The Cuntz algebra $\mathcal{O}\_{\infty}$ is the universal $C^\*$-algebra generated by countably infinitely many isometries $s\_i$ satisfying the relations $s\_i^\*s\_j = \delta\_{ij}$ (there is no condition about the sum $\sum\_i s\_is\_i^\*$ in this case, since the sum would be infinite). 


There is another way to describe this $C^\*$-algebra as a subalgebra of the bounded operators on a Hilbert space as follows: Let $H$ be a separable Hilbert space of infinite dimension. Then we define the Fock space 


$$
\mathcal{F}(H) = \bigoplus\_{n \in \mathbb{N}\_0} H^{\otimes n}\ .
$$


with $H^{\otimes 0} \cong \mathbb{C}$. Any element $v \in H$ defines a creation operator $s\_v(\xi) = v \otimes \xi$ and an annihilation operator $s\_v^\*(w \otimes \xi) = \langle v, w \rangle \xi$, which is zero on $\mathbb{C} = H^{\otimes 0} \subset \mathcal{F}(H)$. The $C^\*$-algebra generated by $s\_v$ and $s\_v^\*$ (i.e. the norm closure) is again $\mathcal{O}\_{\infty}$. After choosing an orthonormal basis $e\_i \in H$, we can identify $s\_i = s\_{e\_i}$. 


My question is:



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> Is there anything known about the von Neumann algebra generated by $\mathcal{O}\_{\infty}$ in this way, i.e. the weak closure or double commutant of $\mathcal{O}\_{\infty}$ in $B(\mathcal{F}(H))$? What type is it ($III\_1$ or just $I\_{\infty}$)? 
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> 
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EDIT: Are there interesting/canonical states on $\mathcal{O}\_{\infty}$ such that the von Neumann algebra associated to the GNS-construction is type $III\_1$?



Von Neumann algebra associated to the infinite Cuntz algebra
The von Neumann algebra $M$ generated by $\mathcal O\_\infty$ is all of $B(\mathcal F(H))$. 


Indeed, if $a$ belongs to its commutant, let me prove that $a$ is a multiple of the identity. First since for all $v \in H$, $s\_v^\* (a \Omega)= a (s\_v^\* \Omega)=0$, we have that $a \Omega=\lambda \Omega$ for some $\lambda \in \mathbb C$. Then for every $\xi \in \oplus\_n H^{\otimes n}$ (finite sum) pick $x \in \mathcal O\_\infty$ such that $x \Omega=\xi$. Then $a(\xi) = x (a\Omega)=\lambda \xi$. This proves $a=\lambda$.


(Let me add that, according to the standard notation, $\Omega$ here denotes some fixed unit vector in $H^{\otimes 0}$).


