Skip to main content

Timeline for Simple $Z^{*}$ algebra

Current License: CC BY-SA 3.0

12 events
when toggle format what by license comment
Nov 30, 2014 at 13:14 comment added Tomasz Kania The operators with separable range contain a copy of $\ell_\infty$ as a subspace. This is not the case for $K(H)$; dual of each separable subspace of $K(H)$ is separable.
Nov 30, 2014 at 13:12 comment added Ali Taghavi @TomekKania what is their difference as two Banach spaces?
Nov 29, 2014 at 17:04 comment added Tomasz Kania No, not even as a Banach space.
Nov 29, 2014 at 16:56 comment added Ali Taghavi @TomekKania Is $K(H)$ as an abstract C* algebra isomorphic to the algebra of countable rank operators on non separable $H$?
Nov 29, 2014 at 15:22 comment added Tomasz Kania Each compact operator has countable spectrum. Consider your favourite C*-algebra that has elements with uncountable spectrum...
Nov 29, 2014 at 15:15 comment added Ali Taghavi @TomekKania Thank you for your interesting answer. Is this example a universal example in the sense that: every $Z^{*}$ algebra has a faithful or irreducible representation in some $\mathcal{K}(H)$? For example consider the $C^{*}$ subalgebra of $B(H)$ generated by all countable rank operators. According to your example it is again a $Z^{*}$ algebra. Can we embedd it in compact operators ? Can we represent it irreducibly in compact operators?
Nov 29, 2014 at 15:10 vote accept Ali Taghavi
Nov 29, 2014 at 12:56 history edited Tomasz Kania CC BY-SA 3.0
added 38 characters in body
Nov 29, 2014 at 12:56 comment added Tomasz Kania Oh yes, you are of course right. Thank you for spotting that.
Nov 29, 2014 at 11:50 comment added Andreas Thom There is no separable example. If $(a_n)_n$ is a dense sequence, then $\sum_n a_n^*a_n/(2^n \|a_n\|^2)$ is a non-zero divisor.
Nov 29, 2014 at 11:28 history edited Tomasz Kania CC BY-SA 3.0
added 2 characters in body
Nov 29, 2014 at 11:21 history answered Tomasz Kania CC BY-SA 3.0