Timeline for Smoothness of orbit of group scheme
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 3, 2021 at 14:10 | comment | added | afh | @JJH I have added another answer with what I believe is a reasonable mild condition to impose in order to have a positive answer to the question. | |
| Nov 24, 2021 at 16:45 | comment | added | afh | Unfortunately I don't know if one can do much without knowing about flatness of $Z$. I will let you know if I happen to come up with something that could help. I am intrigued by the question. Maybe if you post it as a separate question on MO someone more knowledgeable about this could chime in. | |
| Nov 24, 2021 at 15:48 | comment | added | JJH | In fact, the schematic image is what I’m more interested in. But I don’t how to argue the flatness over the base and the reducedness of the fiber. So, I decide to first try to understand if the orbit is locally closed and smooth. Without knowing much about Z, can we actually check these properties fiberwise? | |
| Nov 24, 2021 at 13:05 | comment | added | afh | (Added: for the first step outlined above I think flatness of $Stab$ might be enough) | |
| Nov 24, 2021 at 13:00 | comment | added | afh | You can just check that the morphism $O \to Z$ is flat, unramified and radicial. I believe all of these can be checked fiberwise (flatness by the fiberwise criterion for flatness). If you know the fibers of the scheme theoretic image $Z$ (is this what they call a spherical Schubert cell?), then you can again reduce to looking at a specific fiber, so you are working over a field + you can probably relate to the local situation (affine grassmannian). Sorry for the long comments, I don't know if this helps at all. Note that right away the problem in my example of $G_a$ is that Stab is not smooth. | |
| Nov 24, 2021 at 12:55 | comment | added | afh | First, I would try to show that the stabilizer group scheme is smooth over the base (so you need to check flatness (a local condition), and then the fact that each fiber is smooth, which in char 0 is automatic and in this case it can be related to the local (affine grassmannian situation). Once you know that the stabilizer is smooth, you get that the quotient $O = G/Stab$ is a smooth algebraic space, and it admits a morphism $O \to Z$, where $Z$ is the scheme theoretic image of the morphism $G \to X$. Now you want to show that $O \to Z$ is represented by an open immersion. | |
| Nov 24, 2021 at 12:52 | comment | added | afh | That is an interesting question. I don't know the answer off the top of my head (you are saying that this cannot be extracted from the literature?). My first instinct (if I wanted to prove your question positively) would be to first equip the orbit with smooth structure and then try to show that it is a locally closed subscheme. | |
| Nov 23, 2021 at 22:09 | comment | added | JJH | just in case you know, my real question is: if Beilinson-Drinfeld Schubert cell is locally closed and smooth over the n-power of the curve. When n=1, it is easy to see. What about n>2. | |
| Nov 23, 2021 at 21:17 | comment | added | JJH | What if we assume the fiberwise orbit has the same dimension? Any reference on general theorem which guarantee local closedness and smoothness? Thanks. | |
| Nov 23, 2021 at 20:59 | history | edited | LSpice | CC BY-SA 4.0 |
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| Nov 23, 2021 at 20:56 | history | answered | afh | CC BY-SA 4.0 |