# Von Neumann algebra associated to the infinite Cuntz algebra

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 $III_1$?

• For the infinite Cuntz algebra, every truncated sum of range projections has to be bounded by 1. So there IS a condition about the sum. Jun 13 at 17:13
• @AksharaPrasad Any projection in a $C^*$-algebra is bounded by 1, right? (since 1-p for a projection p is again a projection). Jun 14 at 8:20

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 $H^{\otimes 0}$).

• Snap! Nice proof. Aug 30, 2012 at 12:06
• Do you know about any states on $\mathcal{O}_{\infty}$ which would yield a type $III$-von Neumann algebra? Aug 30, 2012 at 13:45
• I do not know, but I am not very familiar with $\mathcal O_\infty$ (and with $C^*$-algebras in general). Do you have an argument why such a state should exist? Aug 30, 2012 at 18:33
• @Jon: The state constructed in Dima's paper is just the vector state $\langle \cdot \Omega,\Omega\rangle$ restricted to the $C^*$-algebra generated by the operators $s_v + s_v^*$ for $v$ is some real subspace of $H$ (see def 2.3 and remark 2.6 in the paper). I do not believe this is so relevant if one is looking for a state on $\mathcal O_\infty$. Aug 30, 2012 at 19:22
• @Mikael: Not really an argument. But there are states on $\mathcal{O}_n$ for finite $n$ which yield type $III_{1/n}$-factors. Unfortunately, those are built upon the fact, that $\mathcal{O}_n$ contains a core UHF-algebra and a conditional expectation (via the gauge action) onto it. For $\mathcal{O}_{\infty}$ you only get an AF-algebra. Aug 31, 2012 at 7:01

I'm reasonably convinced the algebra generates all of $B(\mathcal{F}(H))$. Note that $1-\sum_{i=1}^\infty s_is_i^*$ converges strongly to the projection $P\_0$ onto $\mathbb{C}$ and similarly the sum $1-\sum_{|\alpha|=n}s_\alpha s_\alpha^*$ over multi-indices converges to the projection $P_n$ onto vectors with highest tensor power $n-1$. Pick vectors $u\in H^{\otimes n}$, $v\in H^{\otimes m}$, then the operators $s_u$, $s_v$ (the obvious generalisations of the $s_i$) are in your algebra and so is the rank one operator $s_vs_u^*P_{n+1}=|v><\ u|$. Finally note you may approximate any rank one in $B(\mathcal{F}(H))$ by these.

If you take the Fock space $$\mathcal F_P$$ associated to a polarization of $$H$$, as in [W, page 480, line 7], and consider the CAR algebra (=algebra generated by creation and annihilation operators) generated by a subspace $$K\subset H$$, as in [W, page 497, line 29], then the von Neumann algebra this generates is (hyperfinite) of type $$III_1$$ [W, Corollary on the bottom of page 500].

Reference:
[W] Wassermann, Operator algebras and conformal field theory

• This is one of the reasons, I was interested in the question. The Cuntz algebra $\mathcal{O}_{\infty}$ has some properties in common with the ($C^*$-algebraic) CAR-algebra - for example both are strongly self-absorbing. But it is purely infinite and therefore contains no finite projection, which in my head "moves it closer to" type $III_1$-factors. Jul 29, 2013 at 8:47
• I'm confused by your remark: didn't you say that $\mathcal O_\infty$ is isomorphic to the $C^*$-algebraic CAR algebra? Now you seem to indicate that they are only similar? Jul 29, 2013 at 11:04
• No, that is not what I said. The CAR-algebra is isomorphic to the UHF-algebra $M_{2^{\infty}}$ (see Example 1.2.6 in Rordam's "Classification of Nuclear, Simple $C^*$-algebras"). I might have said that the methods in our paper apply to the CAR-algebra as well, which is true since both are strongly self-absorbing. Jul 29, 2013 at 12:14
• What might have caused the confusion is that the descriptions in terms of faithful representations on Fock space really look very similar, but note that for $\mathcal{O}_{\infty}$ I need to take the full tensor algebra and not just the antisymmetric part. If I only take the latter and take creation and annihilation ops then I get CAR. Jul 29, 2013 at 12:19
• I see. Thanks for the clarification. Jul 29, 2013 at 14:08

You asked two questions. First about $\mathcal{O}_\infty$ (first paragraph). And second a special realization of $\mathcal{O}_\infty$ (second paragraph, Fock space).

Your answer may depend on the representation, and the answers given by Mikael de la Salle and Ollie are related to the Fock space representation.

But if $\mathcal{O}_\infty$ is embedded in a Hilbert space in such a way that the strong operator sum $1-\sum_{n \ge 1} s_n s_n^*$ projects onto an infinite dimensional Hilbert space $H$ (not a one dimensional), then I can hardly imagine how you could realize all operators in $B(H)$ in the strong operator closure of $\mathcal{O}_\infty$.

To see this, notice that every element in the algebra generated by $s_i$ has the form: $$\sum_{\alpha,\beta \in \mbox{words}} \lambda_{\alpha,\beta} s_\alpha s_\beta^*$$ for scalars $\lambda$s. Almost all of these operators $s_\alpha s_\beta^*$ project vectors onto $H^\bot$, move them to $H^\bot \oplus H$ and move them back to $H^\bot$.

The only exception being $1=s_\emptyset s_\emptyset^*$, which can move vectors from $H$ to $H$.

Conclusion: $\overline{\mathcal{O}_\infty} \neq B(H \oplus H^\bot)$ in this case.