Extension of tracial state to the multiplier algebra
$begingroup$
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 $M(A)$?
operator-theory operator-algebras c-star-algebras
asked Jul 22 '18 at 2:54
mathrookiemathrookie
914512
$endgroup$
add a comment |
$begingroup$
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 $M(A)$?
operator-theory operator-algebras c-star-algebras
asked Jul 22 '18 at 2:54
mathrookiemathrookie
914512
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $M(A)$?
operator-theory operator-algebras c-star-algebras
asked Jul 22 '18 at 2:54
mathrookiemathrookie
914512
$endgroup$
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
operator-theory operator-algebras c-star-algebras
asked Jul 22 '18 at 2:54
mathrookiemathrookie
914512
asked Jul 22 '18 at 2:54
mathrookiemathrookie
914512
edited Jan 12 at 12:06
user42761
asked Jul 22 '18 at 2:54
mathrookiemathrookie
914512
asked Jul 22 '18 at 2:54
mathrookiemathrookie
914512
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
add a comment |
$begingroup$
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
add a comment |
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.
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Jan 12 at 13:00
answered Jul 24 '18 at 21:59
user42761
add a comment |
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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