Some basic question on the parabolic induction

Let $F$ be a local field and $G=GL_n(F)$ and $P=MN$ its parabolic subgroup whose Levis subgroup $M=GL_{n_1}(F) \times GL_{n_2}(F)$ where $n=n_1 +n_2$.

Write $G_1=GL_{n_1}(F)$ and let $P_1=M_1N_1$ its parabolic subgroup whose Levis subgroup $M=GL_{a}(F) \times GL_{b}(F)$ where $n_1=a+b$. Similarly, write $G_2=GL_{n_2}(F)$ and let $P_2=M_2N_2$ its parabolic subgroup whose Levis subgroup $M=GL_{c}(F) \times GL_{d}(F)$ where $n_2=c+d$.

Let $\sigma_a,\sigma_b,\sigma_c,\sigma_d$ be irreducible representation of $GL_{a},GL_{b},GL_{c},GL_{d}$ respectively.

Then I am wondering the relation between $\text{Ind}^G_P(\text{Ind}^{G_1}_{P_1} (\sigma_a \boxtimes \sigma_b) \boxtimes \text{Ind}^{G_2}_{P_2} (\sigma_c \boxtimes \sigma_d))$ and $Ind^G_{P_{a,b,c,d}}(\sigma_a \boxtimes \sigma_b \boxtimes \sigma_c \boxtimes \sigma_d )$ where $P_{a,b,c,d}$ is the parabolic subgroup of $G$ whose Levi part is $GL_a \times GL_b \times GL_c \times GL_d$.

I guess there willbe some relation between these under some ideal condition, for example irreducibility of induced representation etc.

Any comments on this will be highly appreciated.

• The two representations are canonically isomorphic. The general statement is the following: let $P=MN$ be a parabolic of $G$ and $P'=M'N'$ be a parabolic of $M$. Then the preimage of $P^*:=P'N=M'(N'N)$ is another parabolic of $G$ with $M'$ again as its Levi, and we'd like to say $\mathrm{Ind}_{P^*}^G\sigma=\mathrm{Ind}_P^G\mathrm{Ind}_{P'}^M\sigma$ for any (nice) representation of $M'$. This can be checked from definition. To get your situation we have $P=P_{n_1,n_2}$ and $P'=P_{a,b}\times P_{c,d}$. The same statement is true for normalized induction, as the modular characters will match up. – Cheng-Chiang Tsai Jul 14 at 7:55
• @Tsai, Thank you very much! What is the ‘nice’ do you mean? – Monty Jul 14 at 12:02
• I think the canonical isomorphism is best understood via the associativity of tensor products (albeit over rings with many idempotents, but not units), for the non-archimedean case. This was P. Cartier's approach in his Corvallis notes. – paul garrett Jul 14 at 16:44
• @Monty Pardon that I probably shouldn't stress the adjective; it's nothing serious. When $F$ is non-archimedean (and our representation is over a field in which $p$ is invertible), by "nice" representation we definitely like a smooth representation. When $F=\mathbb{R}$ many know which category to work on much better than me. – Cheng-Chiang Tsai Jul 15 at 0:07
• @Tsai, Thank you very much! Your answer helped me a lot! If you don’t mind, would you please see my earlier question? I think you are the very person who can answer my question. My earlier question is below: mathoverflow.net/questions/304707/… – Monty Jul 15 at 5:13