0 引言

设$ \mathcal{A} $是定义在域$ \mathbb{F} $(char$ \mathbb{F} = 0 $)上的有单位元1的结合代数.任取$ a, b \in \mathcal{A} $, 记$ [a, b] = ab-ba $. $ \mathcal{A} $的中心记为$ Z(\mathcal{A}) $(其中$ Z(\mathcal{A}) = \{z\in\mathcal{A}:[z, a] = 0, \forall a\in\mathcal{A}\} $).映射$ f:\mathcal{A}\rightarrow \mathcal{A} $是弱可加映射是指对任意的$ x, y\in\mathcal{A} $, 都存在$ t_{x, y}, s_{x, y}\in \mathbb{F} $使得$ f(x+y) = t_{x, y}f(x)+s_{x, y}f(y) $, 这里$ t_{x, y}, s_{x, y} $是和$ x, y $有关的数.例如, 设$ M_{2}(\mathbb{F}) $是定义在域$ \mathbb{F} $上的满足普通的矩阵加法和矩阵乘法的矩阵代数, 映射$ f:M_{2}(\mathbb{F})\rightarrow M_{2}(\mathbb{F}) $定义为$ f(x) = \|x\|x $(这里$ \|x\| $是任意一种矩阵范数), 则$ f $是弱可加映射.映射$ f:\mathcal{A}\rightarrow \mathcal{A} $是交换的是指对所有的$ a\in \mathcal{A} $有$ [f(a), a] = 0 $.如果交换映射$ f $的形式是$ f(a) = \lambda a+\mu(a) $, 这里$ \lambda\in Z(\mathcal{A}) $并且$ \mu:\mathcal{A}\rightarrow Z(\mathcal{A}) $是一个映射, 则$ f $被称为是正常的.

通常交换映射主要研究的是映射的形式，然而关于交换性的研究最初是关于交换环的.环$ \mathcal{R} $是交换的是指任意的$ a, b\in\mathcal{R} $有$ ab = ba $成立.交换环及相关的中心映射的研究要追溯到1957年E. C. Posner^[1]证明了如下结论:如果一个素环$ \mathcal{R} $有非零中心导子, 则$ \mathcal{R} $必是交换的.这个结果引起了人们对可加的交换映射的研究兴趣. M. Brešar^[2]证明了如果$ F $是一个从von Neumann代数$ M $到自身的可加的交换映射, 则$ F $是正常的.不久, M. Brešar^[3]给出了素环上的可加交换映射的刻画, 后来这个结果被研究者推广到了其他的环和代数上.关于交换映射的更多内容可参考文献[4-9].

映射$ f:\mathcal{A}\rightarrow \mathcal{A} $是$ k $-交换的是指对所有的$ a\in \mathcal{A} $有$ [f(a), a]_{k} = 0 $(这里$ [f(a), a]_{k} = [\cdots[[f(a), a], a], \cdots, a] $, 其中$ a $有$ k $个).映射$ f:\mathcal{A}\rightarrow \mathcal{A} $是中心映射是指对所有的$ a\in \mathcal{A} $有$ [f(a), a]\in Z(\mathcal{A}) $.映射$ f:\mathcal{A}\rightarrow \mathcal{A} $是$ \ast $-中心映射是指对所有的$ a\in \mathcal{A} $有$ [f(a), a^{\ast}]\in Z(\mathcal{A}) $.最近, 文献[10]研究得出在某些三角代数上的$ k $-交换映射是正常的.李岩波等^[11]讨论了广义矩阵代数上的半中心映射, 并通过复杂的计算刻画了它的一般形式.齐霄霏等^[12]刻画了在某些假设下, 每一个可加映射$ f:\mathcal{R}\rightarrow \mathcal{R} $是$ k $-交换映射当且仅当对所有的$ x\in \mathcal{R} $都有$ f(x) = \alpha x+h(x) $, 这里$ \alpha\in Z(\mathcal{R}) $并且$ h $是一个从$ \mathcal{R} $到$ Z(\mathcal{R}) $的可加映射. Shakir Ali^[13]描述了带有对合的素环上的任意一个可加交换映射是$ \ast $-中心映射.

下面举一个明显的交换映射的例子^[14]:

$ f(x) = \lambda_{0}(x)x^{n}+\lambda_{1}(x)x^{n-1}+\cdots+\lambda_{n-1}(x)x+\lambda_{n}(x), \lambda_{i}:\mathcal{A}\rightarrow Z(\mathcal{A}). $

然而, 这个例子不能说明交换映射就一定是这种形式, 因此需要添加一些限制, 如文献[15]证明了在某些假设下, 如果$ q $是双可加交换映射的迹, 则$ q(x) = \lambda x^{2}+\mu(x)x+v(x) $(这里$ \lambda, \mu(x), v(x) $都属于环的中心).一个自然的问题是什么形式的交换映射满足

$ f(x) = \lambda_{0}(x)x+\lambda_{1}(x), \lambda_{i}:\mathcal{A}\rightarrow Z(\mathcal{A}), i = 0, 1. $

本文的目的是考虑在某些假设下, 如果弱可加映射$ f:\mathcal{A}\rightarrow \mathcal{A} $是交换映射, 则对所有的$ x\in \mathcal{A} $是否存在$ \lambda_{0}(x)\in \mathcal{A} $和一个从$ \mathcal{A} $到$ Z(\mathcal{A}) $的映射$ \lambda_{1} $使得$ f(x) = \lambda_{0}(x) x+\lambda_{1}(x) $.由于$ t_{x, y}, s_{x, y} $是和$ x, y $有关的数, 所以这里不能证明$ \lambda_{0}(x)\in Z(\mathcal{A}) $.本文的主要结论如下.

定理0.1  设$ \mathcal{A} $是域(char$ \mathbb{F} = 0 $)上含有单位元1和幂等元$ e $的结合代数, 同时满足$ a\mathcal{A}e\neq 0, a\mathcal{A}(1-e)\neq0, \forall 0\neq a\in \mathcal{A} $.若$ f:\mathcal{A}\rightarrow \mathcal{A} $是一个弱可加交换映射, 且对任意的$ x, y\in \mathcal{A} $, 满足

(a) $ f(x) = 0\Leftrightarrow x = 0 $;

(b) $ x, y $是线性相关的$ \Leftrightarrow f(x), f(y) $是线性相关的.

则存在$ \lambda_{0}(x)\in \mathcal{A} $和$ \lambda_{1}(x)\in Z(\mathcal{A}) $使得$ f(x) = \lambda_{0}(x)x+\lambda_{1}(x) $成立.

1 主要结果的证明

由条件可知$ e\neq 0, 1 $, 记$ e_{1} = e, e_{2} = 1-e $.则$ \mathcal{A} $可以写成$ \mathcal{A} = \mathcal{A}_{11}+ \mathcal{A}_{12}+\mathcal{A}_{21}+\mathcal{A}_{22} $, 这里$ \mathcal{A}_{ij} = e_{i}\mathcal{A}e_{j}(i, j = 1, 2) $.于是任取$ A\in \mathcal{A} $有$ A = \sum_{i, j = 1}^{2} A_{ij} $(这里$ A_{ij}\in\mathcal{A}_{ij} $).

引理1.1^[11]  设$ \mathcal{R} $是一个含有单位元1和幂等元$ e $的环, 并且$ \mathcal{R} $满足$ a\mathcal{R}e = \{0\}\Rightarrow a = 0, a\mathcal{R}(1-e) = \{0\}\Rightarrow a = 0 $, 则$ \mathcal{R} $的中心是

$ Z(\mathcal{R})\! = \! \{z_{11}+z_{22}:z_{11}\!\in\! \mathcal{R}_{11}, z_{22}\!\in\! \mathcal{R}_{22}, z_{11}a_{12} = a_{12}z_{22}, a_{21}z_{11} = z_{22}a_{21}, \forall a_{12}\!\in \!R_{12}, a_{21}\!\in\! R_{21}\}. $

引理1.2^[16]  设$ \mathcal{R}' $是一个含有单位元1和幂等元$ e $的环, 并且$ \mathcal{R}' $满足$ a\mathcal{R}'e = \{0\}\Rightarrow a = 0 $和$ a\mathcal{R}'(1-e) = \{0\}\Rightarrow a = 0 $.任取$ a\in \mathcal{R}' $, 如果对任意的$ x\in \mathcal{R}' $有$ exeae = eaexe $, 则存在$ \lambda \in Z(\mathcal{R}') $使得$ eae = \lambda e $.

以下引理都是在定理0.1的前提条件下提出的, 下面将不再进行说明.

引理1.3  任取$ x, y\in\mathcal{A} $, 则存在全不为零的$ t_{x, y}, s_{x, y}\in \mathbb{F} $使得$ f(x+y) = t_{x, y}f(x)+s_{x, y}f(y) $.

证    明  任取$ x, y\in\mathcal{A} $, 我们来分情况讨论.

(ⅰ)如果$ x $, $ y $中至少有一个为零, 不妨设$ x = 0 $.由于$ f(x) = 0\Leftrightarrow x = 0 $, 所以$ f(0) = 0 $并且

$ f(x+y) = f(y) = f(x)+f(y). $

(ⅱ)如果$ x $, $ y $全不为零, 设$ t_{x, y}, s_{x, y}\in \mathbb{F} $使得

$ f(x+y) = t_{x, y}f(x)+s_{x, y}f(y). $

如果$ x, y $是线性无关的, 则$ x+y\neq 0 $, 所以$ f(x+y)\neq 0 $.于是$ t_{x, y}, s_{x, y} $不全为零.若$ t_{x, y}, s_{x, y} $中有一个为零, 不妨设

$ f(x+y) = t_{x, y}f(x), t_{x, y}\neq 0, $

则$ f(x+y), f(x) $是线性相关的, 从而$ x+y, x $是线性相关的, 与$ x, y $线性无关矛盾, 因此$ t_{x, y}, s_{x, y} $全不为零.

如果$ x, y $是线性相关的, 则$ f(x), f(y) $是线性相关的, 于是存在不全为零的$ a, b\in\mathbb{F} $使得

$ 0 = af(x)+bf(y). $

由于$ f(x)\neq 0, f(y)\neq 0 $, 所以$ a, b $全不为零.于是对任意的$ p\in\mathbb{F} $都有

$ f(x+y) = (t_{x, y}+ap)f(x)+(s_{x, y}+bp)f(y). $

因为$ \mathrm{char}(\mathbb{F}) = 0 $, 则存在$ p\in\mathbb{F} $使得$ t_{x, y}+ap\neq0 $和$ s_{x, y}+bp\neq0 $.因此可以取$ t_{x, y}+ap, s_{x, y}+bp $为新的$ t_{x, y}, s_{x, y} $, 于是$ t_{x, y}, s_{x, y} $全不为零.

引理1.4  $ f(1)\in Z(\mathcal{A}), f(e_{i})\in\mathcal{A}_{11}+\mathcal{A}_{22}(i = 1, 2) $.并且存在$ z_{i}\in Z(\mathcal{A}) $使得$ e_{i}f(1)e_{i} = z_{i}e_{i} $.

证    明  任取$ a\in \mathcal{A} $, 则$ [f(a+1), a+1] = 0 $.由定义可知存在$ t_{a, 1}, s_{a, 1}\in \mathbb{F} $, 使得

$ [f(a+1), a+1] = [t_{a, 1}f(a)+s_{a, 1}f(1), a+1] = s_{a, 1}[f(1), a] = 0. $

因为$ s_{a, 1}\neq0 $, 所以$ [f(1), a] = 0 $, 因此$ f(1)\in Z(\mathcal{A}) $.任取$ a_{11}\in \mathcal{A}_{11} $和$ a_{22}\in \mathcal{A}_{22} $, 有

$ [f(1), a_{11}] = [e_{1}f(1)e_{1}, a_{11}] = 0 $

和

$ [f(1), a_{22}] = [e_{2}f(1)e_{2}, a_{22}] = 0 $

成立.于是$ e_{1}f(1)e_{1}\in Z(\mathcal{A}_{11}) $和$ e_{2}f(1)e_{2}\in Z(\mathcal{A}_{22}) $.即对任意的$ a\in \mathcal{A} $, 有

$ e_{1}f(1)e_{1}ae_{1} = e_{1}ae_{1}f(1)e_{1} $

和

$ e_{2}f(1)e_{2}ae_{2} = e_{2}ae_{2}f(1)e_{2}. $

由引理1.2, 存在$ z_{1}, z_{2}\in Z(\mathcal{A}) $, 使得$ e_{1}f(1)e_{1} = z_{1}e_{1} $和$ e_{2}f(1)e_{2} = z_{2}e_{2} $成立.

因为$ [f(1), e_{1}] = [f(e_{1}+e_{2}), e_{1}] = 0, [f(e_{1}), e_{1}] = 0 $, 则$ [f(e_{2}), e_{1}] = 0 $, 所以$ f(e_{i})\in\mathcal{A}_{11}+\mathcal{A}_{22} $.

引理1.5  任取$ a_{ii}\in \mathcal{A}_{ii} $, 则$ f(a_{ii})\in \mathcal{A}_{11}+\mathcal{A}_{22}, [e_{i}f(a_{ii})e_{i}, a_{ii}] = 0 $.并且存在弱可加映射$ f_{ii}:\mathcal{A}_{ii}\rightarrow Z(\mathcal{A}) $, 使得$ e_{j}f(a_{ii})e_{j} = f_{ii}(a_{ii})e_{j}, 1\leqslant i\neq j\leqslant 2 $.特别地, $ e_{2}f(e_{1})e_{2} = f_{11}(e_{1})e_{2} $和$ e_{1}f(e_{2})e_{1} = f_{22}(e_{2})e_{1} $成立.

证    明  任取$ a_{11}\in \mathcal{A}_{11} $, 则

$ \begin{align*} 0 = [f(a_{11}), a_{11}]& = [e_{1}f(a_{11})e_{1}+e_{1}f(a_{11})e_{2}+e_{2}f(a_{11})e_{1}+e_{2}f(a_{11})e_{2}, a_{11}]\nonumber\\ & = [e_{1}f(a_{11})e_{1}, a_{11}]+[e_{1}f(a_{11})e_{2}, a_{11}]+[e_{2}f(a_{11})e_{1}, a_{11}]\nonumber. \end{align*} $

注意到

$ [e_{1}f(a_{11})e_{1}, a_{11}]\in \mathcal{A}_{11}, [e_{1}f(a_{11})e_{2}, a_{11}]\in \mathcal{A}_{12}, [e_{2}f(a_{11})e_{1}, a_{11}]\in \mathcal{A}_{21}. $

因此, 对任意的$ a_{11}\in \mathcal{A}_{11} $, 有

$ [e_{1}f(a_{11})e_{1}, a_{11}] = [e_{1}f(a_{11})e_{2}, a_{11}] = [e_{2}f(a_{11})e_{1}, a_{11}] = 0. $

由于

$ [e_{1}f(a_{11})e_{2}, a_{11}] = 0, \forall a_{11}\in \mathcal{A}_{11}, $

计算得

$ a_{11}e_{1}f(a_{11})e_{2} = 0, \forall a_{11}\in \mathcal{A}_{11}, $

因此

$ (a_{11}+e_{1})e_{1}f(a_{11}+e_{1})e_{2} = 0, \forall a_{11}\in \mathcal{A}_{11}. $

由引理1.4得$ e_{1}f(e_{1})e_{2} = 0 $, 所以$ e_{1}f(a_{11})e_{2} = 0 $.又由于

$ [e_{2}f(a_{11})e_{1}, a_{11}] = e_{2}f(a_{11})e_{1}a_{11} = 0, $

类似地, 有

$ e_{2}f(a_{11})e_{1} = 0, \forall a_{11}\in \mathcal{A}_{11}. $

因此, 对任意的$ a_{11}\in \mathcal{A}_{11} $, 有

$ f(a_{11})\in \mathcal{A}_{11}+\mathcal{A}_{22}, [e_{1}f(a_{11})e_{1}, a_{11}] = 0. $

同理, 对任意的$ a_{22}\in \mathcal{A}_{22} $, 有

$ f(a_{22})\in \mathcal{A}_{11}+\mathcal{A}_{22}, [e_{2}f(a_{22})e_{2}, a_{22}] = 0. $

下面考虑$ e_{2}f(a_{11})e_{2} $和$ e_{1}f(a_{22})e_{1} $.

因为$ [f(a_{11}+a_{22}), a_{11}+a_{22}] = 0 $, 所以存在非零的$ t_{a_{11}, a_{22}}, s_{a_{11}, a_{22}}\in \mathbb{F} $, 使得

$ t_{a_{11}, a_{22}}[e_{2}f(a_{11})e_{2}, a_{22}]+s_{a_{11}, a_{22}}[e_{1}f(a_{22})e_{1}, a_{11}] = 0. $

于是对任意的$ a_{11}\in \mathcal{A}_{11}, a_{22}\in \mathcal{A}_{22} $, 有

$ t_{a_{11}, a_{22}}[e_{2}f(a_{11})e_{2}, a_{22}] = s_{a_{11}, a_{22}}[e_{1}f(a_{22})e_{1}, a_{11}] = 0. $

由于$ t_{a_{11}, a_{22}}\neq 0, s_{a_{11}, a_{22}}\neq0 $, 所以$ e_{1}f(a_{22})e_{1}\in Z(\mathcal{A}_{11}) $, 并且$ e_{2}f(a_{11})e_{2}\in Z(\mathcal{A}_{22}) $.因此由引理1.2, 存在$ f_{11}(a_{11}), f_{22}(a_{22})\in Z(\mathcal{A}) $, 使得$ e_{2}f(a_{11})e_{2} = f_{11}(a_{11})e_{2} $和$ e_{1}f(a_{22})e_{1} = f_{22}(a_{22})e_{1} $成立.

实际上, 由$ f $的弱可加性, 任取$ a_{ii}, b_{ii}\in \mathcal{A}_{ii}(i = 1, 2) $, 存在非零的$ t_{a_{ii}, b_{ii}}, s_{a_{ii}, b_{ii}}\in \mathbb{F} $, 使得

$ \begin{equation} f_{ii}(a_{ii}+b_{ii})e_{j} = t_{a_{ii}, b_{ii}}f_{ii}(a_{ii})e_{j}+s_{a_{ii}, b_{ii}}f_{ii}(b_{ii})e_{j}, \end{equation} $

(1)

则

$ (f_{ii}(a_{ii}+b_{ii})-t_{a_{ii}, b_{ii}}f_{ii}(a_{ii})-s_{a_{ii}, b_{ii}}f_{ii}(b_{ii}))\mathcal{A}e_{j} = 0. $

由$ \mathcal{A} $的假设可得

$ f_{ii}(a_{ii}+b_{ii}) = t_{a_{ii}, b_{ii}}f_{ii}(a_{ii})+s_{a_{ii}, b_{ii}}f_{ii}(b_{ii}), $

所以$ f_{ii} $是弱可加映射.

引理1.6  任取$ a_{ij}\in \mathcal{A}_{ij}(1\leqslant i\neq j\leqslant2) $, 则下列结论成立:

(1) $ e_{j}f(a_{ij})e_{i} = 0 $;

(2) $ e_{i}f(a_{ij})e_{j} = \frac {t_{e_{i}, a_{ij}}}{s_{e_{i}, a_{ij}}}(e_{i}f(e_{i})e_{i}a_{ij}-a_{ij} e_{j}f(e_{i})e_{j}) = \frac {t_{e_{i}, a_{ij}}}{s_{e_{i}, a_{ij}}}(e _{i}f(e_{i})e_{i}a_{ij}-f_{ii}(e_{i})e_{i}a_{ij}) = \frac {t_{e_{j}, a_{ij}}}{s_{e_{j}, a_{ij}}}(a_{ij}e_{j}f(e_{j})e_{j}-e_{i}f(e_{j})e_{i}a_{ij}) = \frac {t_{e_{j}, a_{ij}}}{s_{e_{j}, a_{ij}}}(a_{ij}e_{j}f(e_{j})e_{j}-f_{jj}(e_{j})e_{i}a_{ij}) $, 这里$ t_{e_{i}, a_{ij}}, s_{e_{i}, a_{ij}}, t_{e_{j}, a_{ij}}, s_{e_{j}, a_{ij}}\in \mathbb{F} $;

(3) $ e_{i}f(a_{ij})e_{i}a_{ij} = a_{ij}e_{j}f(a_{ij})e_{j} $;

(4) 存在弱可加映射$ f_{ij}, g_{ij}:\mathcal{A}\rightarrow Z(\mathcal{A}) $, 使得$ e_{i}f(a_{ij})e_{i} = f_{ij}(a_{ij})e_{i} $和$ e_{j}f(a_{ij})e_{j} = g_{ij}(a_{ij})e_{j} $成立.

证    明  这里只证明$ (i, j) = (1, 2) $的情形, 另一种$ (i, j) = (2, 1) $的情形类似.

任取$ a_{12}\in \mathcal{A}_{12} $, 则$ [f(e_{1}+a_{12}), e_{1}+a_{12}] = 0 $.因此存在非零的$ t_{e_{1}, a_{12}}, s_{e_{1}, a_{12}}\in \mathbb{F} $, 使得

$ \begin{align} &[f(e_{1}+a_{12}), e_{1}+a_{12}] \\ & = t_{e_{1}, a_{12}}[f(e_{1}), a_{12}]+s_{e_{1}, a_{12}}[f(a_{12}), e_{1}] \\ & = t_{e_{1}, a_{12}}e_{1}f(e_{1})e_{1}a_{12}-t_{e_{1}, a_{12}}a_{12}e_{2}f(e_{1})e_{2}+s_{e_{1}, a_{12}}e_{2}f(a_{12})e_{1}-s_{e_{1}, a_{12}}e_{1}f(a_{12})e_{2}. \end{align} $

(2)

因此$ e_{2}f(a_{12})e_{1} = 0 $.所以结论(1)成立, 并且

$ \begin{align*} e_{1}f(a_{12})e_{2}& = \frac {t_{e_{1}, a_{12}}}{s_{e_{1}, a_{12}}}(e_{1}f(e_{1})e_{1}a_{12}-a_{12}e_{2}f(e_{1})e_{2})\nonumber \\ & = \frac {t_{e_{1}, a_{12}}}{s_{e_{1}, a_{12}}}(e_{1}f(e_{1})e_{1}a_{12}-f_{11}(e_{1})e_{1}a_{12})\nonumber. \end{align*} $

同理, 由等式$ [f(e_{2}+a_{12}), e_{2}+a_{12}] = 0 $可得

$ \begin{align*} e_{1}f(a_{12})e_{2}& = \frac {t_{e_{2}, a_{12}}}{s_{e_{2}, a_{12}}}(a_{12}e_{2}f(e_{2})e_{2}-e_{1}f(e_{2})e_{1}a_{12})\nonumber \\ & = \frac {t_{e_{2}, a_{12}}}{s_{e_{2}, a_{12}}}(a_{12}e_{2}f(e_{2})e_{2}-f_{22}(e_{2})e_{1}a_{12})\nonumber. \end{align*} $

因为$ [f(a_{12}), a_{12}] = 0 $, 所以$ e_{1}f(a_{12})e_{1}a_{12} = a_{12}e_{2}f(a_{12})e_{2} $.因此结论(2)和(3)成立.

下证结论(4).任取$ a_{11}\in \mathcal{A}_{11} $, 则$ [f(a_{11}+a_{12}), a_{11}+a_{12}] = 0 $.由引理1.5和结论(1), 可得存在非零$ s_{a_{11}, a_{12}}\in \mathbb{F} $, 使得

$ [f(a_{11}+a_{12}), a_{11}+a_{12}] = [s_{a_{11}, a_{12}}e_{1}f(a_{12})e_{1}, a_{11}]+g(a_{11}, a_{12}), $

这里

$ [s_{a_{11}, a_{12}}e_{1}f(a_{12})e_{1}, a_{11}]\in \mathcal{A}_{11}, g(a_{11}, a_{12})\in \mathcal{A}_{12}. $

因此, 对任意的$ a_{11}\in \mathcal{A}_{11} $有$ [e_{1}f(a_{12})e_{1}, a_{11}] = 0 $成立, 所以$ e_{1}f(a_{12})e_{1}\in Z(\mathcal{A}_{11}) $.

类似地, 任取$ a_{22}\in \mathcal{A}_{22} $, 由等式$ [f(a_{22}+a_{12}), a_{22}+a_{12}] = 0 $可以证明$ e_{2}f(a_{12})e_{2}\in Z(\mathcal{A}_{22}) $.再由引理1.2, 存在$ f_{12}(a_{12}), g_{12}(a_{12})\in Z(\mathcal{A}) $, 使得

$ e_{1}f(a_{12})e_{1} = f_{12}(a_{12})e_{1}, e_{2}f(a_{12})e_{2} = g_{12}(a_{12})e_{2}. $

同理, 由引理1.5可得$ f_{ij}, g_{ij} $是弱可加的, 因此结论(4)成立.

引理1.7  任取$ a_{12}\in \mathcal{A}_{12}, a_{21}\in \mathcal{A}_{21} $, 则

$ a_{21}e_{1}f(a_{12})e_{1} = e_{2}f(a_{12})e_{2}a_{21} $

和

$ a_{12}e_{2}f(a_{21})e_{2} = e_{1}f(a_{21})e_{1}a_{12} $

成立.

证    明  任取$ a_{12}\in \mathcal{A}_{12}, a_{21}\in \mathcal{A}_{21} $, 则$ [f(a_{12}+a_{21}), a_{12}+a_{21}] = 0 $.因此由引理1.6, 存在非零$ t_{a_{12}, a_{21}}, s_{a_{12}, a_{21}}\in \mathbb{F} $, 使得

$ \begin{align*} &[f(a_{12}+a_{21}), a_{12}+a_{21}]\nonumber\\ = &t_{a_{12}, a_{21}}(-a_{21}e_{1}f(a_{12})e_{1}-a_{21}e_{1}f(a_{12})e_{2}+e_{1}f(a_{12})e_{2}a_{21}+e_{2}f(a_{12})e_{2}a_{21})\nonumber\\ &+s_{a_{12}, a_{21}}(e_{1}f(a_{21})e_{1}a_{12}+e_{2}f(a_{21})e_{1}a_{12}-a_{12}e_{2}f(a_{21})e_{1}-a_{12}e_{2}f(a_{21})e_{2})\nonumber\\ = &(t_{a_{12}, a_{21}}e_{1}f(a_{12})e_{2}a_{21}-s_{a_{12}, a_{21}}a_{12}e_{2}f(a_{21})e_{1})+s_{a_{12}, a_{21}}(e_{1}f(a_{21})e_{1}a_{12}-a_{12}e_{2}f(a_{21})e_{2})\nonumber\\ &+t_{a_{12}, a_{21}}(-a_{21}e_{1}f(a_{12})e_{1}+e_{2}f(a_{12})e_{2}a_{21})+(s_{a_{12}, a_{21}}e_{2}f(a_{21})e_{1}a_{12}-t_{a_{12}, a_{21}}a_{21}e_{1}f(a_{12})e_{2})\nonumber. \end{align*} $

于是,

$ a_{21}e_{1}f(a_{12})e_{1} = e_{2}f(a_{12})e_{2}a_{21}, a_{12}e_{2}f(a_{21})e_{2} = e_{1}f(a_{21})e_{1}a_{12}. $

引理1.8  任取$ a_{ij}\in \mathcal{A}_{ij} $, 则$ e_{i}f(a_{ij})e_{i}+e_{j}f(a_{ij})e_{j}\in Z(\mathcal{A})(1\leqslant i\neq j\leqslant 2) $.

证    明  设$ 1\leqslant i\neq j\leqslant 2 $, 任取$ a_{ij}\in \mathcal{A}_{ij}, a_{ji}\in \mathcal{A}_{ji} $, 由引理1.6(4)得到

$ e_{i}f(a_{ij})e_{i} = f_{ij}(a_{ij})e_{i}, e_{j}f(a_{ij})e_{j} = g_{ij}(a_{ij})e_{j}, $

这里$ f_{ij}(a_{ij}), g_{ij}(a_{ij})\in Z(\mathcal{A}) $.由此式和引理1.7得到

$ a_{ji}f_{ij}(a_{ij})e_{i} = g_{ij}(a_{ij})e_{j}a_{ji}, \forall a_{ji}\in \mathcal{A}_{ji}, $

即

$ (g_{ij}(a_{ij})e_{j}-f_{ij}(a_{ij})e_{j})ae_{i} = 0, \forall a\in \mathcal{A}. $

再由$ \mathcal{A} $的假设, 有$ g_{ij}(a_{ij})e_{j} = f_{ij}(a_{ij})e_{j} $, 因此

$ e_{i}f(a_{ij})e_{i}+e_{j}f(a_{ij})e_{j} = f_{ij}(a_{ij})e_{i}+g_{ij}(a_{ij})e_{j} = f_{ij}(a_{ij})\in Z(\mathcal{A}). $

引理1.9  任取$ a_{11}\in \mathcal{A}_{11} $, 则

$ e_{1}f(a_{11})e_{1} = f_{11}(a_{11})e_{1}+\frac {s_{a_{11}, a_{12}}t_{e_{1}, a_{12}}}{t_{a_{11}, a_{12}}s_{e_{1}, a_{12}}}(e_{1}f(e_{1})e_{1}-f_{11}(e_{1})e_{1})a_{11}. $

证    明  任取$ a_{11}\in \mathcal{A}_{11}, a_{12}\in \mathcal{A}_{12} $, 则$ [f(a_{11}+a_{12}), a_{11}+a_{12}] = 0 $.由引理1.5和引理1.6, 存在非零$ t_{a_{11}, a_{12}}, s_{a_{11}, a_{12}}\in \mathbb{F} $, 使得

$ \begin{eqnarray} &&[f(a_{11}+a_{12}), a_{11}+a_{12}]\\ && = t_{a_{11}, a_{12}}e_{1}f(a_{11})e_{1}a_{12}-t_{a_{11}, a_{12}}a_{12}e_{2}f(a_{11})e_{2}-s_{a_{11}, a_{12}}a_{11}e_{1}f(a_{12})e_{2}. \end{eqnarray} $

易知

$ \begin{align*} e_{i}f(a_{ij})e_{j}& = \frac {t_{e_{i}, a_{ij}}}{s_{e_{i}, a_{ij}}}(e_{i}f(e_{i})e_{i}a_{ij}-a_{ij}e_{j}f(e_{i})e_{j}) \nonumber\\ & = \frac {t_{e_{i}, a_{ij}}}{s_{e_{i}, a_{ij}}}(e _{i}f(e_{i})e_{i}a_{ij}-f_{ii}(e_{i})e_{j}a_{ij})\nonumber, \end{align*} $

则

$ \begin{align*} &t_{a_{11}, a_{12}}e_{1}f(a_{11})e_{1}a_{12}-t_{a_{11}, a_{12}}a_{12}e_{2}f(a_{11})e_{2}\nonumber \\ &\quad -s_{a_{11}, a_{12}}a_{11}\Big(\frac {t_{e_{1}, a_{12}}}{s_{e_{1}, a_{12}}}(e_{1}f(e_{1})e_{1}a_{12}-a_{12}e_{2}f(e_{1})e_{2})\Big) = 0 \nonumber. \end{align*} $

因此

$ \begin{align*} &\Big(t_{a_{11}, a_{12}}e_{1}f(a_{11})e_{1}-t_{a_{11}, a_{12}}f_{11}(a_{11})e_{1}-s_{a_{11}, a_{12}}\frac {t_{e_{1}, a_{12}}}{s_{e_{1}, a_{12}}}a_{11}e_{1}f(e_{1})e_{1}\nonumber \\ &\quad +s_{a_{11}, a_{12}}\frac {t_{e_{1}, a_{12}}}{s_{e_{1}, a_{12}}}a_{11}f_{11} (e_{1})e_{1}\Big)e_{1}ae_{2} = 0\nonumber. \end{align*} $

利用假设得

$ e_{1}f(a_{11})e_{1} = f_{11}(a_{11})e_{1}+\frac {s_{a_{11}, a_{12}}t_{e_{1}, a_{12}}}{t_{a_{11}, a_{12}}s_{e_{1}, a_{12}}}(e_{1}f(e_{1})e_{1}-f_{11}(e_{1})e_{1})a_{11}. $

类似地, 利用等式$ [f(a_{22}+a_{21}), a_{22}+a_{21}] = 0 $得, 存在非零$ t_{a_{22}, a_{21}}, s_{a_{22}, a_{21}}\in \mathbb{F} $, 使得如下结论成立.

引理1.10  任取$ a_{22}\in \mathcal{A}_{22} $, 则

$ e_{2}f(a_{22})e_{2} = f_{22}(a_{22})e_{2}+\frac {s_{a_{22}, a_{21}}t_{e_{2}, a_{21}}}{t_{a_{22}, a_{21}}s_{e_{2}, a_{21}}}(e_{2}f(e_{2})e_{2}-f_{22}(e_{2})e_{2})a_{22}. $

下面证明定理0.1.

定理0.1的证明  任取$ x = a_{11}+a_{12}+a_{21}+a_{22}\in \mathcal{A} $, 由$ f $的定义, 存在非零常数$ u, v\in \mathbb{F} $, 使得定理0.1的

$ \begin{align*} f(x)& = f(a_{11}+a_{12}+a_{21}+a_{22})\nonumber\\ & = uf(a_{11}+a_{12})+vf(a_{21}+a_{22})\nonumber\\ & = ut_{a_{11}, a_{12}}f(a_{11})+us_{a_{11}, a_{12}}f(a_{12})+vs_{a_{22}, a_{21}}f(a_{21})+vt_{a_{22}, a_{21}}f(a_{22})\nonumber. \end{align*} $

设

$ \lambda_{0}(x) = \alpha(e_{1}f(e_{1})e_{1}-f_{11}(e_{1})e_{1})+\beta(e_{2}f(e_{2})e_{2}-f_{22}(e_{2})e_{2}), $

其中

$ \alpha = us_{a_{11}, a_{12}}\frac {t_{e_{1}, a_{12}}}{s_{e_{1}, a_{12}}}, \beta = vs_{a_{22}, a_{21}}\frac {t_{e_{2}, a_{21}}}{s_{e_{2}, a_{21}}}. $

令

$ \lambda_{1}(x) = f(x)-\lambda_{0}(x)x, $

显然$ \lambda_{1}(x) $是$ \mathcal{A} $上的映射.因此, 只须验证, 对所有的$ x\in \mathcal{A} $, 有$ \lambda_{1}(x)\in Z(\mathcal{A}) $.实际上, 由引理1.5和引理1.6可得

$ \begin{align*} \lambda_{1}(x) = &f(a_{11}+a_{12}+a_{21}+a_{22})-\lambda_{0}(x)(a_{11}+a_{12}+a_{21}+a_{22})\nonumber\\ = &ut_{a_{11}, a_{12}}f(a_{11})+us_{a_{11}, a_{12}}f(a_{12})+vs_{a_{22}, a_{21}}f(a_{21})+vt_{a_{22}, a_{21}}f(a_{22})-(\alpha(e_{1}f(e_{1})e_{1}\nonumber\\ &-f_{11}(e_{1})e_{1})+\beta(e_{2}f(e_{2})e_{2}-f_{22}(e_{2})e_{2}))(a_{11}+a_{12}+a_{21}+a_{22})\nonumber\\ = &ut_{a_{11}, a_{12}}e_{1}f(a_{11})e_{1}+ut_{a_{11}, a_{12}}e_{2}f(a_{11})e_{2}+vt_{a_{22}, a_{21}}e_{1}f(a_{22})e_{1}+vt_{a_{22}, a_{21}}e_{2}f(a_{22})e_{2}\nonumber\\ &+us_{a_{11}, a_{12}}e_{1}f(a_{12})e_{1}+us_{a_{11}, a_{12}}e_{2}f(a_{12})e_{2}+vs_{a_{22}, a_{21}}e_{1}f(a_{21})e_{1}\nonumber\\ &+vs_{a_{22}, a_{21}}e_{2}f(a_{21})e_{2}+us_{a_{11}, a_{12}}e_{1}f(a_{12})e_{2}+vs_{a_{22}, a_{21}}e_{2}f(a_{21})e_{1}\nonumber\\ &-\alpha(e_{1}f(e_{1})e_{1}-f_{11}(e_{1})e_{1})a_{11}-\alpha(e_{1}f(e_{1})e_{1}\!-\!f_{11}(e_{1})e_{1})a_{12}\!\nonumber\\ = &-\! \beta(e_{2}f(e_{2})e_{2}\!-\!f_{22}(e_{2})e_{2})a_{21}\!-\! \beta(e_{2}f(e_{2})e_{2}\!-\!f_{22}(e_{2})e_{2})a_{22}ut_{a_{11}, a_{12}}e_{1}f(a_{11})e_{1}\nonumber\\ &+ut_{a_{11}, a_{12}}e_{2}f(a_{11})e_{2}+vt_{a_{22}, a_{21}}e_{1}f(a_{22})e_{1}+vt_{a_{22}, a_{21}}e_{2}f(a_{22})e_{2}\nonumber \\ &+us_{a_{11}, a_{12}}e_{1}f(a_{12})e_{1}+us_{a_{11}, a_{12}}e_{2}f(a_{12})e_{2}+vs_{a_{22}, a_{21}}e_{1}f(a_{21})e_{1}+vt_{a_{22}, a_{21}}e_{2}f(a_{21})e_{2}\nonumber\\ &+\Big(us_{a_{11}, a_{12}}\frac {t_{e_{1}, a_{12}}}{s_{e_{1}, a_{12}}}-\alpha\Big)(e_{1}f(e_{1})e_{1}-f_{11} (e_{1})e_{1})a_{12}+\Big(vs_{a_{22}, a_{21}}\frac {t_{e_{2}, a_{21}}} {s_{e_{2}, a_{21}}}\nonumber\\ &-\beta\Big)(e_{2}f(e_{2})e_{2}-f_{22}(e_{2})e_{2})a_{21}-\alpha(e_{1}f(e_{1})e_{1}-f_{11}(e_{1})e_{1})a_{11}\nonumber\\ &-\beta(e_{2}f(e_{2})e_{2}-f_{22}(e_{2})e_{2})a_{22}.\nonumber \end{align*} $

由引理1.8得到

$ \begin{eqnarray} us_{a_{11}, a_{12}}e_{1}f(a_{12})e_{1}+us_{a_{11}, a_{12}}e_{2}f(a_{12})e_{2}+ vs_{a_{22}, a_{21}}e_{1}f(a_{21})e_{1} +vs_{a_{22}, a_{21}}e_{2}f(a_{21})e_{2}\in Z(\mathcal{A}), \end{eqnarray} $

(3)

再由引理1.5和引理1.9, 有

$ \begin{align} &ut_{a_{11}, a_{12}}e_{1}f(a_{11})e_{1}-\alpha(e_{1}f(e_{1})e_{1}-f_{11}(e_{1})e_{1})a_{11}+ut_{a_{11}, a_{12}}e_{2}f(a_{11})e_{2} \\ & = ut_{a_{11}, a_{12}}e_{1}f(a_{11})e_{1}-ut_{a_{11}, a_{12}}\frac {s_{a_{11}, a_{12}}t_{e_{1}, a_{12}}}{t_{a_{11}, a_{12}}s_{e_{1}, a_{12}}}(e_{1}f(e_{1})e_{1}-f_{11}(e_{1})e_{1})a_{11}+ut_{a_{11}, a_{12}}e_{2}f(a_{11})e_{2} \\ & = ut_{a_{11}, a_{12}}f_{11}(a_{11})e_{1}+ut_{a_{11}, a_{12}}f_{11}(a_{11})e_{2}\\ & = ut_{a_{11}, a_{12}}f_{11}(a_{11})\in Z(\mathcal{A}). \end{align} $

(4)

同理, 由引理1.5和引理1.10, 有

$ \begin{align} &vt_{a_{22}, a_{21}}e_{1}f(a_{22})e_{1}-\beta(e_{2}f(e_{2})e_{2}-f_{22}(e_{2})e_{2})a_{22}+vt_{a_{22}, a_{21}}e_{2}f(a_{22})e_{2} \\ & = vt_{a_{22}, a_{21}}e_{1}f(a_{22})e_{1}-vt_{a_{22}, a_{21}}\frac {s_{a_{22}, a_{21}}t_{e_{2}, a_{21}}}{t_{a_{22}, a_{21}}s_{e_{2}, a_{21}}}(e_{2}f(e_{2})e_{2}-f_{22}(e_{2})e_{2})a_{22}+vt_{a_{22}, a_{21}}e_{2}f(a_{22})e_{2} \\ & = vt_{a_{22}, a_{21}}f_{22}(a_{22})e_{1}+vt_{a_{22}, a_{21}}f_{22}(a_{22})e_{2}\\ & = vt_{a_{22}, a_{21}}f_{22}(a_{22})\in Z(\mathcal{A}). \end{align} $

(5)

综合式(3)–(5), 可以得到对任意的$ x\in \mathcal{A} $有$ \lambda_{1}(x)\in Z(\mathcal{A}) $.

综上所述, 定理0.1成立.

从上述证明过程可以看出定理0.1的逆定理是不成立的.因为$ \lambda_{0}(x) $不一定属于$ Z(\mathcal{A}) $.但是可以增加条件, 使逆定理成立.

推论1.1  设$ \mathcal{A} $是域$ \mathbb{F} $(char$ \mathbb{F} = 0 $)上的结合代数, 且含有单位元1和幂等元$ e $. $ \mathcal{A} $满足$ a\mathcal{A}e\neq 0, a\mathcal{A}(1-e)\neq0, $其中$ a\neq 0, a\in \mathcal{A} $.设$ f:\mathcal{A}\rightarrow \mathcal{A} $是弱可加映射, 任取$ x, y\in \mathcal{A} $, 满足:

(a) $ f(x) = 0\Leftrightarrow x = 0 $;

(b) $ x, y $线性相关$ \Leftrightarrow f(x), f(y) $线性相关.

若$ f(x) = \lambda_{0}(x) x+\lambda_{1}(x) $, 这里$ \lambda_{0}(x) = \alpha(e_{1}f(e_{1})e_{1}-f_{11}(e_{1})e_{1})+\beta(e_{2}f(e_{2})e_{2}-f_{22}(e_{2})e_{2}) $, $ \lambda_{1}(x)\in Z(\mathcal{A}) $并且$ \frac{\alpha}{\beta} = \frac{t_{e_{1}, e_{2}}}{s_{e_{1}, e_{2}}} $, 则$ f $是交换映射.

证    明  由引理1.4和1.5, 存在非零$ t_{e_{1}, e_{2}}, s_{e_{1}, e_{2}}\in \mathbb{F} $, 使得

$ \begin{align*} t_{e_{1}, e_{2}}e_{1}f(e_{1})e_{1}& = e_{1}f(1)e_{1}-s_{e_{1}, e_{2}}e_{1}f(e_{2})e_{1}\nonumber\\ & = (z_{1}-s_{e_{1}, e_{2}}f_{22}(e_{2}))e_{1}\in Z(\mathcal{A}_{11})\nonumber \end{align*} $

和

$ \begin{align*} s_{e_{1}, e_{2}}e_{2}f(e_{2})e_{2}& = e_{2}f(1)e_{2}-t_{e_{1}, e_{2}}e_{2}f(e_{1})e_{2}\nonumber\\ & = (z_{2}-t_{e_{1}, e_{2}}f_{11}(e_{1}))e_{2}\in Z(\mathcal{A}_{22})\nonumber. \end{align*} $

令

$ \alpha_{1} = \frac {1}{t_{e_{1}, e_{2}}}(z_{1}-s_{e_{1}, e_{2}}f_{22}(e_{2})), \alpha_{2} = \frac {1}{s_{e_{1}, e_{2}}}(z_{2}-t_{e_{1}, e_{2}}f_{11}(e_{1})). $

可以看出$ \alpha_{1}, \alpha_{2}\in Z(\mathcal{A}) $, 因此$ e_{i}f(e_{i})e_{i} = \alpha_{i}e_{i}\in Z(\mathcal{A}_{ii})(i = 1, 2) $.

从上述讨论可得$ e_{1}f(e_{1})e_{1}-f_{11}(e_{1})e_{1}\in Z(\mathcal{A}_{11}) $, 并且$ e_{2}f(e_{2})e_{2}-f_{22}(e_{2})e_{2}\in Z(\mathcal{A}_{22}) $.所以由引理1.4, 对任意的$ a_{12}\in \mathcal{A}_{12} $, 有

$ \begin{align*} \alpha e_{1}f(e_{1})e_{1}a_{12}+\beta a_{12}f_{22}(e_{2})e_{2}& = (\alpha e_{1}f(e_{1})e_{1}+\beta f_{22}(e_{2})e_{1})a_{12}\nonumber\\ & = (\alpha e_{1}f(e_{1})e_{1}+\beta e_{1}f(e_{2})e_{1})a_{12}\nonumber\\ & = \gamma(t_{e_{1}, e_{2}}e_{1}f(e_{1})e_{1}+s_{e_{1}, e_{2}}e_{1}f(e_{2})e_{1})a_{12}\nonumber\\ & = \gamma e_{1}f(1)e_{1}a_{12} = \gamma a_{12}e_{2}f(1)e_{2}\nonumber\\ & = \gamma a_{12}(t_{e_{1}, e_{2}}e_{2}f(e_{1})e_{2}+s_{e_{1}, e_{2}}e_{2}f(e_{2})e_{2})\nonumber\\ & = a_{12}(\alpha f_{11}(e_{1})e_{2}+\beta e_{2}f(e_{2})e_{2})\nonumber\\ & = \alpha f_{11}(e_{1})e_{1}a_{12}+\beta a_{12}e_{2}f(e_{2})e_{2}\nonumber, \end{align*} $

这里$ \alpha = \gamma t_{e_{1}, e_{2}}, \beta = \gamma s_{e_{1}, e_{2}}, \gamma\in \mathbb{F} $, 即

$ \alpha(e_{1}f(e_{1})e_{1}-f_{11}(e_{1})e_{1})a_{12} = \beta a_{12}(e_{2}f(e_{2})e_{2}-f_{22}(e_{2})e_{2}), \forall a_{12}\in \mathcal{A}_{12}. $

同理可证

$ \alpha a_{21}(e_{1}f(e_{1})e_{1}-f_{11}(e_{1})e_{1}) = \beta(e_{2}f(e_{2})e_{2}-f_{22}(e_{2})e_{2})a_{21}, \forall a_{21}\in \mathcal{A}_{21}. $

由引理1.1可得

$ \lambda_{0}(x) = \alpha(e_{1}f(e_{1})e_{1}-f_{11}(e_{1})e_{1})+\beta(e_{2}f(e_{2})e_{2}-f_{22}(e_{2})e_{2})\in Z(\mathcal{A}). $

由于矩阵代数是一类特殊的代数, 根据定理0.1和弱可加映射的定义可以得到如下推论.

推论1.2  设$ M_{n}(\mathbb{F}) $是定义在$ \mathbb{F} $(char$ \mathbb{F} = 0 $)上的关于矩阵加法和乘法的矩阵代数, 映射$ f:M_{n}(\mathbb{F})\rightarrow M_{n}(\mathbb{F}) $是一个弱可加映射, 且对任意的$ x, y\in M_{n}(\mathbb{F}) $, 满足:

(a) $ f(x) = 0\Leftrightarrow x = 0 $;

(b) $ x, y $线性相关$ \Leftrightarrow f(x), f(y) $线性相关.

如果$ f $是交换映射, 则存在$ \lambda_{0}(x)\in M_{n}(\mathbb{F}) $和$ \lambda_{1}(x)\in \mathbb{F} $使得$ f(x) = \lambda_{0}(x) x+\lambda_{1}(x)I $.

正如前面提到的例子, $ \mathcal{A} = M_{2}(\mathbb{F}) $是一个$ \mathbb{F} $-结合代数且含有单位元$ I $和幂等元$ e = \left( \begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix} \right) $. $ M_{2}(\mathbb{F}) $满足$ aM_{2}(\mathbb{F})e\neq 0, aM_{2}(\mathbb{F})(I-e)\neq0 $, $ \forall 0\neq a\in \mathcal{A} $. $ f(x) = \|x\|x $是一个弱可加映射, 且对任意的$ x, y\in M_{2}(\mathbb{F}) $, 有

(a) $ f(x) = 0\Leftrightarrow x = 0 $;

(b) $ x, y $线性相关$ \Leftrightarrow f(x), f(y) $线性相关.

则$ f $是交换映射, 并且

$ f(x) = \|x\|x = \lambda_{0}(x) x+\lambda_{1}(x)I, $

这里$ \lambda_{0}(x) = \|x\|I\in Z(M_{2}(\mathbb{F})) $, 并且$ \lambda_{1}(x) = 0\in Z(M_{2}(\mathbb{F})) $.