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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]

The Weight-k hyperbolic Laplacian is a differential operator adapted to weight-$k$ transformation laws on the upper half-plane $z=x+iy$, $y>0$. Using the sign consistent with the nonnegative weight-zero hyperbolic Laplacian, and the holomorphic-weight convention, defineHere ${\partial }_z=\frac{1}{2}({\partial }_x-i{\partial }_y)$ and ${\partial }_{\bar{z}}=\frac{1}{2}({\partial }_x+i{\partial }_y)$ are Wirtinger derivatives. The cited paper uses the opposite overall sign, so its eigenvalues must also have their signs reversed.^[ref]

The correction terms let the operator preserve the transformation law $f((az+b)/(cz+d))=(cz+d{)}^{k}f(z)$ for determinant-one integer matrices, as used for modular forms and Maass forms. At $k=0$ it reduces to the ordinary hyperbolic Laplacian. Every holomorphic function is locally annihilated because ${\partial }_{\bar{z}}f=0$; being holomorphic alone does not supply modularity or growth conditions.

For a direct calculation, take $f(z)=y^s$. Its $x$ derivatives vanish, givingThus both $1$ and $y^{1-k}$ are annihilated; when $k=1$ these coincide and a second independent solution depending only on $y$ is $\log y$. These are local operator examples, not claims that all satisfy a modular transformation law. Conventions using a unit-modulus weight multiplier have a different first-derivative correction, so the displayed formula matters as much as the name.