How do you construct elements in $\operatorname{Ind}_P^G\pi$?

Question

Let $G$ be a $p$-adic reductive group, and $P=MN\subseteq G$ a parabolic subgroup. How do you know that the space of the induced representation $\operatorname{Ind}_P^G\pi$ is non-zero? Namey, how do you know there even exists a nonzero map $f: G\to V_\pi$ satisfying the defining property for $f\in\operatorname{Ind}_P^G\pi$? Is there any explicity way to write down each such $f$ in terms of the give data $(G, P, \pi)$?

Answer 1

To have it written explicitly, $\pi$ is a smooth representation of $M$, extended trivially across $N$ to $P$.

Choose an open subgroup $K_M$ of $M$ that is so small that $\pi^{K_M} \ne 0$, a $K_M$-fixed vector $v \ne 0$, and a compact, open subgroup $K$ of $G$ such that $K \cap P$ is contained in $K_M N$, and define $f(p k) = \pi(p)v$ for all $p \in P$ and $k \in K$. The extension by $0$ of $f$ to $G$ belongs to $\operatorname{Ind}_P^G\pi$.

Not only does this procedure produce non-$0$ elements in the space of the induced representation, but, in fact, the elements so produced span the induced-representation space. Since $G/P$ is compact, the argument is essentially the same as the corresponding fact that $C^\infty(\mathscr O)$, where $\mathscr O$ is the ring of integers of the underlying $p$-adic field, is spanned by characteristic functions of balls.