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Let $P$ be a parabolic subgroup of a connected, reductive group $G$ over a $p$-adic field. Let $M$ be a Levi subgroup of $P$, and let $N$ be the unipotent radical of $P$. If $(\pi,V)$ is a smooth, irreducible representation of $M$, extend $\pi$ to a representation of $P$ by making it trivial on $N$, and let $\sigma = \operatorname{Ind}_P^G \pi$, the smooth representation of $G$ obtained by parabolic induction.

By definition, a function $f: G \rightarrow V$ lies in the space of $\sigma$ if the following conditions are met:

  • $f$ is locally constant.

  • $f(mng) = \pi(m)f(g)$ for all $m \in M, n \in N, g \in G$.

  • There exists an open compact subgroup $K$ of $G$, depending on $f$, such that $f(gk) = f(g)$ for all $g \in G$ and $k \in K$.

Is the third condition redundant in this definition? I know in the general case for smooth induction in totally disconnected groups, it is necessary, but I have thought that since $P \backslash G$ is compact, there should be some way to show the third condition from the first two. I haven't been able to do this. I have seen some authors leave out the third condition in the definition of parabolic induction.

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Let $H$ be any subgroup such that $H\backslash G$ is compact. Let $K$ be an open subgroup. Then there are $x_1,...x_n$ such that $G=Hx_1 K \cup...\cup Hx_nK.$ Suppose $f$ satisfies points 1 and 2. Replace $K$ with a smaller $K'$ so that $f(x_i k)=f(x_i)$ for all $i$ and for all $k \in K'.$ Given $a \in G,$ there are $h \in H, i,$ and $k' \in K'$ such that $a=hx_i k'$. Let $k \in K'$. Then $f(ak)=f(h x_i k' k)= \pi(h)f(x_i k' k)=\pi(h)f(x_i k')=f(h x_i k')=f(a).$

So $f$ satisfies point 3.

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