Admissibility of representations Consider a smooth representation $(\pi,V)$ of the group $G=GL_n(\mathbb{Q}_p)$. The representation is said to be admissible if for every open compact subgroup $K \subseteq G$, the space $V^K$ of $K$-fixed vectors is finite dimensional. 
For an open compact subgroup $K$, let $\hat{K}$ be the equivalence classes of irreducible unitary representations, and for $\rho \in \hat{K}$, we can define the $\rho$-isotypic component, denoted $V^{\rho}$, of the representation $\pi$. For instance, if $\rho$ is the trivial representation, then $V^{\rho}=V^K$, and in fact, we can characterize admissibility equivalently as follows: the representation $(\pi,V)$ is admissible if for every open compact subgroup $K \subseteq G$ and every $\rho \in \hat{K}$, the $\rho$-isotypic component $V^{\rho}$ is finite-dimensional.
But this seems like overkill. Do we really need "for every open compact subgroup $K$ and every $\rho \in \hat{K}$"? What if for some fixed open compact subgroup $K$, it holds that for every $\rho \in \hat{K}$, $V^{\rho}$ is finite dimensional. Will that suffice to conclude anything about admissibility of $\pi$? 
 A: You are correct. If there is some compact open subgroup $K$ of $G$ for which the $\rho$-isotypic component $V_\rho$ is finite-dimensional for each irreducible smooth representation $\rho$ of $K$, then $V$ is admissible. To see this, first consider an open normal subgroup $K_1$ of $K$. Then the subspace $V^{K_1}$ of $K_1$-fixed vectors is $K$-stable. The fact that $V$ is $K$-semisimple implies that $V^{K_1}$ is $K$-semisimple as well, and in fact, we have 
$$V^{K_1} = \bigoplus_{\tau\in\widehat{K/K_1}}(V^{K_1})_\tau,$$
where $\tau$ runs over the irreducible smooth (i.e. finite-dimensional) representations of the finite group $K/K_1$. So there are only finitely many isotypic components in the above direct sum. Moreover, we have $(V^{K_1})_\tau=V_\rho$, where $\rho$ is the inflation of $\tau$ to $K$ along the homomorphism $K\to K/K_1$. Thus 
$$\dim((V^{K_1})\tau)=\dim(V_\rho)<\infty$$
by the assumption on $V$. So $V^{K_1}$ is finite-dimensional. Now if $K^\prime$ is any compact open subgroup of $G$ whatsoever, let $K_1$ be an open normal subgroup of $K$ contained in $K^\prime\cap K$. Then $V^{K^\prime}\subseteq V^{K_1}$, and since we have already proved that $V^{K_1}$ is finite-dimensional, we may conclude that $V^{K^\prime}$ is as well.
