2
$\begingroup$

I want to ask some basic two questions on the parabolic induction.

Let $F$ be a local fields.

Let $\chi_1,\chi_2$ be two characters of $GL_1(F)$ and $GL_1 \times GL_1$ be the Levi part of the parabolic subgroup $P$ of $GL_2$. Let $v$ be the character of $GL_2$ composed with the determinant map and the absolute value.

  1. Then which representation is isomorphic to $(\text{Ind}_P^{GL_2} (\chi_1 \boxtimes \chi_2)) \cdot v$ or its quotient? Is it $\text{Ind}_P^{GL_2} (\chi_1|\cdot|\boxtimes \chi_2|\cdot|)$?

  2. Consider $\text{Ind}_{P_{(1,2,1)}}^{GL_4} (\chi_1|\cdot| \boxtimes \text{Ind}_{P}^{GL_2}(\chi_1 \boxtimes \chi_2) \boxtimes \chi_1 |\cdot|^{-1})$. Then is its irreducible quotients isomorphic to $\text{Ind}_{P_{(3,1)}}^{GL_4} (\chi_1 \circ \det_{GL_3}\boxtimes \chi_2))$?

Any comments on this will be highly appreciated!

$\endgroup$
5
  • 1
    $\begingroup$ Dear Monty, the detail doesn't come to my mind right away (nor for the other question), but I highly recommend reading Bernstein-Zelevinsky if your $F$ is non-arch. (I personally don't enjoy reading paper much, but still!). To me it's like: if all rep'ns are semisimple, then things just follow from Mackey theory, and BZ has the trick that bypass most or all of this difficulty. $\endgroup$ Commented Jul 15, 2018 at 22:02
  • 1
    $\begingroup$ Dear Tsai, Thank you for your comments. Then do you think that the things I wrote is right? I have read BZ paper but I couldn’t find the relevant contents with my question. But I will try to read it again. Thank you very much! $\endgroup$
    – Monty
    Commented Jul 17, 2018 at 4:49
  • 1
    $\begingroup$ I think about it a bit. I think the first one should be $\mathrm{Ind}_P^{GL_2}(\chi_1|\cdot\boxtimes \chi_2|\cdot|)$, if so it should follow directly from definition? For the second, this is where we like to know about BZ's result on intertwining. I also don't think that's the correct one, but I would suggest to look at the detail of BZ, at least first 4 sections. $\endgroup$ Commented Jul 17, 2018 at 7:27
  • 1
    $\begingroup$ Dear Tsai, Thank you! I think the first one you wwrote is right! The paper of BZ you meant is math.tau.ac.il/~bernstei/Publication_list/publication_texts/…? If so, I confused it with another paper. Ok. I will read the 4 sections of the paper. Thank you very much! $\endgroup$
    – Monty
    Commented Jul 17, 2018 at 8:05
  • 1
    $\begingroup$ Ouch. I am sorry. Yes I did mean this linked paper. $\endgroup$ Commented Jul 17, 2018 at 8:30

1 Answer 1

2
$\begingroup$

I will address part 2, as part 1 is addressed in the comments. There are also two types of induction: normalized and un-normalized. I will assume you mean normalized induction.

First of all, it is clear that

$$\mathrm{Ind}_{P_{(2,1,1)}}^{GL_4}(\nu\chi_1\boxtimes\mathrm{Ind}_P^{GL_2}(\chi_1\boxtimes\chi_2)\boxtimes\nu^{-1}\chi_1)=\mathrm{Ind}_{B}^G(\nu\chi_1\boxtimes\chi_1\boxtimes\chi_2\boxtimes\nu^{-1}\chi_1).$$ Now, Theorem 1.9 of Zelevinsky "Induced representations of reductive $\mathfrak p$-adic groups. II." tells you the order of the tensor product does not change the irreducible costituentsof your representations. Thus, you might as well consider irreducible constituents of

$$\mathrm{Ind}_{B}^G(\nu^{-1}\chi_1\boxtimes\chi_1\boxtimes\nu\chi_1\boxtimes\chi_2)=\mathrm{Ind}_{P_{(3,1)}}^G(\mathrm{Ind}_B^G(\nu^{-1}\chi_1\boxtimes\chi_1\boxtimes\nu\chi_1)\boxtimes\chi_2).$$ The representation $\mathrm{Ind}_B^G(\nu^{-1}\chi_1\boxtimes\chi_1\boxtimes\nu\chi_1)$ has length $4$, including $\chi_1\circ\det_{GL_3}$, $\chi_1\otimes \mathrm{St}_{GL_3}$ as subquotients. There is more work to be done to determine when the subquotient $\mathrm{Ind}_{P_{(3,1)}}^G(\chi_1\circ\det_{GL_3}\boxtimes\chi_2)$ is irreducible, which involves case work on whether $\chi_2=\nu^{-2}\chi_1,\nu^{-1}\chi_1,\chi_1,\nu\chi_2,\nu^2\chi_1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .