Derivations and 2-Local Derivations on Matrix Algebras and Algebras of Locally Measurable Operators

Abstract

Let ${\mathcal {A}}$ be a unital algebra over ${\mathbb {C}}$ and ${\mathcal {M}}$ be a unital ${\mathcal {A}}$-bimodule. We show that every derivation $D: M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}}),$$n\ge 2,$ can be represented as a sum $D=D_{m}+\overline{\delta },$ where $D_{m}$ is an inner derivation and $\overline{\delta }$ is a derivation induced by a derivation $\delta$ from ${\mathcal {A}}$ into ${\mathcal {M}}.$ If ${\mathcal {A}}$ commutes with ${\mathcal {M}}$, we prove that every 2-local inner derivation $\Delta : M_{n}({\mathcal {A}}) \rightarrow M_{n}({\mathcal {M}})$, $n \ge 2$, is an inner derivation. In addition, if ${\mathcal {A}}$ is commutative and commutes with ${\mathcal {M}},$ then every 2-local derivation $\Delta : M_{n}({\mathcal {A}}) \rightarrow M_{n}({\mathcal {M}})$, $n \ge 2$, is a derivation. Let ${\mathcal {R}}$ be a finite von Neumann algebra of type $\text {I}$ with center $\mathcal {Z}$ and $LS({\mathcal {R}})$ be the algebra of locally measurable operators affiliated with ${\mathcal {R}}.$ We also prove that if the lattice $\mathcal {Z}_{\mathcal {P}}$ of all projections in $\mathcal {Z}$ is atomic, then every derivation $D:{\mathcal {R}}\rightarrow LS({\mathcal {R}})$ is an inner derivation.

1 Introduction

Let ${\mathcal {A}}$ be an algebra over ${\mathbb {C}}$ the field of complex numbers and ${\mathcal {M}}$ be an ${\mathcal {A}}$-bimodule. A linear map $\delta$ from ${\mathcal {A}}$ into ${\mathcal {M}}$ is called a Jordan derivation if $\delta (a^{2})=\delta (a)a+a\delta (a)$ for each a in ${\mathcal {A}}$. A linear map $\delta$ from ${\mathcal {A}}$ into ${\mathcal {M}}$ is called a derivation if $\delta (ab)=\delta (a)b+a\delta (b)$ for each a, b in ${\mathcal {A}}$. Let m be an element in ${\mathcal {M}},$ the map $\delta _{m}:{\mathcal {A}}\rightarrow {\mathcal {M}}, ~a\rightarrow \delta _{m}(a):=ma-am,$ is a derivation. A derivation $\delta :{\mathcal {A}}\rightarrow {\mathcal {M}}$ is said to be an inner derivation when it can be written in the form $\delta =\delta _{m}$ for some m in ${\mathcal {M}}.$ A fundamental result, due to Sakai [18], states that every derivation on a von Neumann algebra is an inner derivation.

An algebra ${\mathcal {A}}$ is called 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 ${\mathcal {R}}$ be a von Neumann algebra. We denote $S({\mathcal {R}})$ and $LS({\mathcal {R}})$, respectively, the algebras of all measurable and locally measurable operators affiliated with ${\mathcal {R}}.$ For a faithful normal semi-finite trace $\tau$ on ${\mathcal {R}},$ we denote the algebra of all $\tau$-measurable operators from $S({\mathcal {R}})$ by $S({\mathcal {R}},\tau )$ (cf. [1, 4, 14]). If ${\mathcal {R}}$ is an abelian von Neumann algebra, then it is $*$-isomorphic to the algebra $L^{\infty }(\Omega )=L^{\infty }(\Omega ,\Sigma ,\mu )$ of all (classes of equivalence of) essentially bounded measurable complex functions on a measurable space $(\Omega ,\Sigma ,\mu )$, and therefore, $LS({\mathcal {R}})=S({\mathcal {R}})\cong L^{0}(\Omega ),$ where $L^{0}(\Omega )=L^{0}(\Omega ,\Sigma ,\mu )$ is a unital commutative regular algebra of all measurable complex functions on $(\Omega ,\Sigma ,\mu ).$ In this case inner derivations on the algebra $S({\mathcal {R}})$ are identically zero, i.e., trivial.

Ber et al. [9] obtain necessary and sufficient conditions for existence of non-trivial derivations on commutative regular algebras. In particular, they prove that the algebra $L^{0}(0,1)$ of all measurable complex functions on the interval (0, 1) admits non-trivial derivations. Let ${\mathcal {R}}$ be a properly infinite von Neumann algebra. Ayupov and Kudaybergenov [4] show that every derivation on the algebra $LS({\mathcal {R}})$ is an inner derivation.

In 1997, $\breve{S}$emrl [17] introduced 2-local derivations and 2-local automorphisms. A map $\Delta : {\mathcal {A}} \rightarrow {\mathcal {M}}$ (not necessarily linear) is called a 2-local derivation if, for every $x, y \in {\mathcal {A}}$, there exists a derivation $D_{x, y}: {\mathcal {A}} \rightarrow {\mathcal {M}}$ such that $D_{x, y}(x)=\Delta (x)$ and $D_{x,y}(y) = \Delta (y)$. In particular, if, for every $x, y \in {\mathcal {A}},$$D_{x, y}$ is an inner derivation, then we call $\Delta$ is a 2-local inner derivation. Niazi and Peralta [15] introduce the notion of weak-2-local derivation (respectively, $^{*}$-derivation) and prove that every weak-2-local $^{*}$-derivation on $M_{n}$ is a derivation. 2-local derivations and weak-2-local derivations have been investigated by many authors on different algebras and many results have been obtained in [3,4,5,6,7,8, 11, 13, 15,16,17, 19].

Let ${\mathcal {H}}$ be a infinite-dimensional separable Hilbert space. In [17] $\breve{S}$emrl shows that every 2-local derivation on $\mathcal {B}({\mathcal {H}})$ is a derivation. Kim and Kim [13] give a short proof of that every 2-local derivation on a finite-dimensional complex matrix algebra is a derivation. Ayupov and Kudaybergenov [3] extend this result to an arbitrary von Neumann algebra. Ayupov et al. [5] prove that if ${\mathcal {R}}$ is a finite von Neumann algebra of type $\text {I}$ without abelian direct summands, then each 2-local derivation on the algebra $LS({\mathcal {R}})=S({\mathcal {R}})$ is a derivation. In the same paper, the authors also show that if ${\mathcal {R}}$ is an abelian von Neumann algebra such that the lattice of all projections in ${\mathcal {R}}$ is not atomic, then there exists a 2-local derivation on the algebra $S({\mathcal {R}})$ which is not a derivation. Zhang and Li [19] construct an example of a 2-local derivation on the algebra of all triangular complex $2\times 2$ matrices which is not a derivation.

Ayupov et al. [5] show that if ${\mathcal {A}}$ is a unital commutative regular algebra, then every 2-local derivation on the algebra $M_{n}({\mathcal {A}}),$$n\ge 2,$ is a derivation. Ayupov and Arzikulov [8] show that if ${\mathcal {A}}$ is a unital commutative ring, then every 2-local inner derivation on $M_{n}({\mathcal {A}}), ~n\ge 2,$ is an inner derivation. Let ${\mathcal {A}}$ be a unital Banach algebra and ${\mathcal {M}}$ be a unital ${\mathcal {A}}$-bimodule. He et al. [11] prove that if every Jordan derivation from ${\mathcal {A}}$ into ${\mathcal {M}}$ is an inner derivation then every 2-local derivation from $M_{n}({\mathcal {A}})$$(n\ge 3)$ into $M_{n}({\mathcal {M}})$ is a derivation.

Throughout this paper, ${\mathcal {A}}$ is an algebra with unit 1 over ${\mathbb {C}}$ and ${\mathcal {M}}$ is a unital ${\mathcal {A}}$-bimodule. We say that ${\mathcal {A}}$commutes with${\mathcal {M}}$ if $am=ma$ for every $a\in {\mathcal {A}}$ and $m\in {\mathcal {M}}$. From now on, $M_{n}({\mathcal {A}})$, for $n \ge 2,$ will denote the algebra of all $n\times n$ matrices over ${\mathcal {A}}$ with the usual operations. By the way, we denote any element in $M_{n}({\mathcal {A}})$ by $(a_{rs})_{n\times n},$ where $r,s\in \{ 1,2,\ldots ,n \};$$E_{ij},$$i,j\in \{ 1,2,\ldots ,n \},$ the matrix units in $M_{n}({\mathbb {C}})$; and $x\otimes E_{ij},$ the matrix whose (i, j)-th entry is x and zero elsewhere. We use $A_{ij}$ for the (i, j)-th entry of $A\in M_{n}({\mathcal {A}})$ and denote $\mathrm{diag}(x_{1},\ldots , x_{n})$ or $\mathrm{diag}(x_{i})$ the diagonal matrix with entries $x_{i}\in {\mathcal {A}}$, $i\in {\{1,2,\ldots ,n\}},$ in the diagonal positions. Particularly, we denote $\mathrm{diag}(x_{i})$ by $\mathrm{diag}(x)$, where $x_{i}=x$ for every $i\in {\{1,2,\ldots ,n\}}.$

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

$$\begin{aligned} \overline{\delta }((a_{ij})_{n\times n})=(\delta (a_{ij}))_{n\times n}, \end{aligned}$$

(1.1)

we obtain a well-defined linear operator from $M_{n}({\mathcal {A}})$ into $M_{n}({\mathcal {M}}),$ where $M_{n}({\mathcal {M}})$ has a natural structure of $M_{n}({\mathcal {A}})$-bimodule. Moreover, $\overline{\delta }$ is a derivation from $M_{n}({\mathcal {A}})$ into $M_{n}({\mathcal {M}}).$ If ${\mathcal {A}}$ is a commutative algebra, then the restriction of $\overline{\delta }$ onto the center of the algebra $M_{n}({\mathcal {A}})$ coincides with the given $\delta .$

In this paper we give characterizations of derivations, 2-local inner derivations and 2-local derivations from $M_{n}({\mathcal {A}})$ into $M_{n}({\mathcal {M}})$. In Sect. 2, we show that a derivation $D: M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}}),$$n\ge 2$, can be decomposed as a sum of an inner derivation and a derivation induced by a derivation from ${\mathcal {A}}$ to ${\mathcal {M}}$ as (1.1), as follows:

$$\begin{aligned} D=D_{B}+\overline{\delta }. \end{aligned}$$

In addition, the representation of the above form is unique if and only if ${\mathcal {A}}$ commutes with ${\mathcal {M}}$. Let ${\mathcal {R}}$ be a finite von Neumann algebra of type $\text {I}$ with center $\mathcal {Z}$ and $LS({\mathcal {R}})$ be the algebra of locally measurable operators affiliated with ${\mathcal {R}}.$ we prove that if the lattice $\mathcal {Z}_{\mathcal {P}}$ of all projections in $\mathcal {Z}$ is atomic, then every derivation $D:{\mathcal {R}}\rightarrow LS({\mathcal {R}})$ is an inner derivation.

In Sect. 3, we consider 2-local inner derivations and 2-local derivations from $M_{n}({\mathcal {A}})$ into $M_{n}({\mathcal {M}})$. For the case that ${\mathcal {A}}$ commutes with ${\mathcal {M}}$, we obtain that every inner 2-local derivation from $M_{n}({\mathcal {A}})$ into $M_{n}({\mathcal {M}})$ is an inner derivation. In addition, if ${\mathcal {A}}$ is commutative, we prove that every 2-local derivation $\Delta : M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}})$, $n\ge 2$, is a derivation. Let ${\mathcal {R}}$ be an arbitrary von Neumann algebra without abelian direct summands. We also show every 2-local derivation $\Delta : {\mathcal {R}}\rightarrow LS({\mathcal {R}})$ is a derivation.

2 Derivations

Let ${\mathcal {A}}$ be an algebra with unit 1 over ${\mathbb {C}}$ and ${\mathcal {M}}$ be a unital ${\mathcal {A}}$-bimodule. Let $D: M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}}),$$n\ge 2,$ be a derivation. Firstly, we define a map $D^{ij}_{rs}: {\mathcal {A}}\rightarrow {\mathcal {M}}$ by

$$\begin{aligned} D^{ij}_{rs}(a)=[D(a\otimes E_{rs})]_{ij},\quad a\in {\mathcal {A}},~i,j,r,s\in {\{1,2,\ldots ,n\}}. \end{aligned}$$

For any $a,b\in {\mathcal {A}}$ and some fixed $m\in {\{1,2,\ldots ,n\}},$ we have

$$\begin{aligned} D^{ij}_{rs}(ab)&=[D(ab\otimes E_{rs})]_{ij}\nonumber \\&=[D((a\otimes E_{rm})(b\otimes E_{ms}))]_{ij}\nonumber \\&=[D(a\otimes E_{rm})(b\otimes E_{ms})]_{ij}+[(a\otimes E_{rm})D(b\otimes E_{ms})]_{ij}\nonumber \\&=\delta _{js}[D(a\otimes E_{rm})]_{im}b+\delta _{ir}a[D(b\otimes E_{ms})]_{mj}, \end{aligned}$$

where $\delta$ is the Kronecker’s delta. It follows that

$$\begin{aligned} D^{ij}_{rs}(ab)=\delta _{js}[D(a\otimes E_{rm})]_{im}b+\delta _{ir}a[D(b\otimes E_{ms})]_{mj}. \end{aligned}$$

(2.1)

For any $m\in {\{1,2,\ldots ,n\}},$ we deduce from the equality (2.1) that

$$\begin{aligned} D^{mm}_{mm}(ab)=D^{mm}_{mm}(a)b+aD^{mm}_{mm}(b), \end{aligned}$$

thus $D^{mm}_{mm}:{\mathcal {A}}\rightarrow {\mathcal {M}}$ is a derivation. We abbreviate the derivation $D^{mm}_{mm}$ by $D^{m}.$ Particularly, we denote the derivation $D^{11}_{11}$ by $D^{1}.$

Theorem 2.1

Every derivation $D: M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}}),$$n\ge 2,$ can be represented as a sum

$$\begin{aligned} D=D_{B}+\overline{\delta }, \end{aligned}$$

(2.2)

where $D_{B}$ is an inner derivation implemented by an element $B\in M_{n}({\mathcal {M}})$ and $\overline{\delta }$ is a derivation of the form (1.1) induced by a derivation $\delta$ from ${\mathcal {A}}$ into ${\mathcal {M}}$. Furthermore, if this representation is unique for every derivation D, then ${\mathcal {A}}$ commutes with ${\mathcal {M}}$ (i.e., $am=ma$ for every $a\in {\mathcal {A}},~m\in {\mathcal {M}}$); and if ${\mathcal {A}}$ commutes with ${\mathcal {M}}$ then this representation is always unique.

Before the proof of Theorem 2.1, we first present the following lemma.

Lemma 2.2

For every $~i,j,r,s,m\in {\{1,2,\ldots ,n\}}$ and every $a\in {\mathcal {A}}$ the following equalities hold:

• (i) $D^{ij}_{rs}=0,$$i\ne r$ and $j\ne s,$

• (ii) $D^{ij}_{rj}(a)=D^{im}_{rm}(a)=D^{im}_{rm}(1)a,$ if $i\ne r,$

• (iii) $D^{ij}_{is}(a)=D^{mi}_{ms}(a)=aD^{mj}_{ms}(1),$ if $j\ne s,$

• (iv) $D^{im}_{jm}(1)=-D^{mj}_{mi}(1),$

• (v) $D^{ij}_{ij}(a)=D^{im}_{im}(1)a-aD^{jm}_{jm}(1)+D^{m}(a).$

Proof

It obviously follows from (2.1) that statements $\mathrm{(i)}$, $\mathrm{(ii)}$ and $\mathrm{(iii)}$ hold. We only need to prove $\mathrm{(iv)}$ and $\mathrm{(v)}.$

$\mathrm{(iv)}$: In the case $i=j,$ we have

$$\begin{aligned} 0&=[D(1\otimes E_{ii})]_{ii}=[D((1\otimes E_{im})(1\otimes E_{mi}))]_{ii}\nonumber \\&=[D((1\otimes E_{im}))(1\otimes E_{mi})]_{ii}+[(1\otimes E_{im})D((1\otimes E_{mi}))]_{ii}\nonumber \\&=D^{im}_{im}(1)+D^{mi}_{mi}(1), \end{aligned}$$

i.e.,

$$\begin{aligned} D^{im}_{im}(1)=-D^{mi}_{mi}(1). \end{aligned}$$

(2.3)

For the case $i\ne j,$ we have

$$\begin{aligned} 0&= D(0)=[D((1\otimes E_{ii})(1\otimes E_{jj}))]_{ij}\nonumber \\&=[D((1\otimes E_{ii}))(1\otimes E_{jj})]_{ij}+[(1\otimes E_{ii})D((1\otimes E_{jj}))]_{ij}\nonumber \\&=[D(1\otimes E_{ii})]_{ij}+[D(1\otimes E_{jj})]_{ij}\nonumber \\&=D^{ij}_{ii}(1)+D^{ij}_{jj}(1), \end{aligned}$$

i.e.,

$$\begin{aligned} D^{ij}_{jj}(1)=-D^{ij}_{ii}(1). \end{aligned}$$

By $\mathrm{(ii)}$, $\mathrm{(iii)}$ and equality (2.3), it follows that

$$\begin{aligned} D^{im}_{jm}(1)=-D^{mj}_{mi}(1). \end{aligned}$$

$\mathrm{(v)}$: By equality (2.1), we have

$$\begin{aligned} D^{ij}_{ij}(a)=D^{im}_{im}(1)a+D^{mj}_{mj}(a), \end{aligned}$$

(2.4)

and

$$\begin{aligned} D^{ij}_{ij}(a)=D^{im}_{im}(a)+aD^{mj}_{mj}(1). \end{aligned}$$

(2.5)

Taking $j=m$ in equality (2.4), we obtain that

$$\begin{aligned} D^{im}_{im}(a)=D^{im}_{im}(1)a+D^{m}(a). \end{aligned}$$

(2.6)

By equalities (2.3), (2.5) and (2.6), it follows that

$$\begin{aligned} D^{ij}_{ij}(a)=D^{im}_{im}(1)a-aD^{jm}_{jm}(1)+D^{m}(a). \end{aligned}$$

The proof is complete. $\square$

Now we are in position to prove Theorem 2.1.

Proof of Theorem 2.1

Let $(a_{rs})_{n\times n}$ be an arbitrary element in $M_{n}({\mathcal {A}})$ and D be a derivation from $M_{n}({\mathcal {A}})$ into $M_{n}({\mathcal {M}})$. For any $i,j\in {\{1,2,\ldots ,n\}},$ it follows from Lemma 2.2 that

$$\begin{aligned}{}[D((a_{rs})_{n\times n})]_{ij}&=\sum ^{n}_{r,s=1}D^{ij}_{rs}(a_{rs})\nonumber \\&=\sum ^{n}_{r=1}D^{ij}_{rj}(a_{rj})+\sum ^{n}_{s=1}D^{ij}_{is}(a_{is})-D^{ij}_{ij}(a_{ij})\nonumber \\&=\sum _{r\ne i}D^{ij}_{rj}(a_{rj})+\sum _{s\ne j}D^{ij}_{is}(a_{is})+D^{ij}_{ij}(a_{ij})\nonumber \\&=\sum _{r\ne i}D^{i1}_{r1}(1)a_{rj}+\sum _{s\ne j}a_{is}D^{1j}_{1s}(1)+D^{i1}_{i1}(1)a_{ij} \nonumber \\&\quad -\,a_{ij}D^{j1}_{j1}(1)+D^{1}(a_{ij})\nonumber \\&=\sum ^{n}_{r=1}D^{i1}_{r1}(1)a_{rj} -\sum ^{n}_{s=1}a_{is}D^{s1}_{j1}(1)+D^{1}(a_{ij})\nonumber \\&=\sum ^{n}_{k=1}\left( D^{i1}_{k1}(1)a_{kj} -a_{ik}D^{k1}_{j1}(1)\right) +D^{1}(a_{ij})\nonumber \\&=\left[ (D^{r1}_{s1}(1))_{n\times n}(a_{rs})_{n\times n}-(a_{rs})_{n\times n}(D^{r1}_{s1}(1))_{n\times n}\right] _{ij}\nonumber \\&\quad +\,\left[ \overline{D^{1}}((a_{rs})_{n\times n})\right] _{ij}, \end{aligned}$$

i.e.,

$$\begin{aligned}{}[D((a_{rs})_{n\times n})]_{ij}&=\left[ \left( D^{r1}_{s1}(1))_{n\times n}(a_{rs})_{n\times n}-(a_{rs})_{n\times n}(D^{r1}_{s1}(1)\right) _{n\times n}\right] _{ij}\nonumber \\&\quad +\left[ \overline{D^{1}}((a_{rs})_{n\times n})\right] _{ij}, \end{aligned}$$

(2.7)

where $(D^{r1}_{s1}(1))_{n\times n}\in M_{n}({\mathcal {M}})$ and $[(D^{r1}_{s1}(1))_{n\times n}]_{ij}=D^{i1}_{j1}(1).$ By equality (2.7), we have

$$\begin{aligned} D((a_{rs})_{n\times n})=\left[ (D^{r1}_{s1}(1))_{n\times n}(a_{rs})_{n\times n}-(a_{rs})_{n\times n}(D^{r1}_{s1}(1))_{n\times n}\right] +\left[ \overline{D^{1}}((a_{rs})_{n\times n})\right] . \end{aligned}$$

We denote $B=(D^{r1}_{s1}(1))_{n\times n}$ and $\delta =D^{1}.$ Therefore, every derivation $D: M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}}),$$n\ge 2,$ can be represented as a sum

$$\begin{aligned} D=D_{B}+\overline{\delta }. \end{aligned}$$

Suppose that $D_{M}$ is an inner derivation from $M_{n}({\mathcal {A}})$ into $M_{n}({\mathcal {M}})$ implemented by an element $M\in M_{n}({\mathcal {M}}),$ and $\overline{\zeta }$ is a derivation of the form (1.1) induced by a derivation $\zeta$ from ${\mathcal {A}}$ into ${\mathcal {M}},$ such that $D_{M}=\overline{\zeta }.$ The first step is to establish the following. $\square$

$\mathbf Claim~1$ If ${\mathcal {A}}$ commutes with ${\mathcal {M}}$, then $D_{M}=\overline{\zeta }=0.$

Proof of Claim 1

If $i\ne j,~i,j\in {\{1,2,\ldots ,n\}},$ we have

$$\begin{aligned} 0=\overline{\zeta }(E_{ij})=D_{M}(E_{ij})=ME_{ij}-E_{ij}M. \end{aligned}$$

It follows that $M_{ji}=0.$ Thus, M has a diagonal form, i.e., $M=\mathrm{diag}(M_{kk}).$ Suppose that $\overline{\zeta }\ne 0,$ then there exists an element $a\in {\mathcal {A}}$ such that $\zeta (a)\ne 0.$ Take $A=\mathrm{diag}(a),$ then $\overline{\zeta }(A)\ne 0.$ On the other hand,

$$\begin{aligned} \overline{\zeta }(A)=D_{M}(A)=\mathrm{diag}(M_{kk})\mathrm{diag}(a)-\mathrm{diag}(a)diag(M_{kk})=0. \end{aligned}$$

This is a contradiction. Thus, $~\overline{\zeta }=0.$$\square$

$\mathbf Claim~2$ If ${\mathcal {A}}$ does not commute with ${\mathcal {M}}$, then there exist $D_{M}$ and $\overline{\zeta },$ such that  $D_{M}=\overline{\zeta }\ne 0.$

Proof of Claim 2

By assumption, we can take $a\in {\mathcal {A}}$ and $m\in {\mathcal {M}}$ such that $ma\ne am.$ We define a derivation $\zeta :{\mathcal {A}}\rightarrow {\mathcal {M}}$ by $\zeta (x)=mx-xm$ for every x in ${\mathcal {A}}.$ We denote $M=\mathrm{diag}(m)\in M_{n}({\mathcal {M}}),$ then $D_{M}$ is an inner derivation from $M_{n}({\mathcal {A}})$ into $M_{n}({\mathcal {M}}).$ Obviously, $D_{M}=\overline{\zeta }~$ and $~\overline{\zeta }(\mathrm{diag}(a))\ne 0.$ Thus, $D_{M}=\overline{\zeta }\ne 0.$

In the following, we show that the representation of the above form is unique if and only if ${\mathcal {A}}$ commutes with ${\mathcal {M}}.$

Case 1 If ${\mathcal {A}}$ commutes with ${\mathcal {M}},$ we suppose that there exists a derivation $D:M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}}),$$n\ge 2,$ which can be represented as $D=D_{B_{1}}+\overline{\delta _{1}}=D_{B_{2}}+\overline{\delta _{2}}.$ This means that $D_{B_{1}}-D_{B_{2}}=\overline{\delta _{2}}-\overline{\delta _{1}}.$ Since $D_{B_{1}}-D_{B_{2}}=D_{B_{1}-B_{2}}$ and $\overline{\delta _{2}}-\overline{\delta _{1}} =\overline{\delta _{2}-\delta _{1}},$ we have $D_{B_{1}-B_{2}}=\overline{\delta _{2}-\delta _{1}}.$ It follows from Claim 1 that $D_{B_{1}-B_{2}}=\overline{\delta _{2}-\delta _{1}}=0.$ i.e., $D_{B_{1}}=D_{B_{2}}$ and $\overline{\delta _{1}}=\overline{\delta _{2}}.$

Case 2 If ${\mathcal {A}}$ does not commute with ${\mathcal {M}},$ by Claim 2, there exist derivations $D_{M}$ and $\overline{\zeta }$ from $M_{n}({\mathcal {A}})$ into $M_{n}({\mathcal {M}}),$$n\ge 2,$ such that $D_{M}=\overline{\zeta }\ne 0.$ Let $D:M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}}),$$n\ge 2,$ be an arbitrary derivation. By hypothesis, D can be represented as $D=D_{B}+\overline{\delta }.$ We have $D=D_{B}+\overline{\delta }=D_{B}+D_{M}- \overline{\zeta }+\overline{\delta }=D_{B+M}+\overline{\delta -\zeta }.$ This means that the derivation D can be represented as $D=D_{B}+\overline{\delta },$ and as $D=D_{B+M}+\overline{\delta -\zeta }$ too. Therefore, the representation of (2.2) is not unique for every derivation D. It follows from Cases 1 and 2 that the representation of (2.2) is unique if and only if ${\mathcal {A}}$ commutes with ${\mathcal {M}}$. The proof is complete. $\square$

As applications of Theorem 2.1, we obtain the following corollaries.

Corollary 2.3

The following statements are equivalent.

• (i) Every derivation $\delta :{\mathcal {A}}\rightarrow {\mathcal {M}}$ is an inner derivation.

• (ii) Every derivation $D:M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}}),$$n\ge 2,$ is an inner derivation.

Proof

If $\delta :{\mathcal {A}}\rightarrow {\mathcal {M}}$ is an inner derivation, by the equality (1.1), obviously, $\overline{\delta }:M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}}),$$n\ge 2,$ is an inner derivation.

$\mathrm{(i)}$ implies $\mathrm{(ii)}$: Let $D:M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}}),$$n\ge 2,$ be an arbitrary derivation. By Theorem 2.1, D can be represented as a sum $D=D_{M}+\overline{\delta },$ where $D_{M}$ is an inner derivation. By hypothesis, $\delta$ is an inner derivation from ${\mathcal {A}}$ into ${\mathcal {M}},$ and therefore, $\overline{\delta }$ is an inner derivation. We know that the sum of two inner derivations is an inner derivation, this means that $D:M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}}),$$n\ge 2,$ is an inner derivation.

$\mathrm{(ii)}$ implies $\mathrm{(i)}$: Suppose that $\delta$ is a derivation from ${\mathcal {A}}$ into ${\mathcal {M}},$ then $\overline{\delta }:M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}}),$$n\ge 2,$ is a derivation. By hypothesis, $\overline{\delta }$ is an inner derivation. then the restriction of $\overline{\delta }$ onto $E_{11}M_{n}({\mathcal {A}})E_{11},$ the subalgebra of $M_{n}({\mathcal {A}}),$ is an inner derivation. This means that $\delta :{\mathcal {A}}\rightarrow {\mathcal {M}}$ is an inner derivation. $\square$

Corollary 2.4

Let ${\mathcal {A}}$ be a commutative unital algebra over ${\mathbb {C}}.$ Then every derivation on the matrix algebra $M_{n}({\mathcal {A}})~(n\ge 2)$ is inner if and only if every derivation on ${\mathcal {A}}$ is identically zero, i.e., trivial.

Let ${\mathcal {R}}$ be a von Neumann algebra. Denote by $S({\mathcal {R}})$ and $LS({\mathcal {R}})$, respectively, the sets of all measurable and locally measurable operators affiliated with ${\mathcal {R}}.$ Then the set $LS({\mathcal {R}})$ of all locally measurable operators with respect to ${\mathcal {R}}$ is a unital $*$-algebra when equipped with the algebraic operations of strong addition and multiplication and taking the adjoint of an operator and $S({\mathcal {R}})$ is a solid $*$-subalgebra in $LS({\mathcal {R}})$. If ${\mathcal {R}}$ is a finite von Neumann algebra, then $S({\mathcal {R}})=LS({\mathcal {R}})$ (see, for example, [1, 4, 14]). Let ${\mathcal {A}}$ be a commutative algebra with unit 1 over ${\mathbb {C}}.$ We denote by $\nabla$ the set $\{e\in {\mathcal {A}}:e^{2}=e\}$ of all idempotents in ${\mathcal {A}}.$ For $e,f\in \nabla$ we set $e\le f$ if $ef=e.$ Equipped with this partial order, lattice operations $e\vee f=e+f-ef,$$e\wedge f=ef$ and the complement $e^{\perp }=1-e,$ the set $\nabla$ forms a Boolean algebra. A nonzero element q from the Boolean algebra $\nabla$ is called an atom if $0\ne e\le q,$$e\in \nabla ,$ imply that $e=q.$ If given any nonzero $e\in \nabla$ there exists an atom q such that $q\le e,$ then the Boolean algebra $\nabla$ is said to be atomic.

Let ${\mathcal {R}}$ be an abelian von Neumann algebra. Theorem 3.4 of [9] implies that every derivation on the algebra $S({\mathcal {R}})$ is inner if and only if the lattice ${\mathcal {R}}_{\mathcal {P}}$ of all projections in ${\mathcal {R}}$ is atomic. If ${\mathcal {R}}$ is a properly infinite von Neumann algebra, in [4] the authors show that every derivation on the algebra $LS({\mathcal {R}})$ is inner (see [4], Theorem 4.6). In the case of ${\mathcal {R}}$ is a finite von Neumann algebra of type $\text {I}$, Theorem 3.5 of [4] shows that a derivation on the algebra $LS({\mathcal {R}})$ is an inner derivation if and only if it is identically zero on the center of ${\mathcal {R}}.$

As a direct application of Corollary 2.3, we obtain the following result.

Corollary 2.5

Let ${\mathcal {R}}$ be a finite von Neumann algebra of type $\text {I}$ with center $\mathcal {Z}.$ Then every derivation D on the algebra $LS({\mathcal {R}})$ is inner if and only if the lattice $\mathcal {Z}_{\mathcal {P}}$ of all projections in $\mathcal {Z}$ is atomic.

Proof

Let ${\mathcal {R}}$ be a finite von Neumann algebra of type $\text {I}$ with center $\mathcal {Z}.$ There exists a family $\{e_{n}\}_{n\in \mathcal {F}},$$\mathcal {F}\subseteq \mathbb {N},$ of central projections from ${\mathcal {R}}$ with $\bigvee \nolimits _{n\in \mathcal {F}}e_{n}=1$ such that the algebra ${\mathcal {R}}$ is $*$-isomorphic with the $C^{*}$-product of von Neumann algebras $e_{n}{\mathcal {R}}$ of type $\text {I}_{n}$, respectively, $n\in \mathcal {F},$ i.e., ${\mathcal {R}} \cong \bigoplus \nolimits _{n\in \mathcal {F}}e_{n}{\mathcal {R}}.$ By Proposition 1.1 of [1], we have that $LS({\mathcal {R}}) \cong \prod \nolimits _{n\in \mathcal {F}}LS(e_{n}{\mathcal {R}}).$

Suppose that D is a derivation on $LS({\mathcal {R}})$ and $\delta$ its restriction onto the center $S(\mathcal {Z}).$ Since $\delta$ maps each $e_{n}S(\mathcal {Z})$ into itself, $\delta$ generates a derivation $\delta _{n}$ on $e_{n}S(\mathcal {Z})$ for each $n\in \mathcal {F}.$ By Proposition 1.5 of [1], $LS(e_{n}{\mathcal {R}})\cong M_{n}(e_{n}S(\mathcal {Z}))$. Let $\overline{\delta }_{n}$ be the derivation on the matrix algebra $M_{n}(e_{n}S(\mathcal {Z}))$ defined as in (1.1). Put

$$\begin{aligned} \overline{\delta }(\{x_{n}\}_{n\in \mathcal {F}}) =\{\overline{\delta }_{n}(x_{n})\},\quad \{x_{n}\}_{n\in \mathcal {F}}\in LS({\mathcal {R}}). \end{aligned}$$

(2.8)

Then the map $\overline{\delta }$ is a derivation on $LS({\mathcal {R}}).$ Lemma 2.3 of [1] implies that each derivation D on $LS({\mathcal {R}})$ can be uniquely represented in the form $D=D_{B}+\overline{\delta },$ where $D_{B}$ is an inner derivation and $\overline{\delta }$ is a derivation given as (2.8).

If D is an arbitrary derivation on $LS({\mathcal {R}})$ and $\delta$ its restriction onto center $S(\mathcal {Z}),$ by Theorem 3.4 of [9], the lattice $\mathcal {Z}_{\mathcal {P}}$ is atomic if and only if $\delta =0.$ We have $\delta =0$ if and only if $\delta _{n}=0$ for each $n\in \mathcal {F}.$ By Corollary 2.3, $\delta _{n}=0$ if and only if $\overline{\delta _{n}}=0$ for each $n\in \mathcal {F}.$ By equality (2.8), $\overline{\delta _{n}}=0$ for each $n\in \mathcal {F}$ if and only if $\overline{\delta }=0.$ Therefore, every derivation on the algebra $LS({\mathcal {R}})$ is inner derivation if and only if the lattice $\mathcal {Z}_{\mathcal {P}}$ of all projections in $\mathcal {Z}$ is atomic. The proof is complete. $\square$

Let ${\mathcal {R}}$ be a properly infinite von Neumann algebra and ${\mathcal {M}}$ be a ${\mathcal {R}}$-bimodule of locally measurable operators. In [10], the authors show that every derivation $D:{\mathcal {R}}\rightarrow {\mathcal {M}}$ is an inner derivation. In the case of ${\mathcal {R}}$ is a finite von Neumann algebra of type $\text {I}$, we obtain the following result.

Theorem 2.6

Let ${\mathcal {R}}$ be a finite von Neumann algebra of type $\text {I}$ with center $\mathcal {Z}.$ If the lattice $\mathcal {Z}_{\mathcal {P}}$ of all projections in $\mathcal {Z}$ is atomic, then every derivation $D:{\mathcal {R}}\rightarrow LS({\mathcal {R}})$ is an inner derivation.

Proof

Choose a central decomposition $\{e_{n}\}_{n\in \mathcal {F}},$$\mathcal {F}\subseteq \mathbb {N},$ of the unity 1 such that $e_{n}{\mathcal {R}}$ is a type $\text {I}_{n}$ von Neumann algebra for each $n\in \mathcal {F}.$ By hypothesis, it is easy to check that $D(e_{n}{\mathcal {R}})\subseteq e_{n}LS({\mathcal {R}})$ for each $n\in \mathcal {F}.$ Thus, we only need to show that the derivation D restricted to $e_{n}{\mathcal {R}}$ is an inner derivation for each $n\in \mathcal {F}.$

Let $e_{n}{\mathcal {R}}$ be a type $\text {I}_{n} ~(n\in \mathcal {F})$ von Neumann algebra with center $e_{n}\mathcal {Z}.$ It is well known that $e_{n}{\mathcal {R}}\cong M_{n}(e_{n}\mathcal {Z}).$ We denote the center of $S(e_{n}{\mathcal {R}})$ by $\mathcal {Z}(S(e_{n}{\mathcal {R}}))$. By Proposition 1.2 of [1], we have $\mathcal {Z}(S(e_{n}{\mathcal {R}}))=S(e_{n}\mathcal {Z}).$ By Proposition 1.5 of [1], $LS(e_{n}{\mathcal {R}})=S(e_{n}{\mathcal {R}})\cong M_{n}(S(e_{n}\mathcal {Z})).$

By assumption, the lattice $\mathcal {Z}_{\mathcal {P}}$ of all projections in $\mathcal {Z}$ is atomic. This implies that the lattice $e_{n} \mathcal {Z}_{\mathcal {P}}$ is also atomic for each $n\in \mathcal {F}.$ Statements $\mathrm{(ii)}$ of Proposition 2.3 and $\mathrm{(vi)}$ of Proposition 2.6 of [9] imply that every derivation $\delta :e_{n}\mathcal {Z}\rightarrow S(e_{n}\mathcal {Z})$ is trivial. By Corollary 2.3, we have that every derivation from $M_{n}(e_{n}\mathcal {Z})$ into $M_{n}(S(e_{n}\mathcal {Z}))$ is inner. The proof is complete. $\square$

3 2-Local Derivations

This section is devoted to 2-local inner derivations and 2-local derivations from $M_{n}({\mathcal {A}})$ into $M_{n}({\mathcal {M}})$. Throughout this section, we always assume that $\Delta : M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}})$ is a 2-local derivation. Firstly, we give the following lemma.

Lemma 3.1

For every 2-local derivation $\Delta : M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}})$, $n\ge 2$, there exists a derivation $D: M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}})$ such that $\Delta (E_{ij})=D(E_{ij})$ for all $i,j\in {\{1,2,\ldots ,n\}}.$ In particular, if $\Delta$ is a 2-local inner derivation, then D is an inner derivation.

Proof

Let $\Delta : M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}})$, $n\ge 2$, be a 2-local derivation. By Theorem 2.1, with the proof similar to the proof of Theorem 3 in [13], it is easy to check that there exists a derivation D such that $\Delta (E_{ij})=D(E_{ij})$ for all $i,j\in {\{1,2,\ldots ,n\}}.$

Let $\Delta$ be an inner 2-local derivation. We define two matrices S, T in $M_{n}({\mathcal {A}})$ by

$$\begin{aligned} S=\sum ^{n}_{i=1}i1\otimes E_{ii},\quad T=\sum ^{n-1}_{i=1}E_{ii+1}. \end{aligned}$$

By assumption, there exists an inner derivation $D: M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}})$ such that

$$\begin{aligned} \Delta (S)=D(S),\quad \Delta (T)=D(T). \end{aligned}$$

Replacing $\Delta$ by $\Delta -D$ if necessary, we may assume that $\Delta (S)=\Delta (T)=0.$ Fixed $i,j\in {\{1,2,\ldots ,n\}},$ by assumption, we can take two elements X,  Y in $M_{n}({\mathcal {M}})$ such that

$$\begin{aligned} \Delta (E_{ij})=XE_{ij}-E_{ij}X,\quad 0=\Delta (S)=XS-SX, \end{aligned}$$

and

$$\begin{aligned} \Delta (E_{ij})=YE_{ij}-E_{ij}Y,\quad 0=\Delta (T)=YT-TY. \end{aligned}$$

It follows from $XS=SX$ that X is a diagonal matrix. We denote X by $\mathrm{diag}(x_{k}).$ The equality $YT=TY$ implies that Y is of the form

$$\begin{aligned} Y=\left[ \begin{array}{llllll} \ y_{1}&{}\quad y_{2}&{}\quad y_{3}&{}\quad \cdot &{}\quad \cdot &{}\quad y_{n}\\ \ 0&{}\quad y_{1}&{}\quad y_{2}&{}\quad \cdot &{}\quad \cdot &{}\quad y_{n-1}\\ \ 0&{}\quad 0&{}\quad y_{1}&{}\quad \cdot &{}\quad \cdot &{}\quad y_{n-2}\\ \ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ \ \cdots &{}\quad \cdots &{}\quad \cdots &{}\quad \cdot &{}\quad y_{1} &{}\quad y_{2}\\ \ 0 &{}\quad 0 &{}\quad \cdots &{}\quad \cdot &{}\quad 0 &{}\quad y_{1}\\ \end{array} \right] . \end{aligned}$$

On the one side, we have

$$\begin{aligned} \Delta (E_{ij})=XE_{ij}-E_{ij}X=\mathrm{diag}(x_{k})E_{ij} -E_{ij}\mathrm{diag}(x_{k})=(x_{i}-x_{j})\otimes E_{ij}. \end{aligned}$$

On the other side, we have

$$\begin{aligned}{}[\Delta (E_{ij})]_{ij}=[YE_{ij}-E_{ij}Y]_{ij}=0. \end{aligned}$$

Therefore, $\Delta (E_{ij})=0.$ The proof is complete. $\square$

Theorem 3.2

Suppose that ${\mathcal {A}}$ commutes with ${\mathcal {M}}.$ Then every 2-local inner derivation $\Delta : M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}})$, $n\ge 2$, is an inner derivation.

Proof

By Lemma 3.1, we may assume that $\Delta (E_{ij})=0$ for all $i,j\in {\{1,2,\ldots ,n\}}.$ For any $A\in M_{n}({\mathcal {A}}),$ we take a pair (j, i), $j,i\in {\{1,2,\ldots ,n\}},$ by assumption, there exists an inner derivation $D_{B}$, such that $\Delta (A)=D_{B}(A)$ and $0=\Delta (E_{ij})=D_{B}(E_{ij})$. We have

$$\begin{aligned} E_{ij}\Delta (A)E_{ij}&=E_{ij}D_{B}(A)E_{ij}\nonumber \\&=D_{B}(E_{ij}AE_{ij})-D_{B}(E_{ij})AE_{ij} -E_{ij}AD_{B}(E_{ij})=D_{B}(E_{ij}AE_{ij})\nonumber \\&=D_{B}(A_{ji}\otimes E_{ij})=D_{B}(\mathrm{diag}(A_{ji}, \ldots ,A_{ji})E_{ij})\nonumber \\&=D_{B}(\mathrm{diag}(A_{ji},\ldots ,A_{ji}))E_{ij} +\mathrm{diag}(A_{ji},\ldots ,A_{ji})D_{B}(E_{ij})\nonumber \\&=(B\mathrm{diag}(A_{ji},\ldots ,A_{ji})-\mathrm{diag}(A_{ji}, \ldots ,A_{ji})B)E_{ij}\nonumber \\&=0, \end{aligned}$$

i.e.,

$$\begin{aligned} E_{ij}\Delta (A)E_{ij}=0. \end{aligned}$$

Therefore,

$$\begin{aligned} E_{ji}(E_{ij}\Delta (A)E_{ij})E_{ji}=E_{jj}\Delta (A)E_{ii}=0, \end{aligned}$$

i.e.,

$$\begin{aligned}{}[\Delta (A)]_{ji}=0, \end{aligned}$$

for every $j,i\in {\{1,2,\ldots ,n\}}.$ Hence $\Delta (A)=0.$ The proof is complete. $\square$

Corollary 3.3

Suppose that ${\mathcal {A}}$ is a unital commutative algebra over ${\mathbb {C}}.$ Then every 2-local inner derivation $\Delta : M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {A}})$, $n\ge 2$, is an inner derivation.

Remark 3.4

The above result is proved in [8]. By comparison, our proof is more simple.

Suppose that ${\mathcal {A}}$ is an algebra over ${\mathbb {C}}$ and $\mathcal {B}$ is a unital subalgebra in ${\mathcal {A}}.$ We denote the commutant of $\mathcal {B}$ by $\mathcal {B}^{\prime }=\{a\in {\mathcal {A}}:ab=ba, {for~every~}b\in \mathcal {B}\}$. Let $\mathcal {C}$ be a submodule in $\mathcal {B}^{\prime }.$ It follows from Theorem 3.2 that

Corollary 3.5

Every 2-local inner derivation $\Delta : M_{n}(\mathcal {B})\rightarrow M_{n}(\mathcal {C})$, $n\ge 2$, is an inner derivation.

Theorem 3.6

Suppose that ${\mathcal {A}}$ is a commutative algebra which commutes with ${\mathcal {M}}$. Then every 2-local derivation $\Delta : M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}})$, $n\ge 2$, is a derivation.

Proof

The proof is similar to the proof of Theorem 4.3 in [5]. We leave it to the reader. $\square$

Corollary 3.7

Suppose that ${\mathcal {A}}$ is a unital commutative algebra over ${\mathbb {C}}.$ Then every 2-local derivation $\Delta : M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {A}})$, $n\ge 2$, is a derivation.

If ${\mathcal {A}}$ is a non-commutative algebra, by Theorem 2.1 every derivation from $M_{n}({\mathcal {A}})$ into $M_{n}({\mathcal {M}})(n\ge 2)$ can be represented as a sum $D=D_{B}+\overline{\delta }.$ In [7], the authors apply this representation of derivation to prove the following result.

Theorem 3.8

([7], Theorem 2.1) Let ${\mathcal {A}}$ be a unital Banach algebra and ${\mathcal {M}}$ be a unital ${\mathcal {A}}$-bimodule. If every Jordan derivation from ${\mathcal {A}}$ into ${\mathcal {M}}$ is a derivation, then every 2-local derivation $\Delta : M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {A}})$, $n\ge 3$, is a derivation.

Theorem 3.9

Let ${\mathcal {A}}$ be a unital Banach algebra and ${\mathcal {M}}$ be a unital ${\mathcal {A}}$-bimodule. If $n\ge 6$ is a positive integer but not a prime number, then every 2-local derivation $\Delta : M_{n}({\mathcal {A}})\rightarrow M_{n}({\mathcal {M}})$ is a derivation.

Proof

Suppose that $n=rk$, where $r\ge 3$ and $k\ge 2.$ Then $M_{n}({\mathcal {A}})\cong M_{r}(M_{k}({\mathcal {A}}))$ and $M_{n}({\mathcal {M}})\cong M_{r}(M_{k}({\mathcal {M}})).$ In [2], the author proves that every Jordan derivation from $M_{k}({\mathcal {A}})$ into $M_{k}({\mathcal {M}})(k\ge 2)$ is a derivation ([2], Theorem 3.1). By Theorem 3.8, the proof is complete. $\square$

Let ${\mathcal {R}}$ be a type $\text {I}_{n}~ (n\ge 2)$ von Neumann algebra with center $\mathcal {Z}$ and $\tau$ be a faithful normal semi-finite trace on ${\mathcal {R}}.$ We denote the centers of $S({\mathcal {R}})$ and $S({\mathcal {R}},\tau )$ by $\mathcal {Z}(S({\mathcal {R}}))$ and $\mathcal {Z}(S({\mathcal {R}},\tau ))$, respectively. By Proposition 1.2 of [1], we have $\mathcal {Z}(S({\mathcal {R}}))=S(\mathcal {Z})$ and $\mathcal {Z}(S({\mathcal {R}},\tau ))=S(\mathcal {Z},\tau _{\mathcal {Z}}),$ where $\tau _{\mathcal {Z}}$ is the restriction of the trace $\tau$ on $\mathcal {Z}.$ By Propositions 1.4 and 1.5 of [1], $S({\mathcal {R}})=LS({\mathcal {R}})\cong M_{n}(S(\mathcal {Z}))$ and $S({\mathcal {R}},\tau )\cong M_{n}(S(\mathcal {Z},\tau _{\mathcal {Z}})).$

As a direct application of Theorem 3.6, we have the following corollary.

Corollary 3.10

Suppose that ${\mathcal {R}}$ is a type $\text {I}_{n}, n\ge 2,$ von Neumann algebra and $\tau$ is a faithful normal semi-finite trace on ${\mathcal {R}}.$ Then we have

• (1)   every 2-local derivation $\Delta : {\mathcal {R}}\rightarrow LS({\mathcal {R}})$ is a derivation;

• (2)   every 2-local derivation $\Delta : {\mathcal {R}}\rightarrow S({\mathcal {R}},\tau )$ is a derivation.

Lemma 3.11

Let $\Delta :{\mathcal {A}}\rightarrow {\mathcal {M}}$ be a 2-local derivation. If there exists a central idempotent e in ${\mathcal {A}}$ which commutates with ${\mathcal {M}},$ then $\Delta (ea)=e\Delta (a),$ for each a in ${\mathcal {A}}.$

Proof

For any $a \in {\mathcal {A}},$ by assumption, there exists a derivation $\delta :{\mathcal {A}}\rightarrow {\mathcal {M}}$ such that: $\Delta (ea)=\delta (ea),$ and $\Delta (a)=\delta (a).$ By assumption, e is a central idempotent in ${\mathcal {A}}$ which commutes with ${\mathcal {M}},$ it follows that $\delta (e)=0.$ Then

$$\begin{aligned} \Delta (ea)=\delta (ea)=\delta (e)a+e\delta (a)=e\delta (a)=e\Delta (a). \end{aligned}$$

The proof is complete. $\square$

Theorem 3.12

Suppose that ${\mathcal {R}}$ is a finite von Neumann algebra of type $\text {I}$ without abelian direct summands. Then every 2-local derivation $\Delta : {\mathcal {R}}\rightarrow S({\mathcal {R}})= LS({\mathcal {R}})$ is a derivation.

Proof

By assumption, ${\mathcal {R}}$ is a finite von Neumann algebra of type $\text {I}$ without abelian direct summands. Then there exists a family $\{P_{n}\}_{n\in F},~F\subseteq \mathbb {N}{\setminus }{1},$ of orthogonal central projections in ${\mathcal {R}}$ with $\sum _{n\in F} P_{n}=1,$ such that the algebra ${\mathcal {R}}$ is $*$-isomorphic with the $C^{*}$-product of von Neumann algebras $P_{n}{\mathcal {R}}$ of type $\text {I}_{n}$, respectively, $n\in F.$ Then

$$\begin{aligned} P_{n}LS({\mathcal {R}})=P_{n}S({\mathcal {R}})=S(P_{n}{\mathcal {R}})\cong M_{n}(P_{n}Z({\mathcal {R}})),\quad n\in F. \end{aligned}$$

By Lemma 3.11, we have $\Delta (P_{n}A)=P_{n}\Delta (A),$ for all $A\in {\mathcal {R}}$ and each $n\in F.$ This implies that $\Delta$ maps each $P_{n}{\mathcal {R}}$ into $P_{n}S({\mathcal {R}}).$ For each $n\in F,$ we define $\Delta _{n}:P_{n}{\mathcal {R}}\rightarrow P_{n}S({\mathcal {R}})$ by

$$\begin{aligned} \Delta _{n}(P_{n}A)=P_{n}\Delta (A),\quad A\in {\mathcal {R}}. \end{aligned}$$

By assumption, it follows that $\Delta _{n}$ is a 2-local derivation from $P_{n}{\mathcal {R}}$ into $P_{n}S({\mathcal {R}})$ for each $n\in F.$ By (1) of Corollary 3.10, we have that $\Delta _{n}$ is a derivation for each $n\in F.$ Since $\sum _{n\in F}P_{n}=1,$ it follows that $\Delta$ is a linear mapping. For any $A,B\in {\mathcal {R}},$ it follows $\Delta _{n}$ is a derivation for each $n\in F$ that

$$\begin{aligned} P_{n}\Delta (AB)&=\Delta _{n}(P_{n}AB)= \Delta _{n}(P_{n}A)P_{n}B+P_{n}A\Delta _{n}(P_{n}B)\nonumber \\&=P_{n}\Delta (A)B+P_{n}A\Delta (B)\nonumber \\&=P_{n}(\Delta (A)B+A\Delta (B)). \end{aligned}$$

By assumption, $\sum _{n\in F}P_{n}=1,$ we get

$$\begin{aligned} \Delta (AB)=\Delta (A)B+A\Delta (B). \end{aligned}$$

Therefore, $\Delta : {\mathcal {R}}\rightarrow S({\mathcal {R}})$ is a derivation. The proof is complete. $\square$

Ayupov et al. [7] have proved the following result. Now we give a different proof.

Theorem 3.13

([7], Theorem 3.1) Let ${\mathcal {R}}$ be an arbitrary von Neumann algebra without abelian direct summands and $LS({\mathcal {R}})$ be the algebra of all locally measurable operators affiliated with ${\mathcal {R}}.$ Then every 2-local derivation $\Delta : {\mathcal {R}}\rightarrow LS({\mathcal {R}})$ is a derivation.

Proof

Let ${\mathcal {R}}$ be an arbitrary von Neumann algebra without abelian direct summands. We know that ${\mathcal {R}}$ can be decomposed along a central projection into the direct sum of von Neumann algebras of finite type $\text {I}$, type $\text {I}_{\infty },$ type $\text {II}$ and type $\text {III}.$ By Lemma 3.11, we may consider these cases separately.

If ${\mathcal {R}}$ is a von Neumann algebra of finite type $\text {I}$, Theorem 3.12 shows that every 2-local derivation from ${\mathcal {R}}$ into LS$({\mathcal {R}})$ is a derivation.

If ${\mathcal {R}}$ is a von Neumann algebra of types $\text {I}_{\infty },$$\text {II}$ or $\text {III},$ then the halving Lemma ([12], Lemma 6.3.3) for type $\text {I}_{\infty }$ algebras and ([12], Lemma 6.5.6) for types $\text {II}$ or $\text {III}$ algebras implies that the unit of ${\mathcal {R}}$ can be represented as a sum of mutually equivalent orthogonal projections $e_{1},e_{2},\ldots ,e_{6}$ in ${\mathcal {R}}.$ It is well known that ${\mathcal {R}}$ is isomorphic to $M_{6}({\mathcal {A}}),$ where ${\mathcal {A}}=e_{1}{\mathcal {R}}e_{1}.$ Further, the algebra $LS({\mathcal {R}})$ is isomorphic to the algebra $M_{6}(LS({\mathcal {A}})).$ Theorem 3.9 implies that every 2-local derivation from ${\mathcal {R}}$ into LS$({\mathcal {R}})$ is a derivation. The proof is complete. $\square$