0 引言

近年来, 环或代数上可导映射, Jordan可导映射以及Lie可导映射的研究受到了国内外学者广泛关注^[1-5].Daif^[1]证明了一类含非平凡幂等元且满足一定条件的环上的可导映射是可加的.Ji^[2]得到了一类Jordan代数上的Jordan可导映射是可加的.Lu^[3]证明了含非平凡幂等元的素环上的Jordan可导映射是可加的, 并在文献[4]中给出了素环上Lie可导映射的结构表达形式.最近, Liu^[5]证明了套代数上的k-Jordan可导映射是可加的.其他相关工作可参见文献[6-17].

设δ, τ:$\mathscr{A}$→$\mathscr{E}$是两个映射(无可加或线性假设).如果∀a, b∈$\mathscr{A}$, 有δ(ab)=δ(a)b+aτ(b), 则称(δ, τ) 是$\mathscr{A}$上的可导映射对.类似地, 如果∀a, b∈A, 分别有δ(a○b)=δ(a)○b+a○τ(b) 和δ([a, b])=[δ(a), b]+[a, τ(b)], 则分别称(δ, τ) 是$\mathscr{A}$上的Jordan可导映射对和Lie可导映射对.其中a○b=ab+ba为a与b的Jordan积，[a, b]=ab-ba为a与b的Lie积.显然, 当δ=τ时, 可导映射对, Jordan可导映射对和Lie可导映射对(δ, τ) 分别是可导映射, Jordan可导映射和Lie可导映射.本文将研究三角代数上的可导映射对的可加性问题.

设$\mathscr{A}$和$\mathscr{B}$是可交换环$\mathscr{R}$上含单位元的代数, $\mathscr{M}$是($\mathscr{A}$, $\mathscr{B}$)-忠实双边模.在通常的矩阵运算下, 称

$ {\rm{Tri}}\left( {\mathscr{A},\mathscr{M},\mathscr{B}} \right) = \left\{ {\left[ {\begin{array}{*{20}{c}} A&M\\ 0&B \end{array}} \right]:A \in \mathscr{A},M \in \mathscr{M},B \in \mathscr{B}} \right\} $

为$\mathscr{R}$上的三角代数^[12].设$\mathscr{U}$=Tri ($\mathscr{A}$, $\mathscr{M}$, $\mathscr{B}$) 是一个三角代数, $\mathscr{E}$是U的双边模, 1_{$\mathscr{A}$}和1_{$\mathscr{B}$}分别为$\mathscr{A}$和$\mathscr{B}$的单位元.记

$ {p_1} = \left[ {\begin{array}{*{20}{c}} {{1_\mathscr{A}}}&0\\ 0&0 \end{array}} \right],{p_2} = \left[ {\begin{array}{*{20}{c}} 0&0\\ 0&{{1_\mathscr{B}}} \end{array}} \right] $

以及

$ {\mathscr{U}_{ij}} = {p_i}\mathscr{U}{p_j},\;{\mathscr{E} _{ij}} = {p_i}\mathscr{E} {p_j}\left( {1 \le i \le j \le 2} \right). $

显然, $\mathscr{U}$=$\mathscr{U}$₁₁⊕$\mathscr{U}$₁₂⊕$\mathscr{U}$₂₂, $\mathscr{E}$=$\mathscr{E}$₁₁⊕$\mathscr{E}$₁₂⊕$\mathscr{E}$₂₁⊕$\mathscr{E}$₂₂.对a∈$\mathscr{E}$₁₁, b∈$\mathscr{E}$₂₂, 如果aU₁₂=0蕴含a=0且U₁₂b=0蕴含b=0, 则称$\mathscr{E}$是U的标准双边模.易验证, 三角代数U本身就是U的一个标准双边模.

1 主要结果与证明

定理1  设$\mathscr{U}$=Tri ($\mathscr{A}$, $\mathscr{M}$, $\mathscr{B}$) 是三角代数, $\mathscr{E}$是U的标准双边模，则由U到$\mathscr{E}$的任一可导映射对(δ, τ) 是可加的.进而, τ是由U到$\mathscr{E}$的可加导子, δ是由U到$\mathscr{E}$的关于τ的可加广义导子.

以下假设(δ, τ) 为由三角代数U到其标准双边模$\mathscr{E}$的可导映射对.即∀a, b∈$\mathscr{U}$, 有

$ \delta \left( {ab} \right) = \delta \left( a \right)b + a\tau \left( b \right). $

(1)

下面通过几个引理来完成定理1的证明.

引理1  δ(0)=0.

证明  δ(0)=δ(00)=δ(0)0+0τ(0)=0.证毕.

引理2  p_jδ(p_i)p_j=0, τ(p_i)=p_iδ(p_i)p_j-p_jδ(p_j)p_i (1≤i≤j≤2).

证明  在式(1) 中取a=b=p_i, 则

$ \delta \left( {{p_i}} \right) = \delta \left( {{p_i}} \right){p_i} + {p_i}\tau \left( {{p_i}} \right). $

(2)

对式(2) 分别左右同乘p_j, 右乘p_i, 左乘p_i右乘p_j得

$ {p_j}\delta \left( {{p_i}} \right){p_j} = {p_i}\tau \left( {{p_i}} \right){p_i} = 0,{p_i}\tau \left( {{p_i}} \right){p_j} = {p_i}\delta \left( {{p_i}} \right){p_j}. $

在式(1) 中取a=p_j, b=p_i, 则由引理1得

$ 0 = \delta \left( {{p_j}{p_i}} \right) = \delta \left( {{p_j}} \right){p_i} + {p_j}\tau \left( {{p_i}} \right). $

(3)

对式(3) 分别右乘p_j, 左乘p_j右乘p_i得

$ {p_j}\tau \left( {{p_i}} \right){p_j} = 0,\;\;\;{p_j}\tau \left( {{p_i}} \right){p_i} = - {p_j}\delta \left( {{p_j}} \right){p_i}. $

所以

$ \begin{array}{l} \tau \left( {{p_i}} \right) = {p_i}\tau \left( {{p_i}} \right){p_i} + {p_i}\tau \left( {{p_i}} \right){p_j} + {p_j}\tau \left( {{p_i}} \right){p_i} + {p_j}\tau \left( {{p_i}} \right){p_j} = \\ \;\;\;\;\;\;\;\;\;\;\;{p_i}\delta \left( {{p_i}} \right){p_j} - {p_j}\delta \left( {{p_j}} \right){p_i}. \end{array} $

证毕.

记r₁=p₁δ(p₁)p₁+p₂δ(p₁)p₁+p₁δ(p₂)p₂, r₂=p₁δ(p₁)p₂+p₂δ(p₂)p₁+p₂δ(p₂)p₂.定义映射Φ, ψ:U→$\mathscr{E}$分别为

$ \mathit{\Phi }\left( a \right) = \delta \left( a \right) - {r_1}a - a{r_2},\;\;\psi \left( a \right) = \tau \left( a \right) + {r_2}\left( a \right) - a{r_2}. $

(4)

引理3  ∀a, b∈$\mathscr{U}$, 有Φ(ab)=Φ(a)b+aψ(b), 且Φ(p_i)=ψ(p_i)=0(i=1, 2).

证明   ∀a, b∈$\mathscr{U}$, 由式(1) 和(4) 式可知

$ \begin{array}{l} \mathit{\Phi }\left( {ab} \right) = \delta \left( {ab} \right) - {r_1}ab - ab{r_2} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\delta \left( a \right)b + a\tau \left( b \right) - {r_1}ab - ab{r_2} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\mathit{\Phi }\left( a \right) + {r_1}a + a{r_2}} \right)b + a\left( {\psi \left( b \right) - {r_2}b + b{r_2}} \right) - {r_1}ab - ab{r_2} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\mathit{\Phi }\left( a \right)b + a\psi \left( b \right). \end{array} $

即(Φ, ψ) 是U的可导映射对.再由式(4) 和引理2可知

$ \begin{array}{l} \mathit{\Phi }\left( {{p_1}} \right) = \delta \left( {{p_1}} \right) - {r_1}{p_1} - {p_1}{r_2} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\delta \left( {{p_1}} \right) - {p_1}\delta \left( {{p_1}} \right){p_1} - {p_2}\delta \left( {{p_1}} \right){p_1} - {p_1}\delta \left( {{p_1}} \right){p_2} = \\ \;\;\;\;\;\;\;\;\;\;\;\;{p_2}\delta \left( {{p_1}} \right){p_2} = 0. \end{array} $

类似可得Φ(p₂)=0及ψ(p_i)=0 (i=1, 2).证毕.

以下讨论可导映射对(Φ, ψ) 的可加性.

引理4  设a_il∈U_il且b_ik∈U_ik (1≤i, l, k≤2).则

(a) ψ(a_il + b_ik) = Φ(a_il + b_ik),

(b) ψ(a_il) =Φ(a_il) ∈ $\mathscr{E}$_il.

证明  (a) 设1≤j≤2且j≠i.则由引理1和引理3, 得

$ 0 = \mathit{\Phi }\left( {{p_j}\left( {{a_{il}} + {b_{ik}}} \right)} \right) = \mathit{\Phi }\left( {{p_j}} \right)\left( {{a_{il}} + {b_{ik}}} \right) + {p_j}\psi \left( {{a_{il}} + {b_{ik}}} \right) = {p_j}\psi \left( {{a_{il}} + {b_{ik}}} \right), $

且

$ \mathit{\Phi }\left( {{a_{il}} + {b_{ik}}} \right) = \mathit{\Phi }\left( {{p_i}\left( {{a_{il}} + {b_{ik}}} \right)} \right) = \mathit{\Phi }\left( {{p_i}} \right)\left( {{a_{il}} + {b_{ik}}} \right) + {p_i}\psi \left( {{a_{il}} + {b_{ik}}} \right) = {p_i}\psi \left( {{a_{il}} + {b_{ik}}} \right). $

从而

$ \psi \left( {{a_{il}} + {b_{ik}}} \right) = {p_j}\psi \left( {{a_{il}} + {b_{ik}}} \right) + {p_i}\psi \left( {{a_{il}} + {b_{ik}}} \right) = \mathit{\Phi }\left( {{a_{il}} + {b_{ik}}} \right). $

(5)

(b) 在式(5) 中, 取b_ik=0, 则由引理3, 得

$ \begin{array}{l} \psi \left( {{a_{il}}} \right) = \mathit{\Phi }\left( {{a_{il}}} \right) = \mathit{\Phi }\left( {{p_i}{a_{il}}{p_l}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\mathit{\Phi }\left( {{p_i}{a_{il}}} \right){p_l} + {p_i}{a_{il}}\psi \left( {{p_l}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\mathit{\Phi }\left( {{p_l}} \right){a_{il}} + {p_i}\psi \left( {{a_i}j} \right)} \right){p_l} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;{p_i}\psi \left( {{a_{il}}} \right){p_l} \in {\mathscr{E}_{il}}. \end{array} $

证毕.

引理5  设a_ij∈U_ij(1≤i≤j≤2)，则

(a)ψ(a₁₂ + a₂₂) =ψ(a₁₂) +ψ(a₂₂),

(b)ψ(a₁₁ + a₂₂) =ψ(a₁₁) +ψ(a₂₂),

(c)ψ(a₁₁ + a₁₂) =ψ(a₁₁) +ψ(a₁₂).

证明  (a) 由引理4(B) 和引理3, 则

$ \begin{array}{l} \psi \left( {{a_{12}}} \right) = \mathit{\Phi }\left( {{a_{12}}} \right) = \mathit{\Phi }\left( {{p_1}\left( {{a_{12}} + {a_{22}}} \right)} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\mathit{\Phi }\left( {{p_1}} \right)\left( {{a_{12}} + {a_{22}}} \right) + {p_1}\psi \left( {{a_{12}} + {a_{22}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;{p_1}\psi \left( {{a_{12}} + {a_{22}}} \right), \end{array} $

以及

$ \begin{array}{l} \psi \left( {{a_{22}}} \right) = \mathit{\Phi }\left( {{a_{22}}} \right) = \mathit{\Phi }\left( {{p_2}\left( {{a_{12}} + {a_{22}}} \right)} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\mathit{\Phi }\left( {{p_2}} \right)\left( {{a_{12}} + {a_{22}}} \right) + {p_2}\psi \left( {{a_{12}} + {a_{22}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;{p_2}\psi \left( {{a_{12}} + {a_{22}}} \right). \end{array} $

从而

$ \psi \left( {{a_{12}} + {a_{22}}} \right) = {p_1}\psi \left( {{a_{12}} + {a_{22}}} \right) + {p_2}\psi \left( {{a_{12}} + {a_{22}}} \right) = \psi \left( {{a_{12}}} \right) + \psi \left( {{a_{22}}} \right). $

类似可得(b) 和(c) 也成立.证毕.

引理6  设a₁₂, b₁₂∈U₁₂, 则ψ(a₁₂+b₁₂)=ψ(a₁₂)+ψ(b₁₂).

证明  由于a₁₂+b₁₂=(a₁₂+p₁)(p₂+b₁₂), 从而由引理4和引理5, 得

$ \begin{array}{l} \begin{array}{*{20}{c}} {\psi \left( {{a_{12}} + {b_{12}}} \right) = \mathit{\Phi }\left( {{a_{12}} + {b_{12}}} \right) = \mathit{\Phi }\left( {{a_{12}} + {p_1}} \right)\left( {{p_2} + {b_{12}}} \right) + \left( {{a_{12}} + {p_1}} \right)\psi \left( {{p_2} + {b_{12}}} \right) = }\\ {\psi \left( {{a_{12}} + {p_1}} \right)\left( {{p_2} + {b_{12}}} \right) + \left( {{a_{12}} + {p_1}} \right)\psi \left( {{p_2} + {b_{12}}} \right)} \end{array}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\psi \left( {{a_{12}}} \right) + \psi \left( {{p_1}} \right)} \right)\left( {{p_2} + {b_{12}}} \right) + \left( {{a_{12}} + {p_1}} \right)\left( {\psi \left( {{p_2}} \right) + \psi \left( {{b_{12}}} \right)} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\psi \left( {{a_{12}}} \right) + \psi \left( {{b_{12}}} \right). \end{array} $

证毕.

引理7  设a_ii, b_ii∈U_ii (i=1, 2), 则

(a)ψ(a₁₁ +b₁₁) =ψ(a₁₁) +ψ(b₁₁),

(b)ψ(a₂₂ +b₂₂) =ψ(a₂₂) +ψ(b₂₂).

证明  (A) ∀c₁₂∈U₁₂, 根据引理4和引理6, 则一方面,

$ \begin{array}{l} \psi \left( {{a_{11}}{c_{12}} + {b_{11}}{c_{12}}} \right) = \psi \left( {{a_{11}}{c_{12}}} \right) + \psi \left( {{b_{11}}{c_{12}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathit{\Phi }\left( {{a_{11}}{c_{12}}} \right) + \mathit{\Phi }\left( {{b_{11}}{c_{12}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathit{\Phi }\left( {{a_{11}}} \right){c_{12}} + {a_{11}}\psi \left( {{c_{12}}} \right) + \;\mathit{\Phi }\left( {{b_{11}}} \right){c_{12}} + {b_{11}}\psi \left( {{c_{12}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\psi \left( {{a_{11}}} \right){c_{12}} + {a_{11}}\psi \left( {{c_{12}}} \right) + \psi \left( {{b_{11}}} \right){c_{12}} + {b_{11}}\psi \left( {{c_{12}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\psi \left( {{a_{11}}} \right) + \psi \left( {{b_{11}}} \right)} \right){c_{12}} + \left( {{a_{11}} + {b_{11}}} \right)\psi \left( {{c_{12}}} \right). \end{array} $

(6)

另一方面,

$ \begin{array}{l} \psi \left( {{a_{11}}{c_{12}} + {b_{11}}{c_{12}}} \right) = \mathit{\Phi }\left( {{a_{11}}{c_{12}} + {b_{11}}{c_{12}}} \right) = \mathit{\Phi }\left( {\left( {{a_{11}} + {b_{11}}} \right){c_{12}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathit{\Phi }\left( {{a_{11}} + {b_{11}}} \right){c_{12}} + \left( {{a_{11}} + {b_{11}}} \right)\psi \left( {{c_{12}}} \right){\rm{ = }}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\psi \left( {{a_{11}} + {b_{11}}} \right){c_{12}} + \left( {{a_{11}} + {b_{11}}} \right)\psi \left( {{c_{12}}} \right). \end{array} $

(7)

比较式(6) 与(7), ∀c₁₂∈U₁₂, 有

$ \left( {\psi \left( {{a_{11}} + {b_{11}}} \right) - \psi \left( {{a_{11}}} \right) - \psi \left( {{b_{11}}} \right)} \right){c_{12}} = 0. $

由于$\mathscr{E}$是U的标准双边模, 从而

$ \psi \left( {{a_{11}} + {b_{11}}} \right) = \psi \left( {{a_{11}}} \right) + \psi \left( {{b_{11}}} \right). $

类似可得(B) 也成立.证毕.

引理8  设a_ij∈U_ij (1≤i≤j≤2), 则ψ(a₁₁+a₁₂+a₂₂)=ψ(a₁₁)+ψ(a₁₂)+ψ(a₂₂).

证明  由引理3和引理4, 则

$ \begin{array}{l} \psi \left( {{a_{11}} + {a_{12}}} \right) = \mathit{\Phi }\left( {{a_{11}} + {a_{12}}} \right) = \mathit{\Phi }\left( {{p_1}\left( {{a_{11}} + {a_{12}} + {a_{22}}} \right)} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathit{\Phi }\left( {{p_1}} \right)\left( {{a_{11}} + {a_{12}} + {a_{22}}} \right) + {p_1}\psi \left( {{a_{11}} + {a_{12}} + {a_{22}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{p_1}\psi \left( {{a_{11}} + {a_{12}} + {a_{22}}} \right), \end{array} $

(8)

以及

$ \begin{array}{l} \psi \left( {{a_{22}}} \right) = \mathit{\Phi }\left( {{a_{22}}} \right) = \mathit{\Phi }\left( {{p_2}\left( {{a_{11}} + {a_{12}} + {a_{22}}} \right)} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathit{\Phi }\left( {{p_{21}}} \right)\left( {{a_{11}} + {a_{12}} + {a_{22}}} \right) + {p_2}\psi \left( {{a_{11}} + {a_{12}} + {a_{22}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;{p_2}\psi \left( {{a_{11}} + {a_{12}} + {a_{22}}} \right). \end{array} $

(9)

从而由式(8) 与(9) 及引理5(c), 可得

$ \begin{array}{l} \psi \left( {{a_{11}} + {a_{12}} + {a_{22}}} \right) = {p_1}\psi \left( {{a_{11}} + {a_{12}} + {a_{22}}} \right) + {p_2}\psi \left( {{a_{11}} + {a_{12}} + {a_{22}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\psi \left( {{a_{11}} + {a_{12}}} \right) + \psi \left( {{a_{22}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\psi \left( {{a_{11}}} \right) + \psi \left( {{a_{12}}} \right) + \psi \left( {{a_{22}}} \right). \end{array} $

证毕.

引理9   ∀a, b∈U, 有ψ(a+b)=ψ(a)+ψ(b), 即ψ是可加的.

证明  设a, b∈U, 则a=a₁₁+a₁₂+a₂₂, b=b₁₁+b₁₂+b₂₂, 其中a_ij, b_ij∈U_ij(1≤i≤j≤2).从而由引理6~8可知

$ \begin{array}{l} \psi \left( {a + b} \right) = \psi \left( {{a_{11}} + {b_{11}} + {a_{12}} + {b_{12}} + {a_{22}} + {b_{22}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\psi \left( {{a_{11}} + {b_{11}}} \right) + \psi \left( {{a_{12}} + {b_{12}}} \right) + \psi \left( {{a_{22}} + {b_{22}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\psi \left( {{a_{11}}} \right) + \psi \left( {{b_{11}}} \right) + \psi \left( {{a_{12}}} \right) + \psi \left( {{b_{12}}} \right) + \psi \left( {{a_{22}}} \right) + \psi \left( {{b_{22}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\psi \left( {{a_{11}} + {a_{12}} + {a_{22}}} \right) + \psi \left( {{b_{11}} + {b_{12}} + {b_{22}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\psi \left( a \right) + \psi \left( b \right). \end{array} $

证毕.

定理1的证明  由式(4) 和引理9可知, τ是可加的.在式(1) 中取a=1, 则∀b∈$\mathscr{U}$, 有

$ \delta \left( b \right) = \tau \left( b \right) + \delta \left( 1 \right)b. $

(10)

从而δ是可加的.在式(10) 中用ab替代b, 则

$ \begin{array}{l} \tau \left( {ab} \right) = \delta \left( {ab} \right) - \delta \left( 1 \right)ab = \\ \;\;\;\;\;\;\;\;\;\;\;\;\delta \left( a \right)b + a\tau \left( b \right) - \delta \left( 1 \right)ab = \\ \;\;\;\;\;\;\;\;\;\;\;\;\left( {\delta \left( a \right) - \delta \left( 1 \right)a} \right)b + a\tau \left( b \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\tau \left( a \right)b + a\tau \left( b \right). \end{array} $

于是τ是可加导子.再由式(1), 从而δ是关于τ的可加广义导子.证毕.

上三角矩阵块代数和套代数都是特殊的三角代数, 而每一个有限维空间上的非平凡套代数都同构于一个上三角矩阵块代数.所以由定理1和文献[13]可得以下推论.

推论1  设$\mathscr{T}$(R) 是含单位的可交换环R上的上三角矩阵块代数, (δ, τ) 是从$\mathscr{T}$到自身的可导映射对.则存在S, T∈$\mathscr{T}$(R) 以及可加导子α: R→ R, 使得∀A∈$\mathscr{T}$(R), 有δ(A)=AS-TA+A_α, τ(A)=AS-SA+A_α, 其中A_α=(α(a_ij)).

对于无限维的情况, 由定理1和文献[14]可得以下推论.

推论2  设X是数域F上的无限维Banach空间, $\mathscr{N}$是X上含非平凡可补元的套, Alg$\mathscr{N}$是相应的套代数, (δ, τ) 是从Alg$\mathscr{N}$到B(X) 的可导映射对.则存在T, S∈Alg$\mathscr{N}$使得∀A∈Alg$\mathscr{N}$, 有δ(A)=AT-SA, τ(A)=AT-TA.