Timeline for Excision in "3264 and all that" by Eisenbud-Harris
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 2 at 23:17 | comment | added | Andarrkor | @JasonStarr I am not getting what you mean by "homomorphisms differ by rational equivalence". Thanks for your patience. | |
| Jan 2 at 23:16 | comment | added | Andarrkor | Let $X=\mathbb{P}^1$, $\mathbb{P}^1 \times \mathbb{P}^1$ with coordinates $[\lambda : \mu], [x:y]$ respectively, take $\Phi = \lbrace x \mu - y \lambda =0 \rbrace$. Then take $t_0=[1:0], t_1= [0:1], s_0=[1:1], s_1=[-1:1]$. Then, $$ \langle \Phi \cap (\lbrace t_0 \rbrace \times X) \rangle - \langle \Phi \cap (\lbrace t_1 \rbrace \times X) \rangle = \langle [0:1] \rangle - \langle [1:0] \rangle $$ $$ \langle \Phi \cap (\lbrace s_0 \rbrace \times X) \rangle - \langle \Phi \cap (\lbrace s_1 \rbrace \times X) \rangle = \langle [1:1] \rangle - \langle [-1:1] \rangle $$ are these equal? | |
| Dec 31, 2023 at 18:51 | comment | added | Jason Starr | The image is exactly the same. The homomorphisms differ by rational equivalence, but the images are equal. | |
| Dec 31, 2023 at 16:41 | comment | added | Andarrkor | But in $Z(X)$ we do not take the quotient yet. $Z(X)$ is the free abelian group of the set of subvarieties, so the map is not well defined, right? We need the image to be exactly the same, not rationally equivalent. | |
| Dec 30, 2023 at 21:48 | comment | added | Jason Starr | The group $\textbf{PGL}_2$ of automorphisms of $\mathbb{P}^1$ is a dense open subset of $\mathbb{P}^3$. Thus, the images under the homomorphism are rationally equivalent, independent of the choice of rational point $(t_0,t_1)$ of $\mathbb{P}^1\times \mathbb{P}^1 \setminus \Delta(\mathbb{P}^1)$. | |
| Dec 30, 2023 at 15:59 | comment | added | Andarrkor | Could you elaborate on this in an answer, please? I don't get why this transitivity implies that you can choose any two distinct rational points. | |
| Dec 30, 2023 at 14:24 | comment | added | Jason Starr | You can choose them to be any two distinct rational points: the automorphism group of the projective line is doubly transitive (even triply transitive). | |
| S Dec 30, 2023 at 10:45 | review | First questions | |||
| Dec 30, 2023 at 11:55 | |||||
| S Dec 30, 2023 at 10:45 | history | asked | Andarrkor | CC BY-SA 4.0 |