Timeline for Postnikov invariants of the Brauer 3-group
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 7, 2020 at 17:00 | answer | added | Qiaochu Yuan | timeline score: 4 | |
| May 16, 2020 at 20:37 | history | edited | John Baez | CC BY-SA 4.0 |
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| May 15, 2020 at 18:44 | comment | added | Jacob Lurie | 2) Br(K) is an Eilenberg-MacLane space K( K^*, 2) (since the Picard and Brauer groups of a separably closed field vanish). Concretely this gives you a formula for pi_*( Br(k) ) in terms of Galois cohomology. It also tells you that Br(k) admits the structure of a topologica/simplicial abelian group (since for an Eilenberg MacLane space Br(K), choosing such a structure is equivalent to choosing a base point, and the structure survives passage to homotopy fixed points when the base point is fixed by G). | |
| May 15, 2020 at 18:39 | comment | added | Jacob Lurie | They have a common explanation. Assume k a field for simplicity, let K be a separable closure, and let G = Gal(K/k). Let Br(k) denote the classifying space for your 2-category (so it's the loop space of what you're denoting by Br(k) ). Then Br(K) carries a continuous action of G, and there's a natural map e from Br(k) to the (continuous) homotopy fixed points Br(K)^hG. The input you need is the following: 1) The map e is a homotopy equivalence (because the construction k -> Br(k) satisfies etale descent). (cont) | |
| May 14, 2020 at 20:34 | comment | added | John Baez | Nice! Is this why you can compute the groups I'm calling $\pi_i$ using Galois cohomology as $H^{3-i}(\mathrm{Gal}(K|k), K^\star)$ where $K$ is the separable closure of $k$? | |
| May 13, 2020 at 20:33 | comment | added | Jacob Lurie | Br(k) is the 0th space of an HZ-module spectrum, so its Postnikov invariants are trivial (the Postnikov tower is noncanonically split). | |
| May 8, 2020 at 22:46 | comment | added | Theo Johnson-Freyd | As you point out, $\mathbf{Br}(k)$ is symmetric, and so the Postnikov invariants are (the destablizations of) stable cohomology operations. This highly constrains the question. | |
| May 8, 2020 at 22:05 | history | asked | John Baez | CC BY-SA 4.0 |