# The norm of the principal series intertwining operator for $\operatorname{GL}_2$

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 $$1 + \operatorname{Vol}(\mathbb Z_p^{\ast}) \frac{p^{-\sigma}}{1- p^{-\sigma}}$$.

• 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. Apr 30, 2021 at 16:04
• Why doesn't the same argument give you $\lVert M\rVert_2^2 \le 1 + \operatorname{Vol}(\mathbb Z_p^\times)\frac{p^{-2\sigma}}{1 - p^{-2\sigma}}$? Apr 30, 2021 at 16:05
• The integrand must be squared and that seems to disrupt things considerably
– D_S
Apr 30, 2021 at 16:06
• 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. Apr 30, 2021 at 16:07

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.