Strong $3$-Commutativity Preserving Maps on Standard Operator Algebras

0.14 MB·

by  Meiyun Liu

STRONG 3-COMMUTATIVITY PRESERVING MAPS ON STANDARD
OPERATOR ALGEBRAS
MEIYUN LIU,JINCHUANHOU
6
1
0 Abstract. LetX beaBanachspaceofdimension≥2overtherealorcomplexfieldFandA
2
astandardoperatoralgebrainB(X). AmapΦ:A→Aissaidtobestrong3-commutativity

n
a preservingif [Φ(A),Φ(B)]3 =[A,B]3 for all A,B∈A, where [A,B]3 is the 3-commutator of
J
A,B defined by [A,B]3 =[[[A,B],B],B]. The main result in this paper is shown that, if Φ
4
2 is a surjective map on A, then Φ is strong 3-commutativity preserving if and only if there
exist a functional h:A→F and a scalar λ∈F with λ4 =1 such that Φ(A)=λA+h(A)I
]
A
for all A∈A.
F
.
h
t
a
m
[ 1. Introduction

1
Let R be a ring (or an algebra over a field F). Then R is a Lie ring (or Lie algebra)
v
6
under the Lie product [A,B] = AB −BA. Recall that a map Φ from A into itself is called a
3
3
commutativity preservingmap if [Φ(A),Φ(B)] = 0 whenever [A,B] = 0for all A, B ∈ R. The
6
0
problemofcharacterizing commutativity preservingmapshadbeenstudiedintensively (see[2,
.
1
0 3,9,13]andthereferencestherein). BellandDailgavetheconceptionofstrongcommutativity
6
1 preserving maps in [1]. A map Φ : R → R is said to be strong commutativity preserving
:
v if [Φ(A),Φ(B)] = [A,B] for any A,B ∈ R. Clearly, strong commutativity preserving maps
i
X
must be commutativity preserving maps, but the inverse is not true generally. The structure
r
a
of linear (or additive) strong commutativity preserving maps has been investigated in [1, 4,
5, 8]. For nonlinear strong commutativity preserving maps, with R being a prime unital ring
containing a nontrivial idempotent element, Qi and Hou in [12] proved that every nonlinear
surjective strong commutativity preserving map Φ :R → R has the form Φ(A)= λA+f(A)
forallA∈ R,whereλ ∈ {−1,1}andf isamapfromRintoZR,thecenterofR. Particularly,
this result is true for maps on factor von Neumann algebras. In [11] the nonlinear surjective
strong commutativity preserving maps on triangular algebras are studied. Recently, Liu in [7]
obtained that a surjective strong commutativity preserving map Φ :A → A on von Neumann
2010 Mathematics Subject Classification. 47B49; 47B47.
Key words and phrases. 3-commutators, standard operator algebras, Banach spaces, preservers.
This work is partially supported by Natural Science Foundation of China (11171249, 11271217).
1
2 MEIYUNLIU,JINCHUANHOU
algebras A without central summands of type I has the form Φ(A) = ZA + f(A) for all
1
A∈ A, where Z ∈ ZA satisfies Z2 = I and f is a map from A into ZA.
For a ring R and a positive integer k, recall that the k-commutator of elements A,B ∈ R
is defined by [A,B]k = [[A,B]k−1,B] with [A,B]0 = A and [A,B]1 = [A,B] = AB − BA;
a map Φ : R → R is said to be strong k-commutativity preserving if [Φ(A),Φ(B)] =
k
[A,B] for all A,B ∈ R. Obviously, strongk-commutativity preservingmaps are usualstrong
k
commutativity preserving maps if k = 1. It seems the study of the problem of characterizing
strong k-commutativity preserving maps was started by [10], in where it is shown that a
nonlinear surjective map on a unital prime ring containing a nontrivial idempotent is strong
2-commutativity preserving if and only if it has the form A 7→ λA + f(A), where λ is an
element in the extended centroid of the ring satisfying λ3 = 1 and f is a map from the
ring into its center. With k increasing, the problem of characterizing strong k-commutativity
preserving maps becomes much more difficult. Let R be a unital prime ring containing a
nontrivial idempotent. It is reasonable to conjecture that a surjective map Φ : R → R is
strong k-commutativity preserving if and only if there exist an element λ in the extended
centroid of R with λk+1 = 1 and a map f : R → ZR such that Φ(A) = λA+f(A) for all
A ∈ R. However, we even do not know whether or not the conjecture is true for k = 3. The
purpose of this paper is to answer the above conjecture affirmatively for the case when k = 3
and the maps act on standard operator algebras on Banach spaces.
Let F be the real field R or the complex field C and X be a Banach space over F. As usual,
denote by B(X) the Banach algebra of all bounded linear operators acting on X. Recall that,
a subalgebra A ⊆ B(X) is called a standard operator algebra if it contains the identity I and
all finite rank operators.
The following is our main results.
Theorem 1.1. Let X be a Banach space over the real or complex field F with dimX ≥ 2
and A ⊆ B(X) be a standard operator algebra. Assume that Φ : A → A is a surjective map.
Then Φ is strong 3-commutativity preserving if and only if there exist a functional h :A → F
and a scalar λ ∈ F with λ4 = 1 such that Φ(A)= λA+h(A)I for all A∈ A.
2. Proof of the main result
Before provingTheorem 1.1, wegive several lemmas. In this section we always assumethat
A is a standard operator algebra on a real or complex Banach space X and F stands for the
realfieldRorthecomplex fieldC,dependingonX is realorcomplex. For x ∈ X andf ∈ X∗,
STRONG 3-COMMUTATIVITY PRESERVING MAPS ON STANDARD OPERATOR ALGEBRAS 3
we denote x⊗f for the rank one operator defined by z 7→ hz,fix, where hz,fi = f(z). Note
that, every operator of rank ≤ 1 can be written in this form.
The first lemma is obvious by the main result in [6].
Lemma 2.1. Let A be a standard operator algebra, and let A , B , C , D ∈ A, i =
i i j j
1,2,...,n, such that n A TB = n C TD for all T ∈ A. If A ,...,A are linearly
i=1 i i i=1 j j 1 n
P P
independent, then each B is a linear combination of D ,...,D . Similarly, if B ,...,B
i 1 m 1 m
are linearly independent, then each A is a linear combination of C ,...,C . In particular,
i 1 m
if ATB = BTA for all T ∈ A, then A and B are linearly dependent.
Lemma 2.2. Let A be a standard operator algebra and A∈ A, if [A,P] = 0 for any rank
3
one idempotent P ∈ A, then A ∈ Z(A)= {λI : λ ∈ F}.
Proof. For any nonzero vector x ∈ X, there exists f ∈ X∗ such that hx,fi = 1. Let
P = x⊗f; then P is a rank one idempotent. By the assumption,
∗
0 = [A,P] = [A,x⊗f] = [A,x⊗f]= Ax⊗f −x⊗A f,
3 3
which implies that Ax is linearly dependent of x. Thus, for any x ∈ X, there exists some
scalar λ ∈ F such that Ax = λ x. It follows that there exists a scalar λ ∈ F such that
x x
A= λI. Hence A∈ Z(A) = {λI :λ ∈F}. (cid:3)
It was proved in [10] that, for k = 2, if an element S satisfies [A,S] = 0 for all A in a
k
prime ring, then S is a central element. But, for k ≥ 3, this result is not true anymore. For
k = 3, we have the following lemma, which reveals one of the main difficulties to solve the
problem of characterizing the maps preserving strong 3-commutativity.
Lemma 2.3. Let A be a standard operator algebra, N (A) = {N | N2 = 0,N ∈ A} and
2
S ∈ A. Then [A,S] = 0 holds for all A ∈ A if and only if there exist a scalar λ ∈ F and an
3
element N ∈ N (A) such that S = λI +N.
2
Proof. To check the “if” part, assume that S = λI+N with N2 =0. It is easily seen that
[A,S] = [A,λI +N] = [A,N] = AN3−3NAN2+3N2AN −N3A= 0
3 3 3
for any A∈ A.
Next we check the “ only if ” part.
Assume that
[A,S] = AS3−3SAS2+3S2AS −S3A = 0 (2.1)
3
for all A∈ A. By Lemma 2.1, we see that I, S, S2 and S3 are linearly dependent.
If I and S are linearly dependent, then S = λI for some λ ∈ F and the proof is done.
So we may suppose that I and S are linearly independent. We assert that I, S and S2 are
linearly dependent. In fact, if I, S and S2 are linearly independent, then, by Lemma 2.1, we
4 MEIYUNLIU,JINCHUANHOU
get S = λI, a contradiction. So there exist two scalars α,β ∈ F such that S2 = αI +βS.
Then
S3 = S(αI +βS) = αS +β(αI +βS) = αβI +(α+β2)S
and it follows from Eq.(2.1) that
0 = AS3−3SAS2 +3S2AS −S3A
= A(αβI +(α+β2)S)−3SA(αI +βS)+3(αI +βS)AS −(αβI +(α+β2)S)A
= αβA+(α+β2)AS −3αSA−3βSAS +3αI +3βSAS −αβA−(α+β2)SA
= (4α+β2)(AS −SA).
Since there always exists A∈ A such that AS −SA6= 0, we must have 4α+β2 = 0.
Let λ = β, N = S −λI, then λ2 = −α, and
2
N2 = S2−2λS +λ2I = S2−βS −αI = 0.
So S = λI +N with N ∈ N (A). The proof of the Lemma 2.3 is completed. (cid:3)
2
Now we are at the position to give our proof of the main theorem.
Proof of Theorem 1.1. The “ if ” part of the theorem is obvious. In the sequel, we
always assume that Φ : A → A is a surjective strong 3-commutativity preserving map. we
will check the “ only if ” part by several steps.
Step 1. For any A,B ∈ A, there exists a scalar λ ∈ F such that Φ(A+B) = Φ(A)+
A,B
Φ(B)+λ I.
A,B
For any A,B and T ∈ A, we have
[Φ(A+B)−Φ(A)−Φ(B),Φ(T)]
3
= [Φ(A+B),Φ(T)] −[Φ(A),Φ(T)] −[Φ(B),Φ(T)]
3 3 3
= [A+B,T] −[A,T] −[B,T] = 0.
3 3 3
By the surjectivity of Φ and Lemma 2.2, above equation implies Φ(A+B)−Φ(A)−Φ(B)∈
{λI : λ ∈ F}. So the assertion in Step 1 is true.
Step 2. Φ(FI)= FI and Φ(FI +N (A))= FI +N (A).
2 2
Φ(FI) = FI is obvious by Lemma 2.2 and the surjectivity of Φ.
If N ∈ N (A) and λ ∈ F, then, for any A ∈A, we have
2
[Φ(A),Φ(λI +N)] = [A,λI +N] = [A,N] = 0.
3 3 3
By the surjectivity of Φ and Lemma 2.3, we see that Φ(λI +N)∈ FI +N (A).
2
On the other hand, if Φ(B)= λI +N for some scalar λ and N ∈ N (A), then
2
[A,B] = [Φ(A),Φ(B)] = [Φ(A),λI +N] = [Φ(A),N] = 0
3 3 3 3
STRONG 3-COMMUTATIVITY PRESERVING MAPS ON STANDARD OPERATOR ALGEBRAS 5
for all A∈ A. By Lemma 2.3 again, we see that B ∈ FI +N (A).
2
Hence Φ(FI +N (A))= FI +N (A), completing the proof of the Step 2.
2 2
Step 3. For any nontrivial idempotent P ∈ A, there exist three scalars λP, µP,µI−P ∈ F
and an element N ∈ N (A) such that Φ(P) = λ P +µ I +N and Φ(I −P) = λ (I −
P 2 P P P P
P)+µI−PI −NP.
Let P ∈ A be a nontrivial idempotent and write P = P and P = I − P. Then A
1 2
can be written as A = A + A + A + A , where A = P AP (i,j ∈ {1,2}). Write
11 12 21 22 ij i j
Φ(P)= S +S +S +S with S ∈ A .
11 12 21 22 ij ij
We check Step 3 by several claims.
Claim 3.1. For any A∈ A, we have [A,Φ(P)] ∈A +A .
3 12 21
Note that, for any A∈ A, we have
[A,P] = PA(I −P)+(I −P)AP ∈ A +A .
2 12 21
Also notice that [A,Q] = [A,Q] holds for any A,Q ∈ A with Q an idempotent. So we have
3
[A,P] = [A,P] = [[A,P] ,P] , which implies that
3 5 3 2
[Φ(A),Φ(P)] = [[Φ(A),Φ(P)] ,P]
3 3 2
holds for any A ∈ A. By the surjectivity of Φ one gets
[A,Φ(P)] = [[A,Φ(P)] ,P] for all A∈ A.
3 3 2
Therefore, for any A we have
[A,Φ(P)] ∈ A +A .
3 12 21
The next claim is one of the key steps for our proof.
Claim 3.2. S = S = 0.
12 21
Taking A = A ∈ A and applying Claim 3.1, we have
11 11
0 = (I −P)[A ,Φ(P)] (I −P)
11 3
= 3S S A S +3S S A S −3S A S S −3S A S S ,
21 11 11 12 22 21 11 12 21 11 11 12 21 11 12 22
which implies that
(S S +S S )A S = S A (S S +S S )
21 11 22 21 11 12 21 11 11 12 12 22
holds for all A ∈ A , that is,
11 11
(S S +S S )PAPS = S PAP(S S +S S ) (2.2)
21 11 22 21 12 21 11 12 12 22
6 MEIYUNLIU,JINCHUANHOU
holdsforallA∈ A. Thus,byLemma2.1, S S +S S andS mustbelinearlydependent.
21 11 22 21 21
It follows that
S S +S S = λ S (2.3)
21 11 22 21 1 21
for some λ ∈ F. Again by Eq.(2.2), we get
1
S S +S S = λ S . (2.4)
11 12 12 22 1 12
Taking A= A ∈ A and applying Claim 3.1, one gets
12 12
0 =P[A ,Φ(P)] P
12 3
=A S S2 +A S S S +A S S S +A S S S
12 21 11 12 21 12 21 12 22 21 11 12 22 21 12
−3S A S S −3S A S S +3S2 A S +3S S A S
11 12 21 11 11 12 22 21 11 12 21 12 21 12 21
and
0 =(I −P)[A ,Φ(P)] (I −P)
12 3
=−S S2 A −S S21S A −S S S A −S2 S A
21 11 12 22 11 12 21 12 21 12 22 21 12
−3S A S S −3S A S2 +3S S 1A S +3S S A S .
21 12 21 12 21 12 22 21 1 12 22 22 21 12 22
Thus
A [S (S2 +S S )+S (S S +S S )]
12 21 11 12 21 22 21 11 22 21
(2.5)
= 3S A (S S +S S )−3(S2 +S S )A S
11 12 21 11 22 21 11 12 21 12 21
and
[S (S2 +S S )+S (S S +S S )]A
21 11 12 21 22 21 11 22 21 12
(2.6)
= 3S A (S2 +S S )−3(S S +S S )A S
21 12 22 21 12 21 11 22 21 12 22
hold for all A ∈ A .
12 12
Similarly, by taking A = A and applying Claim 3.1, one gets
21
0 =P[A ,Φ(P )] P
21 1 3
=S2 S A +S S S A +S S222A +S S S A
11 12 21 11 12 22 21 12 21 12 21 12 21
−3S S A S −3S S A S +3S A S2 +3S A S S
11 12 21 11 12 22 21 11 12 21 11 12 21 12 21
and
0=(I −P)[A ,Φ(P)] (I −P)
21 3
=A S2 S +A S S S +A S S2 +A S S S
21 11 12 21 11 12 22 21 12 22 21 12 21 12
−3S A S S −3S A S S +3S2 A S +3S S )A S ,
22 21 11 12 22 21 12 22 22 21 12 21 12 21 12
STRONG 3-COMMUTATIVITY PRESERVING MAPS ON STANDARD OPERATOR ALGEBRAS 7
which imply that
[S (S S +S S )+S (S2 +S S )]A
11 11 12 12 22 12 22 21 12 21
(2.7)
= 3(S S +S S )A S −3S A (S2 +S S )
11 12 12 22 21 11 12 21 11 12 21
and
A [S (S S +S S )+S (S2 +S S )]
21 11 11 12 12 22 12 22 21 12
(2.8)
= 3S A (S S +S S )−3(S2 +S S )A S
22 21 11 12 12 22 22 21 12 21 12
hold for all A ∈ A . Obviously, Eqs.(2.3) ∼ (2.8) together are equivalent to say that
21 21
PA[S (S2 +S S )+S (S S +S S )] = [3λ S −3(S2 +S S )]AS ,
21 11 12 21 22 21 11 22 21 1 11 11 12 21 21



[S21(S121+S12S21)+S22(S21S11+S22S21)]A(I −P)= S21A[3λ1S22−3(S222+S21S12)],



[S11(S11S12+S12S22)+S12(S222+S21S12)]AP = S12AP[3λ1S11−3(S121+S12S21)],
(I −P)A[S11(S11S12+S12S22)+S12(S222+S21S12)]




 = [3λ1S22−3(S222 +S21S12)](I −P)AS12



hold for all A ∈A.
Thus, by Lemma 2.1, we see that 3λ S −3(S2 +S S ) and P are linearly dependent.
1 11 11 12 21
So
3λ S −3(S2 +S S )= λ P (2.9)
1 11 11 12 21 2
for some λ ∈ F. This entails that
2
[S (S2 +S S )+S (S S +S S )]= λ S , (2.10)
21 11 12 21 22 21 11 22 21 2 21
[S (S S +S S )+S (S2 +S S )] = λ S (2.11)
11 11 12 12 22 12 22 21 12 2 12
and
[3λ S −3(S2 +S S )] = λ (I −P). (2.12)
1 22 22 21 12 2
It is easily checked by Eqs.(2.9) ∼ (2.12) that
4λ 4λ
(λ2− 2)S = 0 and (λ2− 2)S =0.
1 3 21 1 3 12
If λ2− 4λ2 = 0, then we have
1 3
(Φ(P)− λ1I)2
2
= (S +S +S +S − λ1I)2
11 12 21 22 2
= (S2 +S S )+(S S +S S )+(S S +S S )
11 12 21 11 12 12 22 21 11 22 21
+(S2 +S S )−λ S −λ S −λ S −λ S + λ21I .
22 21 12 1 11 1 12 1 21 1 22 4
= (λ S − λ2P)+λ S +λ S +(λ S − λ2(I −P))−λ S
1 11 3 1 12 1 21 1 22 3 1 11
λ2
−λ S −λ S −λ S + 1I
1 12 1 21 1 22 4
= (λ21 − λ2)I = 0.
4 3
8 MEIYUNLIU,JINCHUANHOU
Let N = Φ(P)− λ1I; then Φ(P) = λ1I +N with N2 = 0. But, by Step 2, this implies
P 2 2 P P
P = λI + N for some λ ∈ F and some N ∈ N (A), a contradiction. So we must have
2
λ2− 4λ2 6= 0. Hence S = S = 0 and Φ(P)= S +S .
1 3 12 21 11 22
Next, let us determine S and S .
11 22
Taking A = A and applying Claim 3.1, we have
11
[A ,S ] = A S3 −3S A S2 +3S2 A S −S3 A = P[A ,Φ(P)] P = 0
11 11 3 11 11 11 11 11 11 11 11 11 11 11 3
holds for all A ∈ A . As A is clearly a standard operator algebra in B(PX), by Lemma
11 11 11
2.3, S = λ P + N with N2 = 0 for some λ ∈ F and some N ∈ N (A ). A similar
11 1 1 1 1 1 2 11
argument can show that S = µ (I −P)+N for some µ ∈ F and some N ∈ N (A ).
22 P 2 P 2 2 22
Hence
Φ(P) = λ P +N +µ (I −P)+N = λ P +µ I +N ,
1 1 P 2 P P P
where λ =λ −µ with N = N +N , N2 = 0 and N N = N N = 0.
P 1 P P 1 2 P 1 2 2 1
Similarly, we can get Φ(I −P) = λI−P(I −P)+µI−PI +NI−P for some λI−P,µI−P ∈ F
and NI−P ∈ N2(A).
Claim 3.3. λP = λ(I−P) and NI−P = −NP.
By Step 1 and Claim 3.2, there exists some scalar c∈ F, such that
Φ(I) = Φ(P)+Φ(I −P)+cI
= λPP +µPI +NP +λI−P(I −P)+µI−PI +NI−P +cI (2.13)
= (λP −λI−P)P +(µP +µI−P +c)I +NP +NI−P.
Since Φ(I) ∈ FI by Step 2, there exist two scalars α,t ∈F, with t = λP −λI−P such that
tP +NP +NI−P = αI. (2.14)
So
A(tP +NP +NI−P)= (tP +NP +NI−P)A for all A ∈ A.
In particular, taking A= N gives
P
tNPP +NPNI−P = tPNP +NI−PNP. (2.15)
Recall that N = N +N with N ∈ N (A ), i = 1,2. So tN P = tPN = tN and hence,
P 1 2 i 2 ii P P 1
by Eq.(2.15), we have NPNI−P = NI−PNP. Then, it follows from Eq.(2.14) that
(tP −αI)2 = (NP +NI−P)2 = 2NPNI−P. (2.16)
Note that (NPNI−P)2 = 0 as NPNI−P = NI−PNP. Thus, Eq.(2.16) entails that t = α = 0,
which, by Eq.(2.14), implies that λI−P = λP and NI−P = −NP, as desired.
STRONG 3-COMMUTATIVITY PRESERVING MAPS ON STANDARD OPERATOR ALGEBRAS 9
Step 4. For any idempotent P ∈ A of rank ≤ 2, we have N = 0; For any rank-1
P
idempotents P,Q with PQ = QP = 0, we have λ = λ .
P Q
Assume that P is a rank-1 idempotent. By Step 3, there exist two scalars λ ,µ ∈ F and
P P
anelement N ∈ N (A), suchthatΦ(P)= λ P+µ I+N . Takingthespacedecomposition
P 2 P P P
1 0
 
X = X +˙X so that P has the matrix representation P = . Since N = N +N
1 2 P 1 2
0 0
 
0 0
 
withN ∈ PAP andN ∈ (I−P)A(I−P),wemusthaveN = 0andhenceN =
1 2 1 P
0 S
 
with S2 = 0.
Assume S 6= 0, then there exists rank-1 idempotent P′ ∈ B((I−P)X), such that P′S 6= 0.
2 2
Let P = 0 ⊕ P′, then PP = P P and with respect to the space decomposition X =
2 2 2 2
PX+˙P X+˙(I −P)(I −P )X, we have
2 2
1 0 0 0 0 0
   
P =  0 0 0  and P2 =  0 1 0 .
   
   
0 0 0 0 0 0
   
Let
1 0 0
 
Q = P +P2 =  0 1 0 .
 
 
0 0 0
 
Obviously, Q2 = Q. By Step 1 and Step 3, there exist scalars c,λ ,λ , λ , µ , µ , µ ∈ F
P P2 Q Q P P2
and elements N ,N ,N ∈ N (A) with
Q P P2 2
t t 0 0 0 0 r 0 R
 11 12     11 13 
NQ =  t21 t22 0 , NP =  0 s22 S23  and NP2 =  0 0 0 
     
 0 0 T   0 S S   R 0 R 
   32 33   31 33 
such that
λ Q+µ I +N = Φ(Q) = Φ(P )+Φ(P )+cI
Q Q Q 1 2
= λ P +µ I +N +λ P +µ I +N .
P P P P2 2 P2 P2
10 MEIYUNLIU,JINCHUANHOU
So we have
λ +µ 0 0 t t 0
 Q Q   11 12 
0 λ +µ 0 + t t 0
 Q Q   21 22 
   
 0 0 µ   0 0 T 
 Q   
λ +µ +µ +c 0 0 r 0 R
 P P P2   11 13 
= 0 λ +µ +µ +c 0 + 0 s S ,
 P2 P P2   22 23 
   
 0 0 µ +µ +c   R S S +R 
 P P2   31 32 33 33 
whichentails thatt =t = 0, R = 0, R = 0, S = 0andS = 0. SinceN2 = 0, weget
12 21 13 31 23 32 Q
also t = t = 0. Then, as N2 = 0 and N2 = 0, one gets further that s = r = 0. So N
11 22 P P2 22 11 P
0 0 0
 
0 0 1 0
  ′  
has the form NP =  0 0 0 ; that is S = . However, as P2 = ,
  0 S 0 0
 0 0 S   33   
 33 
′ ′
we get a contradiction P S = SP = 0. Hence we must have N = 0. Moreover,
2 2 P
λ +µ +µ +c = λ +µ =λ +µ +µ +c.
P P P2 Q Q P2 P P2
This forces that λ = λ . It follows that T = 0; that is N = 0, too.
P2 P Q
Note that, for the case when dimX ≥ 3, we also have µ = µ + µ + c and hence
Q P P2
λ = λ .
Q P
If P is an idempotent of rank 2, taking rank-1 idempotent P ,P so that P = P +P , then
1 2 1 2
the above argument shows that N = 0. So, the assertion of Step 4 is true.
P
Step 5. Assume dimX ≥ 3. λ4 = 1 and there exists a functional h : A → F such that
Φ(A)= λA+h(A)I for any A∈ A.
We first show that, if dimX ≥ 3, then there exists a scalar λ ∈ F such that, for any
rank-1 idempotent P ∈ A, Φ(P) = λP + µ I for some µ ∈ F. In fact, for any rank
P P
one idempotent operators P,Q, as dimX ≥ 3, there exists rank-1 idempotent R such that
PR = RP = QR = RQ = 0. It follows from Step 4 that λ = λ = λ . Hence, there exists
P R Q
a scalar λ such that λ = λ for any rank-1 idempotent P.
P
Thus, for any rank-1 idempotents P, Q ∈A, we have
[P,Q] = [Φ(P),Φ(Q)] = [λP,λQ] = λ4[P,Q] .
3 3 3 3
It follows that λ4 = 1 as there are rank-1 idempotents P,Q so that [P,Q] 6= 0.
3
Now, for any A and any rank-1 idempotent P in A, we have
[A,P] = [Φ(A),Φ(P)] = [Φ(A),λP] = λ3[Φ(A),P] ,
3 3 3 3