Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 1 / 3

Write $d=\mathrm{diag}({\varpi }^{-(n-1)},\dots ,1)$ for the diagonal matrix called $\delta$ in this question. If $v\in V^K$ and $k_i=d^{i}k{d}^{-i}\in K_i$, thenThus $\rho (d^i)$ maps the fixed-vector space $V^K$ into $V^{K_i}$. Its inverse $\rho (d^{-i})$ gives the reverse map, so

For an upper-unipotent matrix $u$, inverse conjugation changes each entry above the diagonal byEvery such factor tends to zero as $i\to \infty$. Because $U_0$ is compact, its finitely many coordinate functions are uniformly bounded; because it is open, it contains a neighbourhood of the identity in $U$. Therefore this contraction is uniform on $U_0$ and gives $d^{-i}{U}_0{d}^i\subseteq U_0$ for every sufficiently large $i$. The diagonal group commutes with $d$, soConjugating this inclusion gives $K\subseteq K_i$ for all sufficiently large $i$. This is an eventual inclusion; arbitrary $U_0$ need not give a monotone sequence at every earlier step.

Consequently $V^{K_i}\subseteq V^K$ eventually. Admissibility makes these spaces finite-dimensional, while the isomorphism above makes their dimensions equal. HenceThis proves contraction and admissible fixed spaces by both inclusion and dimension, not by an assumption that conjugate fixed spaces are literally equal from the outset.

Take any $v\in V$. A smooth representation has an open stabilizer, so some sufficiently small congruence subgroup of $B$ fixes $v$. It can be written $K=T_m{U}_m$, where the diagonal entries in $T_m$ belong to $1+{\mathfrak{p}}^m$ and all entries above the diagonal in $U_m$ belong to ${\mathfrak{p}}^m$, with $m\ge 1$. These are compact-open in their respective groups, $T_m$ normalizes $U_m$, and their product is compact-open in $B$.

By part (a), this vector lies in $V^{K_i}$ for every sufficiently large $i$. Given any $u\in U$, the coordinate contraction $d^{-i}u{d}^i\to 1$ eventually puts it in $U_m$. Choose $i$ large enough for both properties. Then $u\in d^i{U}_m{d}^{-i}\subseteq K_i$, so $\rho (u)v=v$. Since $u$ and $v$ were arbitrary,Thus every admissible representation of this upper-triangular group factors through its diagonal quotient. The argument uses finite-dimensional fixed spaces essentially; smoothness alone does not force the unipotent action to be trivial.