4
4.1
C*-algebras
Definitions and examples
4.1.1 Definition. Let A be a Banach algebra. An involution on A is a map
A → A, a 7→ a∗ such that for all a, b ∈ A and λ ∈ C we have:
(i). (λa)∗ = λa∗ and (a + b)∗ = a∗ + b∗ (conjugate linearity);
(ii). (ab)∗ = b∗ a∗ ; and
(iii). (a∗ )∗ = a.
A C*-algebra is a Banach algebra A equipped with an involution such that
the C*-condition holds:
kak2 = ka∗ ak for all a ∈ A.
4.1.2 Remark. We usually write a∗∗ instead of (a∗ )∗ .
4.1.3 Examples.
(i). The zero Banach algebra {0} is a C*-algebra.
(ii). The complex numbers C form a C*-algebra under the usual Banach
space norm kλk = |λ| and the involution λ∗ = λ.
(iii). If X is a topological space, then BC (X) is a C*-algebra under the
involution f ∗ (x) = f (x). In particular, if X is compact, then BC (X) =
C(X) is a C*-algebra.
(iv). If H is a Hilbert space then B(H) is a C*-algebra under the involution
T 7→ T ∗ defined by the property that hT x, yi = hx, T ∗ yi for all x, y ∈ H
(see [FA ]).
(v). If A is a C*-algebra then a C*-subalgebra of A is a Banach subalgebra
B ⊆ A that is closed under the involution; in other words, B is a closed
linear subspace of A such that whenever a, b ∈ A we have ab ∈ A and
a∗ ∈ A. Clearly, any C*-subalgebra of a C*-algebra is a C*-algebra.
4.1.4 Proposition. Let A be a C*-algebra.
(i). If A has an identity element 1 then 1∗ = 1, and if A is non-zero then
k1k = 1.
(ii). If A is unital then a ∈ Inv A ⇐⇒ a∗ ∈ Inv A, and for a ∈ Inv A we
have (a∗ )−1 = (a−1 )∗ .
(iii). ka∗ k = kak for all a ∈ A.
30
(iv). The involution A → A, a 7→ a∗ is continuous.
(v). σ(a∗ ) = σ(a)∗ = {λ : λ ∈ σ(a)} for all a ∈ A.
Proof. (i) We have
1 = (1∗ )∗ = (1∗ 1)∗ = 1∗ 1∗∗ = 1∗ 1 = 1∗ .
Hence k1k2 = k1∗ 1k = k1k; if A 6= {0} then k1k 6= 0 and we can cancel to
obtain k1k = 1.
(ii) If a ∈ Inv A then a∗ (a−1 )∗ = (a−1 a)∗ = 1∗ = 1 and (a−1 )∗ a∗ =
(aa−1 )∗ = 1∗ = 1. Hence a∗ is invertible, with inverse (a−1 )∗ . Conversely, if
a∗ ∈ Inv A then a∗∗ = a ∈ Inv A by the same argument.
(iii) This is trivial for a = 0. If a 6= 0 then kak2 = ka∗ ak ≤ ka∗ k kak,
which on cancelling gives kak ≤ ka∗ k. Hence ka∗ k ≤ ka∗∗ k = kak ≤ ka∗ k, so
we have equality.
(iii)
(iv) If an , a ∈ A with an → a then ka∗n −a∗ k = k(an −a)∗ k = kan −ak → 0
as n → ∞, so the involution is continuous.
(v) Without loss of generality, suppose that A is unital. For any λ ∈ C,
(ii)
(i)
λ ∈ σ(a∗ ) ⇐⇒ λ1−a∗ = (λ1−a)∗ 6∈ Inv A ⇐⇒ λ1−a 6∈ Inv A ⇐⇒ λ ∈ σ(a).
Hence σ(a∗ ) = σ(a)∗ .
4.1.5 Definition. Let A be a C*-algebra.
An element a ∈ A is normal if a commutes with a∗ .
An element a ∈ A is hermitian if a = a∗ .
An element p ∈ A is a projection if p = p∗ = p2 .
If A is unital then an element u ∈ A is unitary if uu∗ = u∗ u = 1 (that is,
if u is invertible and u−1 = u∗ ).
4.1.6 Remark. Clearly, projections are hermitian, and both unitary and
hermitian elements are normal.
4.1.7 Proposition. Let A be a C*-algebra.
(i). If a ∈ A then a∗ a is hermitian.
(ii). If p ∈ A is a non-zero projection then kpk = 1.
(iii). Every a ∈ A may be written uniquely in the form a = h + ik where h
and k are hermitian elements of A, called the real and imaginary parts
of a.
31
Proof. (i) If h = a∗ a then h∗ = (a∗ a)∗ = a∗ a∗∗ = a∗ a = h, so h is hermitian.
(ii) We have kpk2 = kp∗ pk = kp2 k = kpk. Since kpk 6= 0 we may cancel
to obtain kpk = 1.
(iii) We have a = h + ik where h = 21 (a + a∗ ) and k = 2i1 (a − a∗ ), as
′
is easily verified. If h + ik = h′ + ik ′ where
k ′ are hermitian, then
∗ h, h , k,
′
′
′ ∗
′
′
i(k − k) = h − h = (h − h ) = i(k − k) = −i(k − k), hence h = h′ and
k = k ′ , demonstrating uniqueness.
4.1.8 Lemma. If A is a unital C*-algebra and τ ∈ A∗ with kτ k ≤ 1 and
τ (1) = 1, then τ (h) ∈ R for all hermitian elements h ∈ A.
Proof. Suppose that τ (h) = x + iy where x, y ∈ R. Observe that for t ∈ R
we have τ (h + it1) = τ (h) + it = x + i(y + t), so
x2 +(y+t)2 = |τ (h+it)|2 ≤ kh+itk2 = k(h−it)(h+it)k = kh2 +t2 k ≤ kh2 k+t2
and so x2 + 2yt ≤ kh2 k for all t ∈ R. This forces y = 0, so τ (h) = x ∈ R.
4.1.9 Proposition. If A is a unital abelian C*-algebra then τ (a∗ ) = τ (a)
for all a ∈ A and τ ∈ Ω(A).
Proof. If τ ∈ Ω(A) then kτ k = τ (1) = 1 by Lemma 3.1.4. If a ∈ A then
a = h + ik for some hermitian h, k ∈ A by Proposition 4.1.7(iii), so τ (h) and
τ (k) are real by Lemma 4.1.8. Hence
τ (a∗ ) = τ (h − ik) = τ (h) − iτ (k) = τ (h) + iτ (k) = τ (a).
4.1.10 Corollary. Let A be a unital C*-algebra.
(i). If h ∈ A is hermitian then σA (h) ⊆ R.
(ii). If u ∈ A is unitary then σA (u) ⊆ T.
Proof. If a ∈ A and a commutes with a∗ then let C = {a, a∗ }′′ . This is a
closed unital abelian subalgebra of A and σA (a) = σC (a) = {τ (a) : τ ∈ Ω(C)}
by Proposition 1.3.19 and Theorem 3.2.3. The commutant of a self-adjoint
subset of A is self-adjoint (that is, it is closed under the involution), so C is
self-adjoint. Hence C is a unital abelian C*-subalgebra of A.
(i) If a = h is hermitian then τ (a) ∈ R for all τ ∈ Ω(C) by Proposition 4.1.9, which establishes the result.
(ii) If a = u is unitary then by Proposition 4.1.9 we have
|τ (u)|2 = τ (u)∗ τ (u) = τ (u∗ )τ (u) = τ (u∗ u) = τ (1) = 1
for every τ ∈ Ω(C), so σA (u) ⊆ T.
32
4.1.11 Corollary. Let A be a unital C*-algebra and let B ⊆ A be a C*subalgebra with 1 ∈ B. If b ∈ B then σB (b) = σA (b).
Proof. If b is hermitian then σA (b) ⊆ R by Corollary 4.1.10(i). Therefore
σA (b) has no holes, so σB (b) = σA (b) by Theorem 1.3.16.
If b is any element of B ∩ Inv A and if a is the inverse of b in A then b∗ b is
invertible in A with inverse aa∗ . Since b∗ b is hermitian, the first paragraph
shows that b∗ b is invertible in B; so aa∗ ∈ B. Hence aa∗ b∗ ∈ B and baa∗ b∗ =
aa∗ b∗ b = 1, so b is invertible in B with inverse aa∗ b∗ . Hence b ∈ Inv B.
This shows that Inv B = B ∩ Inv A, so σB (b) = σA (b) for any b ∈ B.
4.1.12 Proposition.
If a is a hermitian element of a C*-algebra then kak = r(a).
Proof. Since a = a∗ we have kak2 = ka∗ ak = ka2 k, so for n ≥ 1 we have
n
n
kak2 = ka2 k by induction. Hence, by Theorem 1.3.12,
n
n
r(a) = lim ka2 k1/2 = kak.
n→∞
4.1.13 Definition. If A and B are C*-algebras then a ∗-homomorphism
θ : A → B is a homomorphism of Banach algebras which respects the involution: θ(a∗ ) = θ(a)∗ for all a ∈ A. If A and B are unital and θ(1) = 1 then we
say that θ is unital. An invertible ∗-homomorphism is called a ∗-isomorphism.
4.1.14 Proposition. Let A and B be unital C*-algebras and let θ : A → B
be a unital ∗-homomorphism.
(i). θ(Inv A) ⊆ Inv B, and θ(a)−1 = θ(a−1 ) for a ∈ Inv A.
(ii). For all a ∈ A we have σA (a) ⊇ σB (θ(a)).
(iii). θ is continuous; in fact, kθ(a)k ≤ kak for all a ∈ A.
(iv). If θ is a ∗-isomorphism then θ is an isometry and σA (a) = σB (θ(a))
for all a ∈ A.
Proof. (i) and (ii) are special cases of Proposition 1.5.12.
(iii) Let a ∈ A. We have
kθ(a)k2 = kθ(a)∗ θ(a)k = rB (θ(a)∗ θ(a)) by Proposition 4.1.12
= rB (θ(a∗ a))
≤ rA (a∗ a) by (ii)
= ka∗ ak by Proposition 4.1.12
= kak2 .
33
So kθ(a)k ≤ kak for all a ∈ A. Since θ is a linear map, this shows that θ is
continuous (with kθk ≤ 1).
(iv) If θ is a ∗-isomorphism then θ is a bijective ∗-homomorphism, from
which it is easy to see that θ−1 is also a ∗-homomorphism. By (iii) both θ
and θ−1 are continuous linear maps with norm no greater than 1, so
kak = kθ−1 (θ(a))k ≤ kθ(a)k ≤ kak
for each a ∈ A. Hence we have equality throughout and θ is an isometry. We
showed that σA (a) = σB (θ(a)) for all a ∈ A in Proposition 1.5.12.
4.2
The Stone-Weierstrass theorem
4.2.1 Lemma. If m ∈ N then there is a sequence of polynomials p1 , p2 , p3 , . . .
with real coefficients such that 0 ≤ pn (t) ≤ 1 for n ≥ 1 and 0 ≤ t ≤ 1, and
1
pn → 0 uniformly on [0, 2m
] and
pn → 1 uniformly on [ m2 , 1].
Proof. A calculus exercise shows that if N ∈ N then
1 − (1 − x)N ≤ N x and (1 − x)N ≤
1
Nx
for each x ∈ (0, 1).
n
Let pn (t) = 1 − (1 − tn )m . We have 0 ≤ pn (t) ≤ 1 for t ∈ [0, 1], and
n
sup |pn (t)| = sup 1 − (1 − tn )m ≤ sup (mt)n ≤ 2−n → 0 as n → ∞,
t∈[0,1/2m]
t∈[0,1/2m]
t∈[0,1/2m]
and similarly,
sup |1−pn (t)| =
t∈[2/m,1]
n
sup (1−tn )m ≤
t∈[2/m,1]
1
≤ 2−n → 0 as n → ∞.
n
t∈[2/m,1] (mt)
sup
Let X be a compact Hausdorff space, and let C(X, R) denote the set of
continuous functions X → R. This is a real Banach space under the uniform
norm [FA 1.7.2], and it is easy to check that the pointwise product turns
it into a real Banach algebra. The basic definitions we made for complex
Banach algebras carry over simply by changing C to R, so we may talk of
subalgebras and unital subalgebras of C(X, R).
If A ⊆ C(X, R) (or A ⊆ C(X)), we say that A separates the points of X
if for every pair of distinct points x, y ∈ X, there is some f ∈ A such that
f (x) 6= f (y).
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4.2.2 Theorem (The real-valued Stone-Weierstrass theorem). Let X be a
compact Hausdorff topological space and let A ⊆ C(X, R). If A is a unital
subalgebra of C(X, R) which separates the points of X, then A is uniformly
dense in C(X, R).
Proof. We will write statements such as “f ≤ 1” to mean f (x) ≤ 1 for all
x ∈ X, and “f ≤ 1 on K” to mean that f (x) ≤ 1 for all x ∈ K. We split
the proof into several steps.
(i). If x, y ∈ X with x 6= y then there is a function f ∈ A with f ≥ 0 such
that f (x) = 0 and f (y) > 0.
Since A separates the points of X, there is a function g ∈ A with
g(x) 6= g(y). Let f = (g − g(x)1)2 . Since A is a unital algebra, f ∈ A,
and clearly f ≥ 0. Moreover, f (x) = 0 and f (y) > 0.
(ii). If L is a compact subset of X and x ∈ X \ L then there is some f ∈ A
with 0 ≤ f ≤ 1 such that f (x) = 0 and f (y) > 0 for all y ∈ L.
By (i), for each y ∈ L there is a function fy ∈ A such that fy ≥ 0,
fy (x) = 0 and fy (y) > 0. Since fy is continuous, there is an open set
Uy containing y such that fy > 0 on Uy . As y varies over L, the sets
Uy cover L. Since L is compact, there are y1 , . . . , yk ∈ L such that
Uy1 , . . . , Uyk cover L. Let f0 = fy1 + · · · + fyk and let f = f0 /kf0 k.
Then f ∈ A, 0 ≤ f ≤ 1, f (x) = 0 and f (y) > 0 for y ∈ L.
(iii). If K, L are disjoint compact subsets of X then there is a function f ∈ A
with 0 ≤ f ≤ 1 such that f ≤ 41 on K and f ≥ 43 on L.
Let x ∈ K. By (ii), there is a function gx ∈ A such that 0 ≤ gx ≤ 1,
gx (x) = 0, and gx > 0 on L. Since L is compact, gx (L) is compact,
so is contained in a set of the form [s, t] for some s > 0. Hence there
is m ∈ N (depending on x) such that gx ≥ m2 on L. For x ∈ K, let
1
Ux = {z ∈ X : gx (z) < 2m
}. Since x ∈ Ux , these open sets cover K,
so by compactness there are x1 , . . . , xk ∈ K such that Ux1 , . . . , Uxk
cover K.
By Lemma 4.2.1, for i = 1, . . . , k there is a polynomial p (depending
on i) such that if we write fi = p(gxi ), then 0 ≤ fi ≤ 1, fi ≤ 14 on Uxi
and fi ≥ ( 43 )1/k on L. Since p is a polynomial and gxi ∈ A, we have
fi ∈ A.
Now let f = f1 f2 . . . fk . Then f ∈ A and f has the desired properties.
(iv). If f ∈ C(X, R) then there is g ∈ A such that kf − gk ≤ 43 kf k.
We may assume (by considering f /kf k) that kf k = 1. Let h ∈ A be
the function obtained by applying (iii) to K = {x ∈ X : f (x) ≤ − 41 }
35
and L = {x ∈ X : f (x) ≥ 41 }, and let g = h − 21 . Since 14 ≤ g ≤ 12
on L, we have |f − g| ≤ 34 on L; similarly, |f − g| ≤ 43 on K. Since
− 21 ≤ g ≤ 12 and − 14 ≤ f ≤ 41 on X \ (K ∪ L) we have |f − g| ≤ 34
on X \ (K ∪ L). Hence kf − gk ≤ 43 .
(v). A is uniformly dense in C(X, R).
Let f ∈ C(X, R). By (iv), there is g1 ∈ A with kf − g1 k ≤ 43 kf k.
Applying (iv) to the function f − g1 , we obtain a function g2 ∈ A with
kf − g1 − g2 k ≤ 43 kf − g1 k ≤ ( 43 )2 kf k. Continuing in this manner, we
obtain functions g1 , g2 , g3 , . . . ∈ A such that
kf − (g1 + · · · + gn )k ≤ ( 43 )n kf k → 0 as n → ∞.
Since g1 + · · · + gn ∈ A, this shows that f is in the uniform closure
of A.
4.2.3 Theorem (The complex Stone-Weierstrass theorem). Let X be a compact Hausdorff topological space and let A ⊆ C(X). If A is a unital ∗subalgebra of C(X) which separates the points of X, then A is uniformly
dense in C(X).
Proof. For f ∈ C(X), let f1 (x) = Re(f (x)) and f2 (x) = Im(f (x)). Then
f1 , f2 ∈ C(X, R). Moreover, f1 = 12 (f + f ∗ ) and f2 = 2i1 (f − f ∗ ) are in A
since A is a complex vector space which is closed under the ∗-operation.
Consider the set A1 = {f1 : f ∈ A}. Since A separates the points of X,
if x, y ∈ X with x 6= y then there is f ∈ A such that f (x) 6= f (y). Hence
either f1 (x) 6= f1 (y) or f2 (x) 6= f2 (y). If f1 (x) 6= f1 (y) then f1 is a function
in A1 separating x and y, and if f2 (x) 6= f2 (y) then g = −if = f2 − if1 ∈ A
and f2 = g1 is a function in A1 separating x and y.
This shows that A1 separates the points of X, and it is easy to see that A1
is a unital subalgebra of C(X, R). By the real-valued Stone-Weierstrass theorem, A1 is uniformly dense in C(X, R).
If f ∈ C(X) then f = f1 + if2 . Given ε > 0 there is a function g1 ∈ A1
with kf1 − g1 k∞ < ε/2 and a function g2 ∈ A1 with kf2 − g2 k∞ < ε/2. Hence
g = g1 + ig2 ∈ A and kf − gk∞ ≤ kf1 − g1 k∞ + ki(f2 − g2 )k∞ < ε, so A is
uniformly dense in C(X).
4.3
Abelian C*-algebras and the continuous functional
calculus
We now apply Gelfand’s theory of abelian unital Banach algebras (§3) to
show that, up to isometric ∗-isomorphism, every unital abelian C*-algebra is
of the form C(X) for some compact Hausdorff space X.
36
4.3.1 Theorem (The Gelfand-Naimark theorem). If A is a unital abelian
C*-algebra then the Gelfand representation of A,
γ : A → C(Ω(A)),
a 7→ b
a
is a unital isometric ∗-isomorphism of A onto C(Ω(A)).
Proof. Theorem 3.2.3 tells us that γ is a unital, norm-decreasing homomorphism with kγ(a)k = r(a). Moreover, for a ∈ A and τ ∈ Ω(A) we have
c
a∗ (τ ) = τ (a∗ ) = τ (a) = b
a(τ ) = (b
a)∗ (τ )
by Proposition 4.1.9, so γ(a∗ ) = γ(a)∗ and γ is a ∗-homomorphism. Now a∗ a
is hermitian by Proposition 4.1.7(i), so by Proposition 4.1.12,
kγ(a)k2 = kγ(a)∗ γ(a)k = kγ(a∗ a)k = r(a∗ a) = ka∗ ak = kak2 .
Hence γ is an isometry.
It remains to show that γ is surjective, for which we appeal to the StoneWeierstrass theorem. Recall that Ω(A) is a compact Hausdorff space by
Theorem 3.1.6. Since γ is a unital ∗-homomorphism, its image γ(A) is a
unital ∗-subalgebra of C(Ω(A)). If τ1 , τ2 ∈ Ω(A) with τ1 6= τ2 then there is
a ∈ A with τ1 (a) 6= τ2 (a), hence b
a(τ1 ) 6= b
a(τ2 ) and so γ(A) separates the
points of Ω(A). By the Stone-Weierstrass theorem 4.2.3, γ(A) is dense in
C(Ω(A)). Since γ is a linear isometry its range is closed. Hence
γ(A) = γ(A) = C(Ω(A)).
4.3.2 Corollary. If A is a unital abelian C*-algebra and a, b ∈ A with τ (a) =
τ (b) for all τ ∈ Ω(A), then a = b.
Proof. b
a = bb, so a = b by the Gelfand-Naimark theorem.
4.3.3 Definition. If S is a subset of a C*-algebra A, then we write C ∗ (S)
for the smallest C*-subalgebra of A containing S. In particular, if A is a
unital C*-algebra and a ∈ A then we will write C ∗ (1, a) = C ∗ ({1, a}).
4.3.4 Lemma. Let A be a C*-algebra and let S ⊆ A.
(i). The linear span of the elements of the form an1 1 an2 2 . . . ank k for k ≥ 1,
n1 , . . . , nk ≥ 1 and ai ∈ S or a∗i ∈ S for 1 ≤ i ≤ k is dense in C ∗ (S).
(ii). Suppose that A is unital and that B is another unital C*-algebra. If
A = C ∗ (S) and θ1 , θ2 are two unital ∗-homomorphisms A → B such
that θ1 (a) = θ2 (a) for all a ∈ S, then θ1 = θ2 .
37
Proof. (i) Clearly, any C*-algebra containing S must also contain every element of the given form, hence C ∗ (S) contains the closed linear span of such
elements. Conversely, it is easy to see that the closed linear span of these
elements forms a C*-algebra containing S, so this C*-algebra is equal to
C ∗ (S).
(ii) Since θ1 and θ2 are unital ∗-homomorphisms, they are continuous by
Proposition 4.1.14(iii), and we have θ1 (b) = θ2 (b) for any b = an1 1 an2 2 . . . annk
with ai ∈ S or a∗i ∈ S for 1 ≤ i ≤ k. Since θ1 and θ2 are linear, they agree
on the linear span of such elements, which is dense by (i). Continuous maps
agreeing on a dense subset of their domain are equal, so θ1 = θ2 .
4.3.5 Lemma. Let A be a unital C*-algebra. If a is a normal element of A
then C ∗ (1, a) is a unital abelian C*-algebra.
Proof. Exercise.
We can now prove a generalisation of Proposition 4.1.12.
4.3.6 Corollary. If a is a normal element of a unital C*-algebra A then
r(a) = kak.
Proof. Let B = C ∗ (1, a). By Corollary 1.3.13, r(a) = rA (a) = rB (a). By the
Gelfand-Naimark theorem 4.3.1, if γ : B → C(Ω(B)) is the Gelfand representation of B then kak = kγ(a)k = rB (a) = r(a).
4.3.7 Lemma. Suppose that a is a normal element of a unital C*-algebra, let
Ω = Ω(C ∗ (1, a)) and let γ : C ∗ (1, a) → C(Ω) be the Gelfand representation
of C ∗ (1, a). The map b
a = γ(a) : Ω → C is a homeomorphism of Ω onto σ(a).
Proof. By Lemma 3.1.4 and Proposition 4.1.9, every τ ∈ Ω is a unital ∗homomorphism C ∗ (1, a) → C. If b
a(τ1 ) = b
a(τ2 ) for some τ1 , τ2 ∈ Ω then
τ1 (a) = τ2 (a) and τ1 (1) = 1 = τ2 (1). Taking S = {1, a} in Lemma 4.3.4(ii)
we see that τ1 = τ2 . Hence b
a is injective.
The map b
a is continuous by the definition of the topology on Ω. Moreover, Ω is compact by Theorem 3.1.6 and b
a(Ω) = σ(a) is Hausdorff. By
Lemma 2.1.1, b
a is a homeomorphism onto σ(a).
4.3.8 Lemma. Suppose that X and Y are compact Hausdorff topological
spaces and ψ : X → Y is a homeomorphism. The map ψ t : C(Y ) → C(X),
f 7→ f ◦ ψ is an isometric unital ∗-isomorphism.
Proof. Exercise.
38
4.3.9 Theorem (The continuous functional calculus). Let a be a normal
element of a unital C*-algebra A. There is a unique unital ∗-homomorphism
θa : C(σ(a)) → A such that θa (z) = a where z ∈ C(σ(a)) is the function
z(λ) = λ for λ ∈ σ(a). Moreover, θa is an isometric ∗-isomorphism onto
C ∗ (1, a).
Proof. Suppose that θ1 and θ2 are two unital ∗-homomorphisms C(σ(a)) → A
with θ1 (z) = θ2 (z) = a. Since z(λ) = z(µ) if and only if λ = µ, the
function z separates the points of σ(a), so C ∗ ({1, z}) is a closed separating
unital ∗-subalgebra of C(σ(a)). By the Stone-Weierstrass theorem 4.2.3,
C ∗ ({1, z}) = C(σ(a)), so θ1 = θ2 by Lemma 4.3.4(ii). This proves the
uniqueness statement.
Let Ω = Ω(C ∗ (1, a)), let γ : C ∗ (1, a) → C(Ω) be the Gelfand representation of C ∗ (1, a) and let b
a = γ(a). By the Gelfand-Naimark theorem 4.3.1
and the last two lemmas, γ and (b
a)t : C(σ(a)) → C(Ω) are both isometric
unital ∗-isomorphisms. Consider the map θa : C(σ(a)) → C ∗ (1, a) given by
θa = γ −1 ◦ (b
a)t :
θa
// C ∗ (1, a)
C(σ(a))
?
??
??
??
??
?
t ??
(b
a)
??




 −1

 γ

C(Ω)
Since it is the composition of two isometric ∗-isomorphisms, θa is an isometric
∗-isomorphism onto C ∗ (1, a).
4.3.10 Definition. If a is a normal element of a unital C*-algebra A and
f ∈ C(σ(a)) then we write f (a) for the element θa (f ) ∈ A.
4.3.11 Remark. If p = λ0 + λ1 z + · · · + λn z n is a complex polynomial
on σ(a), then θa (p) is equal to the element p(a) defined in Definition 1.3.7,
because θa is a unital homomorphism with θa (z) = a. So this new notation
is consistent.
4.3.12 Remark. This notation is quite convenient. By way of example, we
use it to restate the fact that θa is a ∗-homomorphism. For any f, g ∈ C(σ(a))
and µ ∈ C, we have
(f + g)(a) = f (a) + g(a), (µf )(a) = µ f (a),
(f g)(a) = f (a)g(a) and f ∗ (a) = f (a)∗
where the functions f + g, µf, f g and f ∗ are given by applying the usual
pointwise operations in the C*-algebra C(σ(a)).
39
4.3.13 Theorem (Spectral mapping for the continuous functional calculus).
If a is a normal element of a unital C*-algebra A and f ∈ C(σ(a)) then
σ(f (a)) = f (σ(a)) = {f (λ) : λ ∈ σ(a)}.
Proof. Since θa is a unital ∗-isomorphism of C(σ(a)) onto C ∗ (1, a), we have
σA (f (a)) = σC ∗ (1,a) (f (a))
by Corollary 4.1.11
= σC ∗ (1,a) (θa (f ))
by the definition of f (a)
= σC(σ(a)) (f )
by Proposition 4.1.14(iv)
= f (σ(a))
by Example 1.3.2(ii).
4.4
Positive elements of C*-algebras
Let us write R+ for the non-negative real numbers.
4.4.1 Definition. Let A be a C*-algebra. If a ∈ A then we say that a is
positive and write a ≥ 0 if a is hermitian with σ(a) ⊆ R+ . The set of positive
elements of A will be denoted by A+ = {a ∈ A : a ≥ 0}.
4.4.2 Example. In the case A = C(X) where X is a compact Hausdorff
space, we have
C(X)+ = {f ∈ C(X) : f (x) ∈ R+ for x ∈ X}
= {f ∈ C(X) : f = f ∗ and kf − tk ≤ t for some t ∈ R+ }
= {g ∗ g : g ∈ C(X)}.
4.4.3 Corollary. Let A be a unital C*-algebra. For every a ∈ A+ there is a
unique element b ∈ A+ with b2 = a.
√
Proof. Since σ(a) ⊆ R+ , the function f : σ(a) → C, t 7→ t is well-defined,
continuous and takes values in R+ . Hence we can define b = f (a), and since
f = f ∗ we have b∗ = f ∗ (a) = f (a) = b so b is hermitian. By Theorem 4.3.13,
σ(b) = f (σ(a)) ⊆ R+ , so b ∈ A+ . Since f 2 = z is the identity function
on σ(a), by Theorem 4.3.9 we have b2 = f (a)2 = (f 2 )(a) = z(a) = a. This
establishes the existence of b ∈ A+ with b2 = a.
To see that b is the unique element with these properties, suppose that
c ∈ A+ with c2 = a. Then ac = c3 = ca, so B = C ∗ ({1, a, c}) is a unital
abelian C*-algebra and b = f (a) ∈ C ∗ ({1, a}) ⊆ B. If τ ∈ Ω(B) then
τ (b)2 = τ (b2 ) = τ (a) = τ (c2 ) = τ (c)2 ;
since b, c ∈ A+ we have τ (b), τ (c) ≥ 0, so τ (b) = τ (c). By Corollary 4.3.2,
b = c.
40
4.4.4 Definition. If A is a unital C*-algebra and a ∈ A+ then we call the
element b ∈ A+ with b2 = a the positive square root of a, and write it as a1/2 .
4.4.5 Lemma. Let A be a unital C*-algebra.
(i). If a is a hermitian element of A then a2 ≥ 0.
(ii). If a is a hermitian element of A then a ≥ 0 if and only if ka − tk ≤ t
for some t ∈ R+ .
(iii). If a, b ∈ A with a ≥ 0 and b ≥ 0 then a + b ≥ 0.
(iv). If a ∈ A with a ≥ 0 and −a ≥ 0 then a = 0.
(v). If a ∈ A and −a∗ a ≥ 0 then a = 0.
Proof. (i) Since σ(a) ⊆ R by Corollary 4.1.10(i), the spectral mapping theorem 1.3.8 gives σ(a2 ) = {λ2 : λ ∈ σ(a)} ⊆ R+ , so a2 ≥ 0.
(ii) If a ≥ 0 then σ(a) ⊆ [0, t] where t = r(a) ∈ R+ . By Theorem 1.3.8
we have σ(a − t) = σ(a) − t ⊆ [−t, 0] and a − t is hermitian, so we have
ka − tk = r(a − t) ≤ t by Proposition 4.1.12.
Conversely, if a = a∗ and ka − tk ≤ t for some t ∈ R+ then σ(a − t) ⊆
[−t, t], so σ(a) = σ(a − t) + t ⊆ [0, 2t] by Theorem 1.3.8. Hence a ≥ 0.
(iii) Clearly, a + b is hermitian, and by (ii) there exist s, t ∈ R+ such that
ka − sk ≤ s and kb − tk ≤ t. By the triangle inequality,
ka + b − (s + t)k ≤ ka − sk + kb − tk ≤ s + t,
so a + b ≥ 0 by (ii).
(iv) We have σ(a) ⊆ R+ and σ(−a) = −σ(a) ⊆ R+ , so σ(a) = {0}. Hence
kak = r(a) = 0 by Proposition 4.1.12, so a = 0.
(v) Using Proposition 4.1.7(iii), write a = h + ik where h and k are
hermitian elements of A. By Proposition 1.3.3(ii), −aa∗ is also positive.
Now a∗ a + aa∗ = 2(h2 + k 2 ), so a∗ a = 2(h2 + k 2 ) − aa∗ ≥ 0 by (i) and (iii).
By (iv), a∗ a = 0, so kak2 = ka∗ ak = 0 and a = 0.
4.4.6 Theorem. If A is a unital C*-algebra then A+ = {a∗ a : a ∈ A}.
Proof. If b ∈ A+ then b = a2 = a∗ a where a = b1/2 , so A+ ⊆ {a∗ a : a ∈ A}.
It remains to show that if a ∈ A then b = a∗ a ≥ 0. By Proposition 4.1.7(i), b is hermitian. Consider the three functions z, z + , z − ∈ C(σ(b))
given by
(
(
0
if λ ≥ 0
λ
if
λ
≥
0
and z − (λ) =
z(λ) = λ, z + (λ) =
−λ if λ < 0
0
if λ < 0
41
Observe that z ± (σ(b)) ⊆ R+ , z = z + − z − and z + z − = 0. Thus, writing b+ =
z + (b) and b− = z − (b), we have b± ≥ 0 by Theorem 4.3.13, and b = b+ − b−
and b+ b− = 0 by Theorem 4.3.9.
Let c = ab− . We have
−c∗ c = −b− a∗ ab− = −b− (b+ − b− )b− = (b− )3 ≥ 0.
By Lemma 4.4.5(v), c = 0, so (b− )3 = 0. Hence σ(b− ) = {0}; since b− is
hermitian, Proposition 4.1.12 shows that b− = 0, so b = b+ ≥ 0.
4.4.7 Definition. If a, b ∈ A then let us write a ≤ b if b − a ≥ 0.
4.4.8 Remark. It follows from parts (iii) and (iv) of Lemma 4.4.5 that the
relation ≤ is a partial order on A.
4.4.9 Lemma. If A is a unital C*-algebra and a ∈ A+ then a ≤ kak1.
Proof. Exercise.
4.4.10 Corollary. Let A be a unital C*-algebra.
(i). If a, b ∈ A with a ≤ b then c∗ ac ≤ c∗ bc for any c ∈ A.
(ii). If a, b ∈ A then b∗ a∗ ab ≤ kakb∗ b.
Proof. (i) Let d = (b − a)1/2 . Then c∗ bc − c∗ ac = c∗ d2 c = (dc)∗ (dc) ≥ 0 by
Theorem 4.4.6.
(ii) We have a∗ a ≤ kak2 1 by Lemma 4.4.9, so b∗ a∗ ab ≤ b∗ (kak2 1)b =
kak2 b∗ b by (i).
4.5
The GNS representation
If V is a vector space, recall that a positive semi-definite sesquilinear form
on V is a mapping (·, ·) : V × V → C such that
(x, x) ≥ 0,
(λx + µy, z) = λ(x, z) + µ(y, z) and (y, x) = (x, y)
for all x, y, z ∈ V and λ, µ ∈ C. If this mapping also satisfies
(x, x) = 0 =⇒ x = 0
then we say that it is a positive definite, and it is then an inner product on V
(see [FA 4.1]).
You will probably have seen the next result proven for inner products,
but perhaps not for positive semi-definite sesquilinear forms. The proof is
basically the same, though.
42
4.5.1 Lemma (The Cauchy-Schwarz inequality). Let V be a vector space
and let (·, ·) : V × V → C be a positive semi-definite sesquilinear form. Then
|(x, y)|2 ≤ (x, x) (y, y)
for all x, y ∈ V .
Proof. For λ ∈ C we have
0 ≤ (λx − y, λx − y) = |λ|2 (x, x) − λ(x, y) − λ(y, x) + (y, y).
Taking λ = t(x, y) for t ∈ R gives
p(t) = t2 |(x, y)|2 (x, x) − 2t|(x, y)|2 + (y, y) ≥ 0 for all t ∈ R.
Observe that p(t) is a polynomial in t. If p(t) is constant then (x, y) = 0
and we are done. Otherwise, |(x, y)|2 6= 0 and since all non-constant linear
polynomials take negative values, p(t) must have degree 2.
Since p(t) ≥ 0 for all t ∈ R, its discriminant “b2 − 4ac” is not positive.
Hence 4|(x, y)|4 − 4|(x, y)|2 (x, x)(y, y) ≤ 0, so |(x, y)|2 ≤ (x, x)(y, y).
4.5.2 Definition. Let A be a unital C*-algebra. We say that a linear functional τ : A → C is positive if τ (a) ≥ 0 for all a ∈ A+ .
4.5.3 Remark. If τ is a positive linear functional on A and a, b ∈ A with
a ≤ b (see Remark 4.4.8) then b − a ≥ 0 so τ (b − a) = τ (b) − τ (a) ≥ 0, so
τ (a) ≤ τ (b). Thus positive linear functionals are order-preserving.
4.5.4 Lemma. Let A be a unital C*-algebra and let τ be a positive linear
functional on A.
(i). τ (a∗ ) = τ (a) for all a ∈ A.
(ii). |τ (a)| ≤ τ (1) kak for all a ∈ A. Hence τ ∈ A∗ and kτ k = τ (1).
Proof. (i) If h is a hermitian element of A then σ(h) ⊆ [−t, ∞) for some
t ≥ 0, so h + t1 ≥ 0. Hence τ (h + t1) = τ (h) + tτ (1) ≥ 0. Since τ (1) ≥ 0, we
must have τ (h) ∈ R. Now if a ∈ A then a = h + ik for hermitian elements h
and k, so τ (a∗ ) = τ (h − ik) = τ (h) − iτ (k) = τ (h) + iτ (k) = τ (a).
(ii) If a ∈ A then a∗ a ≥ 0 by Theorem 4.4.6, so a∗ a ≤ ka∗ ak1 = kak2 1 by
Lemma 4.4.9. Hence τ (a∗ a) ≤ τ (kak2 1) = τ (1)kak2 .
Consider the map (·, ·) : A × A → C given by (a, b) = τ (b∗ a). Using (i) it
is easy to check that this is a positive semi-definite sesquilinear form, so by
the Cauchy-Schwarz inequality we have
|τ (a)|2 = |(a, 1)|2 ≤ (1, 1) (a, a) = τ (1) τ (a∗ a) ≤ τ (1)2 kak2 .
Hence |τ (a)| ≤ τ (1)kak, and the result follows.
43
4.5.5 Lemma. Let A be a unital C*-algebra and let τ ∈ A∗ . Then τ is
positive if and only if kτ k = τ (1).
Proof. If τ is positive then kτ k = τ (1) by Lemma 4.5.4(ii). Conversely,
suppose that kτ k = τ (1). Multiplying τ by a positive constant if necessary,
we may assume that kτ k = τ (1) = 1. By Lemma 4.1.8, τ (a) is real whenever
a is hermitian. If a ≥ 0 and kak ≤ 1 then k1 − ak = r(1 − a) ≤ 1. Hence
|1 − τ (a)| = |τ (1 − a)| ≤ 1.
Since τ (a) is real, this shows that τ (a) ≥ 0 so τ is positive.
4.5.6 Definition. Let A be a unital C*-algebra. A state on A is a positive
linear functional of norm 1. We write S(A) for the set of states of A. Thus
by the preceding lemma,
S(A) = {τ ∈ A∗ : kτ k = τ (1) = 1}.
4.5.7 Remark. If A is an abelian unital C*-algebra then Ω(A) ⊆ S(A)
by Lemma 3.1.4. Thus states generalise the characters in the abelian case.
Moreover, if we turn S(A) into a topological space by giving it the subspace
topology from the weak* topology on A∗ , then S(A) = J1−1 (1) ∩ A∗1 is a
compact Hausdorff space.
4.5.8 Lemma. Let A be a unital C*-algebra. If a is a hermitian element
of A then there is a state τ ∈ S(A) with |τ (a)| = kak.
Proof. Let B = C ∗ (1, a), which is a unital abelian C*-algebra. By Corollary 3.1.11(ii), there is τ0 ∈ Ω(B) with |τ0 (a)| = r(a), and r(a) = kak by
Proposition 4.1.12. Moreover, we have τ0 (1) = 1 = kτ0 k by Lemma 3.1.4.
By the Hahn-Banach theorem [FA 3.6], we can extend τ0 to a functional
τ ∈ A∗ with kτ k ≤ 1. Since τ (1) = τ0 (1) = 1 we have kτ k = τ (1) = 1 so
τ ∈ S(A), and kak = |τ0 (a)| = |τ (a)|.
4.5.9 Lemma. Let A be a unital C*-algebra. For every τ ∈ S(A) there is a
Hilbert space H and a ∗-homomorphism π : A → B(H) such that
τ (a∗ a) ≤ kπ(a)k2 ≤ kak2
for all a ∈ A.
Proof. Consider the mapping (·, ·) : A × A → C defined by
(a, b) = τ (b∗ a),
a, b ∈ A.
It is easy to check that this is a positive semi-definite sesquilinear form.
Observe that if a ∈ A then a∗ a ≤ kak2 1 by Lemma 4.4.9, so
(a, a) = τ (a∗ a) ≤ kak2 τ (1) = kak2 .
44
Moreover, if a, b, c ∈ A then (ab, c) = τ (c∗ ab) = τ ((a∗ c)∗ b) = (b, a∗ c).
By the Cauchy-Schwarz inequality 4.5.1, for each a ∈ A we have
(a, a) = 0 ⇐⇒ (a, b) = 0 for all b ∈ A.
Hence if N = {a ∈ A : (a, a) = 0} then
N = {a ∈ A : (a, b) = 0 for all b ∈ A}.
Using this expression, we can check that N is a left ideal of A. Indeed, if
a, b ∈ N then (a + b, c) = (a, c) + (b, c) = 0 and (λa, c) = λ(a, c) = 0 and
(ca, d) = (a, c∗ d) = 0 for all c, d ∈ A and λ ∈ C, so a + b, λa and ca are in N .
Consider the vector space V = A/N and let us write [a] = a + N ∈ V for
a ∈ A. We define a mapping h·, ·i : V × V → C by
h[a], [b]i = (a, b),
a, b ∈ A.
This is well-defined since if [a1 ] = [a2 ] and [b1 ] = [b2 ] then a2 − a1 ∈ N and
b2 − b1 ∈ N , so (a1 , b1 ) = (a1 , b1 ) + (a2 − a1 , b1 ) + (a2 , b2 − b1 ) = (a2 , b2 ). It
follows that h·, ·i is a positive semi-definite sesquilinear form on V . In fact, it
is positive definite since if h[a], [a]i = 0 then (a, a) = 0 so a ∈ N , i.e. [a] = 0.
Hence (V, h·, ·i) is an inner product space.
Let H be the completion of this inner product space. It is not hard to
show that h·, ·i extends to an inner product on H turning H into a Hilbert
space.
Let L(V ) denote the set of linear maps V → V , and let π0 : A → L(V )
be given by π0 (a)[b] = [ab] for [b] ∈ V . This is well-defined since if [b1 ] = [b2 ]
then b2 − b1 ∈ N , so if a ∈ A then a(b2 − b1 ) = ab2 − ab1 ∈ N (since N is a left
ideal) and so [ab1 ] = [ab2 ]. Moreover, for a, b ∈ A we have b∗ a∗ ab ≤ kak2 b∗ b
by Corollary 4.4.10, so
kπ0 (a)[b]k2 = k[ab]k2 = (ab, ab) = τ (b∗ a∗ ab) ≤ kak2 τ (b∗ b) = kak2 k[b]k2 .
Hence kπ0 (a)[b]k ≤ kak k[b]k, so π0 (a) is a continuous linear map by [FA 1.8.1],
and its operator norm satisfies kπ0 (a)k ≤ kak.
Thus π0 (a) has a unique extension to a map in B(H), written π(a), and
kπ(a)k ≤ kak. This defines a mapping π : A → B(H).
We claim that π is a ∗-homomorphism. Since V is dense in H and π(a)
is continuous for all a ∈ A, it suffices to check these properties on V . For
a, b, c ∈ A and λ ∈ C we have
π(a + b)[c] = [(a + b)c] = [ac + bc] = π(a)[c] + π(b)[c] = π(a) + π(b) [c] and
π(λa)[c] = [λac] = λπ(a)[c] so π is linear,
π(ab)[c] = [abc] = π(a)[bc] = π(a)π(b)[c], so π is a homomorphism, and
hπ(a)[b], [c]i = (ab, c) = (b, a∗ c) = h[b], π(a∗ )[c]i so π(a∗ ) = π(a)∗ .
45
Hence π is a ∗-homomorphism.
It remains to establish the inequality τ (a∗ a) ≤ kπ(a)k2 ≤ kak2 . We have
already remarked that kπ(a)k ≤ kak. Now k[1]k2 = (1, 1) = τ (1) = 1, so
kπ(a)k2 ≥ kπ(a)[1]k2 = k[a]k2 = (a, a) = τ (a∗ a).
We can now show that every C*-algebra is, up to isometric ∗-isomorphism,
a subalgebra of B(H) for some Hilbert space H. To do so, we will piece together all of the representations constructed in Lemma 4.5.9. Since there are
a lot of these representations, we will need to take care of a few technicalities
first.
Let I be an index set. If we are given αi ≥ 0 for each i ∈ I, then we will
say that (αi )i∈I is summable if the set of finite sums
X
αi : J is a finite subset of I
i∈J
P
is bounded above, and then we declare the value
P of i∈I αi to be supremum
of this set. If (αi )i∈I is summable,
P we write i∈I αi < ∞.
Suppose that
i | < ∞. Writing J = {j ∈ I : tj ≥ 0}
i∈I |tP
P ti ∈ R and
P
we can define i∈I ti = j∈J tj − k∈I\J (−tk ). Similarly, if λi ∈ C and
P
P
P
P
i∈I |λi | < ∞ then we can define
i∈I λi =
i∈I Re(λi ) + i
i∈I Im(λi ).
If {Hi : i ∈ I} is a family of Hilbert spaces, then we can define an inner
product space H as follows:
X
2
H = (ξi )i∈I : ξi ∈ Hi for each i ∈ I, and
kξi k < ∞
i∈I
with “pointwise” vector space operations (ξi )i∈I + (ηi )i∈I = (ξi + ηi )i∈I ,
λ(ξi )i∈I = (λξi )i∈I and inner product
X
h(ξi )i∈I , (ηi )i∈I i =
hξi , ηi i.
i∈I
P
It follows from the Cauchy-Schwarz inequality that
i∈I |hξi , ηi i| < ∞ if
(ξi )i∈I , (ηi )i∈I ∈ H, so the inner product is well-defined.
It is not too hard to show that H is then a complete inner product space,
i.e. H is a Hilbert space. We usually write
M
H=
Hi
i∈I
and call H the Hilbertian direct sum of the Hilbert spaces {Hi : i ∈ I}.
46
Suppose that ai ∈ B(Hi ) for each i ∈ I and that supi∈I kai k < ∞. We
may defineP
an operator a ∈ B(H) by a((ξ
P i )i∈I ) =2 (a(ξi ))i∈I for (ξi )i∈I ∈ H.
2
Note that i∈I kai ξi k ≤ (supi∈I kai k) i∈I kξi k for each x ∈ X, so a is a
bounded operator, with kak ≤ supi∈I kai k. In L
fact, it is easy to show that
ai .
kak = supi∈I kai k [exercise]. We will write a =
i∈I
4.5.10 Theorem (The Gelfand-Naimark-Segal theorem). If A is a unital
C*-algebra then A is isometrically ∗-isomorphic to a subalgebra of B(H) for
some Hilbert space H.
Proof. Given τ ∈ S(A), let us write Hτ and πτ : A → B(Hτ ) for the Hilbert
space and the ∗-homomorphism obtained from Lemma 4.5.9. Let
M
M
H=
Hτ and define π(a) =
πτ (a) for a ∈ A.
τ ∈S(A)
τ ∈S(A)
It is easy to see that this defines a ∗-homomorphism π : A → B(H). If a ∈ A
then since kπ(a)k = supτ ∈S(A) kπτ (a)k, we have
sup τ (a∗ a) ≤ kπ(a)k2 ≤ kak2
τ ∈S(A)
by Lemma 4.5.9. However, a∗ a is hermitian so by Lemma 4.5.8,
sup τ (a∗ a) ≥ ka∗ ak = kak2 .
τ ∈S(A)
Putting these inequalities together shows that kπ(a)k = kak for each a ∈ A,
so π is an isometric ∗-homomorphism. In particular, the range of π is a closed
∗-subalgebra of B(H) which is ∗-isomorphic to A.
47