Maass form ($\Delta f=\lambda f$)

In the weight-zero convention, a Maass form for a discrete subgroup $\Gamma \subseteq {\mathrm{SL}}_2(\mathbb{R})$ is a smooth function on the upper half-plane, invariant under $z\mapsto (az+b)/(cz+d)$, satisfying a hyperbolic Laplacian eigenvalue equation and specified growth conditions at the cusps:Additional conditions specify a cuspidal or square-integrable class. The sign of $\Delta$ is a convention and must be stated.^[ref]

The constant function $f=1$ is an invariant eigenfunction with $\lambda =0$ and moderate growth, but is not cuspidal: its average along a horizontal period is nonzero. A Maass cusp form has zero such constant term at each cusp. The eigenfunction condition allows nonholomorphic automorphic forms, connecting arithmetic symmetries with spectral geometry.

A shifted polyharmonic Maass form is a smooth automorphic function that satisfies a generalized eigenfunction equation for a hyperbolic Laplacian. In weight zero, choose the convention $\Delta =-y^2({\partial }_x^2+{\partial }_y^2)$ and requirefor an integer $m\ge 1$, together with automorphy and the specified growth conditions at each cusp. Moderate growth means at most polynomial growth in cusp coordinates. The weight-$k$ version replaces $\Delta$ by the corresponding Weight-k hyperbolic Laplacian and uses the weight-$k$ transformation law.^[ref]

The least possible $m$ is the exact depth. For $m=1$ this is an ordinary eigenfunction condition; higher depth allows a chain of generalized eigenfunctions. It is not merely an equation with a different eigenvalue. The cited paper uses the opposite Laplacian sign, so its eigenvalue parameter is the negative of the one used here.

For an illuminating construction, let a nonholomorphic Eisenstein series $E(z,s)$ satisfy $\Delta E=s(1-s)E$. At a regular value $s=2$, differentiate in $s$. The function $F={\partial }_{s}E{|}_{s=2}$ obeysThe first right-hand side is nonzero, so this gives depth two. Parameter differentiation introduces a generalized eigenfunction while preserving automorphy.^[ref]

In the integer-depth convention, a polyharmonic Maass form is a smooth automorphic function annihilated by a positive integer power of its weight-appropriate hyperbolic Laplacian, with specified growth at the cusps. In weight zero, using $\Delta =-y^2({\partial }_x^2+{\partial }_y^2)$, the equation is ${\Delta }^{m}F=0$. It is the zero-shift case of a shifted polyharmonic Maass form. Exact depth is the least such integer $m$. Some refined conventions also use half-integer depths, so the convention must be stated.^[ref]

A global depth-two example comes from the finite part of a completed nonholomorphic Eisenstein series. Near $s=1$, writewhere $c$ is a nonzero constant and $\Delta E^{\ast }=s(1-s)E^{\ast }$. Comparing constant Laurent coefficients gives $\Delta F=-c$, hence ${\Delta }^{2}F=0$ but $\Delta F\ne 0$. The finite part is automorphic and has moderate growth.^[ref]

This demonstrates why higher depth is substantive: the space contains functions that are not harmonic but become harmonic after one application of the Laplacian. Merely solving a local differential equation is insufficient; automorphy and cusp growth remain part of the definition.

A Maass cusp form is a nonzero Maass form with vanishing constant term of an automorphic form at every cusp. For weight zero on the modular group $\Gamma ={\mathrm{SL}}_2(\mathbb{Z})$, it is a smooth, $\Gamma$-invariant function on the upper half-plane, square-integrable on a fundamental domain with measure $dxdy/y^2$, satisfyingThe integral condition describes the cusp at infinity; other groups require the corresponding condition in coordinates at each cusp.^[ref]

Translation invariance gives a Fourier series. When $\lambda =1/4+r^2$, its decaying modes have the formwhere $K_{ir}$ is a Modified Bessel function of the second kind. The absent $n=0$ term expresses cuspidality. The Fourier coefficients must also satisfy the full automorphic symmetry.^[ref]

The constant function $f=1$ is a useful test: it satisfies the symmetry and has eigenvalue zero, but its horizontal average is $1$, so it is not a Maass cusp form. The extra condition isolates oscillating eigenfunctions connecting arithmetic symmetry with spectral geometry.