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I noticed many papers concerning the theory of smooth representations of connected reductive p-adic groups, omit the mention of the specific parabolic subgroup $P\subseteq G$ used in defining the parabolic induction functor $Ind^G_{P,M}$, and write instead $Ind^G_M$ (where $M$ is some Levi subgroup). My question is why is this justified? are the functors $Ind^G_{P,M}$ and $Ind^G_{Q,M}$ naturally isomorphic for different prabolics $P\neq Q$? If so, how can one show this?

roy yanai
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The functors $Ind^G_P$ and $Ind^G_{P^{op}}$ (I'm assuming you mean normalized parabolic induction) are not the same but they do have the same Jordan--Holder factors. So they are the same in the Grothendieck group. I should also point out that the same is not true for Jacquet modules: $Jac^G_P$ and $Jac^G_{P^{op}}$ do not agree in the Grothendieck group. These issues once caused me a huge headache in a paper I was working on due to some typos I had discovered in the literature relating to this very question. So my challenge to you is to be careful about this in your own work!

Alexander
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  • Thank you so much for the answer. I assume that when you say there are only two choices for parabolic you mean that all parabolics containing a certain Levi are conjugate to either $P$ or $P^{op}$, if so where can this fact be found in the literature? Also could you direct me to a text where the fact that the Jordan-Holder factors agree in such a case is proved? – roy yanai Mar 10 '22 at 09:04
  • @royyanai To be slightly more precise, there are precisely to parabolic subgroups containing a given $L$ as the Levi subgroup of $P$. You can find this as Proposition 14.21 of Borel's book on linear algebraic groups. The intuition, and the naming, come from thinking of describing parabolic subgroups as $P_I$ where $I$ is parabolic subset of roots and so then the opposite is $P_{-I}$. – Alex Youcis Mar 10 '22 at 11:04
  • In general there are more than 2 parabolic subgroups of G containing the Levi subgroup M of P ! For instance, if M is the diagonal torus of GL(n) (e.g. P is a the usual Borel subgroup), then there are n! parabolic subgroups of GL(n) containing M. (For n! is the cardinal of the Weyl group, that is the number of chambers in the standard apartment.) – Paul Broussous Mar 10 '22 at 16:41
  • The fact that the two induced representations share the same Jordan-Holder factors is one of the main result of Bernstein, I. N. ; Zelevinsky, A. V. Annales scientifiques de l'École Normale Supérieure, Série 4, Tome 10 (1977) no. 4, pp. 441-472. (You can get it on NUMDAM.) – Paul Broussous Mar 10 '22 at 16:49
  • @PaulBroussous You are correct, I meant to say that for a given levi $L$ of $P$ there is at most one other parabolic subgroup $P'$ also containing $L$ as a Levi such that $P'\cap P=L$. – Alex Youcis Mar 11 '22 at 01:38
  • @Paul Broussous As far as i can tell the paper you suggested does not contain the desired result. Instead it adresses the properies of induced representations of cuspidal representations, and does not prove the independence of parabolic induction from the specific parabolic used. – roy yanai Mar 11 '22 at 14:45