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Maass forms of weight $0$ with trivial character

A Maass form 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 $k$ with trivial character

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. it 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. it 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 Nathan Ryan on 2019-05-01 11:07:32
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