Chapter 1 Banach Algebras

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

kabk ≤ kak kbk for all a, b ∈ A, λ ∈ C.

For any x, x0, y, y0 ∈ A, we have

kxy − x0y0k = kx(y − y0) + (x − x0)yk ≤ kxk ky − y0k + kx − x0k ky0k 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,b in 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 1l0 are units, then 1l = 1l1l0 = 1l0. Example 1.2. If E is a complex Banach space, then B(E), the set of bounded linear 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 k1lk ≤ k1lk k1lk, which implies that k1lk ≥ 1.

1 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, 0 y ∈ A. Then if Lx = Lx0 , it follows that Lx1l = Lx0 1l and so x = x . Hence x 7→ Lx is an injective map from A into the set of linear operators on A. Now,

kLxyk = kxyk ≤ kxk kyk , for y ∈ A which implies that Lx is bounded, and kLxk ≤ kxk. Put |||x||| = kLxk. Then we have just shown that |||x||| ≤ kxk, for any x ∈ A. On the other hand,

|||x||| = kLxk = sup{kLxyk : kyk ≤ 1} = sup{kxyk : kyk ≤ 1} 1l ≥ kxy0k, where y0 = k1lk kxk = . k1lk

Hence, kxk/k1lk ≤ |||x||| ≤ kxk, for all x ∈ A, which shows that the two norms |||· ||| and k · k are equivalent. Moreover, for any x, y ∈ A,

|||xy||| = kLxyk

= kLxLyk

≤ kLxk kLyk = |||x||||||y||| and so A with norm ||| · ||| is a Banach algebra. To complete the proof, we have |||1l||| = kL1l k = 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 AI by (x, λ)(y, µ) = (xy + µx + λy, λµ).

Department of Mathematics 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 k(x, λ)k = kxk + |λ| . Then AI is a Banach space when equipped with this norm. Furthermore,

k(x, λ)(y, µ)k = k(xy + µx + λy, λµ)k = kxy + µx + λyk + |λµ| ≤ kxkkyk + |µ|kxk + |λ|k + k|λ||µ| = (kxk + |λ|)(kyk + |µ|) = k(x, λ)k k(y, µ)k .

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 k1lk = 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 functions dened on the interval [0, 1] equipped with the sup-norm, namely, , and with multiplication dened pointwise: kfk = sups∈[0,1] |f(s)|

(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 kfk = 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. Let be the Banach space 1 and dene by P A ` (Z) xy (xy)n = m xmyn−m for x = (xn) and y = (yn) in A. Then X X X |(xy)n| ≤ |xm||yn−m| n n m X X = |xm| |yn−m| m n X = |xm| kyk m = kxk kyk .

1 Thus xy ∈ ` (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) in A if there is some z ∈ A such that xz = zx = 1l. Note that if such a z exists, then it is unique; if z0x = xz0 = 1l, then z = z1l = zxz0 = 1lz0 = z0 . z is called the inverse of x, and is written x−1, as usual. 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 algebra 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 with , and put Pn k, . Then y ∈ A kyk < 1 sn = k=0 y n ∈ N (sn) is a Cauchy sequence in A and so converges, since A is complete. Let denote its limit; P∞ k. We claim that is the inverse of 1l . w w = k=0 y w − y Indeed, we have

(1l − y)w = lim(1l − y)sn n = lim(1l − yn+1) = 1l n and

w(1l − y) = lim sn(1l − y) n = lim(1l − yn+1) = 1l n which establishes the claim. Hence, if x ∈ A with k1l − xk < 1, then writing x = 1l − (1l − x) and arguing as above (with y = 1l − x), we see that x is invertible and its inverse −1 is given by the convergent series P∞ 1l k. x k=0( −x) Let . Then for any , we have −1 . Now, x0 ∈ G(A) x ∈ A x = x0x0 x 1l −1 −1 −1 k − x0 xk = kx0 (x0 − x)k ≤ kx0 k kx0 − xk,

Department of Mathematics Banach algebras 5 and so we conclude that if −1 −1 then −1 is invertible with kx − x0k < kx0 k x0 x inverse given by ∞ −1 −1 X k (x0 x) = y k=0 with 1l −1 . Hence is invertible and y = ( − x0 x) x ∞ −1 X −1 k −1 x = [x0 (x0 − x)] x0 . k=0

−1 To see that x 7→ x is continuous on G(A), suppose that x0 ∈ G(A) and that (xn) is a sequence in G(A) such that xn → x0 as n → ∞. Then for all suciently large , −1 −1 and so n kxn − x0k < kx0 k ∞ −1 −1 X −1 k −1 kx − x k = x (x0 − xn) x n 0 0 0 k=1 ∞ X −1 k −1 ≤ kx0 kkx0 − xnk kx0 k k=1 → 0 as n → ∞.

Denition 1.8. Let A be a unital algebra, and let x ∈ A. The spectrum of x is 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 of the 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 f does 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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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, whenever is not a constant, but when is the σA0 (f) = C f σA0 (f) = {α} f 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 1 satises 1l and . Thus, every open set p(x) = 1 + 2 εx k − pk < ε p∈ / A0 containing 1l also contains singular elements of A0.

Example 1.12. Let A denote the algebra of meromorphic functions and let B denote the algebra of entire functions. Clearly B is a subalgebra of A, and both 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. −1 On the other hand, the function z 7→ (z − λ) 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| ≤ kxk}. Proof. For given x ∈ A, x − λ1l is invertible whenever |λ| > kxk. Indeed, for any such λ, (x − λ1l) = −λ1l − x/λ has inverse given by a convergent series expansion in powers of x/λ. Hence σA(x) ⊆ {λ ∈ C : |λ| ≤ kxk}, as claimed. 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 −1 x − λ1l ∈ G(A), i.e., if and only if λ ∈ f (G(A)). It follows that C \ σA(x) = −1 f (G(A)). But it is clear that f : C → A is continuous and so G(A) open −1 implies that f (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

Department of Mathematics 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 → (x − λ1l)−2 ζ 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 (x − (λ + ζ)1l)−1 6ζ(x − λ1l)−1 = ζ 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 functionals on A). Then the map λ 7→ ϕ((x−λ1l)−1) is everywhere dierentiable −1 in C, that is, g(λ) ≡ ϕ((x − λ1l) ) is an entire function. For all suciently large |λ|, we may write

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

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

Proof. (i) Suppose that p has degree n ≥ 1. For any µ ∈ C, let λ1, . . . , λn be 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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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

−1 −1 a2a1a3 . . . an(a1 . . . an) = a1 . . . an(a1 . . . an) = 1l −1 = (a1 . . . an) a1 . . . an −1 = (a1 . . . an) a1a3 . . . ana2, which shows that a2 is invertible. Similarly, it is easy to see that any ai is invertible.) 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) = µ which 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 kxk < 1 and also kyk < 1. Then it follows that kxyk ≤ kxk kyk < 1, and similarly, kyxk < 1. We have seen that this implies that 1l − xy and 1l − yx are both invertible with inverses given, respectively, by a = (1l − xy)−1 = P∞ n and 1l −1 P∞ n. Evidently, n=0(xy) b = ( − yx) = n=0(yx) 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) x | {z } = 1l = 1l − yx + yx = 1l .

Department of Mathematics Banach algebras 9

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 Re α ≥ 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 ∈ N which is not possible. On the other hand, if α ∈ σA(ba) with Re α < 0, then α ∈ 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 position and momentum of a particle are represented by linear operators q and p acting on a Hilbert space and are supposed to satisfy the Heisenberg canonical commutation relation pq − qp = 1l. It therefore follows that q and p cannot 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 n 1/n and let x ∈ A. Then the limit limn→∞ kx k exists and satises

n 1/n n 1/n lim kx k = rA(x) = inf kx k . n→∞ n Proof. Set γ(x) = inf{kxnk1/n : n = 1, 2,... }.

n 1/n We shall show that kx k → γ(x) as n → ∞. Given any ε > 0, let k ∈ N be such that kxkk1/k < γ(x) + ε.

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+ 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, n αk + β αk β 1 = = = + n n n n αk α 1 and so → 1 as n → ∞, that is, → , as n → ∞. Now, n n k kxnk1/n = k(xk)αxβk1/n ≤ kxkkα/n kxkβ/n. and the right hand side converges to kxkk1/k, as n → ∞, which is less than γ(x) + ε. Hence, for all suciently large n, kxnk1/n ≤ kxkkα/n kxkβ/n < γ(x) + ε. On the other hand, γ(x) ≤ kxnk1/n for any n = 1, 2,... and so γ(x) ≤ kxnk1/n < γ(x) + ε for all suciently large n. Thus kxnk1/n converges to γ(x) as n → ∞, i.e., n 1/n m 1/m limn kx k exists and is equal to infm kx k . We must now show that the above limit is equal to rA(x). Recall, rst, that for any y ∈ A, σA(y) ⊆ {λ : |λ| ≤ kyk} and so rA(y) ≤ kyk. n n By proposition 1.14, it follows that {λ : λ ∈ σA(x)} = σA(x ) and so n n n n rA(x ) = rA(x) , for any n ∈ N. But rA(x ) ≤ kx k and so we obtain n n n rA(x) = rA(x ) ≤ kx k n 1/n =⇒ rA(x) ≤ kx k , for all n,

=⇒ rA(x) ≤ γ(x). ∗ We want to show now that rA(x) ≥ γ(x). Let ϕ ∈ A . Then g : λ 7→ −1 ϕ((x−λ1l) ) is analytic on C\σA(x), and so has a Laurent series expansion on {λ : |λ| > rA(x)}. But for |λ| > kxk, we know that g has the expansion ∞ X ϕ(xn) g(λ) = . λn+1 n=0

This must therefore be absolutely convergent in the region {λ : |λ| > rA(x)}. n n+1 Fix λ with |λ| > rA(x). Then, in particular, ϕ(x /λ ) → 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 > 0 such that kxn/λn+1k ≤ κ for all n. Hence kxnk ≤ κ|λ| |λn| and so kxnk1/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.

Department of Mathematics Banach algebras 11

Remark 1.18. We have shown that

n 1/n n 1/n inf kx k = lim kx k = sup{|λ| : λ ∈ σA(x)}. n n 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 in C(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

k cl xk = inf kx + vk = inf kx0k. v∈V x0∼x where cl x denotes the equivalence class of x in X/V . Proof. First recall that X/V is the set of equivalence classes in X determined by the relation x ∼ x0 if and only if x − x0 ∈ V . X/V is a linear space when equipped with the obvious operations; cl x + cl y = cl(x + y) and λ cl x = 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 k · k is a norm.

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If cl x = 0 in X/V , then x ∈ V and so, taking v = −x, we get k cl xk = infv∈V kx+vk = 0. On the other hand, if k cl xk = 0, then there is a sequence (vn) in V such that kx + vnk → 0 as n → ∞. Hence −vn → x in X. Since V is closed, we deduce that x ∈ V and so cl x = 0 in X/V . For λ ∈ C, λ 6= 0, and x ∈ X, k cl λxk = inf kλx + vk = inf kλx + λv0k v∈V v0∈V = |λ| inf kx + v0k = |λ| k cl xk. v0∈V For any x, y ∈ X,

k cl x + cl yk = inf kx + y + vk v∈V = inf kx + y + v + wk v,w∈V ≤ inf kx + vk + ky + wk v,w∈V = k cl xk + k cl yk.

Hence k · k 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 P∞ , then there is such that Pn as . m=1 kymk < ∞ y ∈ Y m=1 ym → y n → ∞ Proof. If is complete and satises P∞ , then it is clear Y (yn) k=1 kykk < ∞ that Pn is a Cauchy sequence and hence converges. k=1 yk

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 ∈ N be such that 1 for all . Now let be such kxn1 − xmk < 2 m > n1 n2 > n1 that 1 for all . Continuing in this way, we obtain a kxn2 − xmk < 4 m > n2 subsequence of such that (xnk ) (xn) 1 kx − x k < , k = 1, 2,... nk nk+1 2k Set for . Then yk = xnk − xnk+1 k = 1, 2,...

∞ ∞ X X kykk = kxnk − xnk+1 k k=1 k=1 ∞ X 1 < < ∞ . 2k k=1

Department of Mathematics Banach algebras 13

By hypothesis, there is some y ∈ Y such that

m X y = lim yk m→∞ k=1

= lim (xn − xn ) + (xn − xn ) + ··· + (xn − xn ) m→∞ 1 2 2 3 m m+1

= lim (xn − xn ). m→∞ 1 m+1 That is, converges in (to ). But if a subsequence of a Cauchy (xnk ) Y xn1 − y sequence 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 k cl xk = inf kx + vk. v∈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 (cl xn) be any sequence in such that P∞ . By denition of the inmum, X/V n=1 k cl xnk < ∞ for each n, there is vn ∈ V such that 1 kxn + vnk < inf kxn + vk + v∈V 2n 1 = k cl x k + . n 2n Hence ∞ ∞ X X 1 kx + v k < k cl xk + < ∞. n n 2n n=1 n=1 Now, X is a Banach space, so, by the previous proposition, it follows that there is y ∈ X such that

n X y = lim (xk + vk). n→∞ k=1 We claim that Pn in . Indeed, k=1 cl xk → cl y X/V

n n X X k cl y − cl xkk = k cl(y − xk)k k=1 k=1 n X = inf ky − xk + vk v∈V k=1

King's College London 14 Chapter 1

n n X X ≤ ky − xk − vkk k=1 k=1 n X = ky − (xk + vk)k k=1 → 0 as n → ∞.

Hence Pn converges in (to ) and so is a Banach space, k=1 cl xk X/V cl y X/V 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 k cl xk = inf kx + vk. v∈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 cl x cl y = cl xy. Furthermore,

k cl x cl yk = k cl xyk = inf kxy + vk v∈V ≤ inf kxy + xw + vy + vw k v,w∈V | {z } ∈ V = inf k(x + v)(y + w)k v,w∈V ≤ inf kx + vk ky + wk v,w∈V = k cl xk k cl yk.

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 k1lk = 1, then

k cl 1lk = inf k1l + vk v∈V

Department of Mathematics Banach algebras 15

≤ k1lk, (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 k cl 1lk = 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 associated 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 J cannot 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 J is 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), which is impossible). This means that ϕ(a) belongs to σA(a) and it follows that |ϕ(a)| ≤ kak. This inequality remains valid when ϕ(a) = 0 and we conclude that ϕ is continuous on A. 0 If A is non-unital, we consider AI instead. Dene the map ϕ : AI → C 0 by ϕ ((a, λ)) = ϕ(a) + λ, (a, λ) ∈ AI . Then it is straightforward to check 0 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 algebra 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) `(b) b = a + b − a . `(a) `(a)

`(b) Since b − a ∈ ker ` = J, we see that A = a + J and hence J is a `(a) C maximal ideal. Now suppose that J is a maximal ideal. Then J is closed and so A/J is 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 A which contains J as a proper subset. (J + aA does not contain 1l since cl a is 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 `0 have the same kernel. Then for any a ∈ A, a − `(a)1l belongs to ker ` = ker `0 and so `0(a) = `(a) since `0(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 invertible, 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 xA is a proper ideal and so is contained in a maximal proper ideal J, say, by Zorn'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 , then, for , the equality 2 would imply A i 6= j eij = 0 that `(eij) = 0. Hence the equality eii = eijeji, applied with i 6= j, would imply 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 Sp A.

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

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

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 ∈ G there 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 each a ∈ 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, Sp A, of a commutative unital Banach algebra 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 Sp A is contained in the unit ball of A∗. To show that Sp A is w∗-closed, we shall show that A∗ \ Sp A is open. Let ∗ . If , we have 1l 1 ∗ , ϕ ∈ A \ Sp A ϕ = 0 0 = ϕ ∈ N(0; { }, 2 ) ⊂ A \ Sp A since `(1l) = 1 for any ` ∈ Sp A. Suppose that ϕ 6= 0. Then there is a, b ∈ A such 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), i e , for suciently small ε > 0, N(ϕ : {a, b, ab}, ε) is contained in A∗ \ Sp A. Hence A∗ \ Sp A is open in A∗, and so Sp A is w∗-closed. Sp A is compact since it is a closed subset of a compact set.

Remark 2.12. The fact that Sp A is w∗-closed can be demonstrated quite ∗ easily using nets. Suppose that (`α) is a net in Sp A 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 ∈ Sp A. Note that ϕ is non-zero since ϕ(1l) = 1.

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

xb(`) = `(x). Then the range of the function on satises xb Sp A

ran xb = σA(x). Furthermore, the map is a homomorphism and b b : A → C(Sp A) for . kxbk∞ ≤ kxk, x ∈ A is called the Gelfand transform. b

Department of Mathematics Gelfand Theory 21

Proof. We have seen already that for any x ∈ A and ` ∈ Sp A we have ; i e, and so the range of satises the inclusion `(x) ∈ σA(x) xb(`) ∈ σA(x) xb . ran xb ⊆ 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 Sp A with ker ` = J. Then x − λ1l ∈ J implies that . Hence and it follows that . `(x) = λ xb(`) = `(x) = λ ran xb = σA(x) It is clear that is a homomorphism; for example, b

xyc(`) = `(xy) = `(x)`(y) for any , , = xb(`) yb(`) x, y ∈ A ` ∈ Sp A and so . Similarly, one sees that is linear. xyc = xb yb b To show that , let be any open set in . We must show xb ∈ C(Sp A) U C that −1 is open in . If −1 , then we are done. So suppose xb (U) Sp A xb (U) = ∅ that −1 . Let be any element of −1 . Then there is such xb (U) 6= ∅ ` xb (U) ζ ∈ U that . Since is open in , there is such that xb(`) = ζ U C ε > 0 Nε(ζ) ≡ {z ∈ C : |z − ζ| < ε} ⊆ U. Let V = N(` : {x}, ε) ≡ {ω ∈ Sp A : |ω(x) − `(x)| < . Then for all , i e, −1 . We ε} ω(x) = xb(ω) ∈ U ω ∈ V ` ∈ V ⊆ xb (U) deduce that −1 is open in and hence is continuous; xb (U) Sp A xb : Sp A → C that is, . (Alternatively, the continuity of can easily be xb( · ) ∈ C(Sp A) xb established using nets, as follows. Suppose that `α → ` in Sp A. Then , by denition of the ∗-topology. In other xb(`α) = `α(x) → `(x) = xb(`) w words, is continuous.) xb Finally, we have that and so it follows ran xb = σA(x) ⊆ {λ : |λ| ≤ kxk} that , for all . Thus for any . |xb(`)| ≤ kxk ` ∈ Sp A kxbk∞ ≤ kxk 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. Then the map is a homeomorphism. ba : Sp A → σA(a) ⊂ C Proof. We know that is a continuous function on with , ba Sp A ran ba = σA(a) i e, is continuous and onto. Now, both and ba : Sp A → σA(a) Sp A σA(a) are compact Hausdor spaces, so we need only show that is injective. ba To see this, suppose that , so that . Using the ba(`1) = ba(`2) `1(a) = `2(a) multiplicativity of `1 and `2, we see that for given N ∈ N and c0, c1, . . . , cN in C N N X n X n `1 cna = `2 cna . n=0 n=0

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 of M2(C) consisting of those elements of the form α β , with , . Then 0 α α β ∈ C

α β 0 1 = α1l + βq, where q = . 0 α 0 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, , for α β . Indeed, for , σA(x) x = 0 α λ ∈ C

α − λ β x − λ1l = 0 α − λ is invertible in M2(C) if and only if λ 6= α. If λ 6= α, then, in fact, (α − λ)−1 −β(α − λ)−2 (x − λ1l)−1 = , 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; Sp A = {`}, where ` is given uniquely by the action `(1l) = 1 and `(q) = 0. The Gelfand transform is the map , 1l 1l . But x 7→ xb α + βq 7→ αb + βqb 1l and so that and we have 1l , for b = 1 qb(`) = `(q) = 0 qb = 0 (α + βq)b = α any , . The transform has kernel , so we see that is α β ∈ C b {βq : β ∈ C} b 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 Sp A ' σA(q), via the homeomorphism , . Indeed, both qb : Sp A → σA(q) ` 7→ qb(`) = 0 Sp A and σA(q) are singleton sets! Alternatively, we can determine the spectrum of x ∈ A using the equality σ (x) = ran x. For x = α β , we have A b 0 α

σA(x) = {xb(`)} = {`(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) ka∗k = kak.

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) ka∗ak = kak2, 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 kak2 = ka∗ak ≤ ka∗k kak and so kak ≤ ka∗k. (If kak = 0, then a = 0 = a∗.) Replacing a by a∗ gives ka∗k ≤ kak and so their equality follows. Now suppose that A is a C∗-algebra with unit 1l. Then, by the C∗- property, k1l∗1lk = k1lk2 and so k1lk = k1lk2 which implies that k1lk = 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∗-algebra with unit, when equipped with the supremum norm and the involution ∗ . Since ∗ 2 it follows that ∗ 2 . f 7→ f = f f f = |f| kf fk∞ = kfk∞

2. Let Ω be a topological space. Cb(Ω), the algebra of continuous bounded complex-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 → C belongs 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; i e , the set {ω ∈ Ω: |f(ω)| ≥ ε} ∗ is compact.) Then C0(Ω) is a C -algebra without a unit. (A unit 2 f ∈ C0(Ω) must satisfy f = f 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 ω0 ∈ Ω \ K. Since Ω is locally compact, there 0 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 ω0 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 compact, non-compact and non-Hausdor. Indeed, let Ω = {0} ∪ N, and dene 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 subcover, 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 −1 n ∈ N; to see this, note that f (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 −1 ∅ and {0}, we conclude that n∈ / f (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 > 0 there is a compact set K, 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,

Department of Mathematics 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 isomorphic 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 kxk denote the operator norm of x (given by sup{kxξk : kξk ≤ 1, ξ ∈ H}) and let ∗ be the operator adjoint. Then B(H), equipped with this structure, is a unital C∗-algebra. (If dim H = 1, then we have the previous example.)

7. Any norm closed subalgebra of B(H) which contains x∗ if it contains x 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 adjoint 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- n dimensional complex Hilbert space C . (This is the same as dening kxk 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 Q consisting of those elements for α Aα (aα) which is nite. Then is a Banach space with respect to supα kaαk A the norm . Furthermore, is a ∗-algebra when k(aα)k = supα kaαk A equipped with component-wise operations; for example, the adjoint of is just ∗ , and the product of with is . It is (aα) (aα) (aα) (bα) (aαbα) 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 unit for , then it is readily see that for any , is a unit for . The A α0 eα0 Aα0 converse is clear.)

King's College London 26 Chapter 3

∗ ∗ The C -algebra A is called the direct sum of the C -algebras {Aα}. If ∞ α 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 symmetric 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. An element 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 1 1 x = (x + x∗) + i (x − x∗). 2 2i 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 − ik so that 1 ∗ and 1 ∗ . In other words, the decomposition h = 2 (x+x ) k = 2i (x−x ) 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 A of 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 ∞ X (it)n u = eith ≡ hn. t n! n=0 (the series converges in A(h) since A(h) is complete). Then by the continuity of the involution ∗ , we see that

n n X (it)k X (−it)k u∗ = lim hk ∗ = lim , hk t n→∞ k! n→∞ k! k=0 k=0

= u−t . By multiplying the series (as in the complex case), we see that

∗ 1l ut ut = u−tut = u0 = . Hence ∗ 2, and so for all . 1 = kut utk = kutk kutk = 1 t ∈ R Let ` ∈ Sp A(h). Then, since ` is continuous,

∞ X (it)n `(u ) = ` hn t n! n=0

Department of Mathematics C∗-algebras 27

∞ X (it)n = `(h)n n! n=0 = eit`(h).

it`(h) Since k`k = 1, we have |`(ut)| ≤ kutk = 1, and hence |e | ≤ 1 for all t ∈ R. It follows that `(h) ∈ R. This holds for all ` ∈ Sp A(h) and so bh is real-valued on . But and therefore . Sp A(h) σA(h)(h) = ran bh σA(h)(h) ⊂ R Now and so and we conclude that . A(h) ⊆ A σA(h) ⊆ σA(h)(h) σA(h) ⊂ R Theorem 3.7. (Gelfand-Naimark) Suppose that A is a commutative unital Banach ∗-algebra. The Gelfand transform is an isometric b : A → C(Sp A) ∗-isomorphism if and only if A is a C∗-algebra. Proof. If is an isometric ∗-isomorphism, then for any , b x ∈ A ∗ ∗ ∗ kx xk = kxdxk∞ = k(dx )xbk∞ 2 = k xb xb k∞ = k |xb| k∞ 2 2 = k xb k∞ = kxk which is precisely the C∗-property (v). So A is a C∗-algebra. (Indeed, C(Sp A) is a C∗-algebra, so if A is isometrically ∗-isomorphic to C(Sp A), 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, bh is real-valued. For any x ∈ A, write x = (x + x∗)/2 + i(x − x∗)/2i. Then x + x∗ (x − x∗) (dx∗)(`) = `(x∗) = ` − i 2 2i x + x∗ x − x∗ = ` − i ` 2 2i x + x∗ x − x∗ = ` + i ` 2 2i = `(x)

= xb(`), using the fact that is real if is symmetric. It follows that is a `(h) h b ∗-homomorphism. To show that is isometric (and hence also an injection) consider again b h = h∗ ∈ A. Then, by the C∗-property of the norm khk2 = kh2k and so khk2n = kh2n k. Therefore

2n 1/2n kbhk∞ = r(h) = lim kh k = khk n→∞ and so is isometric on self-adjoint elements. For any we have b x ∈ A 2 k xb k∞ = k xb xb k∞

King's College London 28 Chapter 3

∗ since is a ∗-homomorphism, = k xdx k∞ , b = kx∗xk , since x∗x is self-adjoint, = kxk2 , by the C∗-property of the norm.

Hence is isometric. b We must now show that is surjective. To see this, we note that is b A complete and since is isometric it follows that is closed in . b ran b C(Sp A) Now, is a ∗-homomorphism and so is a closed ∗-subalgebra of b ran b C(Sp A) which contains the constant function 1 = 1l.b Furthermore, if `1 6= `2, then, by denition, there is such that , that is, . x ∈ A `1(x) 6= `2(x) xb(`1) 6= xb(`2) Thus separates points of . It follows from the Stone-Weierstrass ran b Sp A theorem that . ran b = C(Sp A) 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(Sp A(x)). Let f : σA(x) → C be continuous. Since , it follows that is ran xb = σA(x) f ◦ xb : Sp A(x) → C well-dened and is continuous; i e, . Hence there exists f ◦ xb ∈ C(Sp A(x)) such that , and is unique since is an isomorphism. y ∈ A(x) ⊆ A yb = f ◦xb y b 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 invertible. Then x−1 belongs to the C∗-subalgebra of A generated by 1l, x and x∗ (i e , the closure in A of the set of complex polynomials in 1l, x, x∗). Proof. Suppose rst that x = x∗, and let A denote the unital C∗-algebra generated by x, and B that generated by x and x−1. Evidently B is commutative and . We know that the Gelfand transform A ⊆ B ⊆ A b : B → Bb = C(Sp B) is an (isometric ∗-) isomorphism and, since A is a C∗-subalgebra of B, it follows that (the range of under ) is a ∗-subalgebra of . Ab A b C Bb Let `1, `2 ∈ Sp B and suppose that `1(x) = `2(x). For any ` ∈ Sp B, −1 −1 −1 −1 −1 `(xx ) = `(x)`(x ) = `(1l) = 1. Hence `1(x ) = `1(x) = `2(x) = −1 −1 `2(x ). Since B is generated by x and x , we deduce that `1 = `2. Thus, if then or . That is, separates `1 6= `2 `1(x) 6= `2(x) xb(`1) 6= xb(`2) Ab points of Sp B. By the Stone-Weierstrass theorem, it follows that Ab = Bb 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

Department of Mathematics C∗-algebras 29

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) and so 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 of M2(C) consisting of those 2×2 complex matrices of the form α 0 , , and let be the ∗-subalgebra ( 0 0 ) α ∈ C B C consisting of those matrices of the form α 0 , . Then , and 0 β α, β ∈ C A ⊂ B and are both unital with units given by 1l 1 0 and 1l 1 0 . A B A = ( 0 0 ) B = ( 0 1 ) Note that 1l but 1l 1l in . Now if α 0 , then A ∈ B A 6= B B a = ( 0 0 ) ∈ A 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 via the Gelfand transform . On the other hand, C(Sp A) b : A → C(Sp A) bh : Sp A → σA(h) is a homeomorphism. Dene α : C(Sp A) → C(σA(h)) by α(f) = f ◦bh−1 so that α(f)(λ) = f ◦bh−1(λ)) and, in particular, α(bh)(λ) = λ ∗ for λ ∈ σA(h). Then α is an isometric -isomorphism from C(Sp A) onto . Hence is an isometric ∗-isomorphism. C(σA(h)) α ◦ b : A → C(σA(h)) Let p be a polynomial. Then α ◦ b(p(h)) (λ) = α(p(bh))(λ) = p(α(bh))(λ) = p(α(bh)(λ)) = p(λ) for any λ ∈ σA(h). Theorem 3.13. Let Ω be a compact Hausdor space, and let A be the commutative unital C∗-algebra C(Ω). Then Sp A ' Ω.

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

King's College London 30 Chapter 3 one-one, i.e., it is an identication of Ω as a subset of Sp A. We shall show that this mapping is onto Sp A. To see this, let ` ∈ Sp A, 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 ω such 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 , say. Put {Nω1 ,...,Nωk }

2 2 x = |bω1 | + ··· + |bωk | .

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

`(x) = `(b∗ b ) + ··· + `(b∗ b ) ω1 ω1 ωk ωk = `(b∗ )`(b ) + ··· + `(b∗ )`(b ) ω1 ω1 ωk ωk = 0 since for each . Thus, and on `(bωj ) = 0 j = 1, . . . , k x ∈ ker ` x > 0 Ω. But x > 0 implies that x−1 exists in A = C(Ω), and therefore `(1l) = −1 `(x)`(x ) = 0, which is impossible. We deduce that there is some ω0 ∈ Ω such that . Hence the map is one-one and onto . ` = ϕω0 ϕ : ω 7→ ϕω Sp A It remains to show that this is a homeomorphism. Since both Ω and Sp A are compact, it is enough to show that ϕ is continuous (for then ϕ−1 is automatically also continuous). Let U ⊆ Sp A be a non-empty open set and suppose that −1 , so that . Since is open, ω0 ∈ ϕ (U) ϕω0 ∈ U U there is ε > 0 and a nite set S ⊂ A such that the w∗-neighbourhood . Each is continuous at and so there are open N(ϕω0 : S, ε) ⊆ U x ∈ S ω0 neighbourhoods Vx of ω0 in Ω such that |x(ω) − x(ω0)| < ε for all ω ∈ Vx. Put T . Then is open in and . For any , we V = x∈S Vx V Ω ω0 ∈ V ω ∈ V have , for all , that is, , for all |x(ω) − x(ω0)| < ε x ∈ S |ϕω(x) − ϕω0 (x)| < ε . Hence and thus −1 x ∈ S ϕ(V ) ⊆ N(ϕω0 : S, ε) ω0 ∈ V ⊆ ϕ (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 in . Then, for each , we have ωα → ω Ω x ∈ A = C(Ω) ϕωα (x) = x(ωα) → x(ω) since x is a continuous function on Ω. But x(ω) = ϕω(x), and so we conclude that with respect to the ∗-topology on . ϕωα → ϕω w Sp A

We shall now consider the problem of giving a unit to a C∗-algebra when one does not already exist. Let A be a C∗-algebra, and let B(A) denote the Banach algebra of bounded operators on A. Let Ae denote the normed subalgebra of B(A) consisting of those elements of the form La + α1l, with

Department of Mathematics C∗-algebras 31 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 kLaxk = kaxk ≤ kakkxk, so we have kLak ≤ kak. The map a 7→ La from A into Ae is linear, and since ab 7→ Lab = LaLb, we see that it is an algebra homomorphism. Furthermore,

∗ ∗ ∗ 2 kLaa k = kaa k = ka k = kak2 = kakka∗k which, together with the inequality kLak ≤ kak, implies that kLak = kak. Thus the map a 7→ La is isometric and, in particular, injective. ∗ Dene an involution on Ae by (La + α1l) = La∗ + α 1l. Then the map ∗ a 7→ La of A into Ae is an isometric -homomorphism. If A has a unit, then Lα1l = α1l, so that every element of Ae has the form ∗ La for some a ∈ A. In this case, a 7→ La is an isometric -isomorphism of A onto Ae (and so Ae 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 Ae with codimension 1. We claim that Ae is a unital C∗-algebra. We must show that

∗ 2 k(La + α1l) (La + α1l)k = kLa + α1lk and that Ae is complete. For α ∈ C, and a, x ∈ A, we have

2 2 k(La + α1l)xk = kax + αxk = k(ax + αx)∗(ax + αx)k = k(x∗a∗ + αx∗)(ax + αx)k ∗ ∗ = kx (La + α1l) (La + α1l)xk ∗ ∗ ≤ kx k k(La + α1l) (La + α1l)k kxk ∗ 2 = k(La + α1l) (La + α1l)k kxk .

Therefore, 2 ∗ kLa + α1lk ≤ k(La + α1l) (La + α1l)k (∗) It follows that

2 ∗ kLa + α1lk ≤ k(La + α1l) (La + α1l)k ∗ ≤ k(La + α1l) k kLa + α1lk.

We observe now that if kLa + α1lk = 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

King's College London 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

∗ kLa + α1lk ≤ k(La + α1l) k

= kLa∗ + α 1lk.

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

∗ kLa + α1lk = k(La + α1l) k.

This, together with the inequality (∗), gives

2 ∗ ∗ kLa + α1lk ≤ kLa + α1l La + α1lk ≤ kLa + α1l k kLa + α1lk 2 = kLa + α1lk

∗ 2 ∗ and so kLa + α1l La + α1lk = kLa + α1lk , which is the C -property. It remains to show that Ae is complete. To see this, let us dene φ : Ae → C to be the map φ(La +α1l) = α. Note that φ is a well-dened linear functional on Ae 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., the kernel of φ is a closed subset of B(A). It follows that φ is a continuous linear functional on Ae. Suppose that 1l is a Cauchy sequence in . Then, evidently, (Lan + αn ) Ae 1l is a Cauchy sequence in , that is, is a Cauchy sequence (φ(Lan +αn )) C (αn) in C, and so converges to α, say. Now 1l 1l 1l 1l kLan − Lam k ≤ k(Lan − αn ) − (Lam − αm )k + kαn − αm k | {z } =|αn−αm| and so is a Cauchy sequence in . But (Lan ) B(A) kan − amk = kLan − Lam k implies that (an) is a Cauchy sequence in A. Hence there is a ∈ A such that . It follows that and 1l 1l in , which implies that an → a Lan → La αn → α Ae 1l 1l in , and we deduce that is complete. Lan + αn → La + α Ae Ae ∗ Identifying A with {La : a ∈ A}, via the isometric -isomorphism a 7→ La allows us to regard A as a C∗-subalgebra (and also a closed two-sided ∗-ideal) in Ae, 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 Ae is just A ⊕ C with the involution a ⊕ α 7→ a∗ ⊕ α, multiplication (a ⊕ α)(b ⊕ β) = (ab + αb + βa) ⊕ αβ and norm k(a ⊕ α)k = sup{kax + αxk : x ∈ A, kxk ≤ 1}. The unit is 0 ⊕ 1.

Department of Mathematics C∗-algebras 33

(Note that if A has a unit, then we saw that Ae = A and so, in this case, Ae is 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 k · k 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 Ae, 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

khk = kbhk∞ = sup{|λ| : λ ∈ σ(h)} = |||h||| since inverses are dened algebraically (independently of the norm). In general, for any x ∈ A,

kxk2 = kx∗xk = |||x∗x||| = |||x|||2 =⇒ k · k = ||| · |||.

Remark 3.16. This result also follows from the observation that for h = h∗ ∈ A, khk = 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 generated (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 generated by a in Ae 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 A be 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{kax + αxk : x ∈ K(H), kxk ≤ 1} .

King's College London 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∗-algebra with 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 Ae. 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 kpk = 1 so that

k(a + α1l)ξk = kaξ + αξk = k(ap + αp)ξk ≤ kap + αpk ≤ |||a + α1l||| .

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

ka + α1lk ≤ |||a + α1l||| .

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

kax + αxk ≤ ka + α1lkkxk ≤ ka + α1lk and so, taking the supremum over all such x,

|||a + α||| ≤ ka + α1lk .

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 -subalgebra of 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(Sp A), where Sp A is the space of characters on A. Suppose now that A is a commutative C∗-algebra without a unit. Evidently, Ae is a commutative unital C∗-algebra and so is isomorphic to the C∗-algebra of continuous functions on its character space, Sp Ae. It is clear that any character on A extends uniquely to a character on Ae (by simply assigning it the value 1 on 1l, the unit in Ae). Conversely, every character on Ae 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 Ae has one more character than A, that being the character κ0, say, given by κ0(a) = 0 for all a ∈ A ⊂ Ae (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 Ae. Then K is a compact Hausdor space and Ae 'C(K), by 3.7. Thus A is isometrically ∗-isomorphic to a C∗-subalgebra of C(K) via the Gelfand transform on Ae. Let κ0 ∈ K be the character on Ae as above; ( 0, for a ∈ A ⊂ Ae κ0(a) = 1, a = 1l .

Then, for any , , so that the image of under consists a ∈ A ba(κ0) = 0 A b of functions in C(K) which vanish at κ0 ∈ K. Conversely, suppose that is such that . Let be such that . Now can f ∈ C(K) f(κ0) = 0 x ∈ Ae xb = f x be written as x = a + µ1l, for some a ∈ A and µ ∈ C, and we have

f(κ0) = 0 =⇒ xb(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 Ae 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 Ae 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 ⊂ Ae corresponding ( 0, κ = κ to the continuous function e(κ) = 0 would be a unit for A, b 1, otherwise 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 , such that is unitarily equivalent to L 2 and is unitarily a H α L (σ(a), dµα) a equivalent to multiplication by λ, λ ∈ σ(a); i e , if ζ ∈ H corresponds to L 2 , then corresponds to the map , for f ∈ α L (σ(a), dµα) aζ λ 7→ λf(λ) λ ∈ σ(a).

Proof. Let A be the commutative unital C∗-subalgebra of B(H) generated by . Then we have seen (theorem 3.12) that the map −1 , de- a ψ : x 7→ xb(ba (·) ) 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 Z µξ(f) = f dµξ . σ(a) Dene u : C(σ(a)) → H by uf = φ(f)ξ. Then

kufk2 = (φ(f)∗φ(f)ξ, ξ) = (φ(f ∗f)ξ, ξ) Z 2 = |f| dµξ σ(a) 2 = kfk 2 . L (σ(a),dµξ)

37 38 Chapter 4

Hence u is isometric with a dense range (since ξ is assumed cyclic) and with 2 a dense domain of denition in L (σ(a), dµξ), namely C(σ(a)). Therefore u 2 extends uniquely to dene a unitary operator : L (σ(a), dµξ) → H. For any f ∈ C(σ(a)), we have

(u−1a u f)(λ) = (u−1a φ(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.

2 By continuity, this holds for any f ∈ L (σ(a), dµξ) and gives the required unitary equivalence. If ξ ∈ H does not exist as above (i e , if there is no cyclic vector ξ), then, by Zorn's lemma, there is a family of orthogonal subspaces {Hα} in H, with L , and vectors such that is dense in . α Hα = H ξα ∈ Hα Aξα Hα As above, we construct 2 , for each . Then uα : L (σ(a), dµξα ) → Hα α L gives the required unitary equivalence. u = α uα 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,

Z Z kak (iv) a = λ deλ = lim λ deλ, where the integral is a norm ε↓0 R −kak−ε 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 ∗-algebra generated by . Then , via the Gelfand transform , C a A'C(K) b where , and is real-valued, since is self-adjoint. For given K = Sp A ba(·) a λ ∈ R, dene a function pλ(·) on K by

p (·) = χ (·) , λ {κ∈K:ba(κ)≤λ} i e, if , otherwise . Write pλ(κ) = 1 ba(κ) ≤ λ pλ(κ) = 0 Kλ = {κ ∈ K : ba(κ) ≤ . Since is continuous, it follows that is closed in . λ} ba Kλ 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 n ζ (κ) → pλ(κ) for each κ ∈ K, as n → ∞. Let . The map , , denes a positive linear ξ ∈ H xb 7→ (xξ, ξ) x ∈ A functional on C(K), and so, by the Riesz-Markov representation theorem, there is a regular Borel measure µξ on K such that Z (xξ, ξ) = xb(κ) dµξ(κ) . K n Now, by the dominated convergence theorem, we see that ζ → pλ in 2 n 2 L (K, dµξ). In particular, (ζ ) is L -Cauchy. Since ζ ∈ C(K), there is such that , and so n n . Note that . We have y ∈ A ζ = yb ζ = (y )b kyk ≤ 1 Z n m 2 n m 2 kζ − ζ k 2 = |y − y | dµ (κ) L (K,dµξ) c c ξ K = ((yn − ym)∗(yn − ym)ξ, ξ) = k(yn − ym)ξk2 and so we see that (ynξ) is a Cauchy sequence in H, for each ξ ∈ H. It n follows that s-limn→∞ y exists, and denes a bounded operator which we denote by . Since is real-valued, each n is self-adjoint, and so ∗ . eλ ζ y eλ = eλ Furthermore, since (kynk) is bounded (by 1) we have

2 n n 2n eλ = s-limn→∞ y y = s-limn→∞ y = 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 ζ0 is another element of C(K) such that 0 ≤ ζ0(κ) < 1, for 0 0 κ∈ / Kλ, and ζ (κ) = 1, for κ ∈ K, then we repeat the proof to obtain y ∈ A with 0n 0 strongly as . But both n and 0 n converge to y → eλ n → ∞ (yb ) (yb ) pλ in L2 and so we can consider the combined sequence ( yn, n even y00 = n y0n, n odd.

This is 2-Cauchy, and, as before, there exists 00 00. Hence 00 L eλ = s-lim yn eλ = 0 ; i.e., is independent of the choice of subject to its dening eλ = eλ eλ ζ 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. Then n ζµζλ will also give eµ since (ζµζλ) (κ) → pµ(κ) as n → ∞, for each κ ∈ K. Hence, with the obvious notation, ( etc.) ybλ = ζλ n n eλeµ = s-limn→∞ yλ yµ

King's College London 40 Chapter 4

n by the previous observation = s-limn→∞ yµ ,

= eµ. This proves (ii). To prove (iii): for ξ ∈ H, we have 2 n 2 k(eλ+ε − eλ)ξk = lim k(y − yλ)ξk n→∞ λ+ε Z n n 2 = lim |(ζλ+ε − ζλ )(κ)| dµξ n→∞ K Z 2 = |pλ+ε(κ) − pλ(κ)| dµξ . K

But pλ+ε(κ) → pλ(κ) as ε ↓ 0 for each κ ∈ K, so by the dominated convergence theorem, the integral converges to 0, i e , eλ+ε → eλ strongly as ε ↓ 0. Similarly, one veries that k(1l − eλ)ξk → 0 as λ → ∞ and that keλξk → 0 as λ → −∞, which completes the proof of (iii). In fact, eλξ = 0 for any λ < −kak, and eλξ = ξ for all λ > kak. In order to establish (iv), we rst observe that . Divide the |ba( · )| ≤ kak interval (−kak − ε, kak] into n open-closed intervals, (λj, λj+1], 1 ≤ j ≤ n, of equal length. Then

χ ( · ) = (p − p )( · ). {κ∈K : ba(κ)∈Ij } λj+1 λj Put s (κ) = Pn λ (p − p )(κ). Then it is easy to see that s → a n j=1 j+1 λj+1 λj n b uniformly on K as n → ∞. Thus, for given δ > 0, there is N such that for all n > N, and any ξ ∈ H, Z Z 2 |sn − ba| dµξ < δ dµξ K K = δ(ξ, ξ) = δkξk2. That is, for n > N, and ξ ∈ H, n 2 X 2 λj+1(eλ − eλ ) − a ξ < δ kξk j+1 j j=1 and so we see that the sum converges in norm to a; Z kak a = λ deλ , ε > 0 arbitrary. kak−ε

If λ < kak, then pλ = 0 implies that eλ = 0, and if λ > kak then pλ = 1 implies that eλ = 1l, so we can write Z ∞ 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 Z Z Z 2 3 3 n n a = λ deλ , a = λ deλ , . . . , a = λ deλ

(just square the approximating sum etc.). Now let g(·) be the character- istic function of the interval (−kak − 1, µ]. Then, for any ξ ∈ H, Z g(λ) d(eλξ, ξ) = (eµξ, ξ) , since eλ = 0 for λ < −kak ,

2 = keµξk .

Let (Pn(λ)) be a sequence of polynomials converging pointwise to g(λ) on (−kak − 1, kak], and uniformly bounded on this interval. Then Z 2 keµξk = g(λ) d(eλξ, ξ) Z = lim Pn(λ) d(eλξ, ξ) n

= lim(Pn(a)ξ, ξ). n

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 acting 2 on the Hilbert space L ([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 on such that is unitarily equivalent to L 2 {µα} K H α L (K, dµα) and each a ∈ A is unitarily equivalent to multiplication by a continuous function on K. Proof. The proof is just as before, with , L is the measure K = Sp A µ = α µα on K given by the vectors ζα ∈ Hα such that Aζα is dense in Hα, and L . The unitaries are dened by , for α Hα = H uα : C(K) → Aζα uαba = aζα 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 Ae if A has no unit. Proof. Suppose that σ(a) ⊂ [0, ∞). Let A(a) be the unital C∗-subalgebra of A (or Ae if A does not have a unit) generated by a. Then A(a) 'C(Sp A(a)) and , and so as a function on . Let ran ba = σ(a) ⊂ [0, ∞) ba( · ) ≥ 0 Sp A(a) be the positive square root of . Then there is ∗ f ∈ C(Sp A(a)) ba( · ) h = h ∈ A with and hence 2 which implies that 2 . That is, bh = f bh = ba h = a a ≥ 0 as an element of A(a). To see that, in fact, a ≥ 0 as an element of A, we note 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, kak]. Since σ(a) ⊆ [0, kak], we deduce that 2 bh = lim pn(bh ), in C(Sp A(a)), = lim pn(ba).

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 2 , since is real-valued. σ(a) = ran ba = ran bh ⊂ [0, ∞) bh Corollary 5.3. Let a ∈ A where A is a C∗-algebra without a unit. Then a ≥ 0 in A if and only if a ≥ 0 in Ae. Proof. This follows immediately from the proposition. If a ≥ 0 in A, then certainly a ≥ 0 in Ae. On the other hand, if a ∈ A and a ≥ 0 in Ae, then σ (a) ⊂ [0, ∞). By the proposition, a ≥ 0 in A. Ae 43 44 Chapter 5

Corollary 5.4. Suppose that the element a in a C∗-algebra A satises a ≥ 0 and 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 . But then (where is the Gelfand transform of as σ(a) = {0} ran ba = {0} ba a an element of the commutative unital ∗-algebra it generates), and so C ba = 0 and 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(Sp A(t)), via the Gelfand map, and we have 2 2 where and since ba = sb = bt sb ≥ 0 bt ≥ 0 and . It follows that , and therefore . s ≥ 0 t ≥ 0 sb = bt s = t If is invertible, then , i.e., . Hence on . a 0 ∈/ σ(a) 0 ∈/ ran ba ba > 0 Sp A(a) It follows that on and so , i.e., is invertible. sb > 0 Sp A(a) 0 ∈/ σ(s) s 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 ∗ and . Then if and only if a = a ∈ A kak ≤ 1 a ≥ 0 ba ≥ 0 as a function on Sp A(a), where A(a) is the commutative unital C∗-algebra generated by , i.e., if and only if , i.e., if and only if 1l . a |1−ba( · )| ≤ 1 k −ak ≤ 1 (Note that if and only if .) We have kak ≤ 1 |ba( · )| ≤ 1

h + k k khk + kkk − h − kk 1l − = khk + kkk khk + kkk kkhk − hk + kkkk − kk ≤ khk + kkk khkk(1l − h/khk)k + kkkk(1l − k/kkk)k = khk + kkk ≤ 1 since h/khk ≥ 0 and k/kkk ≥ 0.

h + k h + k Since ≤ 1, we deduce that ≥ 0, and hence h + k ≥ khk + kkk khk + kkk 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 Ae). Suppose that a = x∗x is not positive. Then is a non-positive continuous function and so there is ba ∈ C(Sp A(a)) `0 ∈ such that . Thus there is a neighbourhood of in Sp(A(a)) ba(`0) < 0 N `0 such that for all . Let be such that Sp A(a) ba(`) < 0 ` ∈ N f ∈ C(Sp A(a)) f vanishes on the closed set Sp A(a) \ N, 0 ≤ f ≤ 1 on Sp A(a) and f(`0) = 1 (such an exists, by Urysohn's lemma). Then on and f f baf ≤ 0 Sp A(a) . (f baf)(`0) < 0 Let be such that . Then ∗, and b ∈ A(a) b = f b = b bbab = f baf ≤ 0 implies that bab ≤ 0 (i.e., −bab ≥ 0). Furthermore, bab 6= 0 since `0(bab) = . (f baf)(`0) < 0 Write , with ∗ 1 ∗ and ∗ xb = h + ik h = h = 2 (xb + (xb) ) ∈ A(a) k = k = 1 ∗ . Then 2i (xb − (xb) ) ∈ A(a) (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 ≥ 0 and 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 babc = σ(bab) = {0} and so babc = 0 which implies that bab = 0. This is a 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 ≥ 0 for 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 kak ≤ kbk.

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 −kbk ≤ b( · ) ≤ kbk on Sp(A(b)) and so b ≤ kbk1l. Hence we have −kbk1l ≤ −b ≤ a ≤ b ≤ kbk1l. It follows that 1l 1l and so on . −kbk ≤ a ≤ kbk −kbk ≤ ba( · ) ≤ kbk Sp(A(a)) Hence kak ≤ kbk. 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 denote −1/2 −1/2. Then ∗, is invertible and . Hence −1 b ab c = c c 0 < bc( · ) ≤ 1 bc ≥ 1 on 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 ∈ R with such that on , where, as usual, is the γ > 0 0 < γ ≤ ba Sp A(a) A(a) commutative unital C∗-subalgebra of A generated by a. Hence a ≥ γ1l. But then, b − γ1l = (b − a) + (a − γ1l) is the sum of two positive elements of A and therefore is itself positive. This means that b ≥ γ > 0 on Sp A(b), and so 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) 1 ∗ 1 ∗ . a = 2 (a + a ) + i 2i (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, by looking at , we see that are both . Write 1 1 bh |h|±h ≥ 0 h = 2 (h+|h|)− 2 (|h|−h) and use (i). (iii) Let h = h∗ ∈ A. Then h2 ≥ 0. Suppose that khk < 1. Then 1l − h2 ≥ 0 and so has a positive square root, (1l −h2)1/2. Put u = h+i(1l −h2)1/2. Then one checks that ∗ ∗ 1l, i.e., is unitary. Moreover, 1 ∗ . u u = uu = u h = 2 (u + u ) 1 If khk ≥ 1 then consider αh with α = . Then, as above, we can 2khk write 1 ∗ with unitary. Thus 1 ∗ . For general αh = 2 (v + v ) v h = 2α (v + v ) 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± ≥ 0 and k+k− = 0. Then we have

2 2 2 |h| = h = (k+ − k−) 2 2 = k+ + k− 2 = (k+ + k−) .

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 1 and h+ = 2 (|h|+h) 1 are called the positive and negative parts of . h− = 2 (|h| − h) h Example 5.14. Suppose that p is a non-trivial projection in the unital C∗- algebraA. Set h = cos α1l = cos αp + cos αp⊥, where, say, 0 < α < π/2 and p⊥ = 1l − p. Clearly h = h∗ and khk < 1. Then u = eiα1l and v = eiαp + e−iαp⊥ are distinct unitary elements of A, but we have h = 1 ∗ 1 ∗ . 2 (u + u ) = 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 by changing some of the 's to ∗ . Then and are unitary in and, u uα uα u v A 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 elements 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 kh − khk1lk ≤ khk (in Ae, if A does not have a unit). Indeed,

kh − khk1lk = sup |bh(κ) − khk| κ∈Sp A(h)

= sup(khk − bh(κ)), since ran bh ∈ R and khk = kbhk∞ ≥ bh(κ), κ ≤ khk , if and only if bh(κ) ≥ 0 for all κ, i.e., if and only if σ(h) ⊂ [0, ∞) if and only if h ≥ 0. Now, ∗ and therefore ∗ ∗ . We also have hn = hn h = lim hn = lim hn = h khnk → khk. By the above remark, khn − khnk1lk ≤ khnk. Letting n → ∞, we see that kh − khk1lk ≤ khk 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 kabk ≤ kbak.

Proof. By adjoining a unit, if necessary, we may suppose that A is unital. Since ab = (ab)∗, it follows that

kabk = r(ab), the spectral radius of ab (= kabb k∞), = sup{|λ| : λ ∈ σ(ab)} = sup{|λ| : λ ∈ σ(ba)}, since σ(ab) ∪ {0} = σ(ba) ∪ {0}, ≤ kbak, 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 Ae). Suppose rst that a and b are both invertible. Then a ≤ b implies that 0 ≤ b−1/2ab−1/2 ≤ 1l. Hence,

kb−1/4a1/2b−1/4k2 ≤ kb−1/2a1/2k2, as above, = kb−1/2a1/2a1/2b−1/2k, by the C∗-property, = kb−1/2ab−1/2k ≤ 1.

Therefore kb−1/4a1/2b−1/4k ≤ 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.

Department of Mathematics Positive elements of a C∗-algebra 49

√ Now, as , 1l in , and since the map is uniformly ε ↓ 0 a + ε → a A t 7→√ t √ continuous on , with , say, it follows that [0,M] M = kak+1 √ ba√+ ε− ba → 0 uniformly on , as . In other words, as Sp A(a) ε ↓ 0 k ba + ε − bak∞ → 0 ε ↓ 0, i.e., k(a + ε1l)1/2 − a1/2k → 0 as ε ↓ 0. Similarly, k(b + ε1l)1/2 − b1/2k → 0 in A as ε ↓ 0. Hence b1/2 − a1/2 = 1/2 1/2 limε↓0(b + ε1l) − (a + ε1l) 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 P ∗ , and set 1l −1, where denotes the yλ = x∈λ x x uλ = nλyλ( +nλyλ) nλ number of members of λ. Note that nλyλ ≥ 0 and so (1l +nλyλ) is invertible. We note also that , ∗ and that 1l. uλ ∈ J uλ = uλ 0 ≤ uλ ≤

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 −1 uλ = 1l − (1l + nλyλ) −1 ≤ 1l − (1l + nµyµ)

= uµ. (ii) Let x ∈ J. Then 2 ∗ 2 kx − xuλk = k(x − xuλ) k ∗ = kx(1l − uλ)(1l − uλ)x k ∗ 2 ≤ kx(1l − uλ)x k, since (1l − uλ) ≤ (1l − uλ), 1l −1 ∗ = kx( + nλyλ x k ≤ kx(1l + mx∗x)−1x∗k whenever λ has at least m elements and contains x; since in this case, nλyλ ≥ mx∗x,

≤ kx(1l + mx∗x)−1k kx∗k = k(1l + mx∗x)−1x∗x(1l + mx∗x)−1k1/2kx∗k 1 ≤ kmx∗x(1l + mx∗x)−1k1/2kx∗k since k(1l + mx∗x)−1k ≤ 1, m1/2 kx∗k ≤ m1/2 since kmx∗x(1l + mx∗x)−1k ≤ 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, kuλk ≤ 1 and uλ ≤ uµ whenever λ ≤ µ, λ, µ ∈ Λ.

(ii) For any x ∈ J, kx − xuλk → 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).

0 (ii) For any x ∈ J, kx − uλxk → 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 ,

∗ kx − uλxk = k(x − uλx) k ∗ ∗ = kx − x uλk → 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.

00 (ii) For any a ∈ A, ka−auλk → 0 along Λ, and hence also (as above) ka − uλak → 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 Ae, and then J is a left ideal of Ae.) 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. The point is that even if A is not unital, it still has an approximate unit. The following results show the usefulness of approximate units.

Department of Mathematics 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 = lim uλj . 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 kx − xuλk = k cl xk where k cl xk is the norm of cl x in A/J.

Proof. Let x ∈ A. By denition, k cl xk = infj∈J kx + jk. Put α = k cl xk. Then

2 2 2 α = inf kx + jk ≤ kx − xuλk , since xuλ ∈ J, j∈J 2 ∗ = kx(1l − uλ) x k ∗ ≤ kx(1l − uλ)x k.

∗ ∗ Now, x(1l − uλ)x is decreasing in λ and so kx(1l − uλ)x k 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 kx + jk < α + ε. Then

(α + ε)2 > kx + jk2 = k(x + j)(x∗ + j∗)k ∗ ∗ ≥ k(x + j)(1l − uλ)(x + j )k ∗ ∗ ∗ ∗ = kx(1l − uλ)x + j(1l − uλ)x + x(1l − uλ)j + j(1l − uλ)j k

∗ ∗ for any λ. Now, j(1l − uλ), (1l − uλ)j , j(1l − uλ)j → 0. Furthermore, ∗ 2 kx(1l − uλ)x k → β. It follows that (α + ε) ≥ β for all ε > 0, and therefore α2 ≥ β. Thus α2 = β and

2 2 α ≤ kx − xuλk ∗ ≤ kx(1l − uλ)x k → β = α2.

Hence α = lim kx − xuλk, 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 (cl x)∗ = cl x∗; since J is closed, J = J ∗, and this is a well-dened involution on A/J. Furthermore,

k(cl x)∗k = k cl x∗k = inf kx∗ + jk j∈J = inf kx∗ + j∗k = inf kx + jk j∈J j∈J = k cl xk.

Hence

k cl x∗ cl xk ≤ k cl x∗kk cl xk = k cl xk2.

We must show that equality holds. However, if (uλ) is an approximate unit for J (as a left ideal), we have

∗ 2 ∗ 2 k cl x cl xk = lim kx x(1l − uλ)k ∗ ∗ = lim k(1l − uλ)x xx x(1l − uλ)k ∗ ∗ ≥ lim k(1l − uλ)x x(1l − uλ)(1l − uλ)x x(1l − uλ)k ∗ 2 = lim k(1l − uλ)x x(1l − uλ)k 4 = lim kx(1l − uλ)k = k cl xk4 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 homomorphisms 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 with ∗. Then for all a ∈ A a = a |ba(κ)| ≤ kbak∞ κ ∈ Sp A(a) implies that −kbak∞ ≤ ba( · ) ≤ kbak∞ and so 1l and . Hence (using ), kbak∞ − ba( · ) ≥ 0 ba( · ) + kbak∞ ≥ 0 kbak∞ = kak kak1l − a ≥ 0 and a + kak1l ≥ 0, that is, −kak1l ≤ a ≤ kak1l. It follows that −kak1l ≤ ϕ(a) ≤ kak1l since ϕ(1l) = 1l and ϕ is order preserving. By considering the Gelfand transform of ϕ(a) (= ϕ(a)∗), we see that

−kak1l ≤ ϕd(a) ≤ kak1l and so kϕd(a)k∞ = kϕ(a)k ≤ kak for any a ∈ A with a = a∗.

53 54 Chapter 6

For any x ∈ A, we have

kϕ(x)k2 = kϕ(x)∗ϕ(x)k = kϕ(x∗x)k ≤ kx∗xk = kxk2.

Proposition 6.3. Let ϕ : A → B be a homomorphism, and suppose that ϕ is 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−, where x± ∈ 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 , ∗ ∗ ∗ . Let 1/2 . Then the b ∈ B b ϕ(x+)b − b ϕ(x−)b = b yb b = ϕ(x− ) above equality becomes

1/2 1/2 2 ∗ ϕ(x− x+x− ) − ϕ(x−) = b yb . | {z } = 0

Hence 2 ∗ and we deduce that 2 . But 2 −ϕ(x−) = b yb ϕ(x−) ≤ 0 ϕ(x−) = 2 ϕ(x−) ≥ 0, since ϕ(x−) is self-adjoint. Thus

2 ϕ(x−) = 0 =⇒ ϕ(x−)ϕ(x−) = 0

=⇒ ϕ(x−) = 0, e.g., using the Gelfand map,

=⇒ x− = 0,

−1 since ϕ is 11. Thus a = x+ ≥ 0, as claimed, which shows that ϕ is order preserving. Now, for any x ∈ A, kϕ(x)k ≤ kxk, by the previous proposition. On the other hand, putting y = ϕ(x), we have ϕ−1(y) = x and kϕ−1(y)k ≤ kyk again by the previous proposition. Hence kxk ≤ kϕ(x)k ≤ kxk which gives the required equality kϕ(x)k = kxk, 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.

Department of Mathematics Homomorphisms 55

Theorem 6.5. Let ϕ : A → B be a homomorphism. Then ϕ(A) is a C∗- subalgebra of B.

Proof. Let J = ker ϕ. Then it is clear that J is a closed two-sided ∗-ideal of 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 automorphism of A. The collection of automorphisms of A is denoted Aut A. Evidently, Aut A 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 of a Hilbert space H and a homomorphism π : A → B(H). The representation (H, π) is said to be faithful if ker π = {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 homomorphisms 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 observables 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), where ξ ∈ 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), where ξ, 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) ∈ R whenever 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 ha, bi = ϕ(b∗a) denes a sesquilinear form on A, with ha, ai ≥ 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 kωk = ω(1l).

Proof. For any h = h∗ ∈ A, we have −khk1l ≤ h ≤ khk1l, and so the linearity and positivity of ω give −khkω(1l) ≤ ω(h) ≤ khkω(1l) which yields |ω(h)| ≤ ω(1l)khk. 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)ka∗ak as above, with h = a∗a, = ω(1l)2kak2.

Thus |ω(a)| ≤ ω(1l)kak, so that ω is bounded and kωk ≤ ω(1l). But then we conclude that kωk = ω(1l).

Theorem 7.5. Suppose that ω is a bounded linear functional on a unital C∗- algebra A, satisfying kωk = 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)) for all λ ∈ R. Since ω(1l) = kωk, it follows, in particular, that ω(1l) ∈ R, and therefore |ω(h + iλ1l)| ≥ |β + λω(1l)|. On the other hand,

|ω(h + iλ1l)| ≤ kωk kh + iλ1lk 1/2 = ω(1l) khk2 + λ2 as is readily seen by using the Gelfand transform. Therefore

|β + λω(1l)|2 ≤ ω(1l)2 khk2 + 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 khk ≤ 1. Then

|ω(1l) − ω(h)| = |ω(1l − h)| ≤ kωk k1l − hk = ω(1l)k1l − hk ≤ ω(1l) which implies that ω(h) ≥ 0.

Remark 7.6. Thus, a state ω on a unital C∗-algebra is a positive linear functional with kωk = 1, or equivalently, a linear functional with kωk = 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 kωk = ω(1l) = 1. By the Hahn- Banach theorem, there is a continuous linear functional ρ, say, on B, such that kρk = kωk and ρ A = ω. Hence ρ(1l) = ω(1l) = 1, since 1l ∈ A ⊆ B, and so kρk = kωk = 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 + ik with h = h∗ and k = k∗. Then either h 6= 0 or k 6= 0 (otherwise x = 0). If ω is 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(Sp A(h)). Since h 6= 0 there is 0 such that 0 . Dene on by 0 κ ∈ Sp A(h) bh(κ ) 6= 0 ρ A(h) ρ(a) = ba(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) = bh(κ0) 6= 0. Theorem 7.9. The involution of a C∗-algebra is unique.

Proof. Let A be a C∗-algebra. By considering Ae instead of A, we may assume that A has a unit. Let ∗ and † be involutions on A with respect to which A is a C∗-algebra. Let ω be a state on A. Note that a state ω is any bounded linear functional on A with kωk = 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†), and so ω(x∗ − x†) = 0 for all states ω on A. Since the set of states separates the points 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 continuous. 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∗-algebra A. 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 kxnk → 0 but ω(xn) = 1. By positivity, 7.3, ∗ . Setting 1 ∗ , we have ω(xn) = ω(xn) = 1 yn = 2 (xn + xn) kynk ≤ kxnk → 0 and ω(yn) = 1. Now, by 5.12, each yn can be written as 0 00, where 0 and 00 are given by 0 1 , yn = zn − zn zn ≥ 0 zn ≥ 0 zn = 2 (|yn| + yn) 00 1 . Thus 0 , and 0 00 , zn = 2 (|yn| − yn) kznk ≤ kynk → 0 ω(zn) = ω(yn) + ω(zn) ≥ 1 by positivity of . Putting 0 −1 0 , we have , , ω an = ω(zn) zn an ∈ A an ≥ 0 kank → 0 and ω(an) = 1. By passing to a subsequence, if necessary, we may assume that −n, for each . Set Pn . Then kank ≤ 2 n ∈ N bn = k=1 an bn ≥ 0 for 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. But ω is positive and so ω(b − bn) ≥ 0, that is,

n X ω(b) ≥ ω(bn) = ω(an) = n k=1 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 Ae, the C∗-algebra obtained from A by adjoining a unit. First we need the following lemma.

Lemma 7.11. For any positive functional ω on a C∗-algebra A,

|ω(k)| ≤ kωk1/2ω(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 kk − uλkk → 0 so that ∗ ω(k) = limλ ω(uλk), for any k = k ∈ A. However, Schwarz' inequality gives q 2 2 |ω(uλk)| ≤ ω(uλ)ω(k ) ≤ kωk1/2ω(k2)1/2 since 2 2 for all . The result follows. |ω(uλ)| ≤ kωkkuλk ≤ kωk λ

Department of Mathematics States on a C∗-algebra 61

Theorem 7.12. Suppose that ω is a positive linear functional on a C∗-algebra without unit. For any xed dene 1l , for A µ ≥ kωk ωe(a + λ ) = ω(a) + λµ and . Then is a positive linear functional on such that a ∈ A λ ∈ C ωe Ae . Moreover, all positive extensions of to are of this form for ωe A = ω ω Ae suitable µ ≥ kωk. Proof. It is clear that is a well-dened linear functional on . Suppose ωe Ae that z ∈ Ae with z ≥ 0. Then z = h2 for some h = h∗ ∈ Ae. Writing h = k + α1l, it follows that k = k∗ and α = α. By replacing h by −h, if necessary, we may suppose that α ≥ 0. We have 2 21l ωe(z) = ωe(k + 2αk + α ) = ω(k2) + 2αω(k) + α2µ = ω(k2)1/2 − αµ1/22 + 2αω(k2)1/2µ1/2 + 2αω(k) ≥ 0 by the lemma. Thus is positive. ωe If ρ is a positive linear functional on Ae, then kρk = ρ(1l), so that

ρ(a + λ1l) = ρ(a) + λρ(1l) = ρ(a) + λkρk for any a ∈ A and λ ∈ C. If ρ A = ω, then clearly kρk ≥ kωk and therefore , as above, with . ρ = ωe µ = kρk Corollary 7.13. Let A be a C∗-algebra without a unit, and suppose that ω is a positive linear functional on A with kωk ≤ 1. Then ω has a unique extension to a state on Ae. Proof. Dene on by ωe Ae 1l ωe(a + λ ) = ω(a) + λ , for and . Then, since, , it follows that is positive. But a ∈ A λ ∈ C kωk ≤ 1 ωe 1l and so is a state on . ωe( ) = 1 ωe Ae If ρ is state on Ae such that ρ A = ω, then ρ(a + λ1l) = ρ(a) + λρ(1l) = 1l , which gives the uniqueness of the extension. ω(a) + λ = ωe(a + λ ) 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 Ae. Proof. The formula 1l , for , , denes a ωe(a + λ ) = ω(a) + λkωk a ∈ A λ ∈ C positive linear extension of to . Furthermore, 1l . ω Ae kωek = ωe( ) = kωk Any positive extension ρ of ω to Ae satises kρk = ρ(1l), and so the equality kρk = kωk implies that ρ(1l) = kωk 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 kωk = 1. Remark 7.16. According to the preceding discussion, such a functional then has a unique extension to a positive linear functional, , say, on , with ωe Ae (or, equivalently, with 1l ). We see that is the restriction kωek = 1 ωe( ) = 1 ω to A of a unique state, as earlier dened, on the unital C∗-algebra Ae. In other words, 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, the map x 7→ (π(x)ξ, ξ) is clearly a state on A. So from any given representation 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 (H0, π0, Ω0) is another such triple, then there is a unitary operator u : H0 → H such that uΩ0 = Ω and uπ0(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 hξ, ηi = ω(y∗x), where x ∈ ξ and y ∈ η. It is straightforward to verify that h · , · i is a well- dened sesquilinear form on K, i.e., denes an inner product. Moreover, if 2 , then ∗ , for any . Hence, , and so kξkω ≡ hξ, ξi = 0 ω(x x) = 0 x ∈ ξ x ∈ N ξ = 0 in K = A/N. In other words, k · kω is a norm on K.

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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, 2 kLaξkω = hLaξ, Laξi = 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 kρk = ρ(1l), i.e., |ρ(b)| ≤ ρ(1l)kbk, for b ∈ A. Hence |ω(x∗bx)| ≤ ω(x∗x)kbk. So, with = a∗a, we get |ω(x∗a∗ax)| ≤ ω(x∗x)ka∗ak = ω(x∗x)kak2 = hξ, ξikak2.

That is, kLaξkω ≤ kak kξkω. 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 hLa∗ ξ, ηi = ω(y a x) = ω((ay) x) = hξ, Laηi, where x ∈ ξ and y ∈ η. Let H be the completion of K with respect to the norm k · kω. 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 that Ω is a cyclic vector for the representation (H, π). We note that, for any a ∈ A, hπ(a)Ω, Ωi = hLa cl 1l, cl 1li = ω(a). To establish the uniqueness, up to unitary equivalence, suppose that (H0, π0, Ω0) is another such triple. Dene u : H0 → H by uπ0(a)Ω0 = π(a)Ω. Then

kuπ0(a)Ω0k2 = kπ(a)Ωk2 H H = ω(a∗a) = kπ0(a)Ω0k2 . H0 So u is an isometric linear operator from a dense set in H0 to a dense set in H and thus extends to dene a unitary operator from H0 onto H. This unitary provides the required equivalence; to see this, we consider

uπ0(a)u−1π(b)Ω = uπ0(a)π0(b)Ω0 = uπ0(ab)Ω

Department of Mathematics 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 uπ0(a)u−1 = π(a) for a ∈ A.

Remark 8.3. (H, π, Ω) is called the Gelfand, Naimark, Segal (GNS) representation (or triple) associated with ω on A. Examples 8.4. 1. Let be the state on given by R 1 . Then the GNS ω C([0, 1]) x 7→ 0 x(s) ds triple (H, π, Ω) is given by H = L2([0, 1]), with Lebesgue measure, Ω is the vector 1 in H, and π(x) is given on H by multiplication by the function x ∈ C([0, 1]). 2. Suppose is the state on given, as above, by R 1 . ω C([0, 2]) x 7→ 0 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 . Their direct sum is the representation with L , A (H, π) H = α∈I Hα and action L for . π(a) = α∈I πα(a) 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∗-algebra of operators on a Hilbert space.

Proof. Without loss of generality, we may suppose that A has a unit (if not, we consider Ae instead of A). Let SA denote the set of states on A, and, for each ω ∈ SA, let (Hω, πω, Ωω) be the corresponding GNS representation of . Let be their direct sum, L and L , and A (H, π) H = ω∈S Hω π = ω∈S πω let be the vector L A A Ω ω Ωω Suppose that π(a) = π(b) for some a, b ∈ A. Then 0 = (π(a) − π(b))Ω = L Hence for all . In particular, ω(πω(a) − πω(b))Ωω . πω(a − b)Ωω = 0 ω ∈ SA (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∗-algebra A. 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.

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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 E if 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 1 1 ω = 2 ω1 + 2 ω2. Indeed, suppose that ω = αω0 + (1 − α)ω00 for states ω0, ω00 and 0 < α < 1. Without loss of generality, we may assume that 1 . Then can be 0 < α ≤ 2 ω written as 1 1 , where 0 00 and 00. ω = 2 ω1 + 2 ω2 ω1 = 2αω + (1 − 2α)ω 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 Sp A: i.e., a state ω is pure if and only if ω is a character.

Proof. Let ω ∈ Sp A, and suppose that ω = ω1 + ω2, where ω1 and ω2 are states on A. Let a ∈ A with a = a∗. Then

2 1 2 2 ω(a ) = 2 ω1(a ) + ω2(a ) = ω(a)2, since ω ∈ Sp A, 1 2 = 4 (ω1(a) + ω2(a)) .

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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

2 1 2 2 1 2 2 ω(a ) = 2 ω1(a ) + ω2(a ) ≥ 2 ω1(a) + ω2(a) . Combining this with the earlier inequality gives

1 2 1 2 2 4 (ω1(a) + ω2(a)) ≥ 2 ω1(a) + ω2(a) . This reduces to the inequality

2 0 ≥ (ω1(a) − ω2(a))

∗ 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 and arguing as above, we deduce that

ω(bx) = 0 = ω(b)ω(x) , and so we obtain ω(ax) = ω(x) = ω(a)ω(x) ,

Department of Mathematics 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), and X 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 that . Then so that there is some such that ρ0 ∈/ F ρ0 A 6= ω a ∈ A ρ0(a) 6= ω(a) ∗ Put ε = |ρ0(a) − ω(a)|. Then ε > 0 and the w -neighbourhood Nω = N{ρ0 : 1 which, we recall, is dened as 1 , is {a}, 2 ε} {ρ ∈ SB : |ρ(a) − ρ0(a)| < 2 ε} contained 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 , we have . But and ρ ∈ F ω = αρ1 A+(1−α)ρ2 A ρ1 A ρ2 A are both states on , and is pure (on ). Hence , A ω A 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 of to a state on is pure if and only if is ωe ω Ae ω pure.

Proof. Suppose that ω is not pure on A. Then there are states ω1, ω2 on A with and such that 1 1 on . Evidently, 1 1 ω1 6= ω2 ω = 2 ω1 + 2 ω2 A ωe = 2 ωe1 + 2 ωe2 on , and . Thus is not pure on . Ae ωe1 6= ωe2 ωe Ae Conversely, suppose that ω is pure on A. Then, by the theorem, ω has an extension to a pure state on . But is the unique extension of to a Ae ωe ω state on , so that is this pure extension. Ae ωe Corollary 9.7. The pure states separate points of a unital C∗-algebra A.

King's College London 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 ∗-algebra generated by , and let denote the Gelfand isomorphism C a b between and . Since , is not the zero function in A(a) C(Sp A(a)) a 6= 0 ba , and so there is some such that , i.e., C(Sp A(a)) κ0 ∈ Sp A(a) ba(κ0) 6= 0 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 Ae, 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 bh of h, i.e., there is κ ∈ Sp A(h) such that bh(κ) = λ, i.e., κ(h) = λ. By the theorem, there is a pure state ω on A such that ω A(h) = κ. But then 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 spanned by the unit vector . Clearly, . Let H ξ σ(p) =√{0, 1} η ∈ H be a unit vector orthogonal to ξ, and set ζ = (ξ + η)/ 2. Let ωζ denote 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

1 1 ωζ (p) = (pζ, ζ) = 2 (ξ, ξ) = 2 , but 1 . 2 ∈/ 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∗)

Department of Mathematics Pure States 71

= k cl bk2 k cl ak2 . ω 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) ≥ 0 for 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

S0 = {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 0 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 0 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)0, and suppose that t 6= 0 and t 6= 1l. Dene ρ on A by ρ(a) = (tπ(a)Ω, Ω), a ∈ A.

King's College London 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

ρ( · ) (ω − ρ)( · ) ω( · ) = ρ(1l) + (ω − ρ)(1l) ρ(1l) (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) (π(a)Ω, Ω) | {z } ρ(1l) and so ((t − ρ(1l))π(a)Ω, Ω) = 0, for all a ∈ A. It follows that t = ρ(1l)1l, 0 and hence we have π(A) = C1l.

Denition 9.13. A representation (H, π) of a C∗-algebra A is called irre- 0 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)0 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)H are 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 isomorphic 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 representation with L , and L . Each is irreducible, H = ω∈E Hω π = ω∈E πω (Hω, πω) 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 ω is 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, π0) of A by the assignment π0(a) = π(α(a)) for a ∈ A and consider the triple (H, π0, Ω). Since α(A) = A, Ω is cyclic for π0. Furthermore,

(Ω, π0(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 ∗ = π0(a) = π(α(a)) for all a ∈ A. Suppose that V is another unitary operator on H with the same properties. 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 Aut A. Suppose that ω is 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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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 that U( · ) 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 ∈ Aut A a representation of the group G in Aut A. 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 ≡ 0 ({ 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 P Ω 6= 0 and P Ω 6= Ω. Let ω1 be the state on A given by (P Ω, π(a)P Ω) ω (a) = 1 kP Ωk2 for a ∈ A, and let ω2 be the state given by (QΩ, π(a)QΩ) ω (a) = 2 kQΩk2 for a ∈ A, where P +Q = 1l. Evidently, ω is given as the convex combination

2 2 ω = kP Ωk ω1 + kQΩk ω2 .

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 → P ξ. But then

2 2 2 ω1(an) = (P Ω, π(an)P Ω)/kP Ωk = (P Ω, π(an)Ω)/kP Ωk → (P Ω, ξ)/kP Ωk . However,

2 ω2(an) = (QΩ, π(an)QΩ)/kQΩk 2 = (QΩ, π(an)Ω)/kQΩk → (QΩ, P ξ)/kQΩk2 = 0 .

The equality ω1 = ω2 would then entail that (P Ω, ξ) = 0 for all ξ ∈ H, thus giving P Ω = 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.

Department of Mathematics Pure States 75

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)0 such that

ρ(b∗a) = (π(b)Ω, T π(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 = ω giving ω1 = ω2 which is false. We conclude that ω is extremal invariant, as claimed.

King's College London 76 Chapter 9

Department of Mathematics