Just like the title. I want a simple proof of the statment in the title. $\mathbb{Q}_p$ is the p-adic field. I wonder which module (or vector space) will be chosen as the space for the representation. Is this statement true for arbitarily module/vector space? Thanks!
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2$\begingroup$ Just a remark: as Kevin and Brian's arguments below show, it is smoothness, rather than admissibility, which is the key assumption. (If the representation is finite-dimensional, admissibility will be automatic!) $\endgroup$– EmertonCommented Aug 3, 2010 at 15:42
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What does "admissible" mean for you? Does it imply smoothness (stabilisers are open)? If not then I think the statement might be false (choose some hopelessly discontinuous injection from $\mathbf{Q}_p$ into $\mathbf{C}$ and then consider the induced "natural" representation of $GL(2,\mathbf{Q}_p)$ on $\mathbf{C}^2$; that's definitely irreducible, and the space of vectors fixed by some compact open will definitely be finite-dimensional).
But assuming smoothness too, it is true that any finite-dimensional smooth irreducible representation of $GL(2,\mathbf{Q}_p)$ is 1-dimensional. The proof is: if $V$ is such a thing, then choose a basis for $V$ and for each basis vector choose a compact open subgroup stabilising it. The intersection of these guys is still compact and open, and fixes everything. So the kernel of the representation contains a compact open subgroup. But this is a bit worrying because the kernel is normal. Now use the fact that the normal subgroup generated by matrices $(1,e;0,1)$ and $(1,0;e,1)$ for $e$ small is still the whole of $SL(2,\mathbf{Q}_p)$ to deduce that $SL(2,\mathbf{Q}_p)$ is in the kernel, and now the action has to factor through $GL(2,\mathbf{Q}_p)/SL(2,\mathbf{Q}_p)=\mathbf{Q}_p^\times$. But now Schur's Lemma, which works for smooth irreducible representations, says $V$ is 1-dimensional.
answered Aug 3, 2010 at 13:30
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4$\begingroup$ Here's the proof that ${\rm{SL}}_2$ case (for which Kevin's argument works over any non-arch. local field $k$) implies case of any $k$-split conn'd reductive $G$. Let $G' \rightarrow D(G)$ be the $k$-split simply conn'd central cover of derived gp, so $G'(k) \rightarrow G(k)$ has normal image with abelian quotient. By Schur can replace $G$ with $G'$ to reduce to showing the only finite-dim. irred. adm. rep'n of $G(k)$ is trivial one when $G$ is simply conn'd. The cocharacter lattice of split max. $k$-torus is gen'td by coroots since $G$ is s.c., so $G(k)$ is gen'td by ${\rm{SL}}_2(k)$'s. QED $\endgroup$– BCnrdCommented Aug 3, 2010 at 13:59
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$\begingroup$ to BCnrd: I was just too stupid at that time, being unable to figure out why it is enough to prove only for SL2(k)...your illustration is clear. Thanks! $\endgroup$– AlgSoulCommented Aug 3, 2010 at 15:10