Let $G$ be a split reductive group over $\mathbf{Q}_p$ and assume $G$ has connected center. Let $T$ be a maximal split subtorus of $G$ and $R$ be the roots of $(G,T)$.

Let $\chi : T(\mathbf{Q}_p) \to \mathbf{Z}_p^\times$ be a continuous character and assume $\chi \circ \alpha^\vee \neq 1$ for all $\alpha \in R$.

Question: Do we have $w(\chi)=\chi$ if and only if $w=1$ ?

This is true for $\mathrm{GL}_n$ or $\mathrm{GSp}_{2n}$ and also for unramified characters, but is it in general ?

Remark: If the center of $G$ is not connected there are counterexamples, e.g. $G=\mathrm{SL}_2$ and $\chi : \mathrm{diag}(x,x^{-1}) \mapsto (-1)^{\mathrm{ord}_p(x)}$.

(Edit) Remark: If the center of $G$ is connected and $\alpha \in R$, then $\chi \circ \alpha^\vee \neq 1 \Leftrightarrow s_\alpha(\chi) \neq \chi$.

Edit / Answer: The connectedness of the center is not a sufficient condition. There is a counterexample with $G_2$ : the longest element of its Weyl group $w_0$ acts on $T(\mathbf{Q}_p)$ by $w_0(t)=t^{-1}$ and one can construct a character $\chi$ such that $\chi \circ \alpha^\vee \neq 1$ for all $\alpha \in R$ and $w_0(\chi)=\chi$ as follow. Assume $p \neq 2$ so that $\mathbf{Q}_p^\times \cong \mathbf{Z} \times \left( \mathbf{Z}/p\mathbf{Z} \right)^\times \times \mathbf{Z}_p$ and let $\chi_1 : \mathbf{Q}_p^\times \twoheadrightarrow \mathbf{Z} \to \{\pm 1\}$ and $\chi_2 : \mathbf{Q}_p^\times \twoheadrightarrow \left( \mathbf{Z}/p\mathbf{Z} \right)^\times \to \{\pm 1\}$, then define $\chi : T(\mathbf{Q}_p) \cong \mathbf{Q}_p^\times \times \mathbf{Q}_p^\times \to \{\pm 1\}$ by $\chi(t_1,t_2)=\chi_1(t_1)\chi_2(t_2)$.

  • $\begingroup$ Isn't the unramified $\chi({\rm{diag}}(x,y)) = (-1)^{{\rm{ord}}_p(x/y)}$ an unramified counterexample of GL$_2$? It seems likely that torsion $\chi$ will generally create problems, and that is where the intuition related to Weyl chambers breaks down. $\endgroup$ – user76758 Mar 16 '14 at 6:19
  • $\begingroup$ No, because with your example $\chi \circ \alpha^\vee=1$. More generally, for $\mathrm{GL}_n$ the characters of $T(\mathbf{Q}_p) \cong \mathbf{Q}_p^n$ can be written $\chi = \chi_1 \otimes \dots \otimes \chi_n$ and the genericity condition means that $\chi_i \neq \chi_j$ for all $i \neq j$. Thus $w(\chi)=\chi_{w(1)} \otimes \dots \otimes \chi_{w(n)} \neq \chi$ for $w \neq 1$. $\endgroup$ – Arkandias Mar 16 '14 at 10:14
  • $\begingroup$ Isn't your remark valid for all reductive $G$ and all $\chi\in X(T)$? $\endgroup$ – A Stasinski Mar 17 '14 at 8:47
  • $\begingroup$ Yes, for algebraic character one always have $\chi \circ \alpha^\vee \neq 1 \Leftrightarrow s_\alpha(\chi) \neq \chi$. But for $G=\mathrm{SL}_2$ and $\chi(\mathrm{diag}(x,x^{-1}))=(-1)^{\mathrm{ord}_p(x)}$, one has $\chi \circ \alpha^\vee \neq 1$ but $s_\alpha(\chi)=\chi$. In general in order to prove $\chi \circ \alpha^\vee \neq 1 \Rightarrow s_\alpha(\chi) \neq \chi$, one has to suppose the center of $G$ connected and use the fundamental coweight associated to $\alpha$. $\endgroup$ – Arkandias Mar 17 '14 at 11:06

The answer is false due to torsion $\chi$, such as for PGL$_2$ (center trivial, hence connected!) with $\chi({\rm{diag}}(x,1)) = (-1)^{{\rm{ord}}_p(x)}$. [EDIT: This is wrong, as the OP notes below.] But if we ignore torsion, which amounts to considering continuous $\chi:T(\mathbf{Q}_p) \rightarrow \mathbf{Q}_p$, then the answer is affirmative. Indeed, this reduces [EDIT: not quite, but see comments below] to an analogous question at the level of the dual space ${\rm{Lie}}(T(\mathbf{Q}_p))^{\ast} = {\rm{X}}(T) \otimes_{\mathbf{Q}} \mathbf{Q}_p$, for which it suffices to show that the subspace of $w$-fixed points for any $w \ne 1$ is contained in some root hyperplane. This in turn is a statement of purely linear algebraic nature, so the problem over $\mathbf{Q}_p$ is equivalent to the one over $\mathbf{Q}$, which in turn is equivalent to the one over $\mathbf{R}$, where we may use considerations with Weyl chambers to conclude.

  • $\begingroup$ 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. $\endgroup$ – Arkandias Mar 16 '14 at 16:57
  • $\begingroup$ 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). $\endgroup$ – Arkandias Mar 16 '14 at 17:31
  • $\begingroup$ 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.) $\endgroup$ – user76758 Mar 16 '14 at 17:38

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