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$\DeclareMathOperator\Ind{Ind}$Let $G$ be a reductive group over a $p$-adic local field $F$, and $P=MN$ a parabolic subgroup.

Let $\sigma$ be an irreducible representation of $M(F)$ and consider its unnormalized induced representation $\Ind_P^G(\sigma)$. Let $\pi$ be a subrepresentation of $\Ind_P^G(\sigma)$. For an arbitrary element $v$ in $\sigma$, we can choose $f \in \pi$ such that $f(e)=v$.

Let $U^-$ be the unipotent radical of the opposite parabolic subgroup $P^-$ of $P$. Then I am wondering whether we can choose $U^-$-invariant $f \in \pi$ such that $f(e)=v$? If $\pi$ is the full induced representation, it is possible because $P \cap U^- = \{e\}$. But I don't know whether it holds for proper sub-representation of $\Ind_P^G(\sigma)$.

Furthermore, I am also wondering whether we can choose $f\in \pi$ so that $f(e)=v$ and $f$ has small support near $e$. This also holds when $\pi$ is the full induced representation. But I don't know whether it is possible for proper sub-representation $\pi$.

Thank you very much!

(PS: I am especially think of the case when $\pi$ is the image of the local intertwining operator $M(s) : \Ind_P^G(\sigma\cdot |\det|^s) \to \Ind_P^G( \sigma\cdot |\det|^{-s})$ defined by $M(s)f(s)(g)= \int_{N}f(wng)dn$, where $w$ is the longest Weyl element. If it is difficult to consider general $\pi$, how about the case when $\pi=M(s)(\Ind_P^G(\sigma\cdot |\det|^s) )$?)

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    $\begingroup$ The image of $\operatorname{Res}^G_M \pi \to \sigma$, $f \mapsto f(e)$, is an $M$-submodule, since $\pi(m)f \mapsto f(m) = \sigma(m)f(e)$. So it is all of $V$ unless every vector in $\pi$ vanishes at $e$, which implies that $\pi$ is the $0$ representation, since $\pi(g)f \mapsto f(g)$ for all $f$ in the space of $\pi$ and all $g \in G$. (Also, I encourage you never to use $\nu$ for a vector; I guarantee it will lead to confusion at some point.) $\endgroup$
    – LSpice
    Commented May 22, 2020 at 15:35
  • $\begingroup$ @LSpice, Thank you very much! It helps me a lot! $\endgroup$
    – Monty
    Commented May 22, 2020 at 15:52
  • $\begingroup$ @LSpice, can we choose such $f$ which is either $U^{-}$-invariant or sufficiently small support near $e$? (here, $U^{-}$ is the unipotent radical of the opposite parabolic subgroup $P^{-}$ of $P$) These are possible if $\pi$ is the full induced representation. But I am wondering if it holds for a proper subrepresentation $\pi$. How do u think of it? $\endgroup$
    – Monty
    Commented May 22, 2020 at 16:04
  • $\begingroup$ I don't have a specific counterexample, but I suspect that there's too much flexibility with subrepresentations to insist that they all must contain functions of small support. As for $U^-$-invariance, choose a compact open subgroup $K$ such that $G = P K$, and replace $f$ by $\int_{U^- \cap K} \pi(u^-)f\,\mathrm du^-$. $\endgroup$
    – LSpice
    Commented May 22, 2020 at 16:26
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    $\begingroup$ @LSpice, Thank you very much for your comments. I revised my question! $\endgroup$
    – Monty
    Commented May 22, 2020 at 16:57

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