xy x y = x(y y ) + (x x )y x y y + x x y and so we see that multiplication is jointly continuous.

Transcription

1 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 and for all a, b A, λ C. For any x, x, y, y A, we have λ(ab) = (λa)b = a(λb) ab a b xy x y = x(y y ) + (x x )y x y y + x x y and so we see that multiplication is jointly continuous. The algebra A is said to be commutative (or abelian) if ab = ba for all a,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 1l are units, then 1l = 1l1l = 1l. 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 = 1l 2 and so we have 1l 1l 1l, which implies that 1l 1. 1

2 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 L x denote the linear operator L x : y xy A, y A. Then if L x = L x, it follows that L x 1l = L x 1l and so x = x. Hence x L x is an injective map from A into the set of linear operators on A. Now, L x y = xy x y, for y A which implies that L x is bounded, and L x x. Put x = L x. Then we have just shown that x x, for any x A. On the other hand, x = L x = sup{ L x y : y 1} = sup{ xy : y 1} xy, where y = 1l 1l = x 1l. Hence, x / 1l x x, for all x A, which shows that the two norms and are equivalent. Moreover, for any x, y A, xy = L xy = L x L y L x L y = x y and so A with norm is a Banach algebra. To complete the proof, we have 1l = L 1l = 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 A I as an ideal of codimension one. Proof. Let A I = A C as a linear space, and dene a multiplication in A I by (x, λ)(y, µ) = (xy + µx + λy, λµ). Department of Mathematics

3 Banach algebras 3 It is easily checked that this is associative and distributive. Moreover, the element (0, 1) is a unit for this multiplication: (x, λ)(0, 1) = (x0 + x + λ0, λ1) = (x, λ) = (0, 1)(x, λ). Put (x, λ) = x + λ. Then A I is a Banach space when equipped with this norm. Furthermore, (x, λ)(y, µ) = (xy + µx + λy, λµ) = xy + µx + λy + λµ x y + µ x + λ + λ µ = ( x + λ )( y + µ ) = (x, λ) (y, µ). Hence A I is a Banach algebra with unit. We may identify A with the ideal {(x, 0) : x A} in A I via the isometric isomorphism x (x, 0). We write (x, λ) as (x, λ) = x + λ1l A I. (Compare this with complex numbers a + ib (a, b).) Note that 1l = 1. We will see, later, that an analogous result holds for C -algebras, but more care has to be taken regarding the norm. Examples Consider C([0, 1]), the Banach space of continuous complex-valued functions dened on the interval [0, 1] equipped with the sup-norm, namely, f = sup s [0,1] f(s), and with multiplication dened pointwise: (fg)(s) = f(s) g(s), for s [0, 1]. Then C([0, 1]) is a commutative unital Banach algebra; the constant function 1 is the unit element. 2. As above, but replace [0, 1] by any compact topological space. 3. Let D denote the closed unit disc in C, and let A denote the set of continuous complex-valued functions on D which are analytic in the interior of D. Equip A with pointwise addition and multiplication and the norm f = sup{ f(z) : z D} where D is the boundary of D, that is, the unit circle. (That this is, indeed, a norm follows from the maximum modulus principle.) Then A is complete, and so is a (commutative) unital Banach algebra. A is called the disc algebra. King's College London

4 4 Chapter 1 4. Let A be the Banach space l 1 (Z) and dene xy by (xy) n = m x my n m for x = (x n ) and y = (y n ) in A. Then (xy) n n n = m x m y n m m x m n y n m = m x m y = x y. Thus xy l 1 (Z) and so A is a Banach algebra. Furthermore, A has a unit given by (x n ) = (δ 0n ) = (..., 0, 0, 1, 0, 0,... ) where the 1 appears in the 0 th 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 z x = xz = 1l, then z = z1l = zxz = 1lz = z. 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 x 1 is a continuous map from G(A) to G(A). Proof. First let y A with y < 1, and put s n = n k=0 yk, n N. Then (s n ) is a Cauchy sequence in A and so converges, since A is complete. Let w denote its limit; w = k=0 yk. We claim that w is the inverse of 1l y. Indeed, we have (1l y)w = lim n (1l y)s n and = lim n (1l y n+1 ) = 1l w(1l y) = lim n s n (1l y) = lim n (1l y n+1 ) = 1l which establishes the claim. Hence, if x A with 1l x < 1, then writing x = 1l (1l x) and arguing as above (with y = 1l x), we see that x is invertible and its inverse x 1 is given by the convergent series k=0 (1l x)k. Let x 0 G(A). Then for any x A, we have x = x 0 x 1 0 x. Now, 1l x 1 0 x = x 1 0 (x 0 x) x 1 0 x 0 x, Department of Mathematics

5 Banach algebras 5 and so we conclude that if x x 0 < x then x 1 0 x is invertible with inverse given by ( ) (x 1 0 x) 1 = y k k=0 with y = (1l x 1 0 x). Hence x is invertible and x 1 = k=0 [x 1 0 (x 0 x)] k x 1 To see that x x 1 is continuous on G(A), suppose that x 0 G(A) and that (x n ) is a sequence in G(A) such that x n x 0 as n. Then for all suciently large n, x n x 0 < x and so as n. x 1 n 0. x 1 0 = [ x 1 0 (x 0 x n ) ] k x 1 0 k=1 [ x 1 0 x 0 x n ] k x 1 k=1 0 Denition 1.8. Let A be a unital algebra, and let x A. The spectrum of x is the subset σ A (x) of C given by σ A (x) = {λ C : x λ1l / G(A)}. 0 The resolvent set ρ A (x) of x is the complement of the spectrum of x; ρ A (x) = C \ σ A (x). The spectral radius r A (x) of an element x A is dened as provided σ A (x) is not empty. r A (x) = sup{ λ : λ σ A (x)} Example 1.9. Let A be the unital Banach algebra M n (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, σ A (a) is just the set of eigenvalues of the matrix a. Example 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 σ A (f) is equal to the set of values assumed by f, i.e., σ A (f) = ran f, the range of f. King's College London

6 6 Chapter 1 Example 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 A 0 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 A 0 is dense in A. Then one shows that the product in A 0 extends (by continuity) to a product on A, and that, when equipped with this product, A is a Banach algebra.) Let A 0 = C[x], the algebra of complex polynomials in the indeterminate x, considered as a subalgebra of A = C([0, 1]). When equipped with the supremum norm, A 0 is a unital normed algebra whose completion, by Weierstrass' theorem, is just the unital Banach algebra A. Let f A 0. For any λ C, f λ1l fails to be invertible in A 0, unless f is a constant (i.e., a polynomial of degree zero) not equal to λ. That is, σ A0 (f) = C whenever f is not a constant, but σ A0 (f) = {α} when f is the constant α. In particular, an element of A 0 is invertible in A 0 if and only if it is a non-zero constant. Hence G(A 0 ) = {f C[x] : f = α, α 0}. Thus we see that G(A 0 ) is not an open set in A 0. Indeed, 1l, the constant polynomial 1, belongs to G(A 0 ), of course, but for any ε > 0, the polynomial p(x) = εx satises 1l p < ε and p / A 0. Thus, every open set containing 1l also contains singular elements of A 0. Example 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 f(z) = z, z C. For any λ C, the function z (z λ) 1 is meromorphic, i.e., f λ1l is invertible in A. On the other hand, the function z (z λ) 1 is not entire for any λ C. Thus we see that σ A (f) = C but σ B (f) =. Theorem For any x in a unital Banach algebra A, the spectrum σ A (x) is a non-empty compact subset of C with σ A (x) {λ C : λ x }. Proof. For given x A, x λ1l is invertible whenever λ > x. Indeed, for any such λ, (x λ1l) = λ ( 1l x/λ ) has inverse given by a convergent series expansion in powers of x/λ. Hence σ A (x) {λ C : λ x }, as claimed. In particular, this shows that σ A (x) is bounded. Let f be the map f : λ x λ1l. Then λ / σ A (x) if and only if x λ1l G(A), i.e., if and only if λ f 1 (G(A)). It follows that C \ σ A (x) = f 1 (G(A)). But it is clear that f : C A is continuous and so G(A) open implies that f 1 (G(A)) = C \ σ A (x) is open and so σ A (x) is closed. Hence σ A (x) is a closed, bounded subset of C and therefore compact. We must show that σ A (x) is non-empty. Indeed, suppose the contrary, σ A (x) =. Then (x λ1l) is invertible for all λ C. We claim that the Department of Mathematics

7 Banach algebras 7 map λ (x λ1l) 1 is dierentiable, with derivative (x λ1l) 2, that is, we claim that (x (λ + ζ)1l) 1 (x λ1l) 1 in A as ζ 0 in C (with ζ 0). To see this, note that ζ (x λ1l) 2 (x α1l) 1 (x β1l) 1 = (x α1l) 1{ (x β1l) (x α1l) } (x β1l) 1 = (x α1l) 1 (α β)(x β1l) 1. Therefore (x (λ + ζ)1l) 1 (x λ1l) 1 ζ = (x (λ + ζ)1l) 1 ζ(x λ1l) 1 ζ (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 ϕ((x λ1l) 1 ) is everywhere dierentiable in C, that is, g(λ) ϕ((x λ1l) 1 ) is an entire function. For all suciently large λ, we may write (x λ1l) 1 = ( λ) 1 (1l λ 1 x) 1 = ( λ) 1 (λ 1 x) n ( λ > x ). 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 deduce that g is identically zero on C. But then we have ϕ((x λ1l) 1 ) = 0 for all ϕ A, which implies that (x λ1l) 1 = 0. This is impossible because (x λ1l) 1 (x λ1l) = 1l 0. We conclude that σ A (x). Proposition Let A be a unital Banach algebra, and let a A. (i) σ A (p(a)) = p(σ A (a)) for any complex polynomial p. (ii) If a is invertible, σ A (a 1 ) = σ A (a) 1. Proof. (i) Suppose that p has degree n 1. For any µ C, let λ 1,..., λ n be the n complex roots of the polynomial p( ) µ. Then, for any z C, p(z) µ = α(z λ 1 )... (z λ n ), for some non-zero α C and so p(a) µ1l = α(a λ 1 1l)... (a λ n 1l). Now, if a 1,..., a n are mutually commuting elements of A (i.e., a i a j = a j a i, for any 1 i, j n), then the product a 1... a n is invertible if and King's College London

8 8 Chapter 1 only if each a i is invertible. (If each a i is invertible, then their product is invertible, regardless of the commutativity assumption. Conversely, if the product is invertible, then, for example, we have a 2 a 1 a 3... a n (a 1... a n ) 1 = a 1... a n (a 1... a n ) 1 = 1l = (a 1... a n ) 1 a 1... a n = (a 1... a n ) 1 a 1 a 3... a n a 2, which shows that a 2 is invertible. Similarly, it is easy to see that any a i is invertible.) Suppose that µ σ A (p(a)). Then p(a) µ1l is singular and so therefore is a λ i 1l, for some 1 i n. That is, λ i σ A (a). But p(λ i ) = µ which shows that µ p(σ A (a)). Conversely, suppose that λ σ A (a) and let µ = p(λ). Then, with the notation above, it follows that λ = λ i for some 1 i n, and that p(a) µ1l is singular. Thus p(λ) σ A (p(a)), and the result follows. (ii) Suppose that a A is invertible, i.e., 0 / σ A (a). For any λ C, λ 0, we have a λ1l = a(1l λa 1 ) = aλ(λ 1 1l a 1 ) which implies that a λ1l is singular if and only if a 1 λ 1 1l is singular. Suppose that x and y are elements of a unital Banach algebra A such that x < 1 and also y < 1. Then it follows that xy x y < 1, and similarly, yx < 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 = n=0 (xy)n and b = (1l yx) 1 = n=0 (yx)n. Evidently, and so yax = y(1l + xy + (xy) )x = yx + (yx) 2 + (yx) l + yax = 1l + yx + (yx) = 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 }{{} = 1l = 1l yx + yx = 1l. Department of Mathematics

9 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 For any x, y in a unital Banach algebra A, σ A (xy) {0} = σ A (yx) {0}. Proof. Suppose that λ C, λ 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 σ A (xy) \ {0} = σ A (yx) \ {0}, and the result follows. Remark 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, σ A (ab) {0} = σ A (ba) {0}. On the other hand, we know from 1.14 that σ A (ab) = σ A (1l+ba) = 1+σ A (ba). Therefore σ A (ba) {0} = {0} {1 + σ A (ba)}. This is impossible because σ A (ba) is bounded. (If α σ A (ba) with Re α 0, then 1 + α 0 and belongs to the right hand side of the above equality. So it also belongs to σ A (ba). By induction, we see that α + n belongs to σ A (ba) for any n N which is not possible. On the other hand, if α σ A (ba) with Re α < 0, then α 1 + σ A (ba) and so α 1 σ A (ba). Again, by induction, it follows that α n σ A (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 (Spectral radius formula) Let A be a unital Banach algebra and let x A. Then the limit lim n x n 1/n exists and satises lim n xn 1/n = r A (x) = inf n xn 1/n. Proof. Set γ(x) = inf{ x n 1/n : n = 1, 2,... }. We shall show that x n 1/n γ(x) as n. Given any ε > 0, let k N be such that x k 1/k < γ(x) + ε. King's College London

10 10 Chapter 1 For any n N, write n as n = αk + β where 0 β < k and α, β Z +. (Note that k is xed and α, β depend on n.) Then β/n 0 as n since β is always less than k. Also, 1 = n n = αk + β n = αk n + β n and so αk n 1 as n, that is, α n 1, as n. Now, k x n 1/n = (x k ) α x β 1/n x k α/n x β/n. and the right hand side converges to x k 1/k, as n, which is less than γ(x) + ε. Hence, for all suciently large n, x n 1/n x k α/n x β/n < γ(x) + ε. On the other hand, γ(x) x n 1/n for any n = 1, 2,... and so γ(x) x n 1/n < γ(x) + ε for all suciently large n. Thus x n 1/n converges to γ(x) as n, i.e., lim n x n 1/n exists and is equal to inf m x m 1/m. We must now show that the above limit is equal to r A (x). Recall, rst, that for any y A, σ A (y) {λ : λ y } and so r A (y) y. By proposition 1.14, it follows that {λ n : λ σ A (x)} = σ A (x n ) and so r A (x n ) = r A (x) n, for any n N. But r A (x n ) x n and so we obtain r A (x) n = r A (x n ) x n = r A (x) x n 1/n, for all n, = r A (x) γ(x). We want to show now that r A (x) γ(x). Let ϕ A. Then g : λ ϕ((x λ1l) 1 ) is analytic on C\σ A (x), and so has a Laurent series expansion on {λ : λ > r A (x)}. But for λ > x, we know that g has the expansion g(λ) = n=0 ϕ(x n ) λ n+1. This must therefore be absolutely convergent in the region {λ : λ > r A (x)}. Fix λ with λ > r A (x). Then, in particular, ϕ(x n /λ n+1 ) 0 as n. This holds for any ϕ A, and so, by the uniform boundedness principle, it follows that (x n /λ n+1 ) is a bounded sequence in A, that is, there is κ > 0 such that x n /λ n+1 κ for all n. Hence x n κ λ λ n and so x n 1/n (κ λ ) 1/n λ. Letting n gives γ(x) λ. This holds for any λ with λ > r A (x) and so we deduce that γ(x) r A (x). It follows that r A (x) = γ(x) and the proof is complete. Department of Mathematics

11 Banach algebras 11 Remark We have shown that inf n xn 1/n = lim n x n 1/n = sup{ λ : λ σ A (x)}. The right hand side is purely algebraic in that it involves (non-)existence of an inverse in A, whereas the left hand side involves the norm, that is, the metric aspects of the algebra. We see here an inter-relationship between purely metric and purely algebraic parts of the theory. By enlarging the algebra, the spectrum may change, but the spectral radius will not. Example Let A be the disc algebra. Then each f in A is uniquely determined by its values on the unit circle D = S 1. Thus, A can be regarded as a subalgebra of C(S 1 ), the Banach algebra of continuous complex-valued functions on the circle S 1. This identication of A in C(S 1 ) 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 σ A (g) = D = {λ : λ 1}. On the other hand, considered as an element of C(S 1 ), g is the function g(θ) = e iθ, with the obvious notation. Then (g λ1l) fails to be invertible in C(S 1 ) if and only if λ = 1. Hence σ C(S 1 )(g) = D = {λ : λ = 1}. Note that r A (g) = r C(S 1 )(g) (= 1), as we know should be the case. Theorem (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, σ A (x) 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 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 cl x = inf x + v = inf v V x x 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 x if and only if x x 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 is a norm. King's College London

12 12 Chapter 1 If cl x = 0 in X/V, then x V and so, taking v = x, we get cl x = inf v V x+v = 0. On the other hand, if cl x = 0, then there is a sequence (v n ) in V such that x + v n 0 as n. Hence v n x in X. Since V is closed, we deduce that x V and so cl x = 0 in X/V. For λ C, λ 0, and x X, For any x, y X, Hence is a norm on X/V. cl λx = inf λx + v = inf λx + v V v V λv = λ inf x + v V v = λ cl x. cl x + cl y = inf x + y + v v V = inf x + y + v + w v,w V inf v,w V ( x + v + y + w ) = cl x + cl y. 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 Let Y be a normed space. Then Y is complete if and only if it has the following property: if (y m ) is any sequence in Y such that m=1 y m <, then there is y Y such that n m=1 y m y as n. Proof. If Y is complete and (y n ) satises k=1 y k <, then it is clear that ( n k=1 y k) is a Cauchy sequence and hence converges. Conversely, suppose that Y has the stated property, and let (x n ) be a Cauchy sequence in Y. We construct a subsequence as follows. Let n 1 N be such that x n1 x m < 1 2 for all m > n 1. Now let n 2 > n 1 be such that x n2 x m < 1 4 for all m > n 2. Continuing in this way, we obtain a subsequence (x nk ) of (x n ) such that x nk x nk+1 < 1, k = 1, 2,... 2k Set y k = x nk x nk+1 for k = 1, 2,.... Then y k = k=1 < x nk x nk+1 k=1 k=1 1 2 k <. Department of Mathematics

13 Banach algebras 13 By hypothesis, there is some y Y such that y = lim m y k m k=1 = lim m (x n 1 x n2 ) + (x n2 x n3 ) + + (x nm x nm+1 ) = lim m (x n 1 x nm+1 ). That is, (x nk ) converges in Y (to x n1 y). But if a subsequence of a Cauchy sequence converges, the whole sequence does; i.e., (x n ) converges in Y and we conclude that Y is complete. Proposition 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 cl x = inf x + v. 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 x n ) be any sequence in X/V such that n=1 cl x n <. By denition of the inmum, for each n, there is v n V such that Hence x n + v n < inf v V x n + v n x n + v n < n=1 = cl x n n. ( cl x + 12 ) n <. n=1 Now, X is a Banach space, so, by the previous proposition, it follows that there is y X such that y = lim n k=1 n (x k + v k ). We claim that n k=1 cl x k cl y in X/V. Indeed, cl y n cl x k = cl(y k=1 n x k ) k=1 n = inf y x k + v v V k=1 King's College London

14 14 Chapter 1 y = y n x k k=1 n v k k=1 n (x k + v k ) k=1 0 as n. Hence n k=1 cl x k converges in X/V (to cl y) and so X/V is a Banach space, again by the previous proposition. Denition 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 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 cl x = inf x + v. 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, cl x cl y = cl xy = inf xy + v v V inf v,w V xy + xw + vy + vw }{{} V = inf (x + v)(y + w) v,w V inf x + v y + w v,w V = cl x cl y. Thus A/V is a Banach algebra. If V = A, then A/V is trivial, that is, {0}. If V is proper, then A/V is not equal to {0} and one sees that cl 1l is a unit for A/V. Furthermore, if 1l = 1, then cl 1l = inf 1l + v v V Department of Mathematics

15 Banach algebras 15 1l, (taking v = 0), = 1. However, we know that the unit of a Banach algebra always has norm greater than or equal to 1, so we obtain cl 1l = 1. King's College London

16 16 Chapter 1 Department of Mathematics

17 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 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 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) 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 σ A (a) and it follows that ϕ(a) a. This inequality remains valid when ϕ(a) = 0 and we conclude that ϕ is continuous on A. If A is non-unital, we consider A I instead. Dene the map ϕ : A I C by ϕ ((a, λ)) = ϕ(a) + λ, (a, λ) A I. Then it is straightforward to check that ϕ is a homomorphism and therefore, by the above, is continuous on A I. In particular, its restriction to A in A I is continuous, i.e., ϕ is continuous. 17

18 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 l is a character on A, ker l is a maximal ideal in A, and every maximal ideal has this form for some unique character. Proof. Suppose that l : A C is a character and let J = ker l. We have J A since l is non-zero. Let a / J. Then any b A can be written as b = a l(b) ( l(a) + b a l(b) ). l(a) Since b a l(b) ker l = J, we see that A = Ca + J and hence J is a l(a) 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 l with ker l = J. This association is one-one, since l is uniquely determined by its kernel. Indeed, suppose that l and l have the same kernel. Then for any a A, a l(a)1l belongs to ker l = ker l and so l (a) = l(a) since l (1l) = 1. Proposition 2.6. Any commutative unital Banach algebra possesses at least one character. Department of Mathematics

19 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 = M n (C), with n > 1, and let e ij be the n n matrix all of whose entries are 0 except for the ij-entry which is equal to 1. If l were a character on A, then, for i j, the equality e 2 ij = 0 would imply that l(e ij ) = 0. Hence the equality e ii = e ij e ji, applied with i j, would imply that l(e ii ) = 0 for each i = 1, 2,... n. We would conclude that l(1l) = l(e 11 ) + + l(e nn ) = 0, which is impossible. Hence M n (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 ϕ 1, ϕ 2 A, with ϕ 1 ϕ 2 there exists a A such that ϕ 1 (a) ϕ 2 (a). Let ε 0 = ϕ 1 (a) ϕ 1 (a) /3. Then evidently the neighbourhoods N(ϕ 1 : {a}, ε 0 ) and N(ϕ 2 : {a}, ε 0 ) have empty intersection. In terms of nets, the w -topology is described as the weakest topology on A such that a net (ω α ) ω in A if and only if ω α (a) ω(a) for each a A. King's College London

20 20 Chapter 2 We shall use the following resultthe Banach-Alaoglu theorem. Theorem The closed unit ball of A, the dual of the Banach space A, is w -compact. Proposition 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 ϕ A \ Sp A. If ϕ = 0, we have 0 = ϕ N(0; {1l}, 1 2 ) A \ Sp A, since l(1l) = 1 for any l Sp A. Suppose that ϕ 0. Then there is a, b A such that ϕ(ab) 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) ω(a)ω(b), ie, 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 The fact that Sp A is w -closed can be demonstrated quite easily using nets. Suppose that (l α ) is a net in Sp A converging to ϕ A. Then, by denition of the w -topology, l α (x) ϕ(x) for each x A. But then for any x, y A ϕ(xy) = lim l α (xy) = lim l α (x) l α (y) = ϕ(x) ϕ(y) and it follows that ϕ Sp A. Note that ϕ is non-zero since ϕ(1l) = 1. Theorem Let A be a commutative unital Banach algebra. For x A and l Sp A, dene x : Sp A C by x(l) = l(x). Then the range of the function x on Sp A satises ran x = σ A (x). Furthermore, the map is a homomorphism : A C(Sp A) and is called the Gelfand transform. x x, for x A. Department of Mathematics

21 Gelfand Theory 21 Proof. We have seen already that for any x A and l Sp A we have l(x) σ A (x) ; ie, x(l) σ A (x) and so the range of x satises the inclusion ran x σ A (x). Let λ σ A (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 l be that element of Sp A with ker l = J. Then x λ1l J implies that l(x) = λ. Hence x(l) = l(x) = λ and it follows that ran x = σ A (x). It is clear that is a homomorphism; for example, xy(l) = l(xy) = l(x)l(y) = x(l) ŷ(l) for any x, y A, l Sp A, and so xy = x ŷ. Similarly, one sees that is linear. To show that x C(Sp A), let U be any open set in C. We must show that x 1 (U) is open in Sp A. If x 1 (U) =, then we are done. So suppose that x 1 (U). Let l be any element of x 1 (U). Then there is ζ U such that x(l) = ζ. Since U is open in C, there is ε > 0 such that N ε (ζ) {z C : z ζ < ε} U. Let V = N(l : {x}, ε) {ω Sp A : ω(x) l(x) < ε}. Then ω(x) = x(ω) U for all ω V, ie, l V x 1 (U). We deduce that x 1 (U) is open in Sp A and hence x : Sp A C is continuous; that is, x( ) C(Sp A). (Alternatively, the continuity of x can easily be established using nets, as follows. Suppose that l α l in Sp A. Then x(l α ) = l α (x) l(x) = x(l), by denition of the w -topology. In other words, x is continuous.) Finally, we have that ran x = σ A (x) {λ : λ x } and so it follows that x(l) x, for all l Sp A. Thus x x for any x A. This theorem has a sharper form for C -algebras, as we will see in the next section. Theorem 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 â : Sp A σ A (a) C is a homeomorphism. Proof. We know that â is a continuous function on Sp A with ran â = σ A (a), ie, â : Sp A σ A (a) is continuous and onto. Now, both Sp A and σ A (a) are compact Hausdor spaces, so we need only show that â is injective. To see this, suppose that â(l 1 ) = â(l 2 ), so that l 1 (a) = l 2 (a). Using the multiplicativity of l 1 and l 2, we see that for given N N and c 0, c 1,..., c N in C ( N ) ( N l 1 c n a n = l 2 c n a ). n n=0 Since l 1 and l 2 are continuous and a generates A, it follows that l 1 = l 2. n=0 King's College London

22 22 Chapter 2 Example Let A be the subalgebra of M 2 (C) consisting of those elements of the form ( ) α β 0 α, with α, β C. Then ( ) α β 0 α = α1l + βq, where q = ( ) We notice that q 2 = 0 (i.e., q is nilpotent). Evidently, A is a two-dimensional commutative Banach algebra with unit 1l. We shall compute the spectrum, σ A (x), for x = ( ) α β 0 α. Indeed, for λ C, ( ) α λ β x λ1l = 0 α λ is invertible in M 2 (C) if and only if λ α. If λ α, then, in fact, ( (x λ1l) 1 (α λ) 1 β(α λ) = 2 ) 0 (α λ) 1, which belongs to A. Hence, σ A (x) = σ A (α1l + βq) = {α}. In particular, σ A (q) = σ A (βq) = {0}, but q 0. Now we consider the characters of A. If l is a character, then l(xy) = l(x)l(y) implies that l(q 2 ) = l(q)l(q). But q 2 = 0 and therefore l(q) = 0. Since l(1l) = 1, we nd that l(α1l + βq) = α, for any α, β in C. In other words, there is just one character on A; Sp A = {l}, where l is given uniquely by the action l(1l) = 1 and l(q) = 0. The Gelfand transform is the map x x, α1l + βq α 1l + β q. But 1l = 1 and q(l) = l(q) = 0 so that q = 0 and we have (α1l + βq) = α, for any α, β C. The transform has kernel {βq : β C}, so we see that is not an isomorphism. The algebra A has exactly one maximal ideal, namely, the kernel of l. A is the unital algebra generated by the element q, and so Sp A σ A (q), via the homeomorphism q : Sp A σ A (q), l q(l) = 0. Indeed, both Sp A and σ A (q) are singleton sets! Alternatively, we can determine the spectrum of x A using the equality σ A (x) = ran x. For x = ( α β 0 α ), we have since l(q) = 0. σ A (x) = { x(l)} = {l(x)} = {l(α1l + βq)} = {αl(1l) + βl(q)} = {α} Department of Mathematics

23 Chapter 3 C -algebras Denition 3.1. A Banach -algebra is a Banach algebra A together with an involution a a satisfying : (i) is conjugate linear, i.e., (αa) = ᾱa for α C, a A; (ii) a = a for every a A; (iii) (ab) = b a for any a, b A; (iv) a = a. By (iv), we see that : A A is continuous. If A has a unit 1l, then 1l = 1l 1l = (1l 1l) = 1l = 1l. A C -algebra is a Banach -algebra for which (v) a a = a 2, for all a A. This property (v) is often referred to as the C -property of the norm. Remark 3.2. We note that in a C -algebra, property (iv) follows from the others: indeed, for any a in a C -algebra A, property (v) implies that a 2 = a a a a and so a a. (If a = 0, then a = 0 = a.) Replacing a by a gives a a and so their equality follows. Now suppose that A is a C -algebra with unit 1l. Then, by the C - property, 1l 1l = 1l 2 and so 1l = 1l 2 which implies that 1l = 1. If A does not have a unit, then one can be adjoined but care must be taken not to spoil the C -property, property (v). We will return to this later. 23

24 24 Chapter 3 Examples 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 f f = f. Since f f = f 2 it follows that f f = f Let Ω be a topological space. C b (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 C 0 (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 C 0 (Ω) if and only if for any given ε > 0 there is a compact set in Ω outside of which f is less than ε; ie, the set {ω Ω : f(ω) ε} is compact.) Then C 0 (Ω) is a C -algebra without a unit. (A unit f C 0 (Ω) must satisfy f = f 2 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 C 0 (Ω), g = gf implies that g vanishes on the non-empty open set Ω \ K. Let ω Ω \ K. Since Ω is locally compact, there is a compact set K 1 and an open set G such that ω G K 1. By Urysohn's lemma, there is a continuous map g : Ω C such that g takes the value 1 at ω and vanishes on the closed set Ω \ G. But this means that {ω Ω : f(ω) 0} K 1 and so g C 0 (Ω). This contradicts the fact that g should vanish outside of K, and we conclude that C 0 (Ω) has no unit.) 4. It is quite possible for C 0 (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 n N; to see this, note that f 1 (C \ {f(1)}) is an open set in Ω not containing 1. Since the only open sets in Ω which do not contain 1 are and {0}, we conclude that n / f 1 (C \ {f(1)}) for any n N. In other words, f(n) = f(1) for n N. Suppose now that f C 0 (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

25 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, C 0 (Ω) consists of all functions f : Ω C such that f(1) = f(2) = f(3) = = 0. Evidently, the function e : C 0 (Ω) C given by e(0) = 1, e(1) = e(2) = = 0, is a unit for the C -algebra C 0 (Ω). Of course, C 0 (Ω) is not a particularly interesting C -algebrait is isomorphic to C, via the map f f(0). 5. C, with the obvious structure, is a C -algebra. (This is just example 1, above, when Ω consists of a single point.) 6. We denote the linear space of bounded linear operators on the complex Hilbert space H by B(H). For any x B(H) let x denote the operator norm of x (given by sup{ xξ : ξ 1, ξ H}) and let be the operator adjoint. Then B(H), equipped with this structure, is a unital C -algebra. (If 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. M n (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 M n (C) as an algebra of linear operators on the nitedimensional complex Hilbert space C n. (This is the same as dening x to be the square root of the largest eigenvalue of the positive selfadjoint matrix x x, x M n (C).) This is just example 6 when H is n-dimensional. 9. l (N) and l (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 {A α } of C -algebras, let A denote the subset of the Cartesian product α A α consisting of those elements (a α ) for which sup α a α is nite. Then A is a Banach space with respect to the norm (a α ) = sup α a α. Furthermore, A is a -algebra when equipped with component-wise operations; for example, the adjoint of (a α ) is just (a α), and the product of (a α ) with (b α ) is (a α b α ). It is straightforward to check that, with this structure, A is a C -algebra. Moreover, A is unital if and only if each A α is unital. (If (e α ) is a unit for A, then it is readily see that for any α 0, e α0 is a unit for A α0. The converse is clear.) King's College London

26 26 Chapter 3 The C -algebra A is called the direct sum of the C -algebras {A α }. If α runs over Λ and A α = C, for each α Λ, then A = l (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 p 2 = 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 x = 1 2 (x + x ) + i 1 2i (x x ). We see that (x + x )/2 and (x x )/2i are both self-adjoint elements of A. Conversely, if h and k are self-adjoint of A and x = h + ik, then x = h ik so that h = 1 2 (x+x ) and k = 1 2i (x x ). In other words, the decomposition x = h + ik with h = h, k = k in A is unique. (These are often referred to as the real and imaginary parts of x, respectively.) If A is a unital C -algebra and x A, we shall denote by A(x) the unital C -algebra of A generated by x; that is, A(x) is the closure in 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 σ A (h) R. Proof. Suppose that h A is self-adjoint, and consider the commutative unital C -algebra A(h). For t R, put u t = e ith n=0 (it) n n! (the series converges in A(h) since A(h) is complete). Then by the continuity of the involution, we see that ( n u (it) k ) n t = lim h k = lim, ( it)k h k n k! n k! = u t. k=0 h n. k=0 By multiplying the series (as in the complex case), we see that u t u t = u t u t = u 0 = 1l. Hence 1 = u t u t = u t 2, and so u t = 1 for all t R. Let l Sp A(h). Then, since l is continuous, ( (it) n ) l(u t ) = l h n n! n=0 Department of Mathematics

27 C -algebras 27 = n=0 = e itl(h). (it) n n! l(h) n Since l = 1, we have l(u t ) u t = 1, and hence e itl(h) 1 for all t R. It follows that l(h) R. This holds for all l Sp A(h) and so ĥ is real-valued on Sp A(h). But σ A(h) (h) = ran ĥ and therefore σ A(h)(h) R. Now A(h) A and so σ A (h) σ A(h) (h) and we conclude that σ A (h) R. Theorem 3.7. (Gelfand-Naimark) Suppose that A is a commutative unital Banach -algebra. The Gelfand transform : A C(Sp A) is an isometric -isomorphism if and only if A is a C -algebra. Proof. If is an isometric -isomorphism, then for any x A, x x = x x = (x ) x = x x = x 2 = x 2 = x 2 which is precisely the C -property (v). So A is a C -algebra. (Indeed, C(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 σ A (h) R; that is, ĥ is real-valued. For any x A, write x = (x + x )/2 + i(x x )/2i. Then (x )(l) = l(x ) = l ( x + x i (x x )) 2 2i = l ( x + x ) (x x ) i l ( 2 2i = l ( x + x ) (x x ) ) + i l 2 2i = l(x) = x(l), using the fact that l(h) is real if h is symmetric. It follows that is a -homomorphism. To show that is isometric (and hence also an injection) consider again h = h A. Then, by the C -property of the norm h 2 = h 2 and so h 2n = h 2n. Therefore ĥ = r(h) = lim n h2n 1/2n = h and so is isometric on self-adjoint elements. For any x A we have x 2 = x x King's College London

28 28 Chapter 3 = x x, since is a -homomorphism, = x x, since x x is self-adjoint, = x 2, by the C -property of the norm. Hence is isometric. We must now show that is surjective. To see this, we note that A is complete and since is isometric it follows that ran is closed in C(Sp A). Now, is a -homomorphism and so ran is a closed -subalgebra of C(Sp A) which contains the constant function 1 = 1l. Furthermore, if l 1 l 2, then, by denition, there is x A such that l 1 (x) l 2 (x), that is, x(l 1 ) x(l 2 ). Thus ran separates points of Sp A. It follows from the Stone-Weierstrass theorem that ran = 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 : σ A (x) C be continuous. Since ran x = σ A (x), it follows that f x : Sp A(x) C is well-dened and is continuous; ie, f x C(Sp A(x)). Hence there exists y A(x) A such that ŷ = f x, and y is unique since is an isomorphism. We write y = f(x). In this way, we can dene f(x) as an element of A for any normal (and, in particular, for any self-adjoint) element x A and any function f continuous on the spectrum of x. If f is a polynomial, then f(x) is the obvious polynomial in x. We have seen that the spectrum of an element can depend on the Banach algebra it is considered to belong to. The following is the key result telling us that this is not so for C -algebras. Theorem 3.9. Let A be a unital C -algebra and suppose that x A is invertible. Then x 1 belongs to the C -subalgebra of A generated by 1l, x and x (ie, the closure in A of the set of complex polynomials in 1l, x, x ). Proof. Suppose rst that x = x, and let A denote the unital C -algebra generated by x, and B that generated by x and x 1. Evidently B is commutative and A B A. We know that the Gelfand transform : B B = C(Sp B) is an (isometric -) isomorphism and, since A is a C -subalgebra of B, it follows that  (the range of A under ) is a C -subalgebra of B. Let l 1, l 2 Sp B and suppose that l 1 (x) = l 2 (x). For any l Sp B, l(xx 1 ) = l(x)l(x 1 ) = l(1l) = 1. Hence l 1 (x 1 ) = l 1 (x) 1 = l 2 (x) 1 = l 2 (x 1 ). Since B is generated by x and x 1, we deduce that l 1 = l 2. Thus, if l 1 l 2 then l 1 (x) l 2 (x) or x(l 1 ) x(l 2 ). That is,  separates points of Sp B. By the Stone-Weierstrass theorem, it follows that  = B and so A = B and therefore x 1 A. Now let x A be arbitrary with inverse x 1. Then x x is invertible with inverse x 1 (x 1 ). But x x is self-adjoint and so x 1 (x 1 ) belongs to the C -algebra generated by 1l and x x which is contained in that generated by Department of Mathematics

29 C -algebras 29 1l, x, and x. But then x 1 = x 1 (x 1 ) x = (x x) 1 x is also contained in this last algebra. Corollary Let A B be unital C -algebras with the same unit, and let x A. Then σ A (x) = σ B (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 \ σ B (x) C \ σ A (x) and so σ B (x) σ A (x). But A B implies that σ A (x) σ B (x) and therefore we have σ A (x) = σ B (x). Example Let A be the C -subalgebra of M 2 (C) consisting of those 2 2 complex matrices of the form ( α ), α C, and let B be the C -subalgebra consisting of those matrices of the form ( ) α 0 0 β, α, β C. Then A B, and A and B are both unital with units given by 1l A = ( ) and 1l B = ( ). Note that 1l A B but 1l A 1l B in B. Now if a = ( α ) A, then σ A (a) = {α}, but σ B (a) = {0, α}. Evidently σ A (a) σ B (a). In fact, we have σ A (a) σ B (a) despite the fact that A B. Theorem 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 σ A (h) which maps polynomials in h to the same polynomial on σ A (h). Proof. A is a commutative C -algebra and is isomorphic as a C -algebra to C(Sp A) via the Gelfand transform : A C(Sp A). On the other hand, ĥ : Sp A σ A (h) is a homeomorphism. Dene α : C(Sp A) C(σ A (h)) by α(f) = f ĥ 1 so that α(f)(λ) = f ĥ 1 (λ)) and, in particular, α(ĥ)(λ) = λ for λ σ A (h). Then α is an isometric -isomorphism from C(Sp A) onto C(σ A (h)). Hence α : A C(σ A (h)) is an isometric -isomorphism. Let p be a polynomial. Then ( ) α (p(h)) (λ) = α(p( ĥ))(λ) for any λ σ A (h). = p(α(ĥ))(λ) = p(α(ĥ)(λ)) = p(λ) Theorem 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 ϕ ω : A C by ϕ ω (a) = a(ω), a A. Then, clearly, ϕ ω Sp A. Moreover, if ϕ ω1 = ϕ ω2, then a(ω 1 ) = a(ω 2 ) for all a A. Since A = C(Ω) separates points of Ω, we have ω 1 = ω 2. Thus ω ϕ ω is King's College London