Timeline for The Frobenius at the infinite prime?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 9, 2019 at 14:31 | comment | added | Sylvain JULIEN | @Will Sawin : the Weil group you mention is a semi-direct product of some group with the cyclic group $C_2$, like a dihedral group. Can it thus be seen as some symmetry group of an abstract analogue of a polygon with infinitely many sides? | |
| May 9, 2019 at 11:45 | comment | added | Will Sawin | @LSpice Thanks, good point. | |
| May 9, 2019 at 2:55 | comment | added | LSpice | I remember Tasho Kaletha talking about how some of the computations in his regular supercuspidals paper indicate whether the extension $\mathbb C/\mathbb R$ should be considered as ramified or unramified … but I don't remember which way the verdict fell. | |
| May 9, 2019 at 2:54 | comment | added | Wenzhe | @WillSawin Thank you, do you know of any references about the Weil group (at the infinite prime), and how it determines the Gamma factors? | |
| May 9, 2019 at 2:54 | comment | added | LSpice | @WillSawin, I think that the extension $1 \to \mathbb C^\times \to W_{\mathbb R} \to \mathbb Z/2\mathbb Z \to 1$ is non-split (there's a lift of the non-trivial element of $\mathbb Z/2\mathbb Z$ to an element of $W_{\mathbb R}$ whose square is $-1 \in \mathbb C^\times$, but not one whose square is $1$), so that one shouldn't describe the Weil group as a semi-direct product like that. | |
| May 9, 2019 at 2:50 | comment | added | Will Sawin | The analogue of the Galois group at the infinite prime should be the Weil group $\mathbb C^\times \rtimes \mathbb Z/2$, which acts on the cohomology and describes its Hodge numbers and complex conjugation structure. There may be different ways to decide what part of that counts as inertia and what part counts as Frobenius, but regardless the action of this group determines the Gamma factors. | |
| May 9, 2019 at 2:46 | comment | added | Will Sawin | To whatever extent there is a Frobenius at the infinite prime, it doesn't make sense to consider it a limit of the finite Frobeniuses: In the function field setting, there is an Frobenius at the infinite prime, but it can't be expressed as a limit like this in a reasonable way. | |
| May 9, 2019 at 2:30 | history | asked | Wenzhe | CC BY-SA 4.0 |