Extensions of pure states

Transcription

1 Extensions of pure M. Anoussis 07/ 2016

2 1 C algebras

3 C -algebras Definition Let A be a Banach algebra. An involution on A is a map a a on A s.t. (a + b) = a + b (λa) = λa, λ C a = a (ab) = b a

4 Definition A C -algebra is a Banach algebra with an involution wich satisfies a a = a 2.

5 Examples C(X), for X compact. g = sup x X g(x), g(x) = g(x) B(H), for H Hilbert space T = sup x H, x 1 Tx Tx, y = x, T y

6 Theorem Let A be a C -algebra. Then A is isometrically isomorphic to a closed subalgebra of B(H) for some Hilbert space H.

7 C algebras Definition Let A be a C -algebra. An element a A is selfadjoint if a = a. Definition Let A be a C -algebra. An element a A is positive if it selfadjoint and σ(a) R +.

8 Theorem Let A be a C -algebra and a A. The following are equivalent: a is positive a = b 2 for some positive b A a = b b for some b A If A B(H), ax, x 0, x H

9 Definition Let A be a C -algebra. A linear form on A is positive if f(a a) 0 a A. Lemma Let f be a positive linear form on A. Then 1 f(b a) = f(a b) 2 f(b a) 2 f(a a)f(b b) 3 f(a ) = f(a) 4 f(a) 2 f(e)f(a a)

10 Lemma Let f be a positive linear form on A. Then f is bounded and f = f(e). proof If not take a positive element a n such that, a n 1 and f(a n ) 2 n. Then f( a n 2 n ) 1. Set We have a 2 and a = n=0 a n 2 n.

11 N 1 N f( which is a contradiction. From Lemma n=0 a n 2 ) f( n n=0 a n 2 n ) f(a), f(a) 2 f(e)f(a a) f(e) f a a = f(e) f a 2 f(a) 2 a 2 f(e) f f f(e)

12 Lemma Let f be a bounded linear form on A. Then f is positive iff f = f(e). proof We show that if f = f(e), then f is positive. Replacing f by f/f(e) we may assume f = f(e) = 1. We show first that if a is selfadjoint, then f(a) R. Let a A, a = a with a 1. Set f(a) = γ + iδ. We show that δ = 0. We may assume that δ 0. We have f(a ine) 2 (a ine) 2 = (a + ine)(a ine) = a 2 + n 2 e 1 + n 2

13 We also have f(a ine) 2 = γ + δi in 2 = γ 2 + (δ n) 2 and hence If δ 0, γ 2 + (δ n) n 2 2δn 1 γ 2 δ 2. which is absurde. n 1 γ2 δ 2 2δ

14 Now, let a A, a 0 and a 1. Then e a is selfadjoint and e a 1. Hence, f(e a) 1, and 1 f(a) 1 f(a) 0.

15 Definition Let A be a C -algebra. A state is a linear form on A which is positive and satisfies f(e) = 1.

16 Theorem (GNS) Let f be a state on A. There exists a representation ρ f of A on a Hilbert space H and ξ H, such that f(a) = ρ f (a)ξ, ξ.

17 The set of S(A) of a C -algebra A is a w -compact set of the dual of A. It is convex, hence by the Krein-Milman theorem it has extreme points. Definition A state is pure if it is an extreme point of S(A).

18 Examples C(X), for X compact. A state on C(X) is a probability measure. A pure state is a Dirac measure. B(H) for a Hilbert space H. If ξ H, f(a) = aξ, ξ is a state. These are called vector.

19 Theorem Let A, B be C -algebras such that A B and e A. Let f be a state on A. Then there exists a state g on B which extends f. proof Let g be a Hahn-Banach extension of f. Then g = f = f(e) = g(e) and hence g is a state.

20 Let A, B be C -algebras such that A B and e A. Let f be a state on A and E(f) be the set of of B that extend f. Theorem If f is pure, the extreme points of E(f) are pure on B. proof Let g be an extreme point of E(f). If it is not a pure state of B, there are g 1, g 2 and α (0, 1) such that g = αg 1 + (1 α)g 2. We restrict to A and obtain f = αg 1 A + (1 α)g 2 A. Since f is pure, g 1 A and g 2 A are proportional to f, hence g 1, g 2 belong to E(f).

21 Corollary If f is a pure state on A, then it has an extension which is a pure state on B.

22 Examples Let D be the C -algebra of 2 2 diagonal complex matrices. A linear form on D is of the form (( )) a 0 f = xa + yd 0 d for some x, y C. f is a state if and only if x + y = 1 and xa + yd 0 when a 0 and d 0. That is x 0 and y 0.

23 Examples f is pure iff x = 0 or y = 0. Indeed if f = (1, 0) and f 1 = (x 1, y 1 ), f 2 = (x 2, y 2 ), α (0, 1) are such that f = αf 1 + (1 α)f 2 we obtain (( )) a 0 f = (αx 1 + (1 α)x 2 )a + (αy 1 + (1 α)y 2 )d = a. 0 d It follows that αy 1 + (1 α)y 2 = 0 which implies that y 1 = y 2 = 0 and f 1 = f 2 = f.

24 Examples If f = (x, y), with x 0, y 0, then f = xf 1 + yf 2 where f 1 = (1, 0) and f 2 = (0, 1).

25 Examples Let A be the C -algebra of 2 2 complex matrices. A linear form on A is of the form (( )) a b f = xa + yb + zc + wd c d for some ( x, y, z, ) w C, which is equal to tr(ga) where x z G =. y w

26 Examples f is a state if and only if tr G = 1 and G is hermitian positive semidefinite. (A hermitian positive semidefinite v Av 0, v C 2.) Indeed if G is hermitian positive semidefinite, then f(b B) = tr(gb B) = tr(bgb ). But BGB is hermitian positive semidefinite and hence has positive trace. Hence f(b B) 0. Conversely if f is positive, then f(vv ) 0, v C 2, which implies v Gv 0.

27 Examples Consider f on D defined by f = (1, 0). We show that f has a unique extension to A. Let g be an extension of f to A. Then g is determined by a hermitian positive semidefinite matrix G. We have )) (( )) x y g (( a 0 0 d = tr z w ) ( a 0 0 d = a xa + wd = a for every a and d. Hence x = 1 and w = 0. Now, G is positive semidefinite, and has determinant 0. Thus yz 0 y = z = 0, since y = z. Hence f has a unique extension.

28 Examples Consider f on D defined by f = (1, 1)/2. Then f is a state and is not pure. Let g be the linear form on A defined by (( )) a b g = (a + b + c + d)/2. c d g is determined by the matrix 1 2 ( which is hermitan positive semidefinite with trace 1, hence it is a state. )

29 Examples Let h be the linear form on A defined by (( )) a b h = (a + d)/2. c d Then h is determined by the matrix 1 2 ( which is hermitian positive semidefinite with trace 1, hence it is a state. )

30 Examples g and h both extend f.

31 C algebras Definition Let X be a set. A filter F on X is a collection of subsets of X s.t. X F / F A F, B F A B F A F, A B B F

32 Examples Let X be a topological space and x 0 X. The collection of all neighbourhoods of x 0 is a filter. Let X be a set and x 0 X. Then F = {A X, x 0 A} is a filter. X = N. Set F = {A N : N A is finite}. Then F is a filter.

33 Definition Let X be a set. An ultrafilter F on X is a filter which is maximal. Example Let X be a set and x 0 X. Then F = {A X, x 0 A} is an ultrafilter. These are called principal.

34 Definition The set of all of N is denoted by βn. remark Let F = {A N : N A is finite}. There is not principal ultrafilter that contains F. Hence βn N. There exists a topology on βn such that βn is a compact Hausdorff space N (i.e. the principal ) is dense in βn. The space βn is the Stone-Čech compactification of N.

35 Definition Let (a n ) be a sequence in l and U an ultrafilter in βn. We say that x is a U-limit of (a n ) if for every neighbourhood S of x, the set {n N : a n S} is in U. Proposition Let (a n ) be a sequence in l and U an ultrafilter in βn. Then there exists an x which is the U-limit of (a n ). We denote x = lim U a n.

36 Proposition Let (a n ), (b n ) be sequences in l and U an ultrafilter in βn. Then lim U (a n + b n ) = lim U a n + lim U b n lim U (λa n ) = λ lim U a n for λ C. lim U (a n ) = (lim U (a n )) lim U (a n b n ) = lim U a n lim U b n

37 Theorem The spaces l and C(βN) are isometrically isomorphic. proof If a = (a n ) l, define g a : βn C by g a (U) = lim U a n If g C(βN), define a g l : a g = (a g n ), where a g n = g(u n )

38 Let D be the C -algebra of diagonal operators on l 2. Theorem The C -algebra D is isometrically isomorphic to C(βN). We have the following: Definition Let U be an ultrafilter in βn. Define f U : l C by f U (a n ) = lim U a n

39 Theorem The pure of D are the linear functionals of the form f U, for U βn. a = (a n ) g a g a (U) = f U (a n )

40 The Kadison-Singer problem Let D be the C -algebra of diagonal operators on l 2. Let f be a pure state on D. The Kadison-Singer problem is the following: Problem Does f has a unique extension on B(l 2 )?

41 Let E be the map that sends an operator to its diagonal part. E(A)e n, e n = Ae n, e n E(A)e n, e m = 0, if n m. The map E : B(l 2 ) D is bounded, positive, E 2 = E and E(I) = I. Let f be a state on D. Define: f : B(l2 ) C, by f(a) = f(e(a)). Then f is a state and extends f. So, the question becomes: Problem Let f be a pure state on D. Is f the only extension of f?

42 Definition A diagonal projection is an orthogonal projection on l 2 which lies in D. Note that P is a diagonal projection iff there exists a subset S of N such that P is the orthogonal projection on the subspace of l 2 spanned by {e n : n S}. Example If S = {n = 2k : k N}, then the matrix of P is a diagonal matrix (p nn ) s.t. p nn = 1 if n = 2k and p nn = 0 if n = 2k + 1.

43 conjecture (P1) Let A B(l 2 ) with 0 diagonal and ɛ > 0. There exist r N and r pairwise orthogonal diagonal projections P 1, P 2,..., P r such that r i=1 P i = 1 and r P i AP i ɛ A conjecture (P1 ) i=1 Let A B(l 2 ) and ɛ > 0. There exist r N and r pairwise orthogonal diagonal projections P 1, P 2,..., P r such that r i=1 P i = 1 and r P i AP i E(A) ɛ A i=1

44 conjecture (P2) Let ɛ > 0. There exists r N such that for every A B(l 2 ) with E(A) = 0 there exist r pairwise orthogonal diagonal projections P 1, P 2,..., P r such that r i=1 P i = 1 and r P i AP i ɛ A i=1

45 Theorem P1 P2 proof P2 implies P1. We show P1 P2. Assume not. Then there exists ɛ > 0 such that for every r N, there exists A r B(l 2 ), with E(A r ) = 0 such that if P 1, P 2,..., P r are r pairwise orthogonal diagonal projections with r i=1 P i = 1 then r P i A r P i > ɛ A r. i=1 Rescaling the operators A r, we may assume that A r = 1, r N. Set A = A r. Then A B(l 2 ) and E(A) = 0.

46 It follows from the assumption that there exists s N, and s pairwise orthogonal diagonal projections P 1, P 2,..., P s, s.t. s i=1 P i = 1 and In particular which is absurde. s P i AP i ɛ A. i=1 s P i A s P i ɛ A ɛ A s, i=1

47 Theorem KS is true iff every extension g of a pure state f on D satisfies E(A) = 0 g(a) = 0. proof Let f be a pure state on D and f its canonical extension. Assume g is another extension. We have E(A E(A)) = 0, hence g(a E(A)) = 0. Since g(a E(A)) = g(a) g(e(a)) = g(a) f(e(a)) = g(a) f(a), we obtain g(a) = f(a).

48 Theorem P1 is equivalent to KS. proof We show P1 KS. Let f be a pure state on D and g an extension of f. Let A B(l 2 ) with E(A) = 0. We have to show that g(a) = 0. Let ɛ > 0 and P 1, P 2,..., P r be pairwise orthogonal diagonal projections such that r i=1 P i = 1 and r P i AP i ɛ A. i=1

49 We have g(p i X) g(p i )g(x X) for every X B(l 2 ). Similarly g(xp j ) g(x X)g(P j ) for every X B(l 2 ). Since f is multiplicative, we have: f(p i ) = 0 or f(p i ) = 1 for all i. Since f is a state there exists i 0 s.t. f(p i0 ) = 1 and f(p i ) = 0 for all i i 0. It follows that g(a) = g( r i=1 r P i AP j ) = g(p i0 AP i0 ) ɛ A. j=1

50 conjecture (P3) Let ɛ > 0. There exists r N such that for every n N and every A B(l n 2 ) with E(A) = 0, there exist r pairwise orthogonal diagonal projections such that r i=1 P i = 1 and r P i AP i ɛ A i=1

51 Theorem P3 P2 proof Let ɛ > 0. Let A B(l 2 ) with E(A) = 0. Set Q n the orthogonal projection onto the subspace spanned by the vectors e 1, e 2,..., e n. Then Q n AQ n B(l n 2 ) and E(Q naq n ) = 0. There exist r N such that for every n N there are r pairwise orthogonal diagonal projections P1 n, Pn 2,..., Pn r s.t. r i=1 Pn i = I and r i=1 P n i Q n AQ n P n i ɛ Q n AQ n ɛ A

52 The space {P D : P projection} is stronly compact. By a diagonal argument we find an increasing sequence (n k ) and diagonal projections P 1, P 2,..., P r s.t. P nk i P i for every i = 1, 2,..., r. The projections P i are pairwise orthogonal and r i=1 P i = I. Then and hence r i=1 P nk i Q nk AQ nk P nk i sot r P i AP i i=1 r P i AP i ɛ A i=1

53 C algebras A nest N is a totally ordered family of closed subspaces of a Hilbert space H containing {0} and H, which is closed under intersection and closed span. If N a nest on H, then the nest algebra AlgN is the algebra of all operators T such that T(N) N for all N N. AlgN = {A B(H) : PAP = AP, for all P N }

54 Let {e n } n=1 be an orthonormal basis of H, and set P n = [e k : 1 k n]. We consider the nest N = {P n : n N} {H} {{0}}. Then AlgN consists of the operators of the form:

55

56 Let S be the set of strictly upper triangular matrices, and D the set of diagonal matrices. Proposition The map J J D is a bijection between ideals of AlgN containing S and ideals of D, with inverse I I + S. Proposition Let I be a maximal ideal of D. Then I + S is a a maximal ideal of AlgN.

57 question Is every maximal ideal of AlgN of this form? Theorem Yes proof Let J be a maximal ideal which does not contain S. Let X S, X / J. Then the ideal generated by J and X is AlgN. Let I be the identity operator. We have k I = B + C i XD i i=1 for some B J, k N and C i, D i AlgN. Then Y = k i=1 C ixd i S.

58 Let ɛ > 0. There exist r N and r pairwise orthogonal diagonal projections P 1, P 2,..., P r such that r i=1 P i = 1 and r P i YP i ɛ Y i=1 r r P i (I B)P i = I P i BP i ɛ Y. i=1 i=1 It follows that I J, which is a contradiction.