A question on representation theory of p-adic groups Let $V$ be a complex vector space of infinite dimension and let $(\pi,V)$ be a representation of the $p$-adic group $G:=GL_2(\mathbb{Q}_p)$. From representation theory, we know that if the representation $(\pi,V)$ is both smooth and irreducible, then $(\pi,V)$ is admissible. So now it is natural to think about the inverse question. 
Does there exist a representation $(\pi,V)$ of $G$ which is smooth and admissible but not irreducible? 
This question is not difficult. You can find two smooth and irreducible representations $(\pi_1,V_1)$ and $(\pi_2,V_2)$ of $G$ and give the example $(\pi_1 \oplus \pi_2, V_1 \oplus V_2)$. Let $\pi=\pi_1 \oplus \pi_2$ and $V=V_1 \oplus V_2$, then $(\pi,V)$ is the representation we want for the question. 
But if we change a little on the question, does there exist a representation $(\pi,V)$ of $G$ which is smooth, admissible and indecomposable but not irreducible? I have no idea on this question since the above example does not work in this case.
 A: It is indeed well-known that the category of smooth admissible representations of $G$ (and other reductive $p$-adic groups)
is not semi-simple. The principal series, that is the representations induced from a character of the Borel of $G=GL_2(\mathbb Q_p)$, are always indecomposable, but they may nor be irreducible -- think of the case of the trivial character.
Another nice way to construct an example is with trees. You may know that $G$ acts transitively on the regular infinite tree $T$ of arity $p+1$ (i.e. every vertex has exactly $p+1$ neighbors -- this tree is the Bruhat-Tits tree of $G$). The stabilizer of a vertex is $ZK$ where $Z$ is the center and $K$ a compact maximal subgroup (= a conjugate of $GL_2(\mathbb Z_p)$).
If you take for $W$ the set of functions with finite support on the set of vertices of the tree $T$, then $W$ is a representation of $G$ (since $G$ acts on $T$) which is smooth (stabilizers are intersection of finitely many $K_iZ$, with $K_i$ compact  open), admits a non-trivial $G$-morphism to the trivial representation (the linear form which to a function $f$ on $T$ attaches the sum of its value), but certainly does not contain the trivial representation as sub-representation, because $W$ has no function invariant by $G$ except 0 -- they would be of infinite support.
Small problem, $W$ is not admissible. But now let $D: W \rightarrow W$ be the operator which to a function $f$ attaches the function $g(x)=\sum f(y)$ with $y$ running over the $p+1$ neighbors of $x$, and define $V = W/ (D-(p+1))W$. You can easily check that the trivial representation is still a quotient of $V$, and with some little geometric reasoning on the tree that I leave as an exercise, that the trivial representation is not a sub-representation of $V$, and that $V$ is admissible. Hence a second example of smooth-admissible non-semi-simple reducible representation which shows that the answer to your question is yes. (Actually this example is the dual of the first one).
