Chapter 1

Banach algebras

Whilst we are primarily concerned with C∗-algebras, we shall begin with a

study of a more general class of algebras, namely, Banach algebras. These

are of interest in their own right and, in any case, many of the concepts

introduced in their analysis are needed for that of C∗-algebras. Furthermore,

some feeling for the kind of behaviour that can occur in various Banach

algebras helps one to appreciate how well-behaved C∗-algebras are.

Denition 1.1. A Banach algebra is a complex Banach space A together with

an associative and distributive multiplication such that

λ(ab) = (λa)b = a(λb)

and

‖ab‖ ≤ ‖a‖ ‖b‖

for all a, b ∈ A, λ ∈ C.

For any x, x′, y, y′ ∈ A, we have

‖xy − x′y′‖ = ‖x(y − y′) + (x− x′)y‖ ≤ ‖x‖ ‖y − y′‖+ ‖x− x′‖ ‖y′‖

and so we see that multiplication is jointly continuous.

The algebra A is said to be commutative (or abelian) if ab = ba for all a,bin A, and A is said to be unital if it possesses a (multiplicative) unit (this

is also called an identity). Note that if A has an identity, then it is unique:

since if 1l and 1l ′ are units, then 1l = 1l1l ′ = 1l ′.

Example 1.2. If E is a complex Banach space, then B(E), the set of boundedlinear operators on E is a unital Banach algebra when equipped with the

usual linear structure and operator norm.

If 1l denotes the unit in the unital Banach algebra A, then 1l = 1l2 and

so we have ‖1l‖ ≤ ‖1l‖ ‖1l‖, which implies that ‖1l‖ ≥ 1.

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2 Chapter 1

Lemma 1.3. Let A be a Banach algebra with identity 1l. Then there is a norm

||| · ||| on A, equivalent to the original norm, such that (A, ||| · |||) is a unital

Banach algebra with |||1l ||| = 1.

Proof. For each x ∈ A, let Lx denote the linear operator Lx : y 7→ xy ∈ A,y ∈ A. Then if Lx = Lx′ , it follows that Lx1l = Lx′1l and so x = x′. Hencex 7→ Lx is an injective map from A into the set of linear operators on A.Now,

‖Lxy‖ = ‖xy‖ ≤ ‖x‖ ‖y‖ , for y ∈ A

which implies that Lx is bounded, and ‖Lx‖ ≤ ‖x‖. Put |||x||| = ‖Lx‖. Thenwe have just shown that |||x||| ≤ ‖x‖, for any x ∈ A.

On the other hand,

|||x||| = ‖Lx‖ = sup‖Lxy‖ : ‖y‖ ≤ 1= sup‖xy‖ : ‖y‖ ≤ 1

≥ ‖xy′‖, where y′ =1l

‖1l‖

=‖x‖‖1l‖

.

Hence, ‖x‖/‖1l‖ ≤ |||x||| ≤ ‖x‖, for all x ∈ A, which shows that the two norms

||| · ||| and ‖ · ‖ are equivalent.Moreover, for any x, y ∈ A,

|||xy||| = ‖Lxy‖= ‖LxLy‖≤ ‖Lx‖ ‖Ly‖= |||x||| |||y|||

and so A with norm ||| · ||| is a Banach algebra. To complete the proof, we

have |||1l ||| = ‖L1l‖ = 1.

This lemma allows us to assume that the unit of a unital Banach algebra

has norm 1. In fact, this is often taken as part of the denition of a unital

Banach algebra. If A does not have a unit, then we can adjoin one as

follows.

Lemma 1.4. A Banach algebra A without a unit can be embedded into a unital

Banach algebra AI as an ideal of codimension one.

Proof. Let AI = A ⊕ C as a linear space, and dene a multiplication in AIby

(x, λ)(y, µ) = (xy + µx+ λy, λµ).

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Banach algebras 3

It is easily checked that this is associative and distributive. Moreover, the

element (0, 1) is a unit for this multiplication:

(x, λ)(0, 1) = (x0 + x+ λ0, λ1) = (x, λ) = (0, 1)(x, λ).

Put ‖(x, λ)‖ = ‖x‖ + |λ| . Then AI is a Banach space when equipped with

this norm. Furthermore,

‖(x, λ)(y, µ)‖ = ‖(xy + µx+ λy, λµ)‖= ‖xy + µx+ λy‖+ |λµ|≤ ‖x‖‖y‖+ |µ|‖x‖+ |λ|‖+ ‖|λ||µ|= (‖x‖+ |λ|)(‖y‖+ |µ|)= ‖(x, λ)‖ ‖(y, µ)‖ .

Hence AI is a Banach algebra with unit. We may identify A with the ideal

(x, 0) : x ∈ A in AI via the isometric isomorphism x 7→ (x, 0).

We write (x, λ) as (x, λ) = x + λ1l ∈ AI . (Compare this with complex

numbers a + ib ↔ (a, b).) Note that ‖1l‖ = 1. We will see, later, that

an analogous result holds for C∗-algebras, but more care has to be taken

regarding the norm.

Examples 1.5.

1. Consider C([0, 1]), the Banach space of continuous complex-valued func-

tions dened on the interval [0, 1] equipped with the sup-norm, namely,

‖f‖ = sups∈[0,1] |f(s)|, and with multiplication dened pointwise:

(fg)(s) = f(s) g(s) , for s ∈ [0, 1].

Then C([0, 1]) is a commutative unital Banach algebra; the constant

function 1 is the unit element.

2. As above, but replace [0, 1] by any compact topological space.

3. Let D denote the closed unit disc in C, and let A denote the set of

continuous complex-valued functions on D which are analytic in the

interior of D. Equip A with pointwise addition and multiplication and

the norm

‖f‖ = sup|f(z)| : z ∈ ∂D

where ∂D is the boundary of D, that is, the unit circle. (That this is,

indeed, a norm follows from the maximum modulus principle.) Then

A is complete, and so is a (commutative) unital Banach algebra. A is

called the disc algebra.

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4 Chapter 1

4. LetA be the Banach space `1(Z) and dene xy by (xy)n =∑

m xmyn−mfor x = (xn) and y = (yn) in A. Then∑

n

|(xy)n| ≤∑n

∑m

|xm||yn−m|

=∑m

|xm|∑n

|yn−m|

=∑m

|xm| ‖y‖

= ‖x‖ ‖y‖ .

Thus xy ∈ `1(Z) and so A is a Banach algebra. Furthermore, A has a

unit given by (xn) = (δ0n) = (. . . , 0, 0, 1, 0, 0, . . . ) where the 1 appears

in the 0th position.

Denition 1.6. An element x in a unital Banach algebra A is said to be

invertible (or non-singular) inA if there is some z ∈ A such that xz = zx = 1l.

Note that if such a z exists, then it is unique; if z′x = xz′ = 1l, then

z = z1l = zxz′ = 1lz′ = z′ . z is called the inverse of x, and is written x−1, asusual. Evidently, the set of invertible elements forms a group. Non-invertible

elements are also called singular.

Proposition 1.7. The set G(A) of invertible elements in a unital Banach al-

gebra A is open in A, and the inverse operation x 7→ x−1 is a continuous

map from G(A) to G(A).

Proof. First let y ∈ A with ‖y‖ < 1, and put sn =∑n

k=0 yk, n ∈ N. Then

(sn) is a Cauchy sequence in A and so converges, since A is complete. Let

w denote its limit; w =∑∞

k=0 yk. We claim that w is the inverse of 1l − y.

Indeed, we have

(1l − y)w = limn

(1l − y)sn

= limn

(1l − yn+1) = 1l

and

w(1l − y) = limnsn(1l − y)

= limn

(1l − yn+1) = 1l

which establishes the claim. Hence, if x ∈ A with ‖1l − x‖ < 1, then writing

x = 1l − (1l − x) and arguing as above (with y = 1l − x), we see that x is

invertible and its inverse x−1 is given by the convergent series∑∞

k=0(1l−x)k.

Let x0 ∈ G(A). Then for any x ∈ A, we have x = x0x−10 x. Now,

‖1l − x−10 x‖ = ‖x−10 (x0 − x)‖ ≤ ‖x−10 ‖ ‖x0 − x‖,

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Banach algebras 5

and so we conclude that if ‖x− x0‖ < ‖x−10 ‖−1 then x−10 x is invertible with

inverse given by

(x−10 x)−1 =

( ∞∑k=0

yk)

with y = (1l − x−10 x). Hence x is invertible and

x−1 =∞∑k=0

[x−10 (x0 − x)]kx−10 .

To see that x 7→ x−1 is continuous on G(A), suppose that x0 ∈ G(A) andthat (xn) is a sequence in G(A) such that xn → x0 as n→∞. Then for all

suciently large n, ‖xn − x0‖ < ‖x−10 ‖−1 and so

‖x−1n − x−10 ‖ =

∥∥∥∥ ∞∑k=1

[x−10 (x0 − xn)

]kx−10

∥∥∥∥≤∞∑k=1

[‖x−10 ‖‖x0 − xn‖

]k‖x−10 ‖

→ 0

as n→∞.

Denition 1.8. Let A be a unital algebra, and let x ∈ A. The spectrum of xis the subset σA(x) of C given by

σA(x) = λ ∈ C : x− λ1l /∈ G(A).

The resolvent set ρA(x) of x is the complement of the spectrum of x;

ρA(x) = C \ σA(x).

The spectral radius rA(x) of an element x ∈ A is dened as

rA(x) = sup|λ| : λ ∈ σA(x)

provided σA(x) is not empty.

Example 1.9. Let A be the unital Banach algebra Mn(C) of n × n complex

matrices. For a ∈ A and λ ∈ C, a− λ1l is invertible in A if and only if λ is

not an eigenvalue of a. In other words, σA(a) is just the set of eigenvalues ofthe matrix a.

Example 1.10. Suppose that A is the commutative unital Banach algebra

C[0, 1], and f ∈ A. Then, for λ ∈ C, f − λ1l is invertible in A provided fdoes not take the value λ. Hence σA(f) is equal to the set of values assumed

by f , i.e., σA(f) = ran f , the range of f .

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6 Chapter 1

Example 1.11. A normed algebra is dened in just the same way as a Banach

algebra, except that the completeness of the space is no longer required, i.e.,

the space is merely a normed space rather than a Banach space. However,

if A0 is a normed algebra, then it is not dicult to see that its completion

A, say, is in fact a Banach algebra. (To show this, suppose that A0 is dense

in A. Then one shows that the product in A0 extends (by continuity) to a

product on A, and that, when equipped with this product, A is a Banach

algebra.)

Let A0 = C[x], the algebra of complex polynomials in the indetermi-

nate x, considered as a subalgebra of A = C([0, 1]). When equipped with

the supremum norm, A0 is a unital normed algebra whose completion, by

Weierstrass' theorem, is just the unital Banach algebra A.

Let f ∈ A0. For any λ ∈ C, f − λ1l fails to be invertible in A0, unless

f is a constant (i.e., a polynomial of degree zero) not equal to λ. That is,

σA0(f) = C whenever f is not a constant, but σA0(f) = α when f is the

constant α. In particular, an element of A0 is invertible in A0 if and only

if it is a non-zero constant. Hence G(A0) = f ∈ C[x] : f = α, α 6= 0.Thus we see that G(A0) is not an open set in A0. Indeed, 1l, the constant

polynomial 1, belongs to G(A0), of course, but for any ε > 0, the polynomial

p(x) = 1 + 12εx satises ‖1l − p‖ < ε and p /∈ A0. Thus, every open set

containing 1l also contains singular elements of A0.

Example 1.12. Let A denote the algebra of meromorphic functions and let Bdenote the algebra of entire functions. Clearly B is a subalgebra of A, andboth A and B are unital; the unit being the constant function equal to 1 on

C. Let f : C → C be the function z 7→ f(z) = z, z ∈ C. For any λ ∈ C,the function z 7→ (z − λ)−1 is meromorphic, i.e., f − λ1l is invertible in A.On the other hand, the function z 7→ (z − λ)−1 is not entire for any λ ∈ C.Thus we see that σA(f) = C but σB(f) = ∅.

Theorem 1.13. For any x in a unital Banach algebra A, the spectrum σA(x)is a non-empty compact subset of C with σA(x) ⊆ λ ∈ C : |λ| ≤ ‖x‖.

Proof. For given x ∈ A, x − λ1l is invertible whenever |λ| > ‖x‖. Indeed,

for any such λ, (x − λ1l) = −λ(1l − x/λ

)has inverse given by a convergent

series expansion in powers of x/λ. Hence σA(x) ⊆ λ ∈ C : |λ| ≤ ‖x‖, asclaimed. In particular, this shows that σA(x) is bounded.

Let f be the map f : λ 7→ x − λ1l. Then λ /∈ σA(x) if and only if

x−λ1l ∈ G(A), i.e., if and only if λ ∈ f−1(G(A)). It follows that C\σA(x) =f−1(G(A)). But it is clear that f : C → A is continuous and so G(A) open

implies that f−1(G(A)) = C \ σA(x) is open and so σA(x) is closed. Hence

σA(x) is a closed, bounded subset of C and therefore compact.

We must show that σA(x) is non-empty. Indeed, suppose the contrary,

σA(x) = ∅. Then (x − λ1l) is invertible for all λ ∈ C. We claim that the

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Banach algebras 7

map λ 7→ (x − λ1l)−1 is dierentiable, with derivative (x − λ1l)−2, that is,we claim that

(x− (λ+ ζ)1l)−1 − (x− λ1l)−1

ζ→ (x− λ1l)−2

in A as ζ → 0 in C (with ζ 6= 0).To see this, note that

(x− α1l)−1 − (x− β1l)−1 = (x− α1l)−1

(x− β1l)− (x− α1l)

(x− β1l)−1

= (x− α1l)−1(α− β)(x− β1l)−1.

Therefore

(x− (λ+ ζ)1l)−1 − (x− λ1l)−1

ζ=

(x− (λ+ ζ)1l)−1 6ζ(x− λ1l)−1

6ζ→ (x− λ1l)−2

in A as ζ → 0, since x − (λ + ζ) → x − λ and the taking of the inverse is

continuous. This proves the claim.

Now let ϕ ∈ A∗, the dual space of A (the space of continuous linear func-

tionals on A). Then the map λ 7→ ϕ((x−λ1l)−1) is everywhere dierentiablein C, that is, g(λ) ≡ ϕ((x− λ1l)−1) is an entire function. For all suciently

large |λ|, we may write

(x− λ1l)−1 = (−λ)−1(1l − λ−1x)−1

= (−λ)−1∞∑n=0

(λ−1x)n (|λ| > ‖x‖).

Clearly, the right hand side converges to 0 as |λ| → ∞, and so the same is true

of the left hand side, thus g(λ)→ 0 as |λ| → ∞. By Liouville's theorem, we

deduce that g is identically zero on C. But then we have ϕ((x− λ1l)−1) = 0for all ϕ ∈ A∗, which implies that (x−λ1l)−1 = 0. This is impossible because

(x− λ1l)−1(x− λ1l) = 1l 6= 0. We conclude that σA(x) 6= ∅.

Proposition 1.14. Let A be a unital Banach algebra, and let a ∈ A.

(i) σA(p(a)) = p(σA(a)) for any complex polynomial p.

(ii) If a is invertible, σA(a−1) = σA(a)−1.

Proof. (i) Suppose that p has degree n ≥ 1. For any µ ∈ C, let λ1, . . . , λnbe the n complex roots of the polynomial p( · ) − µ. Then, for any z ∈ C,p(z)−µ = α(z−λ1) . . . (z−λn), for some non-zero α ∈ C and so p(a)−µ1l =α(a− λ11l) . . . (a− λn1l).

Now, if a1, . . . , an are mutually commuting elements of A (i.e., aiaj =ajai, for any 1 ≤ i, j ≤ n), then the product a1 . . . an is invertible if and

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8 Chapter 1

only if each ai is invertible. (If each ai is invertible, then their product is

invertible, regardless of the commutativity assumption. Conversely, if the

product is invertible, then, for example, we have

a2a1a3 . . . an(a1 . . . an)−1 = a1 . . . an(a1 . . . an)−1 = 1l

= (a1 . . . an)−1a1 . . . an

= (a1 . . . an)−1a1a3 . . . ana2,

which shows that a2 is invertible. Similarly, it is easy to see that any ai isinvertible.) Suppose that µ ∈ σA(p(a)). Then p(a) − µ1l is singular and so

therefore is a− λi1l, for some 1 ≤ i ≤ n. That is, λi ∈ σA(a). But p(λi) = µwhich shows that µ ∈ p(σA(a)). Conversely, suppose that λ ∈ σA(a) and let

µ = p(λ). Then, with the notation above, it follows that λ = λi for some

1 ≤ i ≤ n, and that p(a) − µ1l is singular. Thus p(λ) ∈ σA(p(a)), and the

result follows.

(ii) Suppose that a ∈ A is invertible, i.e., 0 /∈ σA(a). For any λ ∈ C, λ 6= 0,we have

a− λ1l = a(1l − λa−1) = aλ(λ−11l − a−1)which implies that a−λ1l is singular if and only if a−1−λ−11l is singular.

Suppose that x and y are elements of a unital Banach algebra A such that

‖x‖ < 1 and also ‖y‖ < 1. Then it follows that ‖xy‖ ≤ ‖x‖ ‖y‖ < 1, andsimilarly, ‖yx‖ < 1. We have seen that this implies that 1l − xy and 1l − yxare both invertible with inverses given, respectively, by a = (1l − xy)−1 =∑∞

n=0(xy)n and b = (1l − yx)−1 =∑∞

n=0(yx)n. Evidently,

yax = y(1l + xy + (xy)2 + . . . )x

= yx+ (yx)2 + (yx)3 + . . .

and so

1l + yax = 1l + yx+ (yx)2 + . . .

= b .

Now suppose that x and y are elements of A such that 1l − xy is invertible,

but otherwise x and y are arbitrary. Let a = (1l − xy)−1, and b = 1l + yax.Is it still true that b is the inverse of 1l − yx? That this is, indeed, still the

case follows by straightforward calculation;

b(1l − yx) = (1l + yax)(1l − yx)

= 1l − yx+ yax− yaxyx= 1l − yx+ y a(1l − xy)︸ ︷︷ ︸

= 1l

x

= 1l − yx+ yx

= 1l .

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A similar calculation shows that (1l − yx)b = 1l. (In fact, this demonstration

that 1l + yax is the inverse of 1l − yx is valid in any ring with unitonly the

motivation was carried out in a Banach algebra.) These observations lead

directly to the following result.

Proposition 1.15. For any x, y in a unital Banach algebra A,

σA(xy) ∪ 0 = σA(yx) ∪ 0.

Proof. Suppose that λ ∈ C, λ 6= 0. Then, using the observation above, we

see that λ1l − xy is invertible if and only if λ(1l − xy/λ) is invertible if and

only if λ(1l − yx/λ) is invertible if and only if λ − yx is invertible. Hence

σA(xy) \ 0 = σA(yx) \ 0, and the result follows.

Remark 1.16. It is a consequence of this proposition, that an identity of the

form ab−ba = 1l cannot possibly hold in any unital Banach algebra. Indeed,

suppose the contrary. Then, by the proposition, σA(ab)∪0 = σA(ba)∪0.On the other hand, we know from 1.14 that σA(ab) = σA(1l+ba) = 1+σA(ba).Therefore σA(ba) ∪ 0 = 0 ∪ 1 + σA(ba). This is impossible because

σA(ba) is bounded. (If α ∈ σA(ba) with Reα ≥ 0, then 1 + α 6= 0 and

belongs to the right hand side of the above equality. So it also belongs to

σA(ba). By induction, we see that α + n belongs to σA(ba) for any n ∈ Nwhich is not possible. On the other hand, if α ∈ σA(ba) with Reα < 0, thenα ∈ 1 + σA(ba) and so α − 1 ∈ σA(ba). Again, by induction, it follows that

α− n ∈ σA(ba) for any n ∈ N which, once again, is not possible.)

This result is of great importance in quantum mechanics where the posi-

tion and momentum of a particle are represented by linear operators q and pacting on a Hilbert space and are supposed to satisfy the Heisenberg canon-

ical commutation relation pq − qp = 1l. It therefore follows that q and pcannot both be bounded linear operators on the Hilbert space. To salvage

this situation, one must consider unbounded operators. (In fact, it turns

out (by a result of von Neumann) that both q and p must be unbounded

operators.)

Theorem 1.17. (Spectral radius formula) Let A be a unital Banach algebra

and let x ∈ A. Then the limit limn→∞ ‖xn‖1/n exists and satises

limn→∞

‖xn‖1/n = rA(x) = infn‖xn‖1/n.

Proof. Set

γ(x) = inf‖xn‖1/n : n = 1, 2, . . . .

We shall show that ‖xn‖1/n → γ(x) as n→∞. Given any ε > 0, let k ∈ Nbe such that

‖xk‖1/k < γ(x) + ε.

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10 Chapter 1

For any n ∈ N, write n as n = αk + β where 0 ≤ β < k and α, β ∈ Z+.

(Note that k is xed and α, β depend on n.) Then β/n→ 0 as n→∞ since

β is always less than k. Also,

1 =n

n=αk + β

n=αk

n+β

n

and soαk

n→ 1 as n→∞, that is,

α

n→ 1

k, as n→∞. Now,

‖xn‖1/n = ‖(xk)αxβ‖1/n

≤ ‖xk‖α/n ‖x‖β/n.

and the right hand side converges to ‖xk‖1/k, as n→∞, which is less than

γ(x) + ε. Hence, for all suciently large n,

‖xn‖1/n ≤ ‖xk‖α/n ‖x‖β/n

< γ(x) + ε.

On the other hand, γ(x) ≤ ‖xn‖1/n for any n = 1, 2, . . . and so

γ(x) ≤ ‖xn‖1/n < γ(x) + ε

for all suciently large n. Thus ‖xn‖1/n converges to γ(x) as n → ∞, i.e.,

limn ‖xn‖1/n exists and is equal to infm ‖xm‖1/m.We must now show that the above limit is equal to rA(x). Recall, rst,

that for any y ∈ A, σA(y) ⊆ λ : |λ| ≤ ‖y‖ and so rA(y) ≤ ‖y‖.By proposition 1.14, it follows that λn : λ ∈ σA(x) = σA(xn) and so

rA(xn) = rA(x)n, for any n ∈ N. But rA(xn) ≤ ‖xn‖ and so we obtain

rA(x)n = rA(xn) ≤ ‖xn‖=⇒ rA(x) ≤ ‖xn‖1/n, for all n,

=⇒ rA(x) ≤ γ(x).

We want to show now that rA(x) ≥ γ(x). Let ϕ ∈ A∗. Then g : λ 7→ϕ((x−λ1l)−1) is analytic on C\σA(x), and so has a Laurent series expansion

on λ : |λ| > rA(x). But for |λ| > ‖x‖, we know that g has the expansion

g(λ) =∞∑n=0

ϕ(xn)

λn+1.

This must therefore be absolutely convergent in the region λ : |λ| > rA(x).Fix λ with |λ| > rA(x). Then, in particular, ϕ(xn/λn+1) → 0 as n → ∞.

This holds for any ϕ ∈ A∗, and so, by the uniform boundedness principle, it

follows that (xn/λn+1) is a bounded sequence in A, that is, there is κ > 0such that ‖xn/λn+1‖ ≤ κ for all n. Hence ‖xn‖ ≤ κ|λ| |λn| and so ‖xn‖1/n ≤(κ|λ|)1/n|λ|. Letting n → ∞ gives γ(x) ≤ |λ|. This holds for any λ with

|λ| > rA(x) and so we deduce that γ(x) ≤ rA(x).It follows that rA(x) = γ(x) and the proof is complete.

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Remark 1.18. We have shown that

infn‖xn‖1/n = lim

n‖xn‖1/n = sup|λ| : λ ∈ σA(x).

The right hand side is purely algebraic in that it involves (non-)existence

of an inverse in A, whereas the left hand side involves the norm, that is,

the metric aspects of the algebra. We see here an inter-relationship between

purely metric and purely algebraic parts of the theory. By enlarging the

algebra, the spectrum may change, but the spectral radius will not.

Example 1.19. Let A be the disc algebra. Then each f in A is uniquely

determined by its values on the unit circle ∂D = S1. Thus, A can be regarded

as a subalgebra of C(S1), the Banach algebra of continuous complex-valued

functions on the circle S1. This identication of A in C(S1) is also norm

preserving (by the maximum modulus principle).

Let g(z) = z, for z ∈ D. Then g ∈ A. Evidently (z − λ1l) fails to be an

invertible analytic function on D if and only if λ ∈ D. Thus we see that

σA(g) = D = λ : |λ| ≤ 1.

On the other hand, considered as an element of C(S1), g is the function

g(θ) = eiθ, with the obvious notation. Then (g−λ1l) fails to be invertible inC(S1) if and only if |λ| = 1. Hence

σC(S1)(g) = ∂D = λ : |λ| = 1.

Note that rA(g) = rC(S1)(g) (= 1), as we know should be the case.

Theorem 1.20. (Gelfand-Mazur) Let A be a unital Banach algebra such that

each non-zero element of A is invertible. Then A ' C.

Proof. For any x ∈ A, σA(x) 6= ∅ and so there is λ ∈ C with x−λ1l /∈ G(A).But then this must mean that x−λ1l = 0, that is x = λ1l for some λ ∈ C.

The next theorem concerns quotient spaces.

Proposition 1.21. Let X be a normed space, and V a closed linear subspace

of X. Then the quotient space X/V is a normed space with respect to the

quotient norm dened by

‖ clx‖ = infv∈V‖x+ v‖ = inf

x′∼x‖x′‖.

where clx denotes the equivalence class of x in X/V .

Proof. First recall thatX/V is the set of equivalence classes inX determined

by the relation x ∼ x′ if and only if x − x′ ∈ V . X/V is a linear space

when equipped with the obvious operations; clx + cl y = cl(x + y) and

λ clx = cl(λx), for x, y ∈ A and λ ∈ C (one can readily check that these

operations are well-dened, i.e., independent of the representatives used).

We must show that ‖ · ‖ is a norm.

King's College London

12 Chapter 1

If clx = 0 in X/V , then x ∈ V and so, taking v = −x, we get ‖ clx‖ =infv∈V ‖x+v‖ = 0. On the other hand, if ‖ clx‖ = 0, then there is a sequence

(vn) in V such that ‖x + vn‖ → 0 as n → ∞. Hence −vn → x in X. Since

V is closed, we deduce that x ∈ V and so clx = 0 in X/V .

For λ ∈ C, λ 6= 0, and x ∈ X,

‖ clλx‖ = infv∈V‖λx+ v‖ = inf

v′∈V‖λx+ λv′‖

= |λ| infv′∈V‖x+ v′‖ = |λ| ‖ clx‖.

For any x, y ∈ X,

‖ clx+ cl y‖ = infv∈V‖x+ y + v‖

= infv,w∈V

‖x+ y + v + w‖

≤ infv,w∈V

(‖x+ v‖+ ‖y + w‖

)= ‖ clx‖+ ‖ cl y‖.

Hence ‖ · ‖ is a norm on X/V .

If X is a Banach space, then so is X/V , as we shall now show. We shall

use the following standard result from Banach space theory.

Proposition 1.22. Let Y be a normed space. Then Y is complete if and only

if it has the following property: if (ym) is any sequence in Y such that∑∞m=1 ‖ym‖ <∞, then there is y ∈ Y such that

∑nm=1 ym → y as n→∞.

Proof. If Y is complete and (yn) satises∑∞

k=1 ‖yk‖ < ∞, then it is clear

that(∑n

k=1 yk)is a Cauchy sequence and hence converges.

Conversely, suppose that Y has the stated property, and let (xn) be a

Cauchy sequence in Y . We construct a subsequence as follows. Let n1 ∈ Nbe such that ‖xn1 − xm‖ < 1

2 for all m > n1. Now let n2 > n1 be such

that ‖xn2 − xm‖ < 14 for all m > n2. Continuing in this way, we obtain a

subsequence (xnk) of (xn) such that

‖xnk − xnk+1‖ < 1

2k, k = 1, 2, . . .

Set yk = xnk − xnk+1for k = 1, 2, . . . . Then

∞∑k=1

‖yk‖ =

∞∑k=1

‖xnk − xnk+1‖

<∞∑k=1

1

2k< ∞ .

Department of Mathematics

Banach algebras 13

By hypothesis, there is some y ∈ Y such that

y = limm→∞

m∑k=1

yk

= limm→∞

(xn1 − xn2) + (xn2 − xn3) + · · ·+ (xnm − xnm+1)

= limm→∞

(xn1 − xnm+1).

That is, (xnk) converges in Y (to xn1−y). But if a subsequence of a Cauchysequence converges, the whole sequence does; i.e., (xn) converges in Y and

we conclude that Y is complete.

Proposition 1.23. Let X be a Banach space and suppose that V is a closed

linear subspace of X. Then X/V is a Banach space with respect to the

quotient norm

‖ clx‖ = infv∈V‖x+ v‖.

Proof. We have already shown that X/V is a normed space. To show that

it is indeed a Banach space we will use the last result. Let (clxn) be any

sequence inX/V such that∑∞

n=1 ‖ clxn‖ <∞. By denition of the inmum,

for each n, there is vn ∈ V such that

‖xn + vn‖ < infv∈V‖xn + v‖+

1

2n

= ‖ clxn‖+1

2n.

Hence∞∑n=1

‖xn + vn‖ <∞∑n=1

(‖ clx‖+

1

2n

)< ∞.

Now, X is a Banach space, so, by the previous proposition, it follows that

there is y ∈ X such that

y = limn→∞

n∑k=1

(xk + vk).

We claim that∑n

k=1 clxk → cl y in X/V . Indeed,

‖ cl y −n∑k=1

clxk‖ = ‖ cl(y −n∑k=1

xk)‖

= infv∈V‖y −

n∑k=1

xk + v‖

King's College London

14 Chapter 1

≤ ‖y −n∑k=1

xk −n∑k=1

vk‖

= ‖y −n∑k=1

(xk + vk)‖

→ 0 as n→∞.

Hence∑n

k=1 clxk converges in X/V (to cl y) and so X/V is a Banach space,

again by the previous proposition.

Denition 1.24. A linear subset V in an algebra A is a left (respectively,

right) ideal if av ∈ V (respectively, va ∈ V ) for all a ∈ A and v ∈ V . V is a

two-sided ideal in A if it is both a left and a right ideal.

We can obtain new algebras by taking suitable quotients, as the next

theorem shows.

Theorem 1.25. Let A be a Banach algebra and suppose that V is a closed

two-sided ideal in A. Then A/V is a Banach algebra with respect to the

quotient norm

‖ clx‖ = infv∈V‖x+ v‖.

If A is unital and V is proper, then A/V is unital. Moreover the identity

of A/V has unit norm.

Proof. We have shown that A/V is a Banach space. Since V is a two-sided

ideal it is easy to see that A/V is an algebra with respect to the multiplication

clx cl y = clxy. Furthermore,

‖ clx cl y‖ = ‖ clxy‖= inf

v∈V‖xy + v‖

≤ infv,w∈V

‖xy + xw + vy + vw︸ ︷︷ ︸∈V

‖

= infv,w∈V

‖(x+ v)(y + w)‖

≤ infv,w∈V

‖x+ v‖ ‖y + w‖

= ‖ clx‖ ‖ cl y‖.

Thus A/V is a Banach algebra.

If V = A, then A/V is trivial, that is, 0. If V is proper, then A/V is

not equal to 0 and one sees that cl 1l is a unit for A/V .Furthermore, if ‖1l‖ = 1, then

‖ cl 1l‖ = infv∈V‖1l + v‖

Department of Mathematics

Banach algebras 15

≤ ‖1l‖, (taking v = 0),

= 1.

However, we know that the unit of a Banach algebra always has norm greater

than or equal to 1, so we obtain ‖ cl 1l‖ = 1.

King's College London

16 Chapter 1

Department of Mathematics

Chapter 2

Gelfand Theory

In this section we shall investigate the interplay between the maximal ideals

of a unital Banach algebra, the multiplicative linear functionals and associ-

ated function spaces.

Denition 2.1. An ideal in an algebra is said to be maximal if it is proper

(i.e., not equal to the whole algebra) and is not contained as a proper subset

of any other proper ideal. Thus, `maximal' is synonymous with `maximal

proper'.

Proposition 2.2. Every maximal ideal in a unital Banach algebra is closed.

Proof. Let J be a maximal ideal in the unital Banach algebra A. Then Jcannot contain any invertible elements, otherwise we would have J = A.Hence J ⊆ A \ G(A). Now, G(A) is open and so A \ G(A) is closed, hence

J ⊆ J ⊆ A \ G(A).

In particular, J 6= A. But J is an ideal containing J , and so J = J since Jis a maximal ideal. That is, J is closed.

Proposition 2.3. Every complex-valued homomorphism on a Banach algebra

is continuous.

Proof. Let A be a Banach algebra, and ϕ : A → C a homomorphism. If

ϕ = 0, then it is certainly continuous. So suppose that ϕ 6= 0 and suppose

also that A is unital. For any a ∈ A, ϕ(a) = ϕ(a1l) = ϕ(a)ϕ(1l) and so

ϕ(1l) = 1. If a ∈ A and ϕ(a) 6= 0, then b = a− ϕ(a)1l belongs to the kernel

of ϕ and therefore is singular (otherwise, 1 = ϕ(bb−1) = ϕ(b)ϕ(b−1), whichis impossible). This means that ϕ(a) belongs to σA(a) and it follows that

|ϕ(a)| ≤ ‖a‖. This inequality remains valid when ϕ(a) = 0 and we conclude

that ϕ is continuous on A.If A is non-unital, we consider AI instead. Dene the map ϕ′ : AI → C

by ϕ′((a, λ)) = ϕ(a) + λ, (a, λ) ∈ AI . Then it is straightforward to check

that ϕ′ is a homomorphism and therefore, by the above, is continuous on AI .In particular, its restriction to A in AI is continuous, i.e., ϕ is continuous.

17

18 Chapter 2

Denition 2.4. A non-zero complex-valued homomorphism on a Banach al-

gebra is called a character (or multiplicative linear functional).

By the last proposition, characters are necessarily continuous.

Theorem 2.5. (Gelfand-Mazur) There is a canonical bijection between the

maximal ideals in a commutative unital Banach algebra A and its characters

given by associating to each character its kernel; i.e., if ` is a character on

A, ker ` is a maximal ideal in A, and every maximal ideal has this form for

some unique character.

Proof. Suppose that ` : A → C is a character and let J = ker `. We have

J 6= A since ` is non-zero. Let a /∈ J . Then any b ∈ A can be written as

b = a`(b)

`(a)+

(b− a `(b)

`(a)

).

Since b − a`(b)

`(a)∈ ker ` = J , we see that A = Ca + J and hence J is a

maximal ideal.

Now suppose that J is a maximal ideal. Then J is closed and so A/Jis a Banach algebra. We claim that the maximality of J implies that every

non-zero element of A/J is invertible. To see this, suppose that cl a is a

non-zero non-invertible element of A/J . Then J + aA is a proper ideal of Awhich contains J as a proper subset. (J + aA does not contain 1l since cl ais non-invertible in A/J .) This contradicts the supposed maximality of J ,and the claim is established.

It then follows that every element in A/J has the form λ cl 1l, for λ ∈ C.Let ϕ : A/J → C denote this isomorphism, and let π denote the canonical

projection π : A→ A/J . Then φ π is a homomorphism from A to C with

kernel equal to J ; for

φ π(ab) = φ(π(ab)) = φ(cl ab)

= φ(cl a cl b) = φ(cl a)φ(cl b)

= (φ π(a))(φ π(b)),

and φ π(a) = 0 if and only if π(a) = 0 if and only if a ∈ J .Thus we have a correspondence between maximal ideals J and characters

` with ker ` = J . This association is one-one, since ` is uniquely determined

by its kernel. Indeed, suppose that ` and `′ have the same kernel. Then

for any a ∈ A, a − `(a)1l belongs to ker ` = ker `′ and so `′(a) = `(a) since

`′(1l) = 1.

Proposition 2.6. Any commutative unital Banach algebra possesses at least

one character.

Department of Mathematics

Gelfand Theory 19

Proof. If all elements of the commutative unital Banach algebra A are in-

vertible, then A ' C and the eecting isomorphism is a character. On the

other hand, if there is some x ∈ A such that x is not invertible, then xAis a proper ideal and so is contained in a maximal proper ideal J , say, byZorn's lemma. (The set of proper ideals containing J is partially ordered by

set-theoretic inclusion, and the union of any increasing family of such ideals

is also a proper ideal containing J (since none of these can contain the

unit). Zorn's lemma states that there exists a maximal such set, i.e., proper

and containing J .) But then we know that J is the kernel of a character on

A.

Without the assumption of commutativity, there may be no characters

at all on an algebra.

Example 2.7. Let A = Mn(C), with n > 1, and let eij be the n × n matrix

all of whose entries are 0 except for the ij-entry which is equal to 1. If `were a character on A, then, for i 6= j, the equality e2ij = 0 would imply

that `(eij) = 0. Hence the equality eii = eijeji, applied with i 6= j, wouldimply that `(eii) = 0 for each i = 1, 2, . . . n. We would conclude that `(1l) =`(e11) + · · ·+ `(enn) = 0, which is impossible. Hence Mn(C), for any n > 1,possesses no characters.

Denition 2.8. The set of characters of a commutative unital Banach algebra

A is called the spectrum (or carrier space, or structure space, or maximal

ideal space) of A, and is denoted SpA.

We recall the denition of the w∗-topology on A∗, the dual of the Banachspace A.

Denition 2.9. The w∗-topology on A∗ is that generated by the neighbour-

hoods

N(ϕ : S, ε) = ω ∈ A∗ : |ω(a)− ϕ(a)| < ε for all a ∈ S

where ϕ ∈ A∗, ε is any positive real number and S is any nite subset of A.Thus a set G in A∗ is open in the w∗-topology if and only if for each ψ ∈ Gthere is some N(ψ : S, ε) as above with N(ψ : S, ε) ⊆ G.

This gives rise to a Hausdor topology on A∗; indeed, for any ϕ1, ϕ2 ∈ A∗,with ϕ1 6= ϕ2 there exists a ∈ A such that ϕ1(a) 6= ϕ2(a). Let ε0 =|ϕ1(a) − ϕ1(a)|/3. Then evidently the neighbourhoods N(ϕ1 : a, ε0) and

N(ϕ2 : a, ε0) have empty intersection.

In terms of nets, the w∗-topology is described as the weakest topology

on A∗ such that a net (ωα)→ ω in A∗ if and only if ωα(a)→ ω(a) for eacha ∈ A.

King's College London

20 Chapter 2

We shall use the following resultthe Banach-Alaoglu theorem.

Theorem 2.10. The closed unit ball of A∗, the dual of the Banach space A,is w∗-compact.

Proposition 2.11. The spectrum, SpA, of a commutative unital Banach al-

gebra A is a w∗-closed subset of the unit ball of A∗, and hence is compact.

Proof. We know from 2.3 that any character is bounded with norm equal to

1, and therefore SpA is contained in the unit ball of A∗.

To show that SpA is w∗-closed, we shall show that A∗ \ SpA is open.

Let ϕ ∈ A∗ \ SpA. If ϕ = 0, we have 0 = ϕ ∈ N(0; 1l, 12) ⊂ A∗ \ SpA,since `(1l) = 1 for any ` ∈ SpA. Suppose that ϕ 6= 0. Then there is a, b ∈ Asuch that ϕ(ab) 6= ϕ(a)ϕ(b). Consider the neighbourhood N(ϕ : a, b, ab, ε)of ϕ in A∗. This consists of all those ω ∈ A∗ such that |ω(a) − ϕ(b)| < ε,|ω(b) − ϕ(b)| < ε and |ω(ab) − ϕ(ab)| < ε. Clearly, if ε > 0 is suciently

small, then for any such ω, we have ω(ab) 6= ω(a)ω(b), ie, for suciently

small ε > 0, N(ϕ : a, b, ab, ε) is contained in A∗ \ SpA. Hence A∗ \ SpAis open in A∗, and so SpA is w∗-closed. SpA is compact since it is a closed

subset of a compact set.

Remark 2.12. The fact that SpA is w∗-closed can be demonstrated quite

easily using nets. Suppose that (`α) is a net in SpA converging to ϕ ∈ A∗.Then, by denition of the w∗-topology, `α(x) → ϕ(x) for each x ∈ A. But

then for any x, y ∈ A

ϕ(xy) = lim `α(xy)

= lim `α(x) `α(y)

= ϕ(x)ϕ(y)

and it follows that ϕ ∈ SpA. Note that ϕ is non-zero since ϕ(1l) = 1.

Theorem 2.13. Let A be a commutative unital Banach algebra. For x ∈ Aand ` ∈ SpA, dene x : SpA→ C by

x(`) = `(x).

Then the range of the function x on SpA satises

ran x = σA(x).

Furthermore, the map is a homomorphism : A→ C(SpA) and

‖x‖∞ ≤ ‖x‖, for x ∈ A.

is called the Gelfand transform.

Department of Mathematics

Gelfand Theory 21

Proof. We have seen already that for any x ∈ A and ` ∈ SpA we have

`(x) ∈ σA(x) ; i e, x(`) ∈ σA(x) and so the range of x satises the inclusion

ran x ⊆ σA(x).Let λ ∈ σA(x). Then x − λ1l is not invertible and so belongs to some

maximal ideal, J , say. (In fact, x−λ1l belongs to the proper ideal A(x−λ1l)which is contained in a maximal ideal, by Zorn's lemma.)

Let ` be that element of SpA with ker ` = J . Then x − λ1l ∈ J implies

that `(x) = λ. Hence x(`) = `(x) = λ and it follows that ran x = σA(x).It is clear that is a homomorphism; for example,

xy(`) = `(xy) = `(x)`(y)

= x(`) y(`) for any x, y ∈ A, ` ∈ SpA,

and so xy = x y. Similarly, one sees that is linear.

To show that x ∈ C(SpA), let U be any open set in C. We must show

that x−1(U) is open in SpA. If x−1(U) = ∅, then we are done. So suppose

that x−1(U) 6= ∅. Let ` be any element of x−1(U). Then there is ζ ∈ U such

that x(`) = ζ. Since U is open in C, there is ε > 0 such that Nε(ζ) ≡ z ∈C : |z − ζ| < ε ⊆ U . Let V = N(` : x, ε) ≡ ω ∈ SpA : |ω(x) − `(x)| <ε. Then ω(x) = x(ω) ∈ U for all ω ∈ V , i e, ` ∈ V ⊆ x−1(U). We

deduce that x−1(U) is open in SpA and hence x : SpA → C is continuous;

that is, x( · ) ∈ C(SpA). (Alternatively, the continuity of x can easily be

established using nets, as follows. Suppose that `α → ` in SpA. Then

x(`α) = `α(x) → `(x) = x(`), by denition of the w∗-topology. In other

words, x is continuous.)

Finally, we have that ran x = σA(x) ⊆ λ : |λ| ≤ ‖x‖ and so it follows

that |x(`)| ≤ ‖x‖, for all ` ∈ SpA. Thus ‖x‖∞ ≤ ‖x‖ for any x ∈ A.

This theorem has a sharper form for C∗-algebras, as we will see in the

next section.

Theorem 2.14. Let A be a commutative unital Banach algebra generated by

the single element a: that is, the set of polynomials in a is dense in A. Thenthe map a : SpA→ σA(a) ⊂ C is a homeomorphism.

Proof. We know that a is a continuous function on SpA with ran a = σA(a),i e, a : SpA → σA(a) is continuous and onto. Now, both SpA and σA(a)are compact Hausdor spaces, so we need only show that a is injective.

To see this, suppose that a(`1) = a(`2), so that `1(a) = `2(a). Using the

multiplicativity of `1 and `2, we see that for given N ∈ N and c0, c1, . . . , cNin C

`1

( N∑n=0

cnan

)= `2

( N∑n=0

cnan

).

Since `1 and `2 are continuous and a generates A, it follows that `1 = `2.

King's College London

22 Chapter 2

Example 2.15. Let A be the subalgebra ofM2(C) consisting of those elements

of the form(α β0 α

), with α, β ∈ C. Then(α β0 α

)= α1l + βq, where q =

(0 10 0

).

We notice that q2 = 0 (i.e., q is nilpotent). Evidently, A is a two-dimensional

commutative Banach algebra with unit 1l. We shall compute the spectrum,

σA(x), for x =(α β0 α

). Indeed, for λ ∈ C,

x− λ1l =

(α− λ β

0 α− λ

)is invertible in M2(C) if and only if λ 6= α. If λ 6= α, then, in fact,

(x− λ1l)−1 =

((α− λ)−1 −β(α− λ)−2

0 (α− λ)−1

),

which belongs to A. Hence, σA(x) = σA(α1l + βq) = α. In particular,

σA(q) = σA(βq) = 0, but q 6= 0.Now we consider the characters of A. If ` is a character, then `(xy) =

`(x)`(y) implies that `(q2) = `(q)`(q). But q2 = 0 and therefore `(q) = 0.Since `(1l) = 1, we nd that `(α1l + βq) = α, for any α, β in C. In other

words, there is just one character on A; SpA = `, where ` is given uniquely

by the action `(1l) = 1 and `(q) = 0.The Gelfand transform is the map x 7→ x, α1l + βq 7→ α1l + βq. But

1l = 1 and q(`) = `(q) = 0 so that q = 0 and we have (α1l + βq) = α, forany α, β ∈ C. The transform has kernel βq : β ∈ C, so we see that is

not an isomorphism.

The algebra A has exactly one maximal ideal, namely, the kernel of `.A is the unital algebra generated by the element q, and so SpA ' σA(q),

via the homeomorphism q : SpA→ σA(q), ` 7→ q(`) = 0. Indeed, both SpAand σA(q) are singleton sets!

Alternatively, we can determine the spectrum of x ∈ A using the equality

σA(x) = ran x. For x =(α β0 α

), we have

σA(x) = x(`) = `(x)= `(α1l + βq)= α`(1l) + β`(q)= α

since `(q) = 0.

Department of Mathematics

Chapter 3

C∗-algebras

Denition 3.1. A Banach ∗-algebra is a Banach algebra A together with an

involution a 7→ a∗ satisfying :

(i) ∗ is conjugate linear, i.e., (αa)∗ = αa∗ for α ∈ C, a ∈ A;

(ii) a∗∗ = a for every a ∈ A;

(iii) (ab)∗ = b∗a∗ for any a, b ∈ A;

(iv) ‖a∗‖ = ‖a‖.

By (iv), we see that ∗ : A→ A is continuous.

If A has a unit 1l, then 1l∗ = 1l∗1l = (1l∗1l)∗ = 1l∗∗ = 1l.

A C∗-algebra is a Banach ∗-algebra for which

(v) ‖a∗a‖ = ‖a‖2, for all a ∈ A.

This property (v) is often referred to as the C∗-property of the norm.

Remark 3.2. We note that in a C∗-algebra, property (iv) follows from the

others: indeed, for any a in a C∗-algebra A, property (v) implies that ‖a‖2 =‖a∗a‖ ≤ ‖a∗‖ ‖a‖ and so ‖a‖ ≤ ‖a∗‖. (If ‖a‖ = 0, then a = 0 = a∗.)Replacing a by a∗ gives ‖a∗‖ ≤ ‖a‖ and so their equality follows.

Now suppose that A is a C∗-algebra with unit 1l. Then, by the C∗-property, ‖1l∗1l‖ = ‖1l‖2 and so ‖1l‖ = ‖1l‖2 which implies that ‖1l‖ = 1.

If A does not have a unit, then one can be adjoined but care must be taken

not to spoil the C∗-property, property (v). We will return to this later.

23

24 Chapter 3

Examples 3.3.

1. Let Ω be a compact space. Then C(Ω) is a commutative C∗-algebrawith unit, when equipped with the supremum norm and the involution

f 7→ f∗ = f . Since f∗f = |f |2 it follows that ‖f∗f‖∞ = ‖f‖2∞.

2. Let Ω be a topological space. Cb(Ω), the algebra of continuous boundedcomplex-valued functions on Ω, is a commutative C∗-algebra with unit,as above.

3. Let Ω be a non-compact, locally compact Hausdor space and let C0(Ω)denote the set of continuous complex-valued functions on Ω vanishing

at innity. (That is, the continuous complex-valued function f : Ω→ Cbelongs to C0(Ω) if and only if for any given ε > 0 there is a compact set

in Ω outside of which |f | is less than ε; ie, the set ω ∈ Ω : |f(ω)| ≥ εis compact.) Then C0(Ω) is a C∗-algebra without a unit. (A unit

f ∈ C0(Ω) must satisfy f = f2 and so can only take the values 0 and

1. The set K = ω ∈ Ω : f(ω) = 1 is necessarily compact and so is a

proper subset of Ω, since Ω is supposed to be non-compact. But then

for any g ∈ C0(Ω), g = gf implies that g vanishes on the non-empty

open set Ω \ K. Let ω′ ∈ Ω \ K. Since Ω is locally compact, there

is a compact set K1 and an open set G such that ω′ ∈ G ⊆ K1. By

Urysohn's lemma, there is a continuous map g : Ω → C such that

g takes the value 1 at ω′ and vanishes on the closed set Ω \ G. But

this means that ω ∈ Ω : f(ω) 6= 0 ⊆ K1 and so g ∈ C0(Ω). This

contradicts the fact that g should vanish outside of K, and we conclude

that C0(Ω) has no unit.)

4. It is quite possible for C0(Ω) to possess a unit when Ω is locally com-

pact, non-compact and non-Hausdor. Indeed, let Ω = 0 ∪ N, anddene a topology on Ω by declaring a non-empty set to be open if it

is equal to 0 or if it contains the element 1. Then the points 1 and

2, for example, cannot be separated by disjoint open sets (since any

open set containing 2 also contains 1) so Ω is non-Hausdor, and 0,1, 1, 2, 1, 3, 1, 4, . . . is an open cover of Ω with no nite sub-

cover, so Ω is non-compact. However,any ω ∈ Ω is contained in the

set ω ∪ 1, which is open and compact, and therefore Ω is locally

compact. Let f : Ω → C be continuous. Then f(n) = f(1) for all

n ∈ N; to see this, note that f−1(C \ f(1)) is an open set in Ω not

containing 1. Since the only open sets in Ω which do not contain 1 are

∅ and 0, we conclude that n /∈ f−1(C \ f(1)) for any n ∈ N. In

other words, f(n) = f(1) for n ∈ N.

Suppose now that f ∈ C0(Ω). Then, by denition, for any given ε > 0there is a compact setK, say, in Ω such that |f(ω)| < ε for all ω ∈ Ω\K.

The only compact sets in Ω are those with a nite number of elements,

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C∗-algebras 25

so for any given ε > 0, we must have |f(n)| < ε for all suciently large

n. However, since f is continuous, we have f(n) = f(1), for all n ∈ N.It follows that f(n) = f(1) = 0, for all n ∈ N. Thus, C0(Ω) consists

of all functions f : Ω → C such that f(1) = f(2) = f(3) = · · · = 0.Evidently, the function e : C0(Ω)→ C given by e(0) = 1, e(1) = e(2) =· · · = 0, is a unit for the C∗-algebra C0(Ω).

Of course, C0(Ω) is not a particularly interesting C∗-algebrait is iso-

morphic to C, via the map f 7→ f(0).

5. C, with the obvious structure, is a C∗-algebra. (This is just example

1, above, when Ω consists of a single point.)

6. We denote the linear space of bounded linear operators on the complex

Hilbert space H by B(H). For any x ∈ B(H) let ‖x‖ denote the

operator norm of x (given by sup‖xξ‖ : ‖ξ‖ ≤ 1, ξ ∈ H) and let ∗

be the operator adjoint. Then B(H), equipped with this structure, is a

unital C∗-algebra. (If dimH = 1, then we have the previous example.)

7. Any norm closed subalgebra of B(H) which contains x∗ if it containsx is a C∗-algebra. This is the typical C∗-algebra, as we will see.

8. Mn(C), the algebra of n × n complex matrices, with ∗ being the ad-

joint is a unital C∗-algebra. The norm is the operator norm obtained

by considering Mn(C) as an algebra of linear operators on the nite-

dimensional complex Hilbert space Cn. (This is the same as dening

‖x‖ to be the square root of the largest eigenvalue of the positive self-

adjoint matrix x∗x, x ∈ Mn(C).) This is just example 6 when H is

n-dimensional.

9. `∞(N) and `∞(Z) equipped with component-wise operations and the

sup-norm are commutative unital C∗-algebras. (These are examples of

2 above.)

10. The set K of compact operators on a Hilbert space H is a C∗-algebra.Note, however, that K has a unit if and only if H is nite-dimensional.

11. Given any family Aα of C∗-algebras, let A denote the subset of

the Cartesian product∏αAα consisting of those elements (aα) for

which supα ‖aα‖ is nite. Then A is a Banach space with respect to

the norm ‖(aα)‖ = supα ‖aα‖. Furthermore, A is a ∗-algebra when

equipped with component-wise operations; for example, the adjoint of

(aα) is just (a∗α), and the product of (aα) with (bα) is (aαbα). It is

straightforward to check that, with this structure, A is a C∗-algebra.Moreover, A is unital if and only if each Aα is unital. (If (eα) is a unitfor A, then it is readily see that for any α0, eα0 is a unit for Aα0 . The

converse is clear.)

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26 Chapter 3

The C∗-algebra A is called the direct sum of the C∗-algebras Aα. Ifα runs over Λ and Aα = C, for each α ∈ Λ, then A = `∞(Λ).

Denition 3.4. An element x in a C∗-algebra A is called self-adjoint (or sym-

metric or hermitian) if x∗ = x. A projection in A is a self-adjoint element p,say, such that p2 = p. An element x ∈ A is called normal if xx∗ = x∗x. Anelement u in a unital C∗-algebra A is called unitary if uu∗ = u∗u = 1l.

Remark 3.5. We can write any x ∈ A as the linear combination

x =1

2(x+ x∗) + i

1

2i(x− x∗).

We see that (x+ x∗)/2 and (x− x∗)/2i are both self-adjoint elements of A.Conversely, if h and k are self-adjoint of A and x = h+ ik, then x∗ = h− ikso that h = 1

2(x+x∗) and k = 12i(x−x

∗). In other words, the decomposition

x = h+ ik with h = h∗, k = k∗ in A is unique. (These are often referred to

as the real and imaginary parts of x, respectively.)

If A is a unital C∗-algebra and x ∈ A, we shall denote by A(x) the

unital C∗-algebra of A generated by x; that is, A(x) is the closure in Aof the ∗-algebra of complex polynomials in x, x∗ and 1l. Clearly, A(x) is

commutative if and only if x is normal.

Proposition 3.6. Let A be a unital C∗-algebra, and let h ∈ A be self-adjoint.

Then σA(h) ⊂ R.

Proof. Suppose that h ∈ A is self-adjoint, and consider the commutative

unital C∗-algebra A(h). For t ∈ R, put

ut = eith ≡∞∑n=0

(it)n

n!hn.

(the series converges in A(h) since A(h) is complete). Then by the conti-

nuity of the involution ∗ , we see that

u∗t = limn→∞

( n∑k=0

(it)k

k!hk)∗ = lim

n→∞

n∑k=0

,(−it)k

k!hk

= u−t .

By multiplying the series (as in the complex case), we see that

u∗tut = u−tut = u0 = 1l .

Hence 1 = ‖u∗tut‖ = ‖ut‖2, and so ‖ut‖ = 1 for all t ∈ R.Let ` ∈ SpA(h). Then, since ` is continuous,

`(ut) = `

( ∞∑n=0

(it)n

n!hn)

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C∗-algebras 27

=

∞∑n=0

(it)n

n!`(h)n

= eit`(h).

Since ‖`‖ = 1, we have |`(ut)| ≤ ‖ut‖ = 1, and hence |eit`(h)| ≤ 1 for all

t ∈ R. It follows that `(h) ∈ R. This holds for all ` ∈ SpA(h) and so h is

real-valued on SpA(h). But σA(h)(h) = ran h and therefore σA(h)(h) ⊂ R.Now A(h) ⊆ A and so σA(h) ⊆ σA(h)(h) and we conclude that σA(h) ⊂ R.

Theorem 3.7. (Gelfand-Naimark) Suppose that A is a commutative unital

Banach ∗-algebra. The Gelfand transform : A → C(SpA) is an isometric∗-isomorphism if and only if A is a C∗-algebra.

Proof. If is an isometric ∗-isomorphism, then for any x ∈ A,

‖x∗x‖ = ‖x∗x‖∞ = ‖(x∗)x‖∞= ‖ x x ‖∞ = ‖ |x |2 ‖∞= ‖ x ‖2∞ = ‖x‖2

which is precisely the C∗-property (v). So A is a C∗-algebra. (Indeed,

C(SpA) is a C∗-algebra, so if A is isometrically ∗-isomorphic to C(SpA),then A must also be a C∗-algebra.)

Conversely, suppose that A is a C∗-algebra. Then for any h = h∗ ∈ A,we know that σA(h) ⊂ R; that is, h is real-valued. For any x ∈ A, writex = (x+ x∗)/2 + i(x− x∗)/2i. Then

(x∗)(`) = `(x∗) = `(x+ x∗

2− i(x− x

∗)

2i

)= `(x+ x∗

2

)− i `

(x− x∗2i

)=

(`(x+ x∗

2

)+ i `

(x− x∗2i

))= `(x)

= x(`),

using the fact that `(h) is real if h is symmetric. It follows that is a∗-homomorphism.

To show that is isometric (and hence also an injection) consider again

h = h∗ ∈ A. Then, by the C∗-property of the norm ‖h‖2 = ‖h2‖ and so

‖h‖2n = ‖h2n‖. Therefore

‖h‖∞ = r(h) = limn→∞

‖h2n‖1/2n = ‖h‖

and so is isometric on self-adjoint elements. For any x ∈ A we have

‖ x ‖2∞ = ‖ x x ‖∞

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28 Chapter 3

= ‖ x∗x ‖∞ , since is a ∗-homomorphism,

= ‖x∗x‖ , since x∗x is self-adjoint,

= ‖x‖2 , by the C∗-property of the norm.

Hence is isometric.

We must now show that is surjective. To see this, we note that A is

complete and since is isometric it follows that ran is closed in C(SpA).Now, is a ∗-homomorphism and so ran is a closed ∗-subalgebra of C(SpA)which contains the constant function 1 = 1l . Furthermore, if `1 6= `2, then,by denition, there is x ∈ A such that `1(x) 6= `2(x), that is, x(`1) 6= x(`2).Thus ran separates points of SpA. It follows from the Stone-Weierstrass

theorem that ran = C(SpA).

Remark 3.8. Let A be a unital C∗-algebra, and let x ∈ A be normal. Then

A(x) is commutative and so A(x) ' C(SpA(x)). Let f : σA(x) → C be

continuous. Since ran x = σA(x), it follows that f x : SpA(x) → C is

well-dened and is continuous; ie, f x ∈ C(SpA(x)). Hence there exists

y ∈ A(x) ⊆ A such that y = f x, and y is unique since is an isomorphism.

We write y = f(x). In this way, we can dene f(x) as an element of A for

any normal (and, in particular, for any self-adjoint) element x ∈ A and any

function f continuous on the spectrum of x. If f is a polynomial, then f(x)is the obvious polynomial in x.

We have seen that the spectrum of an element can depend on the Banach

algebra it is considered to belong to. The following is the key result telling

us that this is not so for C∗-algebras.

Theorem 3.9. Let A be a unital C∗-algebra and suppose that x ∈ A is in-

vertible. Then x−1 belongs to the C∗-subalgebra of A generated by 1l, x and

x∗ (ie, the closure in A of the set of complex polynomials in 1l, x, x∗).

Proof. Suppose rst that x = x∗, and letA denote the unital C∗-algebra gen-erated by x, and B that generated by x and x−1. Evidently B is commutative

and A ⊆ B ⊆ A. We know that the Gelfand transform : B → B = C(SpB)is an (isometric ∗-) isomorphism and, since A is a C∗-subalgebra of B, itfollows that A (the range of A under ) is a C∗-subalgebra of B.

Let `1, `2 ∈ SpB and suppose that `1(x) = `2(x). For any ` ∈ SpB,`(xx−1) = `(x)`(x−1) = `(1l) = 1. Hence `1(x

−1) = `1(x)−1 = `2(x)−1 =`2(x

−1). Since B is generated by x and x−1, we deduce that `1 = `2.Thus, if `1 6= `2 then `1(x) 6= `2(x) or x(`1) 6= x(`2). That is, A separates

points of SpB. By the Stone-Weierstrass theorem, it follows that A = B and

so A = B and therefore x−1 ∈ A.Now let x ∈ A be arbitrary with inverse x−1. Then x∗x is invertible with

inverse x−1(x−1)∗. But x∗x is self-adjoint and so x−1(x−1)∗ belongs to the

C∗-algebra generated by 1l and x∗x which is contained in that generated by

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1l, x, and x∗. But then x−1 = x−1(x−1)∗x∗ = (x∗x)−1x∗ is also contained in

this last algebra.

Corollary 3.10. Let A ⊆ B be unital C∗-algebras with the same unit, and let

x ∈ A. Then σA(x) = σB(x).

Proof. Suppose that x− λ1l is invertible in B. Then this inverse belongs to

the C∗-algebra generated by 1l, x− λ1l and (x− λ1l)∗, which is contained in

A. Hence x−λ1l is invertible in A. It follows that C\σB(x) ⊆ C\σA(x) andso σB(x) ⊇ σA(x). But A ⊆ B implies that σA(x) ⊇ σB(x) and therefore we

have σA(x) = σB(x).

Example 3.11. Let A be the C∗-subalgebra ofM2(C) consisting of those 2×2complex matrices of the form ( α 0

0 0 ), α ∈ C, and let B be the C∗-subalgebraconsisting of those matrices of the form

(α 00 β

), α, β ∈ C. Then A ⊂ B, and

A and B are both unital with units given by 1lA = ( 1 00 0 ) and 1lB = ( 1 0

0 1 ).Note that 1lA ∈ B but 1lA 6= 1lB in B. Now if a = ( α 0

0 0 ) ∈ A, then

σA(a) = α, but σB(a) = 0, α. Evidently σA(a) 6= σB(a). In fact, we

have σA(a) ⊂ σB(a) despite the fact that A ⊂ B.

Theorem 3.12. Let A be a unital C∗-algebra generated by a single normal

element h. Then there is an isometric ∗-isomorphism between A and the

algebra of continuous functions on σA(h) which maps polynomials in h to

the same polynomial on σA(h).

Proof. A is a commutative C∗-algebra and is isomorphic as a C∗-algebra to

C(SpA) via the Gelfand transform : A → C(SpA). On the other hand,

h : SpA→ σA(h) is a homeomorphism. Dene α : C(SpA)→ C(σA(h)) byα(f) = f h−1 so that α(f)(λ) = f h−1(λ)) and, in particular, α(h)(λ) = λfor λ ∈ σA(h). Then α is an isometric ∗-isomorphism from C(SpA) onto

C(σA(h)). Hence α : A→ C(σA(h)) is an isometric ∗-isomorphism.

Let p be a polynomial. Then(α (p(h))

)(λ) = α(p(h))(λ)

= p(α(h))(λ)

= p(α(h)(λ))

= p(λ)

for any λ ∈ σA(h).

Theorem 3.13. Let Ω be a compact Hausdor space, and let A be the com-

mutative unital C∗-algebra C(Ω). Then SpA ' Ω.

Proof. For each ω in Ω, dene ϕω : A → C by ϕω(a) = a(ω), a ∈ A. Then,clearly, ϕω ∈ SpA. Moreover, if ϕω1 = ϕω2 , then a(ω1) = a(ω2) for all a ∈ A.Since A = C(Ω) separates points of Ω, we have ω1 = ω2. Thus ω 7→ ϕω is

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30 Chapter 3

one-one, i.e., it is an identication of Ω as a subset of SpA. We shall show

that this mapping is onto SpA.

To see this, let ` ∈ SpA, and suppose that ` is not of the form ϕω for

any ω ∈ Ω. Then ` − ϕω 6= 0 for all ω ∈ Ω. This means that for each

ω ∈ Ω there is some element aω ∈ A such that `(aω)− ϕω(aω) 6= 0; that is,`(aω) 6= aω(ω). Put bω = aω − `(aω)1l. Then bω 6= 0, indeed, bω(ω) 6= 0, but`(bω) = 0.

Now, bω is continuous on Ω, and so there is a neighbourhood Nω of ωsuch that bω does not vanish on Nω. As ω varies over Ω, we obtain an open

cover Nω : ω ∈ Ω of Ω. Since Ω is compact, there is a nite subcover

Nω1 , . . . , Nωk, say. Put

x = |bω1 |2 + · · ·+ |bωk |2.

Then x ∈ A = C(Ω), and x(ω) > 0 for all ω ∈ Ω. Furthermore,

`(x) = `(b∗ω1bω1) + · · ·+ `(b∗ωkbωk)

= `(b∗ω1)`(bω1) + · · ·+ `(b∗ωk)`(bωk)

= 0

since `(bωj ) = 0 for each j = 1, . . . , k. Thus, x ∈ ker ` and x > 0 on

Ω. But x > 0 implies that x−1 exists in A = C(Ω), and therefore `(1l) =`(x)`(x−1) = 0, which is impossible. We deduce that there is some ω0 ∈ Ωsuch that ` = ϕω0 . Hence the map ϕ : ω 7→ ϕω is one-one and onto SpA.

It remains to show that this is a homeomorphism. Since both Ω and

SpA are compact, it is enough to show that ϕ is continuous (for then ϕ−1

is automatically also continuous). Let U ⊆ SpA be a non-empty open

set and suppose that ω0 ∈ ϕ−1(U), so that ϕω0 ∈ U . Since U is open,

there is ε > 0 and a nite set S ⊂ A such that the w∗-neighbourhoodN(ϕω0 : S, ε) ⊆ U . Each x ∈ S is continuous at ω0 and so there are open

neighbourhoods Vx of ω0 in Ω such that |x(ω) − x(ω0)| < ε for all ω ∈ Vx.Put V =

⋂x∈S Vx. Then V is open in Ω and ω0 ∈ V . For any ω ∈ V , we

have |x(ω)− x(ω0)| < ε, for all x ∈ S, that is, |ϕω(x)− ϕω0(x)| < ε, for allx ∈ S. Hence ϕ(V ) ⊆ N(ϕω0 : S, ε) and thus ω0 ∈ V ⊆ ϕ−1(N(ϕω0 : S, ε))and it follows that ϕ is continuous.

The continuity of ϕ can also be easily seen using nets. Indeed, suppose

that ωα → ω in Ω. Then, for each x ∈ A = C(Ω), we have ϕωα(x) = x(ωα)→x(ω) since x is a continuous function on Ω. But x(ω) = ϕω(x), and so we

conclude that ϕωα → ϕω with respect to the w∗-topology on SpA.

We shall now consider the problem of giving a unit to a C∗-algebra whenone does not already exist. Let A be a C∗-algebra, and let B(A) denote

the Banach algebra of bounded operators on A. Let A denote the normed

subalgebra of B(A) consisting of those elements of the form La + α1l, with

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a ∈ A and α ∈ C, where La is the operator x 7→ ax, x ∈ A, and 1l is the unit

operator in B(A). Then ‖Lax‖ = ‖ax‖ ≤ ‖a‖‖x‖, so we have ‖La‖ ≤ ‖a‖.The map a 7→ La from A into A is linear, and since ab 7→ Lab = LaLb,

we see that it is an algebra homomorphism. Furthermore,

‖Laa∗‖ = ‖aa∗‖ = ‖a∗‖2

= ‖a‖2 = ‖a‖‖a∗‖

which, together with the inequality ‖La‖ ≤ ‖a‖, implies that ‖La‖ = ‖a‖.Thus the map a 7→ La is isometric and, in particular, injective.

Dene an involution on A by (La + α1l)∗ = La∗ + α 1l. Then the map

a 7→ La of A into A is an isometric ∗-homomorphism.

If A has a unit, then Lα1l = α1l, so that every element of A has the form

La for some a ∈ A. In this case, a 7→ La is an isometric ∗-isomorphism of Aonto A (and so A is also a C∗-algebra).

If A does not have a unit, then clearly, identifying A with La : a ∈ A,we see that A is a ∗-subalgebra of A with codimension 1.

We claim that A is a unital C∗-algebra. We must show that

‖(La + α1l)∗(La + α1l)‖ = ‖La + α1l‖2

and that A is complete.

For α ∈ C, and a, x ∈ A, we have

‖(La + α1l)x‖2 = ‖ax+ αx‖2

= ‖(ax+ αx)∗(ax+ αx)‖= ‖(x∗a∗ + αx∗)(ax+ αx)‖= ‖x∗(La + α1l)∗(La + α1l)x‖≤ ‖x∗‖ ‖(La + α1l)∗(La + α1l)‖ ‖x‖= ‖(La + α1l)∗(La + α1l)‖ ‖x‖2.

Therefore,

‖La + α1l‖2 ≤ ‖(La + α1l)∗(La + α1l)‖ (∗)

It follows that

‖La + α1l‖2 ≤ ‖(La + α1l)∗(La + α1l)‖≤ ‖(La + α1l)∗‖ ‖La + α1l‖.

We observe now that if ‖La + α1l‖ = 0, then La + α1l = 0 and therefore

ax = −αx for all x ∈ A. If α were non-zero, this would imply that −a/α is

a left unit for A, and that −a∗/α is a right unit for A, which would mean

that A has a unit. Since A does not have a unit, we must have α = 0. But

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32 Chapter 3

then La = 0 and so it follows that a = 0 (since a 7→ La is injective). The

last inequality above therefore implies that

‖La + α1l‖ ≤ ‖(La + α1l)∗‖= ‖La∗ + α 1l‖.

Replacing a by a∗ and α by α, we deduce that

‖La + α1l‖ = ‖(La + α1l)∗‖.

This, together with the inequality (∗), gives

‖La + α1l‖2 ≤ ‖La + α1l∗La + α1l‖ ≤ ‖La + α1l∗‖ ‖La + α1l‖= ‖La + α1l‖2

and so ‖La + α1l∗La + α1l‖ = ‖La + α1l‖2, which is the C∗-property.

It remains to show that A is complete. To see this, let us dene φ : A→ Cto be the map φ(La+α1l) = α. Note that φ is a well-dened linear functional

on A because La + α1l = Lb + β1l implies that La−b + (α − β)1l = 0 and

hence, as above, a = b and α = β. The kernel of φ is precisely the set

La : a ∈ A. Now, A is complete, and a 7→ La is isometric, and hence

the range La : a ∈ A of this mapping is a closed subset of B(A), i.e., thekernel of φ is a closed subset of B(A). It follows that φ is a continuous linear

functional on A.

Suppose that (Lan + αn1l) is a Cauchy sequence in A. Then, evidently,

(φ(Lan +αn1l)) is a Cauchy sequence in C, that is, (αn) is a Cauchy sequencein C, and so converges to α, say. Now

‖Lan − Lam‖ ≤ ‖(Lan − αn1l)− (Lam − αm1l)‖+ ‖αn1l − αm1l‖︸ ︷︷ ︸=|αn−αm|

and so (Lan) is a Cauchy sequence in B(A). But ‖an − am‖ = ‖Lan − Lam‖implies that (an) is a Cauchy sequence in A. Hence there is a ∈ A such that

an → a. It follows that Lan → La and αn1l → α1l in A, which implies that

Lan + αn1l → La + α1l in A, and we deduce that A is complete.

Identifying A with La : a ∈ A, via the isometric ∗-isomorphism a 7→ Laallows us to regard A as a C∗-subalgebra (and also a closed two-sided ∗-ideal)in A, with codimension 1. We have thus proved the following theorem.

Theorem 3.14. Let A be a C∗-algebra without a unit. Then A is isometrically∗-isomorphic to a C∗-subalgebra of codimension 1 in a unital C∗-algebra.

From the construction, we see that A is just A ⊕ C with the involution

a ⊕ α 7→ a∗ ⊕ α, multiplication (a ⊕ α)(b ⊕ β) = (ab + αb + βa) ⊕ αβ and

norm ‖(a ⊕ α)‖ = sup‖ax + αx‖ : x ∈ A, ‖x‖ ≤ 1. The unit is 0 ⊕ 1.

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(Note that if A has a unit, then we saw that A = A and so, in this case, Ais not the direct sum A⊕ C.)

By virtue of this theorem, it is often possible to assume that there is a

unit in a C∗-algebra under consideration. This may be undesirable, however;

for example, the C∗-algebra of compact operators on an innite-dimensional

Hilbert space does not have a unit, and one may not be willing to step outside

the compact operators for the privilege.

Theorem 3.15. The norm on a C∗-algebra is unique.

Proof. Suppose that ‖ · ‖ and ||| · ||| are two norms on A with respect to

each of which A is a C∗-algebra. Adjoin a unit to A if it does not have one

(note that A, as a set, is just the set of pairs (a, λ) with a ∈ A and λ ∈ C,and so is dened independently of the norm on A. Let h = h∗ ∈ A. Then

‖h‖ = ‖h‖∞ = sup|λ| : λ ∈ σ(h)= |||h|||

since inverses are dened algebraically (independently of the norm).

In general, for any x ∈ A,

‖x‖2 = ‖x∗x‖ = |||x∗x||| = |||x|||2

=⇒ ‖ · ‖ = ||| · |||.

Remark 3.16. This result also follows from the observation that for h = h∗ ∈A, ‖h‖ = sup|`(h)| : ` ∈ Sp(A), where A is the commutative unital C∗-algebra generated by h if A is unital, otherwise it is the commutative unital

C∗-algebra obtained by adjoining a unit to the commutative C∗-algebra gen-erated (in A) by h. The character space of A is dened without reference to

the norm.

Note that because of the uniqueness of the norm, we can assert that for

any element a in a non-unital C∗-algebra A, the unital C∗-algebra generatedby a in A is the same as the C∗-algebra obtained by adjoining a unit to the

C∗-algebra generated by a in A.

Example 3.17. Consider K(H), the C∗-algebra of compact operators on the

Hilbert space H, where we suppose that H is innite-dimensional. Let Abe the unital ∗-subalgebra of B(H) consisting of those operators of the form

a + α1l, with a ∈ K(H), α ∈ C and where 1l denotes the identity in B(H).

Notice that a+ α1l = 0 if and only if a = 0 and α = 0, so that A and K(H)are algebraically ∗-isomorphic. According to the preceeding discussion, A is

a C∗-algebra when equipped with the norm

|||a+ α1l ||| = sup‖ax+ αx‖ : x ∈ K(H), ‖x‖ ≤ 1 .

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34 Chapter 3

With this norm, A is isomorphic to K(H). The question is whether or not

A and K(H) are isomorphic as C∗-algebras when A is equipped with the

operator norm inherited from B(H). If we knew that A were a C∗-algebrawith respect to the operator norm then this would follow from the uniqueness

of the norm in a C∗-algebra. To show this, we must show that A is complete

with respect to the norm in B(H). This can be established by the same

argument as that used above to show the completeness of A. This done, we

conclude that A is precisely K(H), the C∗-algebra obtained by adjoining a

unit to the C∗-algebra K(H).Alternatively, we can see this by showing directly that the operator norm

on A coincides with the above C∗-norm. To see this, let ξ be a unit vector

in H, and let p : H → H be the one-dimensional (orthogonal) projection in

B(H) with pξ = ξ. Then p ∈ K(H) and ‖p‖ = 1 so that

‖(a+ α1l)ξ‖ = ‖aξ + αξ‖ = ‖(ap+ αp)ξ‖≤ ‖ap+ αp‖≤ |||a+ α1l ||| .

Taking the supremum over all such ξ ∈ H, we obtain

‖a+ α1l‖ ≤ |||a+ α1l ||| .

On the other hand, for any x ∈ K(H) with ‖x‖ ≤ 1,

‖ax+ αx‖ ≤ ‖a+ α1l‖‖x‖≤ ‖a+ α1l‖

and so, taking the supremum over all such x,

|||a+ α||| ≤ ‖a+ α1l‖ .

We conclude that the norm on A dened above coincides with the operator

norm induced from B(H), and therefore A = K(H) +C1l is a C∗-subalgebraof B(H), and is the C∗-algebra, K(H), obtained by adjoining a unit to K(H).

We have seen that a commutative, unital C∗-algebra A is isometrically ∗-isomorphic to the C∗-algebra of continuous functions on a compact Hausdor

space, namely, C(SpA), where SpA is the space of characters on A. Supposenow that A is a commutative C∗-algebra without a unit. Evidently, A is

a commutative unital C∗-algebra and so is isomorphic to the C∗-algebraof continuous functions on its character space, Sp A. It is clear that any

character on A extends uniquely to a character on A (by simply assigning

it the value 1 on 1l, the unit in A). Conversely, every character on A denes,

by restriction, a complex-valued homomorphism on A. Such a restriction

Department of Mathematics

C∗-algebras 35

will also be a character provided it is non-zero on A. In other words, we can

say that A has one more character than A, that being the character κ0, say,given by κ0(a) = 0 for all a ∈ A ⊂ A (and, of course, κ0(1l) = 1). This

observation leads to the following result which complements 3.7.

Theorem 3.18. Let A be a commutative C∗-algebra without a unit. Then

there is a locally compact, non-compact, Hausdor space X such that A is

isometrically ∗-isomorphic to C0(X), the C∗-algebra of continuous complex-

valued functions on X vanishing at innity.

Proof. Let K = Sp A. Then K is a compact Hausdor space and A ' C(K),by 3.7. Thus A is isometrically ∗-isomorphic to a C∗-subalgebra of C(K) viathe Gelfand transform on A. Let κ0 ∈ K be the character on A as above;

κ0(a) =

0, for a ∈ A ⊂ A1, a = 1l .

Then, for any a ∈ A, a(κ0) = 0, so that the image of A under consists

of functions in C(K) which vanish at κ0 ∈ K. Conversely, suppose that

f ∈ C(K) is such that f(κ0) = 0. Let x ∈ A be such that x = f . Now x can

be written as x = a+ µ1l, for some a ∈ A and µ ∈ C, and we have

f(κ0) = 0 =⇒ x(κ0) = 0

=⇒ κ0(x) = 0

=⇒ κ0(a) + µκ0(1l) = 0

=⇒ µ = 0, since κ0(a) = 0, for a ∈ A.

Thus x ∈ A, and we see that the Gelfand transform on A maps A onto the

subalgebra of C(K) consisting of those functions which vanish on κ0. (We

could also see this by noting that the kernel of κ0 is a maximal proper ideal

in A which contains A, and hence must equal A, since A itself is a maximal

proper ideal. But the kernel of κ0 acting on C(K), via the Gelfand transform,

is precisely the set of functions above.) Let X = K \κ0. Then X is locally

compact and the map g 7→ g X denes an isometric ∗-isomorphism between

g ∈ C(K) : g(κ0) = 0 and C0(X). Hence A ' C0(X).It remains to verify that X is not compact. If X were compact, then κ0

would be an isolated point of K and the element e ∈ A ⊂ A corresponding

to the continuous function e(κ) =

0, κ = κ0

1, otherwisewould be a unit for A,

contrary to hypothesis. Thus X is not compact.

King's College London

36 Chapter 3

Department of Mathematics

Chapter 4

The Spectral Theorem

We shall present several formulations of the spectral theorem. Each realises

operators as multiplication operators in some sense. This corresponds to

the concept of diagonalisation of matrices.

Theorem 4.1. For any bounded normal operator a on a Hilbert space H there

exists a family µα of real regular Borel measures on σ(a), the spectrum of

a, such that H is unitarily equivalent to⊕

α L2(σ(a), dµα) and a is unitarily

equivalent to multiplication by λ, λ ∈ σ(a); ie, if ζ ∈ H corresponds to

f ∈⊕

α L2(σ(a), dµα), then aζ corresponds to the map λ 7→ λf(λ), for

λ ∈ σ(a).

Proof. Let A be the commutative unital C∗-subalgebra of B(H) generated

by a. Then we have seen (theorem 3.12) that the map ψ : x 7→ x(a (·)−1), de-nes an isometric ∗-isomorphism from A onto C(σ(a)) such that ψ(ax)(λ) =λψ(x)(λ) for x ∈ A and λ ∈ σ(a). Let φ : C(σ(a)) → A denote the inverse

of the map ψ.

Suppose rst that there is a vector ξ ∈ H such that Aξ is dense in H(such a vector ξ is called a cyclic vector for A), and dene µξ to be the

map

f 7→ µξ(f) = (φ(f)ξ, ξ)

for f ∈ C(σ(a)). Then µξ is a continuous positive linear functional on

C(σ(a)). By the Riesz-Markov theorem, there is a (regular Borel) measure

µξ on σ(a) such that

µξ(f) =

∫σ(a)

f dµξ .

Dene u : C(σ(a))→ H by uf = φ(f)ξ. Then

‖uf‖2 = (φ(f)∗φ(f)ξ, ξ) = (φ(f∗f)ξ, ξ)

=

∫σ(a)|f |2dµξ

= ‖f‖2L2(σ(a),dµξ).

37

38 Chapter 4

Hence u is isometric with a dense range (since ξ is assumed cyclic) and with

a dense domain of denition in L2(σ(a), dµξ), namely C(σ(a)). Therefore uextends uniquely to dene a unitary operator : L2(σ(a), dµξ)→ H.

For any f ∈ C(σ(a)), we have

(u−1a u f)(λ) = (u−1aφ(f)ξ)(λ)

=(u−1φ(φ−1(a)f)ξ

)(λ), since φ(φ−1(a)f) = aφ(f),

= (φ−1(a)f)(λ)

= φ−1(a)(λ) f(λ)

= λ f(λ), by the denition of φ−1.

By continuity, this holds for any f ∈ L2(σ(a), dµξ) and gives the required

unitary equivalence.

If ξ ∈ H does not exist as above (ie, if there is no cyclic vector ξ), then,by Zorn's lemma, there is a family of orthogonal subspaces Hα in H, with⊕

αHα = H, and vectors ξα ∈ Hα such that Aξα is dense in Hα.As above, we construct uα : L2(σ(a), dµξα) → Hα, for each α. Then

u =⊕

α uα gives the required unitary equivalence.

The more conventional form of the spectral theorem is the following.

Theorem 4.2. Let a be a bounded self-adjoint operator on a Hilbert space H.Then there is a family eλ : λ ∈ R of projections in H satisfying:

(i) eλ is a strong limit of polynomials in a,

(ii) eλeµ = eµ if µ ≤ λ,

(iii) s-limε↓0 eλ+ε = eλ, s-limλ→−∞ eλ = 0, s-limλ→∞ eλ = 1l,

(iv) a =

∫Rλ deλ = lim

ε↓0

∫ ‖a‖−‖a‖−ε

λ deλ, where the integral is a norm

convergent Stieltjes integral.

Moreover, the family eλ is uniquely determined by (ii), (iii) and (iv).

Proof. To construct such a family eλ, let A be the (commutative) unital

C∗-algebra generated by a. Then A ' C(K), via the Gelfand transform ,where K = SpA, and a(·) is real-valued, since a is self-adjoint. For given

λ ∈ R, dene a function pλ(·) on K by

pλ(·) = χκ∈K:a(κ)≤λ(·) ,

i e, pλ(κ) = 1 if a(κ) ≤ λ, otherwise pλ(κ) = 0. Write Kλ = κ ∈ K : a(κ) ≤λ. Since a is continuous, it follows that Kλ is closed in K.

Department of Mathematics

The Spectral Theorem 39

Let ζ(·) be such that ζ ∈ C(K), 0 ≤ ζ ≤ 1, ζ(κ) = 1 for all κ ∈ Kλ, and

0 ≤ ζ(κ) < 1 for all κ /∈ Kλ. (Such a ζ exists by Urysohn's lemma.) Clearly

ζn(κ)→ pλ(κ) for each κ ∈ K, as n→∞.

Let ξ ∈ H. The map x 7→ (xξ, ξ), x ∈ A, denes a positive linear

functional on C(K), and so, by the Riesz-Markov representation theorem,

there is a regular Borel measure µξ on K such that

(xξ, ξ) =

∫Kx(κ) dµξ(κ) .

Now, by the dominated convergence theorem, we see that ζn → pλ in

L2(K, dµξ). In particular, (ζn) is L2-Cauchy. Since ζ ∈ C(K), there is

y ∈ A such that ζ = y, and so ζn = (yn) . Note that ‖y‖ ≤ 1. We have

‖ζn − ζm‖2L2(K,dµξ)=

∫K|yn − ym|2 dµξ(κ)

= ((yn − ym)∗(yn − ym)ξ, ξ)

= ‖(yn − ym)ξ‖2

and so we see that (ynξ) is a Cauchy sequence in H, for each ξ ∈ H. It

follows that s-limn→∞ yn exists, and denes a bounded operator which we

denote by eλ. Since ζ is real-valued, each yn is self-adjoint, and so eλ = e∗λ.Furthermore, since (‖yn‖) is bounded (by 1) we have

e2λ = s-limn→∞ ynyn = s-limn→∞ y

2n = eλ.

Thus eλ is a projection on H for each λ ∈ R. Moreover, eλ is a strong limit

of polynomials in a because it is a strong limit of a sequence in A, and each

element of A is a norm limit of polynomials in a. This proves (i).Note that if ζ ′ is another element of C(K) such that 0 ≤ ζ ′(κ) < 1, for

κ /∈ Kλ, and ζ′(κ) = 1, for κ ∈ K, then we repeat the proof to obtain y′ ∈ A

with y′n → e′λ strongly as n→∞. But both (yn) and (y′n) converge to pλin L2 and so we can consider the combined sequence

y′′n =

yn, n even

y′n, n odd.

This is L2-Cauchy, and, as before, there exists e′′λ = s-lim y′′n. Hence e′′λ =e′λ = eλ; i.e., eλ is independent of the choice of ζ subject to its dening

requirements.

To prove (ii), we note that if λ ≤ µ, then pλpµ = pµ. Suppose that

ζλ ∈ C(K) determines eλ and that ζµ ∈ C(K) determines eµ, as above. Thenζµζλ will also give eµ since (ζµζλ)n(κ) → pµ(κ) as n → ∞, for each κ ∈ K.

Hence, with the obvious notation, (yλ = ζλ etc.)

eλeµ = s-limn→∞ ynλy

nµ

King's College London

40 Chapter 4

= s-limn→∞ ynµ , by the previous observation

= eµ.

This proves (ii).

To prove (iii): for ξ ∈ H, we have

‖(eλ+ε − eλ)ξ‖2 = limn→∞

‖(ynλ+ε − yλ)ξ‖2

= limn→∞

∫K|(ζnλ+ε − ζnλ )(κ)|2dµξ

=

∫K|pλ+ε(κ) − pλ(κ)|2dµξ .

But pλ+ε(κ) → pλ(κ) as ε ↓ 0 for each κ ∈ K, so by the dominated conver-

gence theorem, the integral converges to 0, i e, eλ+ε → eλ strongly as ε ↓ 0.Similarly, one veries that ‖(1l − eλ)ξ‖ → 0 as λ → ∞ and that ‖eλξ‖ → 0as λ → −∞, which completes the proof of (iii). In fact, eλξ = 0 for any

λ < −‖a‖, and eλξ = ξ for all λ > ‖a‖.In order to establish (iv), we rst observe that |a( · )| ≤ ‖a‖. Divide the

interval (−‖a‖ − ε, ‖a‖] into n open-closed intervals, (λj , λj+1], 1 ≤ j ≤ n,of equal length. Then

χκ∈K : a(κ)∈Ij( · ) = (pλj+1− pλj )( · ).

Put sn(κ) =∑n

j=1 λj+1(pλj+1− pλj )(κ). Then it is easy to see that sn → a

uniformly on K as n → ∞. Thus, for given δ > 0, there is N such that for

all n > N , and any ξ ∈ H,∫K|sn − a|2dµξ < δ

∫Kdµξ

= δ(ξ, ξ)

= δ‖ξ‖2.

That is, for n > N , and ξ ∈ H,∥∥∥∥( n∑j=1

λj+1(eλj+1− eλj )− a

)ξ

∥∥∥∥2 < δ ‖ξ‖2

and so we see that the sum converges in norm to a;

a =

∫ ‖a‖‖a‖−ε

λ deλ , ε > 0 arbitrary.

If λ < ‖a‖, then pλ = 0 implies that eλ = 0, and if λ > ‖a‖ then pλ = 1implies that eλ = 1l, so we can write

a =

∫ ∞−∞

λ deλ

which completes the proof of part (iv).

Department of Mathematics

The Spectral Theorem 41

To see that eλ is unique, we rst note that since eλ − eµ is orthogonal

to eβ − eα whenever (µ, λ]∩ (α, β] = ∅ (since µ ≤ λ implies that eλeµ = eµ)it follows that

a2 =

∫λ deλ , a

3 =

∫λ3 deλ , . . . , a

n =

∫λn deλ

(just square the approximating sum etc.). Now let g(·) be the character-

istic function of the interval (−‖a‖ − 1, µ]. Then, for any ξ ∈ H,∫g(λ) d(eλξ, ξ) = (eµξ, ξ) , since eλ = 0 for λ < −‖a‖ ,

= ‖eµξ‖2.

Let (Pn(λ)) be a sequence of polynomials converging pointwise to g(λ) on

(−‖a‖ − 1, ‖a‖], and uniformly bounded on this interval. Then

‖eµξ‖2 =

∫g(λ) d(eλξ, ξ)

= limn

∫Pn(λ) d(eλξ, ξ)

= limn

(Pn(a)ξ, ξ).

Since (Pn(a)ξ, ξ) is dened independently of eµ, we deduce that eµ is, indeed,uniquely determined by a.

Denition 4.3. The projections eλ are called the spectral projections of the

self-adjoint operator a.

From (i) and (iv), we see that an operator b ∈ B(H) commutes with a if

and only if b commutes with all spectral projections of a.Note also that (i) gives strong convergence; eλ need not belong to A.

Indeed, A may contain no non-trivial projections at all; for example, if A is

the unital C∗-algebra generated by the operator of multiplication by x actingon the Hilbert space L2([0, 1]), then, by Weierstrass' theorem, A ' C[0, 1].(Note that the operator norm in A is the function sup-norm in this example.)

Theorem 4.4. Let A be a commutative unital C∗-algebra of operators on a

Hilbert space H. Then there is a compact Hausdor space K and Borel

measures µα on K such that H is unitarily equivalent to⊕

α L2(K, dµα)

and each a ∈ A is unitarily equivalent to multiplication by a continuous

function on K.

Proof. The proof is just as before, withK = SpA, µ =⊕

α µα is the measure

on K given by the vectors ζα ∈ Hα such that Aζα is dense in Hα, and⊕αHα = H. The unitaries uα : C(K)→ Aζα are dened by uαa = aζα, for

a ∈ A.

King's College London

42 Chapter 4

Department of Mathematics

Chapter 5

Positive elements of a C∗-algebra

The notion of positivity plays a fundamental rôle in the development of the

theory of C∗-algebras.

Denition 5.1. An element a in a C∗-algebra A is said to be positive if and

only if a = h2 for some self-adjoint element h ∈ A. We write a ≥ 0. We

write a ≥ b if and only if a− b ≥ 0.

Proposition 5.2. Let A be a C∗-algebra, and let a ∈ A with a = a∗. Then

a ≥ 0 if and only if σ(a) ⊂ [0,∞), where σ(a) is the spectrum of a considered

as an element of A if A has no unit.

Proof. Suppose that σ(a) ⊂ [0,∞). Let A(a) be the unital C∗-subalgebra ofA (or A if A does not have a unit) generated by a. Then A(a) ' C(SpA(a))and ran a = σ(a) ⊂ [0,∞), and so a( · ) ≥ 0 as a function on SpA(a). Let

f ∈ C(SpA(a)) be the positive square root of a( · ). Then there is h = h∗ ∈ Awith h = f and hence h2 = a which implies that h2 = a. That is, a ≥ 0as an element of A(a). To see that, in fact, a ≥ 0 as an element of A, wenote that as a consequence of Weierstrass' theorem, there is a sequence of

polynomials (pn), say, with pn(0) = 0 such that pn(t) →√t, uniformly on

[0, ‖a‖]. Since σ(a) ⊆ [0, ‖a‖], we deduce that

h = lim pn(h2), in C(SpA(a)),

= lim pn(a).

It follows that h = lim pn(a) ∈ A, and we conclude that a ≥ 0, as required.

Conversely, suppose that a ≥ 0. Then a = h2 for some h = h∗ ∈ A.Hence a = a∗ and the unital C∗-algebra A(a) generated by a is commutative.

It follows that σ(a) = ran a = ran h2 ⊂ [0,∞), since h is real-valued.

Corollary 5.3. Let a ∈ A where A is a C∗-algebra without a unit. Then a ≥ 0in A if and only if a ≥ 0 in A.

Proof. This follows immediately from the proposition. If a ≥ 0 in A, thencertainly a ≥ 0 in A. On the other hand, if a ∈ A and a ≥ 0 in A, thenσA

(a) ⊂ [0,∞). By the proposition, a ≥ 0 in A.

43

44 Chapter 5

Corollary 5.4. Suppose that the element a in a C∗-algebra A satises a ≥ 0and a ≤ 0. Then a = 0.

Proof. We have a ≥ 0 and −a ≥ 0, so that σ(a) ⊂ [0,∞) and σ(−a) ⊂[0,∞). This last inclusion is equivalent to σ(a) ⊂ (−∞, 0]. It follows that

σ(a) = 0. But then ran a = 0 (where a is the Gelfand transform of a as

an element of the commutative unital C∗-algebra it generates), and so a = 0and we conclude that a = 0.

Corollary 5.5. Every positive element of a C∗-algebra has a unique positive

square root; i.e., if A is a C∗-algebra and a ∈ A with a ≥ 0, then there is a

unique s ∈ A with s ≥ 0 and s2 = a. Furthermore, if a is invertible, then so

is the square root s.

Proof. The existence of s ∈ A with s ≥ 0 and s2 = a follows from the

construction in the rst part of the proposition. To prove the uniqueness,

suppose that also t ∈ A with t ≥ 0 and t2 = a. Let A(t) be the commutative

unital C∗-algebra generated by t. Then A(a), the C∗-algebra generated by

a, is contained in A(t), so that s, a ∈ A(a) ⊆ A(t). Now A(t) ' C(SpA(t)),via the Gelfand map, and we have a = s 2 = t 2 where s ≥ 0 and t ≥ 0 since

s ≥ 0 and t ≥ 0. It follows that s = t, and therefore s = t.

If a is invertible, then 0 /∈ σ(a), i.e., 0 /∈ ran a. Hence a > 0 on SpA(a).It follows that s > 0 on SpA(a) and so 0 /∈ σ(s), i.e., s is invertible.

Theorem 5.6. Let A be a C∗-algebra and let h, k ∈ A with h ≥ 0 and k ≥ 0.Then h+ k ≥ 0.

Proof. Suppose a = a∗ ∈ A and ‖a‖ ≤ 1. Then a ≥ 0 if and only if a ≥ 0as a function on SpA(a), where A(a) is the commutative unital C∗-algebragenerated by a, i.e., if and only if |1−a( · )| ≤ 1, i.e., if and only if ‖1l−a‖ ≤ 1.(Note that ‖a‖ ≤ 1 if and only if |a( · )| ≤ 1.) We have∥∥∥∥1l − h+ k

‖h‖+ ‖k‖

∥∥∥∥ =‖ ‖h‖+ ‖k‖ − h− k‖

‖h‖+ ‖k‖

≤ ‖‖h‖ − h‖+ ‖‖k‖ − k‖‖h‖+ ‖k‖

=‖h‖‖(1l − h/‖h‖)‖+ ‖k‖‖(1l − k/‖k‖)‖

‖h‖+ ‖k‖≤ 1 since h/‖h‖ ≥ 0 and k/‖k‖ ≥ 0.

Since

∥∥∥∥ h+ k

‖h‖+ ‖k‖

∥∥∥∥ ≤ 1, we deduce thath+ k

‖h‖+ ‖k‖≥ 0, and hence h+ k ≥

0.

Corollary 5.7. If a ≤ b and b ≤ c in a C∗-algebra A, then a ≤ c in A.

Department of Mathematics

Positive elements of a C∗-algebra 45

Proof. We have c− a = c− b+ b− a ≥ 0 since c− b ≥ 0 and b− a ≥ 0.

Theorem 5.8. Let A be a C∗-algebra, and let a ∈ A. Then a ≥ 0 if and only

if a = x∗x for some x ∈ A.

Proof. If a ≥ 0, then a = h2 for some h = h∗ ∈ A.Conversely, suppose that a = x∗x, for some x ∈ A. Then clearly a =

a∗. To show that a is positive we may suppose that A is unital (if not,

we simply work with A). Suppose that a = x∗x is not positive. Then

a ∈ C(SpA(a)) is a non-positive continuous function and so there is `0 ∈Sp(A(a)) such that a(`0) < 0. Thus there is a neighbourhood N of `0 in

SpA(a) such that a(`) < 0 for all ` ∈ N. Let f ∈ C(SpA(a)) be such that fvanishes on the closed set SpA(a) \N, 0 ≤ f ≤ 1 on SpA(a) and f(`0) = 1(such an f exists, by Urysohn's lemma). Then f af ≤ 0 on SpA(a) and

(f af)(`0) < 0.Let b ∈ A(a) be such that b = f . Then b = b∗, and b a b = f af ≤ 0

implies that bab ≤ 0 (i.e., −bab ≥ 0). Furthermore, bab 6= 0 since `0(bab) =(f af)(`0) < 0.

Write xb = h + ik, with h = h∗ = 12(xb + (xb)∗) ∈ A(a) and k = k∗ =

12i(xb− (xb)∗) ∈ A(a). Then

(xb)∗(xb) = (h− ik)(h+ ik)

= h2 + k2 + ihk − ikh

and

(xb)(xb)∗ = h2 + k2 + ikh− ihk.

Adding, we obtain that (xb)∗(xb) + (xb)(xb)∗ = 2(h2 + k2) ≥ 0 since h2 ≥ 0and k2 ≥ 0. But −(xb)∗(xb) = −bx∗xb = −bab ≥ 0, and so (xb)(xb)∗ =2(h2 + k2) + (−(xb)∗(xb)) ≥ 0, being the sum of two positive elements. So

we have that σ((xb)∗(xb)) ⊂ (−∞, 0] and σ((xb)(xb)∗) ⊂ [0,∞). But for

any y ∈ A, the sets σ(y∗y) and σ(yy∗) dier by at most 0, and so we

conclude that σ((xb)∗(xb)) = 0, i.e., σ(bx∗xb) = σ(bab) = 0. But then

ran bab = σ(bab) = 0 and so bab = 0 which implies that bab = 0. This isa contradiction, and we conclude that x∗x ≥ 0.

Corollary 5.9. Let A be a C∗-algebra and let a ∈ A with a ≥ 0, then b∗ab ≥ 0for any b ∈ A.

Proof. This follows immediately from the theorem; if a = x∗x for some

x ∈ A, then b∗ab = b∗x∗xb = (xb)∗(xb) ≥ 0.

Proposition 5.10. Let a, b be self-adjoint elements of a C∗-algebra A.

(i) Suppose that −b ≤ a ≤ b. Then ‖a‖ ≤ ‖b‖.

King's College London

46 Chapter 5

(ii) Suppose that 0 ≤ a ≤ b and a and b are invertible. Then b−1 ≤a−1.

Proof. By adjoining a unit, if necessary, and looking at Gelfand transforms,

we see that −‖b‖ ≤ b( · ) ≤ ‖b‖ on Sp(A(b)) and so b ≤ ‖b‖1l. Hence we have

−‖b‖1l ≤ −b ≤ a ≤ b ≤ ‖b‖1l .

It follows that −‖b‖1l ≤ a ≤ ‖b‖1l and so −‖b‖ ≤ a( · ) ≤ ‖b‖ on Sp(A(a)).Hence ‖a‖ ≤ ‖b‖. This proves (i).

To prove (ii), we note that 0 ≤ a ≤ b implies that b−1/2ab−1/2 ≤ 1l.

Note that the invertibility of a and b requires that A be unital. Let c denoteb−1/2ab−1/2. Then c = c∗, c is invertible and 0 < c( · ) ≤ 1. Hence c−1 ≥ 1on Sp(A(c)) and so c−1 ≥ 1l. It follows that b1/2a−1b1/2 ≥ 1l and therefore

a−1 ≥ b−1/21lb−1/2 = b−1.

Note that if, in fact, 0 ≤ a ≤ b and if a is invertible, then b is also

invertible. To see this, we note that since a is invertible, there is γ ∈ Rwith γ > 0 such that 0 < γ ≤ a on SpA(a), where, as usual, A(a) is the

commutative unital C∗-subalgebra of A generated by a. Hence a ≥ γ1l. Butthen, b − γ1l = (b − a) + (a − γ1l) is the sum of two positive elements of Aand therefore is itself positive. This means that b ≥ γ > 0 on SpA(b), andso b has an inverse in C(Sp(A(b))), which implies that b is invertible.

Proposition 5.11. Let a and b be elements of a unital C∗-algebra A. If 0 ≤a ≤ b, then a(1l + a)−1 ≤ b(1l + b)−1.

Proof. Using Gelfand transforms, it is evident that both (1l + a) and (1l + b)are invertible. We have

0 ≤ a ≤ b=⇒ 1l + a ≤ 1l + b

=⇒ (1l + b)−1 ≤ (1l + a)−1

=⇒ − (1l + a)−1 ≤ −(1l + b)−1

=⇒ 1l − (1l + a)−1 ≤ 1l − (1l + b)−1

i.e., a(1l + a)−1 ≤ b(1l + b)−1.

Proposition 5.12. Let a ∈ A, where A is a unital C∗-algebra. Then a can be

written as a linear combination of

(i) two self-adjoint elements of A,

(ii) four positive elements of A,

(iii) four unitary elements of A.

Proof. (i) a = 12(a+ a∗) + i 12i(a− a

∗).

Department of Mathematics

Positive elements of a C∗-algebra 47

(ii) Let h = h∗ ∈ A. Let |h| denote the positive square root of h2. Then, bylooking at h, we see that |h|±h are both ≥ 0. Write h = 1

2(h+|h|)− 12(|h|−h)

and use (i).

(iii) Let h = h∗ ∈ A. Then h2 ≥ 0. Suppose that ‖h‖ < 1. Then 1l − h2 ≥ 0and so has a positive square root, (1l−h2)1/2. Put u = h+i(1l−h2)1/2. Thenone checks that u∗u = uu∗ = 1l, i.e., u is unitary. Moreover, h = 1

2(u+ u∗).

If ‖h‖ ≥ 1 then consider αh with α =1

2‖h‖. Then, as above, we can

write αh = 12(v + v∗) with v unitary. Thus h = 1

2α(v + v∗). For general

h ∈ A, the result now follows from part (i).

Remark 5.13. Notice that parts (i) and (ii) remain valid even if A has no

unit. This is because the elements a, a∗, h and |h| all belong to A, if a does.

The positive elements h± = |h| ± h in part (ii) satisfy h+h− = 0 as is

readily computed. The decomposition of a self-adjoint element h into two

such positive elements with disjoint support is unique. This is seen as

follows. Suppose that h = h∗ can be written as h = k+ − k− where k± ≥ 0and k+k− = 0. Then we have

|h|2 = h2 = (k+ − k−)2

= k2+ + k2−

= (k+ + k−)2 .

Now, (k+ + k−) is positive and so, by the uniqueness of the positive square

root, we must have (k+ + k−) = |h|. This, together with the equality h =k+ − k−, implies that 2k+ = |h| + h = 2h+ and 2k− = |h| − h = 2h−which establishes the uniqueness. The positive elements h+ = 1

2(|h|+h) andh− = 1

2(|h| − h) are called the positive and negative parts of h.

Example 5.14. Suppose that p is a non-trivial projection in the unital C∗-algebraA. Set h = cosα1l = cosαp + cosαp⊥, where, say, 0 < α < π/2and p⊥ = 1l − p. Clearly h = h∗ and ‖h‖ < 1. Then u = eiα1l and

v = eiαp + e−iαp⊥ are distinct unitary elements of A, but we have h =12(u+ u∗) = 1

2(v + v∗).As another example, let A be the direct sum of C∗-algebras (Aα), and

let u = (uα) ∈ A, with each uα unitary. Let v = (vα) ∈ A be obtained from

u by changing some of the uα's to u∗α. Then u and v are unitary in Aand,

in general, u 6= v. However, u+ u∗ can also be written as v + v∗.

Proposition 5.15. Let A be a C∗-algebra. Then A+, the set of positive ele-

ments of A, is closed in A.

Proof. Let (hn) be a sequence in A+ such that hn → h in A. We must show

that h ∈ A+. First we observe that if h = h∗, h ∈ A, then h ≥ 0 if and only

King's College London

48 Chapter 5

if ‖h− ‖h‖1l‖ ≤ ‖h‖ (in A, if A does not have a unit). Indeed,

‖h− ‖h‖1l‖ = supκ∈SpA(h)

|h(κ)− ‖h‖|

= supκ

(‖h‖ − h(κ)), since ran h ∈ R and ‖h‖ = ‖h‖∞ ≥ h(κ),

≤ ‖h‖ , if and only if h(κ) ≥ 0 for all κ,

i.e., if and only if σ(h) ⊂ [0,∞) if and only if h ≥ 0.

Now, hn = h∗n and therefore h∗ = limh∗n = limhn = h. We also have

‖hn‖ → ‖h‖. By the above remark, ‖hn − ‖hn‖1l‖ ≤ ‖hn‖. Letting n→∞,

we see that ‖h− ‖h‖1l‖ ≤ ‖h‖ and so, again by the above remark, it follows

that h ≥ 0.

Proposition 5.16. Let a, b ∈ A and suppose that ab is self-adjoint. Then

‖ab‖ ≤ ‖ba‖.

Proof. By adjoining a unit, if necessary, we may suppose that A is unital.

Since ab = (ab)∗, it follows that

‖ab‖ = r(ab), the spectral radius of ab (= ‖ab‖∞),= sup|λ| : λ ∈ σ(ab)= sup|λ| : λ ∈ σ(ba), since σ(ab) ∪ 0 = σ(ba) ∪ 0,≤ ‖ba‖,

as required.

Proposition 5.17. Let 0 ≤ a ≤ b be elements of a C∗-algebra A. Then a1/2 ≤b1/2.

Proof. We may suppose that A is unital (if not, we simply consider A).Suppose rst that a and b are both invertible. Then a ≤ b implies that

0 ≤ b−1/2ab−1/2 ≤ 1l. Hence,

‖b−1/4a1/2b−1/4‖2 ≤ ‖b−1/2a1/2‖2, as above,= ‖b−1/2a1/2a1/2b−1/2‖, by the C∗-property,

= ‖b−1/2ab−1/2‖≤ 1.

Therefore ‖b−1/4a1/2b−1/4‖ ≤ 1 and so 0 ≤ b−1/4a1/2b−1/4 ≤ 1l. That is,

a1/2 ≤ b1/2.In general, for any ε > 0, 0 ≤ a ≤ b implies that 0 ≤ ε1l ≤ a+ε1l ≤ b+ε1l,

and a+ε1l and b+ε1l are invertible. So, by the above argument, we see that

(a+ ε1l)1/2 ≤ (b+ ε1l)1/2, that is, (b+ ε1l)1/2 − (a+ ε1l)1/2 ≥ 0.

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Positive elements of a C∗-algebra 49

Now, as ε ↓ 0, a+ ε1l → a in A, and since the map t 7→√t is uniformly

continuous on [0,M ], withM = ‖a‖+1, say, it follows that√a+ ε−

√a→ 0

uniformly on SpA(a), as ε ↓ 0. In other words, ‖√a+ ε −

√a‖∞ → 0 as

ε ↓ 0, i.e., ‖(a+ ε1l)1/2 − a1/2‖ → 0 as ε ↓ 0.Similarly, ‖(b + ε1l)1/2 − b1/2‖ → 0 in A as ε ↓ 0. Hence b1/2 − a1/2 =

limε↓0(b+ ε1l)1/2 − (a+ ε1l)1/2 belongs to A+, since A+ is closed.

Remark 5.18. In general, the inequality a ≤ b does not imply that a2 ≤ b2.

We shall now turn to the discussion of approximate units. Let J be a left

ideal in a unital C∗-algebra A, and let Λ be the collection of nite subsets

of J ordered by set inclusion ⊆. Then Λ is a directed set. For each λ ∈ Λ,dene yλ =

∑x∈λ x

∗x, and set uλ = nλyλ(1l+nλyλ)−1, where nλ denotes thenumber of members of λ. Note that nλyλ ≥ 0 and so (1l+nλyλ) is invertible.We note also that uλ ∈ J , uλ = u∗λ and that 0 ≤ uλ ≤ 1l.

Theorem 5.19. Let J and (uλ) be as above.

(i) If λ ≤ µ, then uλ ≤ uµ.

(ii) For any x ∈ J , x− xuλ → 0 along the directed set Λ.

Proof. (i) If λ ≤ µ, then clearly nλyλ ≤ nµyµ. Hence

uλ = 1l − (1l + nλyλ)−1

≤ 1l − (1l + nµyµ)−1

= uµ.

(ii) Let x ∈ J . Then

‖x− xuλ‖2 = ‖(x− xuλ)∗‖2

= ‖x(1l − uλ)(1l − uλ)x∗‖≤ ‖x(1l − uλ)x∗‖, since (1l − uλ)2 ≤ (1l − uλ),

= ‖x(1l + nλy−1λ x∗‖

≤ ‖x(1l +mx∗x)−1x∗‖

whenever λ has at leastm elements and contains x; since in this case, nλyλ ≥mx∗x,

≤ ‖x(1l +mx∗x)−1‖ ‖x∗‖= ‖(1l +mx∗x)−1x∗x(1l +mx∗x)−1‖1/2‖x∗‖

≤ 1

m1/2‖mx∗x(1l +mx∗x)−1‖1/2‖x∗‖ since ‖(1l +mx∗x)−1‖ ≤ 1,

≤ ‖x∗‖

m1/2

since ‖mx∗x(1l +mx∗x)−1‖ ≤ 1.

King's College London

50 Chapter 5

The properties of the net (uλ) are the basis for the following denition.

Denition 5.20. An approximate unit for a left ideal J in a C∗-algebra A is

a net (uλ) of elements of J , indexed by Λ, say, satisfying

(i) 0 ≤ uλ, ‖uλ‖ ≤ 1 and uλ ≤ uµ whenever λ ≤ µ, λ, µ ∈ Λ.

(ii) For any x ∈ J , ‖x− xuλ‖ → 0 along Λ.

An approximate unit for a right ideal J is a net (uλ) in J satisfying (i)

together with the following obvious modication of (ii).

(ii)′ For any x ∈ J , ‖x− uλx‖ → 0 along Λ.

Evidently, if J is a left ideal, then J∗ = j∗ : j ∈ J is a right ideal of A.Suppose that (uλ) is an approximate unit for J , then, for any x ∈ J∗,

‖x− uλx‖ = ‖(x− uλx)∗‖= ‖x∗ − x∗uλ‖→ 0, since x∗ ∈ J ,

so that (uλ) is an approximate unit for the right ideal J∗.

An approximate unit for a C∗-algebra A is a net (uλ) in A satisfying

condition (i) together with the following.

(ii)′′ For any a ∈ A, ‖a−auλ‖ → 0 along Λ, and hence also (as above)

‖a− uλa‖ → 0 along Λ.

Theorem 5.21. Suppose that J is a left (resp., right) ideal in a C∗-algebra A.Then J has an approximate unit.

Proof. Let J be a left ideal of A. We may assume, without loss of generality,

that A has a unit. (If not, then we can embed A as a two-sided ideal into

A, and then J is a left ideal of A.) An approximate unit is then constructed

as above.

Now if J is a right ideal, then J∗ is a left ideal and so has an approximate

unit, as above. But then we have seen that this is also an approximate unit

for the right ideal J .

Corollary 5.22. Let A be a C∗-algebra. Then A possesses an approximate

unit.

Proof. This follows immediately from the theorem by taking A = J and

noting that A = A∗.

Of course, if A is unital, then we may take uλ to be 1l, for each λ. Thepoint is that even if A is not unital, it still has an approximate unit. The

following results show the usefulness of approximate units.

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Positive elements of a C∗-algebra 51

Theorem 5.23. Let J be a closed two-sided ideal in a C∗-algebra A. Then

J = J∗, i.e., j ∈ J if and only if j∗ ∈ J .

Proof. Let j ∈ J , and let uλ be an approximate unit for J as a left ideal.

Then j = lim juλ and it follows that j∗ = limuλj∗. But uλj

∗ ∈ J since

uλ ∈ J and J is a two-sided ideal, and, since J is closed, we deduce that

j∗ ∈ J .

Theorem 5.24. Let J be a closed two-sided ideal in a C∗-algebra A and let

(uλ) be an approximate unit for J . Then for any x ∈ A

lim ‖x− xuλ‖ = ‖ clx‖

where ‖ clx‖ is the norm of clx in A/J .

Proof. Let x ∈ A. By denition, ‖ clx‖ = infj∈J ‖x + j‖. Put α = ‖ clx‖.Then

α2 = infj∈J‖x+ j‖2 ≤ ‖x− xuλ‖2, since xuλ ∈ J ,

= ‖x(1l − uλ)2x∗‖≤ ‖x(1l − uλ)x∗‖.

Now, x(1l − uλ)x∗ is decreasing in λ and so ‖x(1l − uλ)x∗‖ is decreasing and

hence converges to its inmum, β, say. It follows that α2 ≤ β.On the other hand, for given ε > 0, there is j ∈ J such that ‖x + j‖ <

α+ ε. Then

(α+ ε)2 > ‖x+ j‖2 = ‖(x+ j)(x∗ + j∗)‖≥ ‖(x+ j)(1l − uλ)(x∗ + j∗)‖= ‖x(1l − uλ)x∗ + j(1l − uλ)x∗ + x(1l − uλ)j∗ + j(1l − uλ)j∗‖

for any λ. Now, j(1l − uλ), (1l − uλ)j∗, j(1l − uλ)j∗ → 0. Furthermore,

‖x(1l − uλ)x∗‖ → β. It follows that (α+ ε)2 ≥ β for all ε > 0, and therefore

α2 ≥ β. Thus α2 = β and

α2 ≤ ‖x− xuλ‖2

≤ ‖x(1l − uλ)x∗‖→ β = α2.

Hence α = lim ‖x− xuλ‖, as required.

Theorem 5.25. Suppose that J is a closed two-sided ideal in a C∗-algebra A.Then A/J is a C∗-algebra.

King's College London

52 Chapter 5

Proof. We know that A/J is a Banach algebra. Dene an involution on

A/J by (clx)∗ = clx∗; since J is closed, J = J∗, and this is a well-dened

involution on A/J . Furthermore,

‖(clx)∗‖ = ‖ clx∗‖ = infj∈J‖x∗ + j‖

= infj∈J‖x∗ + j∗‖ = inf

j∈J‖x+ j‖

= ‖ clx‖.

Hence

‖ clx∗ clx‖ ≤ ‖ clx∗‖‖ clx‖= ‖ clx‖2.

We must show that equality holds. However, if (uλ) is an approximate unit

for J (as a left ideal), we have

‖ clx∗ clx‖2 = lim ‖x∗x(1l − uλ)‖2

= lim ‖(1l − uλ)x∗xx∗x(1l − uλ)‖≥ lim ‖(1l − uλ)x∗x(1l − uλ)(1l − uλ)x∗x(1l − uλ)‖= lim ‖(1l − uλ)x∗x(1l − uλ)‖2

= lim ‖x(1l − uλ)‖4

= ‖ clx‖4

by the previous theorem, and the result follows.

Department of Mathematics

Chapter 6

Homomorphisms

We consider now further interplay between the algebraic structure and the

metric structure of a C∗-algebra. In particular, we shall see that homomor-

phisms are necessarily continuous, isomorphisms are isometric and the range

of a homomorphism is a C∗-algebra.

Denition 6.1. Let A and B be unital C∗-algebras. A homomorphism ϕ :A→ B is a linear map satisfying

(i) ϕ(ab) = ϕ(a)ϕ(b) for all a, b ∈ A;

(ii) ϕ(a∗) = ϕ(a)∗ for all a ∈ A;

(iii) ϕ(1lA) = 1lB.

A linear map ϕ : A→ B is said to be order preserving if a ≥ 0 in A implies

that ϕ(a) ≥ 0 in B.

We will only consider unital C∗-algebras in this chapter.

Proposition 6.2. Let ϕ : A → B be a homomorphism. Then ϕ is order

preserving and norm decreasing. In particular, ϕ is continuous.

Proof. Let a ∈ A, a ≥ 0. Then there is x ∈ A with a = x∗x. Hence

ϕ(a) = ϕ(x∗x) = ϕ(x∗)ϕ(x) = ϕ(x)∗ϕ(x) ≥ 0, which shows that ϕ is order

preserving.

Now let a ∈ A with a = a∗. Then |a(κ)| ≤ ‖a‖∞ for all κ ∈ SpA(a)implies that

−‖a‖∞ ≤ a( · ) ≤ ‖a‖∞and so ‖a‖∞1l − a( · ) ≥ 0 and a( · ) + ‖a‖∞ ≥ 0. Hence (using ‖a‖∞ = ‖a‖),‖a‖1l − a ≥ 0 and a+ ‖a‖1l ≥ 0, that is, −‖a‖1l ≤ a ≤ ‖a‖1l. It follows that−‖a‖1l ≤ ϕ(a) ≤ ‖a‖1l since ϕ(1l) = 1l and ϕ is order preserving.

By considering the Gelfand transform of ϕ(a) (= ϕ(a)∗), we see that

−‖a‖1l ≤ ϕ(a) ≤ ‖a‖1l and so ‖ϕ(a)‖∞ = ‖ϕ(a)‖ ≤ ‖a‖ for any a ∈ A with

a = a∗.

53

54 Chapter 6

For any x ∈ A, we have

‖ϕ(x)‖2 = ‖ϕ(x)∗ϕ(x)‖ = ‖ϕ(x∗x)‖≤ ‖x∗x‖= ‖x‖2.

Proposition 6.3. Let ϕ : A → B be a homomorphism, and suppose that ϕis injective. Then ϕ−1 : ϕ(A) → A is order preserving, and ϕ is norm

preserving (i.e., ϕ is isometric).

Proof. Let y ∈ ϕ(A) be such that y ≥ 0. Then there is a unique a ∈ A such

that y = ϕ(a). We must show that ϕ−1(y) = a ≥ 0.

First we shall show that a = a∗. To see this, observe that ϕ(a∗) =ϕ(a)∗ = y∗ = y. Hence a = a∗, since ϕ is 11. Write a = x+ − x−, wherex± ∈ A are the positive and negative parts of a. Then y = ϕ(x+ − x−) =ϕ(x+)− ϕ(x−). We will show that x− = 0.

For any b ∈ B, b∗ϕ(x+)b− b∗ϕ(x−)b = b∗yb. Let b = ϕ(x1/2− ). Then the

above equality becomes

ϕ(x1/2− x+x

1/2−︸ ︷︷ ︸

=0

)− ϕ(x2−) = b∗yb .

Hence −ϕ(x2−) = b∗yb and we deduce that ϕ(x2−) ≤ 0. But ϕ(x2−) =ϕ(x−)2 ≥ 0, since ϕ(x−) is self-adjoint. Thus

ϕ(x2−) = 0

=⇒ ϕ(x−)ϕ(x−) = 0

=⇒ ϕ(x−) = 0, e.g., using the Gelfand map,

=⇒ x− = 0,

since ϕ is 11. Thus a = x+ ≥ 0, as claimed, which shows that ϕ−1 is orderpreserving.

Now, for any x ∈ A, ‖ϕ(x)‖ ≤ ‖x‖, by the previous proposition. On

the other hand, putting y = ϕ(x), we have ϕ−1(y) = x and ‖ϕ−1(y)‖ ≤ ‖y‖again by the previous proposition. Hence ‖x‖ ≤ ‖ϕ(x)‖ ≤ ‖x‖ which gives

the required equality ‖ϕ(x)‖ = ‖x‖, for any x ∈ A.

Corollary 6.4. Let ϕ : A→ B be an injective homomorphism. Then ϕ(A) is

a C∗-subalgebra of B.

Proof. Evidently, ϕ(A) is a ∗-subalgebra of B. Since ϕ is isometric, it follows

that ϕ(A) is closed in B, and hence is complete.

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Homomorphisms 55

Theorem 6.5. Let ϕ : A → B be a homomorphism. Then ϕ(A) is a C∗-subalgebra of B.

Proof. Let J = kerϕ. Then it is clear that J is a closed two-sided ∗-idealof A. Dene ψ : A/J → B by ψ(cl a) = ϕ(a). It is readily seen that ψ is

a homomorphism from A/J into B, and that ψ is 11. Hence ψ(A/J) is a

C∗-subalgebra of B, i.e., ϕ(A) is a C∗-subalgebra of B.

Denition 6.6. A one-one homomorphism of A onto itself is called an auto-

morphism of A. The collection of automorphisms of A is denoted AutA.Evidently, AutA is a group under composition.

For any unitary u ∈ A, the map a 7→ uau∗, a ∈ A, is an automorphism

of A. Automorphisms of this form are said to be inner. In general, not all

automorphisms are inner. Indeed, in a commutative C∗-algebra, only the

identity automorphism is inner.

Denition 6.7. A representation of a C∗-algebra is a pair (H, π) consisting ofa Hilbert space H and a homomorphism π : A→ B(H). The representation(H, π) is said to be faithful if kerπ = 0.

Thus, a faithful representation (H, π) is 11 and so is isometric. Con-

versely, if π is isometric, then, clearly, (H, π) is a faithful representation.

Denition 6.8. A C∗-algebra is said to be simple if it has no non-trivial closed

two-sided ideals.

Remark 6.9. Since the kernel of any homomorphism ϕ : A → B is a closed

two-sided ideal of A, it follows that for a simple C∗-algebra, all homomor-

phisms are 11, and so are isometric. In particular, all representations are

faithful.

King's College London

56 Chapter 6

Department of Mathematics

Chapter 7

States on a C∗-algebra

To begin with, we shall be mainly concerned with unital C∗-algebras. We

will see later that this is no essential loss of generality.

Denition 7.1. A state on a unital C∗-algebra A is a positive linear functional

ω, say, with ω(1l) = 1.That is, ω : A→ C, such that

(i) ω is linear;

(ii) a ∈ A, a ≥ 0 =⇒ ω(a) ≥ 0;

(iii) ω(1l) = 1.

Condition (iii) is really just a normalization condition.

The term state is borrowed from mathematical physics. The observ-

ables of a physical system are represented by self-adjoint elements of a C∗-algebra and the value ω(a) is supposed to be the expected value of the ob-

servable a in the state ω. To know the expected values of the observables

of the system is to know the state of the system.

Examples 7.2.

1. Let A = B(H), and let ω be any map of the form x 7→ (xξ, ξ), whereξ ∈ H has unit norm. Such a state is called a vector state.

2. Let A = B(H), and ω the map x 7→ α(xξ, ξ) + (1 − α)(xη, η), whereξ, η ∈ H have unit norm, and 0 ≤ α ≤ 1.

3. Let A = C(X), where X is a compact Hausdor space, and let ω be

the map f 7→ f(x), where x is any point in X and f ∈ C(X).

Proposition 7.3. Let ϕ : A → C be a positive linear functional on the C∗-algebra A. Then ϕ(a∗) = ϕ(a), for any a ∈ A. In particular, ϕ(h) ∈ Rwhenever h = h∗ ∈ A. Furthermore, ϕ satises Schwarz' inequality

|ϕ(b∗a)|2 ≤ ϕ(a∗a)ϕ(b∗b)

for any a, b ∈ A.

57

58 Chapter 7

Proof. Let a = h+ ik, h = h∗, k = k∗, and write h = h+−h−, k = k+− k−,with h± ≥ 0 and k± ≥ 0. Then

ϕ(a∗) = ϕ(h+ − h− − i(k+ − k−))

= ϕ(h+)− ϕ(h−)− iϕ(k+) + iϕ(k−)

= (ϕ(h+)− ϕ(h−) + iϕ(k+)− iϕ(k−))−

= ϕ(a),

where we have used ϕ(h±) ≥ 0 and ϕ(k±) ≥ 0.Schwarz' inequality follows immediately from the fact that 〈a, b〉 = ϕ(b∗a)

denes a sesquilinear form on A, with 〈a, a〉 ≥ 0.

Note that the existence of a unit is not required in the above argument.

Theorem 7.4. Let ω be a positive linear functional on a unital C∗-algebra A.Then ω is bounded and ‖ω‖ = ω(1l).

Proof. For any h = h∗ ∈ A, we have −‖h‖1l ≤ h ≤ ‖h‖1l, and so the

linearity and positivity of ω give −‖h‖ω(1l) ≤ ω(h) ≤ ‖h‖ω(1l) which yields

|ω(h)| ≤ ω(1l)‖h‖.For general a ∈ A, we use Schwarz' inequality to obtain

|ω(a)|2 = |ω(1l∗a)|2 ≤ ω(1l∗1l)ω(a∗a)

≤ ω(1l)ω(1l)‖a∗a‖ as above, with h = a∗a,

= ω(1l)2‖a‖2.

Thus |ω(a)| ≤ ω(1l)‖a‖, so that ω is bounded and ‖ω‖ ≤ ω(1l). But then we

conclude that ‖ω‖ = ω(1l).

Theorem 7.5. Suppose that ω is a bounded linear functional on a unital C∗-algebra A, satisfying ‖ω‖ = ω(1l). Then ω is positive.

Proof. We shall rst show that if h = h∗ ∈ A then ω(h) ∈ R. To see this,

write ω(h) = α+ iβ, with α, β ∈ R. Then ω(h+ iλ1l) = α+ i(β+λω(1l)) forall λ ∈ R. Since ω(1l) = ‖ω‖, it follows, in particular, that ω(1l) ∈ R, andtherefore |ω(h+ iλ1l)| ≥ |β + λω(1l)|.

On the other hand,

|ω(h+ iλ1l)| ≤ ‖ω‖ ‖h+ iλ1l‖

= ω(1l)(‖h‖2 + λ2

)1/2as is readily seen by using the Gelfand transform. Therefore

|β + λω(1l)|2 ≤ ω(1l)2(‖h‖2 + λ2

)for all λ ∈ R, which is impossible unless β = 0. Hence ω(h) ∈ R, as claimed.

Department of Mathematics

States on a C∗-algebra 59

Now suppose that h ≥ 0, and, without loss of generality, suppose that

‖h‖ ≤ 1. Then

|ω(1l)− ω(h)| = |ω(1l − h)| ≤ ‖ω‖ ‖1l − h‖= ω(1l)‖1l − h‖≤ ω(1l)

which implies that ω(h) ≥ 0.

Remark 7.6. Thus, a state ω on a unital C∗-algebra is a positive linear func-tional with ‖ω‖ = 1, or equivalently, a linear functional with ‖ω‖ = ω(1l) = 1.

Theorem 7.7. Let A ⊆ B be unital C∗-algebras (with the same unit), and let

ω be a state on A. Then ω has an extension to a state on B, i.e., there is a

state ρ on B such that ρ A = ω.

Proof. Since ω is a state on A, we have ‖ω‖ = ω(1l) = 1. By the Hahn-

Banach theorem, there is a continuous linear functional ρ, say, on B, suchthat ‖ρ‖ = ‖ω‖ and ρ A = ω. Hence ρ(1l) = ω(1l) = 1, since 1l ∈ A ⊆ B,and so ‖ρ‖ = ‖ω‖ = ω(1l) = ρ(1l) = 1. Thus ρ (is positive and) is a state on

B.

Theorem 7.8. The set of states on a unital C∗-algebra A separates the points

of A, i.e., for any a, b ∈ A with a 6= b, there is a state ω on A with ω(a) 6=ω(b).

Proof. Let a, b ∈ A with a 6= b be given. Set x = a− b and write x = h+ ikwith h = h∗ and k = k∗. Then either h 6= 0 or k 6= 0 (otherwise x = 0). If ωis a state, then ω(h) and ω(k) are both real, and so ω(x) 6= 0 if and only if

either ω(h) 6= 0 or ω(k) 6= 0. So the theorem is proved if we can show that

for any h = h∗ ∈ A with h 6= 0, there is a state ω with ω(h) 6= 0.To see this we use the identication A(h) ' C(SpA(h)). Since h 6= 0

there is κ′ ∈ SpA(h) such that h(κ′) 6= 0. Dene ρ on A(h) by ρ(a) = a(κ′)for a ∈ A(h). Clearly, ρ is a state on A(h), and so, by the previous theorem,

has an extension to A. Then ω(h) = ρ(h) = h(κ′) 6= 0.

Theorem 7.9. The involution of a C∗-algebra is unique.

Proof. Let A be a C∗-algebra. By considering A instead of A, we may assume

that A has a unit. Let ∗ and † be involutions on A with respect to which Ais a C∗-algebra. Let ω be a state on A. Note that a state ω is any bounded

linear functional on A with ‖ω‖ = ω(1l) = 1, so that the notion of state is

independent of the involution. But then we know that ω is positive with

respect to both ∗ and †. Hence, for any x ∈ A, ω(x∗) = ω(x) = ω(x†), andso ω(x∗ − x†) = 0 for all states ω on A. Since the set of states separates thepoints of A, we deduce that x∗ = x† for any x ∈ A.

King's College London

60 Chapter 7

We now turn to a discussion of the non-unital case. We have seen (7.4)

that a positive linear functional on a unital C∗-algebra is automatically con-

tinuous. This result is true even if there is no unit, as we shall now show.

Theorem 7.10. Suppose that ω is a positive linear functional on a C∗-algebraA. Then ω is bounded.

Proof. We proceed by contradiction; suppose that ω is not bounded. Then

there is a sequence (xn) in A such that ‖xn‖ → 0 but ω(xn) = 1. By

positivity, 7.3, ω(x∗n) = ω(xn) = 1. Setting yn = 12(xn + x∗n), we have

‖yn‖ ≤ ‖xn‖ → 0 and ω(yn) = 1. Now, by 5.12, each yn can be written

as yn = z′n − z′′n, where z′n ≥ 0 and z′′n ≥ 0 are given by z′n = 12(|yn| + yn),

z′′n = 12(|yn| − yn). Thus ‖z′n‖ ≤ ‖yn‖ → 0, and ω(z′n) = ω(yn) + ω(z′′n) ≥ 1,

by positivity of ω. Putting an = ω(z′n)−1z′n, we have an ∈ A, an ≥ 0,‖an‖ → 0 and ω(an) = 1. By passing to a subsequence, if necessary, we may

assume that ‖an‖ ≤ 2−n, for each n ∈ N. Set bn =∑n

k=1 an. Then bn ≥ 0for each n ∈ N, and there is b ∈ A such that bn → b as n→∞. Since A+ is

closed, it follows that b ≥ 0.

Similarly, we see that for any n ∈ N, b− bn = limm→∞ bm − bn ≥ 0. Butω is positive and so ω(b− bn) ≥ 0, that is,

ω(b) ≥ ω(bn) =n∑k=1

ω(an) = n

for any n ∈ N, which is impossible. The result follows.

We shall now consider the problem of extending a positive functional on

a C∗-algebra A, without unit, to one on A, the C∗-algebra obtained from Aby adjoining a unit. First we need the following lemma.

Lemma 7.11. For any positive functional ω on a C∗-algebra A,

|ω(k)| ≤ ‖ω‖1/2ω(k2)1/2

for all self-adjoint k ∈ A.

Proof. Note that, by 7.10, ω is bounded so the claim of the lemma is mean-

ingful.

Let (uλ) be an approximate unit for A. Then ‖k − uλk‖ → 0 so that

ω(k) = limλ ω(uλk), for any k = k∗ ∈ A. However, Schwarz' inequality gives

|ω(uλk)| ≤√ω(u2λ)ω(k2)

≤ ‖ω‖1/2ω(k2)1/2

since |ω(u2λ)| ≤ ‖ω‖‖u2λ‖ ≤ ‖ω‖ for all λ. The result follows.

Department of Mathematics

States on a C∗-algebra 61

Theorem 7.12. Suppose that ω is a positive linear functional on a C∗-algebraA without unit. For any xed µ ≥ ‖ω‖ dene ω(a + λ1l) = ω(a) + λµ, fora ∈ A and λ ∈ C. Then ω is a positive linear functional on A such that

ω A = ω. Moreover, all positive extensions of ω to A are of this form for

suitable µ ≥ ‖ω‖.

Proof. It is clear that ω is a well-dened linear functional on A. Suppose

that z ∈ A with z ≥ 0. Then z = h2 for some h = h∗ ∈ A. Writing

h = k + α1l, it follows that k = k∗ and α = α. By replacing h by −h, ifnecessary, we may suppose that α ≥ 0. We have

ω(z) = ω(k2 + 2αk + α21l)

= ω(k2) + 2αω(k) + α2µ

=(ω(k2)1/2 − αµ1/2

)2+ 2αω(k2)1/2µ1/2 + 2αω(k)

≥ 0

by the lemma. Thus ω is positive.

If ρ is a positive linear functional on A, then ‖ρ‖ = ρ(1l), so that

ρ(a+ λ1l) = ρ(a) + λρ(1l)

= ρ(a) + λ‖ρ‖

for any a ∈ A and λ ∈ C. If ρ A = ω, then clearly ‖ρ‖ ≥ ‖ω‖ and therefore

ρ = ω, as above, with µ = ‖ρ‖.

Corollary 7.13. Let A be a C∗-algebra without a unit, and suppose that ω is a

positive linear functional on A with ‖ω‖ ≤ 1. Then ω has a unique extension

to a state on A.

Proof. Dene ω on A by

ω(a+ λ1l) = ω(a) + λ ,

for a ∈ A and λ ∈ C. Then, since, ‖ω‖ ≤ 1, it follows that ω is positive. But

ω(1l) = 1 and so ω is a state on A.If ρ is state on A such that ρ A = ω, then ρ(a+ λ1l) = ρ(a) + λρ(1l) =

ω(a) + λ = ω(a+ λ1l), which gives the uniqueness of the extension.

Corollary 7.14. A positive linear functional ω on a C∗-algebra A without unit

has a unique extension to a positive linear functional with the same norm on

A.

Proof. The formula ω(a + λ1l) = ω(a) + λ‖ω‖, for a ∈ A, λ ∈ C, denes apositive linear extension of ω to A. Furthermore, ‖ω‖ = ω(1l) = ‖ω‖.

Any positive extension ρ of ω to A satises ‖ρ‖ = ρ(1l), and so the

equality ‖ρ‖ = ‖ω‖ implies that ρ(1l) = ‖ω‖ and the uniqueness follows.

King's College London

62 Chapter 7

Denition 7.15. A state ω on a C∗-algebra A without a unit is a positive

linear functional on A such that ‖ω‖ = 1.

Remark 7.16. According to the preceding discussion, such a functional then

has a unique extension to a positive linear functional, ω, say, on A, with‖ω‖ = 1 (or, equivalently, with ω(1l) = 1). We see that ω is the restriction

to A of a unique state, as earlier dened, on the unital C∗-algebra A. In otherwords, there are no new features involved in considering positive functionals

(and states) on non-unital C∗-algebras.

Department of Mathematics

Chapter 8

The Gelfand, Naimark, Segal construction

We consider now the construction of a representation of a C∗-algebra A from

a state on A. This will lead to the conclusion that any C∗-algebra can be

realised as a C∗-algebra of operators on a Hilbert space.

Denition 8.1. Let A be a unital C∗-algebra and (H, π) a representation of

A. A vector ξ ∈ H is said to be a cyclic vector (for the representation) if

π(A)ξ is dense in H. If (H, π) has a cyclic vector, then it is called a cyclic

representation.

Now, given any representation (H, π) of A and any unit vector ξ ∈ H, themap x 7→ (π(x)ξ, ξ) is clearly a state on A. So from any given representa-

tion we can easily construct states. The following fundamental construction

establishes the converse.

Theorem 8.2. (Gelfand, Naimark, Segal) Let A be a unital C∗-algebra and

ω a state on A. Then there is a cyclic representation (H, π) of A with unit

cyclic vector Ω ∈ H such that ω(a) = (π(a)Ω,Ω) for all a ∈ A.The triple (H, π,Ω) is unique up to unitary equivalence, i.e., if (H′, π′,Ω′)

is another such triple, then there is a unitary operator u : H′ → H such that

uΩ′ = Ω and uπ′(a)u−1 = π(a) for all a ∈ A.

Proof. Let N = x ∈ A : ω(x∗x)) = 0. Then for any a ∈ A and x ∈ N ,

Schwarz' inequality gives

ω((ax)∗ax) = ω(x∗a∗ax)

≤ ω(x∗x)1/2ω(y∗y)1/2, with y = a∗ax,

= 0

which shows that N is a left ideal in A.Let K = A/N as a vector space, and for ξ, η ∈ K dene 〈ξ, η〉 = ω(y∗x),

where x ∈ ξ and y ∈ η. It is straightforward to verify that 〈 · , · 〉 is a well-

dened sesquilinear form on K, i.e., denes an inner product. Moreover, if

‖ξ‖2ω ≡ 〈ξ, ξ〉 = 0, then ω(x∗x) = 0, for any x ∈ ξ. Hence, x ∈ N , and so

ξ = 0 in K = A/N . In other words, ‖ · ‖ω is a norm on K.

63

64 Chapter 8

We dene an action of A on K by Laξ = cl ax for a ∈ A, where x ∈ ξ.This action is well-dened since N is a left ideal. Furthermore,

‖Laξ‖2ω = 〈Laξ, Laξ〉= ω((ax)∗ax) , x ∈ ξ ,= ω(x∗a∗ax).

Set ρ(b) = ω(x∗bx) for any b ∈ A. Then ρ is a positive linear functional on

A and so ‖ρ‖ = ρ(1l), i.e., |ρ(b)| ≤ ρ(1l)‖b‖, for b ∈ A. Hence |ω(x∗bx)| ≤ω(x∗x)‖b‖. So, with = a∗a, we get

|ω(x∗a∗ax)| ≤ ω(x∗x)‖a∗a‖= ω(x∗x)‖a‖2

= 〈ξ, ξ〉‖a‖2.

That is, ‖Laξ‖ω ≤ ‖a‖ ‖ξ‖ω. Hence La denes a bounded linear operator on

K = A/N . The following relations are readily checked;

La+b = La + Lb ,

Lab = LaLb ,

L1l = 1lK ,

and 〈La∗ξ, η〉 = ω(y∗a∗x) = ω((ay)∗x) = 〈ξ, Laη〉, where x ∈ ξ and y ∈ η.Let H be the completion of K with respect to the norm ‖ · ‖ω. Then H

is a Hilbert space and contains (an isomorphic copy of) K as a dense subset.

Let Ω ∈ K be given by Ω = cl 1l. Then any ξ ∈ K has the form ξ = LxΩ,with x ∈ ξ.

For each a ∈ A, La is a bounded linear map from K into K and so has

a unique bounded linear extension, π(a), say, from H into H. The above

relations remain true, and so we see that π is a representation of A on H.Since K = LxΩ : x ∈ A = π(x)Ω : x ∈ A is dense in H, it follows thatΩ is a cyclic vector for the representation (H, π).

We note that, for any a ∈ A, 〈π(a)Ω,Ω〉 = 〈La cl 1l , cl 1l〉 = ω(a).To establish the uniqueness, up to unitary equivalence, suppose that

(H′, π′,Ω′) is another such triple. Dene u : H′ → H by uπ′(a)Ω′ = π(a)Ω.Then

‖uπ′(a)Ω′‖2H = ‖π(a)Ω‖2H= ω(a∗a)

= ‖π′(a)Ω′‖2H′ .

So u is an isometric linear operator from a dense set in H′ to a dense set

in H and thus extends to dene a unitary operator from H′ onto H. This

unitary provides the required equivalence; to see this, we consider

uπ′(a)u−1π(b)Ω = uπ′(a)π′(b)Ω′ = uπ′(ab)Ω

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The Gelfand, Naimark, Segal construction 65

= π(ab)Ω = π(a)π(b)Ω

for all a, b ∈ A. Since π(b)Ω is dense in H, we deduce that uπ′(a)u−1 = π(a)for a ∈ A.

Remark 8.3. (H, π,Ω) is called the Gelfand, Naimark, Segal (GNS) repre-

sentation (or triple) associated with ω on A.

Examples 8.4.

1. Let ω be the state on C([0, 1]) given by x 7→∫ 10 x(s) ds. Then the GNS

triple (H, π,Ω) is given by H = L2([0, 1]), with Lebesgue measure, Ωis the vector 1 in H, and π(x) is given on H by multiplication by the

function x ∈ C([0, 1]).

2. Suppose ω is the state on C([0, 2]) given, as above, by x 7→∫ 10 x(s) ds.

Then the GNS triple (H, π,Ω) here is exactly the same as for example

1.

3. Let H0 be a Hilbert space and let ξ ∈ H0 be a unit vector. Let ω be

the state on B(H0) given by x 7→ (xξ, ξ), x ∈ H0. Then the GNS triple

(H, π,Ω) is the representation with H = H0, Ω = ξ and π(x) = x for

all x ∈ B(H0). (This follows from the uniqueness.)

Denition 8.5. Let (Hα, πα)α∈I be a collection of representations of a C∗-algebra A. Their direct sum is the representation (H, π) withH =

⊕α∈I Hα,

and action π(a) =⊕

α∈I πα(a) for α ∈ A.

It is easy to see that (H, π) really is a representation.

Theorem 8.6. Any C∗-algebra A is isometrically ∗-isomorphic to a C∗-algebraof operators on a Hilbert space.

Proof. Without loss of generality, we may suppose that A has a unit (if not,

we consider A instead of A). Let SA denote the set of states on A, and, foreach ω ∈ SA, let (Hω, πω,Ωω) be the corresponding GNS representation of

A. Let (H, π) be their direct sum, H =⊕

ω∈SA Hω and π =⊕

ω∈SA πω, andlet Ω be the vector

⊕ω Ωω

Suppose that π(a) = π(b) for some a, b ∈ A. Then 0 = (π(a)− π(b))Ω =⊕ω(πω(a)−πω(b))Ωω . Hence πω(a− b)Ωω = 0 for all ω ∈ SA. In particular,

(πω(a− b)Ωω,Ωω) = 0 i e, ω(a− b) = 0 for all ω ∈ SA. But SA separates the

points of A, and so we conclude that a = b. Thus, π is faithful (injective)

and so A is isometrically ∗-isomorphic to π(A).

Remark 8.7. (H, π) is called the universal representation of the C∗-algebraA. Notice that every state ω on A is realised as a vector state on π(A);ω(a) = (π(a)Ωω,Ωω), a ∈ A, ω ∈ SA. Also, every state on π(A) denes a

state on A, so that every state on π(A) is also given by a vector state.

King's College London

66 Chapter 8

Department of Mathematics

Chapter 9

Pure States

Denition 9.1. A subset E of a linear space is said to be convex if for any

x, y ∈ E, and for all 0 ≤ α ≤ 1, we have αx+ (1− α)y ∈ E,

A point z ∈ E, with E convex, is said to be an extreme point of Eif z = αx + (1 − α)y, with 0 < α < 1, x, y ∈ E, has only the solution

x = y = z; i.e., z is an extreme point of E if it is not a convex combination

of two distinct points of E.

Clearly, if A is a C∗-algebra, then the set of states on A is a convex set

in A∗, the dual of A.

Denition 9.2. The extreme points of the set of states of a C∗-algebra are

called pure states. If a state is not pure, then it called a mixture. (The

terminology is once again taken from theoretical physics.)

One easily sees that a state ω is a mixture if and only if there are states

ω1 6= ω2 such that

ω = 12ω1 + 1

2ω2.

Indeed, suppose that ω = αω′ + (1− α)ω′′ for states ω′, ω′′ and 0 < α < 1.Without loss of generality, we may assume that 0 < α ≤ 1

2 . Then ω can be

written as ω = 12ω1 + 1

2ω2, where ω1 = 2αω′ + (1 − 2α)ω′′ and ω2 = ω′′.Evidently ω1 and ω2 are states on A.

Theorem 9.3. Let A be a commutative unital C∗-algebra. Then the set of

pure states of A is exactly SpA: i.e., a state ω is pure if and only if ω is a

character.

Proof. Let ω ∈ SpA, and suppose that ω = ω1 + ω2, where ω1 and ω2 are

states on A. Let a ∈ A with a = a∗. Then

ω(a2) = 12

(ω1(a

2) + ω2(a2))

= ω(a)2, since ω ∈ SpA,

= 14 (ω1(a) + ω2(a))2 .

67

68 Chapter 9

Now, for any state ρ,

ρ(a)2 = ρ(1l∗a)2

≤ ρ(1l∗1l)ρ(a∗a) , by Schwarz' inequality,

= ρ(a2) , if a = a∗.

Hence, we have

ω(a2) = 12

(ω1(a

2) + ω2(a2))≥ 1

2

(ω1(a)2 + ω2(a)2

).

Combining this with the earlier inequality gives

14 (ω1(a) + ω2(a))2 ≥ 1

2

(ω1(a)2 + ω2(a)2

).

This reduces to the inequality

0 ≥ (ω1(a)− ω2(a))2

and we conclude that ω1(a) = ω2(a) for all a ∈ A with a = a∗. But this

implies that ω1 = ω2 = ω (by 5.12), and we see that ω is pure.

Conversely, suppose that ω is a pure state on A. Suppose that a = a∗ ∈A, 0 ≤ a ≤ 1l, and 0 6= ω(a) 6= 1. For any x ∈ A, set

ω1(x) = ω(ax)/ω(a) ,

ω2(x) = ω((1l − a)x)/ω(1l − a).

Then ω1 and ω2 are states on A, and ω(a)ω1(x)+ω(1l−a)ω2(x) = ω(x), i.e.,ω = ω(a)ω1 + (1 − ω(a))ω2. Since ω is pure, it follows that ω1 = ω2 = ω,i.e., ω(x) = ω1(x) = ω(ax)/ω(a). In other words, ω(ax) = ω(a)ω(x) for all

x ∈ A and for a = a∗ as above.Now suppose that a ≥ 0 and ω(a) = 0. Then

|ω(ax)| = |ω(a1/2a1/2x)|≤ ω(a)1/2ω(x∗ax), by Schwarz' inequality,

= 0

= ω(a)ω(x)

for all x ∈ A. Now suppose that a ≤ 1l and ω(a) = 1. Putting b = 1l − a andarguing as above, we deduce that

ω(bx) = 0 = ω(b)ω(x) ,

and so we obtain

ω(ax) = ω(x) = ω(a)ω(x) ,

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Pure States 69

for all x ∈ A.Thus, combining these three situations, we have

ω(ax) = ω(a)ω(x)

for all x ∈ A and for all a ∈ A with a = a∗ and 0 ≤ a ≤ 1l. By linearity, this

also holds for all 0 ≤ a, and, again by linearity, for all a = a∗, and hence for

arbitrary a ∈ A. Thus, ω is a character.

Corollary 9.4. Let X be a compact Hausdor space. Then X = Sp C(X), andX is the set of pure states on C(X).

Theorem 9.5. Let A ⊆ B be C∗-algebras and suppose that ω is a pure state

on A. Then ω has an extension to a pure state ρ on B.

Proof. Let F = ρ : ρ is a state on B, ρ A = ω. Then we know that F is

non-empty. Evidently F is convex and w∗-closed. (To see this, suppose thatρ0 /∈ F . Then ρ0 A 6= ω so that there is some a ∈ A such that ρ0(a) 6= ω(a)Put ε = |ρ0(a)−ω(a)|. Then ε > 0 and the w∗-neighbourhood Nω = Nρ0 :a, 12ε which, we recall, is dened as ρ ∈ SB : |ρ(a) − ρ0(a)| < 1

2ε, iscontained in the complement of F (every ρ ∈ Nω satises ρ(a) 6= ω(a)).Thus the complement of F is w∗-closed.) It follows that F has extreme

points (by the Krein-Milman theorem). Let ρ be an extreme point of F . We

shall show that ρ is a pure state of B. To see this, suppose the contrary;

ρ = αρ1 + (1− α)ρ2

for 0 < α < 1 and states ρ1 6= ρ2 on B.

Since ρ ∈ F , we have ω = αρ1 A+(1−α)ρ2 A. But ρ1 A and ρ2 Aare both states on A, and ω is pure (on A). Hence ρ1 A = ρ2 A = ω,and it follows that ρ1 and ρ2 belong to F . But ρ is an extreme point of F ,which demands that ρ1 = ρ2. This contradiction implies that ρ is pure, as

claimed.

Corollary 9.6. Suppose that ω is a state on the non-unital C∗-algebra A.Then the unique extension ω of ω to a state on A is pure if and only if ω is

pure.

Proof. Suppose that ω is not pure on A. Then there are states ω1, ω2 on Awith ω1 6= ω2 and such that ω = 1

2ω1 + 12ω2 on A. Evidently, ω = 1

2 ω1 + 12 ω2

on A, and ω1 6= ω2. Thus ω is not pure on A.

Conversely, suppose that ω is pure on A. Then, by the theorem, ω has

an extension to a pure state on A. But ω is the unique extension of ω to a

state on A, so that ω is this pure extension.

Corollary 9.7. The pure states separate points of a unital C∗-algebra A.

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70 Chapter 9

Proof. By the preceding result, we may assume that A has a unit. Let

a ∈ A, with a = a∗ and a 6= 0. As usual, let A(a) denote the commutative

unital C∗-algebra generated by a, and let denote the Gelfand isomorphism

between A(a) and C(SpA(a)). Since a 6= 0, a is not the zero function in

C(SpA(a)), and so there is some κ0 ∈ SpA(a) such that a(κ0) 6= 0, i.e.,κ0(a) 6= 0. But κ0 is a character and so is a pure state on A(a) and hence

has an extension to a pure state ρ, say, on A (or A, if A is non-unital). Hence

ρ(a) = κ0(a) 6= 0. Thus, for any a = a∗ 6= 0 ∈ A, there is a pure state ρ on

A such that ρ(a) 6= 0. The result follows from this.

Corollary 9.8. Let h = h∗ be a self-adjoint element of a unital C∗-algebra A.

Then λ ∈ σ(h) implies that there is a pure state ω on A with ω(h) = λ.

Proof. Note that λ ∈ σ(h) implies that λ is in the range of the Gelfand

transform h of h, i.e., there is κ ∈ SpA(h) such that h(κ) = λ, i.e., κ(h) = λ.By the theorem, there is a pure state ω on A such that ω A(h) = κ. Butthen we have λ = κ(h) = ω(h).

Example 9.9. The converse to the above result is false. Let A be the C∗-algebra B(H), and let p ∈ A be the projection onto the one-dimensional

subspace of H spanned by the unit vector ξ. Clearly, σ(p) = 0, 1. Let

η ∈ H be a unit vector orthogonal to ξ, and set ζ = (ξ + η)/√

2. Let ωζdenote the vector state determined by ζ. Then ωζ is pure on A, since any

vector state on B(H) is pure, (as we will see later) and we have

ωζ(p) = (pζ, ζ) = 12(ξ, ξ) = 1

2 ,

but 12 /∈ σ(p).

Theorem 9.10. Let ω be a state on a unital C∗-algebra A, and let (π,H,Ω)be the associated GNS triple. Suppose that ρ is a positive linear functional

on A with ρ ≤ ω (i.e., ω − ρ is positive). Then there is a unique operator

t ∈ B(H) such that 0 ≤ t ≤ 1l, t commutes with each π(a), a ∈ A, and

ρ(b∗a) = (tπ(a)Ω, π(b)Ω)

for all a, b ∈ A.Conversely, if t ∈ B(H), 0 ≤ t ≤ 1l and t commutes with π(a), for all

a ∈ A, then a 7→ ρ(a) = (tπ(a)Ω,Ω) is a positive linear functional on A with

ρ ≤ ω.

Proof. Let ρ be a given positive linear functional on A with ρ ≤ ω. Let

cl a denote the class of a in A/N , where N is the left-ideal N = x ∈ A :ω(x∗x) = 0. Then we have

|ρ(b∗a)| ≤ ρ(b∗b)ρ(a∗)

≤ ω(b∗b)ω(a∗)

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Pure States 71

= ‖ cl b‖2ω‖ cl a‖2

ω.

Hence ρ denes a bounded sesquilinear form on K = A/N and hence extends

to a bounded sesquilinear form, λ, say, on H, the completion of K. By Riesz'

lemma, there is a unique t ∈ B(H) with

λ(ξ, η) = (tξ, η)

for ξ, η ∈ H. But λ(cl a, cl b) = ρ(b∗a), and so, with b = a, we have

(t cl a, cl a) = ρ(a∗a) ≤ ω(a∗a) = (cl a, cl a). Thus ((1l − t) cl a, cl a) ≥ 0for all a ∈ A. It follows that 0 ≤ t ≤ 1l (since K is dense in H). Further-

more, using cl a = π(a)Ω, we have

ρ(b∗a) = (t cl a, cl b)

= (tπ(a)Ω, π(b)Ω).

Finally, for a, b, c ∈ A,

((tπ(a)− π(a)t)π(c)Ω, π(b)Ω) = ρ(b∗ac)− (tπ(c)Ω, π(a∗b)Ω)

= ρ(b∗ac)− ρ((a∗b)∗c)

= ρ(b∗ac)− ρ(b∗ac)

= 0.

Since Ω is cyclic, it follows that (tπ(a) − π(a)t)π(c)Ω = 0 for all a, c ∈ A,and so tπ(a)−π(a)t = 0 for all a ∈ A; that is, t commutes with all members

of π(A).The proof of the converse is a direct computation.

Denition 9.11. The commutant of a set S of bounded operators on a Hilbert

space H is the set

S′ = y ∈ B(H) : ys = sy for all s ∈ S.

Theorem 9.12. Let ω be a state on a unital C∗-algebra A and (H, π,Ω) the

associated cyclic (GNS) representation of A. Then π(A)′ = C1l if and only

if ω is pure.

Proof. Suppose rst that π(A) = C1l, and let ω = αρ1 + (1 − α)ρ2 with

0 < α < 1. Then (1 − α)ρ2 ≥ 0 implies that αρ1 ≤ ω. Hence, there is

t ∈ π(A)′ such that 0 ≤ t ≤ 1l and αρ1(a) = (tπ(a)Ω,Ω) for all a ∈ A.But π(A) = C1l implies that t = β1l for some 0 ≤ β ≤ 1. Hence αρ1(a) =β(π(a)Ω,Ω) = βω(a), for all a ∈ A. In particular, taking a = 1l, we get

α = β and so ρ1 = ω and we deduce that ω is pure on A.Conversely, suppose that ω is pure. Let 0 ≤ t ≤ 1l, t ∈ π(A)′, and

suppose that t 6= 0 and t 6= 1l. Dene ρ on A by ρ(a) = (tπ(a)Ω,Ω), a ∈ A.

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72 Chapter 9

Then it is easy to see that ρ is a positive linear functional on A and that

ρ ≤ ω.Now, |ρ(a)|2 ≤ ρ(1l)ρ(a∗a), by Schwarz' inequality, and so ρ(1l) = 0 if

and only if ρ(a) = 0 for all a ∈ A. By cyclicity, this is equivalent to t = 0.Hence ρ(1l) 6= 0. Furthermore, (ω − ρ)(a) = ((1l − t)π(a)Ω,Ω) and so a

similar argument (with (1l − t) replacing t) implies that (ω − ρ)(1l) = 0 if

and only if t = 1l. Thus (ω − ρ)(1l) 6= 0. It follows that both ρ( · )/ρ(1l) and(ω − ρ)( · )/(ω − ρ)(1l)) are states on A. But then

ω( · ) = ρ(1l)ρ( · )ρ(1l)

+ (ω − ρ)(1l)(ω − ρ)( · )(ω − ρ)(1l)

expresses ω as a convex combination of states. Since ω is pure, we must have

ρ( · )/ρ(1l) = ω( · ). Thus

(tπ(a)Ω,Ω) = (tΩ,Ω)︸ ︷︷ ︸ρ(1l)

(π(a)Ω,Ω)

and so ((t − ρ(1l))π(a)Ω,Ω) = 0, for all a ∈ A. It follows that t = ρ(1l)1l,and hence we have π(A)′ = C1l.

Denition 9.13. A representation (H, π) of a C∗-algebra A is called irre-

ducible if π(A)′ = C1l.

Thus, the previous theorem says that the GNS representation of a unital

C∗-algebra A induced by a state ω is irreducible if and only if ω is pure.

The point is that if a representation (H, π) is not irreducible, then π(A)′

will contain a non-trivial projection, p, say. In this case, π(A) maps the

subspace pH into pH and (1l − p)H into (1l − p)H, i.e., pH and (1l − p)Hare invariant subspaces. If (H, π) is irreducible, then there are no such non-

trivial invariant subspaces of H (under π(A)).

Example 9.14. Let ρ be the vector state on B(H) given by a unit vector

ξ ∈ H. Then ξ is a cyclic vector for the identity representation π(a) = a,a ∈ B(H), and (π(a)ξ, ξ) = ρ(a), a ∈ B(H). By the uniqueness of the

GNS representation, this representation is (unitarily equivalent to) the GNS

representation. By the remarks above, we conclude that ρ is pure, since

B(H) is irreducible.

Theorem 9.15. Let A be a unital C∗-algebra. Then A is isometrically iso-

morphic to a direct sum of irreducible representations of itself.

Proof. Let E denote the set of pure states of A, and let (H, π) be the represen-tation with H =

⊕ω∈E Hω , and π =

⊕ω∈E πω. Each (Hω, πω) is irreducible,

and (H, π) is faithful since E separates points of A.

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Theorem 9.16. Let ω be a state on a unital C∗-algebra A with associated

GNS triple (H, π,Ω). Suppose that α is an automorphism of A such that ωis invariant under α, i.e., ω(α(a)) = ω(a) for all a ∈ A. Then there is a

unitary operator U on H such that UΩ = Ω and Uπ(a)U∗ = π(α(a)) for all

a ∈ A. Moreover, U is unique.

Proof. Dene the representation (H, π′) of A by the assignment π′(a) =π(α(a)) for a ∈ A and consider the triple (H, π′,Ω).

Since α(A) = A, Ω is cyclic for π′. Furthermore,

(Ω, π′(a)Ω) = (Ω, π(α(a))Ω)

= ω(α(a))

= ω(a)

for all a ∈ A. By the uniqueness of the GNS triple, we deduce that there is

a unitary U on H such that UΩ = Ω and Uπ(a)U∗ = π′(a) = π(α(a)) for alla ∈ A.

Suppose that V is another unitary operator on H with the same proper-

ties. Then

Uπ(a)Ω = Uπ(a)U∗UΩ = Uπ(a)U∗Ω = π(α(a))Ω

= V π(a)V ∗Ω = V π(a)Ω

for all a ∈ A. Since Ω is cyclic, it follows that U = V .

Corollary 9.17. Let A be a C∗-algebra and G a topological group. Suppose

that g 7→ αg, for g ∈ G, is a representation of G in AutA. Suppose that ωis a state on A which is invariant under each αg, i.e., ω(αg(a)) = ω(a) for

each a ∈ A and all g ∈ G. Suppose, further, that for any a, b ∈ A the map

g 7→ ω(b∗αg(a)) is continuous. Then there is a strongly continuous unitary

representation g 7→ U(g) of G on H, where (H, π,Ω) is the GNS triple

associated with ω, satisfying U(g)Ω = Ω for all g ∈ G and U(g)π(a)U(g)∗ =π(αg(a)) for all g ∈ G and a ∈ A. Moreover, the U(g) are unique.

Proof. By the theorem, for each g ∈ G there is a unique unitary operator

U(g) on H satisfying U(g)Ω = Ω and U(g)π(a)U(g)∗ = π(αg(a)) for all

a ∈ A. To see that g 7→ U(g) is a representation of G, we compute

U(g)U(h)π(a)Ω = U(g)π(αh(a))Ω

= π(αg(αh(a)))Ω

= π(αgh(a))Ω

= U(gh)π(a)Ω

for all a ∈ A, g, h ∈ G. Since Ω is cyclic, it follows that U(g)U(h) = U(gh).

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74 Chapter 9

It remains to show that U( · ) is strongly continuous. To see this, let

a, b ∈ A. Then(π(b)Ω, U(g)π(a)Ω) = ω(b∗αg(a))

is continuous in g, by hypothesis. Since each U(g) is unitary, it follows thatU( · ) is weakly continuous on H and therefore strongly continuous.

The uniqueness of the U(g) follows as in the theorem.

Denition 9.18. Let A be a C∗-algebra and g 7→ αg ∈ AutA a representation

of the group G in AutA. A state ω on A is said to be extremal invariant

(with respect to αg) if ω is an extreme point of the convex set ρ state on A :ρ αg = ρ for all g ∈ G .

Corollary 9.19. With the assumptions and notation as above, we have R ≡(U(g) : g ∈ G ∪ π(A) ′ = C1l if and only if ω is extremal invariant.

Proof. Suppose ω is extremal invariant, but R 6= C1l. Let P be a non-trivial

projection in R. Then the cyclicity of ω implies that PΩ 6= 0 and PΩ 6= Ω.Let ω1 be the state on A given by

ω1(a) =(PΩ, π(a)PΩ)

‖PΩ‖2

for a ∈ A, and let ω2 be the state given by

ω2(a) =(QΩ, π(a)QΩ)

‖QΩ‖2

for a ∈ A, where P+Q = 1l. Evidently, ω is given as the convex combination

ω = ‖PΩ‖2ω1 + ‖QΩ‖2ω2 .

Moreover, ω1 and ω2 are each invariant under every αg and one readily sees

that ω1 6= ω2. Indeed, for any xed ξ ∈ H, the cyclicity of Ω implies that

there is a sequence an ∈ A such that π(an)Ω→ Pξ. But then

ω1(an) = (PΩ, π(an)PΩ)/‖PΩ‖2 = (PΩ, π(an)Ω)/‖PΩ‖2 → (PΩ, ξ)/‖PΩ‖2 .

However,

ω2(an) = (QΩ, π(an)QΩ)/‖QΩ‖2

= (QΩ, π(an)Ω)/‖QΩ‖2

→ (QΩ, P ξ)/‖QΩ‖2 = 0 .

The equality ω1 = ω2 would then entail that (PΩ, ξ) = 0 for all ξ ∈ H, thusgiving PΩ = 0, which we know not to be true. We conclude that ω1 6= ω2

and so we have exhibited ω as a convex combination of invariant states.

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This contradicts the supposed extremal invariance of ω and so we must have

R = C1l.For the converse, suppose that R = C1l, but ω is not extremal invariant.

Then there are distinct invariant states ω1 and ω2 and 0 < λ < 1 such that

ω = λω1 + (1 − λ)ω2. Hence ω ≥ λω1 = ρ, say. It follows that there is an

operator T ∈ π(A)′ such that

ρ(b∗a) = (π(b)Ω, Tπ(a)Ω)

for all a, b ∈ A. The invariance of ρ under αg together with the cyclicity of

Ω implies that U(g)∗TU(g) = T which is to say that T commutes with each

U(g). Hence T ∈ R and so T = µ1l for some µ ∈ C. But this implies that ω1

is proportional to ω and hence equal to ω. This in turn means that ω2 = ωgiving ω1 = ω2 which is false. We conclude that ω is extremal invariant, as

claimed.

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