Does every integer map generate a von Neumann algebra of type I?

Question

Consider a map $m: \mathbb{N} \to \mathbb{N}$ (we call it an integer map). Let $E_r$ be the set $m^{-1}(\{r\})$.

Let $H$ be the Hilbert space $\ell^2(\mathbb{N})$ and consider the densely defined operator associated to $m$: $$M : c_{00}(\mathbb{N}) \subset H \to H, \ Me_r = e_{m(r)}, \ \forall r \in \mathbb{N}.$$ It is closable iff its adjoint $M^{\star}$ is densely defined, if (and only if?) the set $E_r$ is finite, $\forall r \in \mathbb{N}$; because then $c_{00}(\mathbb{N}) \subset D(M^{\star})$ as $M^{\star}$ is given by $$ M^{\star}e_r = \sum_{s \in E_r} e_s. $$ Then, let $\overline{M} = M^{\star \star}$ be the closure of $M$. Let $\mathcal{M}$ be the smallest von Neumann algebra that $\overline{M}$ is affiliated with. We can call it $vN(m)$, the von Neumann algebra generated by the integer map $m$.

Question: Is $\mathcal{M}$ a von Neumann algebra of type ${\rm I}$?
If no: What is a counter-example? Is there one with $M$ a bounded operator?

The operator $M$ is bounded iff $\exists k \forall r $, $|E_r|<k$ (then $\Vert M \Vert^2<k$ and $\mathcal{M} = W^{\star}(\overline{M})$).

The von Neumann algebra $\mathcal{M}$ is abelian iff $\overline{M}$ is normal, iff $m$ is bijective (see the proof below), iff $\overline{M}$ is unitary. The map $m$ is a proper injection iff the operator $\overline{M}$ is a proper isometry, only if ${\rm C}^{\star}(\overline{M})$ is the Toeplitz algebra (by Coburn's theorem) and $\mathcal{M}$ of type ${\rm I}$.

So we are reduced to consider non-injective map. The Euler's totient function $\varphi$ is neither injective nor surjective, its associated operator is densely defined and closable. What is $vN(\varphi)$?

Related question: Is $\mathcal{M}$ a hyperfinite von Neumann algebra?

---

Lemma 1: The operator $\overline{M}$ is normal iff $m$ is bijective.
Proof: Observe that $$(MM^{\star}-M^{\star}M)e_r = |E_r|e_r-\sum_{s \in E_{m(r)}} e_s$$ so $\overline{M}$ is a normal iff $\forall r \in \mathbb{N}$, $E_{m(r)} = \{r\}$ and $|E_r| = 1$, iff $m$ is bijective. $\square$

Corollary: The von Neumann algebra $\mathcal{M}$ is abelian iff $m$ is bijective.
Proof: Immediate from Lemma 1 and Kadison-Ringrose 5.6.18 (for the unbounded case). $\square$

Answer 1

I think I have an example: Precisely, I will construct an integer function $m$ such that $M$ is bounded and the algebra $\mathcal{M}$ contains a corner which is the von Neuman algebra completion of a Cuntz algebra $\mathcal{O}_2$, i.e. the von Neuman algebra generated by two element $S_0$ and $S_1$ such that $S_0^* S_0=S_1^* S_1= 1$ and $S_0 S_0^* + S_1 S_1^* =1$.

First, one can replace $\mathbb{N}$ by any infinite countable set. I will construct my function $m$ as a function $\mathbb{N} \coprod \mathbb{N} \rightarrow \mathbb{N} \coprod \mathbb{N}$. You decide that the first component corresponds to odd number and the second to even number if you want a function on $\mathbb{N}$ but I will not do that.

To distinguishes between the two component I will call them respectively $K$ and $I$. the function $m$ is defined as follow (it takes values in $I$ only):

• for $n \in K$, $m(n)=2n \in I$.
• for $n \in I$, $m(n)= \lfloor n/2 \rfloor \in I$.

Elements of $K$ have no pre-image by $m$, $i \in I$ always have two pre-image in $I$: $2i$ and $2i+1$, and the have a third pre-image in $K$ if and only if $i$ is even.

In particular: $MM^*$ is the operator that send all elements of $K$ to zero, multiply by two the odd elements of $I$ and by three the even elements of $I$. We define $P_0$ and $P_1$ to be respectively the orthogonal projection on the set of even and odd element of $I$ and $P=P_0+P_1$ be the projection on all elements of $I$, they can all be writen as polynomial in $MM^*$, hence they belong to $\mathcal{M}$.

The corner I'm going to look at is $P \mathcal{M} P$.

Let $S_0$ and $S_1$ be the endomorphism of $l^2(I)$ defined by $S_0(e_{2n}) = e_n, S_{0}(e_{2n+1})=0, S_1(e_{2n})=0, S_1(e_{2n+1})=e_n$.

$S_0$ and $S_1$ satisfies the Cuntz relation above, and I claim that $P \mathcal{M} P$ is exactly the von Neuman algebra generated by $S_0$ and $S_1$

First, $S_0=M P_0$ and $S_1 = M P_1$ hence they indeed belong to $\mathcal{M}$.

Conversely, in the decomposition $l^2(K \coprod I) = l^2(K) \oplus l^2(I)$, $M$ corresponds to the following $2 \times 2$ matrix: the first column is $(0,0)$ the second column is $(S_0^*,S_0+S_1)$ hence all elements of $\mathcal{M}$ have a matrix decomposition with coefficient in the algebras generated by $S_0$ and $S_1$, in particular the fact their bottom right corner is always in this algebra concludes the proof.