Conditional Expectation for von Neumann algebra.
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Let $M$ be a von Neumann algebra and $T: Mrightarrow mathbbC$ be a finite normal faithful tracial map, s.t., if $phi : M rightarrow A cap A^*$ is a conditional expectation, (A being a weak star closed subalgebra of $M$), $T(phi(x))=T(x) quad forall x in M$.
Does this imply that ,
$T(phi(x))=T(x) quad forall x in L^1(M)$ ?
where $L^1(M)$ is the completion of $M$ in $||.||_1$.
I know that we can extend $phi$ continuously to $L^1(M)$. How can I use this fact to answer my question?
functional-analysis operator-theory operator-algebras von-neumann-algebras
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up vote
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down vote
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Let $M$ be a von Neumann algebra and $T: Mrightarrow mathbbC$ be a finite normal faithful tracial map, s.t., if $phi : M rightarrow A cap A^*$ is a conditional expectation, (A being a weak star closed subalgebra of $M$), $T(phi(x))=T(x) quad forall x in M$.
Does this imply that ,
$T(phi(x))=T(x) quad forall x in L^1(M)$ ?
where $L^1(M)$ is the completion of $M$ in $||.||_1$.
I know that we can extend $phi$ continuously to $L^1(M)$. How can I use this fact to answer my question?
functional-analysis operator-theory operator-algebras von-neumann-algebras
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $M$ be a von Neumann algebra and $T: Mrightarrow mathbbC$ be a finite normal faithful tracial map, s.t., if $phi : M rightarrow A cap A^*$ is a conditional expectation, (A being a weak star closed subalgebra of $M$), $T(phi(x))=T(x) quad forall x in M$.
Does this imply that ,
$T(phi(x))=T(x) quad forall x in L^1(M)$ ?
where $L^1(M)$ is the completion of $M$ in $||.||_1$.
I know that we can extend $phi$ continuously to $L^1(M)$. How can I use this fact to answer my question?
functional-analysis operator-theory operator-algebras von-neumann-algebras
Let $M$ be a von Neumann algebra and $T: Mrightarrow mathbbC$ be a finite normal faithful tracial map, s.t., if $phi : M rightarrow A cap A^*$ is a conditional expectation, (A being a weak star closed subalgebra of $M$), $T(phi(x))=T(x) quad forall x in M$.
Does this imply that ,
$T(phi(x))=T(x) quad forall x in L^1(M)$ ?
where $L^1(M)$ is the completion of $M$ in $||.||_1$.
I know that we can extend $phi$ continuously to $L^1(M)$. How can I use this fact to answer my question?
functional-analysis operator-theory operator-algebras von-neumann-algebras
edited Aug 14 at 14:20
Martin Argerami
116k1071165
116k1071165
asked Aug 14 at 12:01
prince
514
514
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