Timeline for Simple $Z^{*}$ algebra
Current License: CC BY-SA 3.0
12 events
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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 |
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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 |
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Nov 29, 2014 at 11:21 | history | answered | Tomasz Kania | CC BY-SA 3.0 |