Let $G$ be a locally profinite group, and $(\pi, V)$ be a representation of $G$ over a complex vector space $V$. Then the representation is called smooth if, for every $v$ in $V$, there is a compact open subgroup $K$ of $G$ (depending on $v$) such that $\pi(g)v=v$ for all $g$ in $K$.
Let $V^K = \{v \in V \;|\, \pi(g)v=v \quad \forall g \in K \}$ be the space of $\pi(K)$-fixed vectors. The smooth representation is called admissible if the space $V^K$ is finite dimensional for each compact open $K$ of $G$.
I believe the above are standard definitions. My confusion stems from a passage on pg 6 of the paper http://www.math.harvard.edu/~gross/preprints/AdjointGamma5.pdf :
A representation ρ : W → GL(V ) on a finite dimensional complex vector space V is called admissible if ker ρ is open and ρ(Fr) is semisimple.
Here $k$ is a local, nonarchimedean field of characteristic zero, and $W$ is the Weil group of $k$, so $W$ is locally profinite, and Fr is the geometric Frobenius.
Here is my question:
So now if I try to apply the earlier definitions of "admissible", I see that since $V$ is already finite dimensional, I need only show that the representation is smooth? Now it seems that having the kernel of $\rho$ be open is sufficient to get smoothness. So why is the semisimplicity of $\rho(Fr)$ necessary for smoothness and/or admissibility?