On proof of weak operator closed ness of kernel of the representation












1














Let $pi:L^{infty}(X,mu)mapsto B(mathcal{H})$ be a representation of an abelian von Neumann algebra. Where $mu$ is a probability measure and $mathcal{H}$ is a seperable Hilbert space. Prove that $text{ ker }pi$ is weak operator closed.






operator-algebras von-neumann-algebras







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  • What would "weak operator topology" mean on $L^infty$?
    – Martin Argerami
    Nov 25 '18 at 20:06










  • Means thinking $L^{infty}(X,mu)$ sitting inside $B(L^2(X,mu))$
    – mathlover
    Nov 27 '18 at 12:35










  • Ok. And what would "representation" mean here? In terms of continuity?
    – Martin Argerami
    Nov 27 '18 at 15:20










  • Thinking the algebra as multiplication operator in $L^2(X,mu)$
    – mathlover
    Nov 28 '18 at 5:05
















1














Let $pi:L^{infty}(X,mu)mapsto B(mathcal{H})$ be a representation of an abelian von Neumann algebra. Where $mu$ is a probability measure and $mathcal{H}$ is a seperable Hilbert space. Prove that $text{ ker }pi$ is weak operator closed.






operator-algebras von-neumann-algebras







share|cite|improve this question






















  • What would "weak operator topology" mean on $L^infty$?
    – Martin Argerami
    Nov 25 '18 at 20:06










  • Means thinking $L^{infty}(X,mu)$ sitting inside $B(L^2(X,mu))$
    – mathlover
    Nov 27 '18 at 12:35










  • Ok. And what would "representation" mean here? In terms of continuity?
    – Martin Argerami
    Nov 27 '18 at 15:20










  • Thinking the algebra as multiplication operator in $L^2(X,mu)$
    – mathlover
    Nov 28 '18 at 5:05














1












1








1







Let $pi:L^{infty}(X,mu)mapsto B(mathcal{H})$ be a representation of an abelian von Neumann algebra. Where $mu$ is a probability measure and $mathcal{H}$ is a seperable Hilbert space. Prove that $text{ ker }pi$ is weak operator closed.






operator-algebras von-neumann-algebras







share|cite|improve this question













Let $pi:L^{infty}(X,mu)mapsto B(mathcal{H})$ be a representation of an abelian von Neumann algebra. Where $mu$ is a probability measure and $mathcal{H}$ is a seperable Hilbert space. Prove that $text{ ker }pi$ is weak operator closed.






operator-algebras von-neumann-algebras




operator-algebras von-neumann-algebras






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share|cite|improve this question











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share|cite|improve this question










asked Nov 25 '18 at 11:31









mathlover

15119




15119












  • What would "weak operator topology" mean on $L^infty$?
    – Martin Argerami
    Nov 25 '18 at 20:06










  • Means thinking $L^{infty}(X,mu)$ sitting inside $B(L^2(X,mu))$
    – mathlover
    Nov 27 '18 at 12:35










  • Ok. And what would "representation" mean here? In terms of continuity?
    – Martin Argerami
    Nov 27 '18 at 15:20










  • Thinking the algebra as multiplication operator in $L^2(X,mu)$
    – mathlover
    Nov 28 '18 at 5:05


















  • What would "weak operator topology" mean on $L^infty$?
    – Martin Argerami
    Nov 25 '18 at 20:06










  • Means thinking $L^{infty}(X,mu)$ sitting inside $B(L^2(X,mu))$
    – mathlover
    Nov 27 '18 at 12:35










  • Ok. And what would "representation" mean here? In terms of continuity?
    – Martin Argerami
    Nov 27 '18 at 15:20










  • Thinking the algebra as multiplication operator in $L^2(X,mu)$
    – mathlover
    Nov 28 '18 at 5:05
















What would "weak operator topology" mean on $L^infty$?
– Martin Argerami
Nov 25 '18 at 20:06




What would "weak operator topology" mean on $L^infty$?
– Martin Argerami
Nov 25 '18 at 20:06












Means thinking $L^{infty}(X,mu)$ sitting inside $B(L^2(X,mu))$
– mathlover
Nov 27 '18 at 12:35




Means thinking $L^{infty}(X,mu)$ sitting inside $B(L^2(X,mu))$
– mathlover
Nov 27 '18 at 12:35












Ok. And what would "representation" mean here? In terms of continuity?
– Martin Argerami
Nov 27 '18 at 15:20




Ok. And what would "representation" mean here? In terms of continuity?
– Martin Argerami
Nov 27 '18 at 15:20












Thinking the algebra as multiplication operator in $L^2(X,mu)$
– mathlover
Nov 28 '18 at 5:05




Thinking the algebra as multiplication operator in $L^2(X,mu)$
– mathlover
Nov 28 '18 at 5:05










1 Answer
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It's not clear what you are asking. There are two possibilities:




  • You are requiring $pi$ to be wot continuous. In that case, $kerpi=pi^{-1}({0})$ is closed.


  • You are not requiring $pi$ to be wot continuous. In that case the assertion is not true. Take $X=mathbb N$ with the counting measure. Fix a free ultrafilter $omega$, and let $pi:ell^infty(mathbb N)tomathbb C$ be $pi(x)=lim_{ntoomega}x_n$. Then $pi$ is a representation, and $kerpisupset c_0$. But $c_0$ is wot dense in $ell^infty(mathbb N)$.







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    1 Answer
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    1 Answer
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    It's not clear what you are asking. There are two possibilities:




    • You are requiring $pi$ to be wot continuous. In that case, $kerpi=pi^{-1}({0})$ is closed.


    • You are not requiring $pi$ to be wot continuous. In that case the assertion is not true. Take $X=mathbb N$ with the counting measure. Fix a free ultrafilter $omega$, and let $pi:ell^infty(mathbb N)tomathbb C$ be $pi(x)=lim_{ntoomega}x_n$. Then $pi$ is a representation, and $kerpisupset c_0$. But $c_0$ is wot dense in $ell^infty(mathbb N)$.







    share|cite|improve this answer




























      1














      It's not clear what you are asking. There are two possibilities:




      • You are requiring $pi$ to be wot continuous. In that case, $kerpi=pi^{-1}({0})$ is closed.


      • You are not requiring $pi$ to be wot continuous. In that case the assertion is not true. Take $X=mathbb N$ with the counting measure. Fix a free ultrafilter $omega$, and let $pi:ell^infty(mathbb N)tomathbb C$ be $pi(x)=lim_{ntoomega}x_n$. Then $pi$ is a representation, and $kerpisupset c_0$. But $c_0$ is wot dense in $ell^infty(mathbb N)$.







      share|cite|improve this answer


























        1












        1








        1






        It's not clear what you are asking. There are two possibilities:




        • You are requiring $pi$ to be wot continuous. In that case, $kerpi=pi^{-1}({0})$ is closed.


        • You are not requiring $pi$ to be wot continuous. In that case the assertion is not true. Take $X=mathbb N$ with the counting measure. Fix a free ultrafilter $omega$, and let $pi:ell^infty(mathbb N)tomathbb C$ be $pi(x)=lim_{ntoomega}x_n$. Then $pi$ is a representation, and $kerpisupset c_0$. But $c_0$ is wot dense in $ell^infty(mathbb N)$.







        share|cite|improve this answer














        It's not clear what you are asking. There are two possibilities:




        • You are requiring $pi$ to be wot continuous. In that case, $kerpi=pi^{-1}({0})$ is closed.


        • You are not requiring $pi$ to be wot continuous. In that case the assertion is not true. Take $X=mathbb N$ with the counting measure. Fix a free ultrafilter $omega$, and let $pi:ell^infty(mathbb N)tomathbb C$ be $pi(x)=lim_{ntoomega}x_n$. Then $pi$ is a representation, and $kerpisupset c_0$. But $c_0$ is wot dense in $ell^infty(mathbb N)$.








        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 28 '18 at 18:28

























        answered Nov 28 '18 at 17:18









        Martin Argerami

        124k1176174




        124k1176174






























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