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Let $k$ be a nonarchimedean locally compact field, $G$ a connected reductive group over $k$ with minimal parabolic subgroup $P = MN$. Let $\sigma$ be a smooth representation of $M$. In chapter 3 of Casselman's notes on representation theory, he writes "If $P$ is a minimal parabolic subgroup and $\sigma$ is irreducible (hence necessarily finite dimensional)..."

Why does $P$ being a minimal parabolic imply that $\sigma$ is finite dimensional? This is clear if $P$ is a Borel subgroup, because then $M$ is commutative, making $\sigma$ one dimensional. In general, $M$ is a reductive group. This seems to indicate that there are constraints on which reductive groups can occur as Levi subgroups of minimal parabolic subgroups..

Edit: Let $S$ be a maximal split torus of $G$ which is contained in a minimal parabolic $P$. Then the set of roots of $S$ in $M$ is empty. So the question comes down to the following:

If $G$ is a connected, reductive group over $k$ with maximal split torus $S$, and $S$ is contained in the center of $G$, then every smooth irreducible representation of $G$ is finite dimensional.

D_S
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  • This comes from the fact that if $G$ and $S$ are as in your last paragraph, then $G/S$ is anisotropic, whence compact. Then using Schur Lemma and the smoothness of your representation, you are done. – Paul Broussous Jul 03 '18 at 19:06
  • If $T$ is a maximal torus of $G$ containing $S$, then $T/S$ is a maximal torus of $G/S$ which is anisotropic. But why should $G/S$ itself be compact? – D_S Jul 03 '18 at 21:38
  • Any maximal $k$-split torus of $G/S$ is trivial. Hence $G/S$ is an anisotropic reductive group. – Paul Broussous Jul 04 '18 at 03:18

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