Let $G=GL(n, F)$, $B$ be a Borel subgroup and let $B=AN$ be the Langlands decomposition. Let $\nu \in \mathfrak{a}^\*\_{\mathbb{C}}$ be in the positive Weyl chamber. Consider the normalized induced representation $I = I^G\_B(\nu)$. Let $F=\mathbb{R}$ and $J$ be the unique generic irreducible subrepresentation of $I$ which exists thanks to Theorem 6.2 for $F=\mathbb{R}$ of the article by Casselman-Shahidi titled "On irreducibility of standard modules for generic representations". It is known that the intertwining operator
$$
M(w\_0) : I^G\_B(\nu) \rightarrow I^G\_{\bar{B}}(w\_0(\nu)),
$$
corresponding to the longest Weyl group element annihilates every nonzero subrepresentation of $I^G\_B(\nu)$. I was wondering if something more can be said here. For example


1. Is $J$ annihilated by all intertwining operators $M(w)$ where $w\in W$? I am mostly interested in the case of $\nu = \rho$ where $\rho$ is the half sum of positive roots.
2. In the same article the authors have proved analogue of Theorem 6.2 for nonarchimedean field and can be found as Theorem 1. What can we say about the analogue of the above theorem when $F$ is nonarchimedean field?
3. If more general results are known or expected in this direction, a reference would be much appreciated.


Any comments, suggestions or references are welcome. Please feel free to edit if you find something erroneous in the question. Thank you.



Intertwining operators and induced representation
I will answer question 1 for when $\nu=\rho$ and $F$ is $p$-adic (since I am not too familiar with $F=\mathbb R$; I assume a similar argument works for $F=\mathbb R$ as well). It suffices to check that $I\_B^G(\rho)$ is not isomorphic to $I\_B^G(w\rho)$ for $w\ne1\in W$. But this is clear, since
$$\begin{align\*}
\hom\_G(I\_B^G(\rho),1\_G)&\cong\hom\_G(1\_G,I\_B^G(-\rho))\\
&\cong\hom\_G(1\_G,\mathrm{Ind}\_B^G(1\_B))\\
&\cong\hom\_B(1\_B,1\_B)\\
&\cong\mathbb C,
\end{align\*}$$
where $\mathrm{Ind}\_B^G$ denotes naive induction, but
$$\begin{align\*}
\hom\_G(I\_B^G(w\rho),1\_G)&\cong\hom\_G(1\_G,I\_B^G(-w\rho))\\
&\cong\hom\_G(1\_G,\mathrm{Ind}\_B^G(\rho-w\rho))\\
&\cong\hom\_B(1\_B,\rho-w\rho)\\
&=0,
\end{align\*}$$
since $\rho\ne w\rho$.


