Timeline for Weyl group action on continuous characters of the group of $\mathbf{Q}_p$-points of the torus
Current License: CC BY-SA 3.0
7 events
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| Mar 16, 2014 at 17:40 | history | edited | user76758 | CC BY-SA 3.0 |
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| Mar 16, 2014 at 17:38 | comment | added | user76758 | Hmm, I seem to be miscalculating a lot today. The rationalization of the group of $\chi$'s seem to be controlled (in the sense of a short exact sequence) by two copies of the character group: one copy that keeps track of unramified characters and the other which keeps track of ${\rm{Lie}}(\chi)$. Probably a combination of the above "answer" and your argument in the unramified case should then settle the general case up to torsion characters. Please let me know if that leads to some progress. (The torsion aspect sounds like it could be delicate.) | |
| Mar 16, 2014 at 17:31 | comment | added | Arkandias | Furthermore, even in the torsion-free case there is a problem (there is no isomorphism between the group of continuous characters $\chi : T(\mathbf{Q}_p) \to \mathbf{Q}_p^\times$ and $X(T) \otimes_{\mathbf{Z}} \mathbf{Q}_p$ because they do not come from algebraic characters in general). | |
| Mar 16, 2014 at 16:57 | comment | added | Arkandias | For $G=\mathrm{PGL}_2$ and $\chi([\mathrm{diag}(x,1)])=(-1)^{\mathrm{ord}_p(x)}$, one has $(\chi \circ \alpha^\vee)(x) = \chi([\mathrm{diag}(x,x^{-1})]) = \chi([\mathrm{diag}(x^2,1)])=1$ so this does not provide a counterexample. | |
| Mar 16, 2014 at 15:21 | history | edited | user76758 | CC BY-SA 3.0 |
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| Mar 16, 2014 at 13:31 | history | edited | user76758 | CC BY-SA 3.0 |
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| Mar 16, 2014 at 13:26 | history | answered | user76758 | CC BY-SA 3.0 |