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

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). $\square$

• To complete the references, would you add the definition of $\mathbb{C}[\mathbb{N}]$? Is this the sequences with finite support? thank you – Pietro Majer Dec 30 '17 at 7:46
• @PietroMajer: yes. Is there a more usual notation for that? – Sebastien Palcoux Dec 30 '17 at 19:16
• Not that I know :) (I've also seen $\mathbb{C}^{(\mathbb{N})}$, $c_{00}$ and $c_c$) – Pietro Majer Dec 30 '17 at 20:54

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.

• For fun, the translation of your function as a $\mathbb{N} \to \mathbb{N}$ map is: $$n \mapsto \begin{cases} 2(n-1) & \text{if} \;n \;\text{odd} \\ 2 \lfloor n/4 \rfloor & \text{if} \;n \;\text{even} \end{cases}$$ It looks like the Collatz map. I don't know whether the usual Collatz map generates also a ${\rm C}^{\star}$-algebra with a Cuntz algebra corner, but it is interesting. – Sebastien Palcoux Dec 31 '17 at 20:20
• The von Neumann algebra generated by a Cuntz algebra $\mathcal{O}_2$ is possibly of type ${\rm I}$: math.stackexchange.com/q/2108815/84284 – Sebastien Palcoux Jan 1 '18 at 14:44
• @SebastienPalcoux : Hum... You're right I missed that. But I'm optimistic this is fixeable. The representation of $\mathcal{O}_2$ that appear here is rather natural (the two isometry are the map induced $n \mapsto 2n$ and $n \mapsto 2n+1$ on $l^2(\mathbb{N})$, I'm sure one can determine the type of the representation (I didn't have much time to think about it so I'm not sure how). If it is not of type I, one can tweak the construction here to obtain other representations of $\mathcal{O}_2$ or even of $\mathcal{O}_n$ which might be not of type $I$. I'll try to think about it... – Simon Henry Jan 1 '18 at 19:35
• Regarding the resemblance to the Collatz map is not completely a coincidence: My idea was to look at functions that gives rises to interesting dynamics and more importantly where the cardinality of fiber of iterate of the functions are interesting and non trivial (as the corresponding multiplication operators will be elements of the algebra, and having both the operator corresponding to the action of a function and multiplication operator is a good way to get type II things ). I thought about the Collatz map in this role, but the dynamics was just too complicated. – Simon Henry Jan 1 '18 at 19:42
• My question was also motivated by the Collatz map. In your answer, the expected isometries are $S_i^*$, $i=0,1$. The projections $S_i^*S_i$ and $S_iS_i^*$ are Murray-von Neumann equivalents (by definition) whereas the first is a proper sub-projection to the second. – Sebastien Palcoux Jan 1 '18 at 21:27