The problem:
Let $G = \mathrm{GL}_2(\mathbb Q_p)$ and $k$ be an algebraically closed field of characteristic $p.$ Denote by $\overline B$ the subgroup of all lower triangular matrices in $G$ and by $U$ the subgroup of all matrices of the form $\begin{bmatrix} 1 & \star \\ 0 & 1 \end{bmatrix}$. Let $\chi_1, \chi_2 \colon \mathbb Q_p^{\times} \to k^{\times}$ be two smooth characters. Consider the smooth representation
$$ \chi \colon \overline B \to k^\times,\quad \begin{bmatrix} \alpha & 0 \\ \gamma & \delta\end{bmatrix} \mapsto \chi_1(\alpha) \chi_2(\delta).$$
This induces a smooth $G$-representation $\operatorname{Ind}_{\overline B}^G \chi.$ Show that there is a vector space isomorphism:
\begin{align*}\{f \in \operatorname{Ind}_{\overline B}^G \chi \colon \operatorname{Supp} f \subseteq \overline B U\} &\longrightarrow \mathcal C_{\mathrm{cpt}}^{\infty}(\mathbb Q_p, k)\\ f & \longmapsto \left(f \mapsto f \begin{bmatrix}1 & x \\ 0 & 1\end{bmatrix}\right),\end{align*}
where $\mathcal C_{\mathrm{cpt}}^{\infty}(\mathbb Q_p, k)$ is the space of all locally constant, compactly supported functions $\mathbb Q_p \to k.$
I understand that the image of any $f$ in the LHS is a locally constant function. But I cannot see how to show that the image is also compactly supported. Injectivity of the map is obvious. For surjectivity, we can do the following:
Let $\phi$ be any locally constant, compactly supported, $k$-valued map from
$\mathbb Q_p.$
Define $f \colon G \to k$ as follows–
First define $f \colon U \to k$ as
$f\begin{bmatrix} 1 & x \\ 0 & 1\end{bmatrix} = \phi(x)$
Note that $\overline B \cap U = \{1\}.$
Hence we may can extend it to a map $f \colon \overline B U \to k,$
$f \left(T\begin{bmatrix} 1 & x \\ 0 & 1\end{bmatrix}\right) = T\phi(x),\text{ for
all } T \in \overline B.$
Just extend by zero to get a function $f \colon G \to k.$ I need to show that $f$ is fixed by an open subgroup.
Any help will be appreciated.
Best Answer
Given a $f\in\mathrm{Ind}_{\overline B}^G(\chi)$ let $Z:=\{\overline Bg\in\overline B\backslash G:f(g)\ne0\}$, which is well-defined since $f$ is by definition left $\overline B$-invariant. Moreover, $Z\subset \overline B\backslash G$ is a closed subset, so $Z$ is compact. Thus, $Z\cap \overline B\backslash \overline BU$ is also compact, and hence $f|U$ also has compact support.
For the converse, let $g\colon U\to k$ have compact support. Then, we may extend this to $\overline B U$ by left $\overline B$-equivariance (which is well-defined since $\overline B\cap U=1$). Take $N\gg0$ such that the function $g$ is $\mathfrak p^N$-invariant and $\chi_1,\chi_2$ are trivial on $1+\mathfrak p^N$.
For $K_N:=1+M_2(\mathfrak p^N)$, the function $g$ extends to $UK_N\to k$, since $g$ is $U\cap K_N$-invariant. Now, we may further extend $g$ to $\overline BUK_N$ by $\overline B$-equivariance since $\overline B\cap UK_N\subset\begin{pmatrix}1+\mathfrak p^N&*\\&1+\mathfrak p^N\end{pmatrix}$. But $\overline BUK_N=G$. Now it is clear from definition that $f$ is fixed by a compact open subgroup.