Timeline for Equivariant normalization?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 15, 2013 at 16:14 | vote | accept | Jesko Hüttenhain | ||
| Oct 18, 2013 at 9:34 | |||||
| Oct 15, 2013 at 15:50 | comment | added | Jason Starr | The counterexamples are over a positive characteristic field $k$ (say, algebraically closed), where the group scheme $G$ is not uniquely determined by the group $G(k)$. This should never be an issue in characteristic $0$. | |
| Oct 15, 2013 at 14:07 | comment | added | Jesko Hüttenhain | No you're right, I don't see why we would need $G$ to be reductive, or equal to $\mathrm{Gl}_n$. Defining $g.(a/b):=g.a/g.b$ is certainly a valid group action on the fraction field. If $a/b$ is integral over the coordinate ring, applying a group element $g$ to some integral relation it satisfied should yield an integral relation for the translate, because the coefficients remain in $\mathbb C[X]$. However, I might be missing something and my question was directed more @JasonStarr. | |
| Oct 15, 2013 at 14:02 | comment | added | Peter Crooks | You are correct that I imposed no conditions on $G$ other than that it be affine algebraic. Technically, we want an algebraic action $G\times\tilde{X}\rightarrow\tilde{X}$. Such a thing presumably comes from a $\mathbb{C}$-algebra morphism $\mathbb{C}[\tilde{X}]\rightarrow\mathbb{C}[\tilde{X}]\otimes\mathbb{C}[G]$ satisfying the properties of a right-action. I think this should just be what I constructed, but I could be wrong. | |
| Oct 15, 2013 at 12:39 | comment | added | Jesko Hüttenhain | That sounds nice, however I am worried about @JasonStarr's comment (see above) - where in your argument will one use the fact that $G=\mathrm{Gl}_n\mathbb C$ (or that $G$ is reductive, which I suppose will be the most relevant property)? | |
| Oct 15, 2013 at 11:58 | history | answered | Peter Crooks | CC BY-SA 3.0 |