# Lie Triple Derivations on𝒥-Subspace Lattice Algebras

Authors: Ting Wang; Fangyan Lu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We describe the structure of Lie triple derivations on𝒥-subspace lattice algebras. The results can be applied to atomic Boolean subspace lattice algebras and pentagon subspace lattice algebras, respectively.


---

## Body

## 1. Introduction and Preliminaries

Let𝒜 be an associative algebra, and let ℳ be an 𝒜-bimodule. We denote by Z
(
ℳ
,
𝒜
) the center of ℳ relative to 𝒜; that is, Z
(
ℳ
,
𝒜
)
=
{
M
∈
ℳ
:
A
M
=
M
Afor allA
∈
𝒜
}. A linear mapping δ
:
𝒜
→
ℳ is called a 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, where [
A
,
B
]
=
A
B
-
B
A is the usual Lie product. We say that a Lie triple derivation δ is standard if it can be decomposed as a sum of a derivation from 𝒜  to  ℳ and a mapping from 𝒜  to  Z
(
ℳ
,
𝒜
) vanishing on every double commutator. The standard problem, which has been studied for many years, is to find conditions on 𝒜 under which each Lie triple derivation is standard or standard-like. This problem has been investigated for von Neumann algebras in [1], for prime rings in [2], for nest algebras in [3, 4], for TUHF algebras in [5], and for upper triangular algebras in [6]. In this present note, we pursue this line of investigation for 𝒥-subspace lattice algebras.Throughout, all algebras and vector spaces will be over𝔽, where 𝔽 is either the real field ℝ or the complex field ℂ. Given a Banach space X with topological dual X
*, by B
(
X
) we mean the algebra of all bounded linear operators on X. The terms operator on X and subspace of X will mean “bounded linear map of X into itself” and “norm closed linear manifold of X,” respectively. For A
∈
B
(
X
), denote by A
* the adjoint of A. For any nonempty subset L
⊆
X, L
⊥ denotes its annihilator; that is, L
⊥
=
{
f
∈
X
*
:
f
(
x
)
=
0 for all x
∈
L
}. For x
∈
X and f
∈
X
*, the rank one operator x
⊗
f is defined by (
x
⊗
f
)
z
=
f
(
z
)
x for z
∈
X.A familyℒ of subspaces of X is called a subspace lattice on X if it contains (0) and X and is complete in the sense that it is closed under the operations ∨ (closed linear span) and ∧ (set-theoretic intersection). For a subspace lattice ℒ on X, the associated subspace lattice algebra Alg
ℒ is the set of operators on X leaving every subspace in ℒ invariant; that is
(1)
Alg
ℒ
=
{
A
∈
B
(
X
)
:
A
x
∈
L
for
every
x
∈
L
and
for
every
L
∈
ℒ
}
.
Obviously, Alg
ℒ is a unital weakly closed operator algebra. A subalgebra of Alg
ℒ is called a standard subalgebra if it contains all finite-rank operators in Alg
ℒ.Subspace lattice algebras are important and mainly consist of nonself-adjoint operator algebras. Completely distributive subspace lattice algebras, commutative subspace lattice algebras, atomic Boolean subspace lattice algebras, pentagon subspace lattice algebras, and so forth have been widely studied. See, for example, [7–11]. Recently, Panaia in [12] introduced a new class of subspace lattices-𝒥-subspace lattice algebras. Several authors have studied 𝒥-subspace lattice as well as 𝒥-subspace lattice algebras; see, for example, [13–18].Given a subspace latticeℒ on Banach space X, put
(2)
𝒥
(
ℒ
)
=
{
L
∈
ℒ
:
L
≠
(
0
)
,
L
-
≠
X
}
,
where K
-
=
∨
{
M
∈
ℒ
:
M
⊉
K
}. Call ℒ a 𝒥-subspace lattice on X if(1)
∨
{
K
:
K
∈
𝒥
(
ℒ
)
}
=
X,
(2)
∧
{
K
-
:
K
∈
𝒥
(
ℒ
)
}
=
(
0
),
(3)
K
∨
K
-
=
X,  for every  K
∈
𝒥
(
ℒ
),
(4)
K
∧
K
-
=
(
0
),for everyK
∈
𝒥
(
ℒ
).The simplest example of𝒥-subspace lattice is any pentagon subspace lattice 𝒫
=
{
(
0
)
,
K
,
L
,
M
,
X
}. Here K
,
L, and M are subspaces of X satisfying K
∨
L
=
X, K
∧
M
=
(
0
) and L
⊂
M. In this case, K
-
=
M, L
-
=
K, and 𝒥
(
𝒫
)
=
{
K
,
L
}. For further discussion of pentagon subspace lattice see [8, 10]. Another important element of the class of 𝒥-subspace lattice is the atomic Boolean subspace lattice. It follows from [15] that every commutative 𝒥-subspace lattice on a Hilbert space is an atomic Boolean subspace lattice. However, most 𝒥-subspace lattices on Hilbert space are non-commutative.Therefore,𝒥-subspace lattices as well as 𝒥-subspace lattice algebras deserve some attention. In the previous papers [3, 17, 18], we studied algebraic isomorphisms, Jordan isomorphisms, Jordan derivations, and Lie derivations. Here we study Lie triple derivations of 𝒥-subspace lattice algebras. Even for pentagon subspace lattice algebras and atomic Boolean subspace lattice algebras, our results are new.For a subspace latticeℒ, the relevance of 𝒥
(
ℒ
) is due to the following lemma, which is crucial to what follows.Lemma 1 (see [11]).
Ifℒ is a subspace lattice on X, then the rank-one operator x
⊗
f belongs to Alg
ℒ if and only if there exists a subspace K in 𝒥
(
ℒ
) such that x
∈
K and f
∈
K
-
⊥, where K
-
⊥ means (
K
-
)
⊥.From Lemma1, we can see that if ℒ is a 𝒥-subspace lattice, then Alg
ℒ is rich in rank-one operators. Moreover, finite-rank operators in a 𝒥-subspace lattice algebra have nice properties. Given a subspace lattice ℒ, by ℱ
(
ℒ
) we denote the algebra of all finite-rank operators in Alg
ℒ. If K
∈
𝒥
(
ℒ
), then we write ℱ
1
(
K
)
=
{
x
⊗
f
:
x
∈
K
,
f
∈
K
-
⊥
} and ℱ
(
K
)
=
〈
ℱ
1
(
K
)
〉, the linear manifold spanned by ℱ
1
(
K
).Lemma 2 (see [17]).
Letℒ be a 𝒥-subspace lattice on X. Suppose that A is an operator of rank n in ℱ
(
ℒ
). Then A can be written as a sum of n rank-one operators in A
l
g
ℒ.Recalling that a linear mappingδ of an algebra 𝒜 is a local derivation if for every A
∈
𝒜 there is a derivation d
A, depending on A, such that δ
(
A
)
=
d
A
(
A
).Lemma 3 (see [3]).
Letℒ be a 𝒥-subspace lattice on X. Then every local derivation from ℱ
(
ℒ
) to a standard subalgebra of A
l
g ℒ is a derivation.Lemma 4 (see [3]).
Letℒ be a 𝒥-subspace lattice on X. Suppose that 𝒜 is in ℱ
(
ℒ
), S in 𝒜, and C in Z
(
𝒜
). If [
A
,
S
]
=
C, then C
=
0.


## 2. Lie Triple Derivation onℱ
(
ℒ
)

The main result in this section reads as follows.Theorem 5.
Letℒ be a 𝒥-subspace lattice on X and 𝒜 a standard subalgebra of A
l
g
ℒ. Let δ be a Lie triple derivation from ℱ
(
ℒ
) to 𝒜. Then δ is standard.For the proof of the theorem, we need some lemmas. In the following, we keep the notation as in the statement of the theorem. Recalling that the statement means thatδ is the sum of a derivation from ℱ
(
ℒ
) to 𝒜 and a linear mapping from ℱ
(
ℒ
) to Z
(
𝒜
,
ℱ
(
ℒ
)
) vanishing on every double commutator. Here Z
(
𝒜
,
ℱ
(
ℒ
)
)
=
{
A
∈
𝒜
:
A
B
=
B
Afor allB
∈
ℱ
(
ℒ
)
}. From [3, Remark 2.5(i)], we know that Z
(
𝒜
,
ℱ
(
ℒ
)
) is equal to Z
(
𝒜
), the center of 𝒜.Lemma 6.
LetK
∈
𝒥
(
ℒ
) and P be an idempotent operator in ℱ
(
K
). Then there are an operator S in ℱ
(
K
) and a unique operator τ
(
P
) in Z
(
𝒜
) such that δ
(
P
)
=
[
P
,
S
]
+
τ
(
P
).Proof.
SetP
1
=
P and P
2
=
I
-
P. Note that 𝒜 does not necessarily contain I; we understand P
2
B
=
B
-
P
1
B and B
P
2
=
B
-
B
P
1 for B
∈
𝒜.
ForA
11
∈
P
1
ℱ
(
ℒ
)
P
1, we have that
(3)
0
=
δ
(
[
[
P
1
,
A
11
]
,
B
]
)
=
[
[
δ
(
P
1
)
,
A
11
]
,
B
]
+
[
[
P
1
,
δ
(
A
11
)
]
,
B
]
.
Let C
=
[
δ
(
P
1
)
,
A
11
]
+
[
P
1
,
δ
(
A
11
)
]. Since B is arbitrary, we have C
∈
Z
(
𝒜
). Then
(4)
δ
(
P
1
)
A
11
-
A
11
δ
(
P
1
)
+
P
1
δ
(
A
11
)
-
δ
(
A
11
)
P
1
=
C
.
Multiplying this equation by P
1 from both sides, we get
(5)
P
1
δ
(
P
1
)
P
1
A
11
-
A
11
P
1
δ
(
P
1
)
P
1
=
C
P
1
;
that is,
(6)
[
P
1
δ
(
P
1
)
P
1
,
A
11
]
=
C
P
1
.
It follows from the Kleinecke-Shirokov theorem (cf. [19, Problem 232]) that C
P
1 is quasinilpotent. Let L
∈
𝒥
(
ℒ
). Then there exists a scalar η
L
∈
𝔽, such that C
x
=
η
L
x for all x
∈
L. Particularly, C
P
1
x
=
η
L
P
1
x; that is, (
η
L
I
-
C
P
1
)
P
1
x
=
0. This implies that either η
L
=
0 or P
1
x
=
0 since C
P
1 is quasinilpotent. Consequently, we always have C
P
1
x
=
0 for all x
∈
L. Hence C
P
1
=
0 since X
=
∨
{
L
:
L
∈
𝒥
(
ℒ
)
}. Then by (5) we have that
(7)
P
1
δ
(
P
1
)
P
1
A
11
=
A
11
P
1
δ
(
P
1
)
P
1
.
Similarly, forA
22
∈
P
2
ℱ
(
ℒ
)
P
2, by considering [
P
1
,
A
22
], we have that
(8)
P
2
δ
(
P
1
)
P
2
A
22
=
A
22
P
2
δ
(
P
1
)
P
2
.
Now forA
12
∈
P
1
ℱ
(
ℒ
)
P
2, we have that
(9)
δ
(
A
12
)
=
δ
(
[
[
A
12
,
P
1
]
,
P
1
]
)
=
[
[
δ
(
A
12
)
,
P
1
]
,
P
1
]
+
[
[
A
12
,
δ
(
P
1
)
]
,
P
1
]
+
[
[
A
12
,
P
1
]
,
δ
(
P
1
)
]
=
P
1
δ
(
A
12
)
P
2
+
P
2
δ
(
A
12
)
P
1
+
A
12
δ
(
P
1
)
P
1
+
P
1
δ
(
P
1
)
A
12
-
A
12
δ
(
P
1
)
+
δ
(
P
1
)
A
12
.
Multiplying this equation by P
1 from the left side and by P
2 from the right side, we get
(10)
P
1
δ
(
P
1
)
P
1
A
12
=
A
12
P
2
δ
(
P
1
)
P
2
.
Similarly, for A
21
∈
P
2
ℱ
(
ℒ
)
P
1, we have that
(11)
P
2
δ
(
P
1
)
P
2
A
21
=
A
21
P
1
δ
(
P
1
)
P
1
.
Using (7)–(11), it is easy to verify that τ
(
P
)
=
P
1
δ
(
P
1
)
P
1
+
P
2
δ
(
P
1
)
P
2
∈
Z
(
𝒜
). Now let S
=
P
1
δ
(
P
1
)
P
2
-
P
2
δ
(
P
1
)
P
1. Then S
∈
ℱ
(
K
) and δ
(
P
1
)
=
[
P
1
,
S
]
+
τ
(
P
1
). Moreover, by Lemma 4, such τ
(
P
1
) is unique.Lemma 7.
LetA
=
x
⊗
f be in ℱ
(
K
) with K
∈
𝒥
(
ℒ
). Then there are an operator S in ℱ
(
K
) and a unique operator τ
(
A
) in Z
(
𝒜
) such that δ
(
A
)
=
[
A
,
S
]
+
τ
(
A
).Proof.
Iff
(
x
)
≠
0, then the result follows from the linearity and Lemma 6.
We now suppose thatf
(
x
)
=
0. Take y
∈
K such that f
(
y
)
=
1. Obviously, x and y are linearly independent. Set P
=
y
⊗
f. Then by Lemma 6, δ
(
P
)
=
[
P
,
S
1
]
+
τ
(
P
), where S
1
∈
ℱ
(
K
) and τ
(
P
)
∈
Z
(
𝒜
). We associate a new Lie triple derivation as follows:
(12)
Δ
(
T
)
=
δ
(
T
)
-
[
T
,
S
1
]
,
T
∈
ℱ
(
ℒ
)
.
Then Δ
(
P
)
=
τ
(
P
)
∈
Z
(
𝒜
).
LetP
1
=
P and P
2
=
I
-
P. Then
(13)
Δ
(
A
)
=
Δ
(
[
[
A
,
P
1
]
,
P
1
]
)
=
[
[
Δ
(
A
)
,
P
1
]
,
P
1
]
=
P
1
Δ
(
A
)
P
2
+
P
2
Δ
(
A
)
P
1
.
We will show P
1
Δ
(
A
)
P
2
=
0. For this, we first observe that P
1
Δ
(
A
)
P
2
=
y
⊗
h for some h
∈
K
-
⊥. Since
(14)
[
P
1
Δ
(
A
)
P
2
+
P
2
Δ
(
A
)
P
1
,
A
]
=
[
Δ
(
A
)
,
A
]
=
[
Δ
(
A
)
,
[
A
,
P
1
]
]
=
Δ
(
[
A
,
[
A
,
P
1
]
]
)
-
[
A
,
[
Δ
(
A
)
,
P
1
]
]
-
[
A
,
[
A
,
Δ
(
P
1
)
]
]
=
-
[
A
,
[
Δ
(
A
)
,
P
1
]
]
=
-
[
A
,
P
2
Δ
(
A
)
P
1
-
P
1
Δ
(
A
)
P
2
]
=
[
P
2
Δ
(
A
)
P
1
-
P
1
Δ
(
A
)
P
2
,
A
]
,
it follows that [
P
1
Δ
(
A
)
P
2
,
A
]
=
0; that is, h
(
x
)
y
⊗
f
-
x
⊗
h
=
0. Hence h
=
0 since x and y are linearly independent. So P
1
Δ
(
A
)
P
2
=
0.
Now by (13), we have Δ
(
A
)
=
P
2
Δ
(
A
)
P
1. Therefore there exists z
∈
K such that Δ
(
A
)
=
z
⊗
f and f
(
z
)
=
0. Choose g
∈
K
-
⊥ such that g
(
x
)
=
1. Then
(15)
Δ
(
A
)
=
z
⊗
f
=
(
z
⊗
g
)
(
x
⊗
f
)
-
(
x
⊗
f
)
(
z
⊗
g
)
.
Let S
2
=
z
⊗
g. Then Δ
(
A
)
=
[
A
,
-
S
2
], and so δ
(
A
)
=
[
A
,
S
1
-
S
2
]. Thus S
=
S
1
-
S
2 and τ
(
A
)
=
0 are desired.Lemma 8.
SupposeA
=
∑
k
=
1
n
‍
x
k
⊗
f
k
∈
ℱ
(
K
), with K
∈
𝒥
(
ℒ
). Then there are an operator S in ℱ
(
K
) and a unique operator τ
(
A
) in Z
(
𝒜
) such that δ
(
A
)
=
[
A
,
S
]
+
τ
(
A
). Moreover, τ
(
A
)
=
∑
k
=
1
n
‍
τ
(
x
k
⊗
f
k
).Proof.
By [3, Proposition 2.6], there is a matrix unit {
y
i
⊗
g
j
}
i
,
j
=
1
m such that each x
k
⊗
f
k belongs to the algebra
(16)
𝒟
=
{
C
∈
ℱ
(
K
)
:
C
=
∑
i
,
j
=
1
m
‍
λ
i
j
y
i
⊗
g
j
,
λ
i
j
∈
𝔽
}
.
Obviously, 𝒟 is a finite-dimensional Banach algebra which is isomorphic M
m
(
𝔽
) via C
→
[
λ
i
j
]
m
×
m. By [1], δ
|
𝒟 is standard, that is, δ
|
𝒟
=
d
+
h, where d is a derivation from 𝒟 to 𝒜 and h is a linear mapping from 𝒟 to Z
(
𝒜
,
𝒟
) vanishing on every double commutator. By [7, Lemma 10.7], there is an operator T in 𝒜 such that d
(
D
)
=
[
D
,
T
] for all D
∈
𝒟. Consequently, d is a derivation from 𝒟 to ℱ
(
K
). By [7] again, there is an operator S in ℱ
(
K
) such that d
(
D
)
=
[
D
,
S
] for all D
∈
𝒟. Thus for each k, it follows that
(17)
δ
(
x
k
⊗
f
k
)
=
[
x
k
⊗
f
k
,
S
]
+
h
(
x
k
⊗
f
k
)
.
On the other hand, by Lemma7, for each k, there is an operator S
k in ℱ
(
K
) such that
(18)
δ
(
x
k
⊗
f
k
)
=
[
x
k
⊗
f
k
,
S
k
]
+
τ
(
x
k
⊗
f
k
)
.
By (17) and (18), we get
(19)
[
x
k
⊗
f
k
,
T
k
]
=
C
k
,
where T
k
=
S
-
S
k
,
C
k
=
τ
(
x
k
⊗
f
k
)
-
h
(
x
k
⊗
f
k
). Since C
k commutes with x
k
⊗
f
k, it follows from Kleinecke-Shirokov theorem that C
k is quasinilpotent. Moreover, C
k
y
i
=
λ
i
y
i for some λ
i
∈
𝔽 and C
k
y
i
⊗
g
i is quasinilpotent since C
k commutes with y
i
⊗
g
i. This implies that C
k
y
i
=
0
,
i
=
1
,
…
,
m.
Sincex
k
⊗
f
k
=
(
∑
i
=
1
m
‍
y
i
⊗
g
i
)
(
x
k
⊗
f
k
)
(
∑
i
=
1
m
‍
y
i
⊗
g
i
), there are i
,
j such that f
k
(
y
i
)
≠
0, g
j
(
x
k
)
≠
0. Applying (19) to y
i, we get f
k
(
T
k
y
i
)
x
k
-
f
k
(
y
i
)
T
k
x
k
=
C
k
y
i
=
0. So C
k
=
x
k
⊗
g, where g
=
T
k
*
f
k
-
(
f
k
(
T
k
y
i
)
/
f
k
(
y
i
)
)
f
k. Thus g
j
(
x
k
)
y
i
⊗
g
=
(
y
i
⊗
g
j
)
(
x
k
⊗
g
)
=
C
k
(
y
i
⊗
g
j
)
=
0. So g
=
0, and therefore C
k
=
0 and [
x
k
⊗
f
k
,
T
k
]
=
0. Consequently, [
x
k
⊗
f
k
,
S
k
]
=
[
x
k
⊗
f
k
,
S
] and τ
(
x
k
⊗
f
k
)
=
h
(
x
k
⊗
f
k
)
,
k
=
1
,
…
,
n. Thus
(20)
δ
(
A
)
=
∑
k
=
1
n
‍
(
[
x
k
⊗
f
k
,
S
]
+
τ
(
x
k
⊗
f
k
)
)
=
[
A
,
S
]
+
∑
k
=
1
n
‍
τ
(
x
k
⊗
f
k
)
.
By Lemma 4, the sum ∑
k
=
1
n
‍
τ
(
x
k
⊗
f
k
) is independent of the representation of A. Let τ
(
A
)
=
∑
k
=
1
n
‍
τ
(
x
k
⊗
f
k
). The proof is complete.Proof of Theorem5.
LetA
∈
ℱ
(
ℒ
). Then there exists a unique finite family of distinct K
1
,
K
2
,
…
,
K
n in 𝒥
(
ℒ
) such that A
=
A
1
+
A
2
+
·
·
·
+
A
n
,
A
i
∈
ℱ
(
K
i
). By Lemma 8, for each i, there are operators S
i in ℱ
(
K
i
) and τ
(
A
i
) in Z
(
𝒜
) such that δ
(
A
i
)
=
[
A
i
,
S
i
]
+
τ
(
A
i
). Let S
=
S
1
+
S
2
+
·
·
·
+
S
n. Since S
i
A
j
=
A
j
S
i
=
0 if i
≠
j, it follows that
(21)
δ
(
A
)
=
∑
i
=
1
n
‍
δ
(
A
i
)
=
∑
i
=
1
n
‍
(
[
A
i
,
S
i
]
+
τ
(
A
i
)
)
=
[
A
,
S
]
+
∑
i
=
1
n
‍
τ
(
A
i
)
.
Defineτ
(
A
) to be an operator in Z
(
𝒜
) such that δ
(
A
)
=
[
A
,
T
]
+
τ
(
A
) for some T in ℱ
(
ℒ
). By Lemma 8 and the equation above, we see that τ
(
A
) is well defined and τ
(
A
)
=
∑
i
=
1
n
‍
τ
(
A
i
).
Now define a mapping fromℱ
(
ℒ
) to 𝒜 as Δ
(
A
)
=
δ
(
A
)
-
τ
(
A
) for A
∈
ℱ
(
ℒ
). Then Δ is linear since τ is linear. Moreover, for each A
∈
ℱ
(
ℒ
) there is an operator S such that Δ
(
A
)
=
[
A
,
S
]. Consequently, Δ is a local derivation from ℱ
(
ℒ
) to 𝒜. By [3, Proposition 2.6], Δ is a derivation. Thus δ is standard. This completes the proof.Corollary 9.
Letℒ be a 𝒥-subspace lattice and suppose that the dimension of each element in ℱ
(
ℒ
) is infinite. Let δ be a Lie triple derivation from ℱ
(
ℒ
) to itself. Then δ is a derivation.Proof.
By [3, Remark 2.5(iii)], Z
(
ℱ
(
ℒ
)
)
=
0 in this case. Hence by Theorem 5, δ is a derivation.


## 3. Lie Triple Derivations of𝒥-Subspace Lattice Algebras

In this section, we study Lie triple derivations of whole𝒥-subspace lattice algebras. The principal result describes the structure of those mappings.Theorem 10.
Letℒ be a 𝒥-subspace lattice on a Banach space X. Let δ
:
A
l
g
ℒ
→
A
l
g
ℒ be a linear mapping. The following is equivalent.(i)
δ is a Lie triple derivation.
(ii)
For eachK
∈
𝒥
(
ℒ
), there exist an operator T
K in B
(
K
) and a linear functional λ
K
:
A
l
g
ℒ
→
𝔽 vanishing on every double commutator such that δ
(
A
)
x
=
(
T
K
A
-
A
T
K
)
x
+
λ
K
(
A
)
x for all A
∈
A
l
g
ℒ and x
∈
K.Proof.
(
ii
)
⇒
(
i
). This is a straightforward verification.
(
i
)
⇒
(
ii
). Obviously, the restriction of δ to ℱ
(
ℒ
) is a Lie triple derivation. Hence it is standard by Theorem 5. Therefore, there exist a derivation d
:
ℱ
(
ℒ
)
→
Alg
ℒ and a linear mapping τ
:
ℱ
(
ℒ
)
→
Z
(
Alg
ℒ
) vanishing on every double commutator such that δ
(
F
)
=
d
(
F
)
+
τ
(
F
) for every F
∈
ℱ
(
ℒ
).
Fix an elementK
∈
𝒥
(
ℒ
). Take vectors x
K
∈
K and f
K
∈
K
-
⊥ such that f
K
(
x
K
)
=
1. Define a linear mapping T
K
:
K
→
K by
(22)
T
K
x
=
d
(
x
⊗
f
K
)
x
K
for x
∈
K. This is well defined because of x
⊗
f
K
∈
ℱ
(
ℒ
) for x
∈
K.
ForF
∈
ℱ
(
ℒ
) and x
∈
K, by (22) we have
(23)
T
K
F
x
=
d
(
F
x
⊗
f
K
)
x
K
=
d
(
F
)
(
x
⊗
f
K
)
x
K
+
F
d
(
x
⊗
f
K
)
x
K
.
So for all F
∈
ℱ
(
ℒ
) and for all x
∈
K,
(24)
d
(
F
)
x
=
(
T
K
F
-
F
T
K
)
x
.
LetA be in Alg
ℒ. Let x be in K. Then for F
∈
ℱ
(
ℒ
) with F
2
=
F, by (22) and (24), we have
(25)
δ
(
[
[
A
,
F
]
,
F
]
)
x
=
d
(
[
[
A
,
F
]
,
F
]
)
x
+
τ
(
[
[
A
,
F
]
,
F
]
)
x
=
T
K
(
[
[
A
,
F
]
,
F
]
)
x
-
(
[
[
A
,
F
]
,
F
]
)
T
K
x
+
τ
(
[
[
A
,
F
]
,
F
]
)
x
=
(
T
K
A
F
+
T
K
F
A
-
2
T
K
F
A
F
-
A
F
T
K
-
F
A
T
K
+
2
F
A
F
T
K
)
x
+
τ
(
[
[
A
,
F
]
,
F
]
)
x
,
(26)
δ
(
[
[
A
,
F
]
,
F
]
)
x
=
[
[
δ
(
A
)
,
F
]
,
F
]
x
+
[
[
A
,
δ
(
F
)
]
,
F
]
x
+
[
[
A
,
F
]
,
δ
(
F
)
]
x
=
[
[
δ
(
A
)
,
F
]
,
F
]
x
+
[
[
A
,
d
(
F
)
]
,
F
]
x
+
[
[
A
,
F
]
,
d
(
F
)
]
x
=
(
δ
(
A
)
F
-
2
F
δ
(
A
)
F
+
F
δ
(
A
)
+
A
d
(
F
)
F
-
d
(
F
)
A
F
-
F
A
d
(
F
)
+
F
d
(
F
)
A
+
A
F
d
(
F
)
-
F
A
d
(
F
)
-
d
(
F
)
A
F
+
d
(
F
)
F
A
)
x
=
(
+
(
T
K
F
-
F
T
K
)
F
A
)
δ
(
A
)
F
-
2
F
δ
(
A
)
F
+
F
δ
(
A
)
+
A
(
T
K
F
-
F
T
K
)
F
-
(
T
K
F
-
F
T
K
)
A
F
-
F
A
(
T
K
F
-
F
T
K
)
+
F
(
T
K
F
-
F
T
K
)
A
+
A
F
(
T
K
F
-
F
T
K
)
-
F
A
(
T
K
F
-
F
T
K
)
-
(
T
K
F
-
F
T
K
)
A
F
+
(
T
K
F
-
F
T
K
)
F
A
)
x
=
(
δ
(
A
)
F
-
2
F
δ
(
A
)
F
+
F
δ
(
A
)
+
A
T
K
F
-
2
T
K
F
A
F
+
2
F
T
K
A
F
-
2
F
A
T
K
F
+
2
F
A
F
T
K
-
F
T
K
A
-
A
F
T
K
+
T
K
F
A
)
x
.
Comparing these two equations, we get
(27)
(
δ
(
A
)
-
T
K
A
+
A
T
K
)
F
x
=
F
(
T
K
2
δ
(
A
)
F
-
δ
(
A
)
-
2
T
K
A
F
+
2
A
T
K
F
-
T
K
F
A
-
A
T
K
)
x
+
τ
(
[
[
A
,
F
]
,
F
]
)
x
.
Taking f
∈
K
-
⊥ with f
(
x
)
=
1 and then putting F
=
x
⊗
f in the last equation, we get
(28)
(
δ
(
A
)
-
(
T
K
A
-
A
T
K
)
)
x
=
λ
1
(
A
,
x
,
f
)
x
+
τ
(
[
[
A
,
F
]
,
F
]
)
x
,
where λ
1
(
A
,
x
,
f
)
=
f
(
(
2
δ
(
A
)
F
-
δ
(
A
)
-
2
T
K
A
F
+
2
A
T
K
F
-
T
K
F
A
-
A
T
K
)
x
). Since τ
(
[
[
A
,
F
]
,
F
]
) is in the center of Z
(
Alg
ℒ
), it follows from [3, Remark 2.5] that τ
(
[
[
A
,
F
]
,
F
]
)
x
=
λ
2
(
A
,
x
,
f
)
x for some λ
2
(
A
,
x
,
f
)
∈
𝔽. Consequently, (
δ
(
A
)
-
(
T
K
A
-
A
T
K
)
)
x is a scalar multiple of x. Since this holds for each x
∈
K, it follows easily that δ
(
A
)
-
(
T
K
A
-
A
T
K
) viewed as a linear mapping from K to K is a scalar multiple of the identify on K. Namely, there exists a scalar λ
K
(
A
) such that
(29)
(
δ
(
A
)
-
(
T
K
A
-
A
T
K
)
)
x
=
λ
K
(
A
)
x
for each x
∈
K. One can easily see that λ
K
:
Alg
ℒ
→
𝔽 is linear and vanishes on every double commutator.
It remains to verify the boundedness ofT
K. Suppose that {
x
n
}
n
=
1
∞ is a sequence of vectors in K, lim
⁡
n
→
∞
x
n
=
0, and lim
⁡
n
→
∞
T
K
x
n
=
x
0. For any f
∈
K
-
⊥,
(30)
f
(
x
n
)
d
(
x
K
⊗
f
K
)
x
K
=
d
(
(
x
K
⊗
f
)
(
x
n
⊗
f
K
)
)
x
K
=
d
(
x
K
⊗
f
)
x
n
+
(
x
K
⊗
f
)
d
(
x
n
⊗
f
K
)
x
K
=
d
(
x
K
⊗
f
)
x
n
+
f
(
T
K
x
n
)
x
K
.
Taking the limit, we get that f
(
x
0
)
=
0. Since x
0 is in K and f
∈
K
-
⊥ is arbitrary, x
0
=
0. So T
K is closed. The Closed Graph theorem gives the boundedness of T
K, completing the proof.



---

*Source: 969265-2014-01-20.xml*