Uniqueness of faithful (!) tracial states on separable $C^*$-algebra
$begingroup$
Let $A$ be a separable $C^*$-algebra and $Ssubseteq A$ a norm-dense, countable set in $A$. Assume that there are two faithful (meaning that the corresponding GNS-construction gives a faithful representation) tracial states $tau$, $rho$ on $A$. Does this already imply that $tau = rho$?
Background and own thoughts: The GNS-construction gives us two separable Hilbert spaces $cal H_1$ $cal H_2$ which are (by the faithfulness) the closure of $S$ by the norms coming from $tau$ and $rho$. As the Hilbert spaces are separable, there exists an isometric isomorphism $U: cal H_1 rightarrow cal H_2$, so for any $x in S$ we have $rho(x^*x)=leftVert xrightVert _{rho}^{2}=leftVert UxrightVert _{tau}^{2} = tauleft(left(Uxright)^{*}left(Uxright)right)$. Now, if $U$ is implemented by a unitary (or isometry) in $A$, this implies that $rho$ and $tau$ coincide on positive elements of $S$, hence are equal. The question is: Does such a unitary/isometry exist? If not, does uniqueness hold nevertheless?
hilbert-spaces operator-algebras
asked Dec 3 '18 at 9:21
worldreporter14worldreporter14
31318
$endgroup$
add a comment |
$begingroup$
Let $A$ be a separable $C^*$-algebra and $Ssubseteq A$ a norm-dense, countable set in $A$. Assume that there are two faithful (meaning that the corresponding GNS-construction gives a faithful representation) tracial states $tau$, $rho$ on $A$. Does this already imply that $tau = rho$?
Background and own thoughts: The GNS-construction gives us two separable Hilbert spaces $cal H_1$ $cal H_2$ which are (by the faithfulness) the closure of $S$ by the norms coming from $tau$ and $rho$. As the Hilbert spaces are separable, there exists an isometric isomorphism $U: cal H_1 rightarrow cal H_2$, so for any $x in S$ we have $rho(x^*x)=leftVert xrightVert _{rho}^{2}=leftVert UxrightVert _{tau}^{2} = tauleft(left(Uxright)^{*}left(Uxright)right)$. Now, if $U$ is implemented by a unitary (or isometry) in $A$, this implies that $rho$ and $tau$ coincide on positive elements of $S$, hence are equal. The question is: Does such a unitary/isometry exist? If not, does uniqueness hold nevertheless?
hilbert-spaces operator-algebras
asked Dec 3 '18 at 9:21
worldreporter14worldreporter14
31318
$endgroup$
$begingroup$
This is false even in Abelian $C^ast$-algebras. Just take two different probability measures with full support in $mathbb R$.
$endgroup$
– Adrián González-Pérez
Dec 3 '18 at 11:46
$begingroup$
There are some simple $C^ast$-algebras that have the unique trace property , like $C^*_r ( mathbb F_2)$, but that is far from the norm.
$endgroup$
– Adrián González-Pérez
Dec 3 '18 at 11:49
add a comment |
$begingroup$
Let $A$ be a separable $C^*$-algebra and $Ssubseteq A$ a norm-dense, countable set in $A$. Assume that there are two faithful (meaning that the corresponding GNS-construction gives a faithful representation) tracial states $tau$, $rho$ on $A$. Does this already imply that $tau = rho$?
Background and own thoughts: The GNS-construction gives us two separable Hilbert spaces $cal H_1$ $cal H_2$ which are (by the faithfulness) the closure of $S$ by the norms coming from $tau$ and $rho$. As the Hilbert spaces are separable, there exists an isometric isomorphism $U: cal H_1 rightarrow cal H_2$, so for any $x in S$ we have $rho(x^*x)=leftVert xrightVert _{rho}^{2}=leftVert UxrightVert _{tau}^{2} = tauleft(left(Uxright)^{*}left(Uxright)right)$. Now, if $U$ is implemented by a unitary (or isometry) in $A$, this implies that $rho$ and $tau$ coincide on positive elements of $S$, hence are equal. The question is: Does such a unitary/isometry exist? If not, does uniqueness hold nevertheless?
hilbert-spaces operator-algebras
asked Dec 3 '18 at 9:21
worldreporter14worldreporter14
31318
$endgroup$
Let $A$ be a separable $C^*$-algebra and $Ssubseteq A$ a norm-dense, countable set in $A$. Assume that there are two faithful (meaning that the corresponding GNS-construction gives a faithful representation) tracial states $tau$, $rho$ on $A$. Does this already imply that $tau = rho$?
Background and own thoughts: The GNS-construction gives us two separable Hilbert spaces $cal H_1$ $cal H_2$ which are (by the faithfulness) the closure of $S$ by the norms coming from $tau$ and $rho$. As the Hilbert spaces are separable, there exists an isometric isomorphism $U: cal H_1 rightarrow cal H_2$, so for any $x in S$ we have $rho(x^*x)=leftVert xrightVert _{rho}^{2}=leftVert UxrightVert _{tau}^{2} = tauleft(left(Uxright)^{*}left(Uxright)right)$. Now, if $U$ is implemented by a unitary (or isometry) in $A$, this implies that $rho$ and $tau$ coincide on positive elements of $S$, hence are equal. The question is: Does such a unitary/isometry exist? If not, does uniqueness hold nevertheless?
hilbert-spaces operator-algebras
hilbert-spaces operator-algebras
asked Dec 3 '18 at 9:21
worldreporter14worldreporter14
31318
asked Dec 3 '18 at 9:21
worldreporter14worldreporter14
31318
asked Dec 3 '18 at 9:21
worldreporter14worldreporter14
31318
asked Dec 3 '18 at 9:21
worldreporter14worldreporter14
31318
asked Dec 3 '18 at 9:21
worldreporter14worldreporter14
31318
31318
$begingroup$
This is false even in Abelian $C^ast$-algebras. Just take two different probability measures with full support in $mathbb R$.
$endgroup$
– Adrián González-Pérez
Dec 3 '18 at 11:46
$begingroup$
There are some simple $C^ast$-algebras that have the unique trace property , like $C^*_r ( mathbb F_2)$, but that is far from the norm.
$endgroup$
– Adrián González-Pérez
Dec 3 '18 at 11:49
add a comment |
$begingroup$
This is false even in Abelian $C^ast$-algebras. Just take two different probability measures with full support in $mathbb R$.
$endgroup$
– Adrián González-Pérez
Dec 3 '18 at 11:46
$begingroup$
There are some simple $C^ast$-algebras that have the unique trace property , like $C^*_r ( mathbb F_2)$, but that is far from the norm.
$endgroup$
– Adrián González-Pérez
Dec 3 '18 at 11:49
$begingroup$
This is false even in Abelian $C^ast$-algebras. Just take two different probability measures with full support in $mathbb R$.
$endgroup$
– Adrián González-Pérez
Dec 3 '18 at 11:46
$begingroup$
This is false even in Abelian $C^ast$-algebras. Just take two different probability measures with full support in $mathbb R$.
$endgroup$
– Adrián González-Pérez
Dec 3 '18 at 11:46
$begingroup$
There are some simple $C^ast$-algebras that have the unique trace property , like $C^*_r ( mathbb F_2)$, but that is far from the norm.
$endgroup$
– Adrián González-Pérez
Dec 3 '18 at 11:49
$begingroup$
There are some simple $C^ast$-algebras that have the unique trace property , like $C^*_r ( mathbb F_2)$, but that is far from the norm.
$endgroup$
– Adrián González-Pérez
Dec 3 '18 at 11:49
add a comment |
1 Answer
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This fails almost always, unless you have unique trace. For instance let $A=mathbb Coplusmathbb C$. Let $tau(x,y)=(x+y)/2$, $rho(x,y)=x/3+2y/3$.
answered Dec 3 '18 at 15:53
Martin ArgeramiMartin Argerami
127k1182183
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$begingroup$
This fails almost always, unless you have unique trace. For instance let $A=mathbb Coplusmathbb C$. Let $tau(x,y)=(x+y)/2$, $rho(x,y)=x/3+2y/3$.
answered Dec 3 '18 at 15:53
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
This fails almost always, unless you have unique trace. For instance let $A=mathbb Coplusmathbb C$. Let $tau(x,y)=(x+y)/2$, $rho(x,y)=x/3+2y/3$.
answered Dec 3 '18 at 15:53
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
This fails almost always, unless you have unique trace. For instance let $A=mathbb Coplusmathbb C$. Let $tau(x,y)=(x+y)/2$, $rho(x,y)=x/3+2y/3$.
answered Dec 3 '18 at 15:53
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
This fails almost always, unless you have unique trace. For instance let $A=mathbb Coplusmathbb C$. Let $tau(x,y)=(x+y)/2$, $rho(x,y)=x/3+2y/3$.
answered Dec 3 '18 at 15:53
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 3 '18 at 15:53
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 3 '18 at 15:53
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 3 '18 at 15:53
Martin ArgeramiMartin Argerami
127k1182183
127k1182183
add a comment |
add a comment |
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$begingroup$
This is false even in Abelian $C^ast$-algebras. Just take two different probability measures with full support in $mathbb R$.
$endgroup$
– Adrián González-Pérez
Dec 3 '18 at 11:46
$begingroup$
There are some simple $C^ast$-algebras that have the unique trace property , like $C^*_r ( mathbb F_2)$, but that is far from the norm.
$endgroup$
– Adrián González-Pérez
Dec 3 '18 at 11:49