For split groups over a $p$-adic field, every irreducible smooth (complex) representation is either infinite-dimensional or one-dimensional. Is it true for quasisplit groups that split over an unramified extension, or quasisplit groups in general? Unfortunately I don't have a good feel for examples, since the books I consult for learning the representation theory of $p$-adic groups usually stick to groups like $GL_n$.

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    $\begingroup$ @nfdc23 Your comment deserves to be an answer ! $\endgroup$ Apr 5, 2018 at 9:06
  • $\begingroup$ @PaulBroussous, by "rational character" do you mean "abstract homomorphism $G \to \mathbb C^\times$" (or maybe $G \to F^\times$?) or "homomorphism deduced from a rational map of algebraic groups $\boldsymbol G \to \mathrm{GL}_1$"? If the latter, then note that $\boldsymbol G = \mathrm{PGL}_2$ has no such characters (since, for example, the rational characters of $\mathrm{GL}_2$ are all integer powers of the determinant, none of which is trivial on the centre), so that $G^1 = G$; but that the image of $\mathrm{SL}_2(F)$ is an open, normal subgroup. $\endgroup$
    – LSpice
    Apr 8, 2018 at 6:16
  • $\begingroup$ @L.Spice. OK, I deleted my comment. $\endgroup$ Apr 10, 2018 at 12:26

2 Answers 2


See Prop. 3.9 of http://math.stanford.edu/~conrad/JLseminar/Notes/L2.pdf for an optimal affirmative answer (no quasi-split condition needed: any connected reductive group over any non-archimedean local field, with the minimal necessary isotropicity hypotheses).


This is an editted version, as my previous answer had an error.

Every $p$-adic group is totally-disconnected locally compact (tdlc). For the class of tdlc group, every continuous finite dimensional complex representation is through a discrete quotient (that is, the kernel contains an open subgroup). More generally, every continuous homomorphism from a tdlc group to a Lie group (eg $\text{GL}_n(\mathbb{C})$) has this property. To see this, recall that every tdlc group has a compact open subgroup (van Dantzig theorem) and observe that the image of such a group should be a compact (hence closed, hence Lie) tdlc subgroup in the target, so it must be finite.

Let $G$ be the $k$ points of a reductive $k$-algebraic group which has no anisotropic factor. The group that you consider are such, and an affirmative answer to your question will follow from the fact that every discrete quotient of $G$ is abelain. A way to see this is to consider the group generated by all unipotent subgroups of $G$, $G^+$, and to see that every open normal subgroup of $G$ contains $G^+$ and $G/G^+$ is commutative. This is discussed in the work of Borel-Tits and nicely summerized in Margulis book "Discrete subgroups of semisimple Lie groups" Chapter I.1.5. I recommend reading it.

  • $\begingroup$ Only finite dimensional smooth representations have open kernel. Your last paragraph sounds weird. I think the general condition on $G$ in order that all finite dimensional smooth representations are $1$ dimensional is that $G$ has no anisotropic factor. This is true if $G$ is quasi-split. $\endgroup$ Apr 4, 2018 at 9:13
  • $\begingroup$ @PaulBroussous I discuss here finite dim reps. Indeed, I forgot to demand anisotrpoicity - I will edit it. Downvoting is certainly the right thing to do. Thank you. Uri $\endgroup$
    – Uri Bader
    Apr 4, 2018 at 9:32
  • $\begingroup$ Only unramified characters of ${\mathbb Q}_p^\times$ are via ${\mathbb Z}={\mathbb Q}_p^\times /{\mathbb Z}_p^\times$. $\endgroup$ Apr 4, 2018 at 9:33
  • $\begingroup$ It is not necessarily true that the group of rational points of a semisimple group has no (non-trivial) finite quotient, even if the underlying algebraic group is split. For example, the quotient of $\mathrm{PGL}_2(F)$ by the image of $\mathrm{SL}_2(F)$ is the group of square classes $F^\times/(F^\times)^2$. $\endgroup$
    – LSpice
    Apr 8, 2018 at 5:51
  • $\begingroup$ @LSpice yes, I know. But every quotient group is finite abelian when the field is a local field of char 0 and the group is a connected almost simple isotropic. I didn't feel that anybody care about this answer, so left it as it is with this silly mistake. I'll edit... $\endgroup$
    – Uri Bader
    Apr 8, 2018 at 6:09

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