Let $G$ be a locally profinite group. Given an (abstract) representation $V$ of $G$, that is a $\mathbb C$-vector space $V$ together with a group homomorphism $G\rightarrow \operatorname{Aut}_{\mathbb C}(V)$, we define the subspace $$V^{\infty}:=\bigcup_K V^K$$
where the union ranges over all open compact subgroups $K$ of $G$, and $V^K$ is the subset of $V$ consisting of the vectors fixed by all the elements of $K$. It is not difficult to see that $V^{\infty}$ is a $G$-subspace of $V$ and defines a smooth representation of $G$.
Now, given three (abstract) representations $U,V,W$ of $G$ that lie in an exact sequence $$0\rightarrow U \xrightarrow a V \xrightarrow b W \rightarrow 0$$
the restriction of $a$ and $b$ to $U^{\infty}$ and $V^{\infty}$ gives rise to an induced exact sequence $$0\rightarrow U^{\infty} \xrightarrow a V^{\infty} \xrightarrow b W^{\infty}$$
According to the book "The local Langlands conjecture for $\operatorname{GL}(2)$" by Bushnell and Henniart, the map $b$ may indeed not be surjective onto $W^{\infty}$.
I am trying to find an example where such phenomenon appears, but I can't come up with one. Would you be able to provide such an example?
Best Answer
Let $K\subseteq G$ be a compact open subgroup. The group algebra $V = \mathbb C[G]$ is a representation with $V^\infty = \{0\}$. The compact induction $W = \mathbb C[G/K]$ is smooth (because each coset $gK$ is stabilized by $gKg^{-1}$).
Now, you have a canonical surjective morphism $f\colon \mathbb C[G] \to \mathbb C[G/K]$ given by $g\mapsto gK$, such that $f^\infty\colon \mathbb C[G]^\infty = \{0\} \to \mathbb C[G/K]^\infty = \mathbb C[G/K]$ is not surjective.