Timeline for Is the Jordan decomposition of a self-adjoint functional constructive?
Current License: CC BY-SA 3.0
19 events
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| Jun 12, 2015 at 11:46 | answer | added | Simon Henry | timeline score: 2 | |
| Jun 12, 2015 at 0:32 | comment | added | Andre Kornell | David, I am interested in this question because I like to think about C*-algebras in settings where the axiom of choice fails (arxiv.org/abs/1502.01516). I also find it surprising that this canonical decomposition might require the axiom of choice. | |
| Jun 11, 2015 at 22:18 | comment | added | David Handelman | Out of curiosity, why should C*-algebraists be interested in this? We already assume AC, and when it suits us, CH. | |
| Jun 11, 2015 at 15:43 | comment | added | Simon Henry | By the Barr covering theorem and descent theory one can prove that this result hold in any boolean Grothendieck topos for an abstract $C^*$ algebra. So there is a very high chance that it should hold without any form of axiom of choice. At the present time I don't have an explicit proof (and I don't think the boolean hypothesis can be removed). | |
| Jun 11, 2015 at 15:43 | comment | added | Andre Kornell | I edited the question to clarify that I'm asking about abstract C*-algebras. I'm not sure whether it's okay for me to edit the original text, or if I'm supposed to append the clarification, or make it in the comments. | |
| Jun 11, 2015 at 15:33 | history | edited | Andre Kornell | CC BY-SA 3.0 |
+"abstract"
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| Jun 11, 2015 at 15:31 | comment | added | Andre Kornell | I seem to recall that the result is true for all vector functionals. | |
| Jun 11, 2015 at 14:38 | comment | added | Nik Weaver | Mmm, although the proof I have in mind only gets $\phi$ as a difference of positive bounded linear functionals, without the uniqueness or the norm equality. | |
| Jun 11, 2015 at 14:32 | comment | added | Nik Weaver | It may depend on your definition of C*-algebra. I think I can do this if I know that for every self-adjoint $x \in A$ and $\epsilon > 0$ there is a state $f$ with $f(x) \geq \|x\| - \epsilon$. If C*-algebras are defined concretely as algebras of operators then this is trivial using vector states, but if they are defined abstractly maybe you need Hahn-Banach. | |
| Jun 11, 2015 at 13:33 | comment | added | Andre Kornell | Naively, one would expect that a canonical decomposition would not need a lot of choice. | |
| Jun 11, 2015 at 13:31 | comment | added | Andre Kornell | Yes, I think that there is an appeal to the Hahn-Banach theorem, and an appeal to Alaoglu's theorem in both versions. | |
| Jun 11, 2015 at 13:30 | comment | added | Andrej Bauer | Ok. Just speaking off the top of my head here: the Hahn-Banach will be problematic in general and unless you know a proof that avoids it you're out of lack. On the positive side: there ought to be an approximate decomposition where things match up to $\epsilon$ for any $\epsilon > 0$. | |
| Jun 11, 2015 at 13:27 | comment | added | Andre Kornell | Another version of the proof is sketched on Wikipedia: en.wikipedia.org/wiki/… | |
| Jun 11, 2015 at 13:25 | comment | added | Andrej Bauer | Hmm, Google books does not have it scanned. Could you give a general idea of how choice is used? | |
| Jun 11, 2015 at 13:24 | history | edited | Andre Kornell | CC BY-SA 3.0 |
C*-algebras and their Automorphism Groups
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| Jun 11, 2015 at 13:23 | comment | added | Andre Kornell | I am referring to Pedersen's C*-algebras and their Automorphism Groups. | |
| Jun 11, 2015 at 13:10 | comment | added | Andrej Bauer | Can you briefly describe how the axiom of choice is used (or give a more complete reference to "Pedersen's book")? | |
| Jun 11, 2015 at 11:22 | history | edited | Andre Kornell |
measure-theory tag
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| Jun 11, 2015 at 11:16 | history | asked | Andre Kornell | CC BY-SA 3.0 |