Timeline for Description of higher chow groups
Current License: CC BY-SA 4.0
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Jul 31, 2019 at 14:49 | history | edited | Matthias Wendt | CC BY-SA 4.0 |
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Jul 31, 2019 at 10:12 | comment | added | user143777 | My last question remains unanswered: in viewing $CH^2(S,1)$ sitting inside $CH^2(S\times \Delta^1)$ by Bloch, are these cycles null-homologous? That is to say, does $CH^2(S,1)$ live in $CH^2(S\times \Delta^1)_0$? | |
Jul 31, 2019 at 10:08 | comment | added | user143777 | Bloch's identification $\mathcal{O}(Z)^\times = CH^1(Z,1)$ in (viii) is proved in section 6 of his paper. I gather that it should essentially take a function $f \in \mathcal{O}(Z)^\times$ to its graph in $Z \times \Delta^1$, which is what I meant when I described the inverse of the edge map using graphs of functions. | |
Jul 31, 2019 at 10:04 | comment | added | user143777 | Assuming a priori that it is an isomorphism, the inverse of the edge map is as follows, if I got it right. Start with an element represented by $(Z_i,f_i)$ in $\oplus_{x\in S^{(1)}} k(x)^\times$, we can remove enough points on those curves so that $Z_i$ become smooth and $f_i$ have no poles nor zeros on them, so that $f_i \in \mathcal{O}(Z_i)^\times = CH^1(Z_i,1)$ by Bloch (viii). This is then sent to $CH^2(S,1)$ (where we have removed some points from the original $S$) via the displayed map $CH^1(Z_i,1) \rightarrow CH^2(S,1)$ of your answer. | |
Jul 31, 2019 at 8:56 | history | answered | Matthias Wendt | CC BY-SA 4.0 |