Lie Triple Derivations on von Neumann Algebras?

Lei LIU

Abstract Let A be a von Neumann algebra with no central abelian projections.It is proved that if an additive map satisfies δ([[a,b],c])=[[δ(a),b],c]+[[a,δ(b)],c]+[[a,b],δ(c)]for any a,b,c ∈ A with ab=0(resp.ab=P,where P is a fixed nontrivial projection in A),then there exist an additive derivation d from A into itself and an additive mapvanishing at every second commutator[[a,b],c]with ab=0(resp.ab=P)such that δ(a)=d(a)+f(a)for any a ∈ A.

Keywords Derivations,Lie triple derivations,von Neumann algebras

1 Introduction

Let A be an algebra over a field F.Recall that an additive(a linear)map δ from A into itself is called an additive(a linear)derivation if δ(ab)= δ(a)b+aδ(b)for all a,b ∈ A. δ is called an additive(a linear)Lie derivation if δ([a,b])=[δ(a),b]+[a,δ(b)]for all a,b ∈ A,where[a,b]=ab?ba.More generally,δ is called an additive(a linear)Lie triple derivation if δ([[a,b],c])=[[δ(a),b],c]+[[a,δ(b)],c]+[[a,b],δ(c)]for all a,b,c ∈ A.The structures of Lie triple derivations on some operator algebras were intensively studied(see[2,7,9]and references therein).Let M be a von Neumann algebra with no central abelian projections.Miers[9]proved that ifis a linear Lie triple derivation,then there exists an element T∈M and a linear mapwhich annihilates brackets such that L(a)=aT?Ta+f(a)for any a∈M.

In recent years,the local actions of derivations have been studied intensively.One direction is to study the conditions under which derivations of operator algebras can be completely determined by the actions on some elements concerning products.We say that an additive(a linear)mapis derivable at a given point G ∈ A,if δ(ab)= δ(a)b+aδ(b)for all a,b ∈ A with ab=G.This kind of maps were discussed by several authors(see[1,3,4–5,11–12]and references therein).But,so far,there have been few papers on the study of the local actions of Lie triple derivations on operator algebras.We say that an additive(a linear)mapis Lie triple derivable at a given point G ∈ A,if δ([[a,b],c])=[[δ(a),b],c]+[[a,δ(b)],c]+[[a,b],δ(c)]for all a,b,c∈A with ab=G.It is the aim of the present article to investigate the additive(linear)Lie triple derivations on von Neumann algebras with no central abelian projections by the local actions.It is a generalization of the results in[9].

We need some notations and preliminaries about von Neumann algebras.A von Neumann algebra A is a weakly closed,self-adjoint algebra of operators on a Hilbert space H containing the identity I.={z∈A:za=az for all a∈A}is called the center of A.A projection P is called a central abelian projection ifand PAP is abelian.We denotebe the central carrier of a,which is the smallest central projection satisfying Pa=a.It is well known thatis the projection whose range is the closed linear span of{Aa(h):h∈H}.For each self-adjoint operator r∈A,the core of r denoted byis sup{a∈ZA:a=a?,a≤r}.If P ∈A is a projection andwe call P a core-free projection.It is easy to verify thatif and only ifBy[8,Lemma 4],we can say that A is a von Neumann algebra with no central abelian projections if and only if it has a projection P∈A such thatandWe refer the reader to[6]for the theory of von Neumann algebras.

2 Characterizing Lie Triple Derivations by Acting on Zero-Product

In this section,we consider the question of characterizing Lie triple derivations by action at zero product on von Neumann algebras with no central abelian projections.

Theorem 2.1Let A be a von Neumann algebra without central abelian projections.Suppose thatis an additive map satisfying

for all a,b,c∈A with ab=0.Then there exists an additive derivation d from A into itself and an additive mapvanishing at every second commutator[[a,b],c]when ab=0 such that

Note that every linear derivation of a von Neumann is inner(see[10]).By Theorem 2.1,the following corollary is immediate.It is a generalization of Theorem 1 in[9].

Corollary 2.1Let A be a von Neumann algebra without central abelian projections.Suppose thatis a linear map satisfying

for all a,b,c∈A with ab=0.Then there exists an element T∈A and a linear mapvanishing at every second commutator[[a,b],c]when ab=0 such that

Proof of Theorem 2.1By[8,Lemma 4],there is a projection P∈A such thatandIn what follows,we denote P1=P and P2=I?P1.By the definitions of core and central carrier,P2is also a core-free projection andSet Aij=PiAPjfor i,j=1,2.Then A=A11+A12+A21+A22.For an operator aij∈A,we always mean that aij∈Aij.

We shall organize the proof of Theorem 2.1 in a series of claims.

Claim 2.1Let aii∈Aii,i=1,2.If a11b12=b12a22for all b12∈A12,then

For any x11∈A11,x12∈A12,we have a11x11x12=x11x12a22=x11a11x12.Since P2=I,which means that{AP2(h):h∈H}is dense in H,we get a11x11=x11a11,that is,a11∈.By[6,Corollary 5.5.7],we knowSo there exists z1∈ZAsuch that a11=z1P1.

Similarly,we have a22=z2P2,z2∈ZA.It follows that z1b12=a11b12=b12a22=z2b12.Then(z1?z2)P1=0,which implies(z1?z2)AP1=0.Bywe obtain z1=z2.Hence a11+a22∈ZA,the claim holds.

Moreover,for any a12∈A12,since a12P1=0,we have

Multiplying P1from the left side and P2from the right side of the above equation,we arrive at P1δ(P1)a12=a12δ(P1)P2.It follows from Claim 2.1 that P1δ(P1)P1+P2δ(P1)P2∈ ZA.Let E=P1δ(P1)P2? P2δ(P1)P1,and ? = δ? δE,where δEis the inner derivation given by δE(x)=xE?Ex for all x∈A.It is not difficult to verify

and

for any a,b,c∈A with ab=0.

Claim 2.2

Since P2P1=0 and,we have

For any a12∈A12,since P2a12=0,we get

Multiplying the above equation by P1from the left and by P2from the right,we obtain

Claim 2.3

Since a12P1=0 and ?(P1)∈ ZA,we get

Now it suffices to show that P2?(a12)P1=0.Indeed,for any b12∈ A12,x ∈ A,it is easy to check that

This together with P1?(a12)P1=P2?(a12)P2=0 entails thatThis leads toThen P2?(a12)b12=0.Sincewe have P2?(a12)P1=0.Consequently,

Claim 2.4There exist mapssuch that ?(aii)? fi(aii) ∈ Aiifor any aii∈Aii,i=1,2.

Since a11P2=0 and from Claim 2.2,we have

Moreover,for any b22∈A22and x∈A,it is easy to check that

and

Therefore,for any a11∈A11,we have

Similarly,we can define a mapsuch that ?(a22)? f2(a22) ∈ A22for any a22∈A22.So Claim 2.4 is true.

By the definition of d and Claim 2.4,we haveand d(aij)=?(aij)for all

In the following we shall show that d is an additive derivation.

Claim 2.5d is an additive map.

Since d=??f and f=f1+f2,we only need to show that f1and f2are additive maps.

For any a11,b11∈A11,it follows from(2.1)that

and

By(2.3)–(2.5),we get

Similarly,f2is an additive map.

Claim 2.6d(aiibij)=aiid(bij)+d(aii)bijfor any aii∈Aii,bij∈Aij,

Due to bijaii=0,the following equations hold:

With the similar argument in Claim 2.6,we have the following claim.

Claim 2.7d(aijbjj)=aijd(bjj)+d(aij)bjjfor any

Claim 2.8d(aiibii)=aiid(bii)+d(aii)biifor any aii,bii∈Aii,i=1,2.

For any bij∈Aij,we have,from Claim 2.6,that

At the same time,

Comparing the above two equations,we get

Note Pj=I.It follows from the fact that{APj(h):h∈H}is dense in H that d(aiibii)=aiid(bii)+d(aii)bii.

Claim 2.9d(aijbji)=aijd(bji)+d(aij)bjifor any

Since P2a12=0 andwe have

Since d(a)= ?(a)?f(a),?a∈ A,

We shall show f(b21a12?a12b21)=0.Multiplying the above equation by a12to the left side and right side respectively,we obtain the following two equations:

and

Computing(2.7)–(2.8),we get

It follows from Claims 2.6–2.7 that

which combining with the above equation impliesUsing polar decomposition of a12,we havewhich yieldsThis leads toand so

Similarly,we have b21f(b21a12?a12b21)?=0.

Then multiplying(2.6)by f(b21a12?a12b21)?to the right side,we arrive at

Due to Claim 2.8,the following equations hold:

Putting the above two equations into(2.9),we have

Multiplying the equation by f(b21a12?a12b21)?to the left side,we get

which implies f(b21a12?a12b21)=0.So we arrive at

This is equivalent to d(a12b21)=d(a12)b21+a12d(b21)and d(b21a12)=d(b21)a12+b21d(a12),as desired.

By Claims 2.5–2.9,we can conclude that d is an additive derivation.Hence we have δ(a)=?(a)+δE(a)=d(a)+f(a)+δE(a),?a ∈ A.Denote φ(a)=d(a)+δE(a),then δ(a)= φ(a)+f(a),?a∈A.Clearly,φ is an additive derivation on A and f is an additive map from A to

For ab=0,it follows that

3 Characterizing Lie Triple Derivations by Acting on Projection-Product

In this section,we consider the question of characterizing Lie triple derivations by acting at projection-product on von Neumann algebras without central abelian projections.The proof of the following theorem shares the similar outline as that of Theorem 2.1,but it needs different techniques.

Theorem 3.1Let A be a von Neumann algebra without central abelian projections and P be a projection in A withandSuppose thatis an additive map satisfying

for all a,b,c ∈ A with ab=P.Then there exists an additive derivation φ from A into itself and an additive mapvanishing at every second commutator[[a,b],c]when ab=P such that

Note that all linear derivations of von Neumann algebras are inner(see[10]).We have the following corollary.It is a generalization of Theorem 1 in[9].

Corollary 3.1Let A be a von Neumann algebra without central abelian projections and P be a projection in A withandSuppose thatis a linear map satisfying for all a,b,c∈A with ab=P.Then there exists an element T∈A and a linear mapvanishing at every second commutator[[a,b],c]when ab=P such that

Proof of Theorem 3.1We shall use the same symbols with that in Section 2.

For any a12∈A12,sincewe obtain

Multiplying P1from the left side and P2from the right side of the above equation,we arrive at P1δ(P1)a12=a12δ(P1)P2.It follows from Claim 2.1 that P1δ(P1)P1+P2δ(P1)P2∈ ZA.Let E=P1δ(P1)P2?P2δ(P1)P1,and ? = δ?δE,where δEis the inner derivation.It is not difficult to verify that

and

for any a,b∈A with ab=P1.

Now we organize the proof in a series of claims.

Claim 3.1

Since(P1+P2)P1=P1andwe have

For any a12∈A12,since(P1+a12)(P1+P2?a12)=P1,we get

Multiplying the above equation by P1from the left and by P2from the right,we obtain

It follows from Claim 2.1 thatHence

Claim 3.2

Since(P1+a12)P1=P1and,we get

Now,for any b12∈A12,we have

Multiplying the above equation from both side by P2,we arrive at P2?(b12)b12=0.Moreover,it follows that

Multiplying the equation by b12from the right and for the fact P2?(b12)b12=0,we obtain b12?(a12)b12=0.By linearizing,we get b12?(a12)d12+d12?(a12)b12=0 for any b12,d12∈ A12.It is not difficult to check

that is,

As von Neumann algebras are semiprime,we see P2?(a12)b12?(a12)P1=0.Then P2?(a12)P1=0.Consequently,.

Claim 3.3There exists a map f1on A11such that ?(a11)?f1(a11)∈ A11for all a11∈ A11.

First suppose that a11is invertible in A11,i.e.,there existssuch thatSince,we have

Moreover,for any b22∈ A22and x∈A,sinceit is easy to check that

If a11is not invertible in A11,we may find a sufficiently big number n such that nP1?a11is invertible in A11.It follows from the preceding case that P1?(nP1?a11)P2+P2?(nP1?a11)P1=0,and P2?(nP1?a11)P2=ZP2.Since ?(P1)∈ ZA,we also have P1?(a11)P2+P2?(a11)P1=0 and P2?(a11)P2∈ ZAP2.Without loss of generality,we still denote P2?(a11)P2=ZP2.

We define f1:M11→ZMby f1(a11)=Z for all a11∈A11.With the similarly argument as in Claim 2.4,we know f1is well defined.Hence

Claim 3.4There exists a map f2on A22such that ?(a22)?f2(a22)∈ A22for any a22∈ A22.

For any a22∈A22,since(P1+a22)P1=P1,we have

The rest step is similar to the proof of Claim 3.3.

Now,we define two maps f:A→ZAand d:A→A respectively by

and

By the definition of d and Claim 3.4,we have d(P1)=d(P2)=0,d(Aij)?Aij,1≤i,j≤2 and d(aij)=?(aij)for all

In the following we shall show that d is an additive derivation.

Claim 3.5d is an additive map.

The proof is similar to that of Claim 2.5.

Claim 3.6d(a11b12)=a11d(b12)+d(a11)b12for any a11∈A11,b12∈A12.

If a11is invertible in A11,then for any x12∈ A12,we haveIt follows that

Replacing b12withwe have d(a11b12)=a11d(b12)+d(a11)b12.

If a11is not invertible in A11,we may find a sufficiently big number n such that nP1?a11is invertible in A11.Then d((nP1?a11)a12)=(nP1?a11)d(a12)+d(nP1?a11)a12.Clearly,P1is invertible in A11,so we get d(a11b12)=a11d(b12)+d(a11)b12from the above equation.

Claim 3.7d(a21b11)=a21d(b11)+d(a21)b11for any a21∈A21,b11∈A11.

Claim 3.8d(a22b21)=a22d(b21)+d(a22)b21for any a22∈A22,b21∈A21.

Due to(P1+a22?a22b21)(P1+b21)=P1,we compute

that is,d(a22b21)=a22d(b21)+d(a22)b21.

Considering(P1+a12)(P1?b22+a12b22)=P1,we arrive at the following claim.

Claim 3.9d(a12b22)=a12d(b22)+d(a12)b22for any a12∈A12,b22∈A22.

Claim 3.10d(aiibii)=aiid(bii)+d(aii)bii,i=1,2.

It is similar to Claim 2.8.

Claim 3.11d(aijbji)=aijd(bji)+d(aij)bjifor any

Since(a12+P1)P1=P1,we have

Since d(a)=?(a)?f(a),?a∈A,

With the same approach as in Claim 2.9,we can get f(b21a12?a12b21)=0.So we arrive at

This is equivalent to d(b21a12)=d(b21)a12+b21d(a12)and d(a12b21)=d(a12)b21+a12(.b21).Consequently,Claim 3.11 is true.

So we can conclude that d is an additive derivation by Claims 3.5–3.11.Hence we have δ(a)= ?(a)+ δE(a)=d(a)+f(a)+ δE(a), ?a ∈ A.Denote φ(a)=d(a)+ δE(a),then δ(a)= φ(a)+f(a),?a ∈ A.Clearly,φ is an additive derivation on A and f is an additive map from A to ZA.

With the similar argument as in the proof of Theorem 2.1,we can verify the additive map f:A→ZAvanishing at every second commutator[[a,b],c]when ab=P.

AcknowledgementThe author wish to give his thanks to the referees and the editor for their helpful comments and suggestions.