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In his Cuspidal geometry of p-adic groups [J. Anal. Math. 47, 1–36 (1986)], Kazhdan uses a standard (?) result about representations of $p$-adic groups, which I will try to restate here.

Let $F$ be a $p$-adic field and let $G$ denote the $F$-points of a reductive group. We assume G has compact center. Let $R_{\mathbb{Z}}(G)$ denote the Grothendieck group of smooth representations of $G$, set $R(G)=R_{\mathbb{Z}}(G)\otimes_{\mathbb{Z}} \mathbb{C}$, and let $R_I(G)$ denote the subgroup of $R(G)$ generated by representations of $G$ which are parabolically induced.

Fix a special maximal compact $K_0$ in $G$ and let $K \leq K_0$ be a congruence subgroup. Now let $R(G)_K$ (resp. $R_I(G)_K$) denote the subgroup of $R(G)$ (resp. $R_I(G)$) generated by representations which are themselves generated by their $K$-fixed vectors.

Lemma 1 of Kazhdan's paper states that $R(G)_K/R_I(G)_K$ is then finite-dimensional.

Question: Why is this true? Alternatively, is there a good reference for this result that applies to arbitrary reductive groups (as opposed to, say, general linear groups)?

Kazhdan's proof is simply a reference to Proposition 3.8 of the Bernstein-Deligne notes on the Bernstein center. As far as I can tell, this reference allows one to reduce the above question to a single Bernstein block, but I am not sure why it would imply the above Lemma.

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  • $\begingroup$ How about the following? Either the Bernstein block is cuspidal or parabolically induced. If it is induced then the quotient is zero so there is nothing to prove. Otherwise the block is cuspidal, in which case there are only finitely many twists of level $K$. $\endgroup$
    – Kenta Suzuki
    Commented Sep 9 at 22:44
  • $\begingroup$ Hm, maybe I'm misunderstanding what you're saying, but it is the induced blocks that are more complicated, because $R_I(G)$ is only generated by fully induced representations. Take $G=GL_2(F)$ for example: we have $\text{Ind}_B^G(1) = \text{St} + 1$. Here $\text{Ind}_B^G(1)$ is equal to $0$ in the quotient, but $1$ and $\text{St}$ are non-zero (and in fact $1 = - \text{St}$). $\endgroup$
    – bakulator
    Commented Sep 10 at 14:29
  • $\begingroup$ Yes sorry you are right! $\endgroup$
    – Kenta Suzuki
    Commented Sep 10 at 20:52

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Remark: In the Kazhdan paper you are asking about, there is a running assumption of compact center. The result you want holds iff the center is compact.

With this assumption, here is a proof. Both $R(G)$ and $R_I(G)$ have compatible gradings by points $\theta$ in the Bernstein variety. Following the terminology in Bernstein-Deligne-Kazhdan's trace Paley-Wiener paper, say $\theta$ is discrete if $R_I(G)_\theta \subsetneq R(G)_\theta$. By the Proposition in section 3.1 of that paper and our assumption on the center, the discrete $\theta$'s meet each Bernstein component in a finite set. Since each $R(G)_\theta$ is finitely generated, this gives the result.

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  • $\begingroup$ By the way, $R(G)$ is not a ring in this setting. It is just the free abelian group generated by smooth irreps of $G$. $\endgroup$
    – Satan's Minion
    Commented Sep 10 at 18:26
  • $\begingroup$ Thanks for the reference, this looks good! Agreed, I failed to mention the compact center assumption. Also, I don't know why I said ring. Will edit the original question. $\endgroup$
    – bakulator
    Commented Sep 10 at 22:22
  • $\begingroup$ It's my pleasure! $\endgroup$
    – Satan's Minion
    Commented Sep 11 at 1:16

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