Timeline for Wanted: example of a non-algebraic singularity
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 20, 2019 at 18:43 | comment | added | Laurent Moret-Bailly | @TimGrosskreutz: yes. | |
| Nov 20, 2019 at 14:39 | comment | added | user267839 | ...you mean it in the sense, that if $X$ is algebraic, then it's closed subset $Z$ is also algebraic, and $f$ and $g$ are algebraic over $K(Z)=\mathbb{C}(z)$, right? | |
| Nov 20, 2019 at 13:00 | comment | added | Laurent Moret-Bailly | @TimGrosskreutz: If $X$ is algebraic, then $f$ and $g$ must be algebraic over $\mathbb{C}(z)$. | |
| Nov 20, 2019 at 4:04 | comment | added | user267839 | one question about this example: the trick is that since the cross ratio stays invariant under "scanning by lines" and thus it's invariant under trafos of coordinates, we can fully reconstruct $f$ and $g$ from their values on $U \times 0$ independently of the choice of local coordinates,this is the message, right? the only point that I still not understand is why $\dim Z=1$ implies that $f$ and $g$ are already algebraically dependent, if we assume that $X$ is an open subset of an algebraic variety? | |
| Sep 27, 2011 at 23:52 | vote | accept | Anton Geraschenko | ||
| Sep 27, 2011 at 23:52 | comment | added | Anton Geraschenko | Ok, I buy it. I was confused for a bit because I thought you needed to include the information of how to fix $z$ in order to recover $f$ and $g$. But of course the whole point of the cross ratio is that it's the same no matter how you slice it. Thanks for the example! | |
| Sep 26, 2011 at 18:44 | comment | added | Mariano Suárez-Álvarez | Beautiful! ${}$ | |
| Sep 26, 2011 at 18:24 | history | answered | Laurent Moret-Bailly | CC BY-SA 3.0 |