Timeline for Brauer Groups and K-Theory
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 12, 2017 at 13:56 | comment | added | InfiniteLooper | The fact that the super Braur group is given by classes of Clifford Algebras is specific to $\mathbb{R}$ and $\mathbb{C}$. Ultimately it comes from the (ungraded) Brauer groups of $\mathbb{R}$ and $\mathbb{C}$ that are given by (ungraded) Clifford algebra : the non trivial one being the quaternion algebra $\mathbb{H}$. You Can find details at page 115 in the IAS, Quantum fields ans strings, a course for mathematicians. | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 29, 2014 at 16:23 | vote | accept | Kevin Walker | ||
| Jul 28, 2014 at 23:26 | answer | added | Qiaochu Yuan | timeline score: 18 | |
| Jun 28, 2012 at 18:11 | answer | added | Benjamin Antieau | timeline score: 9 | |
| Feb 3, 2012 at 0:07 | answer | added | Charles Rezk | timeline score: 15 | |
| Feb 2, 2012 at 21:31 | comment | added | Kevin Walker | @André: That sounds interesting. Do you know where I can read more about the Brauer group grading? | |
| Feb 2, 2012 at 21:23 | comment | added | André Henriques | K-theory is Brauer-group-graded, and also Z-graded. In other words, it's (Brauer-group x Z)-graded. The surprising thing is that this grading collapses to a cyclic-group grading. If you look at K-theory as a cohomology theory on the category of spaces over X (where X is some arbitrary space), then K-theory will be Brauer(X)-graded, which is no longer a cyclic group. | |
| Feb 2, 2012 at 18:17 | comment | added | Kevin Walker | @Paul: I guess it's because the relation between Brauer groups and Clifford algebras seems contingent to me. Does the relation hold for any field, or is it a special fact about $\mathbb R$ and $\mathbb C$? Is it possible to show that all elements of the Brauer group are represented by Clifford algebras without calculating what the Brauer group is? | |
| Feb 2, 2012 at 17:55 | comment | added | Paul Siegel | Why doesn't the connection with Clifford algebras count as an "a priori" reason? | |
| Feb 2, 2012 at 16:45 | history | asked | Kevin Walker | CC BY-SA 3.0 |