Let $G = \text{PGL}_2(\mathbb{R})$ and let $\Gamma = \text{PGL}_2(\mathbb{Z})$. Let $\varphi : \Gamma \backslash G \rightarrow \mathbb{C}$ be smooth and in $L^2(\Gamma\backslash G)$. Let $\mathfrak{g} \cong \mathfrak{sl}_2(\mathbb{R})$ denote the Lie algebra of $G$ and let $\Delta \in \mathcal{U}(\mathfrak{g}_{\mathbb{C}})$ be the Casimir operator. Assume that $\Delta \varphi = \lambda \varphi$ for some $\lambda \in \mathbb{C}- \mathbb{Z}$. Assume that $\varphi$ is right-$K$-finite, where $K$ is the image of $\text{SO}(2)$ in $G$, or even right-$K$-invariant. Feel free to answer for other analogous triples $(\Gamma, G, K)$.
Is the closed sub-representation $\pi$, generated by $\varphi$ in $L^2(\Gamma \backslash G)$ irreducible? If so, why?
I have the feeling that it would suffice to know that the multiplicity of each irreducible unitary representation $\sigma$ of $K$ in $\pi$, is one. Because, the assumption that $\lambda$ is not an integer, implies that raising and lowering operators in $\mathfrak{g}_{\mathbb{C}}$ define isomorphisms between the (smooth) weight-spaces of $\pi$.
Of course, the one-dimensionality of these weight-spaces would follow if we already knew that $\pi$ was irreducible, by the uniqueness of $K$-fixed vector theorem.