Definition of the hyperbolic Laplace operator

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

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

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

Definition of a Maass form

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

1. ${\displaystyle f(\gamma z)=f(z){\text{ for all }}\gamma \in \Gamma (1),\qquad z\in \mathbb {H} }$
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

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

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

Relation between Maass forms and Dirichlet series

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

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

we have:

${\displaystyle \forall z\in \mathbb {H}$ :\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 ${\displaystyle s\in \mathbb {C} ,y>0}$.

Theorem (Fourier coefficients of Maass forms) — Let ${\displaystyle \lambda \in \mathbb {C} }$ be the eigenvalue of the Maass form ${\displaystyle f}$ corresponding to ${\displaystyle \Delta .}$ There exist ${\displaystyle \nu \in \mathbb {C} }$, unique up to sign, such that ${\textstyle \lambda ={\frac {1}{4}}-\nu ^{2}}$. Then the Fourier coefficients of ${\displaystyle f}$ are

$${\displaystyle {\begin{aligned}a_{n}(y)&=c_{n}{\sqrt {y}}K_{\nu }(2\pi |n|y)\quad c_{n}\in \mathbb {C} &&n\neq 0\\a_{0}(y)&=c_{0}y^{{\frac {1}{2}}-\nu }+d_{0}y^{{\frac {1}{2}}+\nu }\quad c_{0},d_{0}\in \mathbb {C} &&n=0\end{aligned}}}$$

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 ${\displaystyle f:\mathbb {H} \to \mathbb {C} }$ 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 ${\displaystyle f={\tfrac {1}{2}}(f+i(f))+{\tfrac {1}{2}}(f-i(f))}$.

E is a Maass form

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 ${\displaystyle \mathbb {H} \cup \{\infty \}}$.

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 ${\displaystyle z\in \mathbb {H} }$ 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 ${\displaystyle z\in \mathbb {H} }$ 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

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

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 ${\displaystyle C^{\infty }(\mathbb {H} )}$. 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.

Congruence subgroups

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 ${\displaystyle \sigma \infty =c}$.

Maass forms of weight k

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

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

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

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 ${\displaystyle C^{\infty }(\mathbb {H} )}$ by

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

where

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

It can be shown that

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

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

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

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

Definition. A Maass form of weight ${\displaystyle k\in \mathbb {Z} }$ for ${\displaystyle \Gamma }$ is a function ${\displaystyle f\in C^{\infty }(\Gamma \backslash \mathbb {H} ,k)}$ 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 ${\displaystyle f\in C^{\infty }(\Gamma \backslash \mathbb {H} ,k)}$ 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 ${\displaystyle f'\in C^{\infty }(\Gamma '\backslash \mathbb {H} ,k)}$, 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

${\displaystyle \forall z\in \mathbb {H}$ :\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 ${\displaystyle g:\mathbb {H} \to \mathbb {C} }$ 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 ${\displaystyle \Gamma (1)}$.

The spectral problem

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

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

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

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

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

Define ${\displaystyle C(\Gamma \backslash \mathbb {H} ,k)}$ as the space of all cusp forms ${\displaystyle L^{2}(\Gamma \backslash \mathbb {H} ,k)\cap C^{\infty }(\Gamma \backslash \mathbb {H} ,k).}$ Then ${\displaystyle \Delta _{k}}$ operates on ${\displaystyle C(\Gamma \backslash \mathbb {H} ,k)}$ 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 ${\displaystyle \Gamma \backslash \mathbb {H} }$ is compact, the spectral problem simplifies. This is because a discrete cocompact subgroup has no cusps. Here all of the space ${\displaystyle L^{2}(\Gamma \backslash \mathbb {H} ,k)}$ is a sum of eigenspaces.

Embedding into the space L²(Γ \ G)

${\displaystyle G=\mathrm {SL} _{2}(\mathbb {R} )}$ 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 ${\displaystyle L^{2}(\Gamma \backslash \mathbb {H} ,k)}$ can be embedded isometrically into the Hilbert space ${\displaystyle L^{2}(\Gamma \backslash G,k)}$. The isometry is given by the map

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

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)

Definition

Let k be an integer, s be a complex number, and Γ be a discrete subgroup of SL₂(R). A Maass form of weight k for Γ with Laplace eigenvalue s is a smooth function from the upper half-plane to the complex numbers satisfying the following conditions:

• For all ${\displaystyle \gamma =\left({\begin{smallmatrix}a&b\\c&d\end{smallmatrix}}\right)\in \Gamma }$ and all ${\displaystyle z\in \mathbb {H} }$, we have ${\displaystyle f\left({\frac {az+b}{cz+d}}\right)=\left({\frac {cz+d}{|cz+d|}}\right)^{k}f(z).}$
• We have ${\displaystyle \Delta _{k}f=sf}$, where ${\displaystyle \Delta _{k}}$ is the weight k hyperbolic Laplacian defined as
  $${\displaystyle \Delta _{k}=-y^{2}\left({\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}\right)+iky{\frac {\partial }{\partial x}}.}$$

  
• The function ${\displaystyle f}$ is of at most polynomial growth at cusps.

A weak Maass form is defined similarly but with the third condition replaced by "The function ${\displaystyle f}$ has at most linear exponential growth at cusps". Moreover, ${\displaystyle f}$ is said to be harmonic if it is annihilated by the Laplacian operator.

Major results

Let ${\displaystyle f}$ be a weight 0 Maass cusp form. Its normalized Fourier coefficient at a prime p is bounded by p^7/64 + p^−7/64. This theorem is due to Henry Kim and Peter Sarnak. It is an approximation toward Ramanujan-Petersson conjecture.

Higher dimensions

Maass cusp forms can be regarded as automorphic forms on GL(2). It is natural to define Maass cusp forms on GL(n) as spherical automorphic forms on GL(n) over the rational number field. Their existence is proved by Miller, Mueller, etc.

The group GL₂(A)

Let ${\displaystyle R}$ be a commutative ring with unit and let ${\displaystyle G_{R}:=\mathrm {GL} _{2}(R)}$ be the group of ${\displaystyle 2\times 2}$ matrices with entries in ${\displaystyle R}$ and invertible determinant. Let ${\displaystyle \mathbb {A} =\mathbb {A} _{\mathbb {Q} }}$ be the ring of rational adeles, ${\displaystyle \mathbb {A} _{\text{fin}}}$ the ring of the finite (rational) adeles and for a prime number ${\displaystyle p\in \mathbb {N} }$ let ${\displaystyle \mathbb {Q} _{p}}$ be the field of p-adic numbers. Furthermore, let ${\displaystyle \mathbb {Z} _{p}}$ be the ring of the p-adic integers (see Adele ring). Define ${\displaystyle G_{p}:=G_{\mathbb {Q} _{p}}}$. Both ${\displaystyle G_{p}}$ and ${\displaystyle G_{\mathbb {R} }}$ are locally compact unimodular groups if one equips them with the subspace topologies of ${\displaystyle \mathbb {Q} _{p}^{4}}$ respectively ${\displaystyle \mathbb {R} ^{4}}$. Then:

${\displaystyle G_{\text{fin}}:=G_{\mathbb {A} _{\text{fin}}}\cong {\widehat {\prod _{p<\infty }^{K_{p}}}}G_{p}.}$

The right side is the restricted product, concerning the compact, open subgroups ${\displaystyle K_{p}:=G_{\mathbb {Z} _{p}}}$ of ${\displaystyle G_{p}}$. Then ${\displaystyle G_{\text{fin}}}$ locally compact group, if we equip it with the restricted product topology.