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?







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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?







    share|cite|improve this question
























      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?







      share|cite|improve this question














      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?









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 14 at 14:20









      Martin Argerami

      116k1071165




      116k1071165










      asked Aug 14 at 12:01









      prince

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