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## Maass forms

A Maass form (of weight 0) on a subgroup $$\Gamma$$ of $$\GL_{2}(\R)$$ is a smooth, square-integrable, automorphic eigenfunction of the Laplace-Beltrami operator $\Delta$. In other words, $$f\in C^\infty(\mathcal{H}),\quad f\in L^2(\Gamma\backslash{\mathcal H}),\quad f(\gamma z)=f(z)\ \forall\gamma\in\Gamma,\quad (\Delta+\lambda)f(z)=0 \textrm{ for some } \lambda \in \C.$$

## Maass forms of weight $k$

A Maass form $f$ of weight $k$ and multiplier system $v$ on a group $\Gamma$ is a smooth function $f:\mathcal{H} \rightarrow\C$ with the following properties:

1. $f$ transforms according to a unitary weight $$k$$ slash-action: $f|[\gamma,k]( z) = v(\gamma) f(z)$ for all $\gamma \in \Gamma$ where $$f|[\gamma,k]( z) = \exp(-ik\mathrm{Arg}(c z+d)) f(\gamma z)$$ for $\gamma=\begin{pmatrix}a & b \\ c & d \end{pmatrix}$

2. $f$ is an eigenfunction of the corresponding weight $k$ Laplacian $$\Delta_k = y^2\left( \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) - iky\frac{\partial}{\partial x}$$

If the level is equal to 1, the only possible multiplier systems are given by the Dedekind eta function. A compatible multiplier system for weight $k\in \R$ is given by

$v(A) = v_{\eta}^{2k}:=\eta(Az)^{2k}/\eta(z)^{2k}$

or one of its 6 conjugates. One can show that $v(A)$ is well-defined and independent of the point $z$ in the upper half-plane.

## Remark:

One can also consider Maass forms transforming with the usual "holomorphic" weight $k$ slash-action. In this case the corresponding weight $k$ Laplacian will look slightly different: $$\Delta_k = y^2\left( \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) - iky\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right).$$ This convention is usually used in the context of Harmonic weak Maass forms.

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• Review status: reviewed
• Last edited by David Farmer on 2020-07-24 14:13:27
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