Timeline for Absolute Hodge implies Galois invariant?
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 1, 2014 at 16:55 | comment | added | jmc | Sorry, I don't understand your comment. There is no Galois action on Hodge classes. An absolute Hodge cycle in Deligne's sense is a tuple $(s_{dR}, s_{et})$, such that (i) for every complex embedding $\sigma \colon K \to \mathbb{C}$, $s_{dR}$ gives rise to a (rational) Hodge class, and (ii) such that under the comparison of Betti cohomology and etale cohomology, these Hodge classes correspond with $s_{et}$, and (iii) $s_{et}$ is Galois invariant. | |
| Jul 1, 2014 at 8:42 | vote | accept | Lan | ||
| Jul 1, 2014 at 20:13 | |||||
| Jun 30, 2014 at 21:12 | comment | added | Lan | some point unclear for me, it may be a stupid question. I understand Absolute Hodge cycle as those Hodge cycle such that after Galois action it is still a Hodge cyle. But then how to show that the etale component of an absolute Hodge cycle is invariant under the whole Galois group $Gal(\bar{K}|K)$? | |
| Jun 30, 2014 at 10:13 | history | answered | jmc | CC BY-SA 3.0 |