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Jul 15, 2018 at 5:13 comment added Monty @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/…
Jul 15, 2018 at 0:07 comment added Cheng-Chiang Tsai @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.
Jul 14, 2018 at 16:44 comment added paul garrett 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.
Jul 14, 2018 at 12:02 comment added Monty @Tsai, Thank you very much! What is the ‘nice’ do you mean?
Jul 14, 2018 at 7:55 comment added Cheng-Chiang Tsai 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.
Jul 14, 2018 at 7:54 history edited Martin Sleziak CC BY-SA 4.0
typo in the title
Jul 14, 2018 at 7:21 history edited Monty CC BY-SA 4.0
edited title
Jul 14, 2018 at 6:59 history asked Monty CC BY-SA 4.0