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Monty
Monty
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Let $F$ be a number field and $G$ a symplectic group over $F$.

Let $P=MN$ is a maximal parabolic subgroup of $G$ and $W_M=N_G(M)/M$. Since $P$ is maximal, $W_M \simeq S_2$. Let $w$ be a non-trivial element of $W_M$.

Let $\sigma$ is a cuspidal representation of $M$.

Consider the intertwining operator $M_w(s,\sigma) : I(\sigma,s) \to I(w\cdot\sigma,-s)$ defined by $M_w(s,\sigma)(f_s)(g)=\int_{N(F)\backslash N(\mathbb{A})} f_s(wng) dn$, where where $I(\sigma,s)$ is the normalized induced representation induced from $\sigma$.

For $f_s \in I(\sigma,s)$, $M_w(s,\sigma)(f_s)$ has a meromorphic continuation on $\mathbb{C}$. Let $M_{1,w}(s,\sigma)$ be an operator sending $f_s$ to the leading term of $M_w(s,\sigma)(f_s)$ in its Laurant expension. Then it is known that $M_{1,w}(s,\sigma) : I(\sigma,s) \to I(w\cdot\sigma,-s).$

It is also known that $M_{1,w}(-s,w \cdot\sigma) \circ M_{1,w}(\sigma,s)=id$. Since it holds for all $s \in \mathbb{C}$ and cuspidal $\sigma$, if we input $-s, w\cdot \sigma$ instead $s,\sigma$, then we have $M_{1,w}(s,\sigma) \circ M_{1,w}(-s,w\cdot \sigma)=id$. From these two equality, I think that we can deduce $M_{1,w}(\sigma,s)$ is isomorphism.

But in some book, the author says that the intertwing operator may have non-trivial kernel.

Which one is right?

(I am very sorry for explaining in detail the notation here. But the experts or person who are familiar with intertwining operator might easily catch the point I am confusing.

Thank you very much!

Let $F$ be a number field and $G$ a symplectic group over $F$.

Let $P=MN$ is a maximal parabolic subgroup of $G$ and $W_M=N_G(M)/M$. Since $P$ is maximal, $W_M \simeq S_2$. Let $w$ be a non-trivial element of $W_M$.

Let $\sigma$ is a cuspidal representation of $M$.

Consider the intertwining operator $M_w(s,\sigma) : I(\sigma,s) \to I(w\cdot\sigma,-s)$, where $I(\sigma,s)$ is the normalized induced representation induced from $\sigma$.

For $f_s \in I(\sigma,s)$, $M_w(s,\sigma)(f_s)$ has a meromorphic continuation on $\mathbb{C}$. Let $M_{1,w}(s,\sigma)$ be an operator sending $f_s$ to the leading term of $M_w(s,\sigma)(f_s)$ in its Laurant expension. Then it is known that $M_{1,w}(s,\sigma) : I(\sigma,s) \to I(w\cdot\sigma,-s).$

It is also known that $M_{1,w}(-s,w \cdot\sigma) \circ M_{1,w}(\sigma,s)=id$. Since it holds for all $s \in \mathbb{C}$ and cuspidal $\sigma$, if we input $-s, w\cdot \sigma$ instead $s,\sigma$, then we have $M_{1,w}(s,\sigma) \circ M_{1,w}(-s,w\cdot \sigma)=id$. From these two equality, I think that we can deduce $M_{1,w}(\sigma,s)$ is isomorphism.

But in some book, the author says that the intertwing operator may have non-trivial kernel.

Which one is right?

(I am very sorry for explaining in detail the notation here. But the experts or person who are familiar with intertwining operator might easily catch the point I am confusing.

Thank you very much!

Let $F$ be a number field and $G$ a symplectic group over $F$.

Let $P=MN$ is a maximal parabolic subgroup of $G$ and $W_M=N_G(M)/M$. Since $P$ is maximal, $W_M \simeq S_2$. Let $w$ be a non-trivial element of $W_M$.

Let $\sigma$ is a cuspidal representation of $M$.

Consider the intertwining operator $M_w(s,\sigma) : I(\sigma,s) \to I(w\cdot\sigma,-s)$ defined by $M_w(s,\sigma)(f_s)(g)=\int_{N(F)\backslash N(\mathbb{A})} f_s(wng) dn$, where $I(\sigma,s)$ is the normalized induced representation induced from $\sigma$.

For $f_s \in I(\sigma,s)$, $M_w(s,\sigma)(f_s)$ has a meromorphic continuation on $\mathbb{C}$. Let $M_{1,w}(s,\sigma)$ be an operator sending $f_s$ to the leading term of $M_w(s,\sigma)(f_s)$ in its Laurant expension. Then it is known that $M_{1,w}(s,\sigma) : I(\sigma,s) \to I(w\cdot\sigma,-s).$

It is also known that $M_{1,w}(-s,w \cdot\sigma) \circ M_{1,w}(\sigma,s)=id$. Since it holds for all $s \in \mathbb{C}$ and cuspidal $\sigma$, if we input $-s, w\cdot \sigma$ instead $s,\sigma$, then we have $M_{1,w}(s,\sigma) \circ M_{1,w}(-s,w\cdot \sigma)=id$. From these two equality, I think that we can deduce $M_{1,w}(\sigma,s)$ is isomorphism.

But in some book, the author says that the intertwing operator may have non-trivial kernel.

Which one is right?

(I am very sorry for explaining in detail the notation here. But the experts or person who are familiar with intertwining operator might easily catch the point I am confusing.

Thank you very much!

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Monty
Monty
  • 1.8k
  • 9
  • 9

Intertwining operator is not an isomorphism?

Let $F$ be a number field and $G$ a symplectic group over $F$.

Let $P=MN$ is a maximal parabolic subgroup of $G$ and $W_M=N_G(M)/M$. Since $P$ is maximal, $W_M \simeq S_2$. Let $w$ be a non-trivial element of $W_M$.

Let $\sigma$ is a cuspidal representation of $M$.

Consider the intertwining operator $M_w(s,\sigma) : I(\sigma,s) \to I(w\cdot\sigma,-s)$, where $I(\sigma,s)$ is the normalized induced representation induced from $\sigma$.

For $f_s \in I(\sigma,s)$, $M_w(s,\sigma)(f_s)$ has a meromorphic continuation on $\mathbb{C}$. Let $M_{1,w}(s,\sigma)$ be an operator sending $f_s$ to the leading term of $M_w(s,\sigma)(f_s)$ in its Laurant expension. Then it is known that $M_{1,w}(s,\sigma) : I(\sigma,s) \to I(w\cdot\sigma,-s).$

It is also known that $M_{1,w}(-s,w \cdot\sigma) \circ M_{1,w}(\sigma,s)=id$. Since it holds for all $s \in \mathbb{C}$ and cuspidal $\sigma$, if we input $-s, w\cdot \sigma$ instead $s,\sigma$, then we have $M_{1,w}(s,\sigma) \circ M_{1,w}(-s,w\cdot \sigma)=id$. From these two equality, I think that we can deduce $M_{1,w}(\sigma,s)$ is isomorphism.

But in some book, the author says that the intertwing operator may have non-trivial kernel.

Which one is right?

(I am very sorry for explaining in detail the notation here. But the experts or person who are familiar with intertwining operator might easily catch the point I am confusing.

Thank you very much!