Timeline for Lifting morphisms of $p$-divisible groups using Grothendieck-Messing theory
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11 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2016 at 18:58 | comment | added | Keerthi Madapusi | Actually, giving a $p$-divisible group over $\mathrm{Spec}(R_0)$ is equivalent to giving such a compatible system. This takes a bit more work to show. See Lemma 2.4.4 of de Jong, 'Crystalline Dieudonne theory via formal and rigid geometry'. | |
| Jun 10, 2016 at 16:29 | comment | added | Simone | Finally, you can apply G-M as in your comment because giving a $p$-divisible group over $\text{Spf}(R_{0})$ is the same as giving a compatible system of $p$-divisible groups over $\text{Spec}(R_{0}/p^{n})$ for each $n$, right? And this would come from the completeness of $R_{0}$, I guess! | |
| Jun 10, 2016 at 16:26 | comment | added | Simone | I'm not so sure of the fact that giving an isogeny over $R_{0}$ is equivalent to giving a compatible family over all $R_{0}/p^{n}R_{0}$. How would you prove this fact? And finally, if this is true, the problem is solved I think, as using the rigidity of quasi isogenies, one gets that given a quasi isogeny at the level $p$, it can be lifted to a quasi-isogeny at each level without changing the $p^{n}$ factor which makes it become an isogeny. But how do you prove your assertion about isogenies? | |
| Jun 10, 2016 at 15:16 | comment | added | Keerthi Madapusi | As for applying G-M to such a thickening, one just applies it at all finite levels, and observes that giving an actual isogeny (not quasi, but an honest homomorphism) of $p$-divisible groups over $R_0$ is equivalent to giving a compatible family over all $R_0/p^nR_0$. | |
| Jun 10, 2016 at 15:14 | comment | added | Keerthi Madapusi | The PD structure is given simply by the usual divided powers on multiples of $p$ (i.e. $p\mapsto p^n/n!$). When $p\neq 2$, these divided powers tend to $0$, i.e. they're topologically nilpotent. When $p=2$, then the same is true for divided powers of $4$. | |
| Jun 10, 2016 at 8:40 | comment | added | Simone | Thanks for your answers! Dear Keerthi, sorry for this stupid question, but how is it possible to apply Grothendieck-Messing to such a formal divided power thickening? Isn't it necessary to have some local nilpotency conditions on $p$? And moreover, how is the PD structure constructed in this case? Thank you very much! | |
| Jun 9, 2016 at 17:01 | comment | added | Keerthi Madapusi | As David points out, quasi-isogenies always lift over nilpotent thickenings, but S-W are actually applying Grothendieck-Messing directly to the formal divided power thickening $R_0\rightarrow R_0/p$. If $p=2$, there is an issue because the divided powers of $2$ are not nilpotent, but this is okay, since you can first lift to $R_0/4$ by the first remark. | |
| Jun 9, 2016 at 9:20 | comment | added | David Hansen | Sorry, not "subsets"; the quasi-isogenies only give subsets of what you've written after applying $ \otimes \mathbb{Q}$ outside all the Hom groups. | |
| Jun 9, 2016 at 9:18 | comment | added | David Hansen | In the directed limit you've written down, the subsets of quasi-isogenies on the right-hand side are bijectively identified for all n (!) - check out Theorem 1.1.13 here: math.leidenuniv.nl/scripties/wang.pdf | |
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| Jun 9, 2016 at 7:56 | history | asked | user93678 | CC BY-SA 3.0 |