saharon shelah and juris steprans- masas in the calkin algebra without the continuum hypothesis

8/3/2019 Saharon Shelah and Juris Steprans- Masas in the Calkin Algebra without the Continuum Hypothesis

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MASAS IN THE CALKIN ALGEBRA WITHOUT THE CONTINUUM

HYPOTHESIS

SAHARON SHELAH AND JURIS STEPRANS

Abstract. Methods for constructing masas in the Calkin algebra without assuming the ContinuumHypothesis are developed.

1. Introduction

In [1] Anderson began a study of the extension property of C-subalgebras that he continued in [2].

A C

-subalgebra A of a C

-algebra B is said to have the extension property if every pure state on A hasa unique extension to a state on B. In his concluding remarks in [3] Anderson expresses the followingview, In order to make further progress on the extension problem for atomic masas in B(H) it appearsthat a clearer understanding of the structure of the masas in the Calkin algebra would be useful. Asa test question, Anderson asks whether every masa in the Calkin algebra which is generated by itsprojections lifts to a masa in the algebra of all bounded operators on separable Hilbert space. He laterprovides a negative answer to this question assuming the Continuum Hypothesis [3].

The starting point for the present investigation is a potential argument producing the same negativeconclusion as Andersons in [3] but not relying on the Continuum Hypothesis. It will be seen thatthis argument runs into a serious difficulty but, even if it did work, the argument would not producemasas that can be tested for the sort of properties Anderson had in mind. For example, another test

question he asks in [1] is whether every masa in the Calkin algebra that is generated by its projectionsis permutable. This concept will not be explored further here, but it serves to illustrate that the lack ofa classification of masas in the Calkin algebra should not be misunderstood to mean that the structureof these objects can not be further investigated.

Before continuing, some notation will be established. Let H denote a fixed separable Hilbert spacewith inner product x, y and let {en}n be a fixed basis for H. For x H define supp(x) ={i | x, ei = 0}. Define S(H) = {x H | x = 1}. For X N define H(X) to be the sub-space ofH generated by {ei | i X} and define PX to be the orthogonal projection onto H(X). Whenthinking ofH as 2 then the ei will be identified with characteristic functions of singletons. Moreover,PX can be identified with multiplication by the characteristic function of X, an element of

.

Let B(H

) denote the algebra of bounded operators onH

and let C(H

) be the compact operators.For any orthonormal family X H define D(X) to be the subalgebra ofB(H) diagonal with respectto X; in other words, T D(X) if and only if there is a bounded function D : X C such thatthat T(z) =

xXx, zD(x)x. Let C be the Calkin algebra B(H)/C(H) and let : B(H) C be the

quotient map. A masa in a C-algebra is a maximal, abelian, self-adjoint subalgebra.There are two important facts about masas in C relevant to the present investigation. The first is the

following result due to Johnson and Parrott [7]:

Theorem 1.1. IfA B(H) is a masa then (A) is a masa inC.

Research of the first author is supported by the United States-Israel Binational Science Foundation (Grant no. 2002323).

Research of the second author is supported by NSERC of Canada and the Fields Institute. This is paper 931 on Shelahslist.

1


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2 S. SHELAH AND J. STEPRANS

The second is that the masas in B(H) can be characterized as those algebras of the form L() actingon L2() where is a regular probability measure. The following result1 states this more precisely.

Theorem 1.2. If A is a masa inB(H) then there is a compact subset X R and a regular Borelprobability measure on X such thatA is -isomorphic to L() acting on L2() and, moreover, the

-isomorphism is a homeomorphism with respect to the weak operator topology.

The two key cases are provided when is an atomic measure on a countable set or when it is anatomless measure on a set without isolated points. Since L() is not separable, this result on its owndoes not guarantee that the number of masas in B(H) is not greater than 20 . However, given a Borelprobability measure on X, the -isomorphism from L() to B(H) is determined2 by it values onC(X) which is separable. This yields the following.Corollary 1.1. There are no more than 20 masas inB(H).

To begin, it is worth noting why Anderson is concerned in [3] only with masas in C that are generatedby their projections. Let S be the uni-lateral shift defined by S(en) = en+1. Then (S) is normal in other words, (S)(S) = (S)(S) and so there is some masa in

Ccontaining it. On the other

hand, SS = SS so this algebra is not the quotient of any masa in B(H). However, by virtue ofcontaining (S) this algebra will not be generated by its projections. So Anderson showed, assumingthe Continuum Hypothesis, that there is a masa in C which is generated by its projections but is notthe quotient of any masa in B(H).

There is a simple strategy for constructing a masa in C which is not the quotient of any masa inB(H). (See [?] for a description of an argument of Akemann and Weaver using this.) Let A be a familyof almost disjoint subsets ofN of size 20 . For each A A choose a pair of projections Q0A and Q1A suchthat

PAQ0APA = Q0A and PAQ1APA = Q1A

Q0AQ

1A

Q1AQ

0A is not compact.

Note that for any function F : A {0, 1} the familyA(F) =

QF(A)A

| A A

is commuting family of projections. Let Ac(F) be the C-algebra generated by A(F). It is immediatethat the algebras Ac(F) and Ac(F) are distinct if F and F are. If it can be shown that each Ac(F)can be extended to a masa this will complete the proof since it follows that there are 2 2

0 distinct masascontradicting Corollary 1.1 if each of them lifts to a masa in B(H).

Each Ac(F) is clearly abelian, self-adjoint and generated by projections, but extending to a maximalsuch family poses a problem. To see this the following simple fact will be useful: IfA is an abelian C-subalgebra of

Cgenerated by projections thenA is maximal abelian if and only if for every self-adjoint

operator S C \ A there is some projection Q A such that SQ = QS. To see this, let T C \ Abe arbitrary and suppose that T commutes with every element ofA. Let T = A + iB where A and Bare self -adjoint. Each element ofA must be normal since A is an abelian C algebra. Therefore theFuglede Lemma [6] implies that T also commutes with every element ofA and, hence, so do both Aand B. Hence both A and B belong to A and, therefore, so does T.

Therefore, in order to extend Ac(F) to a masa it suffices to add to Ac(F) all self-adjoint T Cwhich commute with each member of Ac(F). The catch is that in order for the extended family to begenerated by projections it is necessary to also add to Ac(F) some projections generating T. Beforepresenting the following example showing that this might not be possible the following definition isneeded.

1See, for example, pages 48 and 53 of [5].2See page 49 of [5].


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MASAS IN THE CALKIN ALGEBRA 3

Definition 1.1. The cardinal p is defined to be the least cardinal of a family of Fof subsets ofN suchthat

A B Ffor any A and B in F there is no infinite X N such that X\ A is finite for all A F.

Theorem 1.3 (Bell [?]). If c = p then Martins Axiom for -centred partial orders holds; in otherwords, given

a partial orderP such thatP = n=1 Pn where eachPn is centred (that is, any finite subfamilyhas a lower bound)

and a family {D} such that < c and each D is a dense subset ofP (that is,for all p Pthere is d D below p)

there is a centred G P such that G D = for each .Example 1.1. Ifp = c then there is a self-adjoint operator A B(H) and an abelian subalgebra C ofC such that

(A) commutes with each member ofC

C is generated by projections ifA ki=1 diPi + K < 1/4 where K is compact and each Pi is a projection and each di R

then there is some j k such that (Pj) does not commute with C ifQ is a projection not commuting with A modulo a compact operator then there is C C such

that (Q) and C do not commute.

In other words, ifC is extended to masa then this will not be generated by projections.

Proof. Let A be any self-adjoint operator such that ess(A), the essential spectrum of A, is [0, 1]. Thefamily C = {(C)}c will be constructed by an induction of length c. Let

ki=1

i

Pi

c and even

enumerate all finite linear combinations of pairwise orthogonal projections in C such that

A ki=1

iPi K 0 both of the sets(t , t + ) ess(API ) and (t , t + ) ess(APJ ) have non-empty interior.

To see this, note first thatki=1

ess(APi) [1/3, 1]


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because otherwise there is an open U [ 13

, 1] such the corresponding spectral projection P has the

property that AP > 13

and PiP is compact for each i k. This contradicts that

A k

i=1 iPi K q then C(zpn) < 2n.

It is immediate from Condition 2 in its definition that P is -centred and that the set of all p such that Bp is dense for any . It will be shown that the set of all p such that p m is also dense foreach m. Given this, let G

Pmeet all these dense sets and define C to be the orthogonal projectiononto the subspaces spanned by {zpn | p G and n p}. It follows from Condition 3 that C and PJ

do not commute modulo a compact set. From Condition 4 it follows that ess(C) = {t}. The definitionof guarantees that C commutes with each C modulo a compact if .

In order to establish the required density assertion it suffices to show that for any p there is someq p such that q = p+1. To this end let p be given. Using a slightly modified version of 4.31 from [?]it is possible to find an orthonormal basis {zn}n such that there are compact operators K, {K}Bpand {Ki}ki=1 as well as functions {X}Bp and {Xi}ki=1 from to {0, 1} and a function : (0, 1)such that

A(zn) = (n)zn + K(zn)

C(zn

) = X(n) + K

(zn

) for

Bp Pi(zn) = Xi(n) + Ki(zn) for 1 i k.

Let = 2p1 and let M be such that K(zn) < , Ki(zn) < and K(zn) < for n > M and

i k and Bp. Using the fact that ess(API) (t , t + ) and ess(APJ) (t , t + ) both havenon-empty interior together with the fact that ess(AC) is nowhere dense for each Bp it followsthat there are I and J greater than M such that

PI (zI) = zI and PJ (zJ) = zJ PI (zJ) = 0 and PJ (zI) = 0 C(ZI) = C(ZJ) = 0 for Bp |(I) t| < and |(J) t| < .

let z = (zI + zJ)/2 and note that 2/3 < PJ (z) < 3/4. Also A(z) tz < 2. Moreover,C(z) = K(z) < 2 for Bp. Hence, it is easy to define z to satisfy Condition 2 and yet be so


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close to z that all these inequalities are still satisfied with z in place of z. Let

p =

Bp, {zpi }p+1i=1

so that z

p+1 = z.Now assume that is odd. It may be assumed that A = and acts on 2 by multiplication

where : N (0, 1) has its range dense in (0, 1). IfQC CQ is not compact for some then thereis nothing to do, so assume that QCCQ is compact for all . Since QQ is not compacta pigeonhole argument produces a < b, > 0 and a sequence {n}n=1 such that

n 2 for each n nP1n (b,1) > nP1n (0,a) > the supports of the n are pairwise disjoint finite sets limn Q(n) = 1

where projections are being identified with characteristic functions thought of as elements of . Let Fbe a free ultrafilter on N and note that for each there is {0, 1} such that limFC(n) = .Using that p = c there is a set X N such that limnX C(n) = for each .Now let {In}n enumerate all intervals with rational endpoints in (0, 1). Then construct Jn, Ain, knand Zn such that:

Jn is a rational interval such that Jn In Zn N is infinite and Zn+1 Zn Z0 = X Ain 1n (0, a) and Ain+1 Ain for i Zn+1 iPAin >

1 nj=0 2j2 for i Zn

kn Zn the image ofAjn under j is disjoint from Jn for every j Zn including kj if j = i then A

kjj is disjoint from the support of ki.

If this can be done, then letting C be the orthogonal projection onto the subspace of 2 spanned by

{kjPAkjj }jit is immediate that CQ QC is not compact. Also immediate is the fact that ess(C) is disjointfrom

j Jj and hence is nowhere dense. Since Z0 = X it follows that C commutes with each C for

.To carry out the induction suppose that Zn is given. Choosing kn to satisfy the last clause is then

easy. Let In+1 =

Li=1 I

i be a partition of In+1 into intervals of length less than 2n3. For one of

these intervals it must be that there is an infinite set Zn+1 and some k

L such that

iPAinP1{In+1\Ik} >

1 n+1j=0

2j2

for every i Zn+1. Then let Ain+1 = Ain 1(In+1 \ Ik) for i Zn+1 and let Jn+1 = Ik. Question 1.1. Is the hypothesis p = c necessary for the example? In other words, is it consistent withset theory that every abelian, self adjoint subalgebra of the Calkin algebra generated by projections canbe extended to a masa generated by projections.

It must be noted at this point that a successful implementation of the strategy outlined before thepreceding example would not signal any progress towards gaining the clearer understanding of the

structure of the masas in the Calkin algebra seen to be useful by Anderson. Since the Axiom of Choiceis invoked to obtain maximality, very little can be said about the structure of the masa produced this


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way. For example, one can still ask whether there is a masa in the Calkin algebra which is not locallythe quotient of a masa in B(H). The following definition makes this precise:

Definition 1.2. A masa A C will be said to be locally the quotient of a masa in B(H) if thereis a non-trivial projection p A, a Hilbert subspace H0 H and a masa A B(H0) such that(A) =

{pap

|a

A}.

It is not possible to say whether or not a masa produced by invoking Zorns Lemma is locally thequotient of a masa in B(H). The next section will provide a method for constructing masas with somecontrol over their structure. In particular, it will be immediate that these masas are locally the quotientof a masa in B(H).

Before continuing, it is worth remarking that Andersons construction of masa in [3] can easily bemodified to produce a masa which is not locally the quotient of a masa in B(H). The idea Andersonexploited is that for any algebra of the form L() there is an operator which commutes with onlycountably many projections in L(). With the assistance of the Continuum Hypothesis, Andersonconstructed a masa in C which contains an uncountable set of projections which he called almost central,namely every element of

Ccommutes with all but countably many of them. Clearly this can not be the

quotient of any L(). It is routine to modify the transfinite construction used by Anderson to ensurethat not only does his masa contain a central family P, but also that for any non-trivial projectionq C the family {qpq | p P} is central. This guarantees that Andersons masa is not locally thequotient of a masa in B(H).

Question 1.2. Is there a masa in the Calkin algebra which is not locally the quotient of a masa inB(H)?

Question 1.3. Is there a masa in the Calkin algebra such that for any non-trivial projection q C thefamily {qpq | p P} is central?

Of course, it has already been noted that a positive answer to Question 1.3 will yield a positive answerto Question 2.1 and, assuming the Continuum Hypothesis, the answer to both is positive. It would alsobe interesting to know whether the two questions are, in fact, equivalent.

2. Masas from almost disjoint families

Definition 2.1. IfIis an ideal on N then define an ideal I on the family of finite, non-empty subsetsofN to be generated by

h [N] 0 and all bounded operators there is some ksuch that

w W P[k,m)(w) < is I-large for all m k.

Theorem 2.1. If there is a strongly separable almost disjoint family then there is a masa in the Calkinalgebra which does not lift to a masa ofB(H).

Proof. IfA is strongly separable then it is possible to choose {wAn }n=0 H(A) such that(1) supp(wAn ) is a finite subset of A for each n

(2) max(supp(wAn )) < min(supp(w

An+1)) for each n

(3) for each W H which is I(A)-large there is some A A such that wAn W for each n


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Now, for each A A normalize each wAn and extend to an orthonormal basis BA,n on the space Csupp(wAn )and let BA =

n=0 BAn . Let A0 be the subalgebra of B(H) generated by

AA

D(BA)

let A1 be the quotient A0/C(H) and let A be the quotient norm closure ofA1. The almost disjointnessofA guarantees that A is abelian as well as self adjoint.

To see that A is maximal abelian let B(H) and suppose that commutes modulo a compactoperator with every member ofA0 but () is not in the quotient norm closure ofA1. Let m = P

mP

m

for any m N. For any m N and > 0 letB(, m) =

x S(H) P

supp(x)m(x) >

and let

Z(, m) = {x S(H) | m(x) m(x), xx > }Claim 2. For every > 0 there is some k = k() such that B(, k) is

I(A

)-small.

Proof. If the conclusion fails for then for any w S(H) define m(w) = min(supp(w)). Then defineB =

w S(H) P{w}P[m(w),)(w) >

and observe that B is I(A)-large. To see this let A A and let k be the least element of N notbelonging to A. Since B(, k + 1) is I(A)-large it is possible to find w B(, k + 1) such that

supp(w) (A {k}) = and then it follows that w + ek B.

Then there is A A such that wAn B for each n. Choose an increasing sequence {yn}n=0 from Nsuch that if mn = m(w

Ayn) and wn = w

Ayn then

PwnP[mn,mn+1)(wn) > /2and supp(wj) [mn, mn+1) = for each n and j = n. Then let Q be the projection onto the spacespanned by {wn}n=0. It then follows that

P[mn,mn+1)(Q Q)(wn) = P[mn,mn+1)(wn) P[mn,mn+1)(wn), wnwn =P[mn,mn+1)(wn) (wn), wnwn =

P{wn}P[mn,mn+1)(wn) + P{wn}(wn) (wn, wnwn = P{wn}P[mn,mn+1)(wn) > /2

contradicting the compactness of Q Q. Claim 3. For every > 0 the set Z(, k(/2)) is I(A)-small.Proof. If the conclusion fails let > 0 witness this. It then follows that W = Z(, k(/2))\B(/2, k(/2))is also I(A)-large. Choose A A such that wAn W for each n and let Q be projection onto the spacespanned by {wAn }n=0 Note that it follows that if n > k(/2) then

Psupp(wAn )(Q Q)(wAn ) = (wAn ), wAn wAn Psupp(wAn )(wAn )

= (wAn ), wAn wAn (wAn ) + (wAn ) Psupp(wAn )(wAn )

j

(wAn

), wAn

wAn

(wAn

)

Psupp(w

An )

(wAn

)

> /2

and this contradicts the compactness of Q Q.


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Note that if k k(/2) then Z(, k) is also I(A)-small. Let B(z) denote the open disk of radius and centre z in the complex plane. Let > and for B(0) and > 0 and m N let

X(,,m,) = {w S(H) \ Z(, m) | |w, m(w) | < }and let

C(, m) =

B(0) | ( > 0)X(,,m,) is I(A)-largeand note that C(, m) is closed. The first thing to notice is that for each k k() the set C(, k) isnot empty. To see this, note that otherwise it is possible to choose for each B(0) a > 0 suchthat X(, , m , ) is I(A)-large. Compactness then yields a finite F B(0) such that

FB()

B(0). Then choose w S(H) \ Z(, m) \FX(, , m , ) yielding a contradiction.

Now choose n C(n, m) such that limn n = and limn n = 0. It will be shown that isunique. To this end suppose that n C(n, m) are chosen such that limn n = and limn n = 0and = . Let > 0 be such that 10 < | |/22 and let M be so large that M k(/2). DefineW to be the set of all w S(H) such that:

w = x0 + x1 such that supp(x0)

supp(x1) =

x0 = x1 = 1/2 x0, M(x0)x0 M(x0) x1, M(x1)x1 M(x1) x0, M(x0) x1, M(x1) P

supp(w)M(w) < .It will first be shown that W isI(A)-large. To see this let A A. Let N be such that |N| < /2 and|N| < /2. Then choose x0 X(,/2, M , N)\(B(/2, M)Z(, M)) such that supp(x0)A = and then choose x1 X(,/2, M , N)\(B(/2, M)Z(, M)) such that supp(x1)(Asupp(x0)) = .Letting w = (x0 + x1)/

2 it is immediate that w

W.

Now let A A be such that each wAn belongs to W and let Q be the projection onto the spacespanned by {wAn }n>M and let wAn = x0n + x1n be the decomposition witnessing that wAn belongs to W.Then for any n > M

Psupp(wAn )(QQ)(wAn ) Psupp(wAn )(Q(Psupp(wAn )(M(wAn ))))Psupp(wAn )(M(x0n)+ M(x1n)) =Q(Psupp(wAn )(M(x0n) + M(x1n))) Psupp(wAn )(M(x0n) + M(x1n))

Q(Psupp(wAn )(M(x0n), x0nx0n+M(x1n), x1nx1n))Psupp(wAn )(M(x0n), x0nx0n+M(x1n), x1nx1n)5 Q(M(x0n), x0nx0n + M(x1n), x1nx1n) M(x0n), x0nx0n M(x1n), x1nx1n 5

Q(Nx

0n +

Nx

1n)

Nx

0n

Nx

1n

9

|N N|

22 9

> 0

and this contradicts the compactness ofQQ. Since each of the C(, m) is compact, the uniquenessof implies that for each > 0 the diametre of C(, m) approaches 0 as m increases to infinity.

It will be shown that for any > 0 there is k N some Q A0 such that k Q + I < . Tosee this, choose k > j(/2) such that C(/2, k) B/3(). For each B+1(0) \ B/3() choose > 0 such that X(/2, , k , ) is I(A)-small. Let E B+1(0) \ B/3 be a finite set such thatEB() B+1(0) \ B/3. For each E choose A I(A) witnessing that X(/2, , k , ) is

I(A)-small. Let A I(A) witness that Z(/3, k) I(A) and let C = A EA.Then for any x H such that supp(x) C = it must be that x, kx k(x) /2 since

x A = . Moreover, |x, k(x) | for each E and since |x, k(x)| it follows that

|x,

k(x)

|

< /2. In other words,

(2.1) kPC(x) PCx <


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for all x. Since PC D(BC) A0 it follows that PCkPC commutes with all members ofD(BC) moduloa compact set. By Theorem 1.1 it follows that (PCkPC) A0. Furthermore, PCkPC and PCkPCare both compact because PC commutes with k modulo a compact operator. Therefore

(2.2) (kI)(x) = (PC+PC)(kI)(PC+PC)(x) = PCkPC(x)+PCkPC(x)+PC(x)+PC(x)

and combining this with Inequality 2.1 yields that

(2.3) (k I)(x) (PCkPC(x) + PC(x)) PCkPC(x) + PC(x) < Since PCkPC+ PC A0 this yields the desired conclusion.

The final thing to show is that A is not the lifting of a masa on Hilbert space. So suppose that : H 2 L2([0, 1]) is an isometry and define A to be the algebra consisting of all 1(Df Mg)where f and g L correspond to multiplication operators. Let = 01 such that 0(x) 2and 1(x) L2([0, 1]). Assume that A = A in order to derive a contradictionProposition 2.1. If {fn} are in L2([0, 1]) and fn > 1 for all n then there is > 0 and a measurableset S

[0, 1] such that there are infinitely many n such that S |fn|

2 > and [0,1]\S |fn|2 > .

Claim 4. If > 0 and U() = {x H | x = 1 and 1(x) > } then U() is not I(A)-large.Proof. If U() is I(A)-large then choose A A such that wAn U() for each n. In other words,1(wAn ) > . Use Proposition 2.1 to find Y N, an infinite set, and S [0, 1] and > 0 such thatS|fn|2 > and

[0,1]\S

|fn|2 > for each n Y. Let M be the projection on L2([0, 1]) defined bymultiplying by the characteristic functions of S and let M = 11 M. Let Y Y be an infinite setsuch that for n Y

kY\{n}

|wAk , M(wAn )|2 < /2

and let Q be projection onto the space spanned by {wAn }nY.

It follows that

(QM M Q)(wAn ) M(wAn ), wAn wAn M(wAn ) /2 =M(wAn ), (wAn )(wAn ) M(wAn ) /2 M1(wAn ), 1(wAn )1(wAn ) M1(wAn ) /2 =

S

|1(wAn )|2

1(wAn ) M1(wAn ) /2

S

|1(wAn )|2

[0,1]\S

|1(wAn )|2

> /2 > 0

In order to conclude that M does not commute, modulo a compact operator, with Q it suffices toshow that for any finite dimensional subspace V L2([0, 1]) the projections onto V of the sequence{M1(w

An ), w

An w

AnM1(w

An )}

n=0 converge to 0. But, since the 1(w

An ) are pairwise orthogonal, their

projections onto V clearly converge to 0. Therefore it suffices to shows that limn PV(M1(wAn )) =

0 where PV is the projection onto V. But it may as well be assumed that V is generated by thecharacteristic functions of a partition R of [0, 1] into finitely many measurable sets. Moreover, there isno harm in refining this partition so that if R R then either R S or R S = . But then

PV(M1(wAn )) =

rR

RM1(wAn )1(wAn ), R =

rR

RS1(wAn )

SR

1(wAn )

and each summand RS1(wAn )SR

1(wAn ) is either equal to 0, if R S = , or it is equal to

R1(wAn ) R 1(w

An ) otherwise. Once again the pairwise orthogonality of the 1(w

An ) implies that

limn RS1(wA

n )SR1(w

A

n ) = 0. Hence limn PV(M1(wA

n )) = 0.However, the commutator of Q and M must be compact because M A = A.


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Now let V() = {n N | (j)0(en) ej2 < }. It will be shown then V() is not I(A)-large forany > 0. If this fails for some > 0 then the complement ofV() is I(A)-large. Let be much smallerthan and let A I(A) witness that U() for is not I(A)-large. It follows that there is some X Asuch that X A is finite and 0(wXn ) ej > for each n and j. Moreover, 0(wXn )2 > 1 2 forall but finitely many n.

It is then possible to choose for all but finitely many n a finite set s(n) such that(2.4) /2 < Ps(n)0(wXn ) < 1 /2.In order to see this, let wXn =

i=o iei. Then first observe that |j| < 1 /2 for each j because

1 i=j

|i|2 + |j|2 i=j

|i|2 + (1 /2)2

and hencei=j |i|2 < (/2)2 and therefore

0(wXn ) ej2 =i=j

|i|2 + (1 j)2 <

contradicting that wXn does not belong to V(). If there is somej such that |j| > /2 then let s(n) = {j}.Otherwise, assume that the support of wXn is disjoint from A and hence 1(wXn ) < . If and aresufficiently small then

i |i|2 > /2. Then let s(n) be a minimal set such that

is(n) |i|2 > /2.

Then, letting j s(n) be arbitraryis(n)

|i|2 =

is(n)\{j}

|i|2 + |j|2 < /2 + (/2)2 < 1 /2

yielding Inequality 2.4.There is then an infinite Y N such that for each n Y such that

m=nis(m) |ei, 0(wXn )|2 < 3

m=n |10 (Ps(n)0(wXn )), wXm|2 < 3for each n. Let : 2 2 be the projection onto the space spanned by {wXn }nY and let C =

nY s(n).

It will be shown that 1PC and do not commute modulo compact.To see this note that

1PC(wXn ) =

10 PC0(w

Xn ) + w1 =

10 Ps(n)0(w

Xn ) + w2 + w1

where w1 < 3 and w2 < 3. Hence 1PC(wXn ) = Ps(n)0(wXn ) 23. On the other hand,1PC(w

Xn ) = (

10 Ps(n)0(w

Xn ) + w2 + w1) = (wXn , 10 Ps(n)0(wXn )wXn ) + w3 + (w1 + w2)

= 0(wXn ), Ps(n)0(wXn )wXn + w3 + (w1 + w2)

where w3 < 3

. So1PC(wXn ) < |0(wXn ), Ps(n)0(wXn )| + 33 = Ps(n)0(wXn )2 + 33

and hence

1PC(wXn ) 1PC(wXn ) 1PC(wXn ) 1PC(wXn ) Ps(n)0(wXn ) 23 (Ps(n)0(wXn )2 + 33)

Since /2 < Ps(n)0(wXn ) < 1 /2, it follows that1PC(wXn ) 1PC(wXn ) > /2 2/4 53 > 0

provided 0 < < 1/4. This contradicts the compactness of the commutator.

Therefore {x + x | supp(x) supp(x) = and x V() and x V()} is also I(A)-large. LetA A be such that for each n there are xn and xn in V() such that


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wAn = (x + x)/

2 supp(x) supp(x) = neither x nor x belongs to U().

Let m(n) and m(n) be the integers satisfying that 0(x) em(n) < and 0(x) em(n) < . Let be projection onto the space spanned by all the wAn and let =

1PM where M =

{m(n)

}nN. It

is routine to check that and do not commute modulo a compact provided that has been chosensufficiently small. Since A this shows that A = A.

It is not known whether it is possible to construct an almost disjoint family which is strongly separablewithout assuming some extra set theoretic axioms. However, there are many models of set theory knownin which there is such a family. Certainly assuming that a = c suffices. This provides an easy way tosee that some axiom like the Continuum Hypothesis is necessary for Andersons construction in [3] if itis to rely on almost central families.

Corollary 2.1. It is consistent that there are no almost central families yet there is a masa inC whichis not the quotient of any masa inB(H).

Proof. Assume that the union of 1 meagre sets never covers the reals and that there is an almostdisjoint family which is strongly separable. (Martins Axiom and 20 > 1 will do.) There is thena masa in C which is not the quotient of one in B(H). Moreover, the space S of all orthonormalsequences is a closed subspace ofH with the product topology. Given any projection P B(H) suchthat (P) = 0 and k the set D(k, P) of all S such that there is some n > k such that thedistance from (n) to the range of P is greater than 1/2 is co-meagre. Hence, given any family Pofprojections of cardinality 1 the intersection

k

PP

D(k, p)

is not empty. Moreover any orthonormal sequence in this intersection yields a projection not commuting

modulo a compact with any projection in P. Hence there are no almost central families in this model. Question 2.1. Is there an almost disjoint family A of subsets ofN such that for every H I(A)+there is at least one X A such that for any n N there is h H such that h X\ n?Question 2.2. If there is an almost disjoint family such as in Question 2.1 does it follow that there isa strongly separable one?

Petr Simon has constructed [4] an almost disjoint family A such that if X I(A I(A))+ thenthere are 20 sets A A such that A X is infinite.Question 2.3. Does there exist an almost disjoint family A such that if X I(A I(A))+ thenthere are 20 sets A

Asuch that A

X is infinite? Does the existence of such a family allow the

construction of a masa in C which is not the quotient of one in B(H)?References

[1] Joel Anderson. A maximal abelian subalgebra of the Calkin algebra with the extension property. Math. Scand.,42(1):101110, 1978.

[2] Joel Anderson. Extensions, restrictions, and representations of states on C-algebras. Trans. Amer. Math. Soc.,249(2):303329, 1979.

[3] Joel Anderson. Pathology in the Calkin algebra. J. Operator Theory, 2(2):159167, 1979.[4] Bohuslav Balcar and Petr Simon. Disjoint refinement. In Handbook of Boolean algebras, Vol. 2, pages 333388. North-

Holland, Amsterdam, 1989.[5] Kenneth R. Davidson. C-algebras by example, volume 6 of Fields Institute Monographs. American Mathematical

Society, Providence, RI, 1996.

[6] Paul Richard Halmos. A Hilbert space problem book, volume 19 of Graduate Texts in Mathematics. Springer-Verlag,New York, second edition, 1982. Encyclopedia of Mathematics and its Applications, 17.


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[7] B. E. Johnson and S. K. Parrott. Operators commuting with a von Neumann algebra modulo the set of compactoperators. J. Functional Analysis, 11:3961, 1972.

Department of Mathematics, Rutgers University, Hill Center, Piscataway, New Jersey, U.S.A.

08854-8019

Current address: Institute of Mathematics, Hebrew University, Givat Ram, Jerusalem 91904, Israel

E-mail address: [email protected]

Department of Mathematics, York University, 4700 Keele Street, Toronto, Ontario, Canada M3J

1P3

E-mail address: [email protected]

saharon shelah and juris steprans- masas in the calkin algebra without the continuum hypothesis

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