Timeline for Why would one "attempt" to define points of a motive as $\operatorname{Ext}^1(\mathbb{Q}(0),M)$?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 28, 2015 at 9:53 | answer | added | Olivier | timeline score: 13 | |
| Sep 28, 2015 at 9:41 | comment | added | Ilias A. | I do think there is too much to say about Kim's deep remarks. | |
| Sep 28, 2015 at 9:31 | comment | added | Ilias A. | For a given scheme $X$ over a field $k=\mathbf{Q}$ or $k=\mathbf{F}_{q}$ , the set of rational points $X(\mathbf{Q})$ are extremely difficult to compute in comparison with $X(\mathbf{F}_{q})$ where we have a "linear" formula involving a particular cohomology theory for schemes. Motives are some how a linearization of the category of schemes as stable homotopy is for topological spaces. The natural question is what can the "linear version" of schemes (motives) can tell us about rational points $X(\mathbf{Q})$. Therefore we look at the relation between X(Q) and $RHOM_{Mot}(\mathbf{Q}(0), M)$. | |
| Sep 28, 2015 at 6:00 | comment | added | Qiaochu Yuan | The set of points of a $\mathbb{Q}$-vector space $V$ is $\text{Hom}(\mathbb{Q}, V)$, so the RHom is the analogous thing in a derived sort of setting. | |
| Sep 28, 2015 at 0:40 | review | First posts | |||
| Sep 28, 2015 at 7:35 | |||||
| Sep 28, 2015 at 0:34 | history | asked | Quinlan Aktaş | CC BY-SA 3.0 |