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$\def\anonabs{\lvert\cdot\rvert}\DeclareMathOperator\GL{GL}\DeclareMathOperator\Ind{Ind}\DeclareMathOperator\SO{SO} $Let $F$ be a $p$-adic local field of characteristic zero.

Let $\chi$ be an unramified character of $F^{\times}$ and $\anonabs$ the absolute value.

Then it is well known that the normalized induced representation $\Ind_{B_2}^{\GL_2}(\chi\anonabs^{1/2} \boxtimes \chi\anonabs^{-1/2})$ has a sub-representation $\chi\circ \det_{\GL_2}$, where $B_2$ is a Borel subgroup of $\GL_2$.

Let $\pi$ be an unramified admissible smooth representation of $\SO_n$.

Then I am wondering that whether the set of all irreducible unramified constituents of $\Ind_{(\GL_1 \times \GL_1) \times \SO_n}^{\SO_{n+4}}((\chi\anonabs^{1/2} \boxtimes \chi\anonabs^{-1/2}) \boxtimes \pi)$ are the same as those of $\Ind_{\GL_2 \times \SO_n}^{\SO_{n+4}}(\chi\circ \det_{\GL_2} \boxtimes \pi)$. If is it true, how does the unramified condition play a role?

(Here, $(\GL_1 \times \GL_1) \times \SO_n$ and $\GL_2 \times \SO_n$ are Levi subgroups of parabolic subgroups of $\SO_{n + 4}$, respectively.)

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    $\begingroup$ Surely you mean to consider parabolic induction (thus requiring you to pick two parabolic subgroups of $\operatorname{SO}_{n + 4}$), not induction directly from the Levi? $\endgroup$
    – LSpice
    Commented Jun 9, 2020 at 17:36
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    $\begingroup$ first, it seems that the one dimensional representation of GL(2) is a quotient of your given induced representation if I am not mistaken. The twisted steinberg representation should be the sub. Second, that assertion is true. Here unramified plays a role because the representation induced from an unramified character has only one unramified component. in the GL2 case in your example, that is the one dimensional representation if the given character is unramified. The steinberg representation is not unramified, and thus the representation induced from St has no unramified component. $\endgroup$
    – Q-Zh
    Commented Jun 9, 2020 at 18:52
  • $\begingroup$ @LSpice, Yes! I mean parabolic induction! $\endgroup$
    – Monty
    Commented Jun 11, 2020 at 0:03
  • $\begingroup$ @QZ0, Thank you very much! Right! I should have said that the one dimension representation is a quotient of the induced representation. For the second, you mean the set of all irreducible unramified constituents of $\pi_1=\text{Ind}_{(GL_1 \times GL_1) \times SO_n}^{SO_{n+4}}((\chi |\cdot|^{1/2} \boxtimes \chi|\cdot|^{-1/2}) \boxtimes \pi)$ are the same as that of $\pi_2=\text{Ind}_{GL_2 \times SO_n}^{SO_{n+4}}(\chi\circ \det_{GL_2} \boxtimes \pi)$. But the set of irreducible ramified representations of $\pi_2$ might be properly contained in that of $\pi_1$. Am I right? $\endgroup$
    – Monty
    Commented Jun 11, 2020 at 15:55

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