Skip to main content
MathOverflow

Return to Answer

added 9 characters in body
Source Link
Kenta Suzuki
Kenta Suzuki
  • 3.1k
  • 1
  • 9
  • 32

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 St_{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 $\chi_2=\nu^{-2}\chi_1,\nu^{-1}\chi_1,\chi_1,\nu\chi_2,\nu^2\chi_1$.

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 $\chi_1\circ\det_{GL_3}$, $\chi_1\otimes St_{GL_3}$ 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 $\chi_2=\nu^{-2}\chi_1,\nu^{-1}\chi_1,\chi_1,\nu\chi_2,\nu^2\chi_1$.

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 $\chi_2=\nu^{-2}\chi_1,\nu^{-1}\chi_1,\chi_1,\nu\chi_2,\nu^2\chi_1$.

Source Link
Kenta Suzuki
Kenta Suzuki
  • 3.1k
  • 1
  • 9
  • 32

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 $\chi_1\circ\det_{GL_3}$, $\chi_1\otimes St_{GL_3}$ 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 $\chi_2=\nu^{-2}\chi_1,\nu^{-1}\chi_1,\chi_1,\nu\chi_2,\nu^2\chi_1$.