Timeline for When are two C*-algebras isomorphic as Banach spaces?
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20 events
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| Feb 26, 2015 at 8:40 | comment | added | weather | This has been answered many times in the comments but is the pair $c_0$ (the null sequences) and the algebra of continuous functions on the closed unit interval not a simple counter example to your initial question? | |
| Feb 6, 2014 at 8:33 | comment | added | Narutaka OZAWA | It is an open problem whether $C^∗_r(F_2)$ is Banach isomorphic to a nuclear $C^*$-algebra (say the CAR algebra $B$ as a model). It is known that (1) the dual of $B$ has the bounded approximation property (because it's AFD), (2) $B$ has a Schauder basis (Junge--Nielsen--Ruan--Xu, Adv. Math. 2004), and (3) $B\cong B\otimes_\alpha B$ for some reasonable tensor product (in this case the spatial tensor product). None of these properties is known for $C^∗_r(F_2)$. | |
| Feb 5, 2014 at 18:05 | comment | added | Norbert | @HannesThiel, so maybe you could organize all this long conversation into the single answer? | |
| Feb 5, 2014 at 17:45 | comment | added | Hannes Thiel | @Tomek: Thank you for pointing out that reference, I was not aware of it. It deals exaclty with the topic I asked about. | |
| Feb 5, 2014 at 17:41 | comment | added | Hannes Thiel | As Taka pointed out, among the simple, separable, non-type I C*-algebras, all the nuclear ones are isomorphic as Banach spaces (even much more is true) but beyond the nuclear case this is no longer the case. This certainly answers my question, thanks. | |
| Feb 5, 2014 at 15:32 | comment | added | Tomasz Kania | Hannes, this ma.utexas.edu/users/rosenthl/pdf-papers/93.pdf survey article of Rosenthal gives a nice overview of the Banach/operator-space structure of C*-algebras. | |
| Feb 4, 2014 at 18:19 | comment | added | Narutaka OZAWA | @BillJohnson: HRS proved that none of Type II and III noncommutative $L^1$ spaces Banach embed into a type I noncommutative $L^1$ space. If $A$ is a type I $C^*$-algebra, then $A^*$ is a predual of type I von Neumann algebra. | |
| Feb 4, 2014 at 16:22 | comment | added | Bill Johnson | Thanks, Nik. Taka, can you give a lazy old man a hint as to why the HRS classification theorem yields that Type I is invariant under Banach space isomorphisms? | |
| Feb 4, 2014 at 8:57 | comment | added | Narutaka OZAWA | By Szankowski (Acta Math 1981), there is a separable (simple, unital, etc.) $C^*$-algebra which does not have the approximation property. Such a $C^*$-algebra cannot be Banach isomorphic to a nuclear one. | |
| Feb 4, 2014 at 8:49 | comment | added | Narutaka OZAWA | A complement to Caleb's remark: Being type I is invariant under a Banach space isomorphism (at least in the separable case). This follows from Haagerup--Rosenthal--Sukochev's classification of noncommutative $L^1$-spaces up to Banach isomorphism. | |
| Feb 4, 2014 at 4:12 | comment | added | Hannes Thiel | @Caleb: Thanks for pointing out the result of Kirchberg. This clarifies the picture a lot. The question now seems to be whether the Banach space structure can detect nuclearity of the C*-algebra. | |
| Feb 4, 2014 at 3:59 | comment | added | Hannes Thiel | @Qiaochu: Thanks for clarifying. Here, two Banach spaces E and F are called isomorphic if there exists a bounded linear map from E to F that is bijective. (The inverse map will automatically be bounded, too). I do not want to assume isometric isomorphism. | |
| Feb 4, 2014 at 3:52 | comment | added | Caleb Eckhardt | Hi Hannes, Kirchberg showed in his subalgebras of CAR algebra paper that all separable nuclear, non Type I C*-algebras are isomorphic as operator spaces (much stronger than just isomorphic as Banach spaces). I think Christensen and others also did some things in this direction for von Neumann algebras: those results are in Pisier's operator space book, but my copy isn't with me. | |
| Feb 4, 2014 at 3:18 | comment | added | Nik Weaver | @Bill: the "simplest" example of a simple C*-algebra (hard not to make puns here) is the algebra $K(H)$ of compact operators on a separable Hilbert space. The irrational rotation algebras $A_\theta \subset B(L^2({\bf T}))$ generated by multiplication by $e^{2\pi i t}$ and rotation by the angle $2\pi\theta$ for irrational $\theta$ are also simple. The CAR algebra is obtained by embedding $M_{2^n}$ in $M_{2^{n+1}}$ via the map $A \mapsto \left[\matrix{A&0\cr 0&A}\right]$ and completing the union $\bigcup_{n=1}^\infty M_{2^n}$; it is also simple. | |
| Feb 4, 2014 at 3:14 | comment | added | Nik Weaver | I assume "isomorphic as Banach spaces" means linear homeomorphism. It's well known that there are non-isomorphic C*-algberas that are isometric as Banach spaces. (Google "C*-algebra not isomorphic to its opposite algebra.) The correct statement is: A and B are isomorphic as C*-algebras if and only if they are completely isometric as operator spaces. | |
| Feb 4, 2014 at 1:36 | comment | added | Bill Johnson | What are some simple and complicated examples of simple separable $C^*$-algebras? | |
| Feb 3, 2014 at 23:45 | comment | added | Qiaochu Yuan | Is an isomorphism of Banach spaces a bounded linear map with bounded linear inverse or an isometric isomorphism? | |
| S Feb 3, 2014 at 23:38 | history | suggested | user5794 | CC BY-SA 3.0 |
some LaTex
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| Feb 3, 2014 at 23:33 | review | Suggested edits | |||
| S Feb 3, 2014 at 23:38 | |||||
| Feb 3, 2014 at 23:01 | history | asked | Hannes Thiel | CC BY-SA 3.0 |