of ∼ gives an alternative approach to Lemma 4.32.) By Lemma 

4.34 A is a Banach space in the quotient norm 

i=1 

[x] = inf{x − z : z ∈ A0} .


To proceed further with the investigation of the structure of A, in the unital case, 

we need the following decomposition theorem for functionals on a C∗-algebra. Recall 

that a continuous functional ω on a C∗-algebra A is selfadjoint when ω(a) = ω(a∗ ), 

or, alternatively, ω(Asa) ⊆ R. The formula 

ω(a) = 1 

2 (ω(a) + ω(a∗ )) + i( 1 

2i (ω(a) − ω(a∗ ))) 

shows that an arbitrary element ω ∈ A∗ is a linear combination of two selfadjoint 

continuous functionals. 

Theorem 4.35. (Hahn-Jordan decomposition) Let A be a unital C ∗ -algebra and 

ω ∈ A ∗ be a selfadjoint functional. There are unique positive linear functionals 

ω+, ω− ∈ A ∗ such that ω = ω+ − ω− and ω = ω+ + ω−. 

Proof. The theorem is trivial when ω = 0 so we consider only the case where 

ω = 0. It follows from Lemma 4.9 that there are states µ1, µ2 ∈ S(A) and a number 

t ∈ [0, 1] such that 

ω −1 ω = tµ1 − (1 − t)µ2 . 

Set ω+ = ωtµ1 and ω− = ω(1 − t)µ2. This give the existence part. To prove 

the uniqueness assume that ω = µ+ − µ− where µ+, µ− are positive functionals on 

A with µ+ + µ− = ω. We must show that ω+ = µ+ and µ− = ω−. Let ǫ > 0. 

Since ω is selfadjoint there is a selfadjoint element a in the unit ball of A such that 

ω(a) > ω − 1 

(1 −a) and note that 0 ≤ b ≤ 1. We have that 

2ǫ2 (check !). Set b = 1 

2 

ω+(1) + ω−(1) = ω+ + ω− = ω < ω(a) + 1 

2 ǫ2 = ω+(a) − ω−(a) + 1 

2 ǫ2 , 

showing that ω+(1 − a) + ω−(1 + a) < 1 

2 ǫ2 or 

ω+(b) + ω−(1 − b) < 1 

4 ǫ2 . 

It follows that 0 ≤ ω+(b) < 1 

inequality, Lemma 1.78, we have that 

and 

4ǫ2 and 0 ≤ ω−(1 − b) < 1 

4ǫ2 . By using the Schwarz 

|ω+(bx)| 2 = |ω+(b 1 

2b 1 

2x)| 2 ≤ ω+(b)ω+(x ∗ bx) 

≤ ω+(b)ω+x 2 < 1 

4 ǫ2ω+x 2 

|ω−((1 − b)x)| 2 = |ω((1 − b) 1 

2(1 − b) 1 

2x)| 2 ≤ ω−(1 − b)ω−(x ∗ (1 − b)x) 

< 1 

4 ǫ2ω−x 2 

for all x ∈ A. Hence |ω+(bx)| ≤ 1 

2ǫω1 2 x and |ω−((1 − b)x)| ≤ 1 

2ǫω1 2 x 

for all x ∈ A. The same argument applies to µ+ and µ−, so we find that also 

|µ+(bx)| ≤ 1 

2ǫω1 2 x and |µ−((1 − b)x)| ≤ 1 

2ǫω1 2 x for all x ∈ A. But since 

ω+ − ω− = µ+ − µ−, we find that 

ω+(x) − µ+(x) = ω+(bx) − µ+(bx) + ω−((1 − b)x) − µ−((1 − b)x) 

and so |ω+(x) − µ+(x)| ≤ 2ǫω 1 

2 x for all x ∈ A. Since ǫ > 0 was arbitrary, we 

see that ω+ = µ+ and hence also ω− = µ−. 

In the following A will denote a unital C ∗ -algebra.


Definition 4.36. An element ω ∈ A ∗ is tracial when ω(xy) = ω(yx), x, y ∈ A. 

A tracial state on A, i.e. a tracial element of S(A), will be called a trace state. The 

set of trace states of A will be denoted by T(A). 

Note that T(A) is a convex weak ∗ -compact subset of A ∗ . 

Lemma 4.37. Let ω ∈ A ∗ . Then the following conditions are equivalent. 

1) ω is tracial. 

2) ω(xx ∗ ) = ω(x ∗ x), x ∈ A. 

3) ω(uxu ∗ ) = ω(x) for all x ∈ A and all unitaries u in A. 

Proof. 1) ⇒ 2) is trivial. 2) ⇒ 3) : When x ≥ 0 we find that ω(uxu∗ ) = 

ω(ux 1 

2x 1 

2u∗ ) = ω(x 1 

2u∗ux 1 

2) = ω(x). Since an arbitrary element x of A is a linear 

combination of 4 positive elements, 3) follows. 3) ⇒ 1) : Assume first that y = u 

is a unitary. Then ω(xy) = ω(xu) = ω(u∗ (ux)u) = ω(ux) = ω(yx), showing that 1) 

holds in this case. The general case follows from the fact that an arbitrary element 

y is a linear combination of 4 unitaries. See Exercise 4.38 below. 

Exercise 4.38. Let y ∈ A. Show that there are 4 unitaries u1, u2, u3, u4 in A 

and scalars λ1, λ2, λ3, λ4 ∈ C such that y = λ1u1 + λ2u2 + λ3u3 + λ4u4. 

(Hint : After a few frustrations the reader will realize that it suffices to show that 

a real-valued continuous function on a compact Hausdorff space, taking values in 

[−1, 1], is the mean of two continuous functions taking values on T. Then remember 

that 2t = t + i √ 1 − t 2 + t − i √ 1 − t 2 , t ∈ [−1, 1]. ) 

We can now easily obtain the Hahn-Jordan decomposition for selfadjoint trace 

functionals. 

Theorem 4.39. Let ω ∈ A ∗ be a selfadjoint tracial functional. There are then 

unique positive tracial functionals ω+, ω− such that ω = ω+ −ω− and ω = ω++ 

ω−. 

Proof. Consider the Hahn-Jordan decomposition ω = ω+ − ω− given by Theorem 

4.35. Let u ∈ A be a unitary and set µ+(·) = ω+(u · u ∗ ), µ−(·) = ω−(u · u ∗ ). 

Since ω is a tracial functional, ω = µ+ − µ−, cf. Lemma 4.37. Furthermore, since 

uxu ∗ = x for all x ∈ A, we have that µ± = ω±. The uniqueness of the 

Hahn-Jordan decomposition implies that µ± = ω±, i.e. ω±(uxu ∗ ) = ω±(x), x ∈ A. 

Since this conclusion holds for all unitaries u in A, we conclude from Lemma 4.37 

that ω± are both tracial. The uniqueness statement follows immediately from the 

uniqueness of the Hahn-Jordan decomposition, Theorem 4.35. 

 

Now we define a closed convex cone A+ in A by 

A+ = {[a] : a ∈ A+} , 

i.e. if we let q : Asa → A be the quotient map, A+ = q(A+). Let u = q(1). 

Lemma 4.40. A is an order unit space with positive cone A+ = q(A+) and order 

unit u = q(1). 

Proof. We must show that {x ∈ A : x ≤ 1} = {x ∈ A : −u ≤ x ≤ u}. 

Assume first that x ≤ 1. Then x = q(a) for some a ∈ Asa. For every z ∈ A0 we 

have that −a − z1 ≤ a − z ≤ a − z1. Hence −a − zu ≤ q(a) = x ≤ a − zu


in A. Since x = inf{a − z : z ∈ A0} ≤ 1 and since A+ is closed, it follows that 

−u ≤ x ≤ u. 

Conversely, assume that −u ≤ x ≤ u. To prove that x ≤ 1, we use the 

Hahn-Banach theorem to find l ∈ A ∗ such that l = 1 and l(x) = x. Set 

ω = l ◦ q ∈ (Asa) ∗ and extend ω to a selfadjoint functional on A by ω(a + ib) = 

ω(a) + iω(b), a, b ∈ Asa. Then ω is a selfadjoint trace functional on A and we write 

ω = ω+ −ω− where ω± are positive trace functionals on A and ω = ω++ω−. 

Since ω± annihilate A0, we can define l± ∈ A ∗ by l± ◦ q = ω±. Note that since A 

has the quotient norm, l± = ω± and l = l ◦ q = ω. Hence l = l+ − l− 

and l = l+ + l−. Since −u ≤ x ≤ u we have that 

−l±(u) ≤ l±(x) ≤ l±(u) . 

Since l±(u) = ω±(1) = ω±, we see that |l±(x)| ≤ l± and hence 

x = l(x) = l+(x) − l−(x) ≤ l+ + l− . 

Since l+ + l− = l = 1 we have that x = l(x) ≤ 1. 

The next aim is to show that the order unit space A is a copy of Aff T(A). Define 

Φ : A → Aff T(A) by Φ(q(a))(ω) = ω(a), ω ∈ T(A), a ∈ Asa. 

Lemma 4.41. Φ is a linear isometry such that Φ(A+) = Aff T(A)+ and Φ(u) = 1. 

Proof. Clearly, Φ is a linear map taking u to the constant function 1 and A+ 

into Aff T(A)+. Let Λ : A → Aff S(A) be the isomorphism of Theorem 4.10, i.e. 

Λ(x)(l) = l(x), x ∈ A, l ∈ S(A). It suffices to show that Φ ◦ Λ −1 : Aff S(A) → 

Aff T(A) is an isometry taking Aff S(A)+ onto Aff T(A)+. To this end, define ψ : 

T(A) → S(A) by ψ(ω)(q(a)) = ω(a), a ∈ Asa, ω ∈ T(A). Then Φ ◦ Λ −1 (f)(ω) = 

f(ψ(ω)), f ∈ Aff S(A), ω ∈ T(A). It therefore suffices to show that ψ is an affine 

homeomorphism. To this end define ϕ : S(A) → T(A) by ϕ(l)(a) = l(q(a)), a ∈ 

Asa, l ∈ S(A). Then ψ and ϕ are both affine and are in fact inverses of each other. 

Indeed, ϕ ◦ ψ(ω)(a) = ψ(ω)(q(a)) = ω(a), ω ∈ T(A), a ∈ Asa, and ψ ◦ ϕ(l)(q(a)) = 

ϕ(l)(a) = l(q(a)), l ∈ S(A), a ∈ Asa. 

 

Corollary 4.42. Let A be a unital C ∗ -algebra. For each f ∈ Aff T(A) there is 

a self-adjoint element a = a ∗ ∈ A such that a = f. (Recall that a(ω) = ω(a), ω ∈ 

T(A).) 

The partial ordering of A∗ sa gives rise to a partial ordering of the selfadjoint trace 

functionals on A in the obvious way; µ1 ≤ µ2 when µ1(a) ≤ µ2(a) for all a ∈ A+. 

Lemma 4.43. Let ω1, ω2 ∈ A ∗ be selfadjoint trace functionals on A. There is then 

a selfadjoint trace functional µ ∈ A ∗ such that µ ≥ ωi, i = 1, 2, and when ν ∈ A ∗ is 

any selfadjoint trace functional on A such that ν ≥ ωi, i = 1, 2, then µ ≤ ν. 

In other words, the supremum of ω1 and ω2 exists in the space of selfadjoint trace 

functionals. 

Proof. By Theorem 4.39 we may write ωi = ω + 

i 

− ω− 

i 

where ω± 

i 

are positive 

trace functionals, i = 1, 2. So by considering µ1 = ω1+ω − 1 +ω− 2 and µ2 = ω2+ω − 1 +ω− 2 

in place of ω1 and ω2, we can assume that ωi ≥ 0, i = 1, 2. We first define µ on A+ 

as 

µ(a) = sup{ω1(x) + ω2(y) : x, y ∈ A+, a ≈ x + y} .


Let s, t ∈ A+. When x, y, x1, y1 ∈ A+ such that s ≈ x + y, t ≈ x1 + y1, we find that 

ω1(x) + ω2(y) + ω1(x1) + ω2(y1) = ω1(x + x1) + ω2(y + y1) ≤ µ(s + t) 

since s + t ≈ (x + x1) + (y + y1). Thus µ(s) + µ(t) ≤ µ(s + t). On the other hand, 

consider x, y ∈ A+ such that s + t ≈ x + y. By definition this means that there is a 

sequence {xn} in A such that ∞ n=1 x∗n xn = s + t, ∞ n=1 xnx∗ n = x + y. But then 

s 1 

2s 1 

2 + t 1 

2t 1 

∞ 

2 = 

n=1 

x ∗ n xn 

so 

 

Lemma 4.31 applies to give us two sequences {vn}, {wn} in A such that s = 

n v∗ nvn, t = 

 

= x+y. Now we apply Lemma 

n w∗ nwn and vkv∗ k +wkw∗ k = xkx∗ k for all k ∈ N. Set s′ = 

n vnv∗ n , t′ = 

n wnw∗ n . Then s ≈ s′ , t ≈ t ′ and s ′ +t ′ = 

k xkx∗ k 

4.31 again to get elements zij ∈ A, i, j ∈ {1, 2}, such that s ′ = z∗ 11z11 + z∗ 12z12, t ′ = 

z∗ 21z21 + z∗ 22z22, x = z11z∗ 11 + z21z∗ 21 and y = z12z∗ 12 + z22z∗ 22. Set as = z11z∗ 11, bs = 

z12z∗ 12 , at = z21z∗ 21 and bt = z22z∗ 22 . Then s ≈ s′ ≈ as + bs, t ≈ t ′ ≈ at + bt, and 

x = as + at, y = bs + bt. Therefore 

ω1(x) + ω2(y) = ω1(as) + ω2(bs) + ω1(at) + ω2(bt) ≤ µ(s) + µ(t) . 

By taking the supremeum over all x, y ∈ A+ with s + t ≈ x + y, we find that 

µ(s + t) ≤ µ(s) + µ(t). It follows that µ(s + t) = µ(s) + µ(t), i.e. µ is additive on 

A+. We can then extend µ to a linear functional on Asa by 

µ(a − b) = µ(a) − µ(b), a, b ∈ A+, 

and then to a selfadjoint functional on A by µ(a+ib) = µ(a)+iµ(b), a, b ∈ Asa. When 

0 ≤ b ≤ 1 we get immediately from the definition of µ that 0 ≤ µ(b) ≤ ω1 + ω2. 

(1+a) ≤ 1 and 

When −1 ≤ a ≤ 1, we have that a = 1 

2 

(1+a) − 1 

2 

(1 −a) where 0 ≤ 1 

2 

0 ≤ 1 

2 (1 − a) ≤ 1, so we conclude that −(ω1 + ω2) ≤ µ(a) ≤ ω1 + ω2. It 

follows that µ is continuous, i.e. that µ ∈ A ∗ . By definition µ is a trace functional 

since ω1 and ω2 are, and it is clear that µ ≥ ωi, i = 1, 2. If ν ∈ A ∗ is a trace 

functional such that ν ≥ ωi, i = 1, 2, then 

ω1(x) + ω2(y) ≤ ν(x + y) = ν(a) 

for all a, x, y ∈ A+ such that a ≈ x + y. Hence µ(a) ≤ ν(a). 

Theorem 4.44. Let A be a unital C ∗ -algebra. Then T(A) is a Choquet simplex. 

Proof. By Theorem 4.26 it suffices to show that the ordered dual Aff T(A) ∗ is 

a lattice. By Lemma 4.41 this is equivalent to A ∗ being a lattice. So let µ1, µ2 ∈ A ∗ . 

The complexifications of the functionals ωi = µi ◦ q, i = 1, 2, are selfadjoint trace 

functionals on A. Let µ be the supremum of ω1 and ω2 among the selfadjoint trace 

functionals on A, cf. Lemma 4.43. We can then define µ ′ ∈ A ∗ by µ ′ (q(x)) = 

µ(x), x ∈ Asa. It is straigthforward to check that µ ′ is the supremum of µ1 and µ2 

in A ∗ . 

4.4. Examples of trace spaces. 

Lemma 4.45. The only trace state on the C∗- algebra Mn of complex n × n 

matrices is given by 

τ (aij) = 1 

n 

aii , (aij) ∈ Mn . 

n 

i=1


Proof. We first check that τ is a trace state. When a = (aij) ∈ Mn we have 

that aa ∗ = ( n 

k=1 aikajk) so that 

τ(aa ∗ ) = 1 

n 

 

|aik| 2 ≥ 0 . 

i,k 

This shows that τ is positive. It is clear that τ(1) = 1 so τ is a state on Mn. Let 

{eij} be the standard system of matrix units in Mn. Then τ(eijekl) = δjkτ(eil) = 

1 

n δjkδil = τ(ekleij). Since arbitrary elements x, y ∈ Mn are linear combinations of 

the matrix units it follows by linearity that τ(xy) = τ(yx), i.e. τ is a trace state. 

Let ω be a trace state on Mn. Then 

ω(ekl) = ω(ekkeklell) = ω(ellekkekl) = δklω(ekk) . 

Since ω(e11) = ω(e1kek1) = ω(ek1e1k) = ω(ekk), k = 1, 2, · · · , n, and 1 = ω(1) = 

ω( n i=1 eii) = n i=1 ω(eii) = nω(e11), it follows that ω = τ. 

From Lemma 4.45 it is easy to give the following description of the tracial state 

space T(F) of the finite dimensional C ∗ -algebra F = Mn1 ⊕Mn2 ⊕· · ·⊕Mnm. Denote 

the trace state of Mni by τi, i = 1, 2, · · · , m. For each vector (t1, t2, · · · , tm) ∈ R m 

we define a linear functional τ(t1,t2,···,tm) : F → C by 

m 

τ(t1,t2,···,tm)(a1, a2, · · · , am) = tiτi(ai) . 

Lemma 4.46. The map (t1, t2, · · · , tm) ↦→ τ(t1,t2,··· ,tm) defines an affine homeomorphism 

from ∆m onto T(F). 

Proof. We leave the reader to check that the map takes ∆m affinely into T(F). 

It is injective because 

i=1 

τ(t1,t2,···,tm)(0, 0, · · · , 0, 1, 0, 0, · · · , 0) 

= ti . 

1 at the i’th entry 

To prove the surjectivity, let ω ∈ T(F) and let pi = (0, 0, · · · , 0, 1, 0, 0, · · · , 0). 

Then 


ti = ω(pi) ∈ [0, 1] and m i=1 ti = ω( 

i pi) = ω(1) = 1, i.e. (t1, t2, · · · , tm) ∈ ∆m. 

We leave the reader to use the uniqueness of the trace state on Mni to confirm that 

ω = τ(t1,t2,···,tm). 

We now consider C ∗ -algebras of the form C(X, A) where X is a compact Hausdorff 

space and A is a unital C ∗ -algebra, cf. Example 1.20. To work with such 

algebras it is convenient to introduce the following notation. When f : X → C is a 

continuous function and a ∈ A we define an element f ⊗ a ∈ C(X, A) by 

We shall need the following lemma. 

f ⊗ a(x) = f(x)a, x ∈ X. 

Lemma 4.47. For every g ∈ C(X, A) and every ǫ > 0 there are finite sets 

{f1, f2, · · · , fn} ⊆ C(X) and {a1, a2, · · · , an} ⊆ A such that g − n 

i=1 fi ⊗ ai ≤ ǫ. 

Proof. By compactness of X and the continuity of g we can find a finite open 

cover {Ui : i = 1, 2, · · · , n} of X and points xi ∈ Ui such that g(xi) − g(x) < ǫ 

for all x ∈ Ui, i = 1, 2, · · · , n. Set ai = g(xi) for all i and let {hi} be a partition of


unity in C(X) subordinate to {Ui : i = 1, 2, · · · , n}, i.e. hi ≥ 0 and supp(hi) ⊆ Ui 

for all i and n 

i=1 hi = 1. Then 

g(x) − 

n 

hi ⊗ ai(x) = 

i=1 

n 

hi(x)g(x) − g(xi) ≤ 

i=1 

n 

hi(x)g(x) − 

i=1 

n 

hi(x)ǫ = ǫ. 

i=1 

n 

hi(x)g(xi) ≤ 

Lemma 4.48. Assume that A is unital and has a unique trace state τ. Then 

T(C(X, A)) is affinely homeomorphic to M1(X). The trace state τµ ∈ T(C(X, A)) 

corresponding to µ ∈ M1(X) is given by 

 

τµ(g) = τ(g(x)) dµ(x), g ∈ C(X, A). 

X 

Proof. It is straightforward to check that the map µ ↦→ τµ is a continuous 

affine map from M1(X) into T(C(X, A)). Since τµ(f ⊗ 1) = 

f dµ, f ∈ C(X), 

X 

we see immediately that the map is injective. To prove that it is also surjective, let 

ρ ∈ T(C(X, A)). The map C(X) ∋ f ↦→ ρ(f ⊗ 1) is a state on C(X), so it is given 

by a regular Borel probability measure µ ∈ M1(X) through the formula 

 

f dµ = ρ(f ⊗ 1), f ∈ C(X). 

X 

We assert that τµ = ρ. To check this it suffices by Lemma 4.47 to check that the 

continuous functionals agree on elements of the form f ⊗a where f ∈ C(X), a ∈ A. 

In fact, since an arbitrary f is a linear combination of 4 positive continuous functions 

it suffices to check the agreement when f ≥ 0. We consider first the case where 

ρ(f ⊗ 1) = 0. Then f dµ = 0 and f must be zero almost everywhere with respect 

to µ. Hence τµ(f ⊗ a) = 

τ(a)f(x) dµ(x) = 0. On the other hand the Schwarz 

X 

inequality gives that 

|ρ(f ⊗ a)| 2 = |ρ((f ⊗ 1)(1 ⊗ a))| 2 ≤ ρ(f 2 ⊗ 1)ρ(1 ⊗ a ∗ a) = ρ(1 ⊗ a ∗ 

a) f 2 dµ = 0 , 

i.e. ρ(f ⊗ a) = 0. Thus the two functionals agree on f ⊗ a in this case. Assume 

next that ρ(f ⊗ 1) = 0. Then χ(a) = ρ(f ⊗ 1) −1 ρ(f ⊗ a), a ∈ A, defines a state on 

A. Let x, y ∈ A. Since (1 ⊗ y)(f ⊗ 1) = (f ⊗ 1)(1 ⊗ y) = f ⊗ y and ρ is a trace on 

C(X, A), we find that 

ρ(f ⊗ 1)χ(xy) = ρ(f ⊗ xy) = ρ((f ⊗ 1)(1 ⊗ x)(1 ⊗ y)) = ρ((1 ⊗ y)(f ⊗ 1)(1 ⊗ x)) 

= ρ((f ⊗ 1)(1 ⊗ y)(1 ⊗ x)) = ρ(f ⊗ yx) = ρ(f ⊗ 1)χ(yx) , 

proving that χ is a trace state on A. By the uniquenes of the trace state of A, we 

conclude that χ = τ. It follows that 

 

ρ(f ⊗ a) = ρ(f ⊗ 1)χ(a) = τ(a) f dµ = τ(f ⊗ a(x)) dµ(x) = τµ(f ⊗ a) . 

X 

X 

i=1 

X


For many C ∗ -algebras the tracial state space is determined completely by the 

K0-group, as we shall now explain. Let G be a pre-ordered abelian group and u an 

order unit in G, i.e. u ∈ G + is an element such that ∀g ∈ G ∃n ∈ N : g ≤ nu. 

A homomorphism ϕ : G → R is positive when ϕ(G + ) ⊆ [0, ∞[. A state on G is a 

positive element ϕ ∈ Hom(G, R) such that ϕ(u) = 1. The set of states of G will be 

denoted by SG. Note that the set of states of G depends not only on the partially 

ordered group structure of G, but also on the choice of order unit. In general there 

are many different order units in G and the corresponding state spaces may be very 

different. The state space SG of G comes equipped with a natural topology, the 

topology of pointwise convergence on G. Sets of the form 

{µ ∈ SG : |µ(gi) − ϕ(gi)| < ǫ, i = 1, 2, · · · , n} 

where ϕ ∈ SG, ǫ > 0 and the finite set {g1, g2, . . ., gn} ⊆ G may vary, form a basis 

for this topology. 

Lemma 4.49. SG is compact in the topology of pointwise convergence on G. 

Proof. Let [0, ∞] be the one-point compactification of [0, ∞[. By Tychonoffs 

theorem 

[0, ∞] 

g∈G 

is compact in the product topology. Recall that 

g∈G [0, ∞] consists of all functions 

f : G → [0, ∞]. We can therefore define Φ : SG → 

g∈G +[0, ∞] by Φ(ϕ)(g) = 

ϕ(g), g ∈ G + . We claim that 

Φ(SG) = {f ∈ 

[0, ∞] : f(g) + f(h) = f(g + h), g, h ∈ G + , f(u) = 1}. (4.9) 

g∈G + 

Obviously, Φ(SG) is contained in the set on the right-hand side. Conversely, if 

f ∈ 

g∈G +[0, ∞] such that f(g) + f(h) = f(g + h), g, h ∈ G + and f(u) = 1, then 

for any g ∈ G + , we can find n ∈ N such that g ≤ nu, i.e. such that g + h = nu for 

some h ∈ G + . Hence f(g) ≤ f(g) + f(h) = f(g + h) = f(nu) = nf(u) = n, proving 

that f(g) < ∞ for all g ∈ G + . It follows that we can define ω : G → R by 

ω(g − h) = f(g) − f(h), g, h ∈ G + . 

It is straightforward to check that ω ∈ SG and that Φ(ω) = f. This proves (4.9). 

Φ is then clearly an affine bijection between SG and {f ∈ 

g∈G +[0, ∞] : f(g) + 

f(h) = f(g + h), g, h ∈ G + , f(u) = 1}. It is straightforward to check that Φ is 

also a homeomorphism when the latter set is given the topology inherited from the 

product topology of 

g∈G +[0, ∞]. Since it is clearly closed in that topology the 

compactness of SG now follows from Tychonoff’s theorem. 

Note that SG is a convex subset of Hom(G, R) and that the topology just defined 

is the relative topology enherited from a locally convex vector space topology on 

Hom(G, R), a topology with a base consisting of sets of the form 

{µ ∈ Hom(G, R) : |µ(gi) − ϕ(gi)| < ǫ, i = 1, 2, · · · , n}, 

where ǫ > 0, ϕ ∈ Hom(G, R) and the finite set {g1, g2, · · · , gn} ⊆ G may vary. Thus 

SG is a compact convex set. 

We now establish an important connection between the tracial state space T(A) 

of a unital C ∗ -algebra A and the state space SK0(A) of K0(A) (with respect to the


order unit [1]). Let ω ∈ T(A). We can then define, for each n ∈ N, a trace functional 

ωn on Mn(A) by 

 

ωn (aij) n 

= ω(aii) . 

In this way we can define a map ω∞ : P(A) → R by ω∞(p) = ωn(p) when p is a 

projection p in Mn(A). The trace property of each ωn ensures that ω∞(p) = ω∞(q) 

whenever p ≈ q. Furthermore, it is clear that ω∞(p ⊕ q) = ω∞(p) + ω∞(q), so we 

see that ω∞ gives rise to a semigroup homomorphism P(A)/≈ → R and hence, by 

the universal property of the Grotendieck construction, to a group homomorphism 

rA(ω) : K0(A) → R. It is straightforward to check that ωn is a positive functional 

on Mn(A), so we conclude that rA(ω)(K0(A) + ) ⊆ [0, ∞[, i.e. rA(ω) is positive. Furthermore, 

rA(ω)([1]) = 1, so we see that rA(ω) ∈ SK0(A). Thus we have produced 

a map 

rA : T(A) → SK0(A) . 

This map will be called the restriction map of A, because it is in some sense obtained 

by restricting ω to projections. 

i=1 

Lemma 4.50. The restriction map rA : T(A) → SK0(A) is continuous and affine. 

Proof. This is really straightforward, and we leave it as an exercise to the 

reader. 

Theorem 4.51. Let A be a unital C ∗ -algebra and A1 ⊆ A2 ⊆ A3 ⊆ · · · a 

sequence of unital C ∗ -subalgebras such that A = 

n An. Assume that the restriction 

map rAn : T(An) → SK0(An) is surjective for all n. It follows that rA : T(A) → 

SK0(A) is surjective. 

Proof. Let s ∈ SK0(A). We must construct ω ∈ T(A) such that rA(ω) = s. 

Let ιn : An → A be the inclusion map. Then s ◦ ιn∗ ∈ SK0(An), so the assumption 

on the rAn’s give us trace states ωn ∈ T(An) such that rAn(ωn) = s ◦ιn∗, n ∈ N. For 

each n we choose a state µn ∈ S(A) extending ωn : An → C. This is possible by the 

Hahn-Banach theorem. Since S(A) is compact there is a weak ∗ -accumulation point 

ω ∈ S(A) for the sequence {µn}. Let x, y ∈ A, ǫ > 0. Since 

n An is dense in A 

there is an N ∈ N and elements a, b ∈ AN such that xy −ab < ǫ and yx−ba < ǫ. 

Furthermore, since ω is an accumulation point for the µn’s, there is a k ≥ N such 

that |ω(xy) − µk(xy)| < ǫ and |ω(yx) − µk(yx)| < ǫ. Since µk is a trace state on AN 

we have that µk(ab) = µk(ba). Thus 

|ω(xy) − ω(yx)| ≤ |µk(xy) − µk(yx)| + 2ǫ ≤ |µk(ab) − µk(ba)| + 4ǫ = 4ǫ . 

Since ǫ > 0 was arbitrary this shows that ω(xy) = ω(yx), i.e. ω is a trace state. 

To show that rA(ω) = s it suffices to take a projection p ∈ Mk(A) for some k ∈ N 

and show that s([p]) = ω(p) , when ω is the extension of ω to a trace functional 

on Mk(A). To this end note that 

n Mk(An) is dense in Mk(A) so that there is an 

m ∈ N and a projection q ∈ Mk(Am) such that [p] = [q] in K0(A), cf. Lemma 3.7. 

For each l ≥ m we have that 

µl(q) = ωl(q) = rAl (ωl)([q]) = s([q]) . 

Since ω is a weak ∗ -accumulation point for the sequence {µn}, this implies that 

ω(q) = s([q]), i.e. that rA(ω)([p]) = s([p]).


Corollary 4.52. Let A be a unital C ∗ -algebra and A1 ⊆ A2 ⊆ A3 ⊆ · · · a 

sequence of unital C ∗ -subalgebras such that A = 

n An. If each An has a trace 

state, then so does A. 

Proof. This follows from a straightforward modification of the proof of Theorem 

4.51 which we leave to the reader. 

Theorem 4.53. Let A be a unital C ∗ -algebra such that the span of projections 

in A is dense in A. Then the restriction map rA : T(A) → SK0(A) is injective. 

Proof. Let ω1, ω2 ∈ T(A) such that rA(ω1) = rA(ω2). For every projection 

p ∈ A we have that ω1(p) = rA(ω1)([p]) = rA(ω2)([p]) = ω2(p). Since A is the closed 

linear span of the projections in A it follows from the linearity and continuity of ω1 

and ω2 that ω1 = ω2. 

Corollary 4.54. Let A be a unital AF-algebra. Then the restriction map rA : 

T(A) → SK0(A) is an affine homeomorphism. 

Proof. By definition a unital AF-algebra is the closure of an increasing sequence 

of finite dimensional unital C ∗ -algebras. By Theorem 4.51, the surjectivity 

of rA : T(A) → SK0(A) follows if we can show that the restriction map of a finite 

dimensional C ∗ -algebra is surjective. This we leave to the reader. The injectivity 

of rA will follow from Theorem 4.53 if we can show that A is the closed linear span 

of its projections. But this is easy : If a = a ∗ ∈ A there is a b = b ∗ in some finite 

dimensional C ∗ -subalgebra of A such that a−b < ǫ. By linear algebra b is a linear 

combination of projections. 

Example 4.55. Let A be the complex vector space of sequences in M2 that 

converge to a diagonal element. More formally, let D2 denote the C∗-subalgebra of 

M2 consisting of the 2 × 2 matrices whose off diagonal elements are 0. Note that 

D2 ≃ C2 . Let 

A = {(xn) ⊆ M2 : lim xn ∈ D2} . 

n→∞ 

Let Ak = {(xn) ∈ A : xn = xk+1, n ≥ k + 1}. Then Ak is a unital C∗-subalgebra of A isomorphic to C 2 ⊕ M2 ⊕ M2 ⊕ · · · ⊕ M2 

 

k times 

). Since A = 

k Ak, this shows that 

A is an AF-algebra. Note that K0(Ak) = Z k+2 as dimension groups and that the 

order unit [1] ∈ K0(Ak) corresponds to (1, 1, 2, 2, · · · , 2). Under this identification 

the inclusion map Ak → Ak+1 induces the map Z k+2 → Z k+3 given by 

(z1, z2, z3, · · · , zk+2) ↦→ (z1, z2, z3, · · · , zk+2, z1 + z2) . 

Thus K0(A) can be identified as the subgroup of Z N consisting of the elements 

(zn) ∈ Z N with the property that zk = z1 + z2 for all large enough k. We seek to 

identify SK0(A). To this end set 

pk = (1, 0, 0, 0, · · · , 0, 

1, 1, 1, · · ·) 

 

k times 

and 

qk = (0, 1, 0, 0, · · · , 0, 

1, 1, 1, · · ·) . 

 

k times 

Then pk, qk ∈ K0(A) and {pk} and {qk} are both decreasing sequences in K0(A). 

It follows that the limits s1 = limk→∞ s(pk) and s2 = limk→∞ s(qk) exist for any state


s ∈ SK0(A). Let rn, n = 3, 4, · · ·, be the element rn = (0, 0, 0, · · · 0, 0, 1, 0, 0, , · · · ) ∈ 

1 at the n’th coordinate 

K0(A). Set sn = s(rn) for any state s ∈ SK0(A). We leave the reader to check that 

s (zn) ∞ 

= snzn, (zn) ∈ K0(A) 

and 

n=1 

s1 + s2 + 2 

∞ 

sn = 1 

n=3 

for all s ∈ SK0(A). Conversely, given any sequence {tn} in [0, 1] such that t1 + t2 + 

2 ∞ n=3 tn = 1, we can define a state s ∈ SK0(A) by 

s (zn) = 

∞ 

i=1 

tizi . 

Then sn = tn for all n as the reader can easily verify. 

In this way we see that SK0(A) may be identified, as a convex set, with 

K = {(sn) ∈ [0, 1] N ∞ 

: s1 + s2 + 2 sn = 1} . 

It is important to realize that this description of SK0(A), as a convex set, does not 

give any description of the topology of SK0(A). In particular, K is not a closed 

subseteq of [0, 1] N so SK0(A) is certainly not affinely homeomorphic to K when K 

is given the relative topology enherited from [0, 1] N . For example, 1 

2 rn ∈ K, n ≥ 3, 

and limn→∞ 1 

2 rn = (0, 0, 0, 0, · · ·) in the topology of [0, 1] N while limn→∞ 1 

2 rn = 

( 1 1 

, , 0, 0, · · ·) in the topology of K coming from the identification with SK0(A). 

2 2 

The topology of K which corresponds to the topology of SK0(A), and which makes 

the above identification of SK0(A) into an affine homeomorphism, is the topology of 

K given by pointwise convergence on K0(A) = {(zn) ∈ Z N : ∃N ∈ N such that zn = 

z1 + z2, n ≥ N}, i.e. (sα ∞ n) → (sn) in this topology if and only if limα i=1 sαi zi 

= 

∞ 

i=1 sizi for all (zn) ∈ K0(A). 

By Corollary 4.54 the above description of SK0(A) is at the same time a description 

of T(A) as a compact convex set. Note that it is by no means apparent that 

T(A) is a Choquet simplex. Let us find Aff K and check that it is. To do this we first 

identify the extreme points of K = SK0(A). Define S1, S2 ∈ SK0(A) by Si (zn) 

= 

zi, i = 1, 2, and for k ≥ 3, define Sk ∈ SK0(A) by Sk (zn) = 1 

2zk, (zn) ∈ K0(A). 

We assert that 

∂eK ⊆ {Sk : k = 1, 2, 3, · · · }. (4.10) 

So let s be an extremal state on K0(A). Assume first that s(rk) = 0 for some k ≥ 3 

and define Pk : K0(A) → K0(A) by Pk (zn) = (0, 0, · · · 0, 0, zk, 0, 0, · · · ). Then 

zk at the k’th coordinate 

0 ≤ Pk(x) ≤ x for all x ∈ K0(A) + and hence 0 ≤ s ◦ Pk ≤ s. Set µ = s − s ◦ Pk. 

Then s = s ◦ Pk + µ , µ(u), s ◦ Pk(u) ∈ [0, 1] (where u = (1, 1, 2, 2, · · ·) is the order 

unit). Note that s ◦Pk(u) = s(2rk) > 0 and that µ(u)+s◦Pk(u) = 1. Now µ(u) = 0 

since otherwise 

s = µ(u)µ(u) −1 µ + s(2rk)s(2rk) −1 s ◦ Pk 

is a convex combination which yields that µ(u) −1 µ = s(2rk) −1 s ◦ Pk because s is 

extremal. This, however, is absurd since µ(rk) = 0 while s ◦ Pk(rk) = s(rk) > 0. 

n=3


Thus µ(u) = 0 as asserted and hence µ = 0. It follows that s = s ◦ Pk and this 

clearly implies that s = Sk. Assume next that s(rk) = 0 for all k ≥ 3. In this case 

the above description of SK0(A) shows that there are numbers s1, s2 ∈ [0, 1] such 

that s (zn) = s1z1 + s2z2, (zn) ∈ K0(A). Note that s1 + s2 = 1. It is easy to see 

that s2 = 0 or s1 = 0 since s is extremal. In the first case s = S1 and in the second 

s = S2. We have established (4.10). The reader may check that the Sk’s are in fact 

extreme points in SK0(A) = K and hence that equality holds in (4.10). We shall 

not need this fact here. 

Note that we have that limk→∞ Sk = 1 

2S1 + 1 

2S2 in SK0(A). Indeed, for (zn) ∈ 

 

K0(A) we have that Sk (zn) = 1 

2zk = 1 

2z1 + 1 

2z2 for all large enough k. Now 

define Φ : Aff SK0(A) → RN by Φ(f) = f(Sk) . Then φ maps Aff SK0(A) into 

{(tk) ∈ R N : limk→∞ tk = 1 

2 (t1 + t2) }. Denote this subspace of R N by X and 

set X+ = {(ti) ∈ X : ti ≥ 0 ∀i}. Then X is an ordered Banach space (in the 

norm (ti) = sup i |ti|) and 1 = (1, 1, 1, · · ·) is an order-unit. Note that Φ takes 

Aff SK0(A)+ into X+ and 1 to 1. If (ti) ∈ X+ we may define f ∈ Aff SK0(A)+ = 

Aff K+ by f (si) = t1s1 +t2s2 +2 ∞ 

n=3 tnsn, (sn) ∈ K. Then Φ(f) = (ti), proving 

that Φ(Aff SK0(A)+) = X+. To prove that Φ is an isometry we shall need the 

following general fact 

When C is a compact convex set and f ∈ Aff C, 

then there is a an element c ∈ ∂eC such that |f(c)| = f. 

(4.11) 

Proof. Proof of (4.11) : Let x ∈ C such that |f(x)| = f. Set h = f if 

f(x) ≥ 0 and h = −f if f(x) < 0. Then h ∈ AffC and h(x) = h = f. 

F = {y ∈ C : h(y) = h} is then a compact convex non-empty subseteq of C. Let 

c be an extreme point of F. We assert that c ∈ ∂eC. To see this let a, b ∈ C, t ∈]0, 1[, 

such that c = ta+(1−t)b. Then h = h(c) = th(a)+(1−t)h(b) ≤ th+(1−t)h = 

h, proving that h(a) = h(b) = h, i.e. that a, b ∈ F. Since c is extreme in F we 

conclude that c = a = b, so that c extreme in C. Since |f(c)| = |h(c)| = h = f, 

we have proved (4.11). 

Given (4.11) and (4.10) it is obvious that Φ(f) = sup k |f(Sk)| = f, f ∈ 

Aff SK0(A), i.e. Φ is is an isometry. Thus we have shown that Aff SK0(A) may 

be identified with X as an order unit space. We leave it as a (not necessarily easy) 

exercise for the reader to check that X has the Riesz interpolation property. 

The simplex K, with the topology coming from the identification with SK0(A), is 

qualitatively different from any example we have considered before. Observe namely 

that ∂eK is not closed in K since limk Sk = 1 

2S1 + 1 

2S2 in K. Therefore K can not 

be affinely homeomorphic to M1(X) for any compact Hausdorff space X. 

Example 4.56. Let A be a UHF-algebra. There is then a sequence A1 ⊆ A2 ⊆ 

A3 ⊆ · · · of unital C∗-subalgebras of A such that each An is ∗- isomorphic to a 

matrix algebra and 

n An = A. By Lemma 4.45 each An has exactly one trace state 

ωn. Note that the uniqueness implies that 

ωn+1|An = ωn . 

We can therefore define ω : 

n An → C by ω|An = ωn, n ∈ An. Since |ω(x)| ≤ x 

for all x ∈ 

n An, we can extend ω by continuity to all of A. Then ω is a trace state 

on A. It follows easily from Lemma 4.45 that ω is the only trace state of A. Thus a


UHF-algebra has exactly one trace state. By Corollary 4.54 there must be one and 

only one state on K0(A). The reader is invited to check this. 

5.1. Ideals. 

5. Ideals and simple C ∗ -algebras 

Definition 5.1. Let A be a C ∗ -algebra. An algebraic ideal in A is a linear 

subspace I ⊆ A such that a ∈ A, x ∈ I ⇒ ax, xa ∈ I. If I is also closed, I is called 

an ideal in A. 

Example 5.2. Let X be a compact Hausdorff space and A = C(X, Mk). Any 

closed subset F ⊆ X defines an ideal IF ⊆ A by 

IF = {f ∈ C(X, Mk) : f(x) = 0, x ∈ F } . 

The map F ↦→ IF defines a bijection between the closed subsets of X and the ideals 

on A = C(X, Mk). When I is an ideal in C(X, Mk) the corresponding closed subset 

of C(X, Mk) is given by 

{x ∈ X : f(x) = 0 ∀f ∈ I} . 

All unproven assertions of this example are left to the reader as exercises. 

For the study of ideals we need the notion of an approximate unit . 

Definition 5.3. Let I be an ideal in the C ∗ -algebra A. An approximate unit in 

I is a net {Eα} of positive elements in I such that 

1) Eα ≤ 1 for all α, 

2) α ≤ β ⇒ Eα ≤ Eβ, 

3) limα Eαx = x for all x ∈ I. 

Lemma 5.4. Let I be an ideal in the C ∗ - algebra A. Then I contains an approximate 

unit. 

Proof. Let U denote the set of finite collections (unordered tuples) α = {a1, a2, · · · , an} 

of elements from I. We introduce an ordering ≤ in U by setting α ≤ β when α = 

{a1, a2, · · · , an}, β = {b1, b2, · · · , bm} and there is an injective map σ : {1, 2, · · · , n} → 

{1, 2, · · · , m} such that ai = bσ(i) ∀i. Then U is a directed set. For α ∈ U as above 

we set 

n 

and 

Fα = 

i=1 

aia ∗ i 

Eα = nFα(1 + nFα) −1 , 

where 1 ∈ Â. Then Eα ∈ I and Eα ≤ 1. To see that Eα ≤ Eβ when α ≤ β, note 

first that Eα = 1 − (1 + nFα) −1 . So to conclude that Eα ≤ Eβ we must show that 

(1 + mFβ) −1 ≤ (1 + nFα) −1 , where Fβ = m i=1 bib∗ i . Since α ≤ β we obviously have 

that nFα ≤ mFβ. But 1 + mFβ ≥ 1 + nFα ≥ 1 implies that X = (1 + nFα) −1 2(1 + 

mFβ)(1 + nFα) −1 

2 ≥ 1. Thus X−1 ≤ 1, i.e. (1 + nFα) 1 

2(1 + mFβ) −1 (1 + nFα) 1 

2 ≤ 1.

5. IDEALS AND SIMPLE C ∗ -ALGEBRAS 129 

This implies that (1 + mFβ) −1 ≤ (1 + nFα) −1 , as needed. To see that limα Eαx = x 

for all x ∈ I, note that 

(Eαai − ai)(Eαai − ai) ∗ = 

i=1 

(1 + nFα) −1 Fα(1 + nFα) −1 = 

n 

(Eα − 1)aia ∗ i(Eα − 1) = 

F 1 

2 

α (1 + nFα) −2 F 1 

2 

α ≤ F 1 

2 

α (1 + nFα) −1 F 1 

2 

α = 

1 

n (1 − (1 + nFα) −1 ) ≤ 1 

1 . 

n 

Thus Eαai −ai2 ≤ 1 

n , i = 1, 2, · · · , n. It follows straightforwardly that limα Eαx = 

x for all x ∈ I. 

 

Theorem 5.5. An ideal I in a C ∗ -algebra A is selfadjoint (i.e. x ∈ I ⇒ x ∗ ∈ I) 

and A/I is a C ∗ -algebra in the quotient norm. 

Proof. Let {Eα} be an approximate unit in I. If x ∈ I, then limα x ∗ Eα−x ∗ = 

limα Eαx − x = 0, i.e. limα x ∗ Eα = x ∗ . Since x ∗ Eα ∈ I for all α, we see that 

x ∗ ∈ I, i.e. I is selfadjoint. In particular, A/I is a ∗-algebra and it suffices to prove 

the C ∗ -identity, i.e. that x + I 2 = xx ∗ + I. Since the reversed inequality is 

trivial, it suffices to show that x + I 2 ≤ xx ∗ + I. This depends on the fact that 

y + I = lim α y − Eαy, y ∈ A . (5.1) 

For all z ∈ I we have that y +z ≥ (1 −Eα)(y +z) for all α. Since limα Eαz = z, 

we see that y +z ≥ lim sup α y −Eαy. Hence y +I ≥ lim sup α y −Eαy. But 

Eαy ∈ I for all α, so clearly lim infα y − Eαy ≥ y + I, proving (5.1). Now we 

find that 

x + I 2 = lim α x − Eαx 2 = lim α (x − Eαx)(x − Eαx) ∗ = 

lim α (1 − Eα)xx ∗ (1 − Eα) ≤ lim α (1 − Eα)xx ∗ = 

lim α xx ∗ − Eαxx ∗ = xx ∗ + I . 

Example 5.6. Let A = C(X, Mk) be as in Example 5.2. If F ⊆ X is a closed 

subset and I = {f ∈ C(X, Mk) : f(x) = 0 ∀x ∈ F }, then A/I is ∗-isomorphic 

to C(F, Mk) and the quotient map can be identified with the restriction map r : 

C(X, Mk) → C(F, Mk) which restricts functions f ∈ C(X, Mk) to F. More formally, 

if q : A → A/I is the quotient map, there is a ∗-isomorphism α : A/I → C(F, Mk) 

such that 

commutes. 


A1 

A r 

C(F, Mk) 

 

q 

 

 

α 

 

 

A/I 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

· · ·


be a sequence of C ∗ -algebras, A = lim 

−→ (An, ϕn) the corresponding inductive limit C ∗ - 

algebra. If I is an ideal in A, 

I = 

n 

ϕ∞,n(ϕ−1 ∞,n (I)) . 

Proof. Set In = ϕ−1 ∞,n (I) and let q : A → A/I be the quotient map. Then 

In = ker (q ◦ ϕ∞,n) and hence q ◦ ϕ∞,n induces an injective ∗-homomorphism qn : 

An/In → A/I in the obvious way, i.e defined such that qn(x + In) = q ◦ ϕ∞,n(x). 

Note that qn is injective and hence and isometry. Let x ∈ I and ǫ > 0. There is an 

n ∈ N and an element a ∈ An such that ϕ∞,n(a) − x < ǫ. Let b be the image of 

a in An/In and note that qn(b) = q ◦ ϕ∞,n(a) = q(ϕ∞,n(a) − x) < ǫ. Since 

qn is isometric this shows that inf{a − z : z ∈ In} < ǫ, i.e. there is an element 

a0 ∈ In such that a −a0 < ǫ. It follows that x −ϕ∞,n(a0) < 2ǫ. Since ǫ > 0 was 

arbitrary, we see that x ∈ 

n 

ϕ∞,n(ϕ−1 ∞,n(I)), proving that I ⊆ 

n 

The reversed inclusion is obvious. 

ϕ∞,n(ϕ −1 

∞,n(I)). 

The essence of this theorem is that an ideal of an inductive limit C ∗ -algebra is 

put together by ideals in the sequence of C ∗ -algebras defining the algebra. This is 

more transparent in the following corollary. 

Corollary 5.8. Let A be a C ∗ -algebra and A1 ⊆ A2 ⊆ A3 · · · an increasing 

sequence of C ∗ -subalgebras of A such that A = 

n An. If I is an ideal in A, 

I = 

An ∩ I . 

Proof. This follows straightforwardly from Theorem 5.7. 

n 

Lemma 5.9. Let F be a finite dimensional C ∗ -algebra and I an ideal in F. There 

is then a central projection q ∈ F such that I = qF. 

Proof. I is a finite dimensional C ∗ -algebra in itself, so I unital. Let q be the 

unit in I. It is then obvious that I = qF. To see that q is central in F, let x ∈ F. 

Since xq ∈ F and q is a unit in I, we have that qxq = xq. If x = x ∗ , this shows that 

qx = (xq) ∗ = (qxq) ∗ = qxq = xq. Hence q commutes with every selfadjoint element 

of F and therefore with every element of F, i.e. q is central. 

 

It is clear that if q is a central projection in any C ∗ - algebra A, then qA is an ideal 

in A. Lemma 5.9 shows that every ideal of a finite dimensional C ∗ -algebra is of this 

form. (The example A = C([0, 1]) shows that this is certainly not a general fact, cf. 

Example 5.6.) We know that a finite dimensional C ∗ -algebra F can be identified with 

a direct sum Mn1⊕Mn2⊕· · ·⊕Mnm of matrix algebras. The center of Mn1⊕Mn2⊕· · ·⊕ 

) = ei, i = 1, 2, · · · , m. 

Mnm is spanned by the projections (0, · · · , 0, 0, 1, 0, 0, · · · , 0 


In other words e1, e2, · · · , em are the minimal non-zero projections in the center of 

Mn1 ⊕ · · · ⊕ Mnm. The most general central projection is therefore of the form 

 

i∈D 

for some subset D ⊆ {1, 2, · · · , m}. So by using Lemma 5.9 we easily reach the 

following conclusion. 

ei


Lemma 5.10. For any subset D ⊆ {1, 2, · · · , m}, define the ideal ID in Mn1 ⊕ 

Mn2 ⊕ · · · ⊕ Mnm by 

ID = {(a1, a2, · · · , am) ∈ Mn1 ⊕ Mn2 ⊕ · · · ⊕ Mnm : ai = 0, i /∈ D} . 

This gives a bijection between the ideals of Mn1 ⊕ · · · ⊕ Mnm and the subsets on 

{1, 2, · · · , m}. 

Proof. Left to the reader. 

Lemma 5.11. Let I be an ideal in the C ∗ - algebra A. If y ∈ I, x ∈ A and 

0 ≤ xx ∗ ≤ y, then x ∈ I. 

Proof. After dividing through with y we may assume that y ≤ 1. By using 

this we prove that 

1 

lim y nx = x . (5.2) 

n→∞ 

This will complete the proof because y 1 

n ∈ I for all n ∈ N. First we observe 

that t(1 − t 1 

n) converges decreasingly to 0 for all t ∈ [0, 1]. By Dinis theorem the 

convergence is uniform on [0, 1], proving that limn→∞ y(1 − y 1 

n) = 0. Since 

y 1 

nx − x 2 = (1 − y 1 

n)xx ∗ (1 − y 1 

n) 

≤ (1 − y 1 

n)y(1 − y 1 

n) ≤ y(1 − y 1 

n) , 

(5.2) follows. 

Now we want to describe the ideals of an AF-algebra A. As one should expect 

the ideals can be read off from the K0(A)-group. 

Definition 5.12. Let G be a partially ordered group. An ideal H in G is a 

subgroup such that H = H ∩ G + − H ∩ G + and 

x ∈ G + , y ∈ H ∩ G + , x ≤ y ⇒ x ∈ H . 

Lemma 5.13. Let A be an AF-algebra and I an ideal in A. Set 

HI = {[e] − [f] : e, f is a projection in Mn(I) for some n ∈ N} 

Then HI is an ideal in K0(A). 

Proof. Note first that HI = HI ∩G + −HI ∩G + by definition. Let x ∈ G + , y ∈ 

HI ∩ G + such that x ≤ y. Then y = [e] − [f] = [p] where e, f and p are projections 

, e, f in Mn(I) and p in Mn(A), for some n. By Lemma 3.27 1) we have that 

e ⊕ 0 ≈ p ⊕ f in M2n(A). Let v ∈ M2n(A) be a partial isometry implementing this 

equivalence. Since M2n(I) is an ideal in M2n(A) and e ⊕ 0 ∈ M2n(I), we see from 

Lemma 5.11 first that v ∈ M2n(I) and then that p, f ∈ Mn(I). Since x ∈ G + there 

is a projection q in some matrix algebra over A such that x = [q]. By increasing n 

we may assume that q ∈ Mn(A). Since [q] ≤ y = [p] there is a projection r ∈ Mn(A), 

orthogonal to q, such that q +r ≈ p in Mn(A), cf. Lemma 3.27 1). Since p ∈ Mn(I), 

it follows from Lemma 5.11 first that q +r ∈ Mn(I) and then that q ∈ Mn(I). Thus 

x = [q] ∈ HI. 

If we let i : I → A be the inclusion, the ideal HI of Lemma 5.13 is i∗ (K0 (I)). 

Let A be an AF-algebra and H is an ideal in K0(A). Set 

IH = {q ∈ A : q = q ∗ = q 2 , [q] ∈ H} .


We assert that span IH is an ideal in A. To see this, write A = 

n An where 

A1 ⊆ A2 ⊆ A3 ⊆ · · · are finite dimensional C ∗ -subalgebras of A. Let Jn = An ∩ IH. 

We assert that span Jn is an ideal in An. To see this, we may assume that 

An = Mn1 ⊕ · · · ⊕ Mnm, 

and let ek ij : i, j = 1, 2, . . ., nk, k = 1, 2, . . ., m be a full set of matrix units in An. 

Set 

D = k ∈ {1, 2, . . ., , m} : e k 

11 ∈ H . 

We prove first that 

span Jn = {(a1, a2, . . .,am) ∈ An| ai = 0, i /∈ D}. (5.3) 

To prove this equality, let q ∈ Jn = An ∩ IH. Write q = (q1, q2, . . ., qm). If qk = 0, 

we have that ek 

11 ≤ [gk] ∈ H which implies that ek 

11 ∈ H. It follows that k ∈ D. 

Thus q ∈ {(a1, a2, . . .,am) ∈ An| ai = 0, i /∈ D} and we conclude that span Jn ⊆ 

{(a1, a2, . . .,am) ∈ An| ai = 0, i /∈ D}. To prove that the reverse inclusion, note that 

every element of {(a1, a2, . . .,am) ∈ An| ai = 0, i /∈ D} is a linear span of projections 

from {(a1, a2, . . .,am) ∈ An| ai = 0, i /∈ D}. It suffices therefore to show that any 

projection p ∈ {(a1, a2, . . .,am) ∈ An| ai = 0, i /∈ D} is in span Jn. To see that this 

is indeed the case, write p = (p1, p2, . . .,pm). If pk = 0, ek 

11 ∈ H and since 

k 

pk ≤ nk e11 , this implies that [pk] ∈ H. Hence pk ∈ IH, and we conclude that 

p ∈ span Jn. 

Note that (5.3) exhibits spanJn as an ideal in An, so to conlcude that span IH 

is an ideal in A it suffices now to show that 

span IH = 

span Jn. (5.4) 

Let q ∈ IH and let ǫ > 0. By Lemma ?? and Lemma 3.14 there is an n and 

a projection p ∈ An such that p and q are Murray-von Neumann equivalent and 

p − q ≤ ǫ. Then p ∈ An ∩ IH, so p ∈ 

n span Jn. Since ǫ > 0 was arbitrary, this 

implies that q ∈ 

n span Jn. It follows that span IH ⊆ 

n span Jn. Since the reverse 

inclsuion is trivial, we have established (5.4). Hence span IH is an ideal in A, as 

asserted. 

Theorem 5.14. Let A be an AF-algebra. The map I ↦→ HI is a bijection between 

the ideals of A and the ideals of K0(A). The inverse is the map H ↦→ span IH. 

Proof. We must show that span IHI = I for every ideal I in A and that 

Hspan = H for every ideal H in K0(A). 

IH 

I ⊆ span IHI : Let x ∈ I ∩ An. Then x is the span a set of projections in 

I ∩ An. Since the involved projections are in IHI , we see that x ∈ span IHI . Since 

I = 

n I ∩ An by Corollary 5.8 

Let first I be an ideal in A. If p ∈ IHI , [p] = [e] − [f], where e, f ∈ Mn(I) 

for some n. It follows that p ⊕ f ≈ e ⊕ 0 in Mn+1(A). Since e ⊕ 0 ∈ Mn+1(I), it 

follows from Lemma 5.11 that p ∈ I. Thus span IHI ⊆ I. To prove the converse 

inclusion, note that I is an AF-algebra in itself by Corollary 5.8 and Lemma 5.10. 

In particular I is the closed linear span of its projections, cf. the proof of Corollary 

4.54. It is therefore enough to prove that every projection e ∈ I is in IHI . But this 

is obvious. Hence span IHI = I. 

n


Let next H be an ideal in K0(A) and take x ∈ Hspan ∩Σ(A). Then x = [e] for 

IH 

some projection e ∈ span IH. By using (??) we may in fact assume that e ∈ span Jn 

for some n ∈ N. But we saw above that every projection of span Jn is in IH, so 

x = [e] for some e ∈ IH, i.e. x ∈ H. Since Hspan IH is generated by Hspan ∩ Σ(A), IH 

we have that Hspan ⊆ H. If x ∈ H ∩ Σ(A), x = [e] for some projection e ∈ A. 

IH 

. 

Then e ∈ IH by definition and x ∈ H span IH . It follows that H ⊆ H span IH 

, with scale Σ(A), then H∩Σ(A) generates H∩K0(A) + as a semigroup. Indeed, if 

x ∈ H∩K0(A) + , there are elements z1, z2, · · · , zn ∈ Σ(A) such that z1+z2+· · ·+zn = 

x. In particular, 0 ≤ zi ≤ x, so zi ∈ H because H is an ideal in K0(A). 

5.2. Simple C ∗ -algebras. In this section we introduce one of the most interesting 

classes of C ∗ -algebras; namely the simple ones. We collect first a series of 

general facts about simple C ∗ -algebras and turn subsequently towards the study of 

simple AF-algebras. 

Definition 5.15. A C ∗ -algebra is simple when A has no non-trivial ideals, i.e. 

when {0} and A are the only ideals in A. 

Lemma 5.16. 1) A is simple if and only if the following condition holds : 

For every pair x, y ∈ A\{0} and every ǫ > 0 there are elements 

a1, a2, · · · , aN, b1, b2, · · · , bN ∈ A 

such that N 

i=1 aixbi − y < ǫ. 

2) When A is unital, A is simple if and only if A contains no non-trivial 

algebraic ideals. 

Proof. 1) : Assume first that the condition holds. If I is a non-zero ideal 

in A and x ∈ I\{0}, then the condition obviously implies that I is dense in A, 

i.e. is all of A. Thus A is simple. Conversely, assume that A is simple and let 

x, y ∈ A\{0}. The closure of span{axb : a, b ∈ A} is clearly an ideal which is 

non-zero because x ∗ xx ∗ x = 0 is in it. By simplicity of A it must be all of A, and 

hence y is in it. The existence of elements a1, a2, · · · , aN, b1, b2, · · · , bN in A such 

that 

i aixbi − y < ǫ is now almost immediate. 2) : If A does not contain 

any non-trivial algebraic ideals it is obviously simple. Conversely, assume that A is 

simple and let I be a non-zero algebraic ideal in A. Let x ∈ I\{0}. By 1) there 

are elements a1, a2. · · · , aN, b1, b2, · · · , bN ∈ A such that N 

i=1 aixbi − 1 < 1. But 

then N 

i=1 aixbi = z is an invertible element in I and hence 1 = zz −1 ∈ I. It follows 

that I = A. 

Example 5.17. A full matrix algebra Mn is simple. This follows from Lemma 

5.9 because 0 and 1 are the only central projections in Mn. (Check !) Now Corollary 

5.8 shows that any UHF-algebra is simple. 

Lemma 5.18. Let A be a simple C ∗ -algebra and ω : A → C a non-zero positive 

trace functional. Then ω is faithful , i.e. ω(a) > 0 for all a ∈ A+\{0}. 

Proof. Set I = {x ∈ A : ω(xx ∗ ) = 0}. Since |ω(ab)| 2 ≤ ω(aa ∗ )ω(bb ∗ ) for all 

a, b ∈ A, we see that I = {x ∈ A : ω(xy) = 0 ∀y ∈ A}. This shows that I is a 

closed subspace of A. Let x ∈ I, a ∈ A. Then ω(xay) = 0 and ω(axy) = ω(xya) = 0 

for all y ∈ A. Thus ax, xa ∈ I, i.e. I is an ideal in A. By assumption ω = 0, so


I = A, i.e. I = {0} by simplicity of A. In particular, ω(a) = ω(a 1 

2a 1 

2) > 0 for all 

a ∈ A+\{0}. 

Lemma 5.19. Let A be a simple C ∗ -algebra. Then Mn(A) is simple. 

Proof. Let {Eα} be an approximate unit in A. For each α we denote by E ij 

α 

the element of Mn(A) whose A-entries are all 0, except for the ij’th where the entry 

is Eα. Let I be a non-zero ideal in Mn(A) and x a positive non-zero element of I. 

If Eii αxEii α = 0 for all α and all i, it follows that Eii 

αx = 0 for all α and all i. Hence 

n 

x = limα i=1 Eii αx = 0, a contradiction. So, for some α and i, Eii αxE ii 

α is a non-zero 

positive element of I with only one non-zero entry, namely the ii’th. Since A is 

simple it follows easily from Lemma 5.16 1) that I contains all elements of Mn(A) 

whose only non-zero element is the ii’th. For each j = i and each b ∈ A, observe 

that 

diag(0, 0, · · · , 0, 0, b, 0, 0, · · · , 0, 0) 

= 

b at the j’th entry 

lim α E ji 

α 

diag(0, 0, · · ·0, 0, b, 0, · · · , 0, 0, 0)E 

b at the i’th entry 

ij 

α ∈ I . 

It follows that diag(a1, a2, · · · , an) ∈ I for all a1, a2, · · · , an ∈ A. In particular, 

Fα = diag(Eα, Eα, · · · , Eα) ∈ I for all α. Since {Fα} is an approximate unit in 

Mn(A), it follows that I = Mn(A). 

Lemma 5.20. Let A be a simple unital C∗- algebra and a ∈ A+ a non-zero 

positive element. There is then a finite set x1, x2, · · · , xN of element in A such that 

N 

= 1 . 

i=1 

xiax ∗ i 

Proof. By Lemma 5.16 there are elements c1, · · · , cn, d1, · · · , dn ∈ A such that 

n i=1 ciadi − 1 < 1. Then n i=1 ciadi is invertible. If we substitute each di with 

 

di j cjadj 

−1 n we get that i=1 ciadi = 1. Now, if x, y are arbitrary elements in A 

we have that xx∗ +y∗y ≥ xy+y ∗x∗ since (xx∗ +yy∗ )−(xy+y ∗x∗ ) = (x−y∗ )(x−y∗ ) ∗ ≥ 

0. Therefore ciadi + d∗ iac∗i ≤ ciac∗ i + d∗i adi for each i and hence 

In particular, 

1 = 1 

2 

n 

ciadi + 

i=1 

n 

d ∗ iac ∗ 1 

i ≤ 

2 

n 

i=1 

h = 1 

2 

n 

i=1 

ciac ∗ i + 

i=1 

n 

i=1 

ciac ∗ i + 

d ∗ 

iadi 

n 

i=1 

d ∗ 

iadi . 

is positive and invertible. Set xi = 1 √ h 

2 −1 2ci, i = 1, 2, · · · , n, and xn+i = 1 

√ h 

2 −1 2d∗ i, i = 

1, 2, · · · , n. Then 2n i=1 xiax∗ i = 1. 

Proposition 5.21. Let A be a unital simple C ∗ - algebra. Then any non-zero 

element of K0(A) + is an order-unit in K0(A). 

Proof. Since [1] is an order unit in K0(A) it suffices to show that if x ∈ 

K0(A) + \{0}, then there are natural numbers n, m = 0 such that n[1] ≤ mx. Let 

x = [e] where e is a non-zero projection in Mn(A). By Lemma 5.19, B = Mn(A)


is simple, so by Lemma 5.20 there are elements x1, x2, · · · , xm 

∈ B such that 

m 

i=1 xiex∗ i = 1B. Set 

⎛ 

⎞ 

x1e x2e . . . xme 

⎜ 0 0 . . . 0 ⎟ 

v = ⎜ 

⎝ . 

⎟ 

. . .. . 

⎠ 

0 0 . . . 0 

. 

Then v ∈ Mm(B) = Mmn(A) and vv ∗ = diag(1B, 0, 0, . . ., 0) ∈ Mm(B). Thus v is a 

partial isometry and v ∗ v must be a projection in Mm(B). Set E = diag(e, e, e, · · · , e) 

and note that vE = v. Hence v ∗ v = Ev ∗ vE ≤ E. It follows that n[1] = 

[diag(1B, 0, 0, · · · , 0)] = [v ∗ v] ≤ [E] in K0(A). Since [E] = m[e], this completes 

the proof. 

Proposition 5.22. Let A be a simple unital C ∗ - algebra. Then either K0(A) + = 

K0(A) or K0(A) + ∩ (−K0(A) + ) = {0}. 

Proof. Assume that K0(A) + ∩ (−K0(A) + ) = {0} and let 0 = x ∈ K0(A) + ∩ 

(−K0(A) + ). By Proposition 5.21 x is an order unit in K0(A), so for any y ∈ K0(A) 

we can find m ∈ N such that y ≤ mx. Since mx ∈ −K0(A) + , this implies that 

y ≤ 0. Thus K0(A) = −K0(A) + , or equivalently, K0(A) = K0(A) + . 

Lemma 5.23. Let A be a simple C ∗ -algebra and q ∈ A a projection. Then qAq 

is a simple C ∗ -algebra. 

Proof. Let x, y ∈ qAq\{0}, ǫ > 0. Since A is simple there are elements 

a ′ 1 , a′ 2 , · · · , a′ N , b′ 1 , b′ 2 , · · · , b′ N 

∈ A such that 

i a′ i xb′ i − y < ǫ. Set ai = qa ′ i q, bi = 

qb ′ i q. Then ai, bi ∈ qAq for, all i and 

i aixbi − y < ǫ. 

We now turn to a study of the simple AF-algebras. 

Theorem 5.24. Let A be an AF-algebra. Then the following conditions are 

equivalent. 

1) A is simple. 

2) K0(A) has no ideals other than {0} and K0(A). 

3) Every non-zero element of K0(A) + is an order unit in K0(A). 

Proof. The equivalence between 1) and 2) is immediate from Theorem 5.14. 2) 

⇒ 3) : Let x ∈ K0(A) + \{0}. Set I = {y ∈ K0(A) : −mx ≤ y ≤ mx for some m ∈ 

N}. It is easily seen that I is an ideal in K0(A), and it is non-zero since x ∈ I. 

Hence I = K0(A) by 2). But this means that x is an order unit in K0(A). 3) ⇒ 2) 

: Let I be a non-zero ideal in K0(A). Then I contains a non-zero element x from 

K0(A) + . By (3) x is a order unit, so for any y ∈ K0(A) there is an n ∈ N such that 

y ≤ nx and −y ≤ nx. Since I is an ideal in K0(A) and 0 ≤ nx − y ≤ 2nx ∈ I, we 

see that y = nx − (nx − y) ∈ I. Hence I = K0(A). 

For many purposes it is very useful to have a criterion for an AF-algebra to be 

simple which is given in terms of its Bratteli diagram. Therefore we introduce the 

following notation and terminology. 

Definition 5.25. Let A1 ⊕A2 ⊕ · · · ⊕An and B1 ⊕B2 ⊕ · · · ⊕Bm be direct sums 

of C ∗ -algebras and 

ϕ : A1 ⊕ A2 ⊕ · · · ⊕ An → B1 ⊕ B2 ⊕ · · · ⊕ Bm


a ∗-homomorphism. For each j ∈ {1, 2, · · · , n}, define ιj : Aj → A1 ⊕ A2 ⊕ · · · ⊕ An 

by 

ιj(a) = (0, 0, · · · , 0, 0, · · · , 0, 0, a, 0, 0, · · · , 0) 

. 

a at the j’th entry 

For each i ∈ {1, 2, · · · , m}, define πi : B1⊕B2⊕· · ·⊕Bm → Bi by πi(a1, a2, · · · , am) = 

ai. Then 

πi ◦ ϕ ◦ ιj : Aj → Bi 

is a ∗-homomorphism which we denote by ϕij. The ∗-homomorphisms ϕij, j = 

1, 2, · · · , n, i = 1, 2, · · · , m, will be called the partial maps of ϕ. 

Note that the partial maps determine ϕ since 

ϕ(a1, a2, · · · , an) = n 

ϕ1j(aj), 

j=1 

n 

ϕ2j(aj), · · · , 

j=1 

n 

ϕmj(aj) 

Definition 5.26. A ∗-homomorphism ϕ : A1 ⊕ A2 ⊕ · · · ⊕ An → B1 ⊕ B2 ⊕ 

· · · ⊕Bm is called mixing when ϕij = 0 for all i, j, i.e. when all its partials maps are 

non-zero. 


A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

· · · 

be a sequence of finite direct sums of matrix algebras, 

An = Mk1 ⊕ Mk2 ⊕ · · · ⊕ Mkmn . 

Assume that for every k ∈ N there is an n > k such that ϕn,k : Ak → An is mixing. 

Then A = lim 

−→ (An, ϕn) is simple. 

Proof. Assume I is a non-zero ideal in A. By Theorem 5.7 ϕ −1 

∞,k (I) must be 

a non-zero ideal in Ak for some k ∈ N. Let qk ∈ Ak be a central projection such 

that qkAk = ϕ −1 

∞,k (I) and choose n1 > k such that ϕn1,k : Ak → An1 is mixing. 

Then ϕn1,k(qk) ∈ ϕ −1 

∞,n1 (I) and qϕn1,k(qk) = 0 for all non-zero central projections 

q in An1 because ϕn1,k is mixing. So Lemma 5.9 implies that ϕ−1 (I) = An1. 

∞,n1 

By assumption there is an n2 > n1 such that ϕn2,n1 : An1 → An2 is mixing. By 

repeating the argument we see that ϕ−1 ∞,n2 (I) = An2, and we get in this way a 

sequence n1 < n2 < n3 · · · in N such that Anm = ϕ−1 ∞,nm (I) for all m. Hence I = A 

by Theorem 5.7. 

In the general, non-unital case, the condition of Lemma 5.27 is only a sufficient, 

but not a necessary condition for a sequence of finite direct sums of matrix algebras 

j=1


to give a simple inductive limit C ∗ -algebra. For example, the Bratteli diagram 

1 1 

 

2 1 

 

3 1 

 

4 1 

. . 

describes such a sequence with limit K, the compact operators on l 2 , see Exercise 

2.27. The limit is simple but the condition of Lemma 5.27 is not satisfied. In the 

unital case, however, the situation is more satisfying. 


A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

· · · 

be a sequence of finite direct sums of matrix algebras. Assume that the connecting 

∗-homomorphisms are unital and injective. Then A = lim(An, 

ϕn) is simple if and 

−→ 

only if there is, for every n ∈ N, an m > n such that ϕm,n is mixing. 

Proof. Assume that A is simple and consider some n ∈ N. Let q1, q2, · · · , qmn 

be the minimal non-zero projections in the center of An. We must show that there 

is a m > n such that qϕm,n(qi) = 0 for all i = 1, 2, · · · , mn, and all non-zero central 

pojections q in Am. This will namely clearly imply that ϕm,n is mixing. Since A is 

simple we can find elements k j 

i , j = 1, 2, · · · , Ni, i = 1, 2, · · · , mn, in A such that 

Ni 

k 

j=1 

j j ∗ 

iϕ∞,n(qi)k i = 1, i = 1, 2, · · · , mn . 

This follows from Lemma 5.20. But then there is an m ∈ N so large that Am contains 

such that 

elements r j 

i 

Ni 

 

j=1 

for all i = 1, 2, · · · , mn. In particular, 

r j j∗ 

iϕm,n(qi)r i − 1 < 1 

Ni 

r 

j=1 

j j∗ 

iϕm,n(qi)r i 

is invertible in Am and hence ϕm,n(qi)q must be non-zero for all non-zero central 

projections in Am.


5.3. Simple AF algebras and simple dimension groups. 

Definition 5.29. A unital C ∗ -algebra A is called approximately divisible when 

the following holds : For any finite subset F ⊆ A, any N ∈ N and any ǫ > 0, there is a 

finite-dimensional unital C ∗ - subalgebra B of A such that B ≃ Mn1⊕Mn2⊕· · ·⊕Mnm, 

where ni ≥ N, i = 1, 2, · · · , m, and a finite set G of A such that gb = bg for all 

g ∈ G, b ∈ B and 

∀f ∈ F ∃g ∈ G : f − g < ǫ . 

Less formally A is approximately divisible if any finite subset approximately 

commutes with a finite dimensional C ∗ -subalgebra which is the sum of matrix algebras 

of arbitrary large dimensions. Maybe the definiton becomes a little more 

transparent if we introduce the following notation. When B is a C ∗ -subalgebra of 

the C ∗ -algebra A , set 

B ′ ∩ A = {a ∈ A : ab = ba ∀b ∈ B} . 

B ′ ∩A is a C ∗ -subalgebra of A and is called the relative commutant of B in A. Then 

A is approximately divisible when the following holds : For any finite subset F ⊆ A, 

any N ∈ N and any ǫ > 0, there is a finite dimensional unital C ∗ -subalgebra B ⊆ A 

such that 

where ni ≥ N, i = 1, 2, · · · , m, and 

B ≃ Mn1 ⊕ Mn2 ⊕ · · · ⊕ Mnm , 

dist(x, B ′ ∩ A) < ǫ, x ∈ F. 

We are aiming to prove that all simple unital AF algebras which are not finite 

dimensional are approximately divisible. For this we need the following lemma. 

Lemma 5.30. Let A ≃ Mm and let B ≃ Mn be a unital C ∗ -subalgebra of A. 

Then 

B ′ ∩ A ≃ M m 

n . 

Proof. Let {eij} and {fij} be full sets of matrix units in A and B, respectively. 

By Lemma 2.6 there is a unitary u ∈ A such that 

ufiju ∗ = 

m 

n 

k=1 

ei+(k−1)n, j+(k−1)n 

for all i, j ∈ {1, 2, · · · , n}. The map x ↦→ uxu ∗ defines a ∗-isomorphism from B ′ ∩ A 

onto C ′ ∩ A where C = uBu ∗ . Note that {ufiju ∗ } form a full set of matrix units 

in C. By substituting C for B and ufiju ∗ for fij, we can therefore assume that 

fij = 

m 

n 

k=1 ei+(k−1)n,j+(k−1)n for all i, j. For each pair k, l ∈ {1, 2, · · · , m 

n 

pkl = 

n 

i=1 

ei+(k−1)n, i+(l−1)n . 

}, set 

It is straightforward, although possibly a little tedious, to check that {pkl} is a set 

of matrix units for a copy of M m 

n inside A and that fijpkl = pklfij for all choices 

of i, j, k, l. In other words, {pkl} are matrix units for a copy of M m 

n inside B′ ∩ A.


To see that {pkl} actually spans all of B ′ ∩ A, we first consider a, b ∈ {1, 2, · · · , m}. 

Straightforward calculations show that 

 

fijeabfji = 

ei+(k−1)n,j+(k−1)neabej+(l−1)n,i+(l−1)n = pkl , 

i,j 

i,j,k,l 

where (k − 1)n + j = a and (l − 1)n + j = b, provided that a and b have the same 

remainder after division by n. If this is not the case we find that 

 

fijeabfji = 

ei+(k−1)n,j+(k−1)neabej+(l−1)n,i+(l−1)n = 0. 

i,j 

i,j,k,l 

In either case 

i,j fijeabfji ∈ span{pkl}. Now, if x ∈ B ′ ∩ A, then 

x = 1 

n 

 

i,j 

fijxfji 

since x commutes with each fij. Thus, by writing 

x = 

λa,beab, 

where λa,b ∈ C for all a, b ∈ {1, 2, · · · , n}, we find that 

x = 1 

λa,bfijeabfji ∈ span{pkl} . 

n 

i,j,a,b 

a,b 

Hence B ′ ∩ A = span{pkl} ≃ M m. 

 

n 

In the following lemma we use the convention that M0 = 0. We remind the 

reader that a ∗-homomorphism ϕ : Mn1 ⊕ · · · ⊕ MnM → Mm1 ⊕ · · · ⊕ MmM is inner 

equivalent to a standard homomorphism given by an M × N matrix S = (sij) with 

entries from N, cf. Lemma 2.6. The matrix S is called the multiplicity matrix for 

ϕ. (Two homomorphisms ϕ, ψ : A → B between C∗-algebras A and B are called 

unitarily equivalent when there is a unitary u ∈ ˆ B such that Ad u ◦ ϕ = ψ.) 

Lemma 5.31. Let ϕ : A = Mn1 ⊕ · · · ⊕ MnN → B = Mm1 ⊕ · · · ⊕ MmM be a 

unital ∗-homomorphism. Then 

ϕ(A) ′ ∩ B ≃ ⊕ 

i,j∈{1,2,···,M}×{1,2,···,N} 

where (sij) is the multiplicity matrix for ϕ. 

Proof. For each i ∈ {1, 2, · · · , M}, let πi : B → Mmi 

ϕ(A) ′ ∩ B ≃ ⊕ M i=1 πi ◦ ϕ(A) ′ ∩ Mmi . 

Msij , 

be the projection. Then 

Fix i and let q1, q2, · · · , qN be the minimal non-zero central projections in A. Then 

pj = πi ◦ ϕ(qj), j = 1, 2, · · · , N, are orthogonal projections with sum 1 in Mmi . Set 

Dij = pjMmipj and Cij = πi ◦ ϕ ◦ ιj(Mnj ), where ιj : Mnj → A is as in Definition 

5.25. Then the map 

x ↦→ (xp1, xp2, · · · , xpN) 

defines a ∗-isomorphism from πi ◦ ϕ(A) ′ ∩ Mmi onto ⊕Nj=1 C′ ij ∩ Dij. Since Cij ≃ 

Mnj and Dij ≃ Msijnj by definition of (sij), we conclude from Lemma 5.30 that 

C ′ ij ∩ Dij ≃ Msij . Hence 

ϕ(A) ′ ∩ B ≃ ⊕ M i=1πi ◦ ϕ(A) ′ ∩ Mmi ≃ ⊕Mi=1 ⊕Nj=1 C′ ij ∩ Dij ≃ ⊕ Msij 

ij 

.



A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

· · · 

be a sequence of finite dimensional C∗-algebras and unital injective ∗-homomorphisms. 

Assume that A = lim(An, 

ϕn) is simple and not finite dimensional. Let S 

−→ n,m = (s n,m 

ij ) 

be the multiplicity matrix for ϕn,m : An → Am, n < m. It follows that 

for all n ∈ N. 

limm→∞ min 

ij sn,m ij 

= ∞ 

Proof. Assume first that there is sequence k1 < k2 < · · · in N such that Akn is 

a matrix algebra for all n. Then A is the inductive limit of the sequence 

A1 

ϕ1 

A2 

ϕ2 

A3 

ϕ3 

· · · 

(and hence A is a UHF algebra). Since A is not finite dimensional we must have 

that limm→∞ dim Akm = ∞. Since Ski,kj is the one-by-one matrix with entry 

S ki,kj 

 

dim Akj 11 = 

, we have that limj→∞ S dim Aki ki,kj 

11 = ∞ for each i. Now, it is a general 

fact that S n,m = S k,m S n,k when n < k < m. Since the ϕj’s are unital and injective, 

so that no row or column is zero in any of the S n,m ’s, it follows that 

min 

ij Sn,m ij 

increases with m −n. Therefore we have that limm→∞ minij S n,m 

ij 

 

= ∞ in this case. 

In the remaining case there is an N ∈ N such that S k,k+1 has at least two rows 

when k ≥ N. By Theorem 5.28 there is a sequence k1 < k2 < · · · in N such that 

min 

ab Ski,kj 

ab ≥ 1 

for all i < j. Since S ki,ki+1 has at least two rows when ki ≥ N, it follows easily that 

min 

ab Ski,kj 

ab 

≥ 2j−i−1 

for all i < j with ki ≥ N. In particular, limj→∞ minab S ki,kj 

ab 

The conclusion that limm→∞ minij S n,m 

ij 

= ∞ when ki ≥ N. 

= ∞ follows now as in the first case. 

Theorem 5.33. Let A be a simple unital AF-algebra which is not finite dimensional. 

Then A is approximately divisible. 

Proof. Let F be a finite subset of A, N ∈ N and ǫ > 0. Write A = 

n An 

where A1 ⊆ A2 ⊆ A3 ⊆ · · · is an increasing sequence of finite dimensional unital 

C∗-subalgebras of A. There is then an n ∈ N and a finite set G ⊆ An such that 

dist(x, G) < ǫ for all x ∈ F. By Lemma 5.31 and Lemma 5.32 there is an m > n 

such that B = A ′ n ∩ Am ≃ Mn1 ⊕ · · · ⊕ Mnd , where ni ≥ N for all i = 1, 2, · · · , d. 

Since G ⊆ B ′ ∩ A, this completes the proof. 

 

Lemma 5.34. Let A be a simple unital AF-algebra which is not finite dimensional. 

Let p be a projection in A and α ∈ [0, 1]. For any ǫ > 0 there is a projection q ∈ A 

such that 

|αω(p) − ω(q)| < ǫ ∀ω ∈ T(A) .


Proof. We may obviously assume that p = 0. Choose N ∈ N such that 1 

N 

Since A is approximately divisible by Theorem 5.33, there is a finite dimensional 

unital C∗-subalgebra B of A such that B ≃ Mn1 ⊕ · · · ⊕ MnL , where ni ≥ N for all 

i, and dist(p, B ′ ∩ A) is as small as we want. By using first Lemma 2.10 and then 

Lemma 2.11 we can produce a projection e ∈ B ′ ∩ A such that p ≈ e. In particular, 

ω(p) = ω(e) > 0, ∀ω ∈ T(A), where the last inequality uses Lemma 5.18 and that 

p = 0. Let {rd ij : i, j = 1, 2, · · · , nd, d = 1, 2, · · · , L} be a full set of matrix units in 

B and set ed = e nd 

j=1 rd jj , d = 1, 2, · · · , L. For each i we choose ji ∈ {1, · · · , ni} 

such that |α − ji 

ni 

| ≤ 1 

ni 

≤ 1 

N 

q = e 

< ǫ. Set 

j1 

k=1 

r 1 kk 

+ e 

j2 

k=1 

r 2 kk 

+ · · · + e 

jL 

k=1 

r L kk . 

Then q is a projection and we claim that it does the job. To check it, note first that 

via er k 1j for all j, k, so that 

for all j, k. It follows that 

er k 11 ≈ erk jj 

ω(er k jj ) = ω(erk 1 

11 ) = ω(ek), ω ∈ T(A), 

nk 

ω( 

for all d = 1, 2, · · · , L. Thus 

jd 

k=1 

er d jd 

kk ) = ω(ed), ω ∈ T(A), 

nd 

|αω(p) − ω(q)| = |αω(e) − ω(q)| = | 

| 

L 

(α − jd 

d=1 

nd 

) ω(ed) 

|ω(e) ≤ 

ω(e) 

L 

αω(ed) − 

d=1 

L 

|α − jd 

d=1 

nd 

| ω(ed) 

ω(e) 

< ǫ 

L 

jd 

nd 

d=1 

L 

d=1 

ω(ed)| = 

ω(ed) 

ω(e) 

= ǫ . 

for all ω ∈ T(A). 

5.4. Alternative proof of Lemma 5.34. In this subsection we present a more 

elementary proof of Lemma 5.34 which, in particular, avoids the notion of approximate 

divisibility. 

Lemma 5.35. Let A = Mn1 ⊕ Mn2 ⊕ · · · ⊕ MnN be a finite direct sum of matrix 

algebras, and set m = mini ni. 

For every t ∈ [0, 1] there is a projection p ∈ A such that |t − τ(p)| ≤ 1 for every 

m 

trace state τ ∈ T(A). 

Proof. Let ωi be the trace state of Mni , i.e. 

ωi = 1 

Tr, 

ni 

where Tr is the usual trace obtained by adding the diagonal entries. Note that for 

each i, there is a projection pi ∈ Mni such that 

|t − ωi(pi)| ≤ 1 

≤ 1 

. (5.5) 

m 

ni 

< ǫ.


Then 

p = (p1, p2, . . .,pN) 

is a projection in A which we claim has the desired property. To check it, let 

τ ∈ T(A) be a trace state. Let qi = (0, 0, . . ., 0, 1, 0, . . ., 0) be the central projection 

of A corresponding to the unit in Mni . Set αi = τ(qi), and note that 

N 

αi = 1. (5.6) 

i=1 

We can define a positive trace functional on Mni 

such that 

Mni ∋ ai ↦→ τ (0, 0, . . ., 0, ai, 0, . . ., 0). 

Since the value of this trace on the unit of Mni is αi, we conclude that τ (0, 0, . . ., 0, ai, 0, . . ., 0) = 

αiωi(ai) for all ai ∈ Mni . Thus, when a = (a1, a2, . . ., aN) ∈ A, we find that 

N 

N 

τ(a) = τ (0, 0, . . .,0, ai, 0, . . ., 0) = αiωi(ai). (5.7) 

i=1 

i=1 

By using (5.5), (5.7), and (5.6) twice, this leads to the estimate 

 

 

N N 

 

 

|t − τ(p)| = αit − αiωi(pi) 

 

≤ 

N 

N 

αi |t − ωi(pi)| ≤ 

i=1 

i=1 

i=1 

i=1 

αi 

1 

m 

= 1 

m . 

Lemma 5.36. Let A be a finite dimensional C∗-algebra and B ⊆ A a unital C∗- subalgebra. Then B ′ ∩ A is a finite dimensional C∗-algebra, so B ′ ∩ A ≃ Mn1 ⊕ 

Mn2 ⊕ · · · ⊕ MnN for some natural numbers ni. Set m = mini ni. 

For every projection p ∈ B and every t ∈ [0, 1] there is a projection q ∈ A such 

that 

|tτ(p) − τ(q)| ≤ 1 

for all trace states τ ∈ T(A). 

m 

(5.8) 

Proof. By Lemma 5.35 there is a projection e ∈ B ′ ∩ A such that 

|t − θ(e)| ≤ 1 

(5.9) 

m 

for all trace states θ ∈ T(B ′ ∩ A). Set q = pe, and note that q is a projection in A. 

To check that q has the desired property, let τ ∈ T(A). If τ(p) = 0, we find that 

0 ≤ τ(q) = τ(pep) ≤ τ(p) = 0, 

i.e. τ(q) = 0. Thus (5.8) holds in this case. Assume therefore that τ(p) > 0. Then 

is a trace state of B ′ ∩ A, 2 and hence 

B ′ ∩ A ∋ x ↦→ 

 

 

 

τ (ep) 

t − 

τ(p) 

τ (xp) 

τ(p) 

≤ 1 

m . 

Since 0 < τ(p) ≤ 1, this implies (5.8). 

2 Check this; it is essential that p ∈ B.


Here is the advertized alternative proof of Lemma 5.34: Let A1 ⊆ A2 ⊆ A3 ⊆ . . . 

be a sequence of unital finite dimensional C∗-subalgebras of A whose union is dense 

in A. By Lemma 3.15 there is an i ∈ N and a projection e ∈ Ai such that e and p 

are Murray-von Neumann equivalent. It follows from Lemma 5.31 and Lemma 5.32 

that there is a j ≥ i such that A ′ i ∩ Aj ≃ Mn1 ⊕ Mn2 ⊕ · · · ⊕ MnN , where 

1 

mini ni 

< ǫ. 

It follows then from Lemma 5.36 that there is a projection q ∈ Aj such that 

|tτ(e) − τ(q)| < ǫ 

for all τ ∈ T (Aj). When ω ∈ T(A) is a trace state of A, its restriction to Aj is a 

trace state of Aj and hence we conclude that 

|tω(e) − ω(q)| < ǫ 

for every trace state ω ∈ T(A). Since ω(e) = ω(p) for every ω ∈ T(A), this proves 

Lemma 5.34. 

For any unital C ∗ -algebra A with a non-empty tracial state space T(A) we can 

define a map 

ρA : K0(A) → Aff T(A) 

by ρA(x)(ω) = rA(ω)(x), ω ∈ T(A), where rA : T(A) → SK0(A) is the restriction 

map introduced in Section 4.4. ρA is obviously a group homomorphism and, 

although it need not be injective, it does carry important information about the 

order structure of K0(A). In fact, as we shall see, it determines the order structure 

completely when A is a simple unital AF-algebra. 

Lemma 5.37. Let A be a simple unital AF-algebra which is not finite dimensional. 

Then ρA(K0(A)) is dense in Aff T(A). 

Proof. Set V = ρA(K0(A)). Then V + V ⊆ V and −V = V since K0(A) is a 

group and ρA a group homomorphism. By Lemma 5.34, αV ⊆ V when α ∈ [0, 1], so 

it follows that V is a closed vector subspace of Aff T(A). To prove that V = Aff T(A) 

it therefore suffices, by the Hahn-Banach theorem, to show that any l ∈ Aff T(A) ∗ 

which satisfies that l(V ) = {0} must be zero. From Lemma 4.9 we know that 

l = t1s1 −t2s2 where t1, t2 ∈ R + and s1, s2 are states on Aff T(A). But then there are 

trace states ωi ∈ T(A) such that si(f) = f(ωi), f ∈ Aff T(A), i = 1, 2, by Theorem 

4.3. Since l vanishes on V and 1 = ρA([1]) ∈ V , we have that l(1) = t1 − t2 = 0, i.e. 

t1 = t2. If t1 = t2 = 0, we must have that s1 − s2 annihilates V . In particular, 

0 = s1(ρA(x)) − s2(ρA(x)) = ρA(x)(ω1) − ρA(x)(ω2) = rA(ω1)(x) − rA(ω2)(x) 

for all x ∈ K0(A). It follows that rA(ω1) = rA(ω2) and hence that ω1 = ω2 by 

Corollary 4.54. Thus s1 = s2, proving that l = 0. 

Lemma 5.38. Let A be a unital AF-algebra. If x ∈ K0(A) is not positive (i.e. if 

x /∈ K0(A) + ), then there is tracial state ω ∈ T(A) such that rA(ω)(x) ≤ 0. 

Proof. Write A = 

n An where A1 ⊆ A2 ⊆ A3 ⊆ · · · is a sequence of unital 

finite-dimensional C ∗ -algebras. Since x is not positive in K0(A) there is an N ∈ N 

and elements xn ∈ K0(An)\K0(An) + such that x = ιn∗(xn), n ≥ N. By identifying 

K0(An) with Z mn for some mn ∈ N and using that xn /∈ Z mn 

+ , it is easy to construct 

a state sn on K0(An) such that sn(xn) < 0. By Corollary 4.54 (or an elementary


consideration) there is a tracial state ωn ∈ T(An) such that rAn(ωn) = sn, n ≥ N. 

Let µn be a state extension of ωn to A and let ω be a weak ∗ - condensation point of the 

sequence {µn} in S(A). As in the proof of Theorem 4.51 it follows that ω ∈ T(A). 

We have then that rA(ω)(x) = limn→∞ rAn(ωn)(xn) = limn→∞ sn(xn) ≤ 0. 

Theorem 5.39. Let A be a simple unital AF-algebra. Then 

K0(A) + = {x ∈ K0(A) : ρA(x) > 0} ∪ {0} . 

Proof. That the last set is contained in the first follows immediately form 

Lemma 5.38 and does not require simplicity of A. To prove the reversed inclusion, 

let p be a non-zero projection in Mn(A) for some n ∈ N. Note that ρA([p])(ω) = 

ωn(p), ω ∈ T(A), where ωn : Mn(A) → C is given by ωn ((aij)) = n 

i=1 ω (aii). Note 

that ωn is a positive trace functional. By Lemma 5.19 Mn(A) is simple and hence 

ωn is faithful by Lemma 5.18. Hence ωn(p) > 0 since p = 0. 

Lemma 5.37 and Theorem 5.39 are not only results on simple AF-algebras, but 

also a result on the structure of simple dimension groups. To make this clear we 

reformulate it as follows. 

Theorem 5.40. Let G be a countable simple dimension group with order unit u. 

Assume that G is not cyclic (i.e. G = Z). Then the map ϕG : G → Aff SG given by 

has dense image in Aff SG and 

ϕG(g)(s) = s(g), s ∈ SG, 

G + = {g ∈ G : ϕG(g) > 0} ∪ {0} . 

Proof. By Theorem 3.45 there is a unital AF-algebra A such that G ≃ K0(A) 

as dimension groups with order units. By Theorem 5.24 A is simple and A can not 

be finite dimensional because then K0(A) ≃ Z. By Corollary 4.54 rA : T(A) → 

SK0(A) is an affine homeomorphism, so it follows from Lemma 5.37 and Theorem 

5.39 that ϕK0(A) : K0(A) → Aff SK0(A) has dense range and K0(A) + = {x ∈ 

K0(A) : ϕK0(A)(x) > 0}∪{0}. The isomorphism G ≃ K0(A) induces an isomorphism 

Aff SG ≃ Aff SK0(A) of order unit spaces in the obvious way such that the diagram 

ϕG 

G ≃ K0(A) 

ϕ K0 (A) 

❄ ❄ 

Aff SG ≃Aff SK0(A) 

commutes. The conclusion follows from this. 

Note that the map ϕG : G → Aff SG of Theorem 5.40 can be defined for any 

partially ordered abelian group G with order unit. Theorem 5.40 tells us that this 

map actually determines the order structure on G when G is a countable simple noncyclic 

dimension group. This is actually true in greater generality; G need not be a 

dimension group to reach this conclusion, it suffices that G is unperforated. At this 

point it is not clear that such a purely algebraic assertion has any signifigance for 

C ∗ -algebras because we have not yet related arbitrary partially ordered unperforated 

groups to C ∗ -algebras. Here we continue by showing that Theorem 5.40 has a natural 

converse, namely the following.


Theorem 5.41. Let G be a torsion free abelian group and ∆ a metrisable Choquet 

simplex. Let ϕ : G → Aff ∆ be a group homomorphism such that ϕ(G) is dense in 

Aff ∆ and contains the constant function 1. Set 

G + = {g ∈ G : ϕ(g) > 0} ∪ {0} . 

When ordered by G + , G becomes a simple non-cyclic dimension group such that 

SG ≃ ∆ for any choice of order unit u in ϕ −1 (1). 

Proof. Since ϕ(G) is dense in Aff ∆, there is a g ∈ G such that 1 −ϕ(g) < 1, 

i.e. G + = {0}. Let g ∈ G + \{0}. For any h ∈ G there is an n ∈ N such that 

ϕ(h) < nϕ(g), and hence h ≤ ng. This proves that g, and hence any non-zero 

element of G + , is an order unit for G. In particular, G = G + − G + (i.e. G is 

directed). Since it is clear from the definition that G + ∩ (−G + ) = {0}, we have 

shown that G is a simple partially ordered abelian group. If g ∈ G and ng ∈ G + 

for some n ∈ N, we have that ng = 0 or nϕ(g) > 0. If ng = 0 we must have that 

g = 0 since G is torsion free and if nϕ(g) > 0 we must have that ϕ(g) > 0. Thus 

g ∈ G + , proving that G is unperforated. To check the Riesz interpolation property, 

let g1, g2, h1, h2 ∈ G + such that gi ≤ hj for all i, j ∈ {1, 2}. Then ϕ(gi) 2ǫ, i = 1, 2, ω ∈ ∆ . 

In case 1) k+ = k + ǫ is an element of Aff ∆ such that k+(ω) − gi(ω) > ǫ and 

hi(ω) − k+(ω) > ǫ for i = 1, 2, ω ∈ ∆. In case 2) k− = k − ǫ is an element of Aff ∆ 

such that k−(ω) − gi(ω) > ǫ and hi(ω) − k−(ω) > ǫ for i = 1, 2, ω ∈ ∆. By the 

density of ϕ(G) we can find d ∈ G such that k± − ϕ(d) < ǫ. Then d interpolates 

the g’s and the h’s. 

Let u ∈ ϕ −1 (1). Define Φ u : ∆ → SG by Φ u (ω)(g) = ϕ(g)(ω). It is clear that 

Φ u is an affine continuous map. It is injective because ϕ(G) is dense in Aff ∆. To 

prove that Φ u is also surjective, let s ∈ SG. If g ∈ ker ϕ, we have that 

−u ≤ ng ≤ u 

for all n ∈ N. Hence −1 ≤ ns(g) ≤ 1 for all n ∈ N, showing that s(g) = 0, i.e. 

ker ϕ ⊆ ker s. We can therefore define ˜s : ϕ(G) → R by ˜s ◦ ϕ = s. If f = ϕ(g), 

for some g ∈ G, and −m m < f < , then −mu ≤ ng ≤ mu in G, and hence 

n n 

|˜s(f)| = |s(g)| ≤ m. 

A simple continuity argument shows now that |˜s(f)| ≤ f for 

n 

all f ∈ ϕ(G). Since ˜s is a group homomorphism, the density of ϕ(G) in Aff ∆ shows 

that ˜s extends by continuity to a group homomorphism Aff ∆ → R which we also 

denote by ˜s. Being a continuous group homomorphism, ˜s must actually be linear. 

Since s is positive on G + and ϕ(G + ) is dense in Aff ∆+, we have that ˜s is positive. 

Since ˜s(1) = s(u) = 1, we have that ˜s is a state on Aff ∆. By Theorem 4.3 there is 

a ω ∈ ∆ such that ˜s(f) = f(ω), f ∈ Aff ∆. Then Φu (ω) = s.


Theorem 5.40 and Theorem 5.41 give a complete description of the (countable) 

non-cyclic simple dimension groups with order unit. Hence they provide us with 

a powerful tool for the study of simple unital AF-algebras. For example, we get 

immediately the following 

Corollary 5.42. Let ∆ be a metrizable Choquet simplex. There is then a simple 

unital AF-algebra A such that T(A) is affinely homeomorphic to ∆. 

Proof. Since ∆ is metrizable, Aff ∆ is separable (by Exercise 4.12). Let {xn} 

be a dense sequence in Aff ∆. We may assume that x1 = 1. Set 

G = spanQ{xn : n ∈ N} 

and let ϕ : G → Aff ∆ be the inclusion map. By Theorem 5.41 there is a simple 

dimension group with order unit (namely, G with order unit 1), such that SG ≃ ∆. 

By Theorem 3.45 there is an AF-algebra A such that K0(A) ≃ G. Since G is simple, 

A is simple by Theorem 5.24. Since T(A) ≃ SK0(A) by Corollary 4.54, the proof 

is complete. 

Remark 5.43. In a weak moment one might think that we could classify nonunital 

AF-algebras by determining the isomorphism class of the AF-algebras obtained 

by adding a unit, and in this way avoid all considerations regarding scales of 

non-unital AF-algebras. However, the implication A + ≃ B + ⇒ B ≃ A fails in 

general, also among AF-algebras. To give an example of this consider the following 

two subsets of the real line : 

X = { 1 

: n = 2, 3, 4, · · · } ∪ {0} , 

n 

and 

Y = { 1 

: n = 1, 2, 3, · · · } . 

n 

Both spaces are locally compact in the relative topology inherited from R, in 

fact X is compact. The abelian C∗-algebras A = C(X) and B = C0(Y ) are both 

AF-algebras. The reader is invited to write down a Bratteli diagram for each. But 

they are certainly not isomorphic since one is unital and the other is not. However, 

: n = 1, 2, 3, · · · } ∪ {0}. 

A + and B + are both isomorphic to C(Z) where Z = { 1 

n 

Example 5.44. The purpose of this example is to give a procedure to produce 

examples of partially ordered abelian groups which are directed and unperforated, 

but do no have the Riesz interpolation property, and hence are not dimension groups. 

Let H be a any non-zero partially ordered torsionfree abelian group which is directed 

and unperforated. Set G = Z ⊕ H with positive semi-group G + = {(z, h) : z > 

0, h ∈ H + \{0}} ∪ {(0, 0)}. We leave the reader to check that G is directed and 

unperforated. To see that G does not have the Riesz interpolation property, let 

h1, h2 ∈ H + such that h1 < h2 (by which we mean that h1 ≤ h2 and h1 = h2). For 

example we could take any non-zero element h1 of H + and choose h2 = 2h1. Set 

x1 = (0, 0), x2 = (0, h1) and y1 = (1, h2), y2 = (2, h2). Then xi ≤ yj in G for all 

i, j. Assume z = (a, b) is some element in G such that xi ≤ z ≤ yj for all i, j. It is 

easy to check that z can not be any of the elements x1, x2, y1 or y2. Therefore, since 

z − x1 ∈ G + , we must have that a > 0. On the other hand, since y1 − z ∈ G + , we 

must also have that a < 1. Since a ∈ Z, this is impossible.


Exercise 5.45. Let A be a C ∗ -algebra and set 

I = {I : I is an ideal in A} . 

I is ordered by inclusion, i.e. I1 ≤ I2 ⇔ I1 ⊆ I2. Show that I is a lattice and that 

for I1, I2 ∈ I. 

I1 ∧ I2 = span I1I2 = I1I2 = I1 ∩ I2, 

I1 ∨ I2 = I1 + I2 = I1 + I2 , 

Exercise 5.46. Let A be a C ∗ -algebra and x ∈ A an arbitrary element of A. 

Show that xAx ∗ is a C ∗ -subalgebra of A. Show next that 

A is simple ⇒ xAx ∗ is simple. 

5.5. On the automorphism group. In this section we shall see how the classification 

result (and the methods leading to it) can be used to get some insight into 

the structure of the automorphism group of AF-algebras. For any C ∗ -algebra A we 

let Aut A denote the set of ∗-automorphism of A, i.e. an element α ∈ Aut A is an 

injective and surjective ∗-homomorphism α : A → A. Aut A is a group under composition; 

when α, β ∈ Aut A, α ◦ β is also an automorphism. Furthermore, AutA is 

a topological group. The norm on the bounded operators on A, considered just as 

a Banach space, gives rise to a metric D on Aut A; viz. 

D(α, β) = sup{α(a) − β(a) : a ≤ 1 } . 

It is not difficult to show (just do it !) that the Aut A is complete with respect to 

this metric. The corresponding topology is called the uniform topology. 

However, there is another topology which is more important, namely the one 

with a basis consisting of all the sets 

Vβ,a,ǫ = {α ∈ Aut A : α(a) − β(a) < ǫ} , 

where β varies over all elements of Aut A, a over all elements of A and ǫ over R + . 

It is straightforward to see that a net {αt} in AutA converges to α ∈ Aut A in this 

topology if and only if limt αt(a) = α(a) for all a ∈ A. It is easy to see that Aut A 

is a topological group in this topology, and unless something else is explicitly stated, 

we will consider Aut A equipped with this topology which we shall refer to as the 

topology of point-wise normconvergence. In case the C∗-algebra A is separable, which 

is the case we shall mostly be concerned with, the topology is actually metrizable, 

i.e. comes from a metric. Indeed, when A is separable we can select a sequence {an} 

which is dense in the unit ball of A, and define a metric d on Aut A by 

∞ 

d(α, β) = 2 −n α(an) − β(an) + α −1 (an) − β −1 (an) . 

n=1 

We leave the reader to check that the topology defined by the metric d is indeed the 

topology of pointwise norm-convergence. 

Exercise 5.47. Prove that the metric d is complete. 

Example 5.48. Let A be the C ∗ -algebra of sequences, {an}, in C (or in another 

C ∗ -algebra if you want) such that limn→∞ an = = 0. Define, for each m ∈ N, an 

automorphism βm of A by 

βm({an}) = (am, a1, a2, a3, a4, · · · , am−1, am+1, am+2, · · · · · ·) .


Then limm→∞ βm(a) = λ(a) for all a = {an} ∈ A, where λ : A → A is the 

endomorphism given by 

λ({an}) = (0, a1, a2, a3, a4, · · · · · ·) . 

Note that λ is certainly not an automorphism (it not surjective!). This example 

shows that Aut A is not complete with respect to a metric d0 given by d0(α, β) = 

∞ 

n=1 2−n α(xn) − β(xn) for some dense sequence {xn} in the unit-ball of A. This 

is why we introduce the additional terms involving the inverses in the definition of d 

above. Note however that d and d0 determine the same topology on Aut A, namely 

the topology of pointwise normconvergence. 

An automorphism α ∈ AutA is called inner when there is a unitary u ∈ A + 

such that α(a) = uau ∗ for all a ∈ A. Of course, when A is unital this happens if 

and only if we can find such a unitary u in A. For any unital algebra B, we let 

U(B) denote the group of unitaries in B, i.e. U(B) = {u ∈ B : uu ∗ = u ∗ u = 1}. 

There is a group homomorphism U(A + ) → Aut A which sends u ∈ U(A + ) to the 

automorphism given by a ↦→ uau ∗ and the inner automorphisms are those which lie 

in the image of this map. We denote the group of inner automorphisms by Inn(A). 

The closure of Inn(A) will be denoted by Inn(A), and an automorphism in Inn(A) 

will be called approximately inner. 

Lemma 5.49. Inn(A) ⊆ {α ∈ Aut A : α∗ = idK0(A) on K0(A)}. 

Proof. It is obvious that the right-hand side defines a subgroup of Aut A, and 

since ueu ∗ ≈ e whenever e is a projection and u a unitary, it is clear that it contains 

Inn(A). We assert that its a closed subgroup. So let {αt} be net in AutA converging 

to α ∈ Aut A, and assume that αt∗ = idK0(A) for all t. Let e, f ∈ Mn(A + ) be 

projections such that [e] − [f] ∈ K0(A). For each automorphism β ∈ Aut A there 

is a unique automorphism β + ∈ Aut A + extending β. In the usual way we may 

furthermore extend β + to an automorphism β + n of Mn(A + ). Onserve that 

lim t αt + n (a) = α+ n (a) 

for all a ∈ Mn(A + ). In particular there are is a t such that αt + n (e) − α+ n (e) < 

1 

3 , αt + n (f) − α+ 1 

n (f) < 3 . But then, by Lemma 3.14, α∗([e] − [f]) = [α + n (e)] − 

[α + n (f)] = [αt + n (e)] − [αt + n (f)] = αt∗([e] − [f]) = [e] − [f] in K00(A + ). It follows 

that {α ∈ Aut A : α∗ = idK0(A) on K0(A)} is a closed subgroup containing 

Inn(A). Consequently it also contains Inn(A). 

Set 

AutΣ(K0(A)) = {γ ∈ Aut K0(A) : γ(Σ(A)) = Σ(A) } . 

AutΣ(K0(A)) is a subgroup of Aut K0(A) and it is clear that the map Aut A ∋ α ↦→ 

α∗ ∈ Aut K0(A) maps into AutΣ(K0(A)). 

Theorem 5.50. Let A be an AF-algebra. Then the following is a short exact 

sequence : 

0 

⊆ α→α∗ 

Inn(A) 

Aut A 

AutΣ(K0(A)) 

Proof. Let first γ ∈ AutΣ(K0(A)). By Theorem 3.35 3) there is an automorphism 

α ∈ Aut A such that α∗ = γ. This shows that the map α ↦→ α∗ is 

surjective. Let next α ∈ Aut A be an automorphism such that α∗ = id on K0(A). 

 

0


There is then, by Theorem 3.35 2) a sequence of unitaries, {un}, in A + such that 

α(a) = limn→∞ unau∗ n for all a ∈ A. But this means that α ∈ Inn(A). This proves 

that the kernel of the map α ↦→ α∗ is contained in Inn(A). By Lemma 5.49 we 

therefore have that this kernel is exactly Inn(A). 

We are going to give an alternative description of the subgroup Inn(A) of Aut A. 

To do this we need some lemmas. 

Lemma 5.51. Let F be a finite dimensional C ∗ -subalgebra of A + and set 

F ′ ∩ A = {x ∈ A : xf = fx , f ∈ F } . 

There is then a surjective positive linear map P : A → F ′ ∩ A such that P(x) = 

x, x ∈ F ′ ∩ A, and P ≤ 2. 

Proof. Let {e d ij : i, j = 1, 2, · · · , nd, d = 1, 2, · · · , N} be a full set of matrix 

units in F and set p = 

d,i ed ii which is a projection in A. Define P : A+ → A + by 

P(a) = (1 − p)a(1 − p) + 

N 

nd 

1 

e 

nd 

d=1 i,j=1 

d ijaedji . (5.10) 

Since ed ∗ d 

ji = eij we see that a ≥ 0 ⇒ P(a) ≥ 0 so that P is a positive linear map. 

For each m, k, l we find that 

and 

N 

e m klP(a) = em nd 

1 

kl e 

nd 

d=1 i,j=1 

d ijaedji P(A)e m kl = 

N 

nd 

1 

e 

nd 

d=1 i,j=1 

d ijaedji emkl = 1 

= 1 

nm 

nm 

nm 

j=1 

nm 

j=1 

e m kj aem jl 

e m kj aem jl . 

Hence P(a) ∈ F ′ ∩ A + for all a ∈ A + . If x ∈ F ′ ∩ A + we find that 

P(x) = (1 − p)x(1 − p) + 

= (1 − p)x + 

d,i 

N 

nd 

1 

e 

nd 

d=1 i,j=1 

d ijxedji e d iix = (1 − p)x + px = x . 

= (1 − p)x + 

N 

nd 

1 

e 

nd 

d=1 i,j=1 

d ijedji x 

In particular, P maps A onto F ′ ∩A. Let a = a ∗ ∈ A with a ≤ 1. Since −1 ≤ a ≤ 1 

in A + , the positivity of P and the fact that P(1) = 1 implies that −1 ≤ P(a) ≤ 1. 

Hence P(a) ≤ 1. If a ∈ A is a general element with a ≤ 1 we write a = a1 +ia2 

where ai = a ∗ i and ai ≤ a ≤ 1, i = 1, 2. It follows that 

P(a) = P(a1) + iP(a2) ≤ P(a1) + P(a2) ≤ 2 , 

proving that P ≤ 2. 

Lemma 5.52. Let A be an AF-algebra and F ⊆ A a finite dimensional C ∗ - 

subalgebra of A. There is then an increasing sequence A1 ⊆ A2 ⊆ · · · of finite 

dimensional C ∗ -subalgebras of A such that A1 = F and A = ∞ 

n=1 An.


Proof. Let B1 ⊆ B2 ⊆ B3 ⊆ · · · be an increasing sequence of finite dimensional 

C∗-subalgebras of A such that A = ∞ n=1 Bn. Let ι : F → A and in : Bn → A denote 

the inclusion maps. By Lemma 3.34 1) there is an m ∈ N and a ∗-homomorphism 

ψ : F → Bm such that ι∗ = im ∗ ◦ ψ∗. By Lemma 3.34 2) there is a unitary u ∈ A + 

such that ι = Adu ◦ im ◦ ψ. It follows that F ⊆ uBmu∗ . Set A1 = F and An = 

uBm+n−2u∗ , n ≥ 2. It straightforward to check that the sequence A1 ⊆ A2 ⊆ · · · 

has the stated properties. 

Lemma 5.53. Let A be an AF-algebra and F ⊆ A a finite dimensional C ∗ - 

subalgebra. It follows that 

is an AF-algebra. 

F ′ ∩ A = {x ∈ A : xf = fx , f ∈ F } 

Proof. We pick an increasing sequence A1 ⊆ A2 ⊆ · · · of finite dimensional 

C∗-subalgebras in A such that F = A1 and A = ∞ n=1 An, cf. Lemma 5.52. Note 

that the map P : A → F ′ ∩ A defined in Lemma 5.51 maps An onto F ′ ∩ An for all 

n. We assert that 

F ′ ∩ A = 

F ′ ∩ An . (5.11) 

n 

This will of course complete the proof. Let x ∈ F ′ ∩A and choose xn ∈ An such that 

limn→∞ xn = x. Then P(x) = limn→∞ P(xn) since P is continuous, cf. Lemma 

5.51. But P(xn) ∈ F ′ ∩ An for all n, so this proves (5.11). 

Lemma 5.54. Let u, v be unitaries in a C ∗ -algebra A such that u −v ≤ µ < 2. 

There is then a continuous path γ : [0, 1] → U(A) such that γ(0) = u, γ(1) = v and 

γ(t) − v ≤ µ, t ∈ [0, 1]. 

Proof. Since uv ∗ − 1 = u − v ≤ µ < 2, the spectrum of uv ∗ is a subset of 

T ∩ {λ ∈ C : |λ − 1| ≤ µ} ⊆ T\{−1}. Let ϕ : T\{−1} →] − 1, 1[ be the continuous 

function such that ϕ(e πit ) = t, t ∈]−1, 1[. Then ϕ(uv ∗ ) = a is a self-adjoint element 

of A with spectrum in 

ϕ({λ ∈ T : |λ − 1| ≤ µ}) 

such that e 2πia = uv ∗ . Define γ(t) = e πita v, t ∈ [0, 1]. Then γ is a continuous 

path of unitaries connecting u to v. Since the spectrum of e 2πita is contained in 

{λ ∈ T : |λ − 1| ≤ µ} for all t, we find that γ(t) − v = e 2πita − 1 ≤ µ for all 

t ∈ [0, 1]. 

Proposition 5.55. Let A be a unital AF-algebra. Then the unitary group U(A) 

is pathconnected. 

Proof. First observe that the unitary group of Mn is pathconnected. Indeed, if 

u is a unitary n × n-matrix there is a unitary w ∈ Mn such that wuw ∗ is a diagonal 

unitary, i.e. is of the form 

diag(λ1, λ2, · · · , λn) 

for some λi ∈ T. Since T is pathconnected there is a path of diagonal unitaries 

connecting this unitary to 1 ∈ Mn. Hence wuw ∗ can be connected to 1 via a 

continuous path γ of unitaries. But then w ∗ γw will be a continuous path of unitaries 

connecting u to 1. It follows straightforwardly that the unitary group of any finite 

dimensional C ∗ -algebra is pathconnected.


Let now 1 ∈ A1 ⊆ A2 ⊆ · · · be a sequence of finite dimensional C∗-subalgebras of A such that A = 

n An. Let u ∈ U(A). For any ǫ we can find n ∈ N and x ∈ An 

such that u − x ≤ ǫ. If just ǫ is small enough, x∗x − 1 = x∗x − u∗u < 1 so 

that w = x(x∗x) −1 

2 is defined. w is then a unitary in An and if ǫ is small enough 

we have that u − w < 2. By Lemma 5.54 there is a continuous path of unitaries 

in U(A) connecting u to w. Since the unitary group of An is pathconnected we also 

have a path of unitaries (in An) connecting w to 1. Putting the two paths together 

we get a continuous path of unitaries in A connecting u to 1. 

Theorem 5.56. Let A be an AF-algebra. Then Inn(A) is the pathconnected 

component of AutA containing idA, i.e. Inn(A) consists of the automorphisms α 

for which there is a continuous path γ : [0, 1] → Aut A such that γ(0) = α and 

γ(1) = idA. 

Proof. Assume first that that there is such a path γ connecting α ∈ Aut A to 

idA. Then, when e is a projection in Mn(A), we get a continuous path, pt = γ(t)(p), 

of projections in Mn(A) connecting α(p) to p. By uniform continuity there is a 

n ∈ N such that pt − ps < 1 

2 

when |t − s| < . By using Lemma 3.14 we conclude 

3 n 

that 

[α(p)] = [p 1 

n 

] = [p 2 

n 

] = [p 3 

n 

] = · · · = [p n−1] 

= [p] 

n 

in K0(A). It follows that α∗ = idK0(A) and hence α ∈ Inn(A) by Theorem 5.50. 

Conversely, let α ∈ Inn(A) and choose an increasing sequence A1 ⊆ A2 ⊆ · · · of 

finite dimensional C∗- algebras with dense union in A. Since α∗ = idK0(A) there is 

for each n a unitary un ∈ U(A + ) such that α|An = Ad un. This follows from Lemma 

3.34 2). We will construct, for each n, a continuous path of unitaries wn t , t ∈ 

[ 1 1 , n+1 n ], in U(A+ ) such that w1 1 = 1, w11 = u 

2 

∗ 1 , wn1 = u 

n 

∗ n−1 , wn1 = u 

n+1 

∗ n , n ≥ 2, and 

Ad wn t ◦ α|An−1 = idAn−1 for all t ∈ [ 1 1 , ]. We do this by induction. Since the 

n+1 n 

argument which starts the induction is the same as the that used for the induction 

step, we only give the latter. So assume that we have constructed wn t , t ∈ [ 1 1 , n+1 n ]. 

Since unau∗ n = un+1au∗ n+1 (= α(a)), for all a ∈ An, we see that u∗ n+1un ∈ A ′ n ∩ A + . 

By combining Lemma 5.53 and Proposition 5.55 we see that the unitary group of 

A ′ n ∩ A+ is pathconnected. So there is a continuous path, vt, t ∈ [ 1 

n+2 , 1 ], of n+1 

= vtu∗ n. It is 

unitaries in A ′ n ∩A + such that v 1 

n+1 

= 1 and v 1 

n+2 

= u ∗ n+1un. Set w n+1 

t 

straigthforward to check that wn+1 has the required properties. Now define a path 

γ : [0, 1] → Aut A of automorphisms by γ(t) = Ad wn 1 1 

t ◦α, t ∈ [ , ], n ∈ N, and 

n+1 n 

γ(0) = idA. Then γ is a continuous path of automorphisms connecting α to idA. 

Thus, for an AF-algebra A, the group Inn(A) is the pathcomponent of Aut A 

containing the identity automorphism. As the next lemma shows, this is actually 

the same as the connected component containing idA. 

Lemma 5.57. Let A be an AF-algebra. The connected component of AutA containing 

idA is the same as the pathconnected component of Aut A containing idA and 

equals Inn(A). 

Proof. For every projection p in A, set 

Autp(A) = {α ∈ AutA : α(p) ≈ p}.


It follows from Lemma 3.27 that {α ∈ Aut A : α∗ = idK0(A)} = 

p Autp(A). 

Note that Autp(A) is a subgroup of AutA : If α(p) ≈v p and β(p) ≈w p, then 

β −1 ◦ α(p) ≈ β −1 (v) β −1 (p) ≈ β −1 (w ∗ ) p. Let α ∈ Autp(A). By Lemma 3.14 the open 

set 

{β ∈ Aut A : α(p) − β(p) < 1 

3 } 

is contained in Autp(A) so we see that Autp(A) is an open subgroup of Autp(A). 

But then Autp(A) is automatically also closed; indeed we can choose representatives, 

say {gi}, for the right Autp(A)-cosets in AutA so that Aut A is the disjoint union 

 

i gi Autp(A). If g0 Autp(A) is the coset containing idA, we have that the comple- 

ment of Autp(A) is 

i=0 gi Autp(A), an open set. Thus Autp(A) is both closed and 

open. Therefore it contains the connected component, Aut 0 A, of Aut A containing 

idA. It follows that Aut 0 A ⊆ 

p Autp(A) = {α ∈ Aut A : α∗ = idK0(A)} = 

Inn(A). The pathconnected component of Aut A containing idA, call it Aut0 A, is 

of course contained in Aut 0 A so by using Theorem 5.50 and Theorem 5.56 we find 

that 

Inn(A) = Aut0 A ⊆ Aut 0 A ⊆ Inn(A) . 

Let us consider some examples. Let’s begin with a matrix algbra A = Mn. 

Lemma 5.58. Every automorphism of Mn is inner, i.e. an arbitrary automorphism 

α ∈ Aut Mn is of the form 

α(x) = uxu ∗ , x ∈ Mn . 

Proof. Let {eij : i, j = 1, 2, · · · , n} be the usual matrix units in Mn. Note that 

α(eii), i = 1, 2, · · · , n, are projections such that α(e11) ≈ α(e22) ≈ · · · ≈ α(enn) 

and 

i α(eii) = 1. It follows that Tr(eii) = 1 for all i. In particular, we find that 

n 

α(e11) ≈ e11. Let v ∈ Mn be a partial isometry such that vv∗ = α(e11), v∗v = e11. 

It is straightforward to check that u = 

i α(ei1)ve1i is a unitary in Mn such that 

ueklu ∗ = α(ekl) for all k, l, proving that α = Ad u. 

It follows from this lemma that the homomorphism Un ∋ u ↦→ Ad u is a surjection. 

It is easy to see that the kernel of this homomorphism is λ1, λ ∈ T, so 

that 

Aut Mn ≃ Un/T . 

On AutMn the uniform topology agrees with the topology of pointwise normconvergence, 

and the above isomorphism is a homeomorphism when Un/T is equipped 

the natural topology. Note that the AutΣ(K0(Mn)) consists of the automorphisms of 

Z which fixes n, i.e. AutΣ(K0(Mn)) is trivial. So for this rather trivial C ∗ -algebra the 

short exact sequence of Theorem 5.50 is degenerate; the quotient is trivial. Consider 

next a finite-dimensional C ∗ -algebra 

F = Mn1 ⊕ Mn2 ⊕ · · · ⊕ MnN . 

Let α ∈ Aut F. To see how α acts on F, let ZF denote the center of F; i.e. 

ZF = {x ∈ F : xy = yx , y ∈ F }. It is straightforward to see that ZF ≃ C N . 

Indeed, set 

pi = (0, 0, · · · , 0, 1, 0, · · · , 0)


where the 1 occurs on the i’th entry. Then pi ∈ ZF, i = 1, 2, · · · , N, and every 

element in ZF is a linear combination of the pi’s. Note that each pi is a minimal 

non-zero projection in ZF. Since also α(pi) must be a minimal non-zero projection 

in ZF, we find that there is a permuation σ of {1, 2, 3, · · · , N} such that 

α(ei) = eσ(i) , i = 1, 2, · · · , N . 

Note that α(eiF) = eσ(i)F for each i. Since eiF = Mni and eσ(i)F = Mn σ(i) we see 

that 

nσ(i) = ni 

for all i = 1, 2, · · · , N. Thus σ is not an arbitrary permutation; indeed, when we 

partition {1, 2, · · ·n} = m 

j=1 Fj such that ni = nk ⇔ i, k ∈ Fj we have that 

σ(Fj) = Fj , j = 1, 2, · · · , m. (5.12) 

Conversely, when σ is a permutation of {1, 2, · · · , N} such that (5.12) holds, we can 

define an automorphism β ∈ Aut F by 

β(a1, a2, · · · , aN) = (aσ(1), aσ(2), · · · , aσ(N)) , (a1, a2, · · · , aN) ∈ F . 

Then β(ei) = eσ(i) for all i. In other words, what we have shown is that there is 

a homomorphism Aut F → {σ ∈ ΣN : σ(Fj) = Fj , j = 1, 2, · · · , m} and this 

homomorphism is not only surjective; it is split surjective. 

Lemma 5.59. α ∈ Aut F is in Inn(F) = Inn(F) if and only α acts trivially on 

ZF. 

Proof. An inner automorphism of F must obviously act trivially on ZF, and by 

continuity the same must be true for an automorphism in Inn(F). If α acts trivially 

on ZF we find that α(ei) = ei and hence α(eiF) = eiF for all i. But eiF = Mni 

and since every automorphism of Mni is inner by Lemma 5.58, there is a unitary 

for each i such that 

ui ∈ Mni 

α(a1, a2, · · · , aN) = (u1a1u ∗ 1 , u2a2u ∗ 2 , · · · , uNaNu ∗ N ) 

for all (a1, a2, · · · , aN) ∈ F. But then u = (u1, u2, · · · , uN) is a unitary in F such 

that α(x) = uxu ∗ . 

So for a finite dimensional C ∗ -algebra F as above, the short exact sequence of 

Theorem 5.50 becomes 

0 

 

InnF 

 

Aut F 

where the quotient map 

 

{σ ∈ ΣN : σ(Fj) = Fj , j = 1, 2, · · · , m} 

0 , 

Aut F → {σ ∈ ΣN : σ(Fj) = Fj , j = 1, 2, · · · , m} 

is split. Note that {σ ∈ ΣN : σ(Fj) = Fj , j = 1, 2, · · · , m} ≃ Σ#F1 × Σ#F2 × 

· · · × Σ#Fm. 

Lemma 5.60. Let H be a dense subgroup of R and set H + = H ∩ R + . Let 

A be a unital AF-algebra such that (K0(A), K0(A) + ) ≃ (H, H + ). Then every 

automorphism of A is approximately inner.


Proof. According to Theorem 5.50 the assertion is equivalent the assertion that 

any scale preserving automorphism of K0(A) is trivial. By Lemma 3.36 an automorphism 

γ of K0(A) is scale preserving if and only if γ([1A]) = [1A]. Using the 

isomorphism (K0(A), K0(A) + ) ≃ (H, H + ) we can consider the γ as an automorphism 

of H instead. By Lemma 5.71 it must have the form γ(x) = tx for some 

t > 0. But since γ fixes a non-zero element; namely the image of [1A], we find that 

t = 1. 

Example 5.61. It should be noted that it crucial that we deal with a unital AFalgebra 

in Lemma 5.60. To see this, let H = Q, ordered in the usual way, and let 

Σ = Q + . By Theorem 3.47 there is an AF-algebra A with (K0(A), K0(A) + , Σ(A)) ≃ 

(Q, Q + , Q + ). Every non-zero element q ∈ Q + defines a scale-preserving automorphism 

of Q by s ↦→ sq, and it is straightforward to use Lemma 5.71 to conclude that 

every scale-preserving automorphism of Q arises this way. So we see that 

AutΣ(K0(A)) ≃ {q ∈ Q : q > 0} . 

The group composition on the right-hand side is given by multiplcation. So A admits 

certainly a large class of automorphisms which are not approximately inner. 

Example 5.62. Let A be a UHF-algebra, i.e. A is the inductive limit of a 

sequence 

Mnk 

ϕk 

−−−→ Mnk+1 , 

where nk|nk+1 and ϕk(a) = diag(a, a, · · · , a). Then K0(A) is isomorphic, as a partially 

ordered abelian group, to the inductive limit of the sequence 

Z z↦→ nk+1 z nk Define µk : Z → R by µk(z) = z 

nk 

−−−−−→ Z. 

and note that 

µk 

Z z↦→ nk+1nk z 

Z 

µk+1 

 

 

R 

commutes. It follows that the µk’s define a homomorphism from K0(A) onto 

HA = { z 

: z ∈ Z, k ∈ N} . 

nk 

This map is easily seen to be an isomorphism of partially ordered groups when HA is 

ordered by HA ∩ R + . Note that HA is dense in R if and only if limk→∞ nk = ∞, i.e. 

if and only if A is not finite dimensional. So we see from Lemma 5.58 and Lemma 

5.60 that every automorphism of a UHF-algebra is approximately inner. 

Example 5.63. Let Aα be the unital AF-algebra with (K0(Aα), K0(Aα) + , [1]) ≃ 

(Z + αZ, (Z + αZ) ∩ R + , 1), cf. Section 0.2. Then every automorphism of Aα is 

approximately inner. 

Example 5.64. Let A be an abelian C ∗ -algebra. Let A + is also abelian and 

hence any inner automorphism of A is trivial. Thus also Inn(A) is trivial. So if in 

addition A is an AF-algebra, we see that the short exact sequence of Theorem 5.50 

is degenerate in a new way; the quotient map is an isomorphism for abelian AFalgebras. 

This does not mean that Aut A is a small group for such an AF-algebra;


only that automorphisms of A are determined by how they act on K0(A). In fact, 

Aut A can easily be enormous in this case. Consider for example the AF-algebra 

C(X) where X = {0} ∪ { 1 

n : n ∈ N}. For any n ∈ N, the symmetric group Σn acts 

by homeomorphisms of X via 

σ( 1 

) = 

k 

 

1 

k 

, k ≥ n + 1, 

, 

, k ∈ {1, 2, · · · , n} 

1 

σ(k) 

σ ∈ Σn. It follows that Aut C(X) contains a copy of Σn for any n, and hence a copy 

of any finite group. But in fact, more is true. Let G be a countable infinite group. 

Then C(X) is isomorphic to 

{f : G → C : ∀ ǫ > 0, ∃ F ⊆ G , λ ∈ C such that #F < ∞, |f(g)−λ| < ǫ , g /∈ F } . 

For each g ∈ G define βg ∈ Aut C(X) by 

βg(f)(h) = f(g −1 h) , h ∈ G . 

Then g ↦→ βg is an injective group homomorphism. So we see that Aut C(X) contains 

a copy of every countable group G. 

5.6. Examples. 

5.6.1. A few facts about continued fractions. Let a0, a1, a2, · · · be a sequence of 

real numbers such that ai = 0, i ≥ 1. For each n, set 

[a0, a1, a2, · · · , an] = a0 + 

a1 + 

a2 + 

i.e. [a0] = a0, [a0, a1] = a0+ 1 

a1 , [a0, a1, a2] = a0+ 1 

a1+ 1 

a 2 

1 

1 

1 

a3 + · · · + 1 

an 

, 

, [a0, a1, a2, a3] = a0+ 1 

a1+ 1 

a 2 + 1 

a 3 

and so on. Consider the case where an ∈ N for all n ∈ N, i.e. the an’s is a sequence 

of natural numbers with ai > 0, i ≥ 1. 

For each n write 

[a0, a1, · · · , an] = pn 

qn 

where pn, qn ∈ N are mutually prime. 

Then 

and 

Lemma 5.65. Write [a1, a2, · · · , an] = p 

q 

pn = a0p + q 

qn = p . 

where p, q ∈ N are mutually prime. 

Proof. We have that [a0, a1, · · · , an] = a0 + [a1, a2, · · · , an] −1 so that pn 

qn = 

a0+ q a0p+q 

= . The conclusion follows because a0p+q and p are mutually prime. 

p p 

and 

Lemma 5.66. 

for k ≥ 2. 

pk = akpk−1 + pk−2 , 

qk = akqk−1 + qk−2 ,


Proof. We shall prove this by induction in k. It is straightforward to check that 

p0 = a0, q0 = 1, p1 = a0a1 + 1, q1 = a1, p2 = a0a1a2 + a0 + a2 and q2 = a1a2 + 1, so 

that the lemma is certainly true for k = 2. Now assume that the lemma holds for 

k < n, regardless of which sequence a0, a1, a2, · · · we are given. By Lemma 5.65 we 

have that pn = a0p + q, qn = p where p, q ∈ N are mutually prime and satisfies that 

p 

q = [a1, a2, · · · , an] . 

By using the induction hypothesis we have that 

p = anu1 + u2 , q = anv1 + v2 

where u1, v1 are mutually prime, u2, v2 are mutually prime and 

and 

u1 

v1 

u2 

v2 

So by using Lemma 5.65 we find that 

and 

= [a1, a2, · · · , an−1] , 

= [a1, a2, · · · , an−2] . 

pn = a0(anu1 + u2) + anv1 + v2 = an(a0u1 + v1) + a0u2 + v2 = anpn−1 + pn−2 

qn = anu1 + u2 = anqn−1 + qn−2 . 

This completes the induction step and hence the proof. 

Note that the conclusion of Lemma 5.66 can be written as the matrix identity 

 

pn qn an 1 pn−1 qn−1 

= 

, (5.13) 

pn−1 qn−1 1 0 pn−2 qn−2 

 

p1 q1 

valid for n ≥ 2. Since det = 1 this shows that 

or 


pn 

qn 

p0 q0 

pnqn−1 − pn−1qn = (−1) n−1 , n ≥ 1 . (5.14) 

pn 

qn 

− pn−2 

qn−2 

− pn−1 

qn−1 

= (−1)n−1 

qnqn−1 

= (−1)n−1 

qnqn−1 

, n ≥ 1 . (5.15) 

+ (−1)n 

qn−1qn−2 

, n ≥ 1 . 

Since the qn’s are increasing (this follows from Lemma 5.66) we find that 

when n is even and 

pn 

qn 

pn 

when n is odd. Hence we get the relations 

for all n ≥ 1. 

p2n−2 

q2n−2 

qn 

< p2n 

q2n 

> pn−2 

qn−2 

< pn−2 

qn−2 

< p2n+1 

q2n+1 

< p2n−1 

q2n−1

Thus we see immediately that 


p2n 

lim 

n→∞ q2n 

p2n−1 

≤ lim 

n→∞ q2n−1 

. (5.16) 

However, it follows from the recursive formulas from Lemma 5.66 that 

k 

[ 

qk ≥ 2 2 ] , k ≥ 1, 

to that limk→∞ qk = ∞. So from (5.15) we conclude that limn→∞ pn 

Therefore the two limits from (5.16) agree and we have that the limit 

qn 

− pn−1 

qn−1 

pn 

lim 

n→∞ qn 

exists. We denote this limit by [a0, a1, a2, a3, · · ·], or if the space permits, by 

a0 + 

a1 + 

a2 + 

Exercise 5.67. Prove that limn→∞ pn 

qn 

1 

1 

a3 + 

1 

1 

a4 + · · · 

is irrational. 

. 

= 0. 

Theorem 5.68. Let t ∈ R + \Q. There is then a unique sequence a0, a1, a2, · · · 

in N such that ai > 0, i ≥ 1, and 

t = [a0, a1, a2, a3, · · ·] = a0 + 

a1 + 

a2 + 

1 

1 

a3 + 

1 

1 

a4 + · · · 

Proof. Set a0 = [t] and define rn ∈ R + recursively by r1 = (t − a0) −1 and 

rn+1 = (rn − [rn]) −1 , n ≥ 1. This is possible because t is irrational. Set an = [rn] 

for n ≥ 1. Then 

1 

t = a0 + 1 

= a0 + 

for all n. Note that 

r1 

when n is even and that 

a1 + 

= a0 + 

a2 + 

1 

a1 + 1 

r2 

1 

1 

a3 + · · · + 1 

= · · · · · · 

rn 

= [a0, a1, · · · , an, rn+1] 

t = [a0, a1, · · · , an, rn+1] ≥ [a0, a1, · · · , an] 

t = [a0, a1, · · · , an, rn+1] ≤ [a0, a1, · · · , an] 

when n is odd. From the first inequality we find that 

t ≥ [a0, a1, a2 · · ·] , 

.


and from the second we find that 

t ≤ [a0, a1, a2 · · ·] . 

This proves the existence part of the statement. To prove uniqueness, let a ′ 0, a ′ 1, a ′ 2, · · · 

= [t] = a0. 

be another sequence in N\{0} such that t = [a ′ 0 , a′ 1 , a′ 2 , · · ·]. Then a′ 0 

Hence [a1, a2, a3, · · ·] = [a ′ 1 , a′ 2 , a′ 3 , · · ·] and consequently, for the same reason, a1 = a ′ 1 

must be the integer part of this common value. Consequently, [a2, a3, a4, · · ·] = 

[a ′ 2, a ′ 3, a ′ 4, · · ·] and thus also a2 = a ′ 2. And so on. We leave it to the reader to 

formalize an induction argument which gives that an = a ′ n for all n. 

Let α ∈ R\Q. We can then define a homomorphism ϕα : Z 2 → R by 

ϕα(z1, z2) = z1 + αz2 . 

Then ϕα is injective because α is irrational. We set G = Z 2 , G + = {(z1, z2) : 

z1 + αz2 ≥ 0}, Gα = ϕα(G) ⊆ R and G + α = {t ∈ ϕα(G) : t ≥ 0}. With these 

definitions ϕα : G → Gα becomes an isomorphism of partially ordered groups. Note 

that Gα is unperforated and totally ordered because R has these properties. Thus 

Gα (and hence also G) is a dimension group. Each Gα contains 1 as an order unit. 

Lemma 5.69. For each α ′ ∈ R\Q there is an element α ∈]0, 1[\Q 

such that 

2 

(Gα ′, G+ 

α ′, 1) ≃ (Gα, G + α , 1) 

as partially ordered groups with order unit (i.e. there is a group isomorphism ψ : 

and ψ(1) = 1. 

Gα ′ → Gα such that ψ(G + 

α ′) = G + α 

Proof. Let β ∈ R\Q. Then (z1, z2) → (z1, −z2) defines an automorphism γ of 

G = Z 2 such that γ(1, 0) = (1, 0) and ϕ−β ◦γ = ϕβ. Hence ϕ−β ◦γ ◦ϕ −1 

β : Gβ → G−β 

is an isomorphism of partially ordered groups with order unit. In short Gβ ≃ G−β 

as partially ordered groups. Let now k ∈ Z. Define γ : Z 2 → Z 2 by γ(z1, z2) = 

(z1 + kz2, z2). Then γ is an automorphism of Z 2 such that ϕβ ◦ γ = ϕβ+k. So as 

above we conclude that Gβ ≃ Gβ+k. These two conclusions imply the lemma. 

Lemma 5.70. Let H be a subgroup of R which is not dense in R. It follows that 

H is cyclic, i.e. there is an element s ∈ H such that H = Zs. 

Proof. We may assume that H = 0 so that H contains some element t > 0. 

Since H is not dense there is an open interval ]a, b[, b > a, such that ]a, b[∩H = ∅. 

Set d = b −a and note that ]0, d[∩H = ∅. It follows that s = inf{t ∈ H : t > 0} ≥ 

d > 0. Then s ∈ H. Indeed, if s /∈ H we can find, by definition of s, two elements 

t1, t2 ∈ H such that s < t2 < t1 < 2s. But then t1 − t2 ∈ H∩]0, s[, a contradiction. 

It follows that Zs ⊆ H. If t is some element in H which is not contained in Zs, we 

have that ks < t < (k + 1)s for some k ∈ Z and then (k + 1)s − t ∈ H∩]0, s[, a 

contradiction. Thus Zs = H. 

Lemma 5.71. Let H be a dense subgroup of R and β : H → R a homomorphism 

such that h ≥ 0 ⇒ β(h) ≥ 0. It follows that there is a t ≥ 0 such that β(g) = 

tg, g ∈ H. 

Proof. Define ˜ β : R → R by 

˜β(t) = sup{β(x) : x ∈ H, x < t} .


Consider s, t ∈ R. When x, y ∈ H such that x < t, y < s, we have that x+y < t+s 

and hence β(x) +β(y) = β(x + y) ≤ ˜ β(s +t). By taking supremum over x and y we 

get that 

˜β(t) + ˜ β(s) ≤ ˜ β(s + t) . (5.17) 

Let instead z ∈ H, z < t+s. Since t−H is dense in R, there is an element x ∈ H 

such that 0 < t −x < (t+s) −z. Then y = z −x = z −t+t−x < z −t+t+s−z = 

s, y ∈ H and x + y = z. Consequently, β(z) = β(x) + β(y) ≤ ˜ β(t) + ˜ β(s), and 

by taking the supremum over z we get the inequality which is reversed to (5.17), 

and hence we conclude that ˜ β is a homomorphism. Note that x ≥ 0 ⇒ ˜ β(x) ≥ 0. 

Take an element h ∈ H such that h > 0. Then ˜ β(h) = th for some unique element 

t ≥ 0. For each n ∈ N we find that n˜ β( 1 

nh) = th, i.e. ˜ β( 1 t h) = h. It follows 

n n 

that ˜ β(qh) = qth for all q ∈ Q. Let x ∈ R be arbitrary. There are sequences 

{xn}, {yn} ⊆ Qh such that the xn’s increases, the yn’s decreases and both have x 

as limit. Then txn = ˜ β(xn) ≤ ˜ β(x) ≤ ˜ β(yn) = tyn for all n, so we conclude that 

˜β(x) = tx. 

Lemma 5.72. Let α, γ ∈]0, 1 

2 [\Q. Then Gα = Z + αZ and Gγ = Z + γZ are 

isomorphic as partially ordered abelian groups with order unit 1 if and only α = γ. 

Proof. Let β : Gα → Gγ be an isomorphism of partially ordered groups such 

that β(1) = 1. Since Gα ≃ Z 2 is not cyclic, it must be dense by Lemma 5.70. 

Then we can apply Lemma 5.71 to conclude that β(h) = th for some t > 0. Since 

β(1) = 1 we have that t = 1, proving that Z + αZ = Z + γZ. Define χ1 : Z → T 

and χ2 : Z → T by χ1(z) = e 2πizα and χ2(z) = e 2πizγ , respectively. Then χ1 and χ2 

are both injective (because α and γ are irrational) and by the conclusion we have 

just obtained their range in T is the same. Hence χ −1 

2 ◦χ1 defines an automorphism 

of Z and must therefore be either the identity of the map z ↦→ −z. It follows that 

e2πiα = e2πiγ or e2πiα = e−2πiγ , i.e. α ±γ ∈ Z. Since α, γ ∈]0, 1[, 

this is only possible 

2 

if α = γ. 

By Theorem 3.45 there is, for each α ∈]0, 1 

that (K0(Aα), K0(Aα) + , [1]) ≃ (Gα, G + α , 1). It follows from Lemma 5.72 that the 

unital AF-algebras Aα and Aβ only are isomorphic if α = β. In the following we 

will describe a Bratteli diagram for the AF-algebra Aα. Let 

α = [0, a1, a2, a3, · · ·] 

2 [\Q, a unital AF-algebra Aα such 

be the continued fraction expansion of α, cf. Theorem 5.68. For each n there is a 

positive homomorphism 

ϕn : Z 2 → Z 2 

given by the matrix 

an 1 

. 

1 0 

Note that we here consider Z 2 as an ordered group with the usual ordering, i.e. 

as a simplex group. For each n ≥ 1, write [0, a1, a2, a3, · · · , an] = pn 

qn where 

pn, qn ∈ N\{0} are mutually prime. Set un = (qn−1 , qn−2) ∈ Z 2 when n ≥ 3. It 

follows from Lemma 5.66 that ϕn is a positive order unit preserving homomorphism 

when n ≥ 3. 

ϕn : (Z 2 , un) → (Z 2 , un+1)


Lemma 5.73. There is an isomorphism of partially ordered groups with order 

unit, 

(Gα, 1) ≃ lim((Z 

−→ 2 , un), ϕn)) . 

Proof. Define θn : Z2 → Z2 to be the homomorphism given by 

It follows from (5.13) that 

−1 pn−1 qn−1 

pn−2 qn−2 

−1 pn−1 qn−1 

pn−2 qn−2 

= (−1) n 

. 

qn−2 −qn−1 

−pn−2 pn−1 

Define ψα : Z 2 → Gα by ψα(z1, z2) = z1α + z2. It follows from (5.13) that we have 

the following commuting diagram 

(Z 2 , un) 

ψn 

θn 

(Z 2 , (0, 1)) ψα 

 

(Z2 , un+1) θn+1 ψα 

 

2 (Z , (0, 1)) 

 

(Gα, 1) 

 

(Gα, 1) 

of positive order-unit preserving maps for each n ≥ 3. Since the θn’s are injective 

we get clearly an injective homomorphism θ : lim 

−→ (Z 2 , ϕn) → Gα which takes the 

order unit coming from the un’s to 1 ∈ Gα. It suffices to check that θ maps the 

positive semigroup in lim(Z 

−→ 2 , ϕn) onto G + α , or alternatively that 

 

First note that 

n 

ψα ◦ θn(Z 2+ ) = G + α . (5.18) 

ψα ◦ θn(z1, z2) = (−1) n (α(qn−2z1 − qn−1z2) + (−pn−2z1 + pn−1z2)) . 

Since (−1) n (αqn−2 − pn−2) ≥ pn−2 

qn−2 qn−2 − pn−2 = 0 and (−1) n (pn−1 − αqn−1) ≥ 

pn−1 − pn−1 = 0 when n is even, we see that we have the inclusion ⊆ in (5.18). 

qn−1 

Conversely, let (z1, z2) ∈ Z2 such that αz1 + z2 > 0. Since limn→∞ pn 

= α, there is 

qn 

an N so large that pn−1z1 + qn−1z2 = qn−1( pn−1 

qn−1 z1 + z2) > 0 for all n > N. Hence 

θn −1 

pn−1 qn−1 

(z1, z2) = 

(z1, z2) ∈ Z 

pn−2 qn−2 

2+ 

for all n > N +1. Since αz1 +z2 = ψα ◦θn ◦θn −1 (z1, z2), we conclude that αz1 +z2 ∈ 

 

n ψα ◦ θn(Z 2+ ) . 

So we see that Aα ≃ lim 

−→ (Fn, λn) where 

Fn = Mqn−2 ⊕ Mqn−1 , n ≥ 3 

and the connecting homomorphisms λn : Fn → Fn+1 is given by the Bratteli diagram 

qn−2 qn−1 

 

 

 

 

 

... 

qn−1 qn 

where there are an arrows from qn−1 to qn. 

 

.


Exercise 5.74. Let A be a unital C ∗ -algebra which is approximately divisible. 

Show that ρA(K0(A)) is a subspace of Aff T(A). Show by example that this is not 

true in general. 

Exercise 5.75. Let A be a C∗-algebra and a ∈ A an element of norm a ≤ 1. 

Show that 

lim 

n→∞ (aa∗ ) 1 

n a = a. 

Exercise 5.76. Let H be an infinite-dimensional separable Hilbert space, and 

denote by L(H) the C ∗ -algebra of bounded operators on H. 

An operator A ∈ L(H) has finite rank when the range space, A(H), of A is finite 

dimensonal. 

a) Let A ∈ L(H) have finite rank. Show that A ∗ has finite rank. 

b) Let F be the set of operators in L(H) of finite rank. Show that F is an 

algebraic ∗-ideal. 

We denote by K the closure K = F of F in L(H). An element of K is a compact 

operator. By solving, or at least reading, the next problem, the reader will understand 

the reason for this name. It is not necessary to solve c) in order to do d)-g) - 

it is only included to explain the name. 

c) Let B be the unit ball in H, i.e. B = {ψ ∈ H : ψ ≤ 1}. Show that 

A ∈ L(H) is compact if and only if A(B) is a compact subset of H. 

d) Show that K is an ideal in L(H) which is not all of L(H). 

e) Let ei, i = 1, 2, 3, . . ., be an orthonormal basis in H, and define πn : Mn (C) → 

L(H) such that 

 

0, k ≥ n + 1, 

πn ((aij))ek = n i=1 aikei, k ≤ n. 

Show that πn is an injective ∗-homomorphism, and that πn (Mn (C)) ⊆ F. 

f) Show that K is an AF-algebra, and find a Bratteli diagram presenting it. 

g) Determine K0(K) as a scaled ordered group. 

Exercise 5.77. Let A be the unital AF-algebra given by the following Bratteli 

diagram: 

• 

 

 

 

 

• • • 

 

 

• 

 

 

 

• • • 

 

 

• 

 

 

 

• • • 

 

 

• 

 

 

 

• • • 

 

 

• 

 

 

 

. . . .


a) Show that A has three non-trivial ideals, and describe the inclusion pattern 

among the ideals in A. 

b) Let I1 and I2 be the maximal non-trivial ideals of A. Show that A/I1 and 

A/I2 are isomorphic AF-algebras. 

c) Let I0 be the minimal non-trivial ideal in A. Show that I0 is isomorphic to a 

well-known simple AF-algebra. 

d) Describe the simplex T(A) of trace-states of A.

CHAPTER 2 

Simple AI-algebras 

163

CHAPTER 3 

The range of the Elliott invariant for simple unital 

AI-algebras 

165

Classifying C∗-algebras

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