Let $H$ be a separable Hilbert space and let $M$ be a [densely defined operator](https://en.wikipedia.org/wiki/Densely_defined_operator) $\mathcal{D}(M) \subset H \to H$. It is closable iff its adjoint $M^{\star}$ is densely defined, and then its closure $\overline{M}$ is $M^{\star \star}$. Let $\mathcal{M}$ be the smallest [von Neumann algebra](https://en.wikipedia.org/wiki/Von_Neumann_algebra) that $\overline{M}$ is [affiliated](https://en.wikipedia.org/wiki/Affiliated_operator) with; it is called the von Neumann algebra generated by $M$. 


**Question 1**: Is there a bounded operator $X \in B(H)$ such that $W^{\star}(X) = \mathcal{M}$? 


In other words: Can a von Neumann algebra generated by a densely defined closable operator, be also generated by a bounded operator? 


**Question 2**: Is there a way to generalize the generation of a von Neumann algebra to any densely defined operator (i.e. non necessarily closable)? 


If an answer to Question 1 gives a process defining $X$ from $M$ and if this process works for any densely defined operator, that would also answer Question 2. 




---


**Some investigations for answering Question 2**


Let first suppose that the operator $M$ is associated to an integer map, i.e. $H = \ell^2(\mathbb{N}^\*)$ and there is a map $m: \mathbb{N}^\* \to \mathbb{N}^\*$ such that $Me\_n = e\_{m(n)}$, with $\mathcal{D}(M) = c\_{00}(\mathbb{N}^\*)$, dense in $H$. 


Assume that $M$ is non-closable iff $\exists n \in \mathbb{N}^\*$ with $m^{-1}(\{ n\})$ infinite. 


First of all, there is the following natural way to associate a bounded operator to $m$: $$Ye\_n = \frac{1}{n}e\_{m(n)}.$$ Unfortunately, if $M$ is closable, it does not generate the same von Neumann algebra than $Y$ in general, because if $m=id$, then $M=I$ and $Y = diag(1/n \ | \ n \in \mathbb{N}^\*)$, so $W^{\star}(M) = \mathbb{C}$ and $W^{\star}(Y) = \ell^{\infty}(\mathbb{N}^\*)$. 


But, there is a way to avoid this problem, by using the operator defined as follows:  

 $$\tilde{M}e\_n = \begin{cases} e\_{m(n)} & \text{if} \;m^{-1}(\{m(n)\}) \;\text{finite} \\ \frac{1}{n}e\_{m(n)} & \text{if} \;m^{-1}(\{m(n)\}) \;\text{infinite} \end{cases}$$ Then $\tilde{M}$ is densely-defined and closable, even if $M$ is non-closable. Moreover, if $M$ is still closable then $\tilde{M} = M$ by construction, so they generate the same von Neumann algebra. 


*Example:* If $m(n)=1$ $\forall n$, then $Me\_n = e\_1$ and $M$ is non-closable, whereas $\tilde{M}e\_n = \frac{1}{n}e\_1$ defines a bounded operator. Note that $\tilde{M}$ is not normal because $\tilde{M}^{\star}\tilde{M}e\_1 = \sum\_n \frac{e\_n}{n}$ and $\tilde{M}\tilde{M}^{\star}e\_1 = \frac{\pi^2}{6}e\_1$. Then, $\mathcal{M}:=W^{\star}(\tilde{M})$ is non-abelian. Now, $\tilde{M}^2 = \tilde{M}$ and $\tilde{M}\tilde{M}^{\star}\tilde{M} = \frac{\pi^2}{6}\tilde{M}$, so $\dim(\mathcal{M}) = 4$. It follows that $\mathcal{M} \simeq M\_2(\mathbb{C})$. 


*Invariance*: Let $\sigma \in S(\mathbb{N}^\*)$ be a permutation, $m\_{\sigma}$ the deformation $\sigma \circ m \circ \sigma^{-1}$. Let ${M}\_{\sigma}$ and $\tilde{M}\_{\sigma}$ be the corresponding operators. Do $\tilde{M}$ and $\tilde{M}\_{\sigma}$ generate isomorphic von Neumann algebras? 


*Hard example*: consider [Conway's game of life](https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life) and let $\mathcal{S}$ be the set of states of the grid with only finitely many alive cells. It is countable infinite, so there is a bijection $b: \mathcal{S} \to \mathbb{N}^\*$. Conway's rule (B3/S23) produces a map $r:\mathcal{S} \to \mathcal{S}$. Let $m$ be the integer map $b \circ r \circ b^{-1} : \mathbb{N}^\* \to \mathbb{N}^\*$. We can then define $\tilde{M}$ as above. Let $\mathcal{M}$ be the von Neumann algebra generated by $\tilde{M}$. Assuming that the above *invariance* is true, $\mathcal{M}$ is independant of the choice of $b$. Thus $\mathcal{M}$ can be called the von Neumann algebra generated by Conway's game of life. *Bonus question*: What is $\mathcal{M}$?  

We can do the same with any other cellular automaton. 


*Conclusion*: the use of $\tilde{M}$ is a way to generalize the generation of a von Neumann algebra to any densely defined operator associated to an integer map. Can we extend it to any densely defined operator (i.e. non necessarily associated to an integer map)? 


*A naive attempt of generalization*: let $H$ be a separable infinite dimensional Hilbert space and let $M$ be any densely defined operator, with maximal domain $\mathcal{D}$. Let $\mathcal{B}$ be a countable basis and $b: \mathcal{B} \to \mathbb{N}^\*$ a bijection. Take $e\_n:= b^{-1}(n)$ and let $K$ be the compact operator defined by $Ke\_n = \frac{1}{n}e\_n$. Let $H\_{\infty}$ be the closure of the subspace of vectors $v \in \mathcal{D}$ such that there is a countable orthonormal basis $\mathcal{B}\_v$ of $\overline{M^{-1}(\mathbb{C}Mv)}$ with $\mathcal{B}\_v \subset \mathcal{D}$ and $$\sum\_{b \in \mathcal{B}\_v} |\langle Mb,Mv \rangle|^2 = \infty.$$ Let $P$ be the orthogonal projection on $H\_{\infty}$. Consider the following operator $$\tilde{M}:=MKP + M(I-P).$$
Is $\tilde{M}$ well-defined? densely-defined? closable? Is the von Neumann algebra generated by $\tilde{M}$ independent of the choice of $\mathcal{B}$ and $b$? If so (...!), this construction would answer Question 2. 


*Invariance*: Would such a von Neumann algebra be invariant replacing $K$ by any positive compact operator $L$ with all eigenvalues distinct and $L^{1+\epsilon}$ trace-class $\forall \epsilon >0$?



The von Neumann algebra generated by a non-closable operator
The answer to Question 1. is positive. Namely, consider the polar decomposition of your operator $M=U|M|$ and define $X:= U f(|M|)$, where $f:[0,\infty) \to [0,\infty)$ is a bounded increasing function, for instance $f(x)= 1 - e^{-x}$ would do. Then the spectral projections of $f(|M|)$ are the same as the spectral projections of $|M|$, so both $U$ and the spectral projections belong to the von Neumann algebra generated by $X$, therefore it is the same as the von Neumann algebra generated by $M$.


Unfortunately, it does not give a clue about Question 2.


