Timeline for Is the union of Cartan subgroups over $k$ dense?
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| Jan 30, 2020 at 20:07 | comment | added | LSpice | As @LaurentMoret-Bailly said, the tori are parameterised by $(G/N_G(T))(k)$. Your argument shows that the tori are usually not parameterised by $G(k)/N_G(T)(k)$, but that is usually only a proper subset of $(G/N_G(T))(k)$. The test for $k$-rationality of the coset of $N_G(T)$ containing $g \in G(\overline k)$ is exactly as you have described. | |
| May 20, 2011 at 15:19 | comment | added | asm | I don't see why they would correspond to the $k$-rational points of $G$/(normalizer of $T$). A conjugate of $T$ can be defined over $k$ even if the conjugating matrix $g$ is not $k$-rational. It is sufficient that $g \phi(g)^{-1}$ is contained in the normalizer of $T$, where $\phi$ denotes the Frobenius. Nevertheless, I believe there are only finitely many since their $k$-conjugacy classes correspond to certain equivalence classes of the Weyl group. | |
| May 6, 2011 at 13:18 | vote | accept | Sophi | ||
| May 6, 2011 at 11:44 | history | answered | Laurent Moret-Bailly | CC BY-SA 3.0 |