Timeline for Universal covering of abelian variety
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 13, 2022 at 21:00 | vote | accept | Desunkid | ||
| Jan 9, 2022 at 11:17 | vote | accept | Desunkid | ||
| Jan 9, 2022 at 18:34 | |||||
| Jan 8, 2022 at 11:19 | comment | added | Johannes Huisman | That exact sequence is not necessarily split. It is like $\lim \mu_{p^\infty}$, where $\mu_{p^\infty}$ is the group of roots of unity in $\mathbf C$ of order a power of $p$. That projective limit is torsion free. Hence, the surjective map from $\lim \mu_{p^\infty}$ to $\mu_{p^\infty}$ does not have a section. | |
| Jan 8, 2022 at 8:49 | comment | added | Desunkid | I think it''s stange for me because we have $0 \rightarrow T_p A \rightarrow B(A) \rightarrow A(\overline{K}) \rightarrow 0$, the map $B(A) \rightarrow A(\overline{K})$ send $(u_0,u_1,...) \mapsto u_0$. It follows that $B(A)=T_p(A) \times A(\overline{K})$ as groups? | |
| Jan 8, 2022 at 7:04 | comment | added | Johannes Huisman | Limits commute with limits. In particular, $\lim(A(\bar K)[p^\infty]\times W)=(\lim A(\bar K)[p^\infty])\times (\lim W)$, and $\lim W=W$ since multiplication by $p$ is a bijection on $W$. | |
| Jan 7, 2022 at 23:10 | comment | added | Desunkid | Thank you for your answer. Can you explain more about $B(A)\cong T_pA\oplus W$? It is really amazing for me! | |
| Jan 7, 2022 at 20:06 | history | answered | Johannes Huisman | CC BY-SA 4.0 |