Suppose that $\mathbb{H}$ is the hyperbolic plane and $\Gamma$ is a discrete group of orientation-preserving isometries of $\mathbb{H}$ such that $M=\Gamma\setminus\mathbb{H}$ is a compact hyperbolic surface.
Given the Laplacian on $M$, by compactness, $L^2(M)$ has an orthonormal basis of Laplacian eigenfunctions $\{\psi_{\lambda_i}\}_{i\geq 0}$ with \begin{align*} \Delta\psi_{\lambda_i} = \lambda_i\psi_{\lambda_i}, \end{align*} and $0=\lambda_0<\lambda_1\leq\ldots\to\infty$. Given one of these eigenfunctions say $\psi_\lambda$, one can lift it to a function $\tilde{\psi}_\lambda$on $\mathbb{H}$ through the covering map.
From what I understand, $\tilde{\psi}_\lambda$ is an eigenfunction of the Laplacian on $\mathbb{H}$. However, I am confused since the spectrum of the Laplacian on $\mathbb{H}$ is $[\frac{1}{4},\infty)$ where as the spectrum of the Laplacian on $M$ is contained in $[0,\infty)$ (and certainly $0$ is an eigenfunction as constants on $M$ have eigenvalue zero). However, would this not imply that (for instance) I can lift $L^2$ eigenfunctions on $M$ with eigenvalue in $[0,\frac{1}{4})$ to eigenfunctions on $\mathbb{H}$ with the same eigenvalue, contradicting the spectrum of $\mathbb{H}$?
I think the solution to this is that $[\frac{1}{4},\infty)$ is the $L^2$ spectrum of the Laplacian on $\mathbb{H}$ and the lifts of these $L^2$ eigenfunctions on $M$ won't necessarily be $L^2$ eigenfunctions on $\mathbb{H}$ (just smooth functions satisfying the eigenvalue equation). Is this true?