Timeline for group actions on blow-ups
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 25, 2013 at 20:46 | answer | added | Daniel Loughran | timeline score: 18 | |
| Feb 25, 2013 at 20:43 | comment | added | user5117 | I guess it depends what kind of answer you want. J.C. Ottem's answer is the correct one, but more simplemindedly, and geometrically, if Z is smooth: Think of the exceptional divisor E as the projectivised normal bundle of Z in X. The differential of the map corresponding to $g \in G$ gives a linear aut. of the normal bundle, which then descends to E. | |
| Feb 25, 2013 at 20:23 | comment | added | Andrew Stout | edit: all $z \in Z$ | |
| Feb 25, 2013 at 20:23 | comment | added | Andrew Stout | isn't a blow-up a proper birational morphism? meaning that on $\bar{X} - \pi^{-1}(Z)$ the group action must be induced canonically via the isomorphism $\bar{X} - \pi^{-1}(Z) \cong X $. Thus, you could extend by acting on fibers in $Z$ $g\pi^{-1}(z) = \pi^{-1}(gz)$ for all $g\in G$ and all $Z\in Z$. well, this is a start anyway. | |
| Feb 25, 2013 at 20:21 | comment | added | J.C. Ottem | Perhaps you can use the fact that $G$ acts on the Rees algebra $R(I)=\bigoplus_{m\ge 0}I_Z^m$ and that $\tilde{X}=Proj R(I)$.. | |
| Feb 25, 2013 at 20:10 | history | asked | blp | CC BY-SA 3.0 |