2
$\begingroup$

Let $H$ be a Hilbert space and $\mathcal{X}\subseteq \mathcal{B}(H)$ be an operator space. We denote $d(T,\mathcal{X})=\inf_{X\in\mathcal{X}}\|T-X\|,\,\,T\in\mathcal{B}(H)$ and $r_{\mathcal{X}}(T)=\sup\left\{|\langle{T \xi,\eta\rangle}|: \|\xi\|=\|\eta\|=1,\,\,\langle{X \xi,\eta\rangle}=0,\,\forall\,X\in\mathcal{X}\right\}.$

We say that $\mathcal{X}$ is hyperreflexive if there exists $k>0$ such that $d(T,\mathcal{X})\leq k \,r_{\mathcal{X}}(T)$ for every $T\in\mathcal{B}(H).$

Fact: Let $\mathcal{A}$ and $\mathcal{B}$ be von Neumann algebras and $\pi\colon \mathcal{A}\to \mathcal{B}$ be a $w^*$-continuous $*$-homomorphism and suppose that $\mathcal{A}$ is hyperreflexive. The kernel $\ker(\pi)$ of the map $\pi$ is an ideal of $\mathcal{A}$ and thus it is of the form $P \mathcal{A} P$ for some central projection $P\in\mathcal{A}.$ By known results it follows thar $\ker(\pi)$ is hyperreflexive.

Search: I have the expectation that an analogus result may hold for kernels of specific derivations, i.e. let $\mathcal{A}$ be a hyperreflexive von Neumann algebra acting on the Hilbert space $H$ and $P\in\mathcal{B}(H)$ be a projection. Consider the inner derivation $\delta(X)=XP-PX,\,X\in\mathcal{A}$ of $\mathcal{A}$ into $\mathcal{B}(H).$ Can we prove that $\ker(\delta)=\mathcal{A}\cap \left\{P\right\}^{\prime}$ is also hyperreflexive?

Maybe it is a hard question but interesting too.

$\endgroup$
5
  • $\begingroup$ What exactly is the known result regarding $\text{ker}(\pi)$ that you referred to? Any central compression of a hyperreflexive vNa is hyperreflexive? (Also, everything here is a vNa, you can define hyperreflexivity for just vNa’s instead of for general operator spaces.) $\endgroup$
    – David Gao
    Commented Sep 6 at 12:15
  • $\begingroup$ Also, if I’m not mistaken, any type II$_\infty$ factor can be generated by finitely many projections. $B(H)$ is hyperreflexive, so your claim would imply any II$_1$ factor with II$_\infty$ commutant is hyperreflexive, which, if I’m not mistaken, is still open according to your comments on one of your previous questions. $\endgroup$
    – David Gao
    Commented Sep 6 at 12:31
  • $\begingroup$ Dear @DavidGao, it is known by Lemma 1.3 on a paper of J. Kraus and David Larson entitle "Reflexivity and distance formulae". Also, any $II_1$ factor $M$ with $II_{\infty}$ commutant $M^\prime$ is hyperreflexive. The open case is if $M^\prime$ is hyperreflexive. $\endgroup$ Commented Sep 6 at 16:41
  • $\begingroup$ Hmm, that would be more challenging. There are plenty of II$_1$ factors that can be generated by finitely many projections, but whether this holds for all II$_1$ factors should still be open, if I’m not mistaken. But this should still mean your conjecture implies a very large class of II$_\infty$ factors with II$_1$ commutants are hyperreflexive, which makes this quite hard. $\endgroup$
    – David Gao
    Commented Sep 6 at 16:46
  • $\begingroup$ Exactly, it is still open. Also, the very large class of those $II_{\infty}$ commutants with $II_1$ commutants, which are hyperreflexive, automatically gives us a very large class of $II_1$ factors satisfying Kadison's similarity problem. $\endgroup$ Commented Sep 6 at 16:55

0

You must log in to answer this question.

Browse other questions tagged .