Linear combination of projections in von Neumann factors

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Contents lists available atScienceDirect

Journal of Mathematical Analysis and Applications

www.elsevier.com/locate/jmaa

Linear combination of projections in von Neumann factors

Stanisław Goldstein^∗, Adam Paszkiewicz^∗

FacultyofMathematicsandComputerScience,UniversityofŁódź,Banacha22,90-238Łódź,Poland

a r t i cl e i n f o a b s t r a c t

Articlehistory:

Received31January2019 Availableonline9April2020 SubmittedbyD.Blecher

Keywords:

Projection Factor

Linearcombination

Itisshownthatanyself-adjointoperatorinaﬁnitediscreteorinﬁnitevonNeumann factor canbewritten as areallinear combinationof 4projections. Ontheother hand,inanytypeII₁algebraandinanytypeII_∞factorthereexistsaself-adjoint operatorthatisnotalinearcombinationof3projections.

1. Introduction

Weproveherefourresults,twopositive andtwonegative:

(1) Any self-adjoint operatorA acting ona ﬁnite-dimensional(complex) Hilbertspace canbe writtenas alinearcombination of 4 projections, with two of thecoeﬃcients chosen arbitrarily from theinterval [2A,∞[.

(2) Anyself-adjoint operatorA in aninﬁnitevon Neumannfactorcanbe writtenas alinearcombination of4 projections,withtwo ofthecoeﬃcientschosenarbitrarilyfromtheinterval]2A,∞[.

(3) Thereisaself-adjointoperatorinanytypeII₁ algebrathatcannotbewrittenasalinearcombination of3 projections.

(4) Thereisaself-adjointoperatorinanytypeII_∞ factorthatcannotbewrittenas alinearcombination of3 projections.

Anyself-adjointoperatoractingina(complex)finite-dimensionalHilbertspacecanbewrittenasalinear combinationofafinitenumberofprojections—this followseasilyfromthespectraltheorem.Whatisnot immediately obvious is whether we canmake the number of projectionsindependent of the dimension of theHilbertspace. IsthisalsopossibleiftheHilbertspaceisinfinitedimensional?Thefirstpositive results

* Correspondingauthors.

E-mailaddresses:stanislaw.goldstein@wmii.uni.lodz.pl(S. Goldstein),adam.paszkiewicz@wmii.uni.lodz.pl(A. Paszkiewicz).

https://doi.org/10.1016/j.jmaa.2020.124135

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in this directionwere obtainedby Fillmore1967 (see [3]), who wasable to get downto 9 projections [4].

Pearcy and Topping[16] reducedit to 8 alreadyin1967, thenthe second-namedauthor to 6 in1980 (see [15]).Using theideasfrom[15], Matsumotoshowedin1984that5 isenough.Alsoin1984Nakamura [12]

provedthatanyHermitian matrixisalinearcombinationof4 projections.

In 1985 the second-named author proved that any self-adjoint operator in B(H) can be written as a linear combination of4 projections, both forfinite and infinite-dimensional HilbertspaceH, with aproof for HermitianmatricesdifferingfromthatofNakamura[12].Thepaper [14] withtheresultswassenttoa journal and gotapositive review, requiringonly minorimprovements, butforsomeinexplicable reasonit hasnotbeenresentto thejournal.Thisfactwentunnoticedbythefirst-named authorwhowascompiling thejointpublicationonlinearcombinationsofprojectionsinvonNeumannalgebras[5] whileonleavefrom his alma mater, and thepaper wasadvertised as acceptedfor publication.Evenworse,some resultsfrom thejointpublication dependontheunpublishedpaper, includingthecriterionforavon Neumannalgebra to beacomplexlinearspanofitsprojections.

In 2016,the ﬁrst-namedauthor met ViacheslavRabanovichat theconference “Groupsand Operators”

in Gothenburg, Sweden. There he learned that [14] has never been published, and that Rabanovich [17]

himself proved that4 projections are suﬃcient in the caseof B(H) with aseparable inﬁnite-dimensional HilbertspaceH.

We decided thatthere are good reasonsfor complementingour earlier paper[5] with this one,dealing with linearcombinations of projectionsin vonNeumann factors. BothNakamura’s [12] andRabanovich’s [17] proofsareelegant, andtheyshowclearlythat4projectionsareenoughforself-adjointoperatorsinH both inﬁnite- andinﬁnite-dimensionalcases.However,thereareseveral reasonsforourpresentation:

(1) Theorem1.1(a)of[5] attributedto[14],andhencedidnotprove,adecompositionofself-adjointoper- atorsinatypeIn factoras linearcombinationsof4projections,withaspeciﬁcformofthecoeﬃcients.

Thatform isnotavailablefrom Nakamura’s[12] but isrequiredtoobtaininTheorem2.1(a)of[5] the decompositionofself-adjointoperatorsinaﬁnitediscretevonNeumannalgebraasalinearcombination of4projectionswith coeﬃcientsinthecenterof thealgebra.Aproof basedontheunpublishedpaper [14] is presentedinTheorem 2.1below.

(2) Theorem 1.1 (a) of [5] attributed to [14] a decomposition of self-adjoint operators inan inﬁnite von Neumannfactorasalinearcombinationof4projections.TheproofpresentedbyRabanovich[17] holds for aseparable Hilbertspace and cannot be easily adaptedto thecase of anarbitrary inﬁnitefactor.

Anindependentproofbasedon[14] ispresentedinTheorem3.5.

(3) Given Theorem2.1andTheorem 3.5,alltheresultsfrom [5] are nowproved.Theresultshavealready been usedbyseveralauthorsthroughout theyears.

Theinterestinthisﬁeldofresearchcontinues(see,forexample,[7]),withmoreattentiondirectedrecently to asituationinC^∗-algebras (see[10]).Nevertheless,asseenbelow,there aresomeopenquestionseven in thevon Neumannalgebracase(see‘Openproblems’attheend ofthepaper).

ForavonNeumannalgebraA wedenotebyAhitsself-adjointpartandbyProjA thelatticeofprojections in A. For a projection E, we denote by AE the reduced von Neumann algebra EAE acting in EH. For A∈ Ah,eA(·) denotesthespectralmeasureofA.TheunitofA isdenotedby1_Aorsimply1,ifthereisno danger of confusion.For projectionsP,Q∈ A, we write P ∼ Q ifthere is apartial isometry U ∈ A such thatP = U^∗U andQ= U U^∗,and thenP Q ifP ∼ R ≤ Q foraprojectionR∈ A, andP ≺ Q ifP Q and P Q.

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2. Finitediscretefactors

LetA befactorof typeI_n,n<∞,andτ the tracialstateonA.Weassume(forconvenience)thatA is representedonaHilbertspaceH insuchawaythatA= B(H).

Theorem2.1. [1.1(a)in[5]] LetA∈ Ahandα = τ (A).Foranyβ,γ≥ 2A,thereareprojectionsP,Q,R,S inA withP ∼ Q andR∼ S suchthat

A = (β + α)P − βQ + (γ + α)R − γS (1)

whenα≥ 0,and

A = βP − (β − α)Q + γR − (γ − α)S (2)

whenα≤ 0.

Proof. (2) followsfrom (1) appliedto−A.Weshow (1).IfA= α1, takeP = Q= 1,R = S = 0. Consider n ≥ 2 and A = α1.Assume thatA = 1, otherwise use (1/A)A. Put A = A˜ − α1. Then A˜ = 0. Let {e1,. . . ,en} bean orthonormal basisin H consistingof eigenvectors of A (andA),˜ and let α1,. . . ,αn be thecorrespondingeigenvaluesofA.˜ Weorder theeigenvalues(andthecorresponding eigenvectors)insuch awaythatforeachj,j = 1,. . . ,n, β_j := α₁+· · · + αj∈ [0,2].ByLemma1in[12] (orLemma3.4below), forany β,γ≥ 2,there existrankoneprojectionsP_j,Q_j inA withP_j,Q_j≤ Ej+ E_j+1,when E_j isarank 1 projectionontothesubspacegeneratedbyej,suchthat

βj(Ej− Ej+1) + αEj =

(β + α)Pj− βQj for j odd, (γ + α)Pj− γQj for j even.

Wehave(notingthatβn= 0 andαn =−βn−1)thatfor1≤ j ≤ n− 1

A = [β1(E1− E2) + αE1] +· · · + [βn−1(En−1− En) + αEn−1] + αEn

= (β + α)(

j odd

Pj)− β(

j odd

Qj) + (γ + α)(

j even

Pj)− γ(

j even

Qj) + αEn.

Ifn isodd,putP =

j oddPj+ En,Q=

j oddQj+ En,R =

j evenPj,S =

j evenQj;ifn iseven, addEn to thesumsovereven indices.(Forn= 2,putR = S = E2.)

3. Inﬁnitefactors

Thefollowinglemmabelongsto mathematicalfolklore.Wegivehereitsproofforcompleteness.

Lemma 3.1. Let A be a von Neumann factor, and let P,Q,R ∈ Proj A with P + Q = R, and R inﬁnite.

ThenR P orR Q.

Proof. Let E,F ∈ Proj A besuch thatE + F = R,E ∼ F ∼ R (see Lemma6.3.3in[8]).Then P∧ E Q∧ F or P ∧ E Q∧ F . AssumeP ∧ E Q∧ F ; then, using Kaplansky’s Parallelogram Law, we get R∼ E = P ∧ E + (E − P ∧ E) Q∧ F + (E ∨ P − P )≤ Q,soR Q.

TheproofinthecasewhenP∧ E Q∧ F isobtainedbyexchangingtherolesofP andQ intheproof above.

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Thenextfourlemmas areessential.

Lemma 3.2. Let A bean inﬁnite factor,and let A∈ Ah.There exists a∈ [−A,A] such that eA

]a− ε,a+ ε[

∼ 1 foreach ε> 0.

Proof. Assume there is no such a. Then, for each x ∈ [−A,A], there is εx > 0 such that eA

]x− ε_x,x+ ε_x[

≺ 1.Fromcompactnessof[−A,A],thereisaﬁnitenumberofintervals]x_n− εxn,x_n+ ε_x_n[ coveringtheinterval[−A,A].Hence,thereisaﬁnitenumberofdisjointintervalsIn,eachcontainedin ]x_n−εxn,x_n+ε_x_n[,withunion[−A,A],sothat1=

p_nforp_n= e_A(I_n)≤ eA

]x_n−εxn,x_n+ε_x_n[

≺ 1, whichcontradicts Lemma3.1.

Notethata satisﬁestheconditionsoftheLemmaifandonlyifitbelongstotheessentialspectrumofA with respecttothelargest closedidealofA (theonegeneratedbyalltheprojectionsnotequivalent to1).

Lemma 3.3. LetA bean inﬁnite factor and letA∈ Ah.Assumethat for each δ > 0,e_A(]− δ,δ[)∼ 1.For any ε> 0, there are:a sequence(εi),i∈ Z of positivenumbers satisfying ε0 =A and

i=0εi ≤ ε, and a sequence(Ei),i∈ Z ofprojections fromA such that Ei ∼ 1 andEiA= AEi foralli,

i∈ZEi = 1 and

−εiE_i≤ AEi≤ εiE_i fori∈ Z.

Proof. Considerthefollowing cases:

I eA({0}) ∼ 1. Choose Ei,i ∈ Z\{0},E₀ so that E₀ ∼ Ei ∼ 1 and

i=0Ei + E₀ = eA({0}). Put E0= E₀ + eA(R\{0}),ε0=A andεi= 0 for i∈ Z\{0}.

II e_A {0}

1. First observe that we can replace the set of indices Z by any other countable set.

Deﬁne, for a sequence (εn),n = 0,1,. . . with ε0 = A,ε1 = ε/2 and εn+1 ≤ εn/2 for n ≥ 2, sets I_n = [−εn,−εn+1[∪]εn+1,ε_n] andD_n = e_A(I_n). Since_∞

n=0D_n ∼ 1 (byLemma3.1),we canchooseε_n so thatD_n = 0 foreachn,byassumptionandagainLemma3.1.There areagaintwocases:

(a) Forinﬁnitelymanyn,sayn∈ M ⊆ N,Dn∼ 1.ThenwecanusethemforformingEi,withE0obtained as 1−

i=0Ei.

(b) IfD_n ∼ 1 forﬁnitelymany(or none)ofindices n,wecanassumethatD_n 1 for eachn = 0.

Now,if1 isσ-finite,wearenecessarilyinasemifinitefactor(allnon-zeroprojectionsinaσ-finitetype III factor are equivalent to 1), and we can use a faithful normal semifinite trace τ on A. We have

_∞

n=1τ (D_n) = τ e_A

]− ε1,ε₁[\{0}

= τ (1) = +∞, by assumption and Lemma 3.1. Now we can find infinitely many disjoint infinite sets Mi ⊆ N such that

n∈Miτ (Dn) = +∞. Thus we cantake Ei=

n∈MiDn,i = 0,with(asbefore)E0= 1−

i=0Ei.

If1 isnotσ-finite,wemayassumethatallprojectionsD_n(n = 0) areinfinite(and≺ 1).Foraprojection F ∈ A,denote by#F thelargest cardinalnumberofamaximalorthogonalfamilyofnon-zeroσ-finite projectionsmajorizedbyF (thiscorresponds directlytothegeneralizeddimensionfunctionofTomiyama inthefactorialcase(see[19]).ByLemma3.1,e_A

]− ε1,ε₁[\{0}

∼ 1.SinceA isafactor,wehave,by Tomiyama[19,Theorem 4anditscorollary] (see alsoBlackadar[1,III.1.7.1])

∞ n=1

#D_n= #

∞ n=1

D_n= #e_A

]− ε1, ε₁[\{0}

= #1.

This means thatthere isno cardinalκ< λ= #1 such that#D_n ≤ κ for n= 1,2,. . . . Hence,λ is a limitcardinalof countablecoﬁnality(see, forexample,[11]),andwecanform asubsequence(D_m_n) of

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(D_n) suchthat#D_m_n  λ.Weusethesubsequenceto build countablymanysets M_i ⊆ N suchthat

n∈Mi#D_m_n= #1 foreachκ.Weﬁnishtheproof asintheσ-ﬁnite case.

Lemma3.4.LetA beavonNeumannalgebra,E,F ∈ Proj A,E⊥ F ,E = U^∗U ,F = U U^∗forsomeU ∈ A (so thatE ∼ F ).Forany α,β ≥ 0 and D∈ A+ satisfying D≤ βE, thereexistP,Q∈ Proj A with P ∼ Q suchthat

(α + β)P− βQ = αE + D − UDU^∗ andthat P,Q≤ E + F .

Proof. If β = 0,thenD = 0 andonecantakeP = Q= E.Hence,we canassumethatβ > 0.Weuse the standardmaterialontherelativeposition oftwoprojectionsinavon Neumannalgebra(see Halmos[6] or Takesaki[18,pp.306–308]).WeconstructP,Q∈ Proj A insuchawaythatP,Q≤ E + F andP ∼ E ∼ Q (inAE+F).Morespeciﬁcally,we representtheoperators as2× 2 matricesoverAE,sothat

A₁₁ A₁₂ A21 A22

= A₁₁+ U A₂₂U^∗+ A₁₂U^∗+ U A₂₁

forAij ∈ AE (notethat1_A_E = E).Then

E =

1 0 0 0

, P =

C² CS CS S²

and Q =

C² CS CS S²

forsuitablychosenC,S,C,S with0≤ C,S≤ 1,0≤ C,S ≤ 1 andC²+ S²= 1,C²+ S² = 1.

Deﬁne ﬁrstH ∈ AE byH := (2D + α1)⁻¹(D + α1) for α > 0 and H := (1/2)1_E forα = 0. Checkthat 0≤ (α + β)⁻¹[D + (α + β)1]H ≤ 1 and 0≤ β⁻¹(β1− D)H ≤ 1.Let C andC be thesquare rootsof the two operators,respectively, and letS = (1− C²)¹²,S = (1− C²)¹². Note thatC,S,C andS belong to theabelianvonNeumannalgebrageneratedbyD andE.Wecheckthat

(α + β)P− βQ =

α1 + D 0

0 −D

,

whichendstheproof.

Theorem 3.5.[1.3(a) in [5]] Let A be aninﬁnite factor, andlet A∈ Ah.For any β,γ > 2A,there exist α∈ R with|α|≤ A and projectionsP,Q,R,S∈ A suchthat, depending onα,(1) or(2) ofTheorem2.1 holds.

Proof. ThecaseA= 0 istrivial.WemayassumethatA= 1,otherwiseuse(1/A)A insteadofA. Let α besuch thateA

]α− ε,α + ε[

∼ 1 foranyε> 0.Itisenoughto consider thecaseα≥ 0:ifα < 0, use

−A insteadofA. Put A = A˜ − α1. Takeε> 0 satisfying2(1+ ε)< β,γ.Let(E_i)_i∈Z and(ε_i)_i∈Z beas in Lemma3.3,forA and˜ ε.

Let U_j ∈ A be any partial isometry satisfyingU_j^∗U_j = E_j,U_jU_j^∗ = E_j+1. Put U =

j∈ZU_j andlet h denotethemappingX → UXU^∗ onA.Note thatU isunitaryandh⁻¹ exists andmapsX toU^∗XU .Let Aj = EjAE˜ j and B0 = (1− α + ε)E0. Use therecurrencerelation Aj = Bj − hBj−1 to deﬁne Bj for all j = 0.Then

(−1 − α)E0≤ A0≤ (1 − α)E0;

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forj ∈ Z,j = 0,

−εjEj≤ Aj ≤ εjEj; ﬁnally,

0≤ Bj≤ 2(1 + ε)Ej. ApplyLemma3.4toobtain

αEj+ Bj− hBj =

(β + α)Pj− βQj for j odd, (γ + α)R_j− γSj for j even.

Then put

P =

j odd

Pj, Q =

j odd

Qj, R =

j even

Rj, S =

j even

Sj.

Note that

A = ˜A + α1 =

j∈Z

(A_j+ αE_j) =

j∈Z

(B_j− hBj+ αE_j)

to endtheproof. 4. TypeII1algebras

LetA beaσ-finitetypeII₁algebraandletτ₀beafixedfaithfulnormaltracialstateonA.Inthissection τ denotesafixed tracethatisascalarmultipleofτ₀.

WewillbeconstructinganoperatorA∈ A+ withthepropertythatwhateverα,β ∈ R+ andγ∈ R,and whatever projectionsP,Q,R∈ A,

A + γR = αP + βQ. (3)

Weshallﬁrstdescribethepropertiesoftheright-handsideof(3),introducingconvenientterminology.

Deﬁnition4.1.Adistributionfunction isanon-decreasing,left-continuousfunctionF : R→ [0,∞[ suchthat lim_t→−∞F (t) = 0.Foradistribution functionF we denote by F⁺ itsright-continuousvariant: F⁺(t):=

lim_stF (s),andbyF^(δ)andF^(δ)+,foranyδ∈ R,thefunctionsF^(δ)(t):= F (t)+F⁺(δ−t) andF^(δ)+(t):=

F⁺(t)+ F (δ− t),respectively.

Forany A∈ Ah,thefunctionFA givenbyFA(t):= τ (eA(]− ∞,t[) isadistributionfunction,calledthe distributionfunctionofA w.r.t.thetraceτ .Notethatlimt→+∞FA(t)= τ (1) andF_A⁺(t)= τ (eA(]− ∞,t])).

Deﬁnition 4.2.A distribution functionF iscalled (α,β)-symmetric (with 0≤ α ≤ β)if it is constanton eachoftheintervals]− ∞,0],]α,β],]α + β,∞[ and F^(α+β)isconstanton]0,α]∪]β,α + β].

F issymmetric ifitis(α,β)-symmetric forsome0≤ α ≤ β.

It is clearthatthenotionsof symmetryjust deﬁneddonotdepend onthechoiceof thetraceτ (being alwaysa scalarmultipleofτ₀).Similarly,ifthedistributionfunctionofanoperatorissymmetric,thenthe distributionfunctionofapositivemultipleofthisoperatorissymmetricaswell.

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Remark 4.3.If the distribution function F_A (w.r.t. τ ) of an operator A ∈ Ah is (α,β)-symmetric, then A∈ A+ and

F_A(0) = 0 (4)

F_A⁺(α + β) = τ (1) (5)

F_A⁺(α) = FA(β), (6)

and

F_A^(α+β)takes on at most three diﬀerent values. (7) Infact,positivityofA and(4) followfromthedistributionF_Abeingconstanton]− ∞,0[.Then(5) and (6) followfrom F_A being constanton]α + β,∞[ and]α,β],respectively.Thedeﬁnitionof(α,β)-symmetry alsoimpliesthatF^(α+β) isconstantandequaltoτ (1) on]− ∞,0]∪]α + β,∞[,andthatF^(α+β)isconstant (andequalto2F_A(β))on]α,β],whichyields (7).

We will now investigateproperties of the distribution function of alinear combination A = αP + βQ of two projections P,Q ∈ A with α,β ≥ 0. Thestructure of such an operator A was described (for real coeﬃcients) byNishio [13]. The symmetry of thespectrum of A follows immediately from Corollary 3of [13],asobservedbyRabanovich,whoaddsafewmorefactsonthesymmetrypropertiesofA insection2of [17].Nevertheless,wehavenotspottedany proof(orevenstatement)ofthemoregeneralfactgivenbelow andconcerningsymmetryof thespectral measureofA, aresultcrucialforthesequel.

Lemma 4.4.Let A be anarbitrary vonNeumann algebra and letP,Q∈ A be twoprojections in ageneric position (i.e. P ∧ Q = P ∧ Q^⊥ = P^⊥ ∧ Q = P^⊥ ∧ Q^⊥ = 0). Then, for any α,β ≥ 0 and any Borel Z ⊆ [0,α + β],

e_{αP +βQ}(Z)∼ eαP +βQ(α + β− Z).

Proof. Weuse2× 2 matrixrepresentationofA over AP (cf. referencesintheproofofLemma3.4).Then

P =

1 0 0 0

and Q =

C² CS CS S²

forsome0≤ C,S≤ 1,C²+ S²= 1.LetV =

0 1

−1 0

. Then

V^∗(αP + βQ)V = αP^⊥+ βQ^⊥= (α + β)1− (αP + βQ), hencethespectralprojections

eαP +βQ(Z) and eαP +βQ(α + β− Z) = e(α+β)1−(αP +βQ)(Z) areunitarilyequivalent.

Corollary 4.5. Let A be an arbitrary von Neumann algebra, and let P,Q∈ Proj A,α,β ∈ R,0 ≤ α ≤ β.

Then

e_{αP +βQ}(]0, α[)∼ eαP +βQ(]β, α + β[) and e_{αP +βQ}(]α, β[) = 0. (8)

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Moreover, if 0< α≤ β,then

e_{αP +βQ}({0}) = P^⊥∧ Q^⊥ and e_{αP +βQ}({α + β}) = P ∧ Q.

If, additionally,α < β,then

eαP +βQ({α}) = P ∧ Q^⊥ and eαP +βQ({β}) = P^⊥∧ Q, while ifα = β,

eαP +βQ({α}) = P ∧ Q^⊥+ P^⊥∧ Q.

(Trivially, for 0 = α < β we have e_{αP +βQ}({0}) = Q^⊥ and e_{αP +βQ}({β}) = Q, and for α = β = 0, e_{αP +βQ}({0})= 1.)

Proof. Let A= αP + βQ.Put(cf. Takesaki[18, pp.306–308)])

P0= P − P ∧ Q − P ∧ Q^⊥, Q0= Q− P ∧ Q − P^⊥∧ Q, (9) and

A0= αP0+ βQ0. (10)

IfP Q= QP ,thenP₀= Q₀= 0.IfP andQ donotcommute,thenP₀and Q₀ arebothnon-zeroand in agenericpositioninAP0∨Q0.Note that

1 = P0∨ Q0+ P∧ Q + P ∧ Q^⊥+ P^⊥∧ Q + P^⊥∧ Q^⊥.

As observed byRabanovichintheﬁrst paragraphof section 2of[17], itfollows from Halmos [6] that the only partsofP andQ that contributeto thepossiblynon-zerovaluesofthespectral measuree_A(·) atthe one-point sets {0},{α},{β} and{α + β} are the ‘commuting’ ones, that is P ∧ Q,P ∧ Q^⊥,P^⊥∧ Q and P^⊥∧Q^⊥.Bytheabove,eA(]0,α[)= eA0(]0,α[), eA(]α,β[)= eA0(]α,β[) andeA(]β,α+β[)= eA0(]β,α+β[).

Note thate_A₀(]α,β[)= 0.This follows,for example,fromtheformula inCorollary 3inNishio [13].Hence also eA(]α,β[)= 0.Therest followsfromLemma4.4.

WereturntoourassumptionthatA isaσ-ﬁnitealgebraoftypeII₁,andτ isascalarmultipleofaﬁnite tracial stateonthealgebra.The followingresult iswell known(see,for example,theproof or Lemma2.5 (iii) in[2]),butwe givehereasimpleproofforcompleteness.

Lemma 4.6.For0≤ B ≤ A and λ∈ R wehave F_A(λ)≤ FB(λ).

Proof. Assume that FA(λ) > FB(λ) for someλ. This means that τ

eB([λ,∞[)

> τ

eA([λ,∞[

, so that there isanon-zeroξ∈

e_B([λ,∞[)∧ eA(]− ∞,λ[)

(H),forwhichAξ,ξ< λξ² andAξ,ξ≥ Bξ,ξ≥ λξ²,acontradiction.

Lemma 4.7. For any P,Q ∈ Proj A in generic position and any 0 ≤ α ≤ β, the distribution function F := F_{αP +βQ} (w.r.t. τ )satisﬁes

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F (0) = 0, (11) F (t) = 1

2τ (1) for t∈]α, β], (12)

F⁺(α + β) = τ (1), (13)

F^(α+β)(t) = τ (1) for t∈ R. (14)

In particular,F is(α,β)-symmetric.

Proof. Since P and Q are in a generic position, also P and Q^⊥ are in a generic position, which implies Q ∼ P ∼ Q^⊥, so thatτ (Q) = (1/2)τ (1).By Lemma4.6, wehave FαP +βQ(0) ≤ F0(0) = 0,which yields (11),andF_{αP +βQ}(β)≤ FβQ(β)= τ (1)− τ(Q)= (1/2)τ (1).ApplyingLemma4.4 withZ =]− ∞,t[ gives

F (t) = τ (eαP +βQ(]− ∞, t[))

= τ (eαP +βQ(]α + β− t, ∞[)) = τ(1) − F⁺(α + β− t) (15) forall t∈ R.Hence(14) and,by(11), F⁺(α + β)= τ (1)− F (0)= τ (1),which shows(13). For(12) note thatby(15) wehaveF⁺(α)= τ (1)− F (β)≥ τ(1)− (1/2)τ(1)= (1/2)τ (1).

Lemma4.8. Forany commuting P,Q∈ Proj A and any 0≤ α ≤ β,thedistributionfunctionF := FαP +βQ

isconstant oneach of theintervals]− ∞,0],]0,α], ]α,β],]β,α + β] and]α + β,∞[. Moreover,

F^(α+β)(t) = 2τ (1)− τ(P ) − τ(Q) for t ∈]0, α]∪]β, α + β]. (16) In particular,F is(α,β)-symmetric.

Proof. Notethat for commutingP and Q, αP + βQ= αP ∧ Q^⊥+ βP^⊥∧ Q+ (α + β)P ∧ Q isa linear combination ofthreemutually orthogonalprojections,whichshowsthatthedistributionfunctionFαP +βQ

hastobe constantoneachoftheintervals]− ∞,0],]0,α],]α,β],]β,α + β] and]α + β,∞[.Inparticular,

τ (P^⊥∧ Q^⊥) =

F (t) for t∈]0, α], F⁺(α + β− t) for t ∈]β, α + β]

and

τ ((P∧ Q)^⊥) =

F (t) for t∈]β, α + β], F⁺(α + β− t) for t ∈]0, α].

Asimplecalculationgivesnow(16).

Theworkdoneso farcanbe summedupinthefollowing:

Theorem 4.9. Let P,Q be projections from A, and let 0 ≤ α ≤ β. Then the distribution function F of A= αP + βQ (w.r.t. τ ) is(α,β)-symmetric,and

F^(α+β)(t) = 2τ (1)− τ(P ) − τ(Q) for t ∈]0, α]∪]β, α + β]. (17)

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Proof. Let usﬁrstextractcommutingandgenericparts ofP andQ.Weput P_c:= P∧ Q + P ∧ Q^⊥, Q_c := P∧ Q + P^⊥∧ Q;

P₀:= P− Pc, Q₀:= Q− Qc.

We consider P₀ and Q₀ in the reduced von Neumann algebra AP0∨Q0. Similarly, we treat P_c and Q_c as elements ofA(P0∨Q0)^⊥.Put F₀:= F_αP₀_+βQ₀ andF_c := F_αP_c_+βQ_c.ByLemmas4.7and4.8,bothF₀andF_c are (α,β)-symmetric,andtheysatisfy,fort∈]0,α]∪]β,α + β],

F₀^(α+β)(t) = τ (P₀∨ Q0) = 2τ (P₀∨ Q0)− τ(P0)− τ(Q0), F_c^(α+β)(t) = 2τ ((P0∨ Q0)^⊥)− τ(Pc)− τ(Qc).

HenceF = F₀+ F_c isalso(α,β)-symmetricandF satisﬁes(17).

Thenextfewlemmas showtheexistenceofapositive operatorA suchthatthedistributionfunctionof A+ γR isnotsymmetric,whateverthechoiceofγ∈ R andprojectionR∈ A.

Lemma 4.10.Choose arbitrary λ > 0 and 0 ≤ Λ < τ (1). If A = B + C,B,C ∈ A+,B < λ and τ (supp C) ≤ Λ, then F_A(λ) ≥ τ(1)− Λ. Consequently, if A = B₁ +· · · + Bm + C₁+· · · + Cn with B_i,C_j ∈ A+,B1+· · · + Bm< λ andτ (supp C₁)+· · · + τ(supp Cn)≤ Λ,thenF_A(λ)≥ τ(1)− Λ.

Proof. Suppose that FA(λ) < τ (1)− Λ. Then τ (eA(] − ∞,λ[)∨ supp C) < (τ (1)− Λ)+ Λ = τ (1), so that e_A([λ,∞[) ∧ (supp C)^⊥ = (e_A(] − ∞,λ[) ∨ supp C)^⊥ = 0, and there is a non-zero vector ξ ∈

e_A([λ,∞[)∧ (supp C)^⊥

(H). Hence Aξ,ξ =Bξ,ξ ≤ Bξ² < λξ², while at the same time

Aξ,ξ≥ λeA([λ,∞[)ξ,ξ= λξ²,acontradiction.

In therestof thissection weassumethat thetrace τ satisﬁesτ (1)≥ 1071.

Remark4.11.Belowyouwillfindafewtechnicallemmasthatleadtotheconstructionofanoperatorthat cannot be written as alinearcombination of three projections. To this aim, webuild in 4.19anoperator A such thatthe distribution function F of A+ γR is not symmetric, whatever the values of γ ∈ R and R∈ Proj A.Thisshows,accordingtoTheorem4.9,thatA+γR isnotalinearcombinationoftwoprojections with positivecoefficients.ThecaseofarbitraryrealcoefficientsisthenobtainedeasilyinTheorem4.20.

Note thattheoperatorA:= B + C + 4E1+ 21E2 constructedinLemma4.19isalinearcombination of 9 projections, withB (constructed inLemma4.15and Corollary4.16)alinearcombinationof 4 mutually orthogonal projections,sayE_B0,. . . ,E_B3, thenC (builtinLemma4.17andCorollary 4.18)of 3 mutually orthogonalprojections,sayEC1,EC2,EC3,andﬁnallywithE1 andE2mutuallyorthogonal.Theconstruc- tionsofB,C andthepairE1,E2areindependentofeachother.Speciﬁcvaluesofτ ontheprojectionswill be usedinprovingCorollary4.22,dealingwithafactoroftypeIn.

Only the values of the trace on the 9 projections (and, of course, the values of the coeﬃcients of the linearcombinations) matterforthevalidityofthelemmas mentionedabove.Thechoiceofallowedvalues, used in Lemmas 4.15and 4.17, wasmade to avoidcomplicated fractions.Theassumption τ (1) ≥ 1701 is needed tomakeroomfortheconstructionoftheseprojections.

Lemma 4.12. LetB,C ∈ A+,B < 1,τ (supp C) < 1, andlet E₁,E₂,R ∈ Proj A withE₁ ⊥ E2. Denote Λi:= τ (Ei),c:= τ (R) andassumethat c> 0,Λ1> Λ2 andthat

A = B + C + λ₁E₁+ λ₂E₂+ γR

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forsome0< λ₁< λ₂< γ.Thenthedistribution functionF ofA satisﬁes:

1. F (t)+ F⁺(γ + λ1− t)≤ 2τ(1)− c− Λ1− Λ2 fort∈]0,λ1];

2. F (t)+ F⁺(γ + λ2− t)≤ 2τ(1)− c− Λ2 fort∈]0,λ2];

3. F (γ)≤ τ(1)− c;

4. F (1)> τ (1)− (c+ 1+ Λ₁+ Λ₂);

5. F (γ + 1)> τ (1)− (1+ Λ1+ Λ2);

6. F (λ1+ 1)> τ (1)− (c+ 1+ Λ2);

7. F (γ + λ₁+ 1)> τ (1)− (1+ Λ₂);

8. F (λ₂+ 1)> τ (1)− (c+ 1);

9. F (γ + λ2+ 1)> τ (1)− 1.

Proof. For3.,useA≥ γR andLemma4.6.For1.and2.,userespectivelyA≥ λ1(E1+ E2)+ γR andA≥ λ₂E₂+γR,Lemma4.6and(17) inTheorem4.9.TheothersresultfromLemma4.10,andrepresentations:4.

A= B+(C +λ1E1+λ2E2+γR);5.A= (B+γR)+(C +λ1E1+λ2E2);6.A= (B+λ1E1)+(C +λ2E2+γR);7.

A= (B+λ1E1+γR)+(C+λ2E2);8.A= (B+λ1E1+λ2E2)+(C+γR);9.A= (B+λ1E1+λ2E2+γR)+C. Corollary4.13.LetB,C∈ A+, B< 1,τ (supp C)< 1,andletE₁,E₂,R∈ Proj A,withE₁⊥ E2,τ (E₁)= 16,τ (E2)= 4 and c:= τ (R)> 0.Let

A = B + C + 4E1+ 21E2+ γR with γ > 21.

ThenthedistributionfunctionF of A satisﬁes:

1. F (t)+ F⁺(γ + 4− t)≤ 2τ(1)− c− 20 for t∈]0,4];in particular F (t)≤ τ(1)− c− 10 or F⁺(γ + 4− t)≤ τ (1)− 10 for anyt∈]0,4];

2. F (t)+ F⁺(γ + 21− t)≤ 2τ(1)− c− 4 fort∈]0,21];inparticularF (t)≤ τ(1)− c− 2 orF⁺(γ + 21− t)≤ τ (1)− 2 for any t∈]0,21];

3. F (γ)≤ τ(1)− c;

4. F (1)> τ (1)− c− 21;

5. F (γ + 1)> τ (1)− 21;

6. F (5)> τ (1)− c− 5;

7. F (γ + 5)> τ (1)− 5;

8. F (22)> τ (1)− c− 1;

9. F (γ + 22)> τ (1)− 1.

Proof. Putλ1= 4,λ2= 21, Λ1= 16,Λ2= 4 andγ > 21 inLemma4.12.

Lemma4.14.If F isadistributionfunctionsatisfying 1.− 9.ofCorollary4.13forγ > 44 andc> 42,then F isnot symmetric.

Proof. AssumeF is symmetric. Weshall consider four cases,oneof whichwould necessarilytakeplace if thefunctionF weresymmetricwithrespect tosome(α,β).

Iβ ∈ [γ,γ + 1[).By4.13.8,

F (β)≥ F (γ) ≥ F (22) > τ(1) − c − 1.

Hence,by4.13.2witht= 11+ γ− β,(5) and(6),either

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F⁺(10)≤ F (11 + γ − β) ≤ τ(1) − c − 2 < F (β) = F⁺(α) or

F⁺(β + 10)≤ τ(1) − 2 < τ(1) = F⁺(β + α).

Thus α≥ 10,andby4.13.6and4.13.7,

F (5) + F⁺(α + β− 5) ≥ F (5) + F (γ + 5) > 2τ(1) − c − 10. (18) Notethatα + β− γ − 2≤ 2β − γ − 2< 2(γ + 1)− γ − 2= γ.Hence,by4.13.1witht= 2,(5) and4.13.3, we haveeither

F (2) + F⁺(α + β− 2) ≤ (τ(1) − c − 10) + τ(1) = 2τ(1) − c − 10 (19) or

F⁺(γ + 2) + F (α + β− γ − 2) ≤ (τ(1) − 10) + (τ(1) − c) = 2τ(1) − c − 10. (20) ByDeﬁnition4.2,thelefthandsidesof(18),(19) and(20) areallequaltothevalueofF^(α+β)on]0,α]∪]β,α+

β],whichyieldsacontradiction.

II β≥ γ + 1. By(6) and4.13.5,F⁺(α)= F (β)> τ (1)− 21> τ (1)− c.By4.13.3,α≥ γ.

By4.13.5,(5) andthe deﬁnitionofsymmetry 4.2,we haveas well21> τ (1)− F (γ + 1)≥ F (α + β)− F⁺(β) = F (α)− F⁺(0); by 4.13.3 and 4.13.5, F (γ + 1)− F⁺(0) ≥ F (γ + 1)− F (γ) ≥ 21. This implies α < γ + 1.

Sinceα≥ γ,wehaveF (α + β− 22)≥ F (2γ − 22)≥ F (γ + 22)> τ (1)− 1,by4.13.9.Thisimplies,again bysymmetryofF and4.13.8,F (2)+ F⁺(α + β−2)= F (22)+ F⁺(α + β−22)> (τ (1)−c−1)+ (τ (1)−1)= 2τ (1)− c− 2,hence

F⁺(2) > τ (1)− c − 2. (21)

In particular, F (3)> τ (1)− c− 10,and by (6) with 4.13.1, F (β)= F⁺(α) ≤ F⁺(γ + 1)≤ τ(1)− 10,so that,by4.13.7,β < γ + 5.

The inequality(21) yieldsalso F (11) > τ (1)− c− 2, so that,by 4.13.2,F⁺(γ + 10)≤ τ(1)− 2. Since γ≤ α < γ +1 andγ + 1≤ β < γ +5,wehaveγ + 10= α + β−t forsomet∈]0,α] witht< γ.Consequently, F (t)+ F⁺(γ + 10)= F (22)+ F⁺(α + β− 22), so thatF⁺(γ + 10)= F (22)+ F⁺(α + β− 22)− F (t)>

(τ (1)− c− 1)+ (τ (1)− 1)− (τ(1)− c)= τ (1)− 2,from 4.13.8and4.13.3.Weobtainedacontradiction.

III β∈ [0,γ/2[ Wehaveα≤ β < γ/2,so thatγ > α + β andF (γ)= τ (1),whichcontradicts4.13.3.

IVβ ∈ [γ/2,γ[ Wehaveα + β≥ γ, sinceby(5) F⁺(α + β)= τ (1) and F (γ)≤ τ(1)− c,by 4.13.3.By 4.13.4, there is a0< < 1 satisfying F () > τ (1)− c− 21.Suppose α + β > γ + 1+ . Choose γ with β < γ < γ. Wehaveγ + 1= α + β− t1,γ = α + β− t2 for some< ti < α,i= 1,2.Usingα≤ β < γ, 4.13.3andthechoiceof,wehaveF (t_i)∈]τ(1)− c− 21,τ (1)− c] fori= 1,2.Ontheotherhand,by4.13.3, F⁺(α + β−t2)= F⁺(γ)≤ τ(1)−c,andF⁺(α + β−t1)= F⁺(γ + 1)> τ (1)−21,by4.13.5.Thiscontradicts c > 42 andthe equalityF (t1)+ F⁺(α + β− t1)= F (t2)+ F⁺(α + β− t2),satisﬁed fort1,t2 ∈]0,α]. We showedthatα + β≤ γ + 1+ < γ + 2,inparticular,by(5),F (γ + 2)= τ (1).

SinceF⁺(γ + 10)= τ (1),by4.13.2witht= 11 wehaveF (11)≤ τ(1)− c− 2.By4.13.8and(6),wehave F⁺(α)= F (β)≥ F (γ/2)≥ F (22)≥ τ(1)− c− 1,so thatα≥ 11.Consequently, F (2)+ F⁺(α + β− 2)= F (5)+ F⁺(α + β− 5).Since γ≤ α + β < γ + 2,wehaveα + β− 2,α + β− 5∈ [γ − 5,γ[⊆]22,γ[,and it followsfrom4.13.3and4.13.8thatF⁺(α + β− 2)− F⁺(α + β− 5)< 1 andF (5)− F (2)< 1.By4.13.6this

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impliesF (2)> τ (1)− c− 6,henceF⁺(γ + 2)≤ τ(1)− 10 by4.13.1with t= 2.Wegotcontradictionwith theequalityF (γ + 2)= τ (1),whichends theproof.

Lemma4.15.Letτ beamultipleofτ0suchthatτ(1)> 17.ThereisanoperatorB∈ A+satisfyingB< 1 andthefollowingcondition: forany operatorC∈ Ah with τ(supp C)≤ 1,thedistribution functionFB+C

(w.r.t. τ) isnotsymmetric.

Proof. IfthedistributionfunctionF ofanoperator(w.r.t.τ)satisﬁes,forsome0= λ₀< λ₁< λ₂< λ₃/2, the inequalitiesF⁺(λi)− F (λi)≥ 4 for i= 0,1,2,F⁺(λ2)≥ τ(1)− 3,F (λ3)< τ(1), then F cannotbe symmetric.Infact,supposethatF is(α,β)−symmetricforsome0≤ α ≤ β,andputδ := α + β.Forδ < λ₃ onehasF⁺(δ)≤ F (λ3)< τ(1), contradicting(5).Ontheotherhand,ifδ≥ λ3,then

F^(δ)+(λ_i)− F^(δ)+(λ_i−1) =

F⁺(λ_i)− F⁺(λ_i−1) +

F (δ− λi)− F (δ − λi−1)

≥

F⁺(λi)− F (λi) +

F (δ− λi)− F (δ − λi−1)

≥ 4 + F (δ − λ2)− τ(1)

≥ 4 + F⁺(λ₂)− τ(1)≥ 1,

fori= 0,1,2 and anyλ₋₁< 0.Consequently,F^(δ)hasatleastthreejumps,soittakesonmorethanthree values,contradicting(7).ThusF isnot(α,β)-symmetric.

Putnow B :=3

i=0λiEi,where Ei aremutuallyorthogonalprojectionsfrom A withsum1_Asuchthat τ(E₀),τ(E₁),τ(E₂) ≥ 5 and τ(E₃) = 2, with λ_i as above, and assume additionally that λ₃ < 1, so thatB < 1.The operatorB + C haseigenspaces E_i(H) for eigenvalues λi, with E_i ≥ Ei∧ (supp C)^⊥, for i = 0,1,2,3. Hence τ(E₀),τ(E₁),τ(E₂) ≥ 4,τ(E₃) ≥ 1 and F_B+C⁺ (λ_i)− FB+C(λ_i) ≥ 4 for i = 0,1,2,F_B+C⁺ (λ3)− FB+C(λ3)> 0.Moreover,byLemma4.10appliedtoτ, F_B+C⁺ (λ2)≥ τ(1)− (τ(E3)+ τ(supp C)) ≥ τ(1)− 3, so thatthe ﬁrst partof the proof applies and, consequently, B has the desired properties.

Corollary 4.16.There existsan operator B ∈ A+ satisfying B< 1 and thefollowing condition: forany operatorC∈ Ah with τ (supp C)≤ 63, thedistributionfunctionof B + C (w.r.t.τ ) isnotsymmetric.

Proof. Putτ = (1/63)τ and build B forτ as inLemma4.15. Forany C with τ (supp C)≤ 63, we have τ(supp C)≤ 1,sothatFB+C is notsymmetric.

Lemma 4.17.There is C ∈ A+ satisfying τ (supp C) < 1 and the following condition: for any operator B ∈ Ah with−· 1A≤ B≤ (1− )· 1A forsome ∈ [0,1],thedistributionfunctionF ofB+ C (w.r.t.

τ ) isnotsymmetric.

Proof. Deﬁne C :=3

i=0λiEi, where λ0 = 0,λ1 = 2, λ2 = 4 and λ3 = 11, and where Ei are mutually orthogonal projections from A with sum 1_A satisfying τ (E1+ E2+ E3)< 1 and τ (Ei)> τ (E3)> 0 for i = 0,1,2. ByLemma4.10, for any B ∈ A+ with B≤ 1, the distributionfunction F := F_B+C of the operatorB + CsatisfiesF⁺(λi+ 1)≥ τ(1)−τ(Ei+1+· · ·+E3)= τ (E0+· · ·+Ei) fori= 0,1,2,3 (anempty sumisthezerooperator).ByLemma4.6,itsatisfiesF (λ_i)≤ τ(E0+· · · + Ei−1) fori= 0,1,2,3.Hence,for anoperatorB as inthestatementofourlemma, wecantakeB := B+ · 1A, andthen thedistribution functionFsatisfiesF(t)= F (t+),sothatF⁺(λ_i+1−)≥ τ(E0+· · ·+Ei),F(λ_i−)≤ τ(E0+· · ·+Ei−1) fori= 0,1,2,3.Thuswehave

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F(t) =

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

0 for t∈] − ∞, −] τ (E₀) for t∈]1 − , 2 − ], τ (E0+ E1) for t∈]3 − , 4 − ], τ (1)− τ(E3) for t∈]5 − , 11 − ], τ (1) for t∈]12 − , ∞[.

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SupposethatF is(α,β)-symmetricforsome0≤ α ≤ β,andletδ := α + β.Ifδ < 11− ,thenF⁺(δ)≤ F(11−)< τ (1).Ifδ≥ 11−,then,foranyt≤ 6,F^(δ)(t)= F(t)+F⁺(δ−t)∈ [F(t)+F⁺(5−),F(t)+ τ (1)]⊆ [F(t)+ τ (1)− τ(E3),F(t)+ τ (1)]. Wehavegot F^(δ)(t₀)< F^(δ)(t₁)< F^(δ)(t₂)< F^(δ)(t₃) for arbitrarily chosen t₀ ≤ −,1− < t₁ ≤ 2− ,3− < t₂≤ 4− ,5− < t₃ ≤ 6, using τ (E₃)< τ (E_i) for i= 0,1,2.Thus,foranyδ∈ R,eitherF⁺(δ)< τ (1) orF^(δ)takesonmorethan3 values,whichcontradicts (7).

Corollary 4.18.ThereisanoperatorC∈ A+ satisfying τ (supp C)< 1 and thefollowingcondition:forany operator B∈ Ah with −1· 1A≤ B ≤ 66· 1A,thedistribution functionofB + C is notsymmetric.

Proof. Take C := 67C, where C is the operator from Lemma 4.17, used with = 1/67. Then, for any B ∈ Ahwith−1· 1A≤ B ≤ 66· 1A,wehaveforB := (1/67)B thatF_B+C isnotsymmetric,sothatF_B+C is notsymmetric,either.

Lemma 4.19.Put A:= B + C + 4E₁+ 21E₂,where E₁,E₂∈ Proj A with E₁⊥ E2,τ (E₁)= 16,τ (E₂)= 4, and where B and C are operators from Corollaries 4.16 and 4.18, respectively. Whatever the choice of projection R∈ A andγ∈ R, thedistributionfunctionof A+ γR is notsymmetric.

Proof. IForR suchthatτ (R)≤ 42 (andanyγ∈ R)wehaveτ (supp(C + 4E₁+ 21E₂+ γR))≤ τ(supp C)+ τ (E₁)+ τ (E₂)+ τ (R)< 1+ 16+ 4+ 42= 63,andCorollary 4.16togetherwith thedeﬁnitionofB imply thatthedistributionfunctionof A+ γR isnotsymmetric.

II For −1 < γ≤ 44, we have−1A ≤ B + 4E1+ 21E2+ γR≤ 66· 1A. Corollary 4.18implies thatthe distributionfunctionofA+ γR = C + (B + 4E1+ 21E2+ γR) isnotsymmetric.

IIIForγ≤ −1 andτ (R)> 42 theoperatorA+ γR isnotpositive.Infact,τ (supp(C + 4E₁+ 21E₂))≤ τ (supp C)+ τ (E₁)+ τ (E₂)= 1+ 4+ 16< τ (R).Hence,for 0 = ξ ∈ (R ∧ (supp(C + 4E1+ 21E₂))^⊥)(H), we have(A+ γR)ξ,ξ=Bξ,ξ+ γξ²≤ (B+ γ)ξ²< (1+ γ)ξ²< 0.

ThusFA+γR(0)> 0,whichbyRemark4.3(4) meansthatA+ γR isnotsymmetric.

IVForγ > 44 andτ (R)> 42,thedistributionfunctionoftheoperatorA+ γR isnotsymmetric.Infact, itsatisﬁes theassumptionsofCorollary4.13,sowe canuseLemma4.14.

It should be noted that if aself-adjoint operator A cannotbe written as a real linear combination of three projections,itcannotbewrittenas acomplexlinearcombinationofthreeprojections,either.Infact, if A = αP + βQ+ γR with α,β,γ ∈ C, then A = (1/2)(A+ A^∗) = (α)P + (β)Q+ (γ)R. Hence, Theorem 4.20, Corollaries 4.21 and 4.22, as well as Theorem 5.3 of the next section, hold for arbitrary combinations ofprojections.

Theorem4.20.LetA beaσ-ﬁnitetypeII₁ algebrawithafaithfulnormaltracialstateτ₀.Forany 0< δ ≤ 1 there exists anoperator A∈ A+ with τ₀(supp A)≤ δ such that A isnot areal linear combination of three projections from A.

Proof. Put τ := (1071/δ)τ0. We will apply Lemma 4.19 with the trace τ , and show that the operator A deﬁned there cannot be of the form αP + βQ+ γR, whatever α,β,γ ∈ R and choice of projections P,Q,R∈ A. Infact: