Let $F$ be a local field and $\pi:F^{\times}=\mathrm{GL}(1, F)\to \mathrm{GL}(2, \mathbb{C})$ be indecomposable non-irreducible admissible (smooth) representation. Here indecomposable mean that it can't be represented as a direct sum of irreducible representations (i.e. not semisimple). Assume that there exists 1-dimensional invariant subspace of $\mathbb{C}^{2}$. How can we show that such representation is isomorphic to $\rho\otimes \chi$, where $$ \rho(a) = \begin{pmatrix} 1 & \log |a| \\ 0 & 1\end{pmatrix} $$ and $\chi:F^{\times} \to \mathbb{C}^{\times}$ a quasicharacter?
By the existence of 1-dimensional space, there exists a basis $\{v_{1}, v_{2}\}$ of $\mathbb{C}^{2}$ such that $\pi(a)$ can be written as $$ \pi(a) = \begin{pmatrix} \chi_{1}(a) & f(a) \\ 0 & \chi_{2}(a) \end{pmatrix}. $$ Then $\pi(a)\pi(b) = \pi(ab) = \pi(b)\pi(a)$ implies that $\chi_{1}, \chi_{2}:F^{\times} \to \mathbb{C}^{\times}$, and $$ f(ab) = \chi_{1}(a)f(b) + \chi_{2}(b)f(a) = \chi_{2}(a)f(b) + \chi_{1}(b)f(a). $$ From the last equation, we have $(\chi_{1}^{-1}(a)-\chi_{2}^{-1}(a))f(a) = (\chi_{1}^{-1}(b)-\chi_{2}^{-1}(b))f(b) = (\chi_{1}^{-1}(1) - \chi_{2}^{-1}(1))f(1)=0$, i.e. $(\chi_{1}(a) - \chi_{2}(a))f(a) =0$ for all $a\in F^{\times}$.
I think we may show that $\chi_{1} = \chi_{2}$ from this, but I can't figure it out now. If we prove $\chi_{1} = \chi_{2}$, then $g(a):= \chi_{1}^{-1}(a)f(a)$ satisfies $g(ab)=g(a) + g(b)$, so that $g$ might be a constant multiple of a log function.