Timeline for homology of abelian variety ?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| May 13, 2010 at 12:50 | history | edited | TOM | CC BY-SA 2.5 |
edited title
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| May 12, 2010 at 18:16 | vote | accept | TOM | ||
| S May 12, 2010 at 18:16 | vote | accept | TOM | ||
| May 12, 2010 at 18:16 | |||||
| May 12, 2010 at 18:16 | vote | accept | TOM | ||
| S May 12, 2010 at 18:16 | |||||
| May 12, 2010 at 18:08 | comment | added | Pete L. Clark | @TOM: Note that the tensor construction is not a case of "base change" in any sense that I know: in particular you start and end with an abelian variety over $\mathbb{C}$. You might want to retitle your question accordingly. | |
| May 12, 2010 at 16:48 | comment | added | BCnrd | The role of $R$ above only matters as $\mathbf{Z}$-module: for finite free $\mathbf{Z}$-module $M$ can similarly define/characterize $A_0 \otimes M$, and to any $m \in M$ we associate $A_0 \rightarrow A_0 \otimes M$ defined by $a \mapsto a \otimes m$ (via def'n of right side). The existence/uniqueness/isom. problems all make sense for any $M$ (recovering above when $M$ is ring), and behavior for direct sum via products reduces us to the case $M = \mathbf{Z}$. Can improve arguments to work with $\mathbf{Z}$ replaced by other (assoc.) rings and $M$ a finite projective (left/right?) module. | |
| May 12, 2010 at 16:46 | answer | added | Torsten Ekedahl | timeline score: 3 | |
| May 12, 2010 at 16:04 | answer | added | Pete L. Clark | timeline score: 3 | |
| May 12, 2010 at 16:02 | comment | added | TOM | you are right ,it is in the isogeny-categoty,which I forget to say. | |
| May 12, 2010 at 15:58 | comment | added | BCnrd | Doesn't make sense to tensor "over Q" against ab. var. (Q doesn't act on ab. var., except in isogeny-category sense that is linguistics). So first use an order in $E$ (& then pass to isog. category): for finite flat $\mathbf{Z}$-alg. $R$, the functor $S \mapsto A_0(S) \otimes_ {\mathbf{Z}} R$ on analytic spaces is rep'td by ab. var. (call it $A_0 \otimes R$), and map $A_0 \rightarrow A_0 \otimes R$ induces ${\rm{H}}_1(A_0) \rightarrow {\rm{H}}_1(A_0 \otimes R)$ whose linearization $R \otimes {\rm{H}}_1(A_0) \rightarrow {\rm{H}}_1(A_0 \otimes R)$ is an isom. Nice exercise with uniformization. | |
| May 12, 2010 at 15:29 | history | asked | TOM | CC BY-SA 2.5 |