Timeline for Motivating the Quotient of an Algebraic Variety
Current License: CC BY-SA 4.0
12 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
May 6, 2019 at 20:54 | vote | accept | user267839 | ||
May 6, 2019 at 15:05 | history | edited | Sean Lawton | CC BY-SA 4.0 |
deleted 75 characters in body; edited title
|
May 6, 2019 at 14:46 | answer | added | Sean Lawton | timeline score: 3 | |
May 6, 2019 at 3:53 | answer | added | user130124 | timeline score: 1 | |
May 6, 2019 at 3:43 | comment | added | user267839 | Another remark: I think that in case of $\mathbb{C}^2/\mathbb{C}^*$ the fact that Zariski topology has the Kolmogorov property en.wikipedia.org/wiki/Kolmogorov_space does it's job. indeed there exist no open set which contains a line $(\lambda x, \lambda y)$ but not the origin $\{(0,0)\}$. But in case of $\mathbb{A}^1/G_m$ it seems to be a bit more subtle: the "point"/orbit $\mathbb{A}^1/G_m - \{0\}$ is open so it can be separated from point $\{0\}$ in Kolmogorov's way. So what fails in case $\mathbb{A}^1/G_m$? | |
May 6, 2019 at 2:34 | comment | added | user267839 | To simplify the problem: I think essentially the same argument that the author gave in his example would show that if we let act the multiplicative variety $G_m$ on $\mathbb{A}^1$. Then the topological quotient $p: \mathbb{A}^1 \to \mathbb{A}^1/G_m$ should also fail to become a variety. Another idea: If we assume that the topological quotient $\mathbb{A}^1/G_m$ has a variety structure, should all orbits /point be closed? Why? If yes, this would lead to the desired contradiction but I don't see an argument why we can make this assumption. | |
May 6, 2019 at 2:18 | comment | added | user267839 | @gcousin: I'm not sure. We consider the underlying topological space endowed with Zariski topology so unless for finite algebraic sets no algebraic set is ever a Hausdorff space. Or do I oversee an aspect? | |
May 6, 2019 at 2:02 | comment | added | gcousin | I imagine the main point is that the topological space underlying a complex variety should be Hausdorff | |
May 6, 2019 at 1:35 | review | Close votes | |||
May 6, 2019 at 20:44 | |||||
May 6, 2019 at 1:23 | history | edited | user267839 | CC BY-SA 4.0 |
added 15 characters in body
|
May 6, 2019 at 1:18 | comment | added | LSpice | This is a good question, but probably not research level. It might be better at MSE. | |
May 6, 2019 at 1:09 | history | asked | user267839 | CC BY-SA 4.0 |