On Completely Singular von Neumann Subalgebras
| dc.creator | Fang, Junsheng | |
| dc.date | 2006-06-26 | |
| dc.date | 2007-08-14 | |
| dc.date.accessioned | 2026-07-07T08:23:29Z | |
| dc.date.available | 2026-07-07T08:23:29Z | |
| dc.description | Let $\M$ be a von Neumann algebra acting on a Hilbert space $\H$, and $\N$ be a singular von Neumann subalgebra of $\M.$ If $\N\tensor\B(\K)$ is singular in $\M\tensor\B(\K)$ for any Hilbert space $\K$, we say $\N$ is \emph{completely singular} in $\M$. We prove that if $\N$ is a singular abelian von Neumann subalgebra or if $\N$ is a singular subfactor of a type $II_1$ factor $\M$, then $\N$ is completely singular in $\M$. For any type $II_1$ factor $\M$, we construct a singular von Neumann subalgebra $\N$ of $\M$ ($\N\neq \M$) such that $\N\tensor\B(l^2(\mathbb{N}))$ is regular (hence not singular) in $\M\tensor \B(l^2(\mathbb{N}))$. If $\H$ is separable, then $\N$ is completely singular in $\M$ if and only if for any $θ\in Aut(\N')$ such that $θ(X)=X$ for all $X\in\M'$, then $θ(Y)=Y$ for all $Y\in\N'$. As an application of this characterization of completely singularity, we prove that if $\M$ is separable (with separable predual) and $\N$ is completely singular in $\M$, then $\N\tensorŁ$ is completely singular in $\M\tensor Ł$ for any separable von Neumann algebra $Ł$. | |
| dc.description | 11 pages, introduction is rewritten | |
| dc.identifier | https://arxiv.org/abs/math/0606649 | |
| dc.identifier | http://arxiv.org/abs/math/0606649 | |
| dc.identifier.uri | http://salesiana.dossiersoluciones.com/handle/123456789/136015 | |
| dc.subject | Operator Algebras | |
| dc.subject | 46L10 | |
| dc.title | On Completely Singular von Neumann Subalgebras | |
| dc.type | text |