设${\cal A}$是$*$-代数, $\xi$是非零复数, $A, B\in{\cal A}$的$\xi$-Jordan $*$-积定义为$A\diamond_{\xi}B = AB+\xi BA^{*},$$1$-Jordan $*$-积通常记为$A\bullet B = AB+BA^{*},$$-1$-Jordan $*$- 积(斜Lie积)通常记为$[A, B]_{*} = AB-BA^{*}.$近年来, 相关的研究吸引了许多作者的注意(参看文献[1-13]). 文献[1]中P. Šemrl在量子函数中首先引入并研究了$-1$-Jordan $*$-积. 文献[2]研究了von Neumann代数(无中心交换投影)到${\cal B(H)}$上的非线性$\xi$-Jordan $*$-可导映射.

在以上文献研究的基础上, 下面将考虑更宽泛的结果: 设${\cal A}$是$*$-代数, 对任意$A, B, C\in{\cal A},$定义$A\diamond_{\xi}B\diamond_{\xi}C: = (A\diamond_{\xi}B)\diamond_{\xi}C$为$\xi$-Jordan $*$-三重积(其中$\diamond_{\xi}$不具有结合性). $\phi:{\cal A\rightarrow A}$是一个映射. 如果$\phi(A\diamond_{\xi}B\diamond_{\xi}C) = \phi(A)\diamond_{\xi}B\diamond_{\xi}C+A\diamond_{\xi}\phi(B)\diamond_{\xi}C+A\diamond_{\xi}B\diamond_{\xi}\phi(C),$那么称$\phi$是$\xi$-Jordan $*$-三重可导映射. 文献[3]得到了因子von Neumann代数上非线性$-1$-Jordan $*$-三重可导映射是可加的$*$-导子. 本文将要研究因子von Neumann代数上非线性$\xi$-Jordan $*$-三重可导映射.

设${\mathbb R}$和${\mathbb C}$分别是实数域和复数域, $\text{i}$是虚数单位. von Neumann代数${\cal A}$是定义在Hilbert空间${\cal H}$上的包含单位算子$I$的弱闭的, 自伴的算子代数. 记${\cal Z(A)}$为${\cal A}$的中心, 若${\cal Z(A)} = {\mathbb C}I,$则${\cal A}$是因子von Neumann代数. 因子von Neumann代数${\cal A}$是素的是指对$A, B\in{\cal A},$如果有$A{\cal A}B = 0,$那么$A = 0$或$B = 0.$

引理1.1^[4]  设${\cal A}$是因子von Neumann代数且$A\in{\cal A}.$若对所有$B\in{\cal A},$有$AB+BA^{*} = 0,$则$A\in\text{i}{\mathbb R}I.$

定理2.1  设${\cal A}$是因子von Neumann代数, $\xi$是非零复数. 非线性映射$\phi:{\cal A\rightarrow A}$满足对所有$A, B, C\in{\cal A},$有$\phi(A\diamond_{\xi}B\diamond_{\xi}C) = \phi(A)\diamond_{\xi}B\diamond_{\xi}C+A\diamond_{\xi}\phi(B)\diamond_{\xi}C+A\diamond_{\xi}B\diamond_{\xi}\phi(C)$当且仅当$\phi$是可加的$*$-导子且对所有$A\in{\cal A},$有$\phi(\xi A) = \xi\phi(A).$

证  设$P_{1}\in{\cal A}$为非平凡投影, 令$P_{2} = I-P_{1}, {\cal A}_{ij} = P_{i}{\cal A}P_{j}, i, j = 1, 2,$则${\cal A} = \sum\limits_{i, j = 1}^{2}{\cal A}_{ij}.$所以对任意$A\in{\cal A},$记$A = \sum\limits_{i, j = 1}^{2}A_{ij},$其中$A_{ij}\in{\cal A}_{ij}.$

当$\xi = -1$时, 文献[3]已证明并得到结论, 下面的证明中假设$\xi\neq -1.$

断言1  $\phi(0) = 0.$

取$A = B = C = 0$得

断言2  $\phi(A_{11}+A_{22}) = \phi(A_{11})+\phi(A_{22}), \forall A_{11}\in{\cal A}_{11}, \forall A_{22}\in{\cal A}_{22}.$

设$T = \phi(A_{11}+A_{22})-\phi(A_{11})-\phi(A_{22}).$由$I\diamond_{\xi}\frac{P_{1}}{1+\xi}\diamond_{\xi}A_{22} = 0$和断言1得

所以$I\diamond_{\xi}\frac{P_{1}}{1+\xi}\diamond_{\xi}T = 0,$由此可得$(1+\xi)T_{11}+T_{12}+\xi T_{21} = 0.$由$\xi\neq 0, -1$得$T_{11} = T_{12} = T_{21} = 0.$类似可得$T_{22} = 0.$因此$\phi(A_{11}+A_{22}) = \phi(A_{11})+\phi(A_{22}).$

断言3  $\phi(A_{12}+A_{21}) = \phi(A_{12})+\phi(A_{21}), \forall A_{12}\in{\cal A}_{12}, \forall A_{21}\in{\cal A}_{21}.$

令$T = \phi(A_{12}+A_{21})-\phi(A_{12})-\phi(A_{21}).$由$I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}A_{12} = 0$和断言1得

所以$I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}T = 0,$由此可得$(1+\xi)T_{11}-(1+\frac{1}{\overline{\xi}})T_{22}+(\xi-\frac{1}{\overline{\xi}})T_{21} = 0.$由$\xi\neq 0, -1$得$T_{11} = T_{22} = 0.$

又由$A_{12}\diamond_{\xi}P_{1}\diamond_{\xi}I = 0$得

所以$T\diamond_{\xi}P_{1}\diamond_{\xi}I = 0,$由此可得$(1+|\xi|^{2})T_{21}+2\xi T_{21}^{*} = 0,$进而$T_{21} = 0.$类似可得$T_{12} = 0.$因此$\phi(A_{12}+A_{21}) = \phi(A_{12})+\phi(A_{21}).$

断言4  设$i, j, k\in\{1, 2\},$若$A_{kk}\in {\cal A}_{kk}, A_{ij}\in {\cal A}_{ij}, i\neq j,$则$\phi(A_{kk}+A_{ij}) = \phi(A_{kk})+\phi(A_{ij}).$

下面只证明$i = k = 1, j = 2,$其他情形同理可得. 令$T = \phi(A_{11}+A_{12})-\phi(A_{11})-\phi(A_{12}).$由$I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}A_{11} = 0$和断言1得

所以$I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}T = 0,$由此可得$T_{12} = T_{21} = T_{22} = 0.$

又由$I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}A_{12} = 0$得

所以$I\diamond_{\xi}\frac{P_{1}-\frac{1}{\overline{\xi}}P_{2}}{1+\xi}\diamond_{\xi}T = 0,$由此可得$T_{11} = 0.$因此$\phi(A_{11}+A_{12}) = \phi(A_{11})+\phi(A_{12}).$

断言5  $\phi(A_{11}+A_{12}+A_{21}) = \phi(A_{11})+\phi(A_{12})+\phi(A_{21}), \phi(A_{12}+A_{21}+A_{22}) = \phi(A_{12})+\phi(A_{21})+\phi(A_{22}), \forall A_{11}\in{\cal A}_{11}, \forall A_{12}\in{\cal A}_{12}, \forall A_{21}\in{\cal A}_{21}, \forall A_{22}\in{\cal A}_{22}.$

令$T = \phi(A_{11}+A_{12}+A_{21})-\phi(A_{11})-\phi(A_{12})-\phi(A_{21}).$由断言3得

所以$I\diamond_{\xi}\frac{P_{2}}{1+\xi}\diamond_{\xi}T = 0,$由此可得$T_{12} = T_{21} = T_{22} = 0.$

又由$I\diamond_{\xi}\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi}\diamond_{\xi}A_{21} = 0$和断言4得

所以$I\diamond_{\xi}\frac{-\frac{1}{\overline{\xi}}P_{1}+P_{2}}{1+\xi}\diamond_{\xi}T = 0,$计算可得$T_{11} = 0.$因此$\phi(A_{11}+A_{12}+A_{21}) = \phi(A_{11})+\phi(A_{12})+\phi(A_{21}),$同理$\phi(A_{12}+A_{21}+A_{22}) = \phi(A_{12})+\phi(A_{21})+\phi(A_{22}).$

断言6  $\phi(A_{ij}+B_{ij}) = \phi(A_{ij})+\phi(B_{ij}), \forall A_{ij}, B_{ij}\in{\cal A}_{ij}, 1\leq i\neq j\leq 2.$

因为$I\diamond_{\xi}\frac{P_{i}+A_{ij}}{1+\xi}\diamond_{\xi}(P_{j}+B_{ij}) = A_{ij}+B_{ij}+\xi A_{ij}^{*}+\xi B_{ij}A_{ij}^{*},$由断言3, 断言4和断言5得

所以$\phi(A_{ij}+B_{ij}) = \phi(A_{ij})+\phi(B_{ij}).$

断言7  $\phi(X_{12}A_{21}+X_{12}B_{21}+\xi A_{12}X_{12}^{*}+\xi B_{12}X_{12}^{*}) = \phi(X_{12}A_{21})+\phi(X_{12}B_{21})+\phi(\xi A_{12}X_{12}^{*})+\phi(\xi B_{12}X_{12}^{*}), \forall X_{12}, A_{12}, B_{12}\in {\cal A}_{12}, \forall A_{21}, B_{21}\in {\cal A}_{21}.$

由断言3和断言6得

所以$\phi(X_{12}A_{21}+X_{12}B_{21}+\xi A_{12}X_{12}^{*}+\xi B_{12}X_{12}^{*}) = \phi(X_{12}A_{21})+\phi(X_{12}B_{21})+\phi(\xi A_{12}X_{12}^{*})+\phi(\xi B_{12}X_{12}^{*}).$

断言8  $\phi$是可加的.

令$A = \sum\limits_{i, i = 1}^{2}A_{ij}, B = \sum\limits_{i, i = 1}^{2}B_{ij}, T = \phi(A+B)-\phi(A)-\phi(B).$任取$X_{12}\in {\cal A}_{12},$由断言5, 断言6和断言7得

由此可得$I\diamond_{\xi}\frac{X_{12}}{1+\xi}\diamond_{\xi}T = 0,$即

对所有$X_{12}\in {\cal A}_{12}$成立. (2.1) 式中$X_{12}$用$\text{i}X_{12}$代替得

(2.1) 式和(2.2) 式相加得对所有$X_{12}\in {\cal A}_{12},$有$X_{12}T = 0$成立, 由${\cal A}$是素的可知$T = 0,$即$\phi$是可加的.

断言9  $\phi$是可加的$*$-导子且$\phi(\xi A) = \xi\phi(A), \forall A\in{\cal A}.$

分以下三种情形讨论.

情形1  $\xi = 1.$

任取$\lambda\in{\mathbb R}, A\in{\cal A},$由断言1得

所以任取$B = B^{*}\in{\cal A}$有$\phi(\text{i}\lambda I)B+B\phi(\text{i}\lambda I)^{*} = 0.$设$B_{1} = \frac{B+B^{*}}{2}, B_{2} = \frac{B-B^{*}}{2\text{i}}$是自伴元, 则对所有$B\in{\cal A}$有$B = B_{1}+\text{i}B_{2},$因此$\phi(\text{i}\lambda I)B+B\phi(\text{i}\lambda I)^{*} = 0.$由引理1.1得$\phi(\text{i}\lambda I)\subseteq\text{i}{\mathbb R}I,$所以

对所有$\lambda\in{\mathbb C}, A\in{\cal A},$由(2.3) 式得

由此可得, 对所有$B = B^{*}\in{\cal A},$有$B\phi(\lambda I) = \phi(\lambda I)B.$设$B_{1} = \frac{B+B^{*}}{2}, B_{2} = \frac{B-B^{*}}{2\text{i}}$是自伴元, 则对所有$B\in{\cal A},$有$B = B_{1}+\text{i}B_{2},$因此$B\phi(\lambda I) = \phi(\lambda I)B.$则$\phi(\lambda I)\subseteq{\mathbb C}I.$所以

由(2.4) 式可得, 存在$\lambda\in{\mathbb C}$使得$\phi(I) = \lambda I.$由

得$\lambda+\overline{\lambda} = 0,$则$\lambda\in\text{i}{\mathbb R}I,$于是存在$\lambda_{1}\in{\mathbb R},$使得

设$P_{i}\in{\cal A}, i = 1, 2$为非平凡投影, 由(2.5) 式得

即

设$i, j = 1, 2, i\neq j,$由(2.6) 式可得

(2.7) 式两边同乘$P_{i}$得

(2.7) 式左乘$P_{i}$右乘$P_{j}$得

(2.7) 式左乘$P_{j}$右乘$P_{i}$得

对所有$A_{ji}\in{\cal A}_{ji},$由(2.5) 式和(2.6) 式得

上式左乘$P_{j}$右乘$P_{i}$得$P_{j}\phi(P_{j})A_{ji}+A_{ji}\phi(P_{j})P_{i} = 0.$结合(2.8) 式可得$P_{j}\phi(P_{j})A_{ji} = 0.$由于${\cal A}$是素的, 所以

由(2.8)–(2.11) 式可得

对所有$A, B\in{\cal A},$由(2.12) 式得

由文献[6]的主要定理得, $\phi$是可加的$*$-导子.

情形2  $|\xi| = 1$且$\xi\neq 1, -1.$

因为

所以

由(2.13) 式得

另一方面

(2.14) 式加(2.15) 式得

(2.16) 式两边取$*$得

若$\phi(I)\neq 0,$则$\xi = 1,$与题设矛盾, 所以

任取$A\in{\cal A},$由(2.17) 式有

由此可得

任取$A, B\in{\cal A},$有

结合(2.18) 式可得

由文献[6]的主要定理得, $\phi$是可加的$*$-导子. 再由(2.18) 式得$\phi(\xi A) = \xi\phi(A), \forall A\in{\cal A}.$

情形3  $|\xi|\neq 1.$

由$\phi(I\diamond_{\xi}\text{i}I\diamond_{\xi}\text{i}I) = \phi(\text{i}I\diamond_{\xi}\text{i}I\diamond_{\xi}I)$得

所以$2\xi\text{i}(\phi(\text{i}I)+\phi(\text{i}I)^{*}) = 0.$由此可得

另一方面, 由$\phi(\text{i}I\diamond_{\xi}\text{i}I\diamond_{\xi}\text{i}I) = -\phi(I\diamond_{\xi}\text{i}I\diamond_{\xi}I)$和(2.19) 式得

由此可得

所以

由(2.19) 式和(2.21) 式可得

另一方面

(2.22) 式加(2.23) 式得

又由(2.19) 式和(2.21) 式得

(2.24) 式加(2.25) 式得$4\text{i}\phi(I) = 0,$则

由(2.20) 式得

任取$A\in{\cal A},$由(2.27) 式得

计算可得

由(2.26) 式得$\phi(A\diamond_{\xi}I\diamond_{\xi}I) = \phi(A)\diamond_{\xi}I\diamond_{\xi}I,$结合(2.28) 式得

另一方面$\phi(I\diamond_{\xi}I\diamond_{\xi}A^{*}) = I\diamond_{\xi}I\diamond_{\xi}\phi(A^{*}).$计算可得

结合(2.29) 式和(2.30) 式得

任取$A\in{\cal A},$由(2.28) 式得

所以

任取$A, B\in{\cal A},$有

另一方面, 由(2.32) 式得

结合(2.33) 式和(2.34) 式得

又由(2.30) 式得$(1+\xi)\phi(AB) = (1+\xi)(\phi(A)B+A\phi(B)).$所以$\phi(AB) = \phi(A)B+A\phi(B).$定理2.1证毕.