Timeline for Constructible sets II (Grothendieck rings)
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 14, 2017 at 9:10 | comment | added | Andrea Ricolfi | No, I do not think that is needed. | |
| May 3, 2017 at 15:44 | comment | added | THC | By the way, for obtaining the first formula, is it needed that the scheme is Noetherian ? (Is this a necessary condition ?) | |
| Apr 26, 2017 at 13:46 | comment | added | Andrea Ricolfi | not sure what you mean by "isomorphic constructible sets". If you find decompositions like in your last comment, then certainly $[S]=[S']$. The converse (stating: if two varieties have the same class then they are piecewise isomorphic) is the "cut and paste conjecture" by Larsen and Lunts, which is false in general. | |
| Apr 26, 2017 at 13:22 | comment | added | THC | I guess one way to do it would be to find appropriate decompositions $S = \coprod_i Z_i$ and $S' = \coprod_jZ_j'$ as above, indexed over the same index set $I$ and such that each $Z_i$ is isomorphic to $Z_i'$. | |
| Apr 26, 2017 at 11:57 | comment | added | THC | Thanks ! Is there some way of expressing that $[S] = [S']$ if $S$ and $S'$ are "isomorphic constructible sets," just as for schemes ? | |
| Apr 25, 2017 at 12:45 | history | answered | Andrea Ricolfi | CC BY-SA 3.0 |