Two questions about Whittaker functions

Question

I am watching the video: Modeling p-adic Whittaker functions, Part I. I have two questions about Whittaker functions in the video.

From 33:00 to 37:00, it is said that after changing of variables, we have \begin{align} W^0(t_{\lambda}) = \int_{U} f^0(w_0 u t_{\lambda}) \psi^{-1}(u) du = \int_{\bar{U}} f^0(\bar{u}) \bar{\psi}^{-1}(t_{\lambda} \bar{u} t_{\lambda}^{-1}) d \bar{u}. \quad (1) \end{align} I tried to prove this formula by letting $\bar{u} = w_0 u t_{\lambda}$. But I didn't get $\int_{\bar{U}} f^0(\bar{u}) \bar{\psi}^{-1}(t_{\lambda} \bar{u} t_{\lambda}^{-1}) d \bar{u}$. How to prove (1)?

I am trying to understand the formula of expressing a Whittaker function as a sum over a crystal. In the case of $SL_2$, $\bar{u} = \left( \begin{matrix} 1 & 0 \\ x & 1 \end{matrix} \right)$. In the video, it is said that if $x \in \mathfrak{o}_F$ (the ring of integers of $F$), then it is done; if $x \not\in \mathfrak{o}_F$, we have a decomposition $\left( \begin{matrix} 1 & 0 \\ x & 1 \end{matrix} \right) = \left( \begin{matrix} x^{-1} & 1 \\ 0 & x \end{matrix} \right) \left( \begin{matrix} 0 & -1 \\ 1 & x^{-1} \end{matrix} \right) $ and spherical vector is constant on "shells". What do "shells" mean? Why we need the decomposition of the matrix $\left( \begin{matrix} 1 & 0 \\ x & 1 \end{matrix} \right)$? Why spherical vector is constant on "shells"? Thank you very much.

Answer 1

In the case of $SL_2$, we have $$ U = \left\{ \left(\begin{matrix} 1 & x \\ 0 & 1 \end{matrix}\right) : x \in F \right\} $$ and \begin{align} & \int_U f^0\left( w_0 u t_{\lambda} \right) \psi^{-1}(u) du \\ & = \int_U f^0\left( \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right) \left(\begin{matrix} 1 & x \\ 0 & 1 \end{matrix}\right) \left(\begin{matrix} a & 0 \\ 0 & b \end{matrix}\right) \right) \psi^{-1}(\left(\begin{matrix} 1 & x \\ 0 & 1 \end{matrix}\right)) dx \\ & = \int_U f^0\left( \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right) \left(\begin{matrix} a & 0 \\ 0 & b \end{matrix}\right) \left(\begin{matrix} a^{-1} & 0 \\ 0 & b^{-1} \end{matrix}\right) \left(\begin{matrix} 1 & x \\ 0 & 1 \end{matrix}\right) \left(\begin{matrix} a & 0 \\ 0 & b \end{matrix}\right) \right) \psi^{-1}(\left(\begin{matrix} 1 & x \\ 0 & 1 \end{matrix}\right)) dx \\ & = \int_U f^0\left( \left(\begin{matrix} b & 0 \\ 0 & a \end{matrix}\right) \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right) \left(\begin{matrix} 1 & a^{-1}bx \\ 0 & 1 \end{matrix}\right) \right) \psi^{-1}(\left(\begin{matrix} 1 & x \\ 0 & 1 \end{matrix}\right)) dx \\ & = \chi_1(b) \chi_2(a) \int_U f^0\left( \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right) \left(\begin{matrix} 1 & a^{-1}bx \\ 0 & 1 \end{matrix}\right) \right) \psi^{-1}(\left(\begin{matrix} 1 & x \\ 0 & 1 \end{matrix}\right)) dx. \end{align} Let $\bar{u} = t_{\lambda}^{-1} u t_{\lambda}$. Then we have \begin{align} & \int_U f^0\left( w_0 u t_{\lambda} \right) \psi^{-1}(u) du \\ & = \chi_1(b) \chi_2(a) \int_U f^0\left( \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right) \left(\begin{matrix} 1 & a^{-1}bx \\ 0 & 1 \end{matrix}\right) \right) \psi^{-1}(\left(\begin{matrix} 1 & x \\ 0 & 1 \end{matrix}\right)) dx \\ & = \chi_1(b) \chi_2(a) \int_U f^0(w_0 \bar{u}) \psi^{-1}(t_{\lambda} \bar{u} t_{\lambda}^{-1}) dx \\ & = \chi_1(b) \chi_2(a) \int_U f^0(w_0 \bar{u}) \psi^{-1}(t_{\lambda} \bar{u} t_{\lambda}^{-1}) \frac{1}{a^{-1}b} d(a^{-1}bx) \\ & = \chi_1(b) \chi_2(a) \int_{\bar{U}} f^0(w_0 \bar{u}) \psi^{-1}(t_{\lambda} \bar{u} t_{\lambda}^{-1}) \frac{1}{a^{-1}b} d(\bar{u}) \\ & = \frac{\chi_1(b) \chi_2(a)}{a^{-1}b} \int_{\bar{U}} f^0(w_0 \bar{u}) \psi^{-1}(t_{\lambda} \bar{u} t_{\lambda}^{-1}) d(\bar{u}). \end{align}

For the second question, if $x \in \mathfrak{o}_F$, then $\left(\begin{matrix} 1 & 0 \\ x & 1 \end{matrix}\right) \in G(\mathfrak{o}_F)=K$. Since $f^{0}$ is invariant under right multiplication of $K$, $f^{0}(w_0 \bar{u}) = f^{0}(w_0)$ in this case.

If $x \not\in \mathfrak{o}_F$, then $|x| > 1$ and $$\left( \begin{matrix} 1 & 0 \\ x & 1 \end{matrix} \right) = \left( \begin{matrix} x^{-1} & 1 \\ 0 & x \end{matrix} \right) \left( \begin{matrix} 0 & -1 \\ 1 & x^{-1} \end{matrix} \right). $$ Therefore $|x^{-1}|<1$ and $\left( \begin{matrix} 0 & -1 \\ 1 & x^{-1} \end{matrix} \right) \in G(\mathfrak{o}_F)$. Therefore \begin{align} & f^0(w_0 \left( \begin{matrix} 1 & 0 \\ x & 1 \end{matrix} \right) ) \\ & = f^0(w_0 \left( \begin{matrix} x^{-1} & 1 \\ 0 & x \end{matrix} \right) \left( \begin{matrix} 0 & -1 \\ 1 & x^{-1} \end{matrix} \right)) \\ & = f^0(w_0 \left( \begin{matrix} x^{-1} & 1 \\ 0 & x \end{matrix} \right)) \\ & = f^0(w_0 \left( \begin{matrix} x^{-1} & 0 \\ 0 & x \end{matrix} \right) \left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right)) \\ & = f^0( \left( \begin{matrix} x & 0 \\ 0 & x^{-1} \end{matrix} \right) w_0 \left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right)) \\ & = \chi_1(x) \chi_2(x^{-1}) f^0( w_0 \left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right)). \end{align}