Von Neumann Algebra Associated to the Infinite Cuntz Algebra – Operator Algebras

c-star-algebrasoa.operator-algebrasvon-neumann-algebras

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:

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 $I_{\infty}$)?

EDIT: Are there interesting/canonical states on $\mathcal{O}_{\infty}$ such that the von Neumann algebra associated to the GNS-construction is type $III_1$?

Best Answer

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 $H^{\otimes 0}$).

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