Let $F$ be a finite degree extension over $\mathbf{Q}\_p$ and consider the locally profinite group $G:=GL\_2(\mathbf{Q}\_p)$. 


**P1**: 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 $V$. 


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.


**P2**: Give an interesting example of a smooth representation of $G$ on a topological $\mathbf{C}$-vector space which is not admissible. 


**added**: Is it possible to construct such representations by inducing over an appropriate closed subgroup of $G$?



non-artificial examples of non-smooth and non-admissible representations of GL_2
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 $G$ *is* 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 $G$. 


The subspace of smooth vectors in $L^2(G)$ is dense, of course.


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.


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 $SL\_n(\mathbb Q\_p)$.


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 *this* induced repn *is* admissible... in contrast to inducing the trivial repn from $K$. 


(Also, of course, *principal series* inducing admissibles on the Levi components are admissible.)


