I am taking a look at the residues of Eisenstein series and have a question about a local computation. Let $k$ be a local field, $G = \operatorname{GL}_2(k)$, and $P$ (resp. $K$) the standard parabolic (resp. maximal) compact subgroup of $G$. Let $I(s)$ be the induced representation consisting of all continuous functions $f: G \rightarrow \mathbb C$ satisfying
$$f( \begin{matrix} t_1 & x \\ 0 & t_2 \end{matrix}g) = |t_1t_2^{-1}|^{(s+1)/2}f(g)$$
for all $g \in G, 0 \neq t_i, x \in \mathbb R$. The intertwining operator $M(s): I(s) \rightarrow I(-s)$ is defined formally by
$$M(s)f(g) = \int\limits_k f(\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} g) dx$$
It converges (I think) for $\operatorname{Re}(s) > 0$, but admits a meromorphic continuation to all $s$ with a pole at $s=0$.
It seems to be the case that when $k = \mathbb R$, $M(1)$ takes $I(1)$ into $I(-1)^K$, which is to say $M(1)f(gk) = M(1)f(g)$ for all $g \in G, k \in K$ (proof below in the answer). This is interesting, because $s=1$ is a pole of the global intertwining operator, given by tensoring the local operators $M(s)$ together. Does $M(1)$ always map $I(1)$ into $I(-1)^K$, when $k$ is any local field?
Proof when $k=\mathbb R$:
Here $K = \operatorname{O}_2(\mathbb R)$. I want to show that $M(1)f(gk) = M(1)f(g)$ for all $g \in G$ and $k \in K$. By the Iwasawa decomposition, I may assume $g=1$. I will also assume $k$ lies in the special orthogonal group, and the case where it does not just consists of looking at a single reflection matrix. Let me go ahead and write
$$k = \begin{pmatrix} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{pmatrix}$$
and
$$M(1)f(k) = \int_{-\infty}^{\infty} f(\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & -x \\ 0 & 1 \end{pmatrix}k) \space dx$$ where I have already changed $x \mapsto -x$. Now using the Iwasawa decomposition, I write
$$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & -x \\ 0 & 1 \end{pmatrix} = pk'$$
where
$$k' = \begin{pmatrix} \frac{x}{\sqrt{x^2+1}} & \frac{1}{\sqrt{x^2+1}} \\ \frac{-1}{\sqrt{x^2+1}} & \frac{x}{\sqrt{x^2+1}} \end{pmatrix}$$
$$p = \begin{pmatrix} \sqrt{x^2+1}^{-1} \\ & \sqrt{x^2+1} \end{pmatrix} \begin{pmatrix} 1 & \frac{x}{x^2+1} \\ & 1 \end{pmatrix}.$$
I also make the substitution $\tan \theta = \frac{1}{x}$, so that
$$M(1)f(k) = \int_{-\infty}^{\infty} (1+x^2)^{-1} f( k'k) \space dx = \int_{-\infty}^{\infty} (1+x^2)^{-1} f( \begin{pmatrix} \cos(\theta + \phi) & -\sin(\theta + \phi) \\ \sin(\theta + \phi) & \cos(\theta + \phi) \end{pmatrix})\space dx$$
$$= 2\int_{0}^{\infty} (1+x^2)^{-1} f( \begin{pmatrix} \cos(\theta + \phi) & -\sin(\theta + \phi) \\ \sin(\theta + \phi) & \cos(\theta + \phi) \end{pmatrix})\space dx.$$
Finally, we differentiate $\theta = \arctan (1/x)$ to get
$$\frac{d\theta}{dx} = - \frac{1}{1+x^2}$$
so that
$$M(1)f(k) = 2\int_{-\pi/2}^{\pi/2}-f( \begin{pmatrix} \cos(\theta + \phi) & -\sin(\theta + \phi) \\ \sin(\theta + \phi) & \cos(\theta + \phi) \end{pmatrix}) \space d\theta $$ $$= 2\int_{-\pi/2}^{\pi/2}-f( \begin{pmatrix} \cos(\theta ) & -\sin(\theta ) \\ \sin(\theta ) & \cos(\theta ) \end{pmatrix}) \space d\theta = M(1)f(1).$$