Timeline for Intuition for the Lefschetz motive (Tate motive)?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 22, 2016 at 12:37 | comment | added | Mikhail Bondarko | $X$ and $L(X)$ are always distinct sheaves. $X$ is usually not a sheaf of groups, whereas $L(X)$ is usually not homotopy invariant. So, no "simple" relations between $M(X)$ and $L(X)$ exist. | |
| Jul 22, 2016 at 9:45 | comment | added | user40276 | I mean putting a smooth scheme $X$ in degree $0$, assuming that you have transfers (or just pick $L (X) $ in degree 0). What's the relation between $M (X)$ and $X$? For $G_m$, $M (G_m, 1) \cong G_m$, as you said it's related by a reduction, but in general (for $X$ different from $G_m$) this seems false. Actually its not even clear that $X$ is effective geometric. Anyway, thank you for your patience until now, I will stop bothering you. | |
| Jul 22, 2016 at 9:34 | comment | added | Mikhail Bondarko | $G_m$ is in degree $0$, yes. I probably don't understand your other questions. | |
| Jul 22, 2016 at 7:28 | comment | added | user40276 | You mean everything is wrong or at least I'm correct in saying $G_m$ is the complex concentrated in degree $0$? So apparently $X$ as an effective motive not even need to be geometric, right? | |
| Jul 22, 2016 at 6:37 | comment | added | Mikhail Bondarko | No, this is totally wrong (if I understand you correctly)!:) | |
| Jul 22, 2016 at 1:13 | comment | added | user40276 | I didn't know about it. I thought that people were too lazy to add the $M (-)$ :) . So, just to make sure I understand. $G_m$ means the complex $L (G_m)$ concentrated in degree $0$? By the way, is it always true that for any smooth scheme $X$, this complex is just the reduced $M (X)$? | |
| Jul 21, 2016 at 23:07 | comment | added | Mikhail Bondarko | Oh yes; $G_m$ may have another meaning: instead of considering the Suslin complex of the variety $G_m$ one can just consider the (homotopy invariant) sheaf with transfers $G_m$. These two motives are closely related: by "taking the quotient by a point". | |
| Jul 21, 2016 at 22:56 | comment | added | user40276 | Sorry, maybe I'm missing something very trivial, but I'm not getting it. See page 30 here for instance uni-due.de/~bm0032/publ/ICTPMotives.pdf . It's claimed that $L = Z(1)[2] = G_m [1]$ (the unreduced $G_m$). However your triangle should work as long as reduction sends distinguished triangles to dist. triangles (I don't know if this is true). I understand your argument, but it's not compatible with the above result (in the link for instance). | |
| Jul 21, 2016 at 22:47 | comment | added | Mikhail Bondarko | I am taking the reduced $G_m$ because the unreduced one cannot be shifted to become L. In particular, $G_m$ has etale cohomology in two distinct degrees! | |
| Jul 21, 2016 at 22:40 | comment | added | user40276 | Whoops! I always exchange signal of the shifting. However I still can't see why are you taking the reduced $G_m$. As I understand the embedding of effective geometric motives into effective motives is given by taking $M (X)$ to $C_{*} Z_{tr} (X)$, so if $L = C_{*}Z_{tr} (G_m) [1]$ its inverse image should be $M (G_m) $ and not the reduced one, right? | |
| Jul 21, 2016 at 22:32 | comment | added | Mikhail Bondarko | Certainly you should consider the motif of $G_m/{1}$ (and not just $G_m$) and $[1]$ here. | |
| Jul 21, 2016 at 22:19 | comment | added | user40276 | Thanks for answer. But I'm not seeing how you are getting $M (\mathbb{G}_m) [-1] \cong Cone (x_{*})$. | |
| Jul 21, 2016 at 21:44 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
added 14 characters in body
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| Jul 21, 2016 at 21:27 | history | answered | Mikhail Bondarko | CC BY-SA 3.0 |