Timeline for Is a C*-algebra with an isomorphic predual a von Neumann algebra?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 20, 2016 at 20:09 | answer | added | Matthew Daws | timeline score: 7 | |
| Jul 20, 2016 at 5:36 | vote | accept | Hannes Thiel | ||
| Jul 19, 2016 at 21:39 | answer | added | Yemon Choi | timeline score: 11 | |
| Jul 19, 2016 at 13:32 | comment | added | Yemon Choi | @jjcale I have a very vague recollection that $L^1({\bf T})/H^1$ (where $H^1$ is Hardy space) has this property: this is explained in Wojtaszczyk's book on Banach space but unfortunately I don't recall who proved this | |
| Jul 17, 2016 at 15:04 | comment | added | Hannes Thiel | To my knowledge, this was first shown in this article: Davis, Johnson: A renorming of nonreflexive Banach spaces, Proc. Amer. Math. Soc. 37 (1973) The result of Davis and Johnson was later generalized, for instance here: Godun: Equivalent norms on nonreflexive Banach spaces, Dokl. Akad. Nauk SSSR 265 (1982) | |
| Jul 17, 2016 at 14:50 | comment | added | jjcale | Can you give a reference to an example of a banach space that is isomorphic to the dual space of a banach space but not isometrically isomorphic to the dual space of any banach space ? | |
| Jul 17, 2016 at 11:12 | comment | added | Andreas Thom | Ok, you are right. | |
| Jul 17, 2016 at 8:07 | comment | added | Hannes Thiel | @AndreasThom At least for general Banch spaces this is not always possible: There are Banach spaces that have a predual but no isometric predual. Thus, there exists a Banach space $Y$ such that $Y$ is isomorphic to some $X^*$, but not isometrically isomorphic to $Z^*$ for any $Z$. My question is whether such a behaviour is possible for C*-algebras. | |
| Jul 17, 2016 at 8:03 | comment | added | Hannes Thiel | @jjcale Yes, by isomorphism I mean isomorphism of Banach spaces, that is, continiuous linear bijections (=linear homeomorphism). | |
| Jul 17, 2016 at 6:25 | comment | added | Andreas Thom | Just change the norm on $X$ according to the duality with the right norm on $A$, and $A$ will be the isometric dual. | |
| Jul 17, 2016 at 6:18 | comment | added | jjcale | Is your definition of "isomorphism" "linear homeomorphism" (see mathoverflow.net/questions/80567/…) ? | |
| Jul 16, 2016 at 19:58 | history | asked | Hannes Thiel | CC BY-SA 3.0 |