Equality of tracial states on dense $C^*$-subalgebra implies equality on generated von Neumann algebra?
$begingroup$
Maybe this is a simple question, but I'm not sure about the following:
Let $cal M$, $cal N$ be von Neumann algebras and $Xsubseteq cal M$ a weakly dense (possibly separable) $C^*$-subalgebra. Let $rho, tau$ be tracial states on $cal N$ where $tau$ is normal and $psi: cal M rightarrow cal N$ a surjective, normal $*$-homomorphism.
The Question: Does $rho (x) = tau circ psi (x)$ for every $xin X$ already imply that $rho = tau circ psi $?
operator-algebras trace von-neumann-algebras
asked Dec 7 '18 at 9:51
worldreporter14worldreporter14
31318
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add a comment |
$begingroup$
Maybe this is a simple question, but I'm not sure about the following:
Let $cal M$, $cal N$ be von Neumann algebras and $Xsubseteq cal M$ a weakly dense (possibly separable) $C^*$-subalgebra. Let $rho, tau$ be tracial states on $cal N$ where $tau$ is normal and $psi: cal M rightarrow cal N$ a surjective, normal $*$-homomorphism.
The Question: Does $rho (x) = tau circ psi (x)$ for every $xin X$ already imply that $rho = tau circ psi $?
operator-algebras trace von-neumann-algebras
asked Dec 7 '18 at 9:51
worldreporter14worldreporter14
31318
$endgroup$
add a comment |
$begingroup$
Maybe this is a simple question, but I'm not sure about the following:
Let $cal M$, $cal N$ be von Neumann algebras and $Xsubseteq cal M$ a weakly dense (possibly separable) $C^*$-subalgebra. Let $rho, tau$ be tracial states on $cal N$ where $tau$ is normal and $psi: cal M rightarrow cal N$ a surjective, normal $*$-homomorphism.
The Question: Does $rho (x) = tau circ psi (x)$ for every $xin X$ already imply that $rho = tau circ psi $?
operator-algebras trace von-neumann-algebras
asked Dec 7 '18 at 9:51
worldreporter14worldreporter14
31318
$endgroup$
Maybe this is a simple question, but I'm not sure about the following:
Let $cal M$, $cal N$ be von Neumann algebras and $Xsubseteq cal M$ a weakly dense (possibly separable) $C^*$-subalgebra. Let $rho, tau$ be tracial states on $cal N$ where $tau$ is normal and $psi: cal M rightarrow cal N$ a surjective, normal $*$-homomorphism.
The Question: Does $rho (x) = tau circ psi (x)$ for every $xin X$ already imply that $rho = tau circ psi $?
operator-algebras trace von-neumann-algebras
operator-algebras trace von-neumann-algebras
asked Dec 7 '18 at 9:51
worldreporter14worldreporter14
31318
asked Dec 7 '18 at 9:51
worldreporter14worldreporter14
31318
asked Dec 7 '18 at 9:51
worldreporter14worldreporter14
31318
asked Dec 7 '18 at 9:51
worldreporter14worldreporter14
31318
asked Dec 7 '18 at 9:51
worldreporter14worldreporter14
31318
31318
add a comment |
add a comment |
1 Answer
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I assume you mean that $rho$ is a state on $mathcal M$.
Unless you assume that $rho$ is normal, the answer is no.
Take $mathcal M=mathcal N=ell^infty(mathbb N)$, $X=c_0$, $psi$ the identity map. Take $tau=0$, and $rho$ the limit along any ultrafilter $omega$. Then $tau$ is normal, $tau=rho$ on $X$, but $taune rho$.
answered Dec 7 '18 at 17:24
Martin ArgeramiMartin Argerami
126k1182181
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1 Answer
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1 Answer
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$begingroup$
I assume you mean that $rho$ is a state on $mathcal M$.
Unless you assume that $rho$ is normal, the answer is no.
Take $mathcal M=mathcal N=ell^infty(mathbb N)$, $X=c_0$, $psi$ the identity map. Take $tau=0$, and $rho$ the limit along any ultrafilter $omega$. Then $tau$ is normal, $tau=rho$ on $X$, but $taune rho$.
answered Dec 7 '18 at 17:24
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
I assume you mean that $rho$ is a state on $mathcal M$.
Unless you assume that $rho$ is normal, the answer is no.
Take $mathcal M=mathcal N=ell^infty(mathbb N)$, $X=c_0$, $psi$ the identity map. Take $tau=0$, and $rho$ the limit along any ultrafilter $omega$. Then $tau$ is normal, $tau=rho$ on $X$, but $taune rho$.
answered Dec 7 '18 at 17:24
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
I assume you mean that $rho$ is a state on $mathcal M$.
Unless you assume that $rho$ is normal, the answer is no.
Take $mathcal M=mathcal N=ell^infty(mathbb N)$, $X=c_0$, $psi$ the identity map. Take $tau=0$, and $rho$ the limit along any ultrafilter $omega$. Then $tau$ is normal, $tau=rho$ on $X$, but $taune rho$.
answered Dec 7 '18 at 17:24
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
I assume you mean that $rho$ is a state on $mathcal M$.
Unless you assume that $rho$ is normal, the answer is no.
Take $mathcal M=mathcal N=ell^infty(mathbb N)$, $X=c_0$, $psi$ the identity map. Take $tau=0$, and $rho$ the limit along any ultrafilter $omega$. Then $tau$ is normal, $tau=rho$ on $X$, but $taune rho$.
answered Dec 7 '18 at 17:24
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 7 '18 at 17:24
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 7 '18 at 17:24
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 7 '18 at 17:24
Martin ArgeramiMartin Argerami
126k1182181
126k1182181
add a comment |
add a comment |
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