2 主要定理及其证明
定理2.1 设 $\mbox{alg}_M\beta, \mbox{alg}_M\gamma$是因子von Neumann代数 $M$中的两个非平凡套子代数, $\varphi:
\mbox{alg}_M\beta\rightarrow\mbox{alg}_M\gamma$是一个线性双射, 且 $\varphi(I)=I, $若对任意的 $A, B\in \mbox{alg}_M\beta$且 $AB=0, $有 $\varphi(A\circ B)=\varphi(A)\circ \varphi(B)$成立, 则 $\varphi$要么是一个同构, 要么是一个反同构.
为证明定理2.1, 我们需要下面几个引理.
引理2.1 设 $\varphi:
\mbox{alg}_M\beta\rightarrow\mbox{alg}_M\gamma$是一个线性双射且 $\varphi(I)=I.$若对任意的 $A, B\in \mbox{alg}_M\beta$且 $AB=0, $有 $ \varphi(A\circ
B)=\varphi(A)\circ \varphi(B), $则 $\varphi$保幂等元.即 $\varphi(E)=\varphi(E)^{2}, $对任意的 $E\in {\cal A}$且 $E=E^{2}.$
证 因为 $E=E^{2}, $所以 $E(I-E)=0, $于是 $ \varphi(E)\circ
\varphi(I-E)=\varphi(E\circ (I-E))=\varphi(0)=0. $又 $\varphi(I)=I, $从而 $\varphi(E)=\varphi(E)^{2}.$
以下总假设 $\varphi: \mbox{alg}_M\beta\rightarrow
\mbox{alg}_M\gamma$是一个线性双射且 $\varphi(I)=I.$若对任意的 $A, B\in \mbox{alg}_M\beta$且 $AB=0, $有 $ \varphi(A\circ
B)=\varphi(A)\circ \varphi(B) $成立.
引理2.2 设 $P\in \beta$是一个固定的非平凡投影, 则对 $A\in \mbox{alg}_M\beta, $有
(1)
$\varphi(P)\varphi(PAP^{\perp})\varphi(P)=\varphi(P^{\perp})\varphi(PAP^{\perp})\varphi(P)=0;$
(2) $\varphi(PAP)=\varphi(P)\varphi(PAP)\varphi(P);$
(3)
$\varphi(P^{\perp}AP^{\perp})=\varphi(P^{\perp})\varphi(P^{\perp}AP^{\perp})\varphi(P^{\perp}).$
证 (1) 设 $A\in \mbox{alg}_M\beta, $则由 $PAP^{\perp}P=0$知
| $
\varphi(PAP^{\perp})=\varphi(PAP^{\perp}\circ
P)=\varphi(PAP^{\perp})\circ \varphi(P),
$ |
上式两边同乘 $\varphi(P), $并由引理2.1得
| $
\begin{equation}
\varphi(P)\varphi(PAP^{\perp})\varphi(P)=0.\end{equation}
$ |
(2.1) |
又 $P^{\perp}PAP^{\perp}=0, $从而 $
\varphi(PAP^{\perp})=\varphi(P^{\perp}\circ
PAP^{\perp})=\varphi(P^{\perp})\circ \varphi(PAP^{\perp}). $上式两边同乘以 $\varphi(P^{\perp}), $并由引理2.1得
| $
\begin{equation}
\varphi(P^{\perp})\varphi(PAP^{\perp})\varphi(P^{\perp})=0.\end{equation}
$ |
(2.2) |
由等式(2.1) 和(2.2) 知, 对任意的 $A\in \mbox{alg}_M\beta, $有
| $
\varphi(P)\varphi(PAP^{\perp})\varphi(P)=\varphi(P^{\perp})\varphi(PAP^{\perp})\varphi(P^{\perp})=0.
$ |
(2) 由 $P^{\perp}PAP=0$得
| $
\begin{equation}
\varphi(P^{\perp})\circ\varphi(PAP)=\varphi(P^{\perp}\circ
PAP)=0.\end{equation}
$ |
(2.3) |
由(2.3) 式及引理2.1得
| $
\varphi(P^{\perp})\varphi(PAP)\varphi(P^{\perp})
=\varphi(P^{\perp})\varphi(PAP)\varphi(P)=\varphi(P)\varphi(PAP)\varphi(P^{\perp})=0.
$ |
综上可知, 对任意的 $A\in \mbox{alg}_M\beta, $有
| $
\varphi(PAP)=\varphi(P)\varphi(PAP)\varphi(P).
$ |
类似地, 我们可以证明(3).证毕.
引理2.3 设 $P\in \beta$是一个固定的非平凡投影, 则对 $A\in \mbox{alg}_M\beta, $有
(1) $\varphi(PAP)=\varphi(P)\varphi(A)\varphi(P),
\varphi(P^{\perp}AP^{\perp})=\varphi(P^{\perp})\varphi(A)\varphi(P^{\perp});$
(2)
$\varphi(PAP^{\perp})=\varphi(P)\varphi(A)\varphi(P^{\perp})+\varphi(P^{\perp})\varphi(A)\varphi(P);$
(3) $\varphi(PA+AP)=\varphi(P)\varphi(A)+\varphi(A)\varphi(P).$
证 (1) 设 $A\in \mbox{alg}_M\beta, $由引理2.2(1) 和(3) 知
| $
\varphi(P)\varphi(PAP^{\perp})\varphi(P)=\varphi(P)\varphi(P^{\perp}AP^{\perp})\varphi(P)=0.
$ |
再利用引理2.2(2) 得
| $
\begin{eqnarray*}
&&\varphi(PAP)=\varphi(P)\varphi(PAP)\varphi(P)+\varphi(P)\varphi(PAP^{\perp})\varphi(P)
+\varphi(P)\varphi(P^{\perp}AP^{\perp})\varphi(P)\\
&=&\varphi(P)\varphi(PAP+PAP^{\perp}+P^{\perp}AP^{\perp})\varphi(P)\\
&=&\varphi(P)\varphi(A)\varphi(P).
\end{eqnarray*}
$ |
同理可得, 对任意的 $A\in \mbox{alg}_M\beta, $有 $
\varphi(P^{\perp}AP^{\perp})=\varphi(P^{\perp})\varphi(A)\varphi(P^{\perp}).
$
(2) 设 $A\in \mbox{alg}_M\beta, $一方面,
$
\varphi(A)=\varphi(PAP+PAP^{\perp}+P^{\perp}AP^{\perp}).
$
另一方面,
$
\varphi(A)=\varphi(P)\varphi(A)\varphi(P)+\varphi(P)\varphi(A)\varphi(P^{\perp})
+\varphi(P^{\perp})\varphi(A)\varphi(P)+\varphi(P^{\perp})\varphi(A)\varphi(P^{\perp}).
$于是, 由(1) 得
| $
\varphi(PAP^{\perp})=\varphi(P)\varphi(A)\varphi(P^{\perp})+\varphi(P^{\perp})\varphi(A)\varphi(P).
$ |
(3) 由(1) 和(2), 并注意到 $PA=PAP, $从而对任意的 $A\in
\mbox{alg}_M\beta, $有
| $
\begin{eqnarray*}
\varphi(PA+AP)
&=&\varphi(2PAP+PAP^{\perp})=2\varphi(PAP)+\varphi(PAP^{\perp})\\
&=&2\varphi(P)\varphi(A)\varphi(P)+\varphi(P)\varphi(A)\varphi(P^{\perp})
+\varphi(P^{\perp})\varphi(A)\varphi(P)\\
&=&\varphi(P)\varphi(A)+\varphi(A)\varphi(P).
\end{eqnarray*}
$ |
证毕.
引理2.4 设 $P\in \beta$是一个固定的非平凡投影, 则对 $A, B\in \mbox{alg}_M\beta, $有
| $
\begin{eqnarray*}&& \varphi(P)\varphi(A)\varphi(P^\bot)\varphi(B)\varphi(P)=0; \
\ \
&&\varphi(P)\varphi(B)\varphi(P^\bot)\varphi(A)\varphi(P)=0;\\
&&\varphi(P^\bot)\varphi(A)\varphi(P)\varphi(B)\varphi(P^\bot)=0;\ \ &&
\varphi(P^\bot)\varphi(B)\varphi(P)\varphi(A)\varphi(P^\bot)=0.\end{eqnarray*}
$ |
证 设 $A, B\in
\mbox{alg}_M\beta, $因为 $PAP^{\perp}PBP^{\perp}=0, $所以
| $
\varphi(PAP^{\perp})\circ\varphi(PBP^{\perp})=\varphi(PAP^{\perp}\circ
PBP^{\perp})=0.
$ |
从而由引理2.3(2),
| $
\begin{eqnarray*}
&&\varphi(P)\varphi(A)\varphi(P^\bot)\varphi(B)\varphi(P)+
\varphi(P^\bot)\varphi(A)\varphi(P)\varphi(B)\varphi(P^\bot)\\
&&+\varphi(P)\varphi(B)\varphi(P^\bot)\varphi(A)\varphi(P)+
\varphi(P^\bot)\varphi(B)\varphi(P)\varphi(A)\varphi(P^\bot)\\
&=&[\varphi(P)\varphi(A)\varphi(P^\bot)+\varphi(P^\bot)\varphi(A)\varphi(P)]\circ [\varphi(P)\varphi(B)\varphi(P^\bot)+\varphi(P^\bot)\varphi(B)\varphi(P)]\\
&=&\varphi(PAP^\bot)\circ \varphi(PBP^\bot)=0.
\end{eqnarray*}
$ |
这说明
| $
\begin{eqnarray}&&
\varphi(P)\varphi(A)\varphi(P^\bot)\varphi(B)\varphi(P)
+\varphi(P)\varphi(B)\varphi(P^\bot)\varphi(A)\varphi(P)=0, \end{eqnarray}
$ |
(2.4) |
| $
\begin{eqnarray}&& \varphi(P^\bot)\varphi(A)\varphi(P)\varphi(B)\varphi(P^\bot)
+\varphi(P^\bot)\varphi(B)\varphi(P)\varphi(A)\varphi(P^\bot)=0.\end{eqnarray}
$ |
(2.5) |
因为 $\varphi$是双射, 我们用 $\varphi(P)\varphi(A)\varphi(P^\bot)$代替(2.4) 和(2.5) 式中的 $\varphi(A), $可得
| $
\varphi(P)\varphi(A)\varphi(P^\bot)\varphi(B)\varphi(P)=0
\mbox{且}
\varphi(P^\bot)\varphi(B)\varphi(P)\varphi(A)\varphi(P^\bot)=0.
$ |
再由(2.4) 和(2.5) 式, 于是对任意的 $A, B\in \mbox{alg}_M\beta, $有
| $
\varphi(P)\varphi(A)\varphi(P^\bot)\varphi(B)\varphi(P)=
\varphi(P^\bot)\varphi(A)\varphi(P)\varphi(B)\varphi(P^\bot)=0
$ |
和
| $
\varphi(P)\varphi(B)\varphi(P^\bot)\varphi(A)\varphi(P)=
\varphi(P^\bot)\varphi(B)\varphi(P)\varphi(A)\varphi(P^\bot)=0.
$ |
证毕.
引理2.5 设 $\beta$是因子von Nuemann代数 $M$中的套, 且 $T, S\in M, $则 $T(\mbox{alg}_M\beta)S=0$当且仅当存在 $P\in\beta$使得 $T=TP^\bot$且 $S=PS$.
证 充分性显然.下面只须证必要性.
设 $P$表示由 ${\cal H}$到 $(\mbox{alg}_M\beta)S{\cal
H}$的闭包上的正交投影, 则 $P\in \beta.$由于 $T(\mbox{alg}_M\beta)S=0, $从而 $TP=0, $即 $T=TP^\bot.$另一方面, 我们有 $P^\bot(\mbox{alg}_M\beta)S=0.$于是 $P^\bot
S=0, $即 $S=PS.$ \证毕.
引理2.6 设 $P\in \beta$是一个固定的非平凡投影, 则对 $T\in \mbox{alg}_M\beta, $有要么 $\varphi(PTP^\bot)=\varphi(P)\varphi(T)\varphi(P^\bot);$要么 $\varphi(PTP^\bot)=\varphi(P^\bot)\varphi(T)\varphi(P).$
证 不妨设 $P\in\beta/\{0, I\}.$令 $
A=\varphi(P)\varphi(T)\varphi(P^\bot)$且 $
B=\varphi(P^\bot)\varphi(T)\varphi(P). $则由引理2.4, 对任意算子 $X\in \mbox{alg}_M\gamma, $都有
| $
\begin{equation}
\varphi(P^\bot)XA=\varphi(P)XB=0.\end{equation}
$ |
(2.6) |
由(2.6) 式和引理2.5, 则存在投影 $P_1\in \gamma$使得
| $
\begin{eqnarray}
\varphi(P^\bot)=\varphi(P^\bot)P_1^\bot, \end{eqnarray}
$ |
(2.7) |
| $
\begin{eqnarray} A=P_1A.\end{eqnarray}
$ |
(2.8) |
存在投影 $P_2\in \gamma$使得
| $
\begin{eqnarray}
\varphi(P)=\varphi(P)P_2^\bot;\end{eqnarray}
$ |
(2.9) |
| $
\begin{eqnarray}B=P_2B. \end{eqnarray}
$ |
(2.10) |
由(2.7) 和(2.9) 式, 则
| $
\varphi(P^\bot)P_1^\bot+\varphi(P)P_2^\bot=\varphi(I)=I.
$ |
对上式两边右乘 $P_1P_2$得 $P_1P_2=0.$如果 $A\neq 0, $则由(2.8) 式知 $P_1\neq0.$于是 $P_2=0.$因此由(2.10) 式得 $B=0.
$同理, 如果 $B\neq 0, $则一定有 $A=0.$这说明 $A=0$或 $B=0.$又 $\varphi(PTP^\bot)=A+B, $从而对任意的 $T\in
\mbox{alg}_M\beta, $必有以下两式
| $
\varphi(PTP^\bot)=\varphi(P)\varphi(T)\varphi(P^\bot) \ \
\mbox{和} \ \
\varphi(PTP^\bot)=\varphi(P^\bot)\varphi(T)\varphi(P)
$ |
之一成立.由于 $M$是因子, 则存在部分等距算子 $V\in
M$使得 $V=PVP^\bot, $从而 $V\in \mbox{alg}_M\beta.$于是由上述讨论可知要么 $\varphi(V)=\varphi(P)\varphi(V)\varphi(P^\bot)$; 要么 $\varphi(V)=\varphi(P^\bot)\varphi(V)\varphi(P).$
当 $\varphi(V)=\varphi(P)\varphi(V)\varphi(P^\bot)$时, 如果存在 $S\in\mbox{alg}_M\beta$使得 $\varphi(PSP^\bot)\neq\varphi(P)\varphi(S)\varphi(P^\bot), $则 $\varphi(PSP^\bot)=\varphi(P^\bot)\varphi(S)\varphi(P).$
另一方面, 由于
$\varphi(V+PSP^\bot)=\varphi(P)\varphi(V+S)\varphi(P^\bot) \
\mbox{和} \ \
\varphi(V+PSP^\bot)=\varphi(P^\bot)\varphi(V+S)\varphi(P)
$之一成立, 并且 $\varphi(V)=\varphi(P)\varphi(V)\varphi(P^\bot), \
\varphi(PSP^\bot)\neq\varphi(P)\varphi(S)\varphi(P^\bot), $则
| $
\varphi(V+PSP^\bot)=\varphi(P^\bot)\varphi(V+S)\varphi(P).
$ |
于是 $\varphi(V)=\varphi(P^\bot)\varphi(V)\varphi(P)=\varphi(P)\varphi(V)\varphi(P^\bot).$这说明 $\varphi(V)=0, $从而 $V=0.$矛盾.所以此时对任意的 $T\in
\mbox{alg}_M\beta, $有
| $
\varphi(PTP^\bot)=\varphi(P)\varphi(T)\varphi(P^\bot).
$ |
同理可得当 $\varphi(V)=\varphi(P^\bot)\varphi(V)\varphi(P)$时, 对任意的 $T\in \mbox{alg}_M\beta, $有
| $
\varphi(PTP^\bot)=\varphi(P^\bot)\varphi(T)\varphi(P).
$ |
因此, 对任意的 $T\in \mbox{alg}_M\beta$和任意投影 $P\in\beta$，要么 $\varphi(PTP^\bot)=\varphi(P)\varphi(T)\varphi(P^\bot);$要么 $\varphi(PTP^\bot)=\varphi(P^\bot)\varphi(T)\varphi(P).$证毕.
引理2.7 设 $P\in \beta$是一个固定的非平凡投影, 则对 $A,
T\in \mbox{alg}_M\beta, $下列之一成立:
(1) $\varphi(APTP^\bot)=\varphi(A)\varphi(PTP^\bot),
\varphi(PTP^\bot A)=\varphi(PTP^\bot)\varphi(A);$
(2) $\varphi(APTP^\bot)=\varphi(PTP^\bot)\varphi(A),
\varphi(PTP^\bot A)=\varphi(A)\varphi(PTP^\bot).$
证 由引理2.6知, 对 $T\in \mbox{alg}_M\beta, $
$\varphi(PTP^\bot)=\varphi(P)\varphi(T)\varphi(P^\bot)$和
| $
\varphi(PTP^\bot)=\varphi(P^\bot)\varphi(T)\varphi(P)
$ |
之一成立.当 $\varphi(PTP^\bot)=\varphi(P)\varphi(T)\varphi(P^\bot)$时, $
\varphi(P^\bot)\varphi(T)\varphi(P)=0. $由引理2.3(1) 知, 对 $T\in
\mbox{alg}_M\beta, $有
| $
\begin{equation}
\varphi(TP)=\varphi(PTP)=\varphi(P)\varphi(T)\varphi(P^\bot)
+\varphi(P^\bot)\varphi(T)\varphi(P)=\varphi(T)\varphi(P).\end{equation}
$ |
(2.11) |
同理可得
| $
\begin{equation}
\varphi(P^{\perp}T)=\varphi(P^{\perp}TP^{\perp})=\varphi(P^{\perp})\varphi(T)\varphi(P^\bot)
+\varphi(P^\bot)\varphi(T)\varphi(P)=\varphi(P^{\perp})\varphi(T).\end{equation}
$ |
(2.12) |
由 $PAP=AP, (PTP^{\perp})AP=0$及(2.11) 式知, 对任意的 $A\in
\mbox{alg}_M\beta, $有
| $
\begin{eqnarray*}
\varphi(APTP^\bot)&=&\varphi(PTP^\bot\circ AP)=\varphi(PTP^\bot)\circ\varphi(AP)\\
&=&\varphi(PTP^\bot)\varphi(AP)+\varphi(AP)\varphi(PTP^\bot)\\
&=&\varphi(A)\varphi(P)\varphi(T)\varphi(P^\bot)\\
&=&\varphi(A)\varphi(PTP^\bot).
\end{eqnarray*}
$ |
又 $P^\bot AP^\bot=P^\bot A, P^\bot A(PTP^{\perp})=0, $由(2.12) 式类似可得
| $
\begin{eqnarray*}
\varphi(PTP^\bot A)&=&\varphi(P^{\perp}A\circ PTP^\bot)=\varphi(P^{\perp}A)\circ\varphi(PTP^\bot)\\
&=&\varphi(PTP^\bot)\varphi(A).
\end{eqnarray*}
$ |
即当 $\varphi(PTP^\bot)=\varphi(P)\varphi(T)\varphi(P^\bot)$时, 结论(1) 成立.
类似可证, 当 $\varphi(PTP^\bot)=\varphi(P^\bot)\varphi(T)\varphi(P)$成立时, 结论(2) 成立.证毕.
定理2.1的证明 由引理2.6, 对任意的 $T\in
\mbox{alg}_M\beta, $
| $
\varphi(PTP^\bot)=\varphi(P)\varphi(T)\varphi(P^\bot) \ \
\mbox{和} \ \ \
\varphi(PTP^\bot)=\varphi(P^\bot)\varphi(T)\varphi(P)
$ |
其中之一成立.
若 $ \varphi(PTP^\bot)=\varphi(P)\varphi(T)\varphi(P^\bot), $则由引理2.7(1), 并注意到 $BP=PBP, $从而对任意的 $A, B\in
\mbox{alg}_M\beta, $
| $
\varphi(AB)\varphi(PTP^\bot)=\varphi(ABPTP^\bot)=\varphi(A)\varphi(BPTP^\bot)
=\varphi(A)\varphi(B)\varphi(PTP^\bot).
$ |
从而对任意 $A, B\in \mbox{alg}_M\beta, $有
| $
\begin{equation}
[\varphi(AB)-\varphi(A)\varphi(B)]\varphi(P)\mbox{alg}_M\gamma\varphi(P^\bot)=0, \end{equation}
$ |
(2.13) |
类似地, 由引理2.7(1) 及 $P^{\perp}A=P^{\perp}AP^{\perp}, $对任意 $A,
B\in \mbox{alg}_M\beta, $有
| $
\begin{equation}
\varphi(P)\mbox{alg}_M\gamma\varphi(P^\bot)[\varphi(AB)-\varphi(A)\varphi(B)]=0.\end{equation}
$ |
(2.14) |
由上面两式和引理2.5知, 存在投影 $Q_1, Q_2\in\gamma$使得
| $
[\varphi(AB)-\varphi(A)\varphi(B)]\varphi(P)=[\varphi(AB)-\varphi(A)\varphi(B)]\varphi(P)Q_1^\bot;
$ |
(2.15) |
| $
\begin{eqnarray}
\varphi(P^\bot)=Q_1\varphi(P^\bot); \end{eqnarray}
$ |
(2.16) |
| $
\begin{eqnarray} \varphi(P)=\varphi(P
)Q_2^\bot;\end{eqnarray}
$ |
(2.17) |
| $
\varphi(P^\bot)[\varphi(AB)-\varphi(A)\varphi(B)]=Q_2\varphi(P^\bot)[\varphi(AB)-\varphi(A)\varphi(B)].
$ |
(2.18) |
由于 $\varphi(PTP^\bot)=\varphi(P)\varphi(T)\varphi(P^\bot), $则对任意的 $T\in
\mbox{alg}_M\beta, $有 $ \varphi(P^\bot)\varphi(T)\varphi(P)=0. $从而 $[\varphi(P)]\in M, [\varphi(P)]$表示由 ${\cal
H}$到 $\varphi(P){\cal H}$上的正交投影.特别地, $
\varphi(P^\bot)Q_1^\bot\varphi(P)=0, $即
| $
\varphi(P^\bot)Q_1^\bot\varphi(P^\bot)=\varphi(P^\bot)Q_1^\bot.
$ |
从而由(2.16) 式知
| $
\varphi(P^\bot)Q_1^\bot=Q_1^\bot\varphi(P^\bot)=0.
$ |
这说明 $Q_1^\bot[\varphi(P^\bot)]=0.$由于 $P$是一个非平凡投影, 则 $[\varphi(P^\bot)]\neq0, $从而 $Q_1^\bot=0.$于是由(2.15) 式得
| $
[\varphi(AB)-\varphi(A)\varphi(B)]\varphi(P)=0.
$ |
(2.19) |
由(2.17) 式, 同上类似地讨论可得 $ \varphi(P)Q_2=Q_2\varphi(P)=0. $这说明 $[\varphi(P)]Q_2=0, $从而 $Q_2=0.$于是由(2.18) 式知,
| $
\varphi(P^\bot)[\varphi(AB)-\varphi(A)\varphi(B)]=0.
$ |
(2.20) |
由于 $P^{\perp}BP^{\perp}PAP^{\perp}=0, $由引理2.2知, 对任意 $A,
B\in \mbox{alg}_M\beta, $有
| $
\begin{eqnarray*}
\varphi(PAP^{\perp}BP^{\perp})
&=& \varphi(P^{\perp}BP^{\perp}\circ PAP^{\perp})=\varphi(P^{\perp}BP^{\perp})\circ \varphi(PAP^{\perp})\\
&=& \varphi(P^{\perp}BP^{\perp})\varphi(PAP^{\perp}). \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{eqnarray*}
$ |
(2.21) |
由于 $PBP^{\perp}PAP=0, $同理可得
| $
\varphi(PAPBP^{\perp})=\varphi(PAP)\varphi(PBP^{\perp}).
$ |
(2.22) |
由等式(2.21)、(2.22) 及引理2.3得
| $
\begin{eqnarray*}
\varphi(P)\varphi(AB)\varphi(P^\bot)
&=&\varphi(PAPBP^\bot)+\varphi(PAP^\bot BP^\bot)\\
&=&\varphi(PAP)\varphi(PBP^\bot)+\varphi(PAP^\bot)\varphi(P^\bot BP^\bot)\\
&=&\varphi(P)\varphi(A)\varphi(P)\varphi(B)\varphi(P^\bot)
+\varphi(P)\varphi(A)\varphi(P^\bot)\varphi(B)\varphi(P^\bot)\\
&=&\varphi(P)\varphi(A)\varphi(B)\varphi(P^\bot).
\end{eqnarray*}
$ |
从而对任意 $A, B\in \mbox{alg}_M\beta, $有
| $
\varphi(P)[\varphi(AB)-\varphi(A)\varphi(B)]\varphi(P^\bot)=0.
$ |
(2.23) |
由(2.19)、(2.20) 及(2.23) 式知, 对任意 $A, B\in \mbox{alg}_M\beta, $有
| $
\varphi(AB)=\varphi(A)\varphi(B).
$ |
如果对任意 $T\in
\mbox{alg}_M\beta, $有 $\varphi(PTP^\bot)=\varphi(P^\bot)\varphi(T)\varphi(P), $对任意 $X\in\mbox{alg}_M\beta^\bot, $我们定义 $\psi(X)=\varphi(JX^*J), $这里 $J$是文[11, 引理2.3]中所定义的共轭线性对合算子.由引理2.7(2) 类似可证
| $
\varphi(AB)=\varphi(B)\varphi(A).
$ |
综上所述 $\varphi$要么是一个同构, 要么是一个反同构.证毕.
2015, Vol. 35 
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