Banach -algebra: Banach algebra with an involution ( 2 = id), which is isometric ( a = a ) anti-isomorphism (antilinear and (ab) = b a ).

Transcription

1 Introduction Banach algebra: algebra over C; Banach space structures are compatible: a + b a + b, λa = λ a ; ab a b. Banach -algebra: Banach algebra with an involution ( 2 = id), which is isometric ( a = a ) anti-isomorphism (antilinear and (ab) = b a ). C -property: a a = a 2. C -algebra is a Banach -algebra satisfying the C -property. Problem: In a C -algebra, 1 + a a is invertible for any a.

2 Examples: C -algebras algebra C(X) of all continuous complex-valued functions on a locally compact Hausdorff space X; algebra B(H) of all bounded operators on a Hilbert space H; Not C -algebras algebra C 1 [0, 1] of functions with continuous derivative; algebra C(0, 1) of continuous functions; algebra l 1 (G) for a countable discrete group G (multiplication is given by the convolution).

3 Unitalization We do not assume C -algebras to have a unit element, but if they have it then 1 = 1 and 1 = 1. Theorem Any non-unital C -algebra A is contained in a unital C -algebra A + as a maximal ideal of codimension 1. Set A + = A C; (a, λ)(bµ) = (ab + λb + µa, λµ); (a, λ) = (a, λ), (a, λ) = sup b 1 ab + λb. Unit: (0, 1). Involution: (a, λ) = (a, λ) (check that is isometry on A + ). Norm is the operator norm. Let B(A) be the algebra of bounded operators on A, L a : A A given by L a (b) = ab. Then A + = {L a + λ 1 : a A, λ C} (the norm on A + is taken from that on B(A); check that a = L a ).

4 C -property for A + : (a, λ) 2 = sup b 1 ab + λb 2 = sup b 1 b a ab + λb a b + λb ab + λ 2 b b sup b 1 a ab + λa b + λab + λ 2 b = (a a + λa + λa, λ 2 ) = (a, λ) (a, λ). The opposite inequality is easier: (a, λ) (a, λ) (a, λ) (a, λ) = (a, λ) 2.

5 Spectrum Let A be a unital Banach algebra, a A. Spectrum: Sp(a) = {λ C : λ 1 a is not invertible}. Resolvent set: Res(a) = C \ Sp(a). Resolvent function: R a (λ) = (λ 1 a) 1 Theorem. Sp(a) ; Sp(a) is a compact set; if λ Sp(a) then λ a ; R a is analytic on Res(a); if p is a polynomial then Sp(p(a)) = p(sp(a)).

6 Let a A. Define its spectral radius by r(a) = sup λ Sp(a) λ. Theorem. r(a) = lim n a n 1/n a ; R a is analytic on {λ C : λ > r(a)}. Theorem. If A is a simple unital abelian Banach algebra then A = C. Proof. Let a A, a λ 1, α Sp(a). Then I = (a α 1)A is a (possibly non-closed) ideal. Any x I is of the form x = (a α 1)b, b A, and cannot be invertible, since a α 1 is not invertible. If 1 x < 1 then x is invertible (use x 1 = 1 + (1 x) + (1 x) ). So, if x I then 1 / I. Moreover, if x I then dist(1, I) 1, hence 1 / I, so I A contradiction with simplicity.

7 For a commutative Banach algebra A we write M A for the set of all non-zero multiplicative linear functionals (characters) of A. Theorem. If ϕ M A then ϕ is continuous of norm 1. Proof. Suppose there exists a A such that a < 1, ϕ(a) = 1. Multiplying by λ, λ = 1, we may assume that ϕ(a) = 1. The series b = n 0 an converges, and a + ab = b. Then 1 + ϕ(b) = ϕ(b) + ϕ(a)ϕ(b) = ϕ(b) contradiction. Hence ϕ 1. But ϕ(1) = 1. Theorem. There is a one-to-one correspondence between M A and the set of maximal ideals of A given by ϕ Ker ϕ. Injectivity: ϕ is determined by its kernel, which has codimension 1, and by ϕ(1) = 1. Surjectivity: If M A is a maximal ideal then dist(1, M) 1, hence 1 / M, so M A, hence M = M. The quotient A/M is a simple unital abelian Banach algebra, hence A/M = C. The quotient homomorphism is a character.

8 M A identified with the set of maximal ideal can be topologized. Let A be the dual Banach space for A, B 1 9A ) its unit ball. Then M A B 1 (A ). The weak topology on B 1 (A ) is determined by the pre-base U aε (ϕ) = {ψ B 1 (A ) : (ψ(a) ϕ(a)) < ε}, a A, ε (0, 2]. M A is obviously Hausdorff: if ψ ϕ then there exists a A such that ψ(a) ϕ(a). By Banach Alaoglu Theorem, B 1 (A ) is compact. If ϕ α ϕ, ϕ α M A, then ϕ is also multiplicative, hence M A is a closed subset of a compact space. For a unital Banach algebra A, M A is a compact Hausdorff space.

9 Non-unital case: If A is a non-unital commutative Banach algebra then A is a maximal ideal in its unitalization A + corresponding to the character ϕ 0 given by ϕ 0 (a + λ 1) = λ. For any other maximal ideal M A + the set M A is an ideal of A of codimension 1. Then the quotient map ϕ : A A/M A is a character, and ϕ(a + λ 1) = ϕ(a) + λ gives a character on A +, which is a unique extension of ϕ. Then there is a one-to-one correspondence between M A and M A + \ {ϕ 0 }, hence M A is locally compact, and M A + is its one-point compactification.

10 Gelfand Transform For a commutative Banach algebra A, a A, ϕ M A, the formula â(ϕ) = ϕ(a) defines a continuous function on M A. Set Γ(a) = â. This gives a Gelfand transform Γ : A C 0 (M A ). Theorem. (1) Γ is a contractive algebra homomorphism; (2) its image separates points in M A. Note that if A is not unital then â(ϕ 0 ) = ϕ 0 (a) = 0 for any a A hence â C 0 (M A ). Proof. (1) Γ(a) = â = sup ϕ MA ϕ(a) ϕ a a. (2) If ψ ϕ then there exists a A such that ψ(a) ϕ(a).

11 Corollary. Let A be a unital commutative Banach algebra, a A. Then TFAE: a is invertible; â is invertible; â(ϕ) 0 for any ϕ M A. Thus Sp(a) = Sp(â) = {ϕ(a) : ϕ M A }, and â = r(a). Proof. If a is invertible then Γ(a) is invertible, and Γ(a) 1 = Γ(a 1 ). If a is not invertible then consider the ideal I = aa. It consists of non-invertibles (with distance from 1 bounded from below by 1), so I A. Hence there exists a maximal ideal M such that I M A. Let ϕ M A be the corresponding character. Then M = Ker ϕ, so â(ϕ) = ϕ(a) = 0, hence â is not invertible. As the range of â = Sp(a), so â = r(a).

12 Now let A be a unital C -algebra (not necessarily commutative). Lemma. Let ϕ be a multiplicative functional on A. Then ϕ(a ) = ϕ(a). (ita) n n=0 Proof. First, let a = a. The series u t = n! and ut = ( ita) n n=0 n! converge, and ut = u 1 t for any t R. By the C -property, u t 2 = ut u t = 1 = 1. Then ϕ(u t ) 1. But ϕ(u t ) = (itϕ(a)) n n=0 n! = e itϕ(a). Hence e t Im ϕ(a) = e itϕ(a) 1 for any t R. So Im ϕ(a) = 0. Now let a be arbitrary. It decomposes as a = b + ic, where b, c are selfadjoint, and a = b ic. As ϕ(b), ϕ(c) are real, the conclusion follows.

13 Theorem. Let A be a commutative C -algebra. Then Γ is an isometric -isomorphism A C 0 (M A ). Proof. If a = a then a 2 = a 2, hence Γ(a) = â = r(a) = lim n a 2n 1/2n = lim n ( a 2n ) 1/2n = a. If a is arbitrary, then a 2 = a a = Γ(a a) = Γ(a) 2. Surjectivity of Γ follows from the Stone Weierstrass Theorem (Γ(A) separates points in M A ).

14 Continuous functional calculus for normals Let A be a C -algebra, a A normal, i.e. a a = aa, and let C (a) A be the C -algebra generated by a (equivalently, the minimal C -subalgebra containing a). Corollary. If C (a) is unital then the Gelfand transform C (a) C(Sp(a)) is an isometric -isomorphism. If C (a) is not unital then the image of the Gelfand transform is C 0 (Sp(a) \ {0}). Indeed, C (a) is commutative, so C (a) = C(X) for some X (with necessary corrections in the non-unital case). Any its character ϕ is determined by ϕ(a) = λ. λ can be the value of ϕ iff λ lies in the range of the function â, which is Sp(a).

15 Continuous functional calculus for normals Let f C(Sp(a)). Then f (a) := Γ 1 (f ) C (a) (or C (a) +, if the latter was not unital; if 0 Sp(a) and f (0) = 0 then f (a) C (a)). As Gelfand transform is an isometry, so if f = lim n p n, where p n = p n (a, a ) are polynomials, then f (a) = lim n p n (a, a ). If a A is normal then (1) Sp(f (a)) = f (Sp(a)); (2) g(f (a)) = (g f )(a) for any function g continuous on f (Sp(a)). Proof: approximation by polynomials. Corollary. In a C -algebra A one has if a A is normal then a = r(a); if a = a then Sp(a) R C; if a a = aa = 1 then Sp(a) S 1 C.

16 Positivity Definition. An element a of a C -algebra is positive if a = a and Sp(a) [0, ). Notation: a 0. Lemma. For any positive a A there exists a unique positive b A such that b 2 = a. Proof: The function f (t) = t is continuous on [0, ), so put b = f (a). If c 2 = a then c = f (c 2 ) = f (a) = b. Lemma. Let a = a. There exist a +, a 0 such that a = a + a and a + a = 0. { { t, if t 0 0, if t 0 Proof: Let f + (t) = 0, if t 0, f (t) = t, if t 0, a ± = f ± (a).

17 Positivity Lemma. Let a = a. TFAE: 1 a 0; 2 a = b 2 for some selfadjoint b; 3 λ 1 a λ for any λ a ; 4 λ 1 a λ for some λ a. Proof. 1 2 was already done, 3 4 is trivial. Remaining implications 2 3 and 4 1 can be checked inside C (a) = C(Sp(a)), where a corresponds to the function f (t) = t (hence λ 1 a corresponds to the function λ t on Sp(a)), positivity for a function f means that f takes only non-negative values, and the norm is the sup-norm. In particular, if sup t Sp(a) λ t λ for some λ > a then Sp(a) [0, ).

18 Corollary. If a, b 0 then a + b 0. Proof: Let λ a, µ b, then λ + µ a + b, and (λ + µ)1 (a + b) λ 1 a + µ 1 b λ + µ. Thus the set of positive elements in a C -algebra is a cone. Theorem. In a C -algebra, a a 0 for any a. Proof uses the following algebraic lemma: Lemma. Let R be a ring, a, b R. 1 ab is invertible iff 1 ba is invertible. Proof: if c = (1 ab) 1 then d = (1 ba) 1, where d = 1 + bca. Corollary. Sp(ab) {0} = Sp(ba) {0}. Proof: If λ 1 ab is (not) invertible and if λ 0 then 1 1 λab is (not) invertible.

19 Positivity Proof of the Theorem. Set b = a a (it is patently selfadjoint) and decompose it as b = b + b, where b +, b 0, b + b = 0. Set c = b 1/2, t = ac. As the square root function is a limit of polynomials that vanish at 0, so b + b = 0 implies cb + = 0. Then t t = ca ac = c(b + b )c = cb c = b, 2 hence t t = b 2 0. Let t = x + iy with x, y selfadjoint. Then t t + tt = (x + iy) (x + iy) + (x + iy)(x + iy) = 2(x 2 + y 2 ) 0. As both t t and t t + tt are positive, their sum is positive: tt 0. Thus, Sp(tt ) [0, ), and Sp(t t) (, 0], so Sp(t t) = {0}. Finally, b 2 = t t = r(t t) = 0, so b = 0.

20 Let a, b be selfadjoints. We write a b if b a 0. Lemma. If a b then x ax x bx for any x. Proof: Let c 0, c 2 = b a. Then x (b a)x = x c 2 x = (cx) (cx) 0. Lemma. Let 0 a b, and let a be invertible. Then b is invertible and b 1 a 1. Invertibility of a means that there exists ε > 0 such that Sp(a) [ε, ). Then b enjoys the same property, hence is invertible. Then 1 b 1/2 ab 1/2 = b 1/2 (b a)b 1/2 0, which means that (a 1/2 b 1/2 ) (a 1/2 b 1/2 ) 1. As Sp(xy) and Sp(yx) coinside up to {0}, we have a 1/2 b 1 a 1/2 = (a 1/2 b 1/2 )(a 1/2 b 1/2 ) 1. Multiplying by a 1/2 both from the left and from the right, we get b 1 a 1.

21 In general, one has to be careful with the partial order. Compare: Theorem. If 0 a b implies that 0 a 2 b 2 for any positive a, b A then A is commutative. Theorem. If 0 a b then a 1/2 b 1/2.

22 Approximate unit Let A be a C -algebra. A set {e λ : λ Λ} A is an approximate unit if Λ is a directed set; e λ 0 and e λ 1 for any λ Λ; e λ e µ if λ µ; lim λ Λ ae λ = lim λ Λ e λ a = a. If A is unital then Λ can contain one element, and 1 can be taken as e λ. Example 1. Let A = C 0 (0, 1] and let Λ = [1, ). Let f λ (t) = { λ t, if t (0, 1 λ ; 1, if t [ 1 λ, 1]. Then {f λ } is an approximate unit. Example 2. Let K(H) be the algebra of compact operators on a separable Hilbert space H, with a fixed orthonormal basis {x n } n N, and let p n K(H) denote the projection onto the linear span of x 1,..., x n. Then {p n } is an approximate unit.

23 Approximate unit Theorem. Any C -algebra A has an approximate unit. If A is separable then it has a countable approximate unit. Proof. Set Λ = {a A : a 0, a < 1} ordered by. We have to show that for any a, b Λ we can find c Λ such that a c, b c. c = a + b is not a right choice. The idea is to re-scale spectra of a and b from [0, 1] to [0, ), then add the rescaled elements together, and then re-scale the sum back. For t 1 t re-scaling forward, we use f (t) =, and for re-scaling backward, we use g(t) = t 1+t = t, g(f (t)) = t. Set x = f (a), y = f (b), z = x + y, c = g(c). As Sp(z) [0, N] for some N R, so c = sup t Sp(z) g(t) < 1, hence N N+1 c Λ. As 1 + x 1 + z, so we have (1 + x) 1 (1 + z) 1, and a = g(c) = 1 (1 + x) 1 1 (1 + z) 1 = c. Symmetrically, b c.

24 Approximate unit Proof (continued). Also, if a, b Λ, a b, then d bd 2 = d (1 b) 2 d d (1 b)d d (1 a)d for any d A. Take d A, d 0. Set a n = g(nd) Λ, h n (t) = t 2 (1 g(nt)) = t2 1+nt t n. Then d a nd = h n (d), hence d a n d = h n Sp(d) 0 as n. Then lim b Λ d bd 2 lim n (sup b Λ;b an d bd 2 ) lim n d(1 a n )d = 0. If d is arbitrary, then d bd 2 = (1 b)d d(1 b) d d bd d 0. Similarly, d db 0 as b Λ. The standard argument shows that if A is separable then one can find a subsequence in Λ with the required properties.

25 Ideals Unless the converse is stated, by an ideal in a C -algebra we always mean a closed, two-sided ideal. Lemma. If J A is an ideal then J = J. Proof. Set B = J J. Then B is a C -algebra, and JJ B. Let {e λ } λ Λ be an approximate unit for B, and let j J. Then lim λ Λ j j e λ 2 = lim λ Λ jj jj e λ e λ (jj jj e λ ) 2 lim λ Λ jj jj e λ = 0. As J is closed and j e λ J then j J. Unlike general rings, for C -algebras one has Lemma. Let J A be an ideal in A, and let I J be an ideal in J. Then I is an ideal in A. Let i I, a A. We need to show that ia I. Assume first that i is positive. Then i = (i 1/2 ) 2, hence i 1/2 a J and i 1/2 i 1/2 a I. The case of arbitrary i I can be done by decomposing i as a linear combination of 4 positive elements.

26 Quotients Let J A be an ideal. Then (J is closed) the quotient Banach algebra A/J is well defined with the norm given by ȧ = inf j J a j, where ȧ = a + J A/J. As J is symmetric, so ȧ = ȧ. Theorem. A/J is a C -algebra. Proof. We need to check the C -property. First, we claim that ȧ = lim λ Λ a ae λ for any approximate unit {e λ } in J. As ae λ J, so we obviously have ȧ a ae λ. For any ε > 0 there exists j J such that a j < ȧ + ε. Then lim λ Λ a ae λ lim λ Λ (a j)(1 e λ ) + j je λ a j < ȧ + ε. As ε is arbitrary, the claim is proved. ȧ ȧ ȧ ȧ = ȧ 2, so itremains to check the opposite inequality: ȧ 2 = lim λ Λ a(1 e λ ) 2 = lim λ Λ (1 e λ )a a(1 e λ ) lim λ Λ a a(1 e λ ) = ȧ ȧ.

27 -homomorphisms Theorem. Let A, B be C -algebras, π : A B a non-zero -homomorphism. Then π = 1 and π(a) is a C -subalgebra in B. If π is injective then π is an isometry. In general, π factorizes as π = π q, where q : A A/ Ker π, and π induced by π is an isometric -isomorphism of A/ Ker π onto π(a). Proof. We prove this for the case when A, B are unital, and π(1) = 1. (The general proof is slightly more technical.) If a A is invertible then π(a) is invertible, hence Sp B (π(a)) Sp A (a) for any a A, therefore, r(π(a)) r(a). Then π(a) 2 = π(a a) = r(π(a a)) r(a a) = a a = a 2, hence π 1. As π(1) = 1, π = 1. Therefore, π is continuous, Ker π is a closed ideal in A, and π factorizes through A/ Ker π with π injective.

28 -homomorphisms Proof (continued). Suppose that π is not an isometry. Then there exists a A such that π(a) < a. Let r = π(a a) < a a = s, and let f C[0, s] be the function such that f (t) = 0 for t [0, r], and f (s) = 1. Then π(f (a a)) = f ( π(a a)) = 0, but as f 0 on Sp(a a), so f (a a) 0 a contradiction with injectivity of π. So, π is an isometry, and π(a) is closed. Corollary. If 1 and 2 are two norms on A, both making A a C -algebra, then 1 = 2. Remark. It is hidden here that A is closed with respect to both norms. If not, there may be many C -norms on a -algebra

29 States A linear functional ϕ : A C is called positive if ϕ(a) 0 for any positive a A. A positive linear functional ϕ is called a state if ϕ = 1. Example 1. Let A = C(X) for a compact Hausdorff X, and let x 0 X. The evaluation map ϕ(f ) = f (x 0 ), f C(X), is a state. Example 2. Let B(H) be the algebra of linear operators on a Hilbert space H, and let ξ H. Then the map ϕ(a) = aξ, ξ, a B(H), is a positive linear functional. It is a state iff ξ = 1. If ϕ is a positive linear functional then the formula a, b = ϕ(b a)defines a positive semi-definite sesquilinear form on A. An immediate corollary is the Cauchy Schwarz inequality ϕ(b a) 2 ϕ(a a)ϕ(b b).

30 Lemma. Let ϕ be a positive linear functional on a C -algebra A. Then ϕ = lim λ Λ ϕ(e λ ) for any approximate unit {e λ } in A. In particular, if A is unital then ϕ = ϕ(1). Proof. We prove this Lemma under the additional assumption that ϕ is bounded. Under this assumption, the set ϕ(e λ ) is increasing and bounded, hence converges to some M = lim λ Λ ϕ(e λ ). Obviously, ϕ M. To prove the opposite inequality, take ε > 0 and find a A such that a 1 and ϕ(a) > ϕ ε. Then ϕ(a) 2 = lim λ Λ ϕ(e λ a) 2 lim λ Λ ϕ(e 2 λ )ϕ(a a) M ϕ (we used here e 2 λ e λ). Thus we have ( ϕ ε) 2 M ϕ. As ε is arbitrary, so ϕ 2 M ϕ, ϕ M. The unital case follows.

31 GNS construction Gelfand Naimark Segal construction Theorem. Let ϕ be a positive linear functional on a C -algebra A. Then there exist a Hilbert space H ϕ, a representation π ϕ of A on H ϕ, and a vector ξ ϕ H ϕ, cyclic for π ϕ, such that ξ ϕ 2 = ϕ and ϕ(a) = π ϕ (a)ξ ϕ, ξ ϕ, a A. Proof. Set N = {a A : ϕ(a a) = 0}, N = {a A : ϕ(b a) = 0 b A}. Both are closed. Due to the Cauchy Schwarz inequality ϕ(b a) 2 ϕ(a a)ϕ(b b), N = N, hence N is a linear subspace in A. Moreover, N is a left ideal: if n N, a, b A then ϕ(b (an)) = ϕ((a b) n) = 0, hence an N. Then we can pass to the quotient Banach space A/N and define a positive definite sesquilinear form ẋ, ẏ = ϕ(y x), where ẋ denotes the class x + N A/N of x A. (It is well-defined and if ẋ, ẋ = 0 then ẋ = 0.)

32 Proof (continued). The inner product ẋ, ẏ defines a norm on A/N. Let H ϕ denote the closure of A/N with respect to this norm. To define π ϕ, let us first define a representation π 0 of A on A/N by π 0 (a)ẋ = (ax) (well-defined because N is a left ideal). π 0 is obviously linear and π 0 (ab) = π 0 (a)π 0 (b) for any a, b A. It is also symmetric: π 0 (a ) ẋ, ẏ = ẋ, π 0 (a )ẏ = ẋ, (a y) = ϕ((a y) x) = ϕ(y ax) = π o (a)ẋ, ẏ, hence π 0 (a ) = π 0 (a). Continuity of π 0 follows from the estimate π 0 (a) 2 = sup ẋ 1 π 0 (a)ẋ 2 = sup ẋ 1 ϕ(x a ax) sup ẋ 1 ϕ(x x) a a = sup ẋ 1 ẋ 2 a 2 (we use here a a a a 1). By continuity, π 0 extends to a representation π ϕ on the completion H ϕ of A/N. If A is unital then set ξ ϕ = 1. It is cyclic: π ϕ (A)ξ ϕ = A/N is dense in H ϕ. Also ϕ = ϕ(1) = ξ ϕ 2. Finally, π ϕ (a)ξ ϕ, ξ ϕ = ϕ(1 a1) = ϕ(a). If A is not unital then the proof is more technical: one has to prove that, for an approximate unit {e λ } the limit lim λ Λ ė λ exists, and then set ξ ϕ = lim λ Λ ė λ. We skip the details.

33 Lemma. Let ϕ be a linear functional on a C -algebra A such that ϕ = 1 = lim λ Λ ϕ(e λ ) for some approximate unit {e λ }. Then ϕ is a state. Proof. First, we reduct this to the unital case. Let ϕ : A + C be a Hahn Banach extension of ϕ to A +, and let ϕ(1) = α C. As ϕ = ϕ = 1, so α 1. As 2e λ 1 1, so ϕ(2e λ 1) 1. Passing to the limit, we get 2 α 1. Hence α = 1. Now we may assume that A is unital, and that ϕ = 1 = ϕ(1). Claim. If a A is selfadjoint then ϕ(a) R. The proof follows from the claim. Let 0 a 1. Then 2a 1 1, hence ϕ(2a 1) 1. As ϕ(a) is real, so we get 1 2ϕ(a) 1 1, hence ϕ(a) 0. Positivity is proved (modulo the Claim).

34 Proof of the Claim. Let a = a, a = 1. Take n N. a + in1 2 = (a in1)(a + in1) = a 2 + n 2 1 = n Similarly, a in1 2 = n Then ϕ(a ± in1) = ϕ(a) ± in n As this holds for any n N, so ϕ(a) [ 1, 1], hence it is real. Lemma. Let B A be a unital C -subalgebra of a unital C -algebra A (with common unit), ϕ a state on B. Then there exists a state ϕ on A, that extends ϕ. Proof. Let ϕ be a Hahn Banach extension of ϕ. As ϕ = 1 = ϕ(1), so ϕ is a state. Corollary. Let a A be selfadjoint. Then there exists a state ϕ on A such that ϕ(a) = a. Proof. If A is unital then set B = C (a) + A. Either r(a) or r(a) belongs to Sp(a), and one of the two formulas ϕ(f ) = f (±r(a)), f C(Sp(a)), gives a required state.

35 Theorem (Gelfand Naimark). Any C -algebra A is isometrically -isomorphic to a closed -subalgebra of B(H) for some Hilbert space H. If A is separable then H can be taken separable. Proof. Let S(A) denote the set of all states on A. For any ϕ S(A), apply the GNS construction. Then π = ϕ S(A) π ϕ is a representation of A on the Hilbert space ϕ S(A) H ϕ. Take a A. Let ϕ S(A) satisfy ϕ(a a) = a a. Then π(a) 2 π ϕ (a) 2 π ϕ (a)ξ ϕ 2 = π ϕ (a a)ξ ϕ, ξ ϕ = ϕ(a a) = a a. On the other hand, π is a -homomorphism, hence π = 1, therefore π(a) = a for any a A. If A is separable then one can use a dense sequence {a n } in A, and the corresponding sequence {ϕ n } of states such that ϕ n (a na n ) = a n 2. Set π = n N π ϕn.

36 Let A be a C -algebra, H a Hilbert space, π : A B(H) a representation. π : A B(H) is algebraically irreducible if is has no proper invariant subspace. It is topologically irreducible if it has no proper closed invariant subspace. For a set S B(A) we denote by A! its commutant in B(H), i.e. S! = {a B(H) : as = sa s S}. TFAE: π(a)! = C 1; π is algebraically irreducible; π is topologically irreducible. Proof requires Borel functional calculus in B(H), the weak topology on B(H), and is based on Kaplansky density Theorem. Representation π is non-degenerate if π(a)h = {π(a)ξ : a A, ξ H} is dense in H. Irreducible non-degenerate: if π(a)h differs from H then it is an invariant subspace.

37 Let J A be an ideal, π : J B(H) a non-degenerate representation. Then π uniquely extends to a representation π of A on H. π is irreducible iff π is irreducible. Proof. For a A, j J, ξ H, set π(a)(π(j)ξ) := π(aj)ξ. It may happen that π(j 1 )ξ 1 = π(a 2 )ξ 2. Let {e λ } be an approximate unit in J. Then π(aj 1 )ξ 1 = lim λ Λ π(ae λ j 1 )ξ 1 = lim λ Λ π(ae λ )π(j 1 )ξ 1 = lim λ Λ π(ae λ )π(j 2 )ξ 2 = π(aj 2 )ξ 2, so π is well defined (and uniquely defined) on the dense set π(a)h. It remains to extend it by continuity to H, and the estimate π(a)(π(j)ξ) = lim λ Λ π(ae λ ) π(j)ξ a π(j)ξ shows that π is continuous. If π has an invariant subspace then π has it as well. Let us check the opposite. Let L H be a closed invariant subspace for π. Then π(a)l = π(a)π(j)l = π(j)l L.

38 Let π : A B(H) be an irreducible representation, J A an ideal. Then π J is irreducible. Proof. Suppose the contrary: there exists a closed subspace L H invariant under π(j), i.e. π(j)l L. Then π(a)π(j)l L. Finitedimensional C -algebras Let A be a finitedimensional C -algebra. Then it is unital. Proof. As A is finitedimensional, so the approximate unit has an accumulation point. Obviously, this is the unit element.

39 Finitedimensional C -algebras First, let us specify the Gelfand Naimark Theorem. For a finitedimensional C -algebra A there exists a faithful representation π : A B(H) on a finitedimensional Hilbert space H. Proof. To show injectivity of π = π ϕ, we don t need the exact equality π ϕ (a) = a. It suffices to have the estimate π ϕ (a) > 1/2 a. For each a A find an open set U a S(A) of states satisfying this estimate. As A is finitedimensional, so S(A) is compact, hence we can find a finite cover by the sets of the form U a. Pick one state in each U a and take a direct sum of GNS representations with these states. Now it is easy to decompose a faithful finitedimensional representation into a direct sum of irreducible ones. Let π be one of those, and let H be the corresponding finitedimensional Hilbert space.

40 Lemma. If π is irreducible and H finitedimensional then π(a) contains a minimal projection of rank 1. By projections we mean selfadjoint projections. Proof. As 1 π(1) is a projection, so one can find a minimal non-zero projection p = π(a) in π(a) (minimal with respect to the patial order ). Consider the set pπ(a)p B(H). Suppose there is b A such that pπ(b)p = π(aba) is not a scalar, i.e. has at least two eigenvalues, λ 1 and λ 2. Let f C(Sp(aba)) satisfy f (λ 1 ) = 1, f (λ) = 0 for all other eigenvalues λ, and let g = 1 f. Then π(f (aba)) = f (π(aba)) and π(g(aba)) = g(π(aba)) are projections with π(f (aba)) + π(g(aba)) = π(a) = p, both non-trivial. This contradicts to minimality of p. Thus, pπ(a)p consists only of scalars. Suppose that rk p > 1. Take two orthonormal vectors ξ, η Im p. Then π(c)ξ, η = pπ(c)pξ, η = λ ξ, η = 0 for any c A, where λ is some scalar, hence π(a)ξ is an invariant space contradiction with irreducibility.

41 Lemma. If π is irreducible and H finitedimensional then π(a) = B(H). Proof. Let ξ H. Irreducibility implies that for any η H there exists c A such that π(c)ξ = η (otherwise the linear subspace π(a)ξ would be proper). Now let p = π(a) be a minimal projection of rank 1 in π(a), and let ξ be a unit vector in the image of p. Then pω = ω, ξ ξ. If π(c)ξ = η then π(ca)ω = ω, ξ η. For ζ H let b A satisfy π(b)ζ = ξ. Then π(cab)ω = ω, ζ η. The operators of this form span B(H). Corollary. If A is a finitedimensional C -algebra then there exists a surjective -homomorphism ρ : A M n, where M n is the matrix algebra B(H) of operators on an n-dimensional Hilbert space H.

42 Theorem. A finitedimensional C -algebra A is isometrically -isomorphic to M n1 M nk. Proof. Let ρ : A M n for some n a surjective -homomorphism, and let e B = Ker ρ be the unit of the ideal B. If a A then a = eae + (1 e)ae + ea(1 e) + (1 e)a(1 e), with the first 3 summands being from B (where 1 is the unit of A). Note that as e is the unit of B, so ea(1 e) = (1 e)ae = 0 for any a A. Thus A = eae (1 e)a(1 e) (the sum is obviously a direct sum), where eae = B. Define a -homomorphism ϕ : A B M n by ϕ(a) = (eae, ρ(a)). As dim(1 e)a(1 e) = dim M n, so it suffices to check injectivity of ϕ. If ρ(a) = 0 then a B, hence a = eae, which is zero.

43 Any two non-zero -homomorphisms ϕ, ψ : M n M n are unitarily equivalent, i.e. ψ = Ad U ϕ for some unitary U M n. If m < n then any -homomorphism M n M m is a zero map. If n < m < 2n then there is no unital -homomorphisms from M n to M m, but there is exactly one (up to unitary equivalence) -homomorphism. If m = 2n then there are exactly two non-trivial -homomorphisms M n M m : one is non-unital and takes rank 1 projections to rank 1 projections; another is unital and takes rank 1 projections to rank 2 projections. If kn m < (k + 1)n then there are exactly k non-trivial -homomorphisms M n M m, taking rank 1 projections to rank k projections. Thus any non-trivial -homomorphism M n M m is characterized by a single positive integer: the rank of the image of rank 1 projections. If it is trivial then the corresponding integer iz zero.

44 Let A = M n1 M nk, B = M m1 M ml be two finitedimensional C -algebras, ϕ : A B a -homomorphism. Let α i : M ni A and β j : B M mj be the obvious inclusion and projection maps respectively. Then β j ϕ α i : M ni M mj is characterized by an integer N ij, and ϕ is characterized by the set of these numbers. This can be represented as a picture, where A is depicted by k points labeled by the dimensions n 1,..., n k, B is depicted by l points labeled by the dimensions m 1,..., m l, and each point labeled by n i is connected to the point labeled by m j by N ij arrows. A C -algebra A is an AF algebra if it has C -subalgebras A 1 A 2 A such that n=1 A n is dense in A. Drawing each inclusion A n A n+1 by points and arrows allows to characterize AF algebras by diagrams, called Bratteli diagrams.

45 The algebra K = K(H) of compact operators on a separable Hilbert space H is the closure of the union of matrix algebras M n = p n Kp n, n N, where p n denotes the projection onto the linear span of the first n vectors of a fixed orthonormal basis of H, with the inclusion M n M n+1 as the upper left corner. This corresponds to the Bratteli diagram Its unitalization K + is the closure of n=1 A n, where A n = Cp n + p + nk + p n = C Mn with the Bratteli diagram

46 Let K be the Cantor set, K = n=0 X n, where X n is the union of 2 n intervals of length 1/3 n. Let A n C(K ) consists of functions that are locally constant on X n. Then A n = C 2 n, and n=0 A n is dense in C(K ). Here is the corresponding diagram:

47 CAR algebra Let H be a separable Hilbert space with a fixed decomposition H = H 1 H 2 with H 1 = H2. Let each of these two Hilbet subspaces is also a direct sum H i = H i,1 H i,2, etc. Let A 1 consist of scalar operators on H, A 2 consist of 2 2 matrices with scalar entries with respect to the decomposition H 1 H 2, etc. Then A n = M2 n 1. The inclusion A n A n+1 is given by a ( ) a 0 0 a. The closure of n=1 A n is a C -algebra called the CAR algebra with the Bratteli diagram Lemma. If two AF algebras A and B have the same Bratteli diagrams then they are -isomorphic.

48 Lemma. Linear combinations of projections are dense in any AF algebra. Proof. Any element of a matrix algebra (hence, of a finitedimensional C -algebra) is a linear combination of projections. Corollary. C[0, 1] is not an AF algebra. Caution: C -subalgebras of AF algebras need not be AF. Let X be a compact metric space, then there exists a continuous surjection K X, which induces an injective -homomorphism C(X) C(K ).

49 Instead of matrix algebras, one may use other C -algebras as building blocks. Let C([0, 1]; M m ) be the C -algebra of all continuous functions on [0, 1] taking values in M m. Even such C -algebras may have non-trivial subalgebras. Example: Dimension-drop algebras. Set D m = {f C([0, 1]; M m ) : f (0) = 0, f (1) is diagonal}. This is a C -algebra. If time allows, later we shall see that its K -groups are torsion. A C -algebra is called an AI algebra if it contains C -subalgebras A 1 A 2... A such that each A n is a C -subalgebra of C([0, 1]; M m ) for some m, and if n=1 A n is dense in A.

50 For a C -algebra A, let M n (A) be the set of all n n matrices with entries from A. Let π : A B(H) be a faithful -representation on a Hilbert space H. Set H = H H (n times). Then there is a canonical -isomorphism M n (B(H)) = B( H). Define π : M n (A) B( H) by π((a ij ) n i,j=1 ) = (π(a ij)) n i,j=1. Then π is a faithful -representation of M n (A). Set (a ij ) n i,j=1 = π((a ij) n i,j=1 ). Then M n(a) can be considered as a norm-closed -subalgebra of B( H), hence it is a C -algebra. As all C -norms on a C -algebra are the same, the norm on M n (A) doesn t depend on choice of π (provided it is faithful). Similarly, For a (locally) compact Hausdorff X, the set of all continuous A-valued functions on X (vanishing at infinity) is a C -algebra C(X; A) with the norm f = sup x X f (x). This starts the theory of tensor products of C -algebras a very interesting topic, which is beyond our short course. A good reference is the book by N. Brown and N. Ozawa.

51 Toeplitz algebra Let L 2 (S) be the Hilbert space of square-integrable functions on the unit circle S C, with the basis {e n } n Z, e n = z n, z C. Let H 2 (S) L 2 (S) be the subspace spanned by e 0, e 1, e 2,..., and let L (S) be the C -algebra of essentially bounded measurable functions. For g L (S), set M g (f ) = gf, f L 2 (S). M g = g. Define an operator T g on H 2 (S) by T g h = PM g h, where P denotes the projection onto H 2 (S). In particular, T z e n = e n+1, so T z is the right shift on H 2 (S). For an operator T on a Hilbert space H define its essential norm T ess as its distance to the set K(H) of compact operators (which is the norm on the quotient C -algebra B(H)/K(H)). Lemma. Let g L (S). Then T g = T g, and T g = T g ess = g.

52 Proof. The first statement is trivial. As regards the norms, one has, trivially, T g ess T g M g = g. To prove the opposite estimate, take ε > 0 and find p = N k= N a kz k L 2 (S) such that p 2 = 1 and gp 2 > g ε (here 2 is the Hilbert space norm on L 2 (S)). Note that z n p H 2 (S) for n > N. Let gp = k= b ke e. Then T g (z n p) = P(z n gp) = k= n b ke k, and there exists N 0 > N such that T g (z n p) 2 > gp 2 ε for any n N 0. Thus, for n > N 0 one has T g (z n p) > g 2ε. Let ξ i = z 3Ni p H 2 (S). These vectors have length 1 and are orthogonal to each other. Let k K(H). Then T g k sup i (T g k)ξ i = lim sup i T g ξ i g 2ε, hence T g ess g.

53 Let H (S) = H 2 (S) L (S). Lemma. Let g L (S), h H (S). Then H 2 (S) isinvariant for M h, and T g T h = T gh, T h T g = T hg. Proof. If f H 2 (S) then T h f = P(hf ) = hf H 2 (S) (as both are series in z n for n 0). T g T h (f ) = T g (hf ) = P(ghf ) = T gh (f ); T h T g = (T g T h ) = T gh = T hg. Lemma. If g L (S) then T z T g T g T z has rank 1. Proof. T z T g T g T z = (PM z PM g PM z M g ) H 2 S = (PM zp )M g H 2 S, and PM z P has rank 1. Corollary. If g L (S), f C(S) then T f T g T fg and T g T f T fg are compact operators.

54 Toeplitz algebra Set T = {T f + k : f C(S), k K(H 2 (S))}. Lemma. T is a C -algebra. Proof. T is closed under the operations of a -algebra, e.g. (T f1 + k 1 )(T f2 + k 2 ) = T f1 f 2 + k for some compact operator k, so it remains to check that it is norm-closed. Let {T fn + k n } n N be a Cauchy sequence. Then f n f m = T fn T fm ess (T fn + k n ) (T fm + k m ), hence the sequence {f n } is a Cauchy sequence as well, thus it has a limit f = lim n f n C(S). Then lim n T fn = T f, and therefore the sequence {k n } is also a Cauchy sequence. As compact operators are norm-closed, so it also has a limit k = lim n k n. Thus lim n T fn + k n = T f + k T. Another way to define the Toeplitz algebra is to take the minimal unital C -subalgebra of B(H 2 (S)) containing T z. Then it is an excercise to check that all compact operators lie in it.

55 The subset K = K(H 2 (S)) T is an ideal. Lemma. The quotient T /K is -isomorphic to C(S). Proof. Note that the quotient is commutative, hence it is C(X) for some X to be determined. There is a map q : T C(S) given by q(t f + k) = f. It is obviously a -homomorphism. As T f = T f ess = f, so Ker q = K. It remains to check surjectivity of q. Note that there is a map s : C(S) T, given by s(f ) = T f. This map is not a -homomorphism, but it is continuous, and q s = id C(S). Recall that an operator is Fredholm if it is invertible modulo the compact operators. If f C(S) never equals 0 then the winding number wind f is well-defined as that of the curve f (S) around 0. T f is Fredholm iff 0 is not in the range of f. In this case, ind T f = wind f. Proof. Homotopy invariance of both ind and wind show that it suffices to check this equality for f (z) = z n, where it is easy.

56 Lemma. There is no -homomorphism C(S) T right-inverse to q. Proof. Suppose σ : C(S) T is a -homomorphism. Commutativity implies that σ(f ) is normal for any f C(S), hence ind σ(f ) = 0. But if q σ = id C(S) then ind σ(f ) = wind f. The ideal K, the quotient C(S) and the Toeplitz algebra can be organized into a short exact sequence 0 K T C(S) 0. Short exact sequences are often called extensions of the quotient by the ideal. The extension given by the Toeplitz algebra plays an important role in the index theory and in the K -theory of C -algebras.

57 Group C -algebras Let G be a countable discrete group, and let π : G U(H) be its unitary representation on a Hilbert space H. It can be extended to a -representation of the group ring C[G] on H. Then a π = π(a), a C[G], defines a semi-norm on C[G], which becomes a norm after we pass to the quotient C[G]/ Ker π. This norm is obviously a C -norm. The completion of C[G]/ Ker π with respect to π gives a C -algebra C π(g) of G determined by π. Example. Let θ be the trivial representation, which sends all group elements to 1. Then C θ (G) = C is just scalars.

58 The reduced group C -algebra Example. Let λ be the (left) regular representation of G on H = l 2 (G) given by λ(g)( h G β h δ h ) = h G β gδ gh, where δ h l 2 (G) are delta functions (orthonormal basis), and β h C are coefficients. The corresponding C -algebra Cλ (G) is called the reduced group C -algebra (sometimes denoted by Cr (G)). Note that λ( g G α g g) λ( g G α g g)δ e 2 = g G α g δ g = g G α g 2, where e G denotes the neutral element, hence Ker λ = {0} and the reduced group C -algebra is a completion of the group ring without taking quotients. The full group C -algebra If {π γ } γ Γ is a family of unitary representations of G (labeled by elements of a set Γ) then one can take π = γ Γ π γ. Then π = sup γ Γ πγ. If we take all representations of G we get the universal, or full group C -algebra C (G).

59 Example. If G is finite then there is no need to complete, so C (G) = C r (G) = C[G]. Example. Let G = Z with a generating element a Z. Let x = n Z α na n C[Z]. Any unitary representation of Z decomposes as a direct integral of irreducible representations, all of which are known: all irreducible representations of Z are one-dimensional, and are determined by π t (a) = e 2πit, t [0, 1]. Therefore, x u = sup t [0,1] π t (x) = n Z α ne 2πint. Hence the map ϕ : C (Z) C(S) given by ϕ(x)(t) = n Z α ne 2πint is an isometry (that s why it is well defined not only on C[Z], but also on C (Z)). An appropriate choice of vectors in l 2 (Z) shows that x λ = sup t [0,1] π t (x).

60 Sometimes the universal norm u coincides with the reduced (or regular) norm λ, but in general u λ. The identity map on C[G] extends to a continuous map C (G) C r (G), which is always surjective, but in general this map can have a non-trivial kernel, and the algebras C (G) and C r (G) may be very different. In particular, if G = F 2 is the free group on two generators then C (F 2 ) has infinitely many ideals of finite codimension, while C r (F 2 ) is simple. Theorem. The surjective -homomorphism C (G) C r (G) is a -isomorphism iff G is an amenable group. Proof: difficult. Cf. book by N. Brown and N. Ozawa.

61 Let A be a C -algebra, G a countable discrete group acting on A by automorphisms due to a homomorphism α : G Aut(A). The triple (A, G, α) is called a C -dynamical system. A covariant representation of (A, G, α) is a pair (π, u), where π : A B(H) is a -representation of A and u : G U(H) is a unitary representation of G on the same Hilbert space H, such that u g π(a)u g = π(α g (a)) for any a A, g G. Similarly to group rings, let A[G] be the algebra of all finite sums a = g G a gg, where a g A. Multiplication on A[G] is defined using the twisted convolution. If b = g G b gg then ab = g,h G a ggb h h = g,h G a g(gb h g 1 )gh = g,h G a gα g (b h )gh = γ G ( g G a gα g (b g 1 γ ))γ. Also a = g G (a gg) = g G α 1 g (a g)g 1. A covariant representation of (A, G, α) yields a -representation of A[G], and, conversely, a -representation of A[G], when restricted on A and on G (if A is unital), gives a covariant representation of (A, G, α).

62 Crossed product C -algebras Let π : A B(H) be a -representation of A, and let λ be the left regular representation of G. Form the Hilbert space l 2 (G; H) of all H-valued square-summable functions x on G with the norm x 2 2 = g G x(g) 2 and define a covariant representation ( π, u) of (A, G, α) on this Hilbert space by ( π(a)x)(g) = π(αg 1 (a))(x(g)); (u g x)(h) = x(g 1 h), g, h G. Define a norm on A[G] by a = sup σ σ(a), where the supremum is taken over all covariant representations of the C -dynamical system (A, G, α). This is correct, as a g G a g <. The (full) crossed product C -algebra A α G is defined as the completion of A[G] with respect to this norm.

63 Example. C α G = C (G). Example. If α is trivial then C(X) α G = C(X; C (G)). In particular, if X = S, G = Z then C(S) α Z = C(T 2 ), where T 2 is a two-dimensional torus. Irrational rotation algebras Let the action α of Z on C(S) be defined by α n (f )(z) = f (e 2πiθn z), n Z, where θ [0, 1] is irrational. (This action is induced by the action by irrational rotation on the circle.) Then A α Z = A θ is called an irrational rotation algebra, or a noncommutative torus. It can be described in terms of generating elements and relations as follows: Let u C(S) denote the identity function u(z) = z, z S, and let v denote a generator of Z. They can be considered as unitary elements of C(S) α Z, and they satisfy the relation vu = e 2πiθ uv. Two unitaries with this relation determine A θ as a certain universal C -algebra, like two commuting unitaries determine C(T 2 ). A θ is simple. (Difficult. Cf. the book by K. Davidson.)

64 By a projection in a C -algebra A we always mean a selfadjoint idempotent (p = p = p 2 ). Two projections, p, q A, are Murray von Neumann equivalent (p MN q) if there exists v A such that p = v v, q = vv ; unitarily equivalent (p u q) if there exists a unitary u A + such that p = u qu; homotopy equivalent (p h q) if there exists a continuous path of projections that connects p and q. There may be several other notions of equivalence, but they usually reduce to these. For example: Lemma. If q = zpz 1 for some z A + then p u q. Proof. It follows from zp = qz and z q = pz that pz z = z qz = z zp, and as p commutes with z z, so it commutes with (z z) 1/2. Set u = z(z z) 1/2. Then u is unitary and upu = z(z z) 1/2 p(z z) 1/2 z = zp(z z) 1 z = qz(z z) 1 z = q.

65 Lemma. Let v v = p and vv = q be projections. Then v = vv v and v = v vv. Proof. v vv v 2 = v(1 p) 2 = (1 p)v v(1 p) = (1 p)p(1 p) = 0. Lemma. If p MN q and 1 p MN 1 q then p u q. Proof. By assumption, there exist v and w such that v v = p, vv = q, w w = 1 p, ww = 1 q. Note that v w = v q(1 q)w = 0, similarly, w v = 0, hence (v + w) (v + w) = v v + w w = p + 1 p = 1. Similarly, (v + w)(v + w) = 1. Thus u = v + w is unitary, and upu = (v + w)v v(v + w) = vv vv = q 2 = q. Let v be the right shift on l 2. Then v v = 1, vv = 1 e 1, where e 1 is the projection onto the first coordinate. This shows that the Murray von Neumann equivalence doesn t imply unitary equivalence.

66 Lemma. Let p A be a projection, and let V = {q A : q is a projection and p q < 1}. There exists a continuous map q u q from V to the unitary group of A + such that u p = 1 and u q pu q = q. Proof. Set v p = 2p 1, v q = 2q 1, z q = v q v p + 1. Then qz q = q(2q 1)(2p 1)+q = 2qp = (2q 1)(2p 1)p+p = z q p. It remains to check invertibility of z q, which follows from the estimate zq 2 1 = 1 2 v qv p + 1 = 1 2 v q(v p v q ) 1 2 v p v q = p q < 1. Corollary. If p q < 1 then p h q. Proof. As u p = 1, so u q lies in the path connected component of 1 in the unitary group, and there is a path u(t) connecting 1 with u q. Then p(t) = u(t)pu(t) gives a required path of projections.

67 Lemma. If p h q then p u q. Proof. Divide the homotopy path connecting p with q by intermediate points p = p 0, p 1, p 2,..., p n = q such that p i+1 p i < 1. All projections p i are unitarily equivalent. Lemma. (1) If p MN q then ) (2) If p u q then ( p ( p h ( q ) ). u ( q ( Proof. (1) If p = v v, q = vv, then set u = v 1 q 1 p v ). An easy calculation shows that u is unitary in M 2 (A) and that u ( p ) u = ( q ). (2) Let q = upu with u A + unitary. The unitary group of a C -algebra need not to be path-connected, but elements of the form ( ) u 0 0 u lie in the connected component of 1M2 (A + ) = ( ) The required path of unitaries is given by w t = ( ) ( u 0 cos t sin t ) ( 1 0 ) ( cos t sin t 0 1 sin t cos t 0 u sin t cos t ). ). Then wt ( p the path connecting the projections. ) w t is

68 Idea: consider projections not in A, but in matrices of arbitrary size over A. Advantages: All equivalences for projections become the same; One can take direct sums of projections. Set M (A) = n N M n (A), and let V (A) denote the set of equivalence classes of projections )] in M (A). Addition [p] + [q] := is well-defined and makes V (A) [( p 0 0 q an abelian semigroup with the trivial element [0]. Example. If A is C, M n or K then the equivalence class of a projection in M k (A) is determined by its rank, hence V (A) = {0, 1, 2...}. Example. If A = B(H) with H infinitedimensional then, besides projections of finite rank, there is one more equivalence class consisting of infinite rank projections, so V (A) = {0, 1, 2,..., }. Example. If A = C(X) then V (A) can be identified with isomorphism classes of locally trivial vector bundles over X.

69 Stable equivalence: Two projections, p, q M (A) are stably equivalent if there exists a projection r such that [p + r] = [q + r]. Consider all pairs ([p], [q]), [p], [q] V (A). Two such pairs, ([p], [q]) and ([p ], [q ]) are equivalent if p + q and p + q are stably equivalent. The set of equivalence classes of such pairs becomes an abelian group with ([p], [q]) + ([p ] + [q ]) = ([p] + [p ], [q] + [q ]), ([p], [q]) = ([q], [p]), and with the neutral element ([p], [p]). If a C -algebra A is unital then define K 0 (A) as the group of equivalence classes of pairs of projections. If ϕ : A B is a -homomorphism of C -algebras then it extends to a -homomorphism ϕ : M n (A) M n (B), and ϕ(p) is a projection in M n (B), hence K 0 is a covariant functor (even for non-unital -homomorphisms). We write ϕ : K 0 (A) K 0 (B).

70 K 0 (C) = K 0 (M n ) = Z; K 0 (B(H)) = 0. Non-unital C -algebras may have no non-trivial projections at all (cf. C 0 (R 2 )). If A is not unital then there is a quotient map θ : A + C with the kernel A. Set K 0 (A) = Ker θ. If B is non-unital, ϕ : A B a -homomorphism, then the ϕ composition K 0 (A) K 0 (B + θ ) K 0 (C) is zero map, hence ϕ (K 0 (A)) K 0 (B), and K 0 is a functor for non-unital C -algebras as well.

71 Continuity of K 0 Lemma. Let A 1 A 2 A with n N A n dense in A. Then K 0 (A) = lim K 0 (A n ). Idea of proof. If p is a projection in A (or in M k (A + )) then, for any ε > 0 there exists some selfadjoint x in A n (or in M k (A + )) for some k, such that p x < ε. This x is not a projection, but x x 2 < 3ε, hence Sp x consists of two distinct pieces one close to 0, and another one close to 1. Applying a function f which equals 0 on (, 1/2) and 1 on (1/2, ) (which is continuous on Sp x), we obtain a projection f (x), which is also close to p (up to 4ε). As this procedure is continuous, it can be applied to homotopies of projections as well.