Timeline for A $C^*$-algebra with the bidual $B(H).$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 2 at 9:07 | comment | added | Karimbergen Kudaybergenov | Thank you very much! | |
| Aug 1 at 16:48 | comment | added | David Gao | @JamieGabe This is even true if $A^{\ast\ast} \equiv B(H)$ means the two are isometrically isomorphic as Banach spaces, since a result of Kadison implies that $A^{\ast\ast}$ and $B(H)$ are isometrically isomorphic as Banach spaces iff they are isomorphic as von Neumann algebras. Also, separability assumption is not necessary, as $A^{\ast\ast} \cong B(H)$ (just isometrically as Banach spaces) implies $A^\ast \cong S_1(H)$ isometrically. As $S_1(H)$ is separable, $A$ must be separable. | |
| Aug 1 at 15:41 | comment | added | Jamie Gabe | I don't understand what $A^{**} \equiv B(H)$ means. If $A$ is a separable $C^\ast$-algebra such that $A^{\ast \ast} \cong B(H)$ (isomorphic as $C^*$-algebras), then $A$ must be simple (since $B(H)$ is a factor) and type I (since $B(H)$ is a type I von Neumann algebra). The only simple separable type I $C^\ast$-algebras are $M_n(\mathbb C)$ and $K(H)$, so one must have $A \cong K(H)$. But $K(H)$ embeds (as a $C^*$-algebra) into $B(H)$ in many different ways. | |
| Aug 1 at 6:37 | comment | added | David Gao | @JochenWengenroth Also, the OP is assuming $A$ is a $C^\ast$-algebra, and any isomorphism in the category of $C^\ast$-algebras is automatically isometric. | |
| Aug 1 at 6:33 | comment | added | David Gao | @JochenWengenroth The corresponding result is false for $K(H)$, since $(K(H) + \mathbb{C})^{\ast\ast}$ has a one-dimensional summand whereas $B(H)$ does not. | |
| Aug 1 at 6:22 | comment | added | Karimbergen Kudaybergenov | Thanks a lot. Of course, I mean isometrically isomorphic when say coincide. | |
| Aug 1 at 6:07 | comment | added | Jochen Wengenroth | What do you mean by coincide? Isomorphic or isometrically isomorphic? The commutative $C^*$-algebras $c$ of all convergent sequences and $c_0$ of all null sequences both have $\ell^\infty$ as the bidual. | |
| Aug 1 at 5:39 | history | edited | Karimbergen Kudaybergenov | CC BY-SA 4.0 |
added 5 characters in body
|
| S Aug 1 at 4:45 | review | First questions | |||
| Aug 1 at 6:17 | |||||
| S Aug 1 at 4:45 | history | asked | Karimbergen Kudaybergenov | CC BY-SA 4.0 |