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I am trying to make sure the meaning of the meromorphic continuation of the intertwining operators.

Assume we deal with a non-Archimedean field $F$ and just consider $G= SL_2$, for simplicity. We fix a maximal split torus $T$. Let $B_1$ and $B_2$ be two Borel subgroups, containing $T$. Then for any quasi-character $\chi : F^{\times} \to \mathbb{C}^{\times}$, we can consider the intertwining operator $M(\chi) : Ind_{B_1}^G\chi \to Ind_{B_2}^G\chi$, which is given by integral \begin{align*} (Mf)(g)=\int_{N_1 \cap N_2\backslash N_2}f(ng)dn. \end{align*} Classical results claim the intertwining operator has meromorphic continuation. But I got confused with the exact meaning of this.

I think it might mean that $M$ takes holomorphic sections to meromorphic sections and all of these meromorphic sections share a common denominator. To be more detailed, it is given as follows. View all of the quasi-characters of $F^{\times}$ as copies of $\mathbb{C}/\frac{2\pi i\mathbb{Z}}{ln(q)}$, indexed by all of the characters $\eta$ of $O^{\times}$. We concentrate then, on each fixed $\eta$. To put this in a more classical statement, for any $f \in Ind_{B_1}^G\eta$, we consider the associated standard section $f^{(s)}$ of $Ind_{B_1}^G\eta_s, \eta_s=\eta|\centerdot|^s$. For any fixed g, we consider $(Mf^{(s)})(g)$ as a function of $s$. Then there should exist some $a(s,\eta)\in \mathbb{C}[q^s,q^{-s}]$, such that $(Mf^{(s)})(g)a(s,\eta)$ is holomorphic for any $g\in G$ and $f\in Ind_{B_1}^G\eta$.

Is this understanding correct? And is there any reference that states the exact definition of the meromorphic continuation?

If this understanding is correct, then I would like to know how to see this intuitively. Since for any $f \in Ind_{B_1}^G\eta$, $f$ is assumed to be right $K$-finite, then we can just deal with finite many $g\in G$. So we have a common denominator for fixed $f$ and all $g \in G$. Then how could we have a common denominator for all $f \in Ind_{B_1}^G\eta$?

I am trying to make sure the meaning of the meromorphic continuation of the intertwining operators.

Assume we deal with a non-Archimedean field $F$ and just consider $G= SL_2$, for simplicity. We fix a maximal split torus $T$. Let $B_1$ and $B_2$ be two Borel subgroups, containing $T$. Then for any quasi-character $\chi : F^{\times} \to \mathbb{C}^{\times}$, we can consider the intertwining operator $M(\chi) : Ind_{B_1}^G\chi \to Ind_{B_2}^G\chi$, which is given by integral \begin{align*} (Mf)(g)=\int_{N_1 \cap N_2\backslash N_2}f(ng)dn. \end{align*} Classical results claim the intertwining operator has meromorphic continuation. But I got confused with the exact meaning of this.

I think it might mean that $M$ takes holomorphic sections to meromorphic sections and all of these meromorphic sections share a common denominator. To be more detailed, it is given as follows. View all of the quasi-characters of $F^{\times}$ as copies of $\mathbb{C}/\frac{2\pi i\mathbb{Z}}{ln(q)}$, indexed by all of the characters $\eta$ of $O^{\times}$. We concentrate then, on each fixed $\eta$. To put this in a more classical statement, for any $f \in Ind_{B_1}^G\eta$, we consider the associated standard section $f^{(s)}$ of $Ind_{B_1}^G\eta_s, \eta_s=\eta|\centerdot|^s$. For any fixed g, we consider $(Mf^{(s)})(g)$ as a function of $s$. Then there should exist some $a(s,\eta)\in \mathbb{C}[q^s,q^{-s}]$, such that $(Mf^{(s)})(g)a(s,\eta)$ is holomorphic for any $g\in G$ and $f\in Ind_{B_1}^G\eta$.

Is this understanding correct? And is there any reference that states the exact definition of the meromorphic continuation?

If this understanding is correct, then I would like to know how to see this intuitively. Since for any $f \in Ind_{B_1}^G\eta$, $f$ is assumed to be right $K$-finite, then we can just deal with finitely many $g\in G$. So we have a common denominator for fixed $f$ and all $g \in G$. Then how could we have a common denominator for all $f \in Ind_{B_1}^G\eta$?

I am trying to make sure the meaning of the meromorphic continuation of the intertwining operators.

Assume we deal with a non-Archimedean field $F$ and just consider $G= SL_2$, for simplicity. We fix a maximal split torus $T$. Let $B_1$ and $B_2$ be two Borel subgroups, containing $T$. Then for any quasi-character $\chi : F^{\times} \to \mathbb{C}^{\times}$, we can consider the intertwining operator $M(\chi) : Ind_{B_1}^G\chi \to Ind_{B_2}^G\chi$, which is given by integral \begin{align*} (Mf)(g)=\int_{N_1 \cap N_2\backslash N_2}f(ng)dn. \end{align*} Classical results claim the intertwining operator has meromorphic continuation. But I got confused with the exact meaning of this.

I think it might mean that $M$ takes holomorphic sections to meromorphic sections and all of these meromorphic sections share a common denominator. To be more detailed, it is given as follows. View all of the quasi-characters of $F^{\times}$ as copies of $\mathbb{C}/\frac{2\pi i\mathbb{Z}}{ln(q)}$, indexed by all of the characters $\eta$ of $O^{\times}$. We concentrate then, on each fixed $\eta$. To put this in a more classical statement, for any $f \in Ind_{B_1}^G\eta$, we consider the associated standard section $f^{(s)}$ of $Ind_{B_1}^G\eta_s, \eta_s=\eta|\centerdot|^s$. For any fixed g, we consider $(Mf^{(s)})(g)$ as a function of $s$. Then there should exist some $c(s,\eta)\in \mathbb{C}[q^s,q^{-s}]$, such that $(Mf^{(s)})(g)c(s,\eta)$ is holomorphic for any $g\in G$ and $f\in Ind_{B_1}^G\eta$.

Is this understanding correct? And is there any reference that states the exact definition of the meromorphic continuation?

If this understanding is correct, then I would like to know how to see this intuitively. Since for any $f \in Ind_{B_1}^G\eta$, $f$ is assumed to be right $K$-finite, then we can just deal with finite many $g\in G$. So we have a common denominator for fixed $f$ and all $g \in G$. Then how could we have a common denominator for all $f \in Ind_{B_1}^G\eta$?

I am trying to make sure the meaning of the meromorphic continuation of the intertwining operators.

Assume we deal with a non-Archimedean field $F$ and just consider $G= SL_2$, for simplicity. We fix a maximal split torus $T$. Let $B_1$ and $B_2$ be two Borel subgroups, containing $T$. Then for any quasi-character $\chi : F^{\times} \to \mathbb{C}^{\times}$, we can consider the intertwining operator $M(\chi) : Ind_{B_1}^G\chi \to Ind_{B_2}^G\chi$, which is given by integral \begin{align*} (Mf)(g)=\int_{N_1 \cap N_2\backslash N_2}f(ng)dn. \end{align*} Classical results claim the intertwining operator has meromorphic continuation. But I got confused with the exact meaning of this.

I think it might mean that $M$ takes holomorphic sections to meromorphic sections and all of these meromorphic sections share a common denominator. To be more detailed, it is given as follows. View all of the quasi-characters of $F^{\times}$ as copies of $\mathbb{C}/\frac{2\pi i\mathbb{Z}}{ln(q)}$, indexed by all of the characters $\eta$ of $O^{\times}$. We concentrate then, on each fixed $\eta$. To put this in a more classical statement, for any $f \in Ind_{B_1}^G\eta$, we consider the associated standard section $f^{(s)}$ of $Ind_{B_1}^G\eta_s, \eta_s=\eta|\centerdot|^s$. For any fixed g, we consider $(Mf^{(s)})(g)$ as a function of $s$. Then there should exist some $a(s,\eta)\in \mathbb{C}[q^s,q^{-s}]$, such that $(Mf^{(s)})(g)a(s,\eta)$ is holomorphic for any $g\in G$ and $f\in Ind_{B_1}^G\eta$.

Is this understanding correct? And is there any reference that states the exact definition of the meromorphic continuation?

If this understanding is correct, then I would like to know how to see this intuitively. Since for any $f \in Ind_{B_1}^G\eta$, $f$ is assumed to be right $K$-finite, then we can just deal with finite many $g\in G$. So we have a common denominator for fixed $f$ and all $g \in G$. Then how could we have a common denominator for all $f \in Ind_{B_1}^G\eta$?