Timeline for Grothendieck-Messing theory
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 14, 2012 at 12:25 | comment | added | Keerthi Madapusi | Given any isogeny $g:A\to B$, and lifts $\widetilde{A}$ and $\widetilde{B}$ over $R$, $f=pg$ lifts uniquely to a map $\widetilde{f}:\widetilde{A}\to\widetilde{B}$. The main point is that $K=\ker(\widetilde{B}\to B)$ is killed by $p$ (viewed as group schemes over $R$). This is because you can use the divided powers to define a logarithm map $K\into Lie(\widetilde{B})$; see II.2.2 in Messing, The Crystals associated to Barsotti-Tate groups. From this, you can deduce the result from the argument in Lemma 1.1.3 in Katz, Serre-Tate Local Moduli. | |
| Oct 12, 2012 at 12:27 | comment | added | Kskvrt | I would like to follow up on my previous comment. If indeed $\mathbb{D}(A)_R=\mathbb{D}(A) \otimes_{W(k)}R$, then there is a big problem. Let $f:A\rightarrow A$ be an isogeny which factors by multiplication by $p$: $f=pg$, where $g$ is another isogeny. Lifting $f$ to $R$ is the same as lifting in a compatible way with $f$ two Hodge filtrations. But the map induced by $f$ on $\mathbb{D}(A)_R$ is just the zero map because $f$ factors by $p$. So the compatibility condition disappears completely, and lifting $f$ reduces to lifting two copies of $A$. That looks unreasonable to me. | |
| Oct 12, 2012 at 12:18 | comment | added | Kskvrt | There's something I don't quite understand: To an abelian variety over $k=\overline{\mathbb{F}_p}$, we can associate a Dieudonné module $\mathbb{D}(A)$ which is a free $W(k)$-module. Now, according to Grothendick-Messing theory, for every locally nilpotent divided power extension $R/k$, there is also a free $R$-module $\mathbb{D}(A)_R$. What is the connection between $\mathbb{D}(A)$ and $\mathbb{D}(A)_R$ ? My first thought was that $\mathbb{D}(A)_R=\mathbb{D}(A)\otimes_{W(k)} R$, where $R$ is considered a $W(k)$-module via the map $W(k)\rightarrow k \rightarrow R$. Is this true ? | |
| Oct 12, 2012 at 11:11 | comment | added | Keerthi Madapusi | By $\mathbb{D}(A)$ I meant the Dieudonne crystal of $A$. | |
| Oct 12, 2012 at 1:04 | comment | added | S. Carnahan♦ | You may use the universal property of tensor product to transfer multiplication by $p$ to the right side. | |
| Oct 11, 2012 at 19:53 | comment | added | Kskvrt | Yes, the map induced on $\mathbb{D}(A)$ is multiplication by $p$, but what about the map induced on $\mathbb{D}(A)_R$ ? | |
| Oct 11, 2012 at 19:37 | comment | added | Keerthi Madapusi | The map induced by $f$ on $\mathbb{D}(A)$ is just multiplication by $p$! Why do you find this so unbelievable? | |
| Oct 11, 2012 at 18:05 | history | asked | Kskvrt | CC BY-SA 3.0 |