7
$\begingroup$

Let $H$ be a separable Hilbert space and let $M$ be a 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 that $\overline{M}$ is affiliated 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 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 $\forall \epsilon >0$?

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I have just edited some investigations for answering Question 2. What do you think? $\endgroup$ – Sebastien Palcoux Jan 4 '18 at 16:03
  • $\begingroup$ So far I don't have anything constructive to say; let me think about it for some time. $\endgroup$ – Mateusz Wasilewski Jan 5 '18 at 8:07
2
$\begingroup$

Only a comment on Question 2: It is already not easy to define the von Neumann algebra generated by a family of closable operators, see e.g. https://projecteuclid.org/euclid.cmp/1103899047 Def. 2.5 and Rem. 2.7.

Closability in this context appears naturally for the following reason: Assume that you have some (possibly non-closed) unbounded operator $M$ on some dense domain $\mathcal D$. To get towards defining a corresponding von Neumann algebra $\mathcal M$, I presume you would first try to compute a formal adjoint $M^*$ and I think you will agree that $M^*$ should also be densely defined. But then you already obtain that $M$ is closable.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.