Timeline for Non-Faithfully Flat Quantum Homogeneous Spaces
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 29, 2012 at 14:11 | vote | accept | Réamonn Ó Buachalla | ||
| Aug 18, 2012 at 17:49 | answer | added | anonymous | timeline score: 2 | |
| Jul 22, 2012 at 14:43 | history | edited | Réamonn Ó Buachalla | CC BY-SA 3.0 |
added 5 characters in body
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| Jul 22, 2012 at 9:47 | comment | added | Réamonn Ó Buachalla | To clarify: For the reasons MTS gave, I am most interested in the surjective case, certainly not injective. | |
| Jul 22, 2012 at 6:27 | comment | added | user22479 | @MTS: Thanks for clarifying, sorry for my initial confusion. | |
| Jul 22, 2012 at 3:19 | comment | added | MTS | Interesting question - I guess I've only really thought about examples that were faithfully flat, so I don't have one off the top of my head. I will think about it. | |
| Jul 22, 2012 at 2:54 | comment | added | MTS | No, I meant to ask if the map was surjective. In that case one would regard $H$ as a "quantum subgroup" of $G$, and the algebra $M$ is a coideal subalgebra of $G$. If $G$ was the function algebra of a group, and $H$ the function algebra of the subgroup, $\pi$ would be the restriction map, and $M$ would be the subalgebra of functions invariant under translation by the subgroup, i.e. the function algebra of the corresponding homogeneous space; hence the term "quantum homogeneous space." | |
| Jul 22, 2012 at 2:09 | comment | added | user22479 | @MTS: Despite the "G" and "H" notation, they are rings and not "geometric" objects, so the map $\pi$ is analogous in the algebraic geometry setting to the map between coordinate rings rather than their spectra. So rather than ask if $\pi$ is surjective, you probably meant to ask if it is assumed to be injective (for which the answer is probably "yes"...which the OP can confirm or not). | |
| Jul 22, 2012 at 1:50 | comment | added | Réamonn Ó Buachalla | No. I would even be interested in cases where $\pi$ is just a bialgebra map. | |
| Jul 22, 2012 at 1:38 | comment | added | MTS | Do you require $\pi$ to be surjective? | |
| Jul 22, 2012 at 0:26 | history | asked | Réamonn Ó Buachalla | CC BY-SA 3.0 |