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.


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.