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Let $G=GL_n(F)$, where $F$ is a p-adic local field, $U$ be the upper triangular maximal unipotent group, and $\theta$ a character of $U$. Then a Theorem of Kazhdan says that for any irreducible smooth representation $(\pi,V)$ of $G$, we have $$\dim V_{U,\theta} $$ has finite dimension, where $V_{U,\theta}$ is the twisted Jacquet functor. This is Theorem 5.21 of "Bernstein-Zelevinski, Representations of $GL(n,F)$, where $F$ is a non-archimedean local field". In fact, in Theorem 5.21, the above dimension is bounded by $n!$.

My question is: is this theorem true for more general groups? If it is, where can I find a proof?

Thanks.

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Yes - this statement is true for any irreducible representation of a split reductive p-adic group $G$, where $U$ is the unipotent radical of the Borel (a Borel subgroup is a minimal algebraic subgroup $B\subset G$ such that $G/B$ is proper; $G$ is split if such a $B$ is defined over your base field). An example is the preimage of $U$ for the standard embedding of $O(n, F), Sp(n, F),$ etc. in $GL(n).$ The source I know this from is Karl Rumelhart's transcription of Bernstein's Harvard course (although the notes only include full proofs for $GL_n$). The general sequence of arguments this follows from is as follows:

  1. an irreducible representation $V$ of a $p$-adic group $G$ is admissible (has finite-dimensional subspace $V^K$ of vectors fixed by any open compact $K\subset G$, and is generated over $G$ by $V^K$ for $K$ sufficiently fine).
  2. If $P$ is a parabolic subgroup of $G$ (such that $G/P$ is compact), $U\subset P$ is its unipotent radical and $L=P/U$ (always a reductive group, called the Levi) then the Jacquet restriction $V_{U,\theta}$ is admissible as a representation of $L$.
  3. An admissible representation of a torus $T$ (i.e., $T(F)$ for $T$ a connected reductive commutative algebraic group) is finite-dimensional. Now just take $T=B/U$.

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  • $\begingroup$ Thanks for your answer. But I have a question on part (2) above. If $\theta$ is trivial, I know that the standard Jacquet functor $V_U$ is an admissible representation of the Levi. But, if $\theta$ is not trivial, the twisted Jacquet functor $V_{U,\theta}$ is not even a representation of the full Levi. Think about the GL(3) case. Let $P$ be the parabolic of type (2,1), i.e., the Levi is $GL(2)\times GL(1)$, with $GL(2)$ at the left upper corner. If $\theta$ is the character defined on the (2,3) position of the unipotent, then $V_{U,\theta}$ is just a representation of the mirabolic $\endgroup$ – user64433 Mar 25 '15 at 3:52
  • $\begingroup$ of the mirabolic of the upper $GL(2)$, not the full Levi. Also, I cannot find the statement which says the twisted Jacquet functor is admissible. Do you know where I can find this? $\endgroup$ – user64433 Mar 25 '15 at 3:54
  • $\begingroup$ Ah sorry, thought you were twisting by a character of the Levi (which would not affect the dimension... your current question makes a lot more sense). $\endgroup$ – Dmitry Vaintrob Mar 25 '15 at 4:55
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Decided to add a new answer instead of editing the old one (which answers a different question). Look at the proof in Bernstein-Zelevinsky. It generalizes to arbitrary split reductive groups. Here's a somewhat simplified (and coarser) argument:

Note that by passing to the contragredient, it suffices to prove your statement for twisted invariants $V^{U,\Theta}$. Now $V$ injects into $i\circ r(V)$ for $r$ the Jacquet restriction and $i$ the induction, both with respect to the Borel. The space $i\circ r(V)$ is canonically the global sections of an equivariant (topological) vector bundle $\mathcal{V}$ on $G/B$ with finite-dimensional fibers isomorphic to $r(V)$, of some dimension $d$. Now observe that $G/B$ has finitely many $U$-orbits, $O_1,\dots, O_N$, and on each orbit, the twisted invariants $\Gamma(\mathcal{V}, O_i)^{U,\Theta}$ are $\le d$-dimensional. Now we have an embedding of $i\circ r(V)$ in $\oplus_i \Gamma(\mathcal{V}, O_i)$ and we are done by left exactness of taking invariants.

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  • $\begingroup$ This argument uses only quasisplitness, not splitness, right? $\endgroup$ – LSpice Jul 10 '17 at 18:11

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