Equality of tracial states on dense $C^*$-subalgebra implies equality on generated von Neumann algebra?












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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 $?










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    2












    $begingroup$


    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 $?










    share|cite|improve this question









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      2












      2








      2





      $begingroup$


      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 $?










      share|cite|improve this question









      $endgroup$




      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






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      asked Dec 7 '18 at 9:51









      worldreporter14worldreporter14

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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 $taune rho$.






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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 $taune rho$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              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 $taune rho$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                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 $taune rho$.






                share|cite|improve this answer









                $endgroup$



                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 $taune rho$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 7 '18 at 17:24









                Martin ArgeramiMartin Argerami

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                126k1182181






























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