Timeline for Topological K-theory of Riemann surface
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 8, 2022 at 20:37 | comment | added | user39380 | @DanRamras Thanks! the definition of the pairing has been added | |
| Mar 8, 2022 at 20:24 | vote | accept | CommunityBot | ||
| Mar 7, 2022 at 16:47 | history | edited | Dan Ramras | CC BY-SA 4.0 |
further corrections as per comments
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| Mar 7, 2022 at 16:36 | comment | added | Dan Ramras | @Tyrone Thanks - I definitely wrote this too early in the morning... | |
| Mar 7, 2022 at 16:19 | comment | added | Tyrone | Your formula for $\Sigma M^g$ is not correct (it should be $\Sigma M^g\simeq (\bigvee_{2g}S^2)\vee S^3$). For an explicit homotopy equivalence choose a basis for $H^1M^g$, represented as maps $M^g\rightarrow S^1\simeq K(\mathbb{Z},1)$. Suspend and use the suspension coordinate to add the maps together with the suspension of the pinching map $M^g\rightarrow S^2$. The result is a homology equivalence $\Sigma M^g\simeq\bigvee_{2g} S^2\vee S^3$. (you should also have $[M^g,U]\cong\bigoplus_{2g}\pi_1U$ (in accordance with your first paragraph). | |
| Mar 7, 2022 at 15:35 | comment | added | Dan Ramras | The pairing the OP is talking about is described here mathoverflow.net/questions/417368/… | |
| Mar 7, 2022 at 15:30 | history | edited | Dan Ramras | CC BY-SA 4.0 |
removed a mistaken paragraph about ring structure
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| Mar 7, 2022 at 15:29 | comment | added | Dan Ramras | Hmmm... I guess I was thinking of the map $ku\to H\mathbb{Z}$... I was a little worried that this seemed too simple... Anyway, I think from the OP's other recent post, Euler pairing means something else, so I'll remove this part of the answer. | |
| Mar 7, 2022 at 15:28 | comment | added | Lennart Meier | One might, however, use the map $KU \to \prod_{i\in \mathbb{Z}} \Sigma^{2i} H\mathbb{Q}$ of ring spectra. | |
| Mar 7, 2022 at 15:24 | comment | added | Denis Nardin | What map $H\mathbb{Z}→KU$? I do not think there's such a map of spectra, let alone of ring spectra (if there were, $KU$ would have trivial k-invariants, which they are not). | |
| Mar 7, 2022 at 15:17 | history | answered | Dan Ramras | CC BY-SA 4.0 |