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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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    I would expect that the enveloping von Neumann algebras of isomorphic C*-algebras are isomorphic. – 2017-01-23
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    related: http://mathoverflow.net/questions/105922/von-neumann-algebra-associated-to-the-infinite-cuntz-algebra – 2017-01-23

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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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    It was really nice. – 2017-01-24
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    Is there an example of type ${\rm III}$? I came here after reading https://mathoverflow.net/a/289569/34538 – 2017-12-31
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    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). – 2017-12-31
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    @MartinArgerami: What kind vn-algebra(s) do you suggest if it is replaced then the question faces positive answer. – 2018-05-11
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    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). – 2018-05-11