Timeline for Reference request - conjugacy classes over local fields
Current License: CC BY-SA 4.0
12 events
when toggle format | what | by | license | comment | |
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Feb 17, 2020 at 14:27 | comment | added | Sasha | @LaurentMoret-Bailly Thank you! | |
Feb 16, 2020 at 16:55 | comment | added | Laurent Moret-Bailly | More generally, if $F$ is a henselian valued field, $X$ a homogeneous space under an algebraic $F$-group $G$, and the stabilizers are smooth, the $G(F)$-orbits are open and closed in $X(F)$. This follows from Proposition 3.4.1 in this paper: note that if $x\in X(F)$ the corresponding orbit map $G\to X$ is a $G_x$-torsor over $X$. | |
Feb 16, 2020 at 14:09 | comment | added | Sasha | @MikhailBorovoi OK, thank you, I will think about that. | |
Feb 16, 2020 at 13:03 | comment | added | Mikhail Borovoi | However, I think that all will work also in char p, if the (scheme-theoretical) stabilizer of $x$ in $G$ is a smooth connected reductive group. | |
Feb 16, 2020 at 12:57 | comment | added | Mikhail Borovoi | In char 0, since $X$ is homogeneous, for any $x\in X(F)$ the map $$\phi_x\colon G\to X, \quad g\mapsto g\cdot x$$ is smooth, and hence the map on $F$-points $G(F)\to X(F)$ is open by the implicit function theorem, and hence the $G(F)$-orbit of $x$ is open. | |
Feb 16, 2020 at 12:51 | comment | added | Mikhail Borovoi | Serre (GC, III.4.4, Thm. 5) assumes that $F$ is perfect when proving that there are finitely many $G(F)$-orbits in $X(F)$. | |
Feb 15, 2020 at 21:32 | comment | added | Sasha | @MikhailBorovoi Do you know if something goes wrong if $F$ has char. p? I think of $G(F)$ etc. as analytic manifolds, so don't have a feeling of what can go wrong in char. p | |
Feb 15, 2020 at 21:27 | comment | added | Sasha | @MikhailBorovoi Dear Michael, the first fact I knew, but your next comment helped me understand better how to think about this rational situation. Thank you! | |
Feb 15, 2020 at 20:06 | comment | added | Mikhail Borovoi | Now it remains to show that if $F$ is a local field and $X$ is a homogeneous space of $G$, then the orbits of $G(F)$ in $X(F)$ are closed in the analytical topology. Proof: the number of orbits is finite (see Serre, Galois Cohomology), and they are open, hence the complement of each orbit is open, hence each orbit is closed. | |
Feb 15, 2020 at 19:58 | comment | added | Mikhail Borovoi | I don't know such a reference. However, I would suggest to ask a reference for the following assertion: the $G(\overline F)$-conjugacy class of any semisimple element in $G(\overline F)$ is Zariski-closed. The tags in such a question should include invariant-theory. | |
Feb 15, 2020 at 16:52 | history | edited | YCor |
edited tags; edited tags
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Feb 15, 2020 at 16:24 | history | asked | Sasha | CC BY-SA 4.0 |