TITLE: local Langlands and the Jacquet module
QUESTION [14 upvotes]: Let $G = GL_n(F)$ be the general linear group over a finite extension $F$ of $Q_p$. This question could be posed for a larger class of groups, but let us stay with $Gl_n$ for the moment. 
Let $\pi$ be a smooth irreducible complex representation of $G$. Let $P \subset G$ be a  parabolic subgroup and $P =MN$ a Levi-decomposition. The Jacquet module $\pi_N$ of $\pi$ is by definition the module of $N$-coinvariants in $\pi$. 
Via the local Langlands correspondence $\pi$ corresponds to a Weil-Deligne representation $\sigma_\pi$. Furthermore, in the cases where $\pi_N$ is irreducible, the representation $\pi_N$ has a corresponding Weil-Deligne representation via local Langlands for $M$.
My question is, does the operation $\pi \to \pi_N$ from $G$-representations to $M$-representations has a "satisfactory" interpretation on the Galois side, via local Langlands for $G$ and $M$?
A point of caution is that $\pi_N$ need not be $M$-irreducible, so it does not go directly into local Langlands for $M$.

REPLY [9 votes]: Let us consider the simple case: $G=GL_2(F)$, $n=2$. (cf. ''The local langlands conjecture for $GL_2(F)$''   C.J. Bushnell and G.Henniart)
In order to tell the story, first we need to give some definitions. Clearly we only need to consider the non-cuspidal case. Let $\chi=\chi_1\otimes \chi_2$ be the character of $T$, we denote $ \chi^{\omega}=\chi_2\otimes \chi_1$, we define $\pi_{\chi}=Ind_B^G(\delta_B^{-\frac{1}{2}}\otimes \chi)$ where $\delta_B$ is the modular function of the group $B$ i,e $\delta_B(tn)=||t_2t_1^{-1}||$ for $t=diag(t_1,t_2)$, $n\in N$, we write $\phi\circ det$, $\phi \cdot St_G$  two other kind of principal series for  $GL_2(F)$.
Now we arrive to  write the Jacquet functor $J: Rep(G) \longrightarrow Rep(T); (\pi, V) \longrightarrow
(\pi_N, V_N)$.
(1) For $\chi_1\chi_2^{-1}\neq ||.||^{\pm}$, $\pi=\pi_{\chi}$ is irreducible,   then
$\pi_N=\delta_B^{-\frac{1}{2}}\otimes \chi \oplus \delta_B^{-\frac{1}{2}}\otimes \chi^{\omega}$.
(2) $\pi=\phi\circ det$, then $\pi_N=\phi\otimes \phi$.
(3) $\pi=\phi \cdot St_G$, then $\pi_N=||.||\phi\otimes ||.||^{-1}\phi$.
We recall some result about local langlands correspondance for general linear  group. We denote $\mathcal{G}_2(F)$ to be the set of equivalence classes of 2-dimensional Frobenius semisimple, Deligne representation of the Weil group $\mathcal{W}_F$; also  $\mathcal{A}_2(F)$ to be the set of equivalence classes of irreducible smooth representations of $GL_2(F)$.  The local langlands correspondance tell us that there is a natural bijective map $l_2$ between $\mathcal{A}_2(F)$ and $\mathcal{G}_2(F)$. The naturality often involves some compatibility conditions. ( For detail one should see the article of Borel in Corvallis).
Assume $\pi$ is irreducible, lying in $\mathcal{A}_2(F)$, we denote $l_2(\pi)=(\rho,W,\mathbf{n})$.
(1) if $\pi=\pi_{\chi}$,  then $\rho=\chi_1 \oplus \chi_2$ and $\mathbf{n}=0$, here we regard $\chi_i$ as the representation of Weil group $\mathcal{W}_F$.
(2) if $\pi=\phi\circ det$, then $\rho=||.||^{-\frac{1}{2}}\phi \oplus ||.||^{\frac{1}{2}}\phi$ and $\mathbf{n}=0$.
(3) if $\pi=\phi \cdot St_G$, then $\rho=||.||^{-\frac{1}{2}}\phi \oplus ||.||^{\frac{1}{2}}\phi$, but in this case
$\mathbf{n}\neq 0$.
Finally we comme to the question that Arno asks. We translate directly  ''the Jacquet functor'' to the Galois side via the local langlands correspondence.
$J: \mathcal {G}_2(F) \longrightarrow \mathcal{G}_1(F)^{\otimes 2}$. More precisely,  the result is outlined as follows:
(1) $J\big((\pi_{\chi}, \mathbf{n}=0)\big)=(\delta_B^{-\frac{1}{2}}\otimes \chi) \oplus (\delta_B^{-\frac{1}{2}}\otimes \chi^{\omega})$;
(2) $J\big((\phi\circ det, \mathbf{n}=0)\big)=\phi\otimes \phi$.
(3) $J\big((\phi \cdot St_G, \mathbf{n}=0)\big)= ||.||\phi\otimes ||.||^{-1}\phi$.
Remark:  for general case,  we  take $\pi \in Irr_{\mathbb{C}}(G)$, one knows $\pi_N$ has finite length and is admissible  as the representation over its Levi subgroup $M$, although we don't even know  it is  semi-simple or not.