non-artificial examples of non-smooth and non-admissible representations of GL_2 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:58:57Z http://mathoverflow.net/feeds/question/109048 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109048/non-artificial-examples-of-non-smooth-and-non-admissible-representations-of-gl-2 non-artificial examples of non-smooth and non-admissible representations of GL_2 Hugo Chapdelaine 2012-10-07T10:08:29Z 2012-10-07T19:19:25Z <p>Let $F$ be a finite degree extension over $\mathbf{Q}_p$ and consider the locally profinite group $G:=GL_2(\mathbf{Q}_p)$. </p> <p><strong>P1</strong>: Give an interesting example (non-artificial one, i.e., one that arises in real life for a representation theorist) of a non-smooth representation $\rho$ of $G$ on a topological $\mathbf{C}$-vector space $V$. </p> <p>So here I would like the representation $\rho:G\rightarrow Aut(V)$ to be at least continuous in the following sense: for every $v\in V$ I want the orbit map $\pi^v:G\rightarrow V$, $g\mapsto \pi(g)(v)$ to be continuous.</p> <p><strong>P2</strong>: Give an interesting example of a smooth representation of $G$ on a topological $\mathbf{C}$-vector space which is not admissible. </p> <p><strong>added</strong>: Is it possible to construct such representations by inducing over an appropriate closed subgroup of $G$?</p> http://mathoverflow.net/questions/109048/non-artificial-examples-of-non-smooth-and-non-admissible-representations-of-gl-2/109059#109059 Answer by paul garrett for non-artificial examples of non-smooth and non-admissible representations of GL_2 paul garrett 2012-10-07T13:33:18Z 2012-10-07T19:19:25Z <p>The "smoothness" prevents taking Hilbert-space completions in general, for example. That is, for example, with $G=GL_n(F)$ for a $p$-adic field $F$ and $n\ge 1$, the Hilbert space $V=L^2(G)$ with right translation by $G$ <em>is</em> a continuous representation in the strong topology but is not smooth. In fact, this would be the case for any unimodular totally-disconnected (non-discrete) group $G$. </p> <p>The subspace of smooth vectors in $L^2(G)$ is dense, of course.</p> <p>The smooth vectors in $L^2(G)$ are a natural example of a smooth repn that is (too big to be) admissible. Not hard to check.</p> <p>Edit: and responding to the further-question above, note that the other good answer about action of G on compactly-supported (complex-valued) functions on the Bruhat-Tits building (or tree, for $SL_2(\mathbb Q_p)$), is indeed the (compactly-supported, smooth) induced representation of the trivial representation on the Iwahori subgroup, up to the whole $SL_n(\mathbb Q_p)$.</p> <p>Crazily enough, if one lifts a "cuspidal" repn from $SL_n(\mathbb F_p)$ to $K=SL_n(\mathbb Z_p)$, and then induces that to $SL_n(\mathbb Q_p)$, a finite direct sum of supercuspidal repns is obtained, so <em>this</em> induced repn <em>is</em> admissible... in contrast to inducing the trivial repn from $K$. </p> <p>(Also, of course, <em>principal series</em> inducing admissibles on the Levi components are admissible.)</p> http://mathoverflow.net/questions/109048/non-artificial-examples-of-non-smooth-and-non-admissible-representations-of-gl-2/109064#109064 Answer by Joël for non-artificial examples of non-smooth and non-admissible representations of GL_2 Joël 2012-10-07T14:08:49Z 2012-10-07T14:08:49Z <p>Another example, this time of a natural representation which is smooth, but not admissible. The group $G$ acts naturally on its Bruhat-Tits tree $X$. Let $C(X)$ be the space of complex-valued functions on (the set of vertices of) $X$ with finite support. Then $G$ has a natural representation on $C(X)$, which is smooth (the stabilizer of a point is a maximal compact subgroup is a compact maximal, the stabilizer of a function is an intersection of finitely many such maximal compact subgroup hence is open) but not admissible (for example, the space of invariants in $C(X)$ by a compact maximal is the underlying space of the unramified (or spherical) Hecke algebra of $G$, which has infinite dimension over $\mathbb C$.</p>