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$\DeclareMathOperator\GL{GL}$Let $(\pi,V)$ be an infinite dimensional irreducible admissible representation of $\GL_2(\mathbb{Q}_p)$. Let us fix an element $v_0\in V$ and define a vector space $$V_{v_0} := \left\{\pi\begin{pmatrix}a& \\ &1\end{pmatrix}v_0,\;\text{s.t.}\;a\in \GL_1(\mathbb{Q}_p)\right\}.$$

We define the representation \begin{align}\GL_1(\mathbb{Q}_p)&{}\to \operatorname{Aut}(V_{v_0}),\\ a&{}\mapsto \pi\begin{pmatrix}a& \\ &1\end{pmatrix}.\end{align} It is irreducible and admissible since $$\begin{pmatrix}\mathbb{Z}_p^{\times}& \\ &1\end{pmatrix}\subseteq GL_2(\mathbb{Z}_p).$$

On the one hand applying Schur's lemma for admissible representations (Lemma 4.2.4 of "Automorphic forms and representations" of Bump) then the representation of $\GL_1(\mathbb{Q}_p)$ should satisfy that $$\pi\begin{pmatrix}a& \\ &1\end{pmatrix}v_0 = \chi(a)v_0,$$ where $\chi$ is a character of $\GL_1(\mathbb{Q}_p)$.

On the other hand, let us suppose that $\pi\simeq \operatorname{Ind}_{P_{\GL_2}}^{\GL_2}\xi$, where $\xi$ is a character defined on the diagonal elements of $\GL_2(\mathbb{Q}_p)$ such that $\pi$ is irreducible. We denote their Satake parameters by $\alpha_1,\;\alpha_2$ and the Whittaker functional by $W(\cdot)$. Theorem 4.6.5 of "Automorphic forms and representations" of Bump states that $$W\left(\pi\begin{pmatrix}p^k& \\ &1\end{pmatrix}v_0\right) = W(v_0)\frac{\alpha_1^{k+1}-\alpha_2^{k+1}}{\alpha_1-\alpha_2}.$$ The function $$p^k\to \frac{\alpha_1^{k+1}-\alpha_2^{k+1}}{\alpha_1-\alpha_2},$$ is not a character for $(p)$. This contradicts the previous Schur's lemma argument. Where is the mistake in those computations?

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    $\begingroup$ Hi! Why should your representation of $GL_1(\mathbb{Q}_p)$ be irreducible? $\endgroup$
    – Aurel
    Commented Nov 15, 2021 at 11:02
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    $\begingroup$ I don’t understand your first claim. What is $V_{v_0}$ supposed to be? As currently written, it isn’t a vector space. Even once that is clarified, I don’t get the irreducibility argument. What does the statement about units tell you? $\endgroup$ Commented Nov 15, 2021 at 11:15
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    $\begingroup$ You seem to be claiming that a cyclic representation (i.e. generated by a single element) is always irreducible. This is clearly false: think about the regular representation of $C_2$ (the cyclic group of order $2$). $\endgroup$
    – Aurel
    Commented Nov 15, 2021 at 11:16
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    $\begingroup$ Are you assuming $v_0$ is a spherical vector? That might be the issue: the spherical vector is scaled by a character when acted on by compact diagonal elements, but not the full torus. The Whittaker formula (a special case of the Casselman-Shalika-Shintani formula) tells you how the full torus acts. $\endgroup$ Commented Nov 15, 2021 at 11:18
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    $\begingroup$ The set $V_{v_0}$ as you define it is still not a vector space (not stable under linear combinations!). In my comments I interpreted it as the vector space generated by the set your wrote. $\endgroup$
    – Aurel
    Commented Nov 15, 2021 at 11:19

1 Answer 1

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$\DeclareMathOperator\GL{GL}$Let me try to clarify. The formula for the Whittaker functional in Theorem 4.6.5 of "Automorphic forms and representations" of Bump states that $$W\left(\pi\begin{pmatrix}p^k& \\ &1\end{pmatrix}v_0\right) = W(v_0)\frac{\alpha_1^{k+1}-\alpha_2^{k+1}}{\alpha_1-\alpha_2},$$ where $v_0$ is a $\GL_2(\mathbb{Z}_p)$-stable vector in the induced representation.

Your claim is then that the cyclic representation generated by $\GL_1(\mathbb{Q}_p)$-translates of $v_0$ is always irreducible as a $\GL_1(\mathbb{Q}_p)$ representation. As Aurel points out in the comments, this is certainly false in general. In the case related to the above formula, the fact that $v_0$ is $\GL_2(\mathbb{Z}_p)$-stable does imply that $\mathbb{Z}_p^\times$ acts by a character since $$\begin{pmatrix}\mathbb{Z}_p^{\times}& \\ &1\end{pmatrix}\subseteq GL_2(\mathbb{Z}_p),$$ but this says nothing about the action of $$ \begin{pmatrix}p^{k}& \\ &1\end{pmatrix} $$ for $k>0$. Those values are determined by the formula.

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