Timeline for Hodge--Tate weights of an abelian surface
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 16, 2019 at 7:06 | comment | added | user138661 | @PiotrAchinger but then the original comment was slightly sloppily worded: "Hodge-Tate weights...they are...". Do you agree? | |
| May 15, 2019 at 20:21 | comment | added | Piotr Achinger | I'm claiming that $\binom{g}{i} \binom{g}{n-i}$ is the multiplicity of $\mathbf{C}_p(-i)$ in the Hodge-Tate decomposition of $H^n(X_{\bar K}, \mathbf{Q}_p)$. This follows from the fact that $H^n = \bigwedge^n H^1$ as Galois representations, and that $H^1$ has Hodge-Tate weights $0$ and $1$, both with multiplicity $g$. | |
| May 15, 2019 at 17:32 | comment | added | user138661 | @PiotrAchinger do you mean that the multiplicity of the Hodge--Tate weight $i$ is the product of binomial coefficients? or am I misunderstanding the terminology? | |
| May 15, 2019 at 17:10 | comment | added | Piotr Achinger | By Hodge-Tate comparison for abelian varieties (proved by Tate), Hodge-Tate weights are the same as Hodge numbers, so for $X$ an abelian variety of dim $g$ they are $\binom{g}{i} \binom{g}{n-i}$ in degree $n$. These are never all $\leq 1$ (I guess this is what you mean by distinct Hodge-Tate weights) unless $g=1$. | |
| May 15, 2019 at 14:20 | review | First posts | |||
| May 15, 2019 at 14:25 | |||||
| May 15, 2019 at 14:16 | history | asked | user140654 | CC BY-SA 4.0 |