论文标题
在非交换性$ l^p $ - 空间上分离地图
Surjective separating maps on noncommutative $L^p$-spaces
论文作者
论文摘要
令$ 1 \ leq p <\ infty $，让$ t \ colon l^p（{\ mathcal m}）\ to l^p（{\ Mathcal n}）$为非交换性$ l^p $ -spaces之间的有界地图。如果$ t $是两种二$（即，对于任何$ x，y \ in l^p（{\ nathcal m}）$，以至于$ x^*y = xy^*= 0 $，我们有$ t（x）^*t（x）^*t（y）= t（x）t（x）t（x）t（y）t（y）^*= 0 $） m} _1 \ Mathop {\ oplus} \ limits^\ infty {\ Mathcal M} _2 $，$ {\ Mathcal n} = {\ Mathcal n} _1 _1 _1 _1 \ Mathop {\ oplus} l^p（{\ Mathcal m} _1）\ to l^p（{\ Mathcal n} _1）$，$ t_2 \ colon l^p（{\ Mathcal M} _2）\ to l^p（{\ Mathcal n} _2）分解和$ T_2 $具有反向直接的Yeadon类型分解。我们进一步表明，在这种情况下，$ t^{ - 1} $是分开的。接下来，我们证明，对于任何$ 1 \ leq p <\ infty $（分别$ 1 \ leq p \ not = 2 <\ infty $），这是分隔地图$ t $ t \ colon l^p（{\ mathcal m}）\与l^p（{\ nathcal n}）$ nisect $ s $ s^1 $ byd（IF n of）;分解$ {\ MATHCAL M} = {\ MATHCAL M} _1 \ MATHOP {\ oplus} \ limits^\ infty {\ Mathcal M} _2 $ $ {\ MATHCAL M} _2 $是亚均匀的。
Let $1\leq p<\infty$ and let $T\colon L^p({\mathcal M})\to L^p({\mathcal N})$ be a bounded map between noncommutative $L^p$-spaces. If $T$ is bijective and separating (i.e., for any $x,y\in L^p({\mathcal M})$ such that $x^*y=xy^*=0$, we have $T(x)^*T(y)=T(x)T(y)^*=0$), we prove the existence of decompositions ${\mathcal M}={\mathcal M}_1\mathop{\oplus}\limits^\infty{\mathcal M}_2$, ${\mathcal N}={\mathcal N}_1 \mathop{\oplus}\limits^\infty{\mathcal N}_2$ and maps $T_1\colon L^p({\mathcal M}_1)\to L^p({\mathcal N}_1)$, $T_2\colon L^p({\mathcal M}_2)\to L^p({\mathcal N}_2)$, such that $T=T_1+T_2$, $T_1$ has a direct Yeadon type factorisation and $T_2$ has an anti-direct Yeadon type factorisation. We further show that $T^{-1}$ is separating in this case. Next we prove that for any $1\leq p<\infty$ (resp. any $1\leq p\not=2<\infty$), a surjective separating map $T\colon L^p({\mathcal M})\to L^p({\mathcal N})$ is $S^1$-bounded (resp. completely bounded) if and only if there exists a decomposition ${\mathcal M}={\mathcal M}_1 \mathop{\oplus}\limits^\infty{\mathcal M}_2$ such that $T|_{L^p({\tiny {\mathcal M}_1})}$ has a direct Yeadon type factorisation and ${\mathcal M}_2$ is subhomogeneous.
