Are automorphic forms eigenfunctions?

Question

The Eisenstein series are eigenfunctions of the non-Euclidean Laplacian operator. Are automorphic forms in general by definition eigenfunctions of certain operators?

Answer 1

Yes and no. :) It is true that the official literature on this may create a misleading impression, akin to saying that a `periodic function' on $\mathbb R$ must be an eigenfunction for $d^2/dx^2$. In the automorphic case, it is implicit that (most) discussions of "$\Gamma$-periodic functions'' really refer to special $\Gamma$-periodic functions, akin to exponentials (or sine and cosine) among periodic functions on the real line.

The difference in the situations is that the individual special periodic functions in the automorphic context are much subtler things than in the periodic-on-$\mathbb R$ context.

But, yes, in fact, the orthodox definition of automorphic form/function requires $\mathfrak z$-finiteness, where $\mathfrak z$ is the center of the enveloping algebra of the Lie algebra $\mathfrak g$ of the real Lie group $G_\infty$ that is the archimedean points of the ambient algebraic group. This slightly evasive generalization of ``simultaneous eigenfunction for $\mathfrak z$'' allows finite linear combinations of eigenfunctions, as well as derivatives of Eisenstein series with regard to the spectral parameter `$s$'.

But, not, not everything in $L^2(\Gamma\backslash G)$, nor in $L^2(G_k\backslash G_{\mathbb A})$ is an eigenfunction for those operators.

The spectral decomposition theorem is that everything in $L^2$ is a superposition of eigenfunctions (meaning to not only use genuine $L^2$ eigenfunctions, but also non-$L^2$ eigenfunctions such as Eisenstein series).