2-Local derivations on matrix algebras and algebras of measurable operators

TL;DR

This paper proves that 2-local derivations on matrix algebras over certain Banach algebras and on algebras of measurable operators affiliated with von Neumann algebras are actual derivations, extending known results.

Contribution

It establishes that 2-local derivations are derivations on matrix algebras over Banach algebras satisfying specific conditions and on algebras of measurable operators.

Findings

2-local derivations on M_n(A) are derivations for na3 3

2-local derivations on algebras of measurable operators are derivations

Results apply to von Neumann algebras without abelian summands

Abstract

Let \(\mathcal{A}\) be a unital Banach algebra such that any Jordan derivation from \(\mathcal{A}\) into any \(\mathcal{A}\)-bimodule \(\mathcal{M}\) is a derivation. We prove that any 2-local derivation from the algebra $M_n(\mathcal{A})$ into $M_n(\mathcal{M})$ $(n\geq 3)$ is a derivation. We apply this result to show that any 2-local derivation on the algebra of locally measurable operators affiliated with a von Neumann algebra without direct abelian summands is a derivation.

Equations

$$\Delta (x^2)=\Delta (x)x+x\Delta (x)$$

$$x=\sum _{i,j=1}^n{x}_{ij}{e}_{ij},x_{ij}\in \mathcal{A},i,j\in \overline{1,n}.$$

$$\overline{\delta }(x)=\sum _{i,j=1}^n\delta (x_{ij})e_{ij},x_{ij}\in \mathcal{A},i,j\in \overline{1,n}$$

$$D=ad(a)+\overline{\delta },$$

$$u=\sum _{i=1}^n\frac{1}{2^i}{e}_{i,i},v=\sum _{i=2}^n{e}_{i-1,i}.$$

$$a=\left(\begin{array}[]{ccccccc}a_{1}&a_{2}&a_{3}&.&\ldots&a_{n}\\ 0&a_{1}&a_{2}&.&\ldots&a_{n-1}\\ 0&0&a_{1}&.&\ldots&a_{n-2}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0&0&\ldots&.&a_{1}&a_{2}\\ 0&0&\ldots&.&0&a_{1}\end{array}\right).$$

$$\Delta (u)=D(u),\Delta (v)=D(v),$$

$$\Delta (e_{i,j})=[h,e_{i,j}]+\overline{\delta }(e_{ij}),\Delta (u)=[h,u]+\overline{\delta }(u).$$

$$\Delta (e_{i,j})=b{e}_{i,j}-e_{i,j}b,$$

$$\Delta (e_{i,j})=h{e}_{i,j}-e_{i,j}h=b{e}_{i,j}-e_{i,j}b.$$

$$h{e}_{i,j}-e_{i,j}h=(h_i-h_j)e_{i,j}$$

$$[b{e}_{i,j}-e_{i,j}b{]}_{i,j}=0,$$

$$e_{i,j}\Delta (x)e_{i,j}$$

$${\Delta }_{i,j}(x)=e_{i,i}\Delta (x)e_{j,j}$$

$$\Delta (x)=[h,x]+\overline{\delta }(x),\Delta (u)=[h,u]+\overline{\delta }(u),$$

$$e_{k,k}\Delta (x)e_{k,k}=\Delta (x_{k,k})$$

$$\Delta (x_{k,k})$$

$$e_{k,k}\Delta (x)e_{k,k}$$

$$e_{j,i}\Delta (x)e_{i,j}=\Delta (e_{j,i}x{e}_{i,j})$$

$$e_{j,i}\Delta (x)e_{i,j}$$

$$\Delta (e_{j,i}x{e}_{i,j})$$

$${\Delta }_{i,i}(x+y)$$

$${\Delta }_{i,i}(x)$$

$${\Delta }_{i,i}(y)$$

$${\Delta }_{i,i}(x+y)={\Delta }_{i,i}(x)+{\Delta }_{i,i}(y).$$

$$x=\left(\begin{array}[]{cccccc}x_{1}&0&0&\ldots&0\\ 0&x_{2}&0&\ldots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ 0&0&\ldots&x_{n-1}&0\\ 0&0&\ldots&0&x_{n}\end{array}\right).$$

$$\Delta (x)=[a,x]+\overline{\delta }(x),\Delta (y)=[a,y]+\overline{\delta }(y).$$

$$0$$

$$a_{k,k}{x}_{k,k}-x_{k,k}{a}_{k,k}+\delta (x_{k,k})=0$$

$$a_{i,k}=a_{k,i}=0$$

$$\Delta (x{)}_{k,k}=a_{k,k}{x}_{k,k}-x_{k,k}{a}_{k,k}+\delta (x_{k,k})=\Delta (y{)}_{k,k}=0.$$

Full text

1. Introduction

Let $\mathcal{A}$ be an associative algebra over $\mathbb{C}$ the field of complex numbers and let $\mathcal{M}$ be an $\mathcal{A}$-bimodule. A linear map $D$ from $\mathcal{A}$ to $\mathcal{M}$ is called a derivation if $D(xy)=D(x)y+xD(y)$ for all $x,y\in\mathcal{A}.$ If it satisfies a weaker condition $D(x^{2})=D(x)x+xD(x)$ for every $x\in\mathcal{A}$ then it is called a Jordan derivation. It is easy to verify that each element $a\in\mathcal{M}$ implements a derivation $D_{a}$ from $\mathcal{A}$ into $\mathcal{M}$ by $D_{a}(x)=ax-xa,\,x\in\mathcal{A}.$ Such derivations $D_{a}$ are called inner derivations.

In 1990, Kadison [12] and Larson and Sourour [15] independently introduced the concept of local derivation. A linear map $\Delta:\mathcal{A}\to\mathcal{M}$ is called a local derivation if for every $x\in\mathcal{A}$ there exists a derivation $D_{x}$ (depending on $x$) such that $\Delta(x)=D_{x}(x).$ It would be interesting to consider under which conditions local derivations automatically become derivations. Many partial results have been done in this problem. In [12] Kadison shows that every norm-continuous local derivation from a von Neumann algebra $M$ into a dual $M$-bimodule is a derivation. In [11] Johnson extends Kadison’s result and proves every local derivation from a $C^{\ast}$-algebra $\mathcal{A}$ into any Banach $\mathcal{A}$-bimodule is a derivation.

Similar problems for local derivations on algebras of measurable operators $S(M)$ and locally measurable operators $LS(M),$ affiliated with a von Neumann algebra $M,$ have been considered in [4] and [9]. Namely, it was proved that if $M$ is a von Neumann algebra without abelian direct summand then every local derivation on $LS(M)$ is a derivation. Moreover, for abelian von Neumann algebras $M$ necessary and sufficient condition are given in [5] for $S(M)=LS(M)$ to admit local derivations which are not derivations (see for details the survey [4, Section 5]).

In 1997, Šemrl [17] initiated the study of so-called 2-local derivations and 2-local automorphisms on algebras. Namely, he described such maps on the algebra $B(H)$ of all bounded linear operators on an infinite dimensional separable Hilbert space $H$.

In the above notations, map $\Delta:\mathcal{A}\to\mathcal{M}$ (not necessarily linear) is called a 2-local derivation if, for every $x,y\in\mathcal{A},$ there exists a derivation $D_{x,y}:\mathcal{A}\to\mathcal{M}$ such that $D_{x,y}(x)=\Delta(x)$ and $D_{x,y}(y)=\Delta(y).$

Afterwards local derivations and 2-local derivations have been investigated by many authors on different algebras and many results have been obtained in [1, 2, 3, 5, 12, 14, 17].

Recall that an algebra $\mathcal{A}$ is called a regular (in the sense of von Neumann) if for each $a\in\mathcal{A}$ there exists $b\in\mathcal{A}$ such that $a=aba.$ Let $M_{n}(\mathcal{A})$ be the algebra of all $n\times n$ matrices over a unital commutative regular algebra $\mathcal{A}.$ In [5], we prove that every 2-local derivation on $M_{n}(\mathcal{A}),$ $n\geq 2,$ is a derivation. We applied this result to a description of 2-local derivations on the algebras of measurable operators $S(M)$ and locally measurable operators $LS(M)$ affiliated with a type I finite von Neumann algebra $M$. Further this result was extended to type I∞ von Neumann algebras: it was proved that in this case every 2-local derivations on the algebra of locally measurable operators is a derivation (see [4, Theorem 6,7]). Moreover in [5] we also gave necessary and sufficient conditions for a commutative regular algebra, in particular for the algebra $S(M)$ of measurable operators affiliated with an abelian von Neumann algebra $M$, to admit 2-local derivations which are not derivations. In [3] we considered a unital semi-prime Banach algebra $\mathcal{A}$ with the inner derivation property and proved that any 2-local derivation on the algebra $M_{2^{n}}(\mathcal{A}),$ $n\geq 2,$ is a derivation. We have applied this result to $AW^{\ast}$-algebras and proved that any 2-local derivation on an arbitrary $AW^{\ast}$-algebra is a derivation. In [10], W. Huang, J. Li and W. Qian, have characterized derivations and 2-local derivations from $M_{n}(\mathcal{A})$ into $M_{n}(\mathcal{M}),n\geq 2,$ where $\mathcal{A}$ is a unital algebra over $\mathbb{C}$ and $\mathcal{M}$ is a unital $\mathcal{A}$-bimodule. They considered a unital Banach algebra such that any Jordan derivation from the algebra $\mathcal{A}$ into any $\mathcal{A}$-bimodule $\mathcal{M}$ is an inner derivation and proved that any 2-local derivation from the algebra $M_{n}(\mathcal{A})$ into $M_{n}(\mathcal{M})$ $(n\geq 3)$ is a derivation, when $\mathcal{A}$ is commutative and commutes with $\mathcal{M}.$

In the present paper we shall consider matrix algebras over unital (non commutative in general) Banach algebras and describe 2-local derivations from $M_{n}(\mathcal{A})$ into $M_{n}(\mathcal{M})$, where $\mathcal{A}$ is a unital Banach algebra such that any Jordan derivation from the algebra $\mathcal{A}$ into any $\mathcal{A}$-bimodule $\mathcal{M}$ is a derivation. The main result of Section 2 asserts that under the above conditions every 2-local derivation from the algebra $M_{n}(\mathcal{A})$ into $M_{n}(\mathcal{M})$ $(n\geq 3)$ is a derivation.

In Section 3, we apply the main result of the previous section to algebras of locally measurable operators affiliated with von Neumann algebras. Namely, we extend all above mentioned results from [3, 4, 5, 10] and prove that for an arbitrary von Neumann algebra $M$ without abelian direct summands every 2-local derivation on each subalgebra $\mathcal{A}$ of the algebra $LS(M)$, such that $M\subseteq\mathcal{A},$ is a derivation. A similar result for local derivation is obtained in [9, Theorem 1] (see also [4, Theorem 5.5]).

2. 2-local derivations on matrix algebras

If $\Delta:\mathcal{A}\rightarrow\mathcal{M}$ is a 2-local derivation, then from the definition it easily follows that $\Delta$ is homogenous. At the same time,

[TABLE]

for each $x\in\mathcal{A}.$ This means that additive (and hence, linear) 2-local derivation is a Jordan derivation.

In [8] Brešar suggested various conditions on an algebra $\mathcal{A}$ under which any Jordan derivation from $\mathcal{A}$ into any $\mathcal{A}$-bimodule $\mathcal{M}$ is a derivation.

In the present paper we shall consider algebras with the following property:

(J): any Jordan derivation from the algebra $\mathcal{A}$ into any $\mathcal{A}$-bimodule $\mathcal{M}$ is a derivation.

Therefore, in the case of algebras with the property (J) in order to prove that a 2-local derivation $\Delta:\mathcal{A}\rightarrow\mathcal{M}$ is a derivation it is sufficient to prove that $\Delta:\mathcal{A}\rightarrow\mathcal{M}$ is additive.

Throughout this paper, $\mathcal{A}$ is a unital Banach algebra over $\mathbb{C},$ $\mathcal{M}$ is an $\mathcal{A}$-bimodule with $\mathbf{1}x=x\mathbf{1}=x$ for all $x\in\mathcal{M},$ where $\mathbf{1}$ is the unit element of $\mathcal{A}.$

The following theorem is the main result of this section.

Theorem 2.1**.**

Let $\mathcal{A}$ be a unital Banach algebra with the property (J), $\mathcal{M}$ be a unital $\mathcal{A}$-bimodule and let $M_{n}(\mathcal{A})$ be the algebra of all $n\times n$-matrices over $\mathcal{A},$ where $n\geq 3.$ Then any 2-local derivation $\Delta$ from $M_{n}(\mathcal{A})$ into $M_{n}(\mathcal{M})$ is a derivation.

The proof of Theorem 2.1 consists of two steps. In the first step we shall show additivity of $\Delta$ on the subalgebra of diagonal matrices from $M_{n}(\mathcal{A}).$

Let $\{e_{i,j}\}_{i,j=1}^{n}$ be the system of matrix units in $M_{n}(\mathcal{A}).$ For $x\in M_{n}(\mathcal{A})$ by $x_{i,j}$ we denote the $(i,j)$-entry of $x,$ where $1\leq i,j\leq n.$ We shall, if necessary, identify this element with the matrix from $M_{n}(\mathcal{A})$ whose $(i,j)$-entry is $x_{i,j},$ other entries are zero, i.e. $x_{i,j}=e_{i,i}xe_{j,j}.$

Each element $x\in M_{n}(\mathcal{A})$ has the form

[TABLE]

Let $\delta:\mathcal{A}\to\mathcal{M}$ be a derivation. Setting

[TABLE]

we obtain a well-defined linear operator $\overline{\delta}$ from $M_{n}(\mathcal{A})$ into $M_{n}(\mathcal{M}).$ Moreover $\overline{\delta}$ is a derivation from $M_{n}(\mathcal{A})$ into $M_{n}(\mathcal{M}).$

It is known [10, Theorem 2.1] that every derivation $D$ from $M_{n}(\mathcal{A})$ into $M_{n}(\mathcal{M})$ can be represented as a sum

[TABLE]

where $\textrm{ad}(a)$ is an inner derivation implemented by an element $a\in M_{n}(\mathcal{M}),$ while $\overline{\delta}$ is the derivation of the form (2.1) generated by a derivation $\delta$ from $\mathcal{A}$ into $\mathcal{M}.$

Consider the following two matrices:

[TABLE]

It is easy to see that an element $x\in M_{n}(\mathcal{M})$ commutes with $u$ if and only if it is diagonal, and if an element $a\in M_{n}(\mathcal{M})$ commutes with $v,$ then $a$ is of the form

[TABLE]

A result, similar to the following one, was proved in [5, Lemma 4.4] for matrix algebras over commutative regular algebras.

Further in Lemmata 2.2–2.5 we assume that $n\geq 2.$

Lemma 2.2**.**

For every $2$-local derivation $\Delta$ from $M_{n}(\mathcal{A})$ into $M_{n}(\mathcal{M})$ there exists a derivation $D$ such that $\Delta|_{\mbox{sp}\{e_{i,j}\}_{i,j=1}^{n}}=D|_{\mbox{sp}\{e_{i,j}\}_{i,j=1}^{n}},$ where $\mbox{sp}\{e_{i,j}\}_{i,j=1}^{n}$ is the linear span of the set $\{e_{i,j}\}_{i,j=1}^{n}.$

Proof.

Take a derivation $D$ from $M_{n}(\mathcal{A})$ into $M_{n}(\mathcal{M})$ such that

[TABLE]

where $u,v$ are the elements from (2.3). Replacing $\Delta$ by $\Delta-D$, if necessary, we can assume that $\Delta(u)=\Delta(v)=0.$

Let $i,j\in\overline{1,n}.$ Take a derivation $D=\textrm{ad}(h)+\overline{\delta}$ of the form (2.2) such that

[TABLE]

Since $\Delta(u)=0$ and $\overline{\delta}(u)=0,$ it follows that $[h,u]=0,$ and therefore $h$ has a diagonal form, i.e. $h=\sum\limits_{s=1}^{n}h_{s}e_{s,s},\,h_{s}\in\mathcal{A},\,s\in\overline{1,n}.$

In the same way, but starting with the element $v$ instead of $u$, we obtain

[TABLE]

where $b$ has the form (2.4), depending on $e_{i,j}.$ So

[TABLE]

Since

[TABLE]

and

[TABLE]

it follows that $\Delta(e_{i,j})=0.$

Now let us take a matrix $x=\sum\limits_{i,j=1}^{n}\lambda_{i,j}e_{i,j}\in M_{n}(\mathbb{C}).$ Then

[TABLE]

i.e. $e_{i,j}\Delta(x)e_{i,j}=0$ for all $i,j\in\overline{1,n}.$ This means that $\Delta(x)=0.$ The proof is complete. ∎

Further in Lemmata 2.3–2.8 we assume that $\Delta$ is a 2-local derivation from $M_{n}(\mathcal{A})$ into $M_{n}(\mathcal{M})$ such that $\Delta|_{\mbox{sp}\{e_{i,j}\}_{i,j=1}^{n}}=0.$

Let $\Delta_{i,j}$ be the restriction of $\Delta$ onto $\mathcal{A}_{i,j}=e_{i,i}M_{n}(\mathcal{A})e_{j,j},$ where $1\leq i,j\leq n.$

Lemma 2.3**.**

$\Delta_{i,j}$* maps $\mathcal{A}_{i,j}$ into itself.*

Proof.

Let us show that

[TABLE]

for all $x\in\mathcal{A}_{i,j}.$

Take $x=x_{i,j}\in\mathcal{A}_{i,j},$ and consider a derivation $D=\textrm{ad}(h)+\overline{\delta}$ of the form (2.2) such that

[TABLE]

where $u$ is the element from (2.3). Since $\Delta(u)=0$ and $\overline{\delta}(u)=0,$ it follows that $[h,u]=0,$ and therefore $h$ has a diagonal form. Then $\Delta(x)=(h_{i}-h_{j})e_{ij}+\delta(x_{ij})e_{ij}.$ This means that $\Delta(x)\in\mathcal{A}_{i,j}.$ The proof is complete. ∎

Lemma 2.4**.**

Let $x=\sum\limits_{i=1}^{n}x_{i,i}$ be a diagonal matrix. Then

[TABLE]

for all $k\in\overline{1,n}.$

Proof.

Take a derivation $D=\textrm{ad}(a)+\overline{\delta}$ of the form (2.2) such that

$\Delta(x)=[a,x]+\overline{\delta}(x)$ and $\Delta(x_{k,k})=[a,x_{k,k}]+\overline{\delta}(x_{kk}).$

Using equality (2.5), we obtain that

[TABLE]

Since $x$ is a diagonal matrix, we get

[TABLE]

Thus $e_{k,k}\Delta(x)e_{k,k}=\Delta(x_{k,k}).$ The proof is complete. ∎

Lemma 2.5**.**

Let $x=x_{i,i}\in\mathcal{A}_{i,i}.$ Then

[TABLE]

for every $j\in\{1,\cdots,n\}.$

Proof.

For $i=j$ we have already proved (see Lemma 2.4).

Suppose that $i\neq j.$ For an arbitrary element $x=x_{i,i}\in\mathcal{A}_{i,i}$ , consider $y=x+e_{j,i}xe_{i,j}\in\mathcal{A}_{i,i}+\mathcal{A}_{j,j}.$ Take a derivation $D=\textrm{ad}(a)+\overline{\delta}$ such that

$\Delta(y)=[a,y]+\overline{\delta}(y)$ and $\Delta(v)=[a,v]+\overline{\delta}(v),$

where $v$ is the element from (2.3). Since $\Delta(v)=0$ and $\overline{\delta}(v)=0,$ it follows that $a$ has the form (2.4). By Lemma 2.4 we obtain that

[TABLE]

The proof is complete. ∎

Further in Lemmata 2.6–2.13 we assume that $n\geq 3.$

Lemma 2.6**.**

$\Delta_{i,i}$* is additive for all $i\in\overline{1,n}.$*

Proof.

Let $i\in\overline{1,n}.$ Since $n\geq 3,$ we can take different numbers $k,s$ such that $(k-i)(s-i)\neq 0.$

For arbitrary $x,y\in\mathcal{A}_{i,i}$ consider the diagonal element $z\in\mathcal{A}_{i,i}+\mathcal{A}_{k,k}+\mathcal{A}_{s,s}$ such that $z_{i,i}=x+y,\,z_{k,k}=x,\,z_{s,s}=y.$ Take a derivation $D=\textrm{ad}(a)+\overline{\delta}$ such that

$\Delta(z)=[a,z]+\overline{\delta}(z)$ and $\Delta(v)=[a,v]+\overline{\delta}(v),$

where $v$ is the element from (2.3). Since $\Delta(v)=0$ and $\overline{\delta}(v)=0,$ it follows that $a$ has the form (2.4). Using Lemmata 2.4 and 2.5 we obtain that

[TABLE]

Hence

[TABLE]

The proof is complete. ∎

As it was mentioned in the beginning of the section any additive 2-local derivation is a Jordan derivation. Since $\mathcal{A}_{i,i}\cong\mathcal{A}$ has the property (J), Lemma 2.6 implies the following result.

Lemma 2.7**.**

$\Delta_{i,i}$* is a derivation for all $i\in\overline{1,n}.$*

Denote by $\mathcal{D}_{n}(\mathcal{A})$ the set of all diagonal matrices from $M_{n}(\mathcal{A}),$ i.e. the set of all matrices of the following form

[TABLE]

Let us consider a derivation $\overline{\Delta_{1,1}}$ of the form (2.1). By Lemmata 2.4 and 2.5 we obtain that

Lemma 2.8**.**

$\Delta|_{\mathcal{D}_{n}(\mathcal{A})}=\overline{\Delta_{1,1}}|_{\mathcal{D}_{n}(\mathcal{A})}$* and $\overline{\Delta_{1,1}}|_{\mbox{sp}\{e_{i,j}\}_{i,j=1}^{n}}=0.$*

Now we are in position to pass to the second step of our proof. In this step we show that if a 2-local derivation $\Delta$ satisfies the following conditions

$\Delta|_{\mathcal{D}_{n}(\mathcal{A})}\equiv 0$ and $\Delta|_{\mbox{sp}\{e_{i,j}\}_{i,j=1}^{n}}\equiv 0,$

then it is identically equal to zero.

Below in the five Lemmata we shall consider 2-local derivations which satisfy the latter equalities.

We denote by $e$ the unit of the algebra $\mathcal{A}.$

Lemma 2.9**.**

Let $x\in M_{n}(\mathcal{A}).$ Then $\Delta(x)_{k,k}=0$ for all $k\in\overline{1,n}.$

Proof.

Let $x\in M_{n}(\mathcal{A}),$ and fix $k\in\overline{1,n}.$ Since $\Delta$ is homogeneous, we can assume that $\|x_{k,k}\|<1,$ where $\|\cdot\|$ is the norm on $\mathcal{A}.$ Take a diagonal element $y$ in $M_{n}(\mathcal{A})$ with $y_{k,k}=e+x_{k,k}$ and $y_{i,i}=0$ otherwise. Since $\|x_{k,k}\|<1,$ it follows that $e+x_{k,k}$ is invertible in $\mathcal{A}.$ Take a derivation $D=\textrm{ad}(a)+\overline{\delta}$ of the form (2.2) such that

[TABLE]

Since $y\in\mathcal{D}_{n}(\mathcal{A})$ we have that $0=\Delta(y)=[a,y]+\overline{\delta}(y),$ and therefore

[TABLE]

for all $i\neq k.$ Thus

[TABLE]

and

[TABLE]

for all $i\neq k.$ The above equalities imply that

[TABLE]

The proof is complete. ∎

Lemma 2.10**.**

Let $x$ be a matrix with $x_{k,s}=e.$ Then $\Delta(x)_{k,s}=0.$

Proof.

We have

[TABLE]

Thus

[TABLE]

This means that $\Delta(x)_{k,s}=0.$ The proof is complete. ∎

Lemma 2.11**.**

Let $k,s$ be numbers such that $k\neq s$ and let $x$ be a matrix with $x_{k,s}=e.$ Then $\Delta(x)_{s,k}=0.$

Proof.

Take a diagonal element $y$ such that $y_{k,k}=x_{s,k}$ and $y_{i,i}=\lambda_{i}e$ otherwise, where $\lambda_{i}\,\,(i\neq k)$ are distinct numbers with $|\lambda_{i}|>\|x_{s,k}\|.$ Take a derivation $D=\textrm{ad}(a)+\overline{\delta}$ such that

$\Delta(x)=[a,x]+\overline{\delta}(x)$ and $\Delta(y)=[a,y]+\overline{\delta}(y).$

Then

[TABLE]

Thus $a_{i,j}=0$ for all $i\neq j,$ i.e. $a$ is a diagonal element. Since

[TABLE]

it follows that $a_{k,k}=a_{s,s}.$ Finally,

[TABLE]

The proof is complete. ∎

Lemma 2.12**.**

Let $k\neq s$ and let $x,$ $y$ be matrices with $x_{i,j}=y_{i,j}$ for all $(i,j)\neq(s,k).$ Then $\Delta(x)_{k,s}=\Delta(y)_{k,s}.$

Proof.

Take a derivation $D=\textrm{ad}(a)+\overline{\delta}$ such that

$\Delta(x)=[a,x]+\overline{\delta}(x)$ and $\Delta(y)=[a,y]+\overline{\delta}(y).$

Then

[TABLE]

The proof is complete. ∎

Lemma 2.13**.**

Let $k\neq s.$ Then $\Delta(x)_{k,s}=0.$

Proof.

Take a matrix $y$ with $y_{s,k}=e$ and $y_{i,j}=x_{i,j}$ otherwise. By Lemma 2.11 we have that $\Delta(y)_{k,s}=0.$ Further Lemma 2.12 implies that

[TABLE]

The proof is complete. ∎

Now we are in position to prove Theorem 2.1.

Proof of Theorem 2.1. Let $\Delta$ be a 2-local derivation from $M_{n}(\mathcal{A})$ into $M_{n}(\mathcal{M}),$ where $n\geq 3.$ By Lemma 2.2 there exists a derivation $D$ such that $\Delta|_{\mbox{sp}\{e_{i,j}\}_{i,j=1}^{n}}=D|_{\mbox{sp}\{e_{i,j}\}_{i,j=1}^{n}}.$ Consider a 2-local derivation $\Theta=\Delta-D.$ Since $\Theta$ is equal to zero on $\mbox{sp}\{e_{i,j}\}_{i,j=1}^{n},$ by Lemma 2.8 we obtain that $\Theta|_{\mathcal{D}_{n}(\mathcal{A})}=\overline{\Theta_{11}}|_{\mathcal{D}_{n}(\mathcal{A})},$ where $\overline{\Theta_{11}}$ is the derivation defined by (2.1). As in Lemma 2.8 we have that

$\left(\Theta-\overline{\Theta_{11}}\right)|_{\mbox{sp}\{e_{i,j}\}_{i,j=1}^{n}}\equiv 0$ and $\left(\Theta-\overline{\Theta_{11}}\right)|_{\mathcal{D}_{n}(\mathcal{A})}\equiv 0.$

Now for an arbitrary element $x\in M_{n}(\mathcal{A}),$ by Lemmata 2.9 and 2.13 we obtain that $\left(\Theta-\overline{\Theta_{11}}\right)(x)_{k,s}=0$ for all $k,s.$ Thus $\left(\Theta-\overline{\Theta_{11}}\right)(x)=0,$ i.e., $\Theta=\overline{\Theta_{11}}.$ So, $\Delta=\overline{\Theta_{11}}+D$ is a derivation. The proof is complete. $\Box$

3. An application to 2-local derivations on algebras of locally measurable operators

In this section we apply Theorem 2.1 to the description of 2-local derivations on the algebra of locally measurable operators affiliated with a von Neumann algebra and on its subalgebras.

In [8, Corollary 3.11] it was proved that if an associative algebra (ring) $\mathcal{A}$ contains a noncommutative simple subalgebra (subring) $\mathcal{A}_{0}$ which contains the unit of $\mathcal{A}$, then every Jordan derivation from $\mathcal{A}$ into any $\mathcal{A}$-bimodule is a derivation, i.e. $\mathcal{A}$ satisfies the property (J). In particular, if there exists a subalgebra $\mathcal{A}_{0}$ of $\mathcal{A}$ which is isomorphic to $M_{n}(\mathbb{C})$ ($n\geq 2$) and contains the unit of $\mathcal{A},$ then $\mathcal{A}$ has the property (J).

Let $M$ be a von Neumann algebra and denote by $S(M)$ the algebra of all measurable operators and by $LS(M)$ the algebra of all locally measurable operators affiliated with $M$ (see for example [16, 18]).

Theorem 3.1**.**

Let $M$ be an arbitrary von Neumann algebra without abelian direct summands and let $LS(M)$ be the algebra of all locally measurable operators affiliated with $M.$ Then any 2-local derivation $\Delta$ from $M$ into $LS(M)$ is a derivation.

Proof.

Let $z$ be a central projection in $M.$ Since $D(z)=0$ for an arbitrary derivation $D,$ it is clear that $\Delta(z)=0$ for any $2$-local derivation $\Delta$ from $M$ into $LS(M).$ Take $x\in M$ and let $D$ be a derivation from $M$ into $LS(M)$ such that $\Delta(zx)=D(zx),\Delta(x)=D(x)$. Then we have $\Delta(zx)=D(zx)=D(z)x+zD(x)=z\Delta(x).$ This means that every 2-local derivation $\Delta$ maps $zM$ into $zLS(M)\cong LS(zM)$ for each central projection $z\in M.$ So, we may consider the restriction of $\Delta$ onto $zM.$ Since an arbitrary von Neumann algebra without abelian direct summands can be decomposed along a central projection into the direct sum of von Neumann algebras of type I${}_{n},n\geq 2,$ type I${}_{\infty},$ type II and type III, we may consider these cases separately.

If $M$ is a von Neumann algebra of type I${}_{n},$ $n\geq 2,$ [10, Corollary 3.12] implies that any 2-local derivation from $M$ into $LS(M)\equiv S(M)$ is a derivation.

Let the von Neumann algebra $M$ have one of the types I${}_{\infty},$ II or III. Then the halving Lemma [13, Lemma 6.3.3] for type I∞-algebras and [13, Lemma 6.5.6] for type II or III algebras, imply that the unit of the algebra $M$ can be represented as a sum of mutually equivalent orthogonal projections $e_{1},e_{2},e_{3}$ from $M.$ Then the map $x\mapsto\sum\limits_{i,j=1}^{3}e_{i}xe_{j}$ defines an isomorphism between the algebra $M$ and the matrix algebra $M_{3}(\mathcal{A}),$ where $\mathcal{A}=e_{1,1}Me_{1,1}.$ Further, the algebra $LS(M)$ is isomorphic to the algebra $M_{3}(LS(\mathcal{A})).$ Moreover, the algebra $\mathcal{A}$ has same type as the algebra $M,$ and therefore contains a subalgebra isomorphic to $M_{3}(\mathbb{C}).$ This means that the algebra $\mathcal{A}$ satisfies the property (J). Therefore Theorem 2.1 implies that any 2-local derivation from $M$ into $LS(M)$ is a derivation. The proof is complete. ∎

Taking into account that any derivation on an abelian von Neumann algebra is trivial, Theorem 3.1 implies the following result (cf. [2, Theorem 2.1] and [3, Theorem 3.1]).

Corollary 3.2**.**

Let $M$ be an arbitrary von Neumann algebra. Then any 2-local derivation $\Delta$ on $M$ is a derivation.

For each $x\in LS(M)$ set $s(x)=l(x)\vee r(x),$ where $l(x)$ is the left and $r(x)$ is the right support of $x.$

Lemma 3.3**.**

Let $\mathcal{B}$ be a subalgebra of $LS(M)$ such that $M\subseteq\mathcal{B}$ and let $\Delta:\mathcal{B}\to LS(M)$ be a 2-local derivation such that $\Delta|_{M}\equiv 0.$ Then $\Delta\equiv 0.$

Proof.

Let us first take an arbitrary element $x\in\mathcal{B}\cap S(M).$ Let $|x|=\int\limits_{0}^{\infty}\lambda\,de_{\lambda}$ be the spectral resolution of $|x|.$ Since $x\in S(M),$ it follows that $e_{n}^{\perp}$ is a finite projection for a sufficiently large $n.$ Take a derivation $D_{x,xe_{n}}$ such that $\Delta(x)=D_{x,xe_{n}}(x)$ and $\Delta(xe_{n})=D_{x,xe_{n}}(xe_{n}),$ $n\in\mathbb{N}.$ Since $xe_{n}\in M,$ it follows that $\Delta(xe_{n})=0$ for all $n\in\mathbb{N}.$ We have

[TABLE]

Let $\mathcal{D}$ be a dimension function on the lattice $P(M)$ of all projections from $M$ (see [18]). Using [6, Lemma 4.3] we obtain that

[TABLE]

and therefore $\Delta(x)=0.$

Now let take an element $x\in\mathcal{B}.$ By the definition of locally measurable operator there exists a sequence $\{z_{n}\}$ of central projections in $M$ such that $z_{n}\uparrow\mathbf{1}$ and $xz_{n}\in S(M)$ for all $n\in\mathbb{N}$ (see [16]). Taking into account the previous case we obtain that

[TABLE]

i.e., $z_{n}\Delta(x)=0$ for all $n\in\mathbb{N}.$ Hence $\Delta(x)=0.$ The proof is complete. ∎

Theorem 3.4**.**

(cf. [4, Theorem 5.5]). Let $M$ be an arbitrary von Neumann algebra without abelian direct summands and let $\mathcal{B}$ be a subalgebra of $LS(M)$ such that $M\subseteq\mathcal{B}.$ Then any 2-local derivation $\Delta$ on $\mathcal{B}$ is a derivation.

Proof.

By Theorem 3.1 the restriction $\Delta|_{M}$ of $\Delta,$ is a derivation from $M$ into $LS(M).$ By [6, Theorem 4.8] the derivation $\Delta|_{M}$ can be extended to a derivation from $\mathcal{B}$ into $LS(M),$ which we denote by $D.$ Since the 2-local derivation $\Delta-D$ is equal to zero on $M,$ Lemma 3.3 implies that $\Delta\equiv D.$ The proof is complete. ∎

Remark 3.5*.*

As it was mentioned in the introduction, the paper [5] gives necessary and sufficient conditions on a commutative regular algebra to admit 2-local derivations which are not derivations. In particular, for an arbitrary abelian von Neumann algebra $M$ with a non atomic lattice of projections $P(M)$ the algebras $S(M)$ and $LS(M)$ always admit a 2-local derivation which is not a derivation.

A complete description of derivations on the algebra $LS(M)$ for type I von Neumann algebras $M$ is given in [4, Section 3]). Moreover, for general von Neumann algebras every derivation on the algebra $LS(M)$ is inner, provided that $M$ is a properly infinite von Neumann algebra [4, 7]. But for type II1 von Neumann algebra $M$ description of structure of derivations on the algebra $S(M)\equiv LS(M)$ is still an open problem (see [4]). In this connection it should be noted that Theorem  3.4 is one of the first results on 2-local derivations without information on the general form of derivations on these algebras.