I am trying to prove
Let $(V,\pi)$ be a smooth representation of $GL_n(F)$. Then $V$ is one dimensional or infinite dimensional.
My attempt:
Let $(V,\pi)$ be a smooth representation of $GL_n(F)$ of finite dimension. Let $\{v_1,\dots, v_m\}$ be a basis of $V$. Since $V$ is smooth, for any $v_i$ there exists an open compact subgroup $K_i$ of $GL_n(F)$ that stabilises $v_i$. So $K=\bigcap K_i$ is again an open compact subgroup of $GL_n(F)$ and moreover it stabilises all $v_1,\cdots, v_n$ and so all $V$. Hence $Ker \pi$ has an open compact subgroup. Moreover, since $Ker \pi$ is normal and a conjugate of $K$ is in $GL_n(\mathcal{O}_F)$, we have \begin{equation*} K^g \leq Ker\pi \cap GL_n(\mathcal{O}_F). \end{equation*}
My claim is: $SL_n(F)\subseteq Ker(F)$.
The idea that I have $SL_n(F)$ is generate:d by the matrices of the form $(e_{ij}(x))_{r,s}$ that have $1$ on the diagonal, $x$ in the entry $i,j$ and $0$ otherwise. Since $Ker \pi$ is open and $(e_{ij}(0))_{r,s}=Id$ is in $Ker \pi$, that is open, for $x$ with "big" valuation $(e_{ij}(x))_{r,s}$ is in the kernel. So the irreducible representation $(V,\pi)$ of $GL_n(F)$ is also an irreducible representation of $GL(F)/SL_n(F)\cong F^\times$ and so it is one dimensional.
I think that I am not so far from the conclusion, but I have some problem. I don't know how to use the fact that there is a compact subgroup in the Kernel that is normal.
