Is there a known bound on the norm of the standard intertwining operator for the principal series of $G = \operatorname{GL}_2(\mathbb Q_p)$?

Background: For a character $\chi = (\chi_1,\chi_2)$ of the standard torus $T$ in $G$, extended to a character of the standard Borel subgroup $B = TU$ we can define the induced representation $I(\chi) = \operatorname{Ind}_B^G \chi$ to be the Banach space of all measurable functions $f: G \rightarrow \mathbb C$ satisfying $f(tng) = \chi(t)\delta_B^{1/2}(t)f(g)$ and the norm condition $$\lVert f\rVert^2 = \int\limits_K \lvert f(k)\rvert^2 dk < \infty$$ where $K = \operatorname{GL}_2(\mathbb Z_p)$ (usually, one works with the subspace $I(\chi)^{\infty}$ of locally constant functions in $I(\chi)$). If $w = \begin{pmatrix} & 1 \\ -1 \end{pmatrix}$, the standard intertwining operator $M: I(\chi) \rightarrow I(w(\chi))$ is densely defined on $I(\chi)^{\infty}$ by $$Mf(g) = \int\limits_N f(wng)dn.$$ This converges for $\sigma = \operatorname{Re}(\chi_1\chi_2^{-1}) > 0$, but can be extended to make sense for most $\chi_1$, $\chi_2$ by a process of analytic continuation.

What I want to do:

I want to compute, or get an upper bound on, the norm $\lVert M\rVert = \sup\limits_{\lVert f\rVert=1} \lVert Mf\rVert$. Actually, I'm trying to show that the global intertwining operator for $\operatorname{GL}_2(\mathbb A_{\mathbb Q})$, densely defined as a product of local intertwining operators, is a bounded linear operator. The norm of the global intertwining operator should be a convergent product of the norms of the local operators.

Computations for $\lVert{-}\rVert_1$:

If we use the $1$-norm instead of the $2$-norm, I can get an estimate for $\lVert M\rVert_1$. There may be a way to modify these calculations for the $2$-norm, but I haven't been able to do it. For $f$ locally constant, \begin{gather*} \lVert Mf\rVert = \int\limits_K \lvert Mf(k)\rvert dk \leq \int\limits_K \int\limits_N \lvert f(wnk)\rvert dn\,dk= \int\limits_{N(\mathbb Z_p)} \int\limits_K \lvert f(wnk)\rvert dk\,dn + \int\limits_{n \not\in N(\mathbb Z_p)} \int\limits_K \lvert f(wnk)\rvert dk\,dn \\ =\int\limits_{N(\mathbb Z_p)} \int\limits_K \lvert f(k)\rvert dk\,dn + \int\limits_{x \in \mathbb Q_p - \mathbb Z_p} \int\limits_K \left\lvert f(w \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix}k)\right\rvert dk\,dx = \lVert f\rVert_1 + \int\limits_{x \in \mathbb Q_p - \mathbb Z_p} \int\limits_K \left\lvert f(w \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix}k)\right\rvert dk\,dx. \end{gather*} This last integral is \begin{gather*} \int\limits_{x \in \mathbb Q_p - \mathbb Z_p} \int\limits_K \left\lvert f( \begin{pmatrix} -x^{-1} & 1 \\ & -x\end{pmatrix} \begin{pmatrix} 1 \\ x^{-1} & 1 \end{pmatrix} k)\right\rvert dk\,dx = \int\limits_{x \in \mathbb Q_p - \mathbb Z_p} \int\limits_K \lvert x\rvert^{-\sigma} \lvert f(k)\rvert dk\,d^{\ast}x = \sum\limits_{k=1}^{\infty} \lVert f\rVert_1 \int\limits_{\lvert x\rvert = p^k} \lvert x\rvert^{-\sigma}d^{\ast}x \\ = \lVert f\rVert_1 \operatorname{Vol}(\mathbb Z_p^{\ast}) \frac{p^{-\sigma}}{1- p^{-\sigma}}. \end{gather*}

So we see that for $\sigma = \operatorname{Re}(\chi_1\chi_2^{-1}) > 0$, the $1$-norm $\lVert M\rVert_1$ is bounded by $1 + \operatorname{Vol}(\mathbb Z_p^{\ast}) \frac{p^{-\sigma}}{1- p^{-\sigma}}$.

  • $\begingroup$ I did some tidying in the TeX. At one point, you said that you're bounding $\lVert f\rVert_1$ when I'm pretty sure you're bounding $\lVert M\rVert_1$. Later, you say you've accomplished a bound on $\lVert M f\rVert_1$, which is true, but the point seemed to be the resulting bound on $\lVert M\rVert_1$. I changed both accordingly. $\endgroup$
    – LSpice
    Apr 30, 2021 at 16:04
  • $\begingroup$ Why doesn't the same argument give you $\lVert M\rVert_2^2 \le 1 + \operatorname{Vol}(\mathbb Z_p^\times)\frac{p^{-2\sigma}}{1 - p^{-2\sigma}}$? $\endgroup$
    – LSpice
    Apr 30, 2021 at 16:05
  • 2
    $\begingroup$ The integrand must be squared and that seems to disrupt things considerably $\endgroup$
    – D_S
    Apr 30, 2021 at 16:06
  • $\begingroup$ Ah, right, the very first integral is $\int_K \left(\int_N \cdots\right)^2$. I wasn't paying attention to where that square went. $\endgroup$
    – LSpice
    Apr 30, 2021 at 16:07

1 Answer 1


Having a bound on $\|M\|_1$, you get by duality a bound on $\|M\|_\infty$ and then you get a bound on $\|M\|_p$ for every $1\leq p\leq\infty$ by the Riesz–Thorin theorem.


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