MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F$ be a locally compact non-archimedean field and $G_{n}$ the locally profinite group $GL(n,F)$. Let $\Gamma_{n,k}$ be the subgroup of $G_{n}$ whose elements are the matrices of the form $$ \begin{pmatrix} A_{1,1} & A_{1,2} \\ \\ A_{2,1} & A_{2,2} \\ \end{pmatrix} $$ where $A_{1,1}\in GL(n-1,O_{F})$, $A_{1,2}\in M_{n-1,1}(O_{F})$, $A_{2,1}\in M_{1,n-1}(p_{F}^{k})$ and $A_{2,2}\in 1+p_{F}^{k}$. Here, $p_F$ denotes the maximal ideal in the ring of integers $O_F$ of $F$.

Let $(\pi,V)$ be a generic representation of $G_{n}$. We know that the space $V^{\Gamma_{n,k}}$ of fixed vectors is non-zero for $k$ large enough. Moreover, if $c(\pi) = \min\{ k\in\mathbb{N} : V^{\Gamma_{n,k}}\neq 0 \}$ ($c(\pi)$ is the conductor of $\pi$) then $\dim(V^{\Gamma_{n,c(\pi)}})=1$. Reference: Jacquet, Piatetski-Shapiro, Shalika, "Conducteur des représentations du groupe linéaire", Math. Ann. 256 (1981).

My question concerns replacing the subgroups $\Gamma_{n,k}$ by small subgroups $P_{n,k}$ whose elements are the upper-triangular matrices mod $p_{F}^{k}$. More precisely, if $\varphi:GL(n,O_{F})\longrightarrow GL(n,O_{F}/p_{F}^{k})$ is the morphism of reduction mod $p_{F}^{k}$, define $P_{n,k}=\varphi^{-1}(B)$, where $B$ is the standard Borel subgroup of $GL(n,O_{F}/p_{F}^{k})$.

It is clear that $V^{P_{n,k}}\neq 0$ for $k$ large enough. Denote $u(\pi) = \min\{ k\in\mathbb{N} : V^{P_{n,k}}\neq 0 \}$.

Question 1: Is it true that $\dim(V^{P_{n,u(\pi)}})=1$?

Question 2: If that is false for a generic representation, does it hold for only a supercuspidal representation of $GL(n,F)$?

share|cite|improve this question
Please include all hyphens present in existing tags (otherwise you create new ones). And, please do not include the number given after the tag. The 15 you used as a tag is the number of times the tag 'p-adic-groups' was used already; thus it makes no sense to include this. – user9072 Jun 13 '13 at 12:12

To my knowledge, it is not known whether $Ind_{P_{n,k}}^{GL_n(o)} 1$ decomposes with single multiplicity.

This is certainly necessary by Frobenius reciprocity $$ dim Hom_{P_{n,k}}( 1 , Res_{P_{n,k}} \pi) = dim Hom_{GL_n(F)}( Ind^{GL_n(F)} Ind_{P_{n,k}}^{GL_n(o)} 1, \pi).$$

For more information, see Parabolic induction GL(n,Zp)

share|cite|improve this answer
Thank you very much for your answer and also for the link. – Rajkarov Jun 13 '13 at 20:12

The group $P_{n,k}$ is not "smaller" than $\Gamma_{n,k}$ as claimed in the question. The $A_{n,n}$ for $\Gamma_{n,k}$ must be in $1+\mathfrak{p}^k$ whereas for $P_{n,k}$ it can be arbitrary element of $\mathcal{O}^\times$. Regardless, the question makes sense. The case $n=2$ is treated in Casselman's paper on the method of Atkin and Lehner. He shows that the answer to Question 1 is positive, and he defines the conductor that way.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.