Timeline for Closed reductive sub-orbits
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 6, 2012 at 20:13 | comment | added | Jason Starr | @Florian -- Thank you for pointing out the mistake. I gave another example where $X$ is affine. | |
| Jul 6, 2012 at 19:52 | history | edited | Jason Starr | CC BY-SA 3.0 |
added 845 characters in body
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| Jul 6, 2012 at 18:59 | history | edited | Jason Starr | CC BY-SA 3.0 |
Corrected mistake
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| Jul 6, 2012 at 16:37 | comment | added | Florian Eisele | Okay, on second thought, the $G$ in my previous comment is not reductive. So I guess that isn't a counter-example either. | |
| Jul 6, 2012 at 16:27 | comment | added | Florian Eisele | Technically this counter-example doesn't fit the question as the OP asks for $X$ to be affine. I'd suggest taking the group of affine transformations $G=\mathbb G_a^n \rtimes GL_n$, letting $X=\mathbb A^n$ be affine $n$-space and taking $H=GL_n$. Then the orbit of any point $0\neq x \in X$ under $H$ is $X-\{ 0 \}$ (i. e. not closed), but $G\cdot x = X$, because $G$ is transitive. | |
| Jul 6, 2012 at 15:46 | comment | added | Jim Humphreys | I guess similar reasoning applies whether the Zariski or complex topology is used here. From the algebraic viewpoint, a simple algebraic group of dimension 3 is acting on its flag variety (a homogeneous space) while a standard maximal torus has only two fixed points (corresponding to the Weyl group). | |
| Jul 6, 2012 at 15:06 | history | answered | Jason Starr | CC BY-SA 3.0 |