Timeline for Why is Maycock's Brauer group of graded C*-algebras connected while Moutuou's is not?
Current License: CC BY-SA 4.0
6 events
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| Jul 28, 2022 at 4:41 | comment | added | Jonathan Beardsley | I should mention that there is a real geometric relationship between the C*-algebra picture and Donovan-Karoubi one given by going between C*-algebras with spectrum X and some kind of Hilbert bundle over X. I don't understand this story well, but it's not just some coincidental isomorphism. | |
| Jul 28, 2022 at 4:38 | comment | added | Jonathan Beardsley | @QiaochuYuan yes. That's all true. Maycock's classifies Morita classes of complex C*-algebras and Moutuou's does the same for real C*-algebras. However, both also look a lot like the Donovan-Karoubi Brauer groups except that Maycock's is missing the $H^0(X;\mathbb{Z}/2)$ that it should have. | |
| Jul 27, 2022 at 22:26 | history | edited | LSpice | CC BY-SA 4.0 |
Names of references
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| Jul 27, 2022 at 22:13 | comment | added | Qiaochu Yuan | The first one looks complex and the second one looks real to me. In general a Brauer group should look something like $H^2(X, \mathcal{O}_X^{\times})$, yes? If we take real-valued functions we get something that looks like $H^2(X, \mathbb{R}^{\times}) \cong H^2(X, \mathbb{Z}_2)$ and if we take complex-valued functions we get something that looks like $H^2(X, \mathbb{C}^{\times}) \cong H^3(X, \mathbb{Z})$. Then these are related by a complexification map corresponding to the Bockstein. | |
| Jul 27, 2022 at 21:01 | history | edited | Jonathan Beardsley | CC BY-SA 4.0 |
extended the main question
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| Jul 27, 2022 at 20:54 | history | asked | Jonathan Beardsley | CC BY-SA 4.0 |