Maass wave form

The group

${\displaystyle G:=\mathrm {SL} _{2}(\mathbb {R} )=\left\{{\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}\in M_{2}(\mathbb {R} ):ad-bc=1\right\}}$

operates on the upper half plane

${\displaystyle {\mathcal {H}}=\{z\in \mathbb {C} :\operatorname {Im} (z)>0\}}$

by fractional linear transformations:

${\displaystyle {\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}\cdot z:={\frac {az+b}{cz+d}}.}$

It can be extended to an operation on ${\displaystyle {\mathcal {H}}\cup \{\infty \}\cup \mathbb {\mathbb {R} } }$ by defining:

${\displaystyle {\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}\cdot z:={\begin{cases}{\frac {az+b}{cz+d}}&{\text{if }}cz+d\neq 0,\\\infty &{\text{if }}cz+d=0,\end{cases}}}$

${\displaystyle {\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}\cdot \infty :=\lim _{\operatorname {Im} (z)\to \infty }{\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}\cdot z={\begin{cases}{\frac {a}{c}}&{\text{if }}c\neq 0\\\infty &{\text{if }}c=0\end{cases}}}$

The Radon measure

${\displaystyle d\mu (z):={\frac {dxdy}{y^{2}}}}$

defined on ${\displaystyle {\mathcal {H}}}$ is invariant under the operation of ${\displaystyle \mathrm {SL} _{2}(\mathbb {R} )}$.

Let ${\displaystyle \Gamma }$ be a discrete subgroup of ${\displaystyle G}$. A fundamental domain for ${\displaystyle \Gamma }$ is an open set ${\displaystyle F\subset {\mathcal {H}}}$, so that there exists a system of representatives ${\displaystyle R}$ of ${\displaystyle \Gamma \backslash {\mathcal {H}}}$ with

${\displaystyle F\subset R\subset {\overline {F}}{\text{ and }}\mu ({\overline {F}}\setminus F)=0.}$

A fundamental domain for the modular group ${\displaystyle \Gamma (1):=\mathrm {SL} _{2}(\mathbb {Z} )}$ is given by

${\displaystyle F:=\left\{z\in {\mathcal {H}}\mid \left|\operatorname {Re} (z)\right|<{\frac {1}{2}},|z|>1\right\}}$

(see Modular form).

A function ${\displaystyle f:{\mathcal {H}}\to \mathbb {C} }$ is called ${\displaystyle \Gamma }$-invariant, if ${\displaystyle f(\gamma z)=f(z)}$ holds for all ${\displaystyle \gamma \in \Gamma }$ and all ${\displaystyle z\in {\mathcal {H}}}$.

For every measurable, ${\displaystyle \Gamma }$-invariant function ${\displaystyle f:{\mathcal {H}}\to \mathbb {C} }$ the equation

${\displaystyle \int _{F}f(z)\,d\mu (z)=\int _{\Gamma \backslash {\mathcal {H}}}f(z)\,d\mu (z),}$

holds. Here the measure ${\displaystyle d\mu }$ on the right side of the equation is the induced measure on the quotient ${\displaystyle \Gamma \backslash {\mathcal {H}}.}$

The hyperbolic Laplace operator on ${\displaystyle {\mathcal {H}}}$ is defined as

${\displaystyle \Delta :C^{\infty }({\mathcal {H}})\to C^{\infty }({\mathcal {H}}),}$

${\displaystyle \Delta =-y^{2}\left({\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}\right)}$

A Maass form for the group ${\displaystyle \Gamma (1):=\mathrm {SL} _{2}(\mathbb {Z} )}$ is a complex-valued smooth function ${\displaystyle f}$ on ${\displaystyle {\mathcal {H}}}$ satisfying

1. ${\displaystyle f(\gamma z)=f(z){\text{ for all }}\gamma \in \Gamma (1),\qquad z\in {\mathcal {H}}}$
2. ${\displaystyle {\text{there exists }}\lambda \in \mathbb {C} {\text{ with }}\Delta (f)=\lambda f}$
3. ${\displaystyle {\text{there exists }}N\in \mathbb {N} {\text{ with }}f(x+iy)={\mathcal {O}}(y^{N}){\text{ for }}y\geq 1}$

If

${\displaystyle \int _{0}^{1}f(z+t)dt=0{\text{ for all }}z\in {\mathcal {H}}}$

we call ${\displaystyle f}$ Maass cusp form.

Let ${\displaystyle f}$ be a Maass form. Since

${\displaystyle \gamma :={\begin{pmatrix}1&1\\0&1\\\end{pmatrix}}\in \Gamma (1)}$

we have:

${\displaystyle \forall z\in {\mathcal {H}}:\qquad f(z)=f(\gamma z)=f(z+1).}$

Therefore ${\displaystyle f}$ has a Fourier expansion of the form

${\displaystyle f(x+iy)=\sum _{n=-\infty }^{\infty }a_{n}(y)e^{2\pi inx},}$

with coefficient functions ${\displaystyle a_{n},n\in \mathbb {Z} .}$

It is easy to show that ${\displaystyle f}$ is Maass cusp form if and only if ${\displaystyle a_{0}(y)=0\;\;\forall y>0}$.

We can calculate the coefficient functions in a precise way. For this we need the Bessel function ${\displaystyle K_{v}}$.

Definition: The Bessel function ${\displaystyle K_{v}}$ is defined as

${\displaystyle K_{s}(y):={\frac {1}{2}}\int _{0}^{\infty }e^{-{\frac {y(t+t^{-1})}{2}}}t^{s}{\frac {dt}{t}},\qquad s\in \mathbb {C} ,y>0.}$

The integral converges locally uniformly absolutely for ${\displaystyle y>0}$ in ${\displaystyle s\in \mathbb {C} }$ and the inequality

${\displaystyle K_{s}(y)\leq e^{-{\frac {y}{2}}}K_{\operatorname {Re} (s)}(2)}$

holds for all ${\displaystyle y>4}$.

Therefore, ${\displaystyle |K_{s}|}$ decreases exponentially for ${\displaystyle y\to \infty }$. Furthermore, we have ${\displaystyle K_{-s}(y)=K_{s}(y)}$ for all ${\displaystyle s\in \mathbb {C} ,y>0}$.

Proof: We have

${\displaystyle \Delta (f)=\left({\frac {1}{4}}-\nu ^{2}\right)f.}$

By the definition of the Fourier coefficients we get

${\displaystyle a_{n}(y)=\int _{0}^{1}f(x+iy)e^{-2\pi inx}dx}$

for ${\displaystyle n\in \mathbb {Z} .}$

Together it follows that

${\displaystyle {\begin{aligned}\left({\frac {1}{4}}-\nu ^{2}\right)a_{n}(y)&=\int _{0}^{1}\left({\frac {1}{4}}-\nu ^{2}\right)f(x+iy)e^{-2\pi inx}dx\\[4pt]&=\int _{0}^{1}(\Delta f)(x+iy)e^{-2\pi inx}dx\\[4pt]&=-y^{2}\left(\int _{0}^{1}{\frac {\partial ^{2}f}{\partial x^{2}}}(x+iy)e^{-2\pi inx}dx+\int _{0}^{1}{\frac {\partial ^{2}f}{\partial y^{2}}}(x+iy)e^{-2\pi inx}dx\right)\\[4pt]&{\overset {(1)}{=}}-y^{2}(2\pi in)^{2}a_{n}(y)-y^{2}{\frac {\partial ^{2}}{\partial y^{2}}}\int _{0}^{1}f(x+iy)e^{-2\pi inx}dx\\[4pt]&=-y^{2}(2\pi in)^{2}a_{n}(y)-y^{2}{\frac {\partial ^{2}}{\partial y^{2}}}a_{n}(y)\\[4pt]&=4\pi ^{2}n^{2}y^{2}a_{n}(y)-y^{2}{\frac {\partial ^{2}}{\partial y^{2}}}a_{n}(y)\end{aligned}}}$

for ${\displaystyle n\in \mathbb {Z} .}$

In (1) we used that the nth Fourier coefficient of ${\textstyle {\frac {\partial ^{2}f}{\partial x^{2}}}}$ is ${\displaystyle (2\pi in)^{2}a_{n}(y)}$ for the first summation term. In the second term we changed the order of integration and differentiation, which is allowed since f is smooth in y . We get a linear differential equation of second degree:

${\displaystyle y^{2}{\frac {\partial ^{2}}{\partial y^{2}}}a_{n}(y)+\left({\frac {1}{4}}-\nu ^{2}-4\pi n^{2}y^{2}\right)a_{n}(y)=0}$

For ${\displaystyle n=0}$ one can show, that for every solution ${\displaystyle f}$ there exist unique coefficients ${\displaystyle c_{0},d_{0}\in \mathbb {C} }$ with the property ${\displaystyle a_{0}(y)=c_{0}y^{{\frac {1}{2}}-\nu }+d_{0}y^{{\frac {1}{2}}+\nu }.}$

For ${\displaystyle n\neq 0}$ every solution ${\displaystyle f}$ has coefficients of the form

${\displaystyle a_{n}(y)=c_{n}{\sqrt {y}}K_{v}(2\pi |n|y)+d_{n}{\sqrt {y}}I_{v}(2\pi |n|y)}$

for unique ${\displaystyle c_{n},d_{n}\in \mathbb {C} }$. Here ${\displaystyle K_{v}(s)}$ and ${\displaystyle I_{v}(s)}$ are Bessel functions.

The Bessel functions ${\displaystyle I_{v}}$ grow exponentially, while the Bessel functions ${\displaystyle K_{v}}$ decrease exponentially. Together with the polynomial growth condition 3) we get ${\displaystyle f:a_{n}(y)=c_{n}{\sqrt {y}}K_{v}(2\pi |n|y)}$ (also ${\displaystyle d_{n}=0}$) for a unique ${\displaystyle c_{n}\in \mathbb {C} }$. Q.E.D.

Even and odd Maass forms: Let ${\displaystyle i(z):=-{\overline {z}}}$. Then i operates on all functions ${\displaystyle f:{\mathcal {H}}\to \mathbb {C} }$ by ${\displaystyle i(f):=f(i(z))}$ and commutes with the hyperbolic Laplacian. A Maass form ${\displaystyle f}$ is called even, if ${\displaystyle i(f)=f}$ and odd if ${\displaystyle i(f)=-f}$. If f is a Maass form, then ${\displaystyle {\tfrac {1}{2}}(f+i(f))}$ is an even Maass form and ${\displaystyle {\tfrac {1}{2}}(f-i(f))}$ an odd Maass form and it holds that ${\displaystyle f={\tfrac {1}{2}}(f+i(f))+{\tfrac {1}{2}}(f-i(f))}$.

Let

${\displaystyle f(x+iy)=\sum _{n\neq 0}c_{n}{\sqrt {y}}K_{\nu }(2\pi |n|y)e^{2\pi inx}}$

be a Maass cusp form. We define the L-function of ${\displaystyle f}$ as

${\displaystyle L(s,f)=\sum _{n=1}^{\infty }c_{n}n^{-s}.}$

Then the series ${\displaystyle L(s,f)}$ converges for ${\textstyle \Re (s)>{\frac {3}{2}}}$ and we can continue it to a whole function on ${\displaystyle \mathbb {C} }$.

If ${\displaystyle f}$ is even or odd we get

${\displaystyle \Lambda (s,f):=\pi ^{-s}\Gamma \left({\frac {s+\varepsilon +\nu }{2}}\right)\Gamma \left({\frac {s+\varepsilon -\nu }{2}}\right)L(s,f).}$

Here ${\displaystyle \varepsilon =0}$ if ${\displaystyle f}$ is even and ${\displaystyle \varepsilon =-1}$ if ${\displaystyle f}$ is odd. Then ${\displaystyle \Lambda }$ satisfies the functional equation

${\displaystyle \Lambda (s,f)=(-1)^{\varepsilon }\Lambda (1-s,f).}$

The non-holomorphic Eisenstein-series is defined for ${\displaystyle z=x+iy\in {\mathcal {H}}}$ and ${\displaystyle s\in \mathbb {C} }$ as

${\displaystyle E(z,s):=\pi ^{-s}\Gamma (s){\frac {1}{2}}\sum _{(m,n)\neq (0,0)}{\frac {y^{s}}{|mz+n|^{2s}}}}$

where ${\displaystyle \Gamma (s)}$ is the Gamma function.

The series converges absolutely in ${\displaystyle z\in {\mathcal {H}}}$ for ${\displaystyle \Re (s)>1}$ and locally uniformly in ${\displaystyle {\mathcal {H}}\times \{\Re (s)>1\}}$, since one can show, that the series

${\displaystyle S(z,s):=\sum _{(m,n)\neq (0,0)}{\frac {1}{|mz+n|^{s}}}}$

converges absolutely in ${\displaystyle z\in {\mathcal {H}}}$, if ${\displaystyle \Re (s)>2}$. More precisely it converges uniformly on every set ${\displaystyle K\times \{\Re (s)\geq \alpha \}}$, for every compact set ${\displaystyle K\subset {\mathcal {H}}}$ and every ${\displaystyle \alpha >2}$.

We only show ${\displaystyle \mathrm {SL} _{2}(\mathbb {Z} )}$-invariance and the differential equation. A proof of the smoothness can be found in Deitmar or Bump. The growth condition follows from the Fourier expansion of the Eisenstein series.

We will first show the ${\displaystyle \mathrm {SL} _{2}(\mathbb {Z} )}$-invariance. Let

${\displaystyle \Gamma _{\infty }:=\pm {\begin{pmatrix}1&\mathbb {Z} \\0&1\\\end{pmatrix}}}$

be the stabilizer group ${\displaystyle \infty }$ corresponding to the operation of ${\displaystyle \mathrm {SL} _{2}(\mathbb {Z} )}$ on ${\displaystyle {\mathcal {H}}\cup \{\infty \}}$.

Proposition. E is ${\displaystyle \Gamma (1)}$-invariant.

Proof. Define:

${\displaystyle {\tilde {E}}(z,s):=\sum _{\gamma \in \Gamma _{\infty }\backslash \Gamma }\Im (\gamma z)^{s}.}$

(a) ${\displaystyle {\tilde {E}}}$ converges absolutely in ${\displaystyle z\in {\mathcal {H}}}$ for ${\displaystyle \Re (s)>1}$ and ${\displaystyle E(z,s)=\pi ^{-s}\Gamma (s)\zeta (2s){\tilde {E}}(z,s).}$

Since

${\displaystyle \gamma ={\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}\in \Gamma (1)\Longrightarrow \Im (\gamma z)={\frac {\Im (z)}{|cz+d|^{2}}},}$

we obtain

${\displaystyle {\tilde {E}}(z,s)=\sum _{\gamma \in \Gamma _{\infty }\backslash \Gamma }\Im (\gamma z)^{s}=\sum _{(c,d)=1{\bmod {\pm }}1}{\frac {y^{s}}{|cz+d|^{2s}}}.}$

That proves the absolute convergence in ${\displaystyle z\in {\mathcal {H}}}$ for ${\displaystyle \operatorname {Re} (s)>1.}$

Furthermore, it follows that

${\displaystyle \zeta (2s){\tilde {E}}(z,s)=\sum _{n=1}^{\infty }n^{-s}\sum _{(c,d)=1{\bmod {\pm }}1}{\frac {y^{s}}{|cz+d|^{2s}}}=\sum _{n=1}^{\infty }\sum _{(c,d)=1{\bmod {\pm }}1}{\frac {y^{s}}{|ncz+nd|^{2s}}}=\sum _{(m,n)\neq (0,0)}{\frac {y^{s}}{|mz+n|^{2s}}},}$

since the map

${\displaystyle {\begin{cases}\mathbb {N} \times \{(x,y)\in \mathbb {Z} ^{2}-\{(0,0)\}:(x,y)=1\}\to \mathbb {Z} ^{2}-\{(0,0)\}\\(n,(x,y))\mapsto (nx,ny)\end{cases}}}$

is a bijection (a) follows.

(b) We have ${\displaystyle E(\gamma z,s)=E(z,s)}$ for all ${\displaystyle \gamma \in \Gamma (1)}$.

For ${\displaystyle {\tilde {\gamma }}\in \Gamma (1)}$ we get

${\displaystyle {\tilde {E}}({\tilde {\gamma }}z,s)=\sum _{\gamma \in \Gamma _{\infty }\backslash \Gamma }\Im ({\tilde {\gamma }}\gamma z)^{s}=\sum _{\gamma \in \Gamma _{\infty }\backslash \Gamma }\Im (\gamma z)^{s}={\tilde {E}}(z,s).}$

Together with (a), ${\displaystyle E}$ is also invariant under ${\displaystyle \Gamma (1)}$. Q.E.D.

Proposition. E is an eigenform of the hyperbolic Laplace operator

We need the following Lemma:

Lemma: ${\displaystyle \Delta }$ commutes with the operation of ${\displaystyle G}$ on ${\displaystyle C^{\infty }({\mathcal {H}})}$. More precisely for all ${\displaystyle g\in G}$ we have: ${\displaystyle L_{g}\Delta =\Delta L_{g}.}$

Proof: The group ${\displaystyle \mathrm {SL} _{2}(\mathbb {R} )}$ is generated by the elements of the form

${\displaystyle {\begin{pmatrix}a&0\\0&{\frac {1}{a}}\\\end{pmatrix}},a\in \mathbb {R} ^{\times };\quad {\begin{pmatrix}1&x\\0&1\\\end{pmatrix}},x\in \mathbb {R} ;\quad S={\begin{pmatrix}0&-1\\1&0\\\end{pmatrix}}.}$

One calculates the claim for these generators and obtains the claim for all ${\displaystyle g\in \mathrm {SL} _{2}(\mathbb {R} )}$. Q.E.D.

Since ${\displaystyle E(z,s)=\pi ^{-s}\Gamma (s)\zeta (2s){\tilde {E}}(z,s)}$ it is sufficient to show the differential equation for ${\displaystyle {\tilde {E}}}$. We have:

${\displaystyle \Delta {\tilde {E}}(z,s):=\Delta \sum _{\gamma \in \Gamma _{\infty }\backslash \Gamma }\Im (\gamma z)^{s}=\sum _{\gamma \in \Gamma _{\infty }\backslash \Gamma }\Delta \left(\Im (\gamma z)^{s}\right)}$

Furthermore, one has

${\displaystyle \Delta \left(\Im (z)^{s}\right)=\Delta (y^{s})=-y^{2}\left({\frac {\partial ^{2}y^{s}}{\partial x^{2}}}+{\frac {\partial ^{2}y^{s}}{\partial y^{2}}}\right)=s(1-s)y^{s}.}$

Since the Laplace Operator commutes with the Operation of ${\displaystyle \Gamma (1)}$, we get

${\displaystyle \forall \gamma \in \Gamma (1):\quad \Delta \left(\Im (\gamma z)^{s}\right)=s(1-s)\Im (\gamma z)^{s}}$

and so

${\displaystyle \Delta {\tilde {E}}(z,s)=s(1-s){\tilde {E}}(z,s).}$

Therefore, the differential equation holds for E in ${\displaystyle \Re (s)>3}$. In order to obtain the claim for all ${\displaystyle s\in \mathbb {C} }$, consider the function ${\displaystyle \Delta E(z,s)-s(1-s)E(z,s)}$. By explicitly calculating the Fourier expansion of this function, we get that it is meromorphic. Since it vanishes for ${\displaystyle \Re (s)>3}$, it must be the zero function by the identity theorem.

The nonholomorphic Eisenstein series has a Fourier expansion

${\displaystyle E(z,s)=\sum _{n=-\infty }^{\infty }a_{n}(y,s)e^{2\pi inx}}$

where

${\displaystyle {\begin{aligned}a_{0}(y,s)&=\pi ^{-s}\Gamma (s)\zeta (2s)y^{s}+\pi ^{s-1}\Gamma (1-s)\zeta (2(1-s))y^{1-s}\\a_{n}(y,s)&=2|n|^{s-{\frac {1}{2}}}\sigma _{1-2s}(|n|){\sqrt {y}}K_{s-{\frac {1}{2}}}(2\pi |n|y)&&n\neq 0\end{aligned}}}$

If ${\displaystyle z\in {\mathcal {H}}}$, ${\displaystyle E(z,s)}$ has a meromorphic continuation on ${\displaystyle \mathbb {C} }$. It is holomorphic except for simple poles at ${\displaystyle s=0,1.}$

The Eisenstein series satisfies the functional equation

${\displaystyle E(z,s)=E(z,1-s)}$

for all ${\displaystyle z\in {\mathcal {H}}}$.

Locally uniformly in ${\displaystyle x\in \mathbb {R} }$ the growth condition

${\displaystyle E(x+iy,s)={\mathcal {0}}(y^{\sigma })}$

holds, where ${\displaystyle \sigma =\max(\operatorname {Re} (s),1-\operatorname {Re} (s)).}$

The meromorphic continuation of E is very important in the spectral theory of the hyperbolic Laplace operator.

For ${\displaystyle N\in \mathbb {N} }$ let ${\displaystyle \Gamma (N)}$ be the kernel of the canonical projection

${\displaystyle \mathrm {SL} _{2}(\mathbb {Z} )\to \mathrm {SL} _{2}(\mathbb {Z} /N\mathbb {Z} ).}$

We call ${\displaystyle \Gamma (N)}$ principal congruence subgroup of level ${\displaystyle N}$. A subgroup ${\displaystyle \Gamma \subseteq \mathrm {SL} _{2}(\mathbb {Z} )}$ is called congruence subgroup, if there exists ${\displaystyle N\in \mathbb {N} }$, so that ${\displaystyle \Gamma (N)\subseteq \Gamma }$. All congruence subgroups are discrete.

Let

${\displaystyle {\overline {\Gamma (1)}}:=\Gamma (1)/\{\pm 1\}.}$

For a congruence subgroup ${\displaystyle \Gamma ,}$ let ${\displaystyle {\overline {\Gamma }}}$ be the image of ${\displaystyle \Gamma }$ in ${\displaystyle {\overline {\Gamma (1)}}}$. If S is a system of representatives of ${\displaystyle {\overline {\Gamma }}\backslash {\overline {\Gamma (1)}}}$, then

${\displaystyle SD=\bigcup _{\gamma \in S}\gamma D}$

is a fundamental domain for ${\displaystyle \Gamma }$. The set ${\displaystyle S}$ is uniquely determined by the fundamental domain ${\displaystyle SD}$. Furthermore, ${\displaystyle S}$ is finite.

The points ${\displaystyle \gamma \infty }$ for ${\displaystyle \gamma \in S}$ are called cusps of the fundamental domain ${\displaystyle SD}$. They are a subset of ${\displaystyle \mathbb {Q} \cup \{\infty \}}$.

For every cusp ${\displaystyle c}$ there exists ${\displaystyle \sigma \in \Gamma (1)}$ with ${\displaystyle \sigma \infty =c}$.

Let ${\displaystyle \Gamma }$ be a congruence subgroup and ${\displaystyle k\in \mathbb {Z} .}$

We define the hyperbolic Laplace operator ${\displaystyle \Delta _{k}}$ of weight ${\displaystyle k}$ as

${\displaystyle \Delta _{k}:C^{\infty }({\mathcal {H}})\to C^{\infty }({\mathcal {H}}),}$

${\displaystyle \Delta _{k}=-y^{2}\left({\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}\right)+iky{\frac {\partial }{\partial x}}.}$

This is a generalization of the hyperbolic Laplace operator ${\displaystyle \Delta _{0}=\Delta }$.

We define an operation of ${\displaystyle \mathrm {SL} _{2}(\mathbb {R} )}$ on ${\displaystyle C^{\infty }({\mathcal {H}})}$ by

${\displaystyle f_{||k}g(z):=\left({\frac {cz+d}{|cz+d|}}\right)^{-k}f(gz)}$

where

${\displaystyle z\in {\mathcal {H}},g={\begin{pmatrix}\ast &\ast \\c&d\\\end{pmatrix}}\in \mathrm {SL} _{2}(\mathbb {R} ),f\in C^{\infty }({\mathcal {H}}).}$

It can be shown that

${\displaystyle (\Delta _{k}f)_{||k}g=\Delta _{k}(f_{||k}g)}$

holds for all ${\displaystyle f\in C^{\infty }({\mathcal {H}}),k\in \mathbb {Z} }$ and every ${\displaystyle g\in \mathrm {SL} _{2}(\mathbb {R} )}$.

Therefore, ${\displaystyle \Delta _{k}}$ operates on the vector space

${\displaystyle C^{\infty }(\Gamma \backslash {\mathcal {H}},k):=\{f\in C^{\infty }({\mathcal {H}}):f_{||k}\gamma =f\forall \gamma \in \Gamma \}}$.

Definition. A Maass form of weight ${\displaystyle k\in \mathbb {Z} }$ for ${\displaystyle \Gamma }$ is a function ${\displaystyle f\in C^{\infty }(\Gamma \backslash {\mathcal {H}},k)}$ that is an eigenfunction of ${\displaystyle \Delta _{k}}$ and is of moderate growth at the cusps.

The term moderate growth at cusps needs clarification. Infinity is a cusp for ${\displaystyle \Gamma ,}$ a function ${\displaystyle f\in C^{\infty }(\Gamma \backslash {\mathcal {H}},k)}$ is of moderate growth at ${\displaystyle \infty }$ if ${\displaystyle f(x+iy)}$ is bounded by a polynomial in y as ${\displaystyle y\to \infty }$. Let ${\displaystyle c\in \mathbb {Q} }$ be another cusp. Then there exists ${\displaystyle \theta \in \mathrm {SL} _{2}(\mathbb {Z} )}$ with ${\displaystyle \theta (\infty )=c}$. Let ${\displaystyle f':=f_{||k}\theta }$. Then ${\displaystyle f'\in C^{\infty }(\Gamma '\backslash {\mathcal {H}},k)}$, where ${\displaystyle \Gamma '}$ is the congruence subgroup ${\displaystyle \theta ^{-1}\Gamma \theta }$. We say ${\displaystyle f}$ is of moderate growth at the cusp ${\displaystyle c}$, if ${\displaystyle f'}$ is of moderate growth at ${\displaystyle \infty }$.

Definition. If ${\displaystyle \Gamma }$ contains a principal congruence subgroup of level ${\displaystyle N}$, we say that ${\displaystyle f}$ is cuspidal at infinity, if

${\displaystyle \forall z\in {\mathcal {H}}:\quad \int _{0}^{N}f(z+u)du=0.}$

We say that ${\displaystyle f}$ is cuspidal at the cusp ${\displaystyle c}$ if ${\displaystyle f'}$ is cuspidal at infinity. If ${\displaystyle f}$ is cuspidal at every cusp, we call ${\displaystyle f}$ a cusp form.

We give a simple example of a Maass form of weight ${\displaystyle k>1}$ for the modular group:

Example. Let ${\displaystyle g:{\mathcal {H}}\to \mathbb {C} }$ be a modular form of even weight ${\displaystyle k}$ for ${\displaystyle \Gamma (1).}$ Then ${\displaystyle f(z):=y^{\frac {k}{2}}g(z)}$ is a Maass form of weight ${\displaystyle k}$ for the group ${\displaystyle \Gamma (1)}$.

Let ${\displaystyle \Gamma }$ be a congruence subgroup of ${\displaystyle \mathrm {SL} _{2}(\mathbb {R} )}$ and let ${\displaystyle L^{2}(\Gamma \backslash {\mathcal {H}},k)}$ be the vector space of all measurable functions ${\displaystyle f:{\mathcal {H}}\to \mathbb {C} }$ with ${\displaystyle f_{||k}\gamma =f}$ for all ${\displaystyle \gamma \in \Gamma }$ satisfying

${\displaystyle \|f\|^{2}:=\int _{\Gamma \backslash {\mathcal {H}}}|f(z)|^{2}d\mu (z)<\infty }$

modulo functions with ${\displaystyle \|f\|=0.}$ The integral is well defined, since the function ${\displaystyle |f(z)|^{2}}$ is ${\displaystyle \Gamma }$-invariant. This is a Hilbert space with inner product

${\displaystyle \langle f,g\rangle =\int _{\Gamma \backslash {\mathcal {H}}}f(z){\overline {g(z)}}d\mu (z).}$

The operator ${\displaystyle \Delta _{k}}$ can be defined in a vector space ${\displaystyle B\subset L^{2}(\Gamma \backslash {\mathcal {H}},k)\cap C^{\infty }(\Gamma \backslash {\mathcal {H}},k)}$ which is dense in ${\displaystyle L^{2}(\Gamma \backslash {\mathcal {H}},k)}$. There ${\displaystyle \Delta _{k}}$ is a positive semidefinite symmetric operator. It can be shown, that there exists a unique self-adjoint continuation on ${\displaystyle L^{2}(\Gamma \backslash {\mathcal {H}},k).}$

Define ${\displaystyle C(\Gamma \backslash {\mathcal {H}},k)}$ as the space of all cusp forms ${\displaystyle L^{2}(\Gamma \backslash {\mathcal {H}},k)\cap C^{\infty }(\Gamma \backslash {\mathcal {H}},k).}$ Then ${\displaystyle \Delta _{k}}$ operates on ${\displaystyle C(\Gamma \backslash {\mathcal {H}},k)}$ and has a discrete spectrum. The spectrum belonging to the orthogonal complement has a continuous part and can be described with the help of (modified) non-holomorphic Eisenstein series, their meromorphic continuations and their residues. (See Bump or Iwaniec).

If ${\displaystyle \Gamma }$ is a discrete (torsion free) subgroup of ${\displaystyle \mathrm {SL} _{2}(\mathbb {R} )}$, so that the quotient ${\displaystyle \Gamma \backslash {\mathcal {H}}}$ is compact, the spectral problem simplifies. This is because a discrete cocompact subgroup has no cusps. Here all of the space ${\displaystyle L^{2}(\Gamma \backslash {\mathcal {H}},k)}$ is a sum of eigenspaces.

${\displaystyle G=\mathrm {SL} _{2}(\mathbb {R} )}$ is a locally compact unimodular group with the topology of ${\displaystyle \mathbb {R} ^{4}.}$ Let ${\displaystyle \Gamma }$ be a congruence subgroup. Since ${\displaystyle \Gamma }$ is discrete in ${\displaystyle G}$, it is closed in ${\displaystyle G}$ as well. The group ${\displaystyle G}$ is unimodular and since the counting measure is a Haar-measure on the discrete group ${\displaystyle \Gamma }$, ${\displaystyle \Gamma }$ is also unimodular. By the Quotient Integral Formula there exists a ${\displaystyle G}$-right-invariant Radon measure ${\displaystyle dx}$ on the locally compact space ${\displaystyle \Gamma \backslash G}$. Let ${\displaystyle L^{2}(\Gamma \backslash G)}$ be the corresponding ${\displaystyle L^{2}}$-space. This space decomposes into a Hilbert space direct sum:

${\displaystyle L^{2}(\Gamma \backslash G)=\bigoplus _{k\in \mathbb {Z} }L^{2}(\Gamma \backslash G,k)}$

where

${\displaystyle L^{2}(\Gamma \backslash G,k):=\left\{\phi \in L^{2}(\Gamma \backslash G)\mid \phi (xk_{\theta })=e^{ik\theta }F(x)\forall x\in \Gamma \backslash G\forall \theta \in \mathbb {R} \right\}}$

and

${\displaystyle k_{\theta }={\begin{pmatrix}\cos(\theta )&-\sin(\theta )\\\sin(\theta )&\cos(\theta )\\\end{pmatrix}}\in SO(2),\theta \in \mathbb {R} .}$

The Hilbert-space ${\displaystyle L^{2}(\Gamma \backslash {\mathcal {H}},k)}$ can be embedded isometrically into the Hilbert space ${\displaystyle L^{2}(\Gamma \backslash G,k)}$. The isometry is given by the map

${\displaystyle {\begin{cases}\psi _{k}:L^{2}(\Gamma \backslash {\mathcal {H}},k)\to L^{2}(\Gamma \backslash G,k)\\\psi _{k}(f)(g):=f_{||k}\gamma (i)\end{cases}}}$

Therefore, all Maass cusp forms for the congruence group ${\displaystyle \Gamma }$ can be thought of as elements of ${\displaystyle L^{2}(\Gamma \backslash G)}$.

${\displaystyle L^{2}(\Gamma \backslash G)}$ is a Hilbert space carrying an operation of the group ${\displaystyle G}$, the so-called right regular representation:

${\displaystyle R_{g}\phi :=\phi (xg),{\text{ where }}x\in \Gamma \backslash G{\text{ and }}\phi \in L^{2}(\Gamma \backslash G).}$

One can easily show, that ${\displaystyle R}$ is a unitary representation of ${\displaystyle G}$ on the Hilbert space ${\displaystyle L^{2}(\Gamma \backslash G)}$. One is interested in a decomposition into irreducible subrepresentations. This is only possible if ${\displaystyle \Gamma }$ is cocompact. If not, there is also a continuous Hilbert-integral part. The interesting part is, that the solution of this problem also solves the spectral problem of Maass forms. (see Bump, C. 2.3)