Part of some generic representation is also generic?

$$\DeclareMathOperator\SL{SL}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\GL{GL}$$Let $$F$$ be local field of characteristic zero and $$(W,\langle,\rangle)$$ be a $$2n$$-dimensional symplectic space over $$F$$.

Let $$X,X^*$$ be maximal totally isotropic subspaces of $$W$$, which are dual with respect to $$\langle,\rangle$$.

Let $$e_1,\dotsc,e_n$$ be a basis of $$X$$ and $$f_1,\dotsc,f_n$$ be a basis of $$X^*$$. Let $$B_{n}=M_nU_n$$ be the parabolic subgroup of $$\Sp(W)$$ stabilizing the flag $$\langle e_1\rangle \subset \langle e_1, e_2\rangle \subset \dotsb \subset\langle e_1, e_2, \dotsc, e_{n}\rangle.$$

Let $$\{ \chi_i \}$$ be characters of $$F^{\times}$$ and $$U_n$$ the unipotent radical of $$B_n$$.

Let $$\pi$$ be the irreducible generic unramified representation of $$\Sp(W)$$ that is a subquotient of $$\operatorname{Ind}_{B_{n}}^{\Sp(W)} (\chi_1 \otimes \dotsb \chi_{n} )$$. (Here, the induction is normalized.)

Then for any $$1 \le i \le n$$, let $$W_i=\left$$ and $$B_{i}$$ be the parabolic subgroup of $$\Sp(W_i)$$ stabilizing $$\left$$. Let $$\pi_i=\operatorname{Ind}_{B_{i}}^{\Sp(W_i)} \chi_i$$.

Then I am wondering whether the irreducible unramified constituent of $$\pi_i$$ is also generic. Is it true? And I also guess $$\pi_i$$ is irreducible.

Let $$\pi$$ be the irreducible generic unramified representation of $$Sp(W)$$ that is a subquotient of $$Ind(\chi_1, \dots, \chi_n)$$.
I think the key here is to realise that this does not exist for all values of $$\chi$$. If the induction is irreducible (which is true for "sufficiently general" tuples of unramified characters $$(\chi_1, \dots, \chi_n)$$), then it is generic and unramified. However, when the induction is reducible, there is usually no subquotient which is unramified and generic at the same time; usually there will be a unique "largest" factor which is the generic one, and a unique "smallest" factor which is the unramified one. For instance, if $$n = 1$$ and $$\chi_1 = |\cdot|^{1/2}$$, then the induction has 2 factors, the trivial rep and the Steinberg; the former is unramified but not generic, and the latter is generic but not unramified!
If you take $$n = 1$$ and $$\chi_1 = 1$$, then the representation is again reducible and the two factors are "the same size" in some sense -- they are representations of $$SL_2$$ which are conjugate in $$GL_2$$. In this case, exactly one of the two factors is generic, and exactly one of the two is spherical. However, which one has which property depends on how you interpret the definitions! There are two conjugacy classes of hyperspecial max compacts in $$SL_2(F)$$ [hence two meanings of "unramified"] and two conj classes of Whittaker characters [hence two meanings of "generic"]. So the question is still not well-defined enough to be answerable.
If you work with "sufficiently general" characters that the induced rep of $$G$$ is irreducible, then the $$\pi_i$$ are all irreducible too. Maybe that's not a very interesting statement, but I'm struggling to find an interpretation of the question which makes sense in the reducible cases.
• Dear professor, I am very sorry for late reply because I just saw your comment. Thank you very much for such very nice comment. I learned much from your comment. Since I am considering $\pi$, the local unramified component of a global tempered representation, the exponents of the standard module of $\pi$ are zero and so they are in ‘general position’ in the sense of Lemma 2.8 of the paper hazu.hr/~tadic/33-LMT-gen-dual.pdf ? Nov 6, 2021 at 17:54
• I am wondering that the term “general position” in the paper is exactly what you mean ‘sufficiently general characters’ that the induced rep of G is irreducible? But the Lemma 2.8 assumes first that the standard module of $\pi$ is generic. Without knowing $\pi$ is generic, can we know whether $\pi$ is irreducible or not? If you don’t mind, may I ask you that where is the definition of “sufficiently general characters” of a standard module which ensures its irreducibility? Thank you very much! Nov 6, 2021 at 18:00