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Let $H$ and $K$ be two Hilbert spaces. Let us consider two pairs of isometries $(x_1,x_2)$ in $B(H)$ and $(y_1,y_2)$ in $B(K)$ satisfying in $$x_1x_1^*+x_2x_2^*=1_H~~~,~~~y_1y_1^*+y_2y_2^*=1_K$$

Cuntz proved that the C*-algebra generated by $x_1,x_2$ is the same as the C*-algebra generated by $y_1,y_2$.

Q. Is the von Neumann algebra generated by $x_1,x_2$ the same as the von Neumann algebra generated by $y_1,y_2$?

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    $\begingroup$ I would expect that the enveloping von Neumann algebras of isomorphic C*-algebras are isomorphic. $\endgroup$
    – user42761
    Jan 23, 2017 at 8:41
  • $\begingroup$ related: mathoverflow.net/questions/105922/… $\endgroup$
    – Jonas Meyer
    Jan 23, 2017 at 20:52

1 Answer 1

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No, they don't have to be isomorphic. Let $\mathcal O_2\subset B(H)$ be an irreducible representation. Then $$ \mathcal O_2''=B(H). $$ Now let $A\subset B(H)$ be a unital, separable, nuclear C$^*$-algebra such that $A''$ is a II$_1$-factor $M$. For instance, we can take $A=UHF(2^\infty)$. Then Kirchberg-Phillips implies that $\mathcal O_2\simeq A\otimes \mathcal O_2$. This means that via this isomorphism we can find isometries $y_1,y_2\in A\otimes\mathcal O_2$ such that $y_1y_1^*+y_2y_2^*=I$, and $C^*(y_1,y_2)=A\otimes\mathcal O_2$. If we consider the irreducible representation $\mathcal O_2\subset B(H)$ as above, we have $$ \{y_1,y_2\}''=C^*(y_1,y_2)''=(A\otimes \mathcal O_2)''= A''\bar\otimes\mathcal O_2''=M\otimes B(H). $$

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  • $\begingroup$ It was really nice. $\endgroup$
    – ABB
    Jan 24, 2017 at 7:50
  • $\begingroup$ Is there an example of type ${\rm III}$? I came here after reading mathoverflow.net/a/289569/34538 $\endgroup$
    – Sebastien Palcoux
    Dec 31, 2017 at 20:31
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    $\begingroup$ Yes. The example above depends on $A$ having a type II$_1$-representation. But you can instead take $A$ with a type III-representation. You can still use $UHF(2^\infty)$ with an appropriate state (think of $A$ as $\bigotimes_nM_{2}(\mathbb C)$ and use the state to get a type III Powers factor). $\endgroup$
    – Martin Argerami
    Dec 31, 2017 at 22:10
  • $\begingroup$ @MartinArgerami: What kind vn-algebra(s) do you suggest if it is replaced then the question faces positive answer. $\endgroup$
    – ABB
    May 11, 2018 at 20:17
  • $\begingroup$ I'm not sure if there is such thing. Basically you are looking for a C$^*$-algebra such that all of its infinite representations give the same von Neumann algebra; I'll be surprised if such a thing exists (although, of course, I don't trust my intuition that much). $\endgroup$
    – Martin Argerami
    May 11, 2018 at 22:48

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