1

This is a more specific version of my last question. Let $H$ be an infinite dimensional Hilbert space. Is the Calkin algebra $Q(H)=B(H)/K(H)$ linearly homeomorphic to a dual space in general? If not, is it for any particular $H$? I suspect that for separable $H,$ it is not. This intuition comes from the fact that $\ell^\infty/ c_0$ (which can be isometrically embedded as the diagonal operators) is not isomorphic to a dual space.

As noted in my last question, such a homeomorphism is never *-isomorphic, but for my purposes, this does not matter.

As I noted in this answer, if we have a Banach space $F$ s.t. $F^*$ is linearly homeomorphic to $Q(H),$ then $F$ corresponds to a minimal closed, point separating subspace of $Q(H)^*.$ This then corresponds to a minimal locally convex Hausdorff topology on $Q(H)$ (namely $\sigma(Q(H),F)$) whose continuous dual is Banach in the canonical way.

Miles Gould
  • 1,284
  • Before giving this some thought, you're implying that the Calkin algebra is not a von Neumann algebra, which I'm sure is true, but what is a quick way to see this? – Just dropped in Feb 24 '25 at 17:42
  • @Justdroppedin I'm not sure how the standard proof goes, but we know its projections don't form a lattice, a requirement for von Neumann algebras. – Miles Gould Feb 24 '25 at 17:51
  • @OnurOktay I don’t think that need be the case. – Miles Gould Feb 25 '25 at 23:12

0 Answers0