Extension of tracial state to the multiplier algebra












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If $A$ is a $C^*$ algebra, I know the fact: If $tau$ is a state on $A$, we can extend it to a state on $M(A)$, by continuity of $tau$ and where $M(A)$ is the multiplier algebra of $A$.



If $tau$ is a tracial state on $A$, can we extend $tau$ to get a tracial state on $M(A)$?










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  • $begingroup$
    I haven't checked the details but I will proceed as follows: First $M(A)$ has a coarser topology that the norm topology, that is the strong topology given by seeing $M(A)$ as the space of adjointable operators over $A$ as a Hilbert $C^*$-module. Then, extend $tau$ by strong continuity and use the fact that the tracial identity is preserved.
    $endgroup$
    – Adrián González-Pérez
    Jul 23 '18 at 11:05
















2












$begingroup$


If $A$ is a $C^*$ algebra, I know the fact: If $tau$ is a state on $A$, we can extend it to a state on $M(A)$, by continuity of $tau$ and where $M(A)$ is the multiplier algebra of $A$.



If $tau$ is a tracial state on $A$, can we extend $tau$ to get a tracial state on $M(A)$?










share|cite|improve this question











$endgroup$












  • $begingroup$
    I haven't checked the details but I will proceed as follows: First $M(A)$ has a coarser topology that the norm topology, that is the strong topology given by seeing $M(A)$ as the space of adjointable operators over $A$ as a Hilbert $C^*$-module. Then, extend $tau$ by strong continuity and use the fact that the tracial identity is preserved.
    $endgroup$
    – Adrián González-Pérez
    Jul 23 '18 at 11:05














2












2








2





$begingroup$


If $A$ is a $C^*$ algebra, I know the fact: If $tau$ is a state on $A$, we can extend it to a state on $M(A)$, by continuity of $tau$ and where $M(A)$ is the multiplier algebra of $A$.



If $tau$ is a tracial state on $A$, can we extend $tau$ to get a tracial state on $M(A)$?










share|cite|improve this question











$endgroup$




If $A$ is a $C^*$ algebra, I know the fact: If $tau$ is a state on $A$, we can extend it to a state on $M(A)$, by continuity of $tau$ and where $M(A)$ is the multiplier algebra of $A$.



If $tau$ is a tracial state on $A$, can we extend $tau$ to get a tracial state on $M(A)$?







operator-theory operator-algebras c-star-algebras






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edited Jan 12 at 12:06







user42761

















asked Jul 22 '18 at 2:54









mathrookiemathrookie

914512




914512












  • $begingroup$
    I haven't checked the details but I will proceed as follows: First $M(A)$ has a coarser topology that the norm topology, that is the strong topology given by seeing $M(A)$ as the space of adjointable operators over $A$ as a Hilbert $C^*$-module. Then, extend $tau$ by strong continuity and use the fact that the tracial identity is preserved.
    $endgroup$
    – Adrián González-Pérez
    Jul 23 '18 at 11:05


















  • $begingroup$
    I haven't checked the details but I will proceed as follows: First $M(A)$ has a coarser topology that the norm topology, that is the strong topology given by seeing $M(A)$ as the space of adjointable operators over $A$ as a Hilbert $C^*$-module. Then, extend $tau$ by strong continuity and use the fact that the tracial identity is preserved.
    $endgroup$
    – Adrián González-Pérez
    Jul 23 '18 at 11:05
















$begingroup$
I haven't checked the details but I will proceed as follows: First $M(A)$ has a coarser topology that the norm topology, that is the strong topology given by seeing $M(A)$ as the space of adjointable operators over $A$ as a Hilbert $C^*$-module. Then, extend $tau$ by strong continuity and use the fact that the tracial identity is preserved.
$endgroup$
– Adrián González-Pérez
Jul 23 '18 at 11:05




$begingroup$
I haven't checked the details but I will proceed as follows: First $M(A)$ has a coarser topology that the norm topology, that is the strong topology given by seeing $M(A)$ as the space of adjointable operators over $A$ as a Hilbert $C^*$-module. Then, extend $tau$ by strong continuity and use the fact that the tracial identity is preserved.
$endgroup$
– Adrián González-Pérez
Jul 23 '18 at 11:05










1 Answer
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$begingroup$

Let $tau$ be a state on $M(A)$ such that its restriction to $A$ satisfies $tau(ab) = tau(ba)$ for all $a,b in A$. For $x,y in M(A)$ we then have
$$
tau(xy) = s-limtau((xe_lambda)(e_lambda y)) = s-lim tau(e_lambda yx e_lambda) = tau(yx),
$$

where $(e_lambda)_lambda$ is an approximate unit for $A$. Note that $e_lambda to 1$ strictly and hence $e_lambda x to x$ strictly. Since $tau$ is strictly continuous, the result follows.



Note that in the case $A$ is separable, $A$ is a hereditary C*-subalgebra of $M(A)$, namely $A = overline{hM(A)h}$, where $h in A$ is strictly positive. In that case every state on $A$ has a unique extension to $M(A)$ given by
$$
tau(x) = lim tau(e_lambda x e_lambda) qquad (x in M(A)).
$$



Edit: For the argumet to work one needs that $tau$ is strictly continuous on $M(A)$. I will check if this is automatic and then update the answer.






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    $begingroup$

    Let $tau$ be a state on $M(A)$ such that its restriction to $A$ satisfies $tau(ab) = tau(ba)$ for all $a,b in A$. For $x,y in M(A)$ we then have
    $$
    tau(xy) = s-limtau((xe_lambda)(e_lambda y)) = s-lim tau(e_lambda yx e_lambda) = tau(yx),
    $$

    where $(e_lambda)_lambda$ is an approximate unit for $A$. Note that $e_lambda to 1$ strictly and hence $e_lambda x to x$ strictly. Since $tau$ is strictly continuous, the result follows.



    Note that in the case $A$ is separable, $A$ is a hereditary C*-subalgebra of $M(A)$, namely $A = overline{hM(A)h}$, where $h in A$ is strictly positive. In that case every state on $A$ has a unique extension to $M(A)$ given by
    $$
    tau(x) = lim tau(e_lambda x e_lambda) qquad (x in M(A)).
    $$



    Edit: For the argumet to work one needs that $tau$ is strictly continuous on $M(A)$. I will check if this is automatic and then update the answer.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Let $tau$ be a state on $M(A)$ such that its restriction to $A$ satisfies $tau(ab) = tau(ba)$ for all $a,b in A$. For $x,y in M(A)$ we then have
      $$
      tau(xy) = s-limtau((xe_lambda)(e_lambda y)) = s-lim tau(e_lambda yx e_lambda) = tau(yx),
      $$

      where $(e_lambda)_lambda$ is an approximate unit for $A$. Note that $e_lambda to 1$ strictly and hence $e_lambda x to x$ strictly. Since $tau$ is strictly continuous, the result follows.



      Note that in the case $A$ is separable, $A$ is a hereditary C*-subalgebra of $M(A)$, namely $A = overline{hM(A)h}$, where $h in A$ is strictly positive. In that case every state on $A$ has a unique extension to $M(A)$ given by
      $$
      tau(x) = lim tau(e_lambda x e_lambda) qquad (x in M(A)).
      $$



      Edit: For the argumet to work one needs that $tau$ is strictly continuous on $M(A)$. I will check if this is automatic and then update the answer.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $tau$ be a state on $M(A)$ such that its restriction to $A$ satisfies $tau(ab) = tau(ba)$ for all $a,b in A$. For $x,y in M(A)$ we then have
        $$
        tau(xy) = s-limtau((xe_lambda)(e_lambda y)) = s-lim tau(e_lambda yx e_lambda) = tau(yx),
        $$

        where $(e_lambda)_lambda$ is an approximate unit for $A$. Note that $e_lambda to 1$ strictly and hence $e_lambda x to x$ strictly. Since $tau$ is strictly continuous, the result follows.



        Note that in the case $A$ is separable, $A$ is a hereditary C*-subalgebra of $M(A)$, namely $A = overline{hM(A)h}$, where $h in A$ is strictly positive. In that case every state on $A$ has a unique extension to $M(A)$ given by
        $$
        tau(x) = lim tau(e_lambda x e_lambda) qquad (x in M(A)).
        $$



        Edit: For the argumet to work one needs that $tau$ is strictly continuous on $M(A)$. I will check if this is automatic and then update the answer.






        share|cite|improve this answer











        $endgroup$



        Let $tau$ be a state on $M(A)$ such that its restriction to $A$ satisfies $tau(ab) = tau(ba)$ for all $a,b in A$. For $x,y in M(A)$ we then have
        $$
        tau(xy) = s-limtau((xe_lambda)(e_lambda y)) = s-lim tau(e_lambda yx e_lambda) = tau(yx),
        $$

        where $(e_lambda)_lambda$ is an approximate unit for $A$. Note that $e_lambda to 1$ strictly and hence $e_lambda x to x$ strictly. Since $tau$ is strictly continuous, the result follows.



        Note that in the case $A$ is separable, $A$ is a hereditary C*-subalgebra of $M(A)$, namely $A = overline{hM(A)h}$, where $h in A$ is strictly positive. In that case every state on $A$ has a unique extension to $M(A)$ given by
        $$
        tau(x) = lim tau(e_lambda x e_lambda) qquad (x in M(A)).
        $$



        Edit: For the argumet to work one needs that $tau$ is strictly continuous on $M(A)$. I will check if this is automatic and then update the answer.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 12 at 13:00

























        answered Jul 24 '18 at 21:59







        user42761





































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