Countable decomposable von Neumann algebra

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Does countable decomposable von Neumann algebra will necessary imply that hilbert space has to be separable where the von Neumann algebra acts, if not give a counterexample.




operator-algebras von-neumann-algebras




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    Does countable decomposable von Neumann algebra will necessary imply that hilbert space has to be separable where the von Neumann algebra acts, if not give a counterexample.




    operator-algebras von-neumann-algebras




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      Does countable decomposable von Neumann algebra will necessary imply that hilbert space has to be separable where the von Neumann algebra acts, if not give a counterexample.




      operator-algebras von-neumann-algebras




      share|cite|improve this question













      Does countable decomposable von Neumann algebra will necessary imply that hilbert space has to be separable where the von Neumann algebra acts, if not give a counterexample.





      operator-algebras von-neumann-algebras





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      edited Jul 16 at 11:32
























      asked Jul 16 at 5:09









      mathlover

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          Counterexample: $mathbb C,Isubset B(ell^2(mathbb R))$. It is a one-dimensional subalgebra, so not only it is countably decomposable, it is finitely decomposable, even "one-decomposable".



          This example is non-degenerate.






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            Counterexample: $mathbb C,Isubset B(ell^2(mathbb R))$. It is a one-dimensional subalgebra, so not only it is countably decomposable, it is finitely decomposable, even "one-decomposable".



            This example is non-degenerate.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              Counterexample: $mathbb C,Isubset B(ell^2(mathbb R))$. It is a one-dimensional subalgebra, so not only it is countably decomposable, it is finitely decomposable, even "one-decomposable".



              This example is non-degenerate.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Counterexample: $mathbb C,Isubset B(ell^2(mathbb R))$. It is a one-dimensional subalgebra, so not only it is countably decomposable, it is finitely decomposable, even "one-decomposable".



                This example is non-degenerate.






                share|cite|improve this answer













                Counterexample: $mathbb C,Isubset B(ell^2(mathbb R))$. It is a one-dimensional subalgebra, so not only it is countably decomposable, it is finitely decomposable, even "one-decomposable".



                This example is non-degenerate.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 17 at 4:57









                Martin Argerami

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