4
$\begingroup$

We assume that $(\pi,V)$ is an admissible, irreducible and infinite-dimensional representation of $GL_2(\mathbb{Q_p})$. In the proof of existence and uniqueness of Kirillov model, the key step is that we give the subspace $V_0 \subseteq V$ defined by \begin{equation*} V_0:= \left \{ v \in V \Bigg| \int_{p^{-n}\mathbb{Z_p}} e_p(-u)\pi \left( \begin{pmatrix} 1&u\\ 0&1 \end{pmatrix} \right) \cdot v\;du=0,\;\text{for all big}\;n \right \} \end{equation*} and prove that $\dim(V/V_0)=1$. Here $e_p$ is the nontrivial additive character from $\mathbb{Q_p}$ to $S^1$. Now we fix a vector $v \in V$ and for each positive integer $n$, we define the integral $I(n)$ as follows: \begin{equation*} I(n):=\int_{p^{-n}\mathbb{Z_p}} e_p(-u)\pi \left( \begin{pmatrix} 1&u\\ 0&1 \end{pmatrix} \right) \cdot v\;du. \end{equation*} We want to show that there exists a big positive integer $N$, s.t. for arbitrary $n>N$, we have $I(n)=I(n+1)=I(n+2)=\cdots$. In other word, we want to prove the stability of the given integral.

My idea is following: Using the substitution rule and the topological structure of $p^{-n}\mathbb{Z_p}$, for any positive integer $n$, we have \begin{equation*} I(n+1)=\left(\sum_{j=0}^{p-1} e_p\left(-\frac{j}{p^{n+1}}\right)\pi \left( \begin{pmatrix} 1&\frac{j}{p^{n+1}}\\ 0&1 \end{pmatrix} \right)\right) I(n). \end{equation*} Now if there exists some big positive integer $N$ s.t. $I(N)=0$, then immediately we have $0=I(n)=I(n+1)=I(n+2)=\cdots$. But if the value is not $0$, I have almost no idea. Can we also get the stability of the integral from the above equation? Or maybe we need more relation between $I(n)$ and $I(n+1)$?

$\endgroup$
  • $\begingroup$ Are you sure that the proof need the value of I(n) to be stable in all cases ? $\endgroup$ – Paul Broussous Nov 16 '17 at 12:44
  • $\begingroup$ Computations with the Kirillov model show that you can find $v\in V$ such that $I(n)$ is not constant for $n\geqslant N$, $N$ large enough. $\endgroup$ – Paul Broussous Nov 21 '17 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.