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Inclusions and positive cones of von Neumann 8 algebras 0 0 2 Yoh Tanimoto n Graduate School of Mathematical Sciences a J University of Tokyo, Komaba, Tokyo, 153-8914, JAPAN 8 e-mail: [email protected] 2 ] A Abstract O We consider cones in a Hilbert space associated to two von Neumann . algebrasanddeterminewhenonealgebraisincludedintheother. Ifacone h isassocated toavonNeumannalgebra, theJordanstructureisnaturally t a recovered from it and we can characterize projections of the given von m Neumann algebra with thestructure in some special situations. [ 1 1 Introduction v 9 The natural positive cone P♮ =∆41M+ξ0 plays a significant role in the theory 5 of von Neumann algebras (see, for example, [1, 5]) where M is a von Neumann 2 4 algebra, ξ0 is a cyclic separating vector for M and ∆ is the Tomita-Takesaki 1. modular operator associated to ξ0. Among them, the result of Connes [6] is of particular interest which characterized the natural positive cones with their 0 8 geometric properties called selfpolarity, facial homogeneity and orientability, 0 and showed that if two von Neumann algebras M and N share a same cone, : then there is a central projection q of M such that N =qM⊕q⊥M′. Connes v i usedtheLiealgebrawithaninvolutionofthelineartransformationgroupofP♮ X in his paper. r Inthepresentpaper,insteadofP♮,westudyP♯ =M ξ ,whichholdsmore a + 0 informations of M, for example, the subalgebra structure. In the second section, we study what occurs when N ξ ⊂ P♯ where N is + 0 another von Neumann algebra. We consider first the case when ξ is not cyclic 0 for N and then assume the cyclicity. It turns out that in the latter case N is included in M except the part where ξ is tracial. 0 Inthethirdsection,wecharacterizecentralprojectionsofMintermsofP♯. A projectionp is in M∩M′ if and only if p and its orthogonalcomplement p⊥ preserve P♯. In the fourth and fifth sections, the Jordan structure on P♯ is studied. We can recoverthe lattice structure of projections and the operator norm from the order structure of P♯. Then we can define the square operation on P♯. 1 Inthe finalsection,using the Jordanstructure,acharacterizationofprojec- tionsinMisobtainedwhenthe modularautomorphismwithrespecttoξ acts 0 ergodically. The result of the second section has an easy application to the theory of half-sided modular inclusions [12, 2]. Let {U(t)} be a one-parameter group of unitaryoperatorswithageneratorH whichkillsξ . AssumethatMisafactor 0 of type III (or more generally a properly infinite algebra). It is easy to see 1 that U(t)MU(t)∗ ⊂ M for t ≥ 0 if and only if U(t) preserves P♯ for t ≥ 0. A similarresultforP♮ and{e−tH}hasbeenobtainedby Borcherswithadditional conditions on H [4]. Davidson has obtained conditions for {U(t)} to generate a one-parameter semigroup of endomorphisms [7]. The relations with the modular group have been shown to be important in his study. 2 Inclusions of positive cones LetMbeavonNeumannalgebraactingonaHilbertspaceHandξ beacyclic 0 separating vector for M. We denote the modular group by ∆it, the modular conjugation by J, modular automorphism by σ and the canonical involution t by S =J∆12. The positive cone associated to ξ0 is denoted by P♯ =M+ξ0. Suppose there is another von Neumann algebra N such that N ξ ⊂ P♯. + 0 We can define a positive contracitve map α from N into M as follows. Lemma 1 Fora∈N thereistheuniquepositiveelementα(a)∈Msatisfying + aξ =α(a)ξ . In addtition, α is contracitive on M . 0 0 + proof Bytheassumption,wehaveaξ ∈P♯. Recallthatforavectoraξ inP♯ 0 0 there is a positive linear operator α(a) affiliated to M such that aξ = α(a)ξ 0 0 [11]. Since kakI−a ispositive, wehave(kakI−a)ξ ∈P♯. This implies, forevery 0 y ∈M′, hα(a)yξ ,yξ i = hα(a)ξ ,y∗yξ i 0 0 0 0 = haξ ,y∗yξ i 0 0 ≤ kakhξ ,y∗yξ i=kakkyξ k2. 0 0 0 Hence α(a) is bounded and in M. 2 We can easily see that α extends to N by linearity. Since α is contractive on N , α is bounded on N . + sa Lemma 2 The map α maps every projection to a projection. proof Take a projection e∈N. Note that, since α maps N into M and is + + contractive, we have α(e)≥α(e)2. 2 Recall that, by the definition of α, we have α(e)ξ = eξ . We calculate as 0 0 follows. hα(e)2ξ ,ξ i = hα(e)ξ ,α(e)ξ i 0 0 0 0 = heξ ,eξ i 0 0 = heξ ,ξ i 0 0 = hα(e)ξ ,ξ i. 0 0 Thisimpliesthath α(e)−α(e)2 ξ ,ξ i=0. Aswenotedabove,α(e)−α(e)2 0 0 must be positive, hen(cid:0)ce the vecto(cid:1)r α(e)−α(e)2 12 ξ must vanish. By the 0 separating property of ξ0, we see α(e)(cid:0)=α(e)2. 2 (cid:1) Recall that a linear mapping φ which preserves every anticommutator is called a Jordan homomorphism: φ(xy+yx)=φ(x)φ(y)+φ(y)φ(x). Now we show the following lemma. The proofofit is essentially takenfrom[9]. Lemma 3 The map α is a Jordan homomorphism. proof Leteandf bemutuallyorthogonalprojectionsinN. Thene+f,α(e), α(f) and α(e)+α(f) are projections. We see the range of α(e) and the range ofα(f) aremutually orthogonalbecauseif not, thenthe sumα(e)+α(f)could not be a projection. This implies that α(e)α(f)=α(f)α(e)=0. In particular, α maps the positive (resp. negative) part of a self-adjoint element x to the positive (reps. negative) part of α(x). From this we see that α is contractive on N . sa Next suppose we have commuting projections e,f ∈N. Remark that, since ef ≤e, positivity of α assures α(ef)≤α(e). Recalling that in this case ef and e are projections, we see the range of α(ef) is included in the range of α(e). Thus we have α(ef)α(e)=α(ef). Now noting e−ef and f are mutually orthogonalprojections, we have 0=α(e−ef)α(e)=α(e)α(f)−α(ef). Hence α preserves products of commuting projections. Since every self-adjoint element in a von Neumann algebra is a uniform limit of linear combinations of mutually orthogonal projections, and since α is continuous in norm on N , α preserves products of commuting self-adjoint sa elements. In particular, α perserves the square of self-adjoint elements. 3 Thisimpliesthat,firstly,αpreservesJordanproductsofself-adjointelements ab+ba=(a+b)2−a2−b2. This shows α(ab+ba) = α (a+b)2 −α(a2)−α(b2) = α((cid:0)a+b)2−(cid:1)α(a)2−α(b)2 = α(a)α(b)+α(b)α(a). Secondly, α preserves squares of arbitrary elements (a+ib)2 = a2+i(ab+ ba)−b2: α (a+ib)2) = α a2+i(ab+ba)−b2 (cid:0) (cid:1) = α((cid:0)a2)+iα(ab+ba)−α(cid:1)(b2) = α(a)2+i(α(a)α(b)+α(b)α(a))−α(b)2 = (α(a)+iα(b))2. Finally, α preserves Jordan products of arbitrary elements xy+yx = (x+ y)2−x2−y2: α(xy+yx) = α (x+y)2 −α(x2)−α(y2) = α((cid:0)x+y)2−(cid:1)α(x)2−α(y)2 = α(x)α(y)+α(y)α(x). This completes the proof. 2 Here we need the following result on Jordan homomorphisms of Jacobson and Rickart [8]. Proposition 4 Suppose φ is a unital Jordan homomorphism from an algebra A into B. Suppose further that A has a system of matrix units. Then there is a central idempotent g of the algebra generated by φ(A) such that φ(·)g is homomorphic and φ(·)(I −g) is antihomomorphic. Note thatevery vonNeumann algebraN decomposes into the commutative part, the I parts, the II part, and the properly infinite part. On the first one n 1 α causes no problemand on the remaining parts we can apply Proposition4 to the case in which φ = α, A = N, B = M. Examining the proof, we see if φ is self-adjoint, then e is an centralprojection of α(N)′′ (the argumenthere is due to Kadison [9]). Next, we show the normality of α. Lemma 5 The map α is a normal linear mapping from N into M. proof We only have to show that for any normal functional ϕ on M the functional ϕ◦α on N is normal. Note that, since M has a separating vector ξ , we may assume ϕ(·)=h·η ,η i for some η ,η ∈H. 0 1 2 1 2 4 Recall that a linear functional on a von Neumann algebra is normal if and only if it is continuous on every bounded set in the weak operator topology. Now suppose that we have a WOT-converging bounded net x → x in i N. Obviously {x ξ } converges to xξ weakly. By the definition of α, we i 0 0 see {α(x )ξ } converges to α(x)ξ weakly. We have, for any y ,y ∈M′, i 0 0 1 2 hα(x )y ξ ,y ξ i = hy α(x )ξ ,y ξ i i 1 0 2 0 1 i 0 2 0 ∗ = hα(x )ξ ,y y ξ i i 0 1 2 0 → hα(x)ξ ,y∗y ξ i 0 1 2 0 = hα(x)y ξ ,y ξ i. 1 0 2 0 First we assume {x } is a net of self-adjoint elements. Then for arbitrary i η ,η ∈ H the convergence hα(x )η ,η i → hα(x)η ,η i holds since {x } is a 1 2 i 1 2 1 2 i bounded net, α is contractive on N , and ξ is cyclic for M′. sa 0 ThenwecanobtaintheconvergenceforarbitraryboundedWOT-converging net {x } since we have the decomposition i x +x∗ x −x∗ x = i i +i i i i 2 2i and each part of the net is self-adjoint or antiself-adjoint, bounded and WOT- converging. 2 We combine this lemma and the proposition of Jacobsonand Rickart to get the following. Lemma 6 Thereisanormalhomomorphism β andnormalantihomomorphism γ of N into M such that α(x)=β(x)+γ(x) and the the range of β and γ are mutually orthogonal. In addition, there are central projections e,f ∈ N and a central projection g ∈ α(N)′′ such that α(e ·)g = β(·) is an isomorphism of Ne and α(f ·)g⊥ = γ(·) is an antiisomorphism of Nf. proof We know from Proposition 4 that there is a central projection g ∈ α(N)′′ suchthat β(·)=α(·)g is a homomorphismof N and γ(·)=α(·)g⊥ is an antihomomorphism of Nf. Then just take e as the support of β and f as the support of γ. Since α is normal, so are β and γ and the definitions of e and f are legitimate. 2 Lemma 7 The von Neumann algebra Nf is finite. proof Let Nh be the properly infinite part of Nf. We have g⊥α(xy) = g⊥α(y)α(x) =α(y)g⊥α(x) for x,y ∈Nh. 5 Again take x,y ∈Nh. By the definiton of α, we have ⊥ ⊥ g xyξ = g α(xy)ξ 0 0 = α(y)g⊥α(x)ξ 0 ⊥ ⊥ g xyξ ,ξ = α(y)g α(x)ξ ,ξ 0 0 0 0 (cid:10) (cid:11) (cid:10) ⊥ ∗ (cid:11) = g α(x)ξ ,α(y )ξ 0 0 (cid:10) ⊥ ∗ (cid:11) = g xξ ,y ξ 0 0 = (cid:10)yg⊥xξ ,ξ (cid:11). 0 0 (cid:10) (cid:11) Since Nh is properly infinite, there is a sequence of isometries {v } ⊂ Nh n such that v v∗ → 0 in SOT-topology (That they are isometries means v∗v = n n n n h). Now hγ(h)ξ ,ξ i = g⊥hξ ,ξ 0 0 0 0 = (cid:10)g⊥v∗v ξ(cid:11),ξ n n 0 0 = (cid:10)v g⊥v∗ξ ,ξ (cid:11) n n 0 0 ≤ (cid:10)hv v∗ξ ,ξ i→(cid:11)0. n n 0 0 But since γ(h) is a projectionin α(N)′′ ⊂M and since ξ is separating for M, 0 γ(h) must be zero. Recalling that h is a subprojection of f and that f is the support of γ, we see that h=0. 2 Theorem 8 Let M and N be von Neumann algebras and ξ is a cyclic sepa- 0 rating vector for M. Suppose N ξ ⊂P♯. + 0 Then we have two disjoint possibilities: 1. The von Neumann algebra M has a subalgebra M such that M ξ = 1 1+ 0 N ξ . + 0 2. For any subalgebra M of M, its “sharpened cone” M ξ cannot coin- 2 2+ 0 cide with N ξ and N has a finite ideal N such that there is a subalgebra + 0 1 of M which is isomorphic to the direct sum of N and Nopp. 1 1 proof Suppose that e and f defined above are mutually orthogonal. Then let us define M =α(N). Since we have ef =0, it decomposes as follows. 1 α(N) = α N[e+e⊥][f +f⊥] = α(cid:0)N[ef⊥+fe⊥+e⊥(cid:1)f⊥] = β(cid:0)Nef⊥ +γ Nfe⊥ , (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) by noting that Ne⊥f⊥ is the kernel of α. Since the range of β and γ are mutually orthogonal, and since e and f are central projections, α(N) is a direct sum of β Nef⊥ and γ Nfe⊥ . (cid:0) (cid:1) (cid:0) (cid:1) 6 Let a be a positive element of N. Then we have aξ = α(a)ξ 0 0 = β(ae)ξ +γ(af)ξ 0 0 ⊥ ⊥ = β(aef )ξ +γ(afe )ξ . 0 0 Conversely it is easy to see that for b ∈ α(N) there is a ∈ N such that + + α(a) = b, hence we have aξ = bξ . This completes the proof of the claimed 0 0 equality M ξ =N +ξ . 1+ 0 0 Next, we assume that ef 6= 0. Note that Nef is noncommutative since by the definitionofβ andγ the commutativepartofN is lefttoβ. Inparticularg is a nontrivial centralprojection in α(Nef)′′. By Lemma 7, Nef is finite. One can easily see that α(Nef)′′ is a subalgebra of M which decomposes into the direct sum of β(Nef) and γ(Nef) where the latter is isomorphic to (Nef)opp. What remainsto proveis thatfor anysubalgebraM ofM we cannothave 2 the equality (Nef) ξ =M ξ . To see this impossibility, recall that + 0 2+ 0 M ξ ={Aξ |A is a closed positive operator affiliated to M}, + 0 0 since ξ is a separating vector for M [11]. Similarly we have 0 M ξ ={Aξ |A is a closed positive operator affiliated to M }. 2+ 0 0 2 Now suppose aξ ∈ M ξ for a positive element a of Nef. By the above 0 2+ 0 remark,wehaveapositiveoperatorAaffiliatedtoM suchthataξ =α(a)ξ = 2 0 0 Aξ . Then for y ∈M′ we have 0 α(a)yξ =yα(a)ξ =yAξ =Ayξ , 0 0 0 0 hence A is bounded and α(a) = A. This implies α(a) ∈ M and α(Nef) ⊂ 2 M . But by Proposition 4 α(Nef) generates β(Nef) ⊕ γ(Nef). We have 2 β(Nef)⊕γ(Nef)⊂M . 2 Wewillshowthatthisleadstoacontradiction. Bytheobservationabovewe see that M ξ contains vectors of the form gaξ ,g⊥bξ where a,b∈(Nef) . 2+ 0 0 0 + Suppose the contrary that gaξ ∈ (Nef) ξ . By the argument similar to 0 + 0 the above one, there is a self-adjoint positive operator A affiliated to Nef such that Aξ = gaξ . Then g⊥Aξ = 0. Noting that f is the support of γ and 0 0 0 that ξ is separating for M, we see g⊥e ξ = γ(e )ξ cannot vanish for any 0 A 0 A 0 nontrivial projection e of Nef. A There area spectralprojectione ofA, a positivescalarǫ andy ∈M′ such A that A≥ǫe and hγ(e )yξ ,yξ i>0. Remark that A A 0 0 ⊥ ⊥ g (A−ǫe )ξ ∈ g (Nef) ξ A 0 + 0 ⊂ g⊥(Nef) ξ + 0 = γ(Nef) ξ . + 0 7 Then we have ⊥ 0 = hyg Aξ ,yξ i 0 0 = hg⊥Aξ ,y∗yξ i 0 0 ⊥ ∗ ⊥ ∗ = hg (A−ǫe )ξ ,y yξ i+hg ǫe ξ ,y yξ i A 0 0 A 0 0 ≥ hg⊥ǫe ξ ,y∗yξ i A 0 0 = hyγ(ǫe )ξ ,yξ i A 0 0 = ǫhγ(e )yξ ,yξ i A 0 0 > 0. This contradiction completes the proof of that (Nef) ξ 6=M ξ . 2 + 0 2+ 0 Ifwefurtherassumethe cyclicityofξ forN,wehaveastrongerresult. For 0 the proof of it, we need the following lemma. This can be found, for example in [3], but here we present another simple proof. Lemma 9 IfA⊂B isaproperinclusion ofvon Neumannalgebras onaHilbert space K and if ζ is a common cyclic separating vector, then B cannot be finite. proof Suppose the contrary, that B is finite. Then A must be finite, too. Hence there is a faithful trace τ on B. Since ζ is separating for B, there is a vector η such that τ(x) = hxη,ηi by the Radon-Nikodym type theorem. Since τ is faithful, η must be separating for B. WecanseethatηiscyclicforBasfollows. Denotetheorthogonalprojection onto Bη by p. By separation verified above, we have B′η = K. On the other hand, by assumption, Bζ = B′ζ = K. By the general theory of equivalence of projections, p∼I in B. But recalling that B is finite, we see that p=I, i.e., η is cycic. Bythesamereasoning,ηiscyclicseparatingtracialforA. Thenthemodular conjugationsJAandJBwithrespecttoηmustcoincideandwehavetherequired equation. A′ ⊃B′ =JBBJB =JABJA ⊃JAAJA =A′. This contradicts the assumption that the inclusion A⊂B is proper. 2 Theorem 10 Let M andN bevon Neumannalgebras andξ bea vectorcyclic 0 separating for M and cyclic for N. Suppose N ξ ⊂P♯. + 0 Then we have the following. 1. The vector ξ is also separating for N. 0 2. There is a central projection e in N such that Ne⊂M. 3. The vector e⊥ξ is tracial for Ne⊥. 0 4. Je⊥Ne⊥Je⊥ ⊂M. 8 In particular, N and Ne⊕Je⊥Ne⊥Je⊥ share the same positive cone PN♯ where Ne⊕Je⊥Ne⊥Je⊥ ⊂M. proof First we show that the induction by g realizes β(·) = gα(·). For arbitrary x,y ∈N we have gxyξ = gα(xy)ξ 0 0 = gα(x)α(y)ξ 0 = gα(x)yξ 0 = α(x)gyξ . 0 Taking it into consideration that ξ is cyclic for N, we see that gx = gα(x) = 0 α(x)g. But, since this holds for arbitrary x ∈ N, in particular for self-adjoint elements. If x=x∗, then we have gx=α(x)g =(gα(x))∗ =(gx)∗ =xg. Since this equation is linear for x, we see that g ∈N′ and gx=gα(x). Now recallthat we havedecomposedαinto a normalhomomorphismβ and anormalantihomomorphismγ. We againdenote the supportofβ bye andthe support of γ by f. LetNh be the properlyinfinite part. By Lemma 7the intersectionofh and f is trivial. Thus we have ghxξ =hgα(hx)ξ =hα(hx)ξ =hxξ , 0 0 0 0 for x∈N. Cyclicity of ξ tells us that gh=h. Then for hx∈Nh we get that 0 α(hx)=ghx=hx. In other words, α maps identically on Nh. In particular, α is decomposed by h, that is, we have ⊥ ⊥ hα(h )=α(h)α(h )=0, since α maps orthogonal projections to orthogonalprojections. Notethathξ iscyclicforNhsinceξ iscyclicforN. Thevectorhξ isalso 0 0 0 separating for Nh since Nh=α(Nh)⊂M and ξ is separating for M. 0 For the proof of remaining part of the theorem, we may assume N is finite. Recall that g⊥ commutes with N. Take x,y ∈N and let us calculate ⊥ ⊥ ⊥ ⊥ ∗ hxyg ξ ,g ξ i = hg yξ ,g x ξ i 0 0 0 0 = hg⊥α(y)ξ ,g⊥α(x∗)ξ i 0 0 ⊥ ⊥ = hg α(x)α(y)ξ ,g ξ i 0 0 = hg⊥α(yx)ξ ,g⊥ξ i 0 0 ⊥ ⊥ = hg yxξ ,g ξ i 0 0 = hyxg⊥ξ ,g⊥ξ i 0 0 9 This shows that g⊥ξ is a tracial vector for Ng⊥. By assumption, ξ is cyclic 0 0 forN,henceg⊥ξ iscyclicforNg⊥. Inaddition,itisalsoseparatingasfollows. 0 If xg⊥ξ =0 for some x∈Ng⊥, then for any y ∈Ng⊥ we have 0 xyg⊥ξ 2 = hy∗x∗xyg⊥ξ ,g⊥ξ i 0 0 0 (cid:13)(cid:13) (cid:13)(cid:13) = hxyy∗x∗g⊥ξ0,g⊥ξ0i ≤ kyk2hxx∗g⊥ξ ,g⊥ξ i 0 0 = kyk2hx∗xg⊥ξ ,g⊥ξ i 0 0 = 0, then the cyclicity implies the separation by g⊥ξ . 0 Now Ng⊥ has the canonical conjugation Jg⊥ defined as (the closure of) Jg⊥ :g⊥H∋xξ0 7−→x∗ξ0 ∈g⊥H. On Ng⊥ we have the canonical antihomomorphism ⊥ ∗ ⊥ Ng ∋x7−→Jg⊥x Jg⊥ ∈Ng . In our situation the composition of the induction by g⊥ and this antihomo- morphismcoincide with the compositionof α and the induction byg⊥. In fact, for any elements x,y,z ∈Ng⊥ we have hJg⊥(xg⊥)∗g⊥Jg⊥yg⊥ξ,zg⊥ξ0i = hz∗g⊥ξ0,x∗y∗g⊥ξ0i = hyxz∗g⊥ξ ,g⊥ξ i 0 0 = hz∗yxg⊥ξ ,g⊥ξ i 0 0 = hg⊥yxξ ,zg⊥ξ i 0 0 = hg⊥α(yx)ξ ,zg⊥ξ i 0 0 = hg⊥α(x)α(y)ξ ,zg⊥ξ i 0 0 = hg⊥α(x)yξ ,zg⊥ξ i 0 0 = hg⊥α(x)yξ ,zg⊥ξ i. 0 0 The cyclicity of g⊥ξ0 shows that g⊥α(x)=Jg⊥(xg⊥)∗Jg⊥. Summing up, we get the following formula for α: α(x) = gα(x)+g⊥α(x) = gx+Jg⊥g⊥x∗Jg⊥. Note that gξ is cyclic separating for Ng. In fact, the cyclicity comes from 0 theassumptionofξ ’scyclicityandseparatingpropertycanbeseenbyobverving 0 Ng =gα(N)⊂M and by separating property of ξ for M. 0 10