Question on the commutant of a von Neumann algebra Let $\mathfrak{R}$ be a von Neumann algebra acting on a complex Hilbert space $\mathcal{H}$.
A projection $P\in\mathfrak{R}$ is said to be cyclic if there exists $x\in\mathcal{H}$ such that $\text{Ran} P=\overline{\mathfrak{R}'x}$. I asked myself in which cases one-dimensional projections are cyclic (if $\mathfrak{R}$ contains some of them). Surely, if $\mathfrak{R}=\mathcal{B}(\mathcal{H})$, all the one-dimensional projections are cyclic since $\mathfrak{R}'=\mathbb{C}I$. But this is the only case? In other words, can $\mathfrak{R}\subsetneq\mathcal{B}(\mathcal{H})$ (so $\mathfrak{R}'\supsetneq\mathbb{C}I$) such that $\overline{\mathfrak{R}'x}=\mathbb{C}Ix=\langle\cdot|x\rangle x$ for some $x\in\mathcal{H}$? In some way, I search for a $\mathfrak{R}$ whose commutant is non-trivial but acts trivially on a proper one-dimensional subspace of $\mathcal{H}$. I tried to reason by contradiction but I did not find any clue: is it possible that, if $\mathfrak{R}'\supsetneq\mathbb{C}I$, there necessarily exists $A\in\mathfrak{R}'$ such that $Ax=y$, with $x,y$ linearly independent?
 A: A one-dimensional projection $P$ of a von Neumann algebra $\mathfrak{R}$ on a Hilbert space $\mathcal{H}$ will always be cyclic.  
Indeed, let $x$ be a non-zero element of $\mathrm{Ran} P$, so $\mathrm{Ran} P = \mathbb{C}x$.  To see that $P$ is cyclic we must show that $\overline{\mathfrak{R}'x}=\mathbb{C}x$.  Clearly, $\mathbb{C}x\subseteq \overline{\mathfrak{R}'x}$, since $x=Ix$ and $I\in\mathfrak{R}'$. For the other direction, $\overline{\mathfrak{R}'x}\subseteq \mathbb{C}x$, let $y\in\mathcal{H}$ be the limit of a sequence of the form $A_1x,\,A_2x,\,\dotsc$, where $A_1,A_2,\dotsc\in \mathfrak{R}'$; we must show that $y\in \mathbb{C}x\equiv \mathrm{Ran}P$, that is, $Py=y$.  Note that since $P$ is an element of $\mathfrak{R}$ and $A_n\in \mathfrak{R}'$, we have $PA_n=A_nP$ for all $n$, and so $Py=P(\lim_n A_n x)=\lim_nPA_nx=\lim_n A_nPx=\lim_n A_nx=y$. Whence $y\in \mathbb{C}x$.
(More generally, given an element $y$ of $\mathcal{H}$ there will always be a projection $P$ of $\mathfrak{R}$ with range $\overline{\mathfrak{R}'y}$, and this will be the least projection $Q$ of $\mathfrak{R}$ with $Qy=y$.)
