Timeline for Full free product of $B(\mathcal H_i)$
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2017 at 17:45 | comment | added | Caleb Eckhardt | It's probably the same answer. $B(K)$ has a unique closed 2-sided ideal when $K$ is infinite dimensional. The full free product would have more than one such ideal if $H_1$ and $H_2$ are both infinite dimensional--just take representations that alternately kill/preserve the compacts in each factor. If one of the Hilbert spaces is finite dimensional and the other is infinite dimensional, I'm not sure (my knowledge of free products is shallow) but I would guess it is not isomorphic to some $B(K)$ in that case as well | |
| Jan 26, 2017 at 14:23 | comment | added | Chris Ramsey | @CalebEckhardt Any ideas for the infinite dimensional case? | |
| Jan 26, 2017 at 13:36 | comment | added | Caleb Eckhardt | I think it is obviously not isomorphic to some $B(K)$ when the $H_i$ are finite dimensional, but not one dimensional. In that case the free product will be infinite dimensional and norm separable which can't happen for any $B(K)$ | |
| Jan 26, 2017 at 4:26 | history | edited | Chris Ramsey | CC BY-SA 3.0 |
Mistakes, mistakes, mistakes
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| Jan 26, 2017 at 4:15 | comment | added | Chris Ramsey | Yes, of course! The identification would only hold if the vector states were faithful, but that's impossible. I'll alter the question accordingly. | |
| Jan 26, 2017 at 3:33 | comment | added | Caleb Eckhardt | I'm probably missing something, but I don't see why the isomorphism you stated is true in the reduced case. For example if $H_1$ and $H_2$ are 2-dimensional and $n=2$, then it seems that the left hand side will be norm-separable and the right hand side will not be norm-separable. | |
| Jan 25, 2017 at 20:53 | history | asked | Chris Ramsey | CC BY-SA 3.0 |