Maass waveform

Maaß's forms or Maaß's wave forms are examined in the theory of automorphic forms , a branch of mathematics. In the classical sense, Maaß's forms are complex-valued , smooth functions of the upper half-level , which have a similar transformation behavior under the operation of a discrete subgroup of on the upper half-level as that of the module forms . They are natural forms of the hyperbolic Laplacian on and meet certain growth conditions in the peaks of a fundamental domain of . In contrast to the modular forms, Maaß's forms do not have to be holomorphic . They were first examined by Hans Maaß in 1949. ${\displaystyle \mathbb {H} }$ ${\displaystyle \Gamma }$ ${\displaystyle SL_{2}(\mathbb {R} )}$ ${\displaystyle \Delta }$ ${\displaystyle \mathbb {H} }$ ${\displaystyle \Gamma }$

General

The special linear group

${\displaystyle G:=SL_{2}(\mathbb {R} )={\Bigg \{}{\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}\in M_{2}(\mathbb {R} )\,\mid \,ad\,-\,bc=1{\Bigg \}}}$

operates on the upper half plane through the Möbius transformations ${\displaystyle \mathbb {H} =\{z\in \mathbb {C} \mid \operatorname {Im} (z)>0\}}$

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

This operation can be expanded to an operation on by defining: ${\displaystyle \mathbb {H} \cup \{\infty \}\cup \mathbb {\mathbb {R} } }$

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

${\displaystyle {\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}.\infty :=\lim \limits _{\operatorname {Im} (z)\rightarrow \infty }{{\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}.z}={\begin{cases}{\frac {a}{c}}&,{\text{ falls }}c\neq 0\\\infty &,{\text{ falls }}c=0\end{cases}}}$

The upper half level is through ${\displaystyle \mathbb {H} }$

${\displaystyle \mathrm {d} \mu :={\frac {\mathrm {d} x\,\mathrm {d} y}{y^{2}}}}$

a radon measure invariant under the operation given. ${\displaystyle SL_{2}(\mathbb {R} )}$

Be a discrete subgroup of . A fundamental domain zu is an open subset , so that a representative system of exists with ${\displaystyle \Gamma }$ ${\displaystyle G}$ ${\displaystyle \Gamma }$ ${\displaystyle F\subset \mathbb {H} }$ ${\displaystyle R}$ ${\displaystyle \Gamma \setminus \mathbb {H} }$

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

A fundamental area for the module group is given by ${\displaystyle \Gamma (1):=SL_{2}(\mathbb {Z} )}$

${\displaystyle D:=\{z\in \mathbb {H} \,\mid \,|Re(z)|<{\frac {1}{2}}\,,\,|z|<1\}}$

(see module form ). A function is called -invariant if holds for each and every . For every measurable invariant function then applies ${\displaystyle f\colon \mathbb {H} \to \mathbb {C} }$ ${\displaystyle \Gamma }$ ${\displaystyle f(\gamma .z)=f(z)}$ ${\displaystyle \gamma \in \Gamma }$ ${\displaystyle z\in \mathbb {H} }$ ${\displaystyle \Gamma }$ ${\displaystyle f\colon \mathbb {H} \to \mathbb {C} }$

${\displaystyle \int _{F}f\,\mathrm {d} \mu =\int _{\Gamma \setminus \mathbb {H} }f\,\mathrm {d} \mu }$,

where that on the right side of the equation represents the measure induced on the quotient . ${\displaystyle \mathrm {d} \mu }$

Classic Maass waveforms

Definition of the hyperbolic Laplace operator

The hyperbolic Laplacian operator on the half plane is defined by ${\displaystyle \mathbb {H} }$

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

With

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

This corresponds precisely to the (generalized) Laplace operator or Laplace-Beltrami operator with regard to the hyperbolic metric on the hyperbolic level . ${\displaystyle \mathbb {H} }$

Definition of a Maass waveform

A Maaßsche waveform to group is a smooth function on , so ${\displaystyle \Gamma (1):=SL_{2}(\mathbb {Z} )}$ ${\displaystyle f}$ ${\displaystyle \mathbb {H} }$

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

Also applies

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

then one calls a Maaß tip shape. ${\displaystyle f}$

Relationship between Maass waveforms and Dirichlet series

Now be a Maaß waveform. Then because of ${\displaystyle f}$ ${\displaystyle \gamma :={\begin{pmatrix}1&1\\0&1\\\end{pmatrix}}\in \Gamma (1)}$

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

Thus has a Fourier expansion of the shape ${\displaystyle f}$

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

with coefficient functions One can check that there is a Maaß peak shape if and only if applies. These coefficient functions can be specified precisely; the K-Bessel function is required for this. ${\displaystyle a_{n}\,,n\in \mathbb {N} .}$ ${\displaystyle f}$ ${\displaystyle \forall y>0:\ a_{0}(y)=0}$

Definition: The K-Bessel function is defined by ${\displaystyle y>0}$

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

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

${\displaystyle K_{s}(y)\leq e^{-y/2}K_{\operatorname {Re} (s)}(2),}$if . ${\displaystyle y>4}$

Thus falls exponentially in terms of amount for . Furthermore applies for all , . ${\displaystyle K_{s}}$ ${\displaystyle y\to \infty }$ ${\displaystyle K_{-s}(y)=K_{s}(y)}$ ${\displaystyle s\in \mathbb {C} }$ ${\displaystyle y>0}$

Theorem: Fourier coefficients of a Maaß waveform

Let be the eigenvalue of the Maaß waveform with respect to . Let the complex number be unique except for the sign . Then applies to the Fourier coefficient functions of ${\displaystyle \lambda \in \mathbb {C} }$ ${\displaystyle f}$ ${\displaystyle \Delta }$ ${\displaystyle \nu \in \mathbb {C} }$ ${\displaystyle \lambda ={\tfrac {1}{4}}-\nu ^{2}}$ ${\displaystyle f}$

${\displaystyle a_{n}(y)=c_{n}{\sqrt {y}}K_{\nu }(2\pi |n|y)\,\,\,,\,\,\,c_{n}\in \mathbb {C} ,}$

if . Is , then applies ${\displaystyle n\neq 0}$ ${\displaystyle n=0}$

${\displaystyle a_{0}(y)=c_{0}y^{{\frac {1}{2}}-\nu }+d_{0}y^{{\frac {1}{2}}+\nu }}$with . ${\displaystyle c_{0},\,d_{0}\in \mathbb {C} }$

Proof: It applies . According to the definition of Fourier coefficients, for ${\displaystyle \Delta (f)=({\tfrac {1}{4}}-\nu ^{2})f}$ ${\displaystyle n\in \mathbb {Z} }$

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

Together it follows for : ${\displaystyle n\in \mathbb {Z} }$

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

In (1) it was used for the first summand that the -th Fourier coefficient of is exact , since we are allowed to differentiate Fourier series term by term. In the second summand, the order of integration and differentiation was changed, which is allowed, since it is continuously differentiable in y as often as desired and one integrates via a compact. The following linear differential equation of the second order results: ${\displaystyle n}$ ${\displaystyle {\frac {\partial ^{2}f}{\partial x^{2}}}}$ ${\displaystyle (2\pi in)^{2}a_{n}(y)}$ ${\displaystyle f}$

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

For one can show that there are unique coefficients for every solution of this differential equation , so that . ${\displaystyle n=0}$ ${\displaystyle f}$ ${\displaystyle c_{0},d_{0}\in \mathbb {C} }$ ${\displaystyle f(y)=c_{0}y^{{\frac {1}{4}}-\nu }+d_{0}y^{{\frac {1}{4}}+\nu }}$

For each solution of the above differential equation is of the form ${\displaystyle n\neq 0}$ ${\displaystyle f}$

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

for unambiguous , where is the K-Bessel function and the I-Bessel functions (see O. Forster). ${\displaystyle c_{n},d_{n}\in \mathbb {C} }$ ${\displaystyle K_{v}(s)}$ ${\displaystyle I_{v}(s)}$

Since the I-Bessel function grows exponentially and the K-Bessel function falls exponentially, it follows with requirement 3) the at most polynomial growth of ${\displaystyle f:}$

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

(so ) for a clear one ${\displaystyle d_{n}=0}$ ${\displaystyle c_{n}\in \mathbb {C} .\,\,\,\square }$

Even and odd Maaß waveforms: Let . Then operates on all functions of the upper half-level via . It is easy to calculate that with mixed up. We call a Maaß waveform even if and odd if . If a Maaß waveform is an even Maaß waveform and an odd Maaß waveform, and it applies . ${\displaystyle i(z):=-{\overline {z}}}$ ${\displaystyle i}$ ${\displaystyle f}$ ${\displaystyle \mathbb {H} }$ ${\displaystyle i(f):=f(i(z))}$ ${\displaystyle \Delta }$ ${\displaystyle i}$ ${\displaystyle f}$ ${\displaystyle i(f)=f}$ ${\displaystyle i(f)=-f}$ ${\displaystyle f}$ ${\displaystyle {\tfrac {1}{2}}(f+i(f))}$ ${\displaystyle {\tfrac {1}{2}}(f-i(f))}$ ${\displaystyle f={\tfrac {1}{2}}(f+i(f))+{\tfrac {1}{2}}(f-i(f))}$

Theorem: L function of a Maaß waveform

Be a Maaß tip shape. We define the so-called L-function of as ${\displaystyle f(x+iy)=\sum _{n\neq 0}c_{n}{\sqrt {y}}K_{\nu }(2\pi |n|y)e^{2\pi inx}}$ ${\displaystyle f}$

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

Then the series converges for and you can get to an entire function to continue. ${\displaystyle L(s,f)}$ ${\displaystyle \operatorname {Re} (s)>{\tfrac {3}{2}}}$ ${\displaystyle \mathbb {C} }$

Is even or odd, that's how you define ${\displaystyle f}$

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

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

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

Proof:

Be a Maaß tip shape. First, let's see how quickly the Fourier coefficients of grow. ${\displaystyle f}$ ${\displaystyle f}$

Assertion: It applies ${\displaystyle c_{n}={\mathcal {O}}(n^{\frac {1}{2}})}$

Proof: Since there is a Maaß point shape, exist such that the inequality holds. If and is conjugated to modulo , it is easy to calculate that it is true. Since is invariant , for : ${\displaystyle f}$ ${\displaystyle C,N>0}$ ${\displaystyle y>1}$ ${\displaystyle |f(x+iy)|\leq Cy^{N}}$ ${\displaystyle y<{\tfrac {1}{2}}}$ ${\displaystyle \omega \in D}$ ${\displaystyle z=x+iy}$ ${\displaystyle \Gamma (1)}$ ${\displaystyle \operatorname {Im} (\omega )\leq {\tfrac {1}{y}}}$ ${\displaystyle f}$ ${\displaystyle \Gamma (1)}$ ${\displaystyle y<{\tfrac {1}{2}}}$

${\displaystyle |f(x+iy)|\leq Cy^{-N}}$.

So it applies to the estimate ${\displaystyle y<{\frac {1}{2}}}$

${\displaystyle |c_{n}|{\sqrt {y}}|K_{\nu }(2\pi |n|y)\leq \int _{0}^{1}|f(x+iy)|\mathrm {d} x\leq Cy^{-N}}$.

For and thus applies ${\displaystyle n\in \mathbb {Z} \,,|n|>2}$ ${\displaystyle y={\tfrac {1}{|n|}}}$

${\displaystyle |c_{n}|\leq Cr^{N+{\frac {1}{2}}}|K_{\nu }(2\pi )|^{-1}}$.

With this we find a constant such that for each one applies: ${\displaystyle D\geq C}$ ${\displaystyle n\neq 0}$

${\displaystyle |c_{n}|\leq Dr^{N+{\frac {1}{2}}}|K_{\nu }(2\pi )|^{-1}}$

But now the K-Bessel function falls exponentially and is a Maaß tip shape. Together it follows that on the fundamental domain of is restricted and thus on . So we can repeat the above proof with and get for one , so . ${\displaystyle f}$ ${\displaystyle f}$ ${\displaystyle \Gamma (1)}$ ${\displaystyle \mathbb {H} }$ ${\displaystyle N=0}$ ${\displaystyle |c_{n}|\leq Kn^{\frac {1}{2}}}$ ${\displaystyle K<\infty }$ ${\displaystyle c_{n}={\mathcal {O}}(n^{\frac {1}{2}})\,.\,\,\square }$

The series thus converges for . ${\displaystyle L(s,f)}$ ${\displaystyle \operatorname {Re} (s)>{\tfrac {3}{2}}}$

To prove the second part of the theorem, we still need the Mellin transform of . ${\displaystyle K_{\nu }}$

For the integral converges ${\displaystyle \operatorname {Re} (s)>|\operatorname {Re} (\nu )|}$

${\displaystyle \int _{0}^{\infty }K_{\nu }(y)y^{s}{\frac {\mathrm {d} y}{y}}}$

absolutely and it applies

${\displaystyle \int _{0}^{\infty }K_{\nu }(y)y^{s}{\frac {\mathrm {d} y}{y}}=2^{s-2}\Gamma ({\frac {s+\nu }{2}})\Gamma ({\frac {s-\nu }{2}})}$.

If it is even or odd, it follows from the uniqueness of the Fourier coefficients for all . ${\displaystyle f}$ ${\displaystyle a_{n}=(-1)^{\epsilon }}$ ${\displaystyle n\in \mathbb {N} }$

Be straight. The odd case works in a similar way and is therefore not shown here. Then: ${\displaystyle f}$ ${\displaystyle f}$

${\displaystyle {\begin{aligned}\int _{0}^{\infty }f(iy)y^{s-{\frac {1}{2}}}{\frac {\mathrm {d} y}{y}}&=2\int _{0}^{\infty }\sum _{n=1}^{\infty }c_{n}y^{s-1}K_{\nu }(2\pi ny)\,\mathrm {d} y\\&=2\sum _{n=1}^{\infty }c_{n}\int _{0}^{\infty }y^{s-1}K_{\nu }(2\pi ny)\mathrm {d} y\\&=2\sum _{n=1}^{\infty }c_{n}(2\pi n)^{-s}\int _{0}^{\infty }y^{s-1}K_{\nu }(y)\mathrm {d} y\\&=2(2\pi )^{-s}\left(\sum _{n=1}^{\infty }c_{n}n^{-s}\right)2^{s-2}\Gamma \left({\frac {s+\nu }{2}}\right)\Gamma \left({\frac {s-\nu }{2}}\right)\\&={\frac {1}{2}}\pi ^{s}\Gamma \left({\frac {s+\nu }{2}}\right)\Gamma \left({\frac {s-\nu }{2}}\right)L(s,f)\\&={\frac {1}{2}}\Lambda (s,f)\end{aligned}}}$

The interchanging of the order of integral and sum is shown, for example, with majorized convergence, whereby one makes use of the fact that for the K-Bessel function : ${\displaystyle y>4}$

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

One also shows that for falls exponentially. ${\displaystyle f(iy)}$ ${\displaystyle y\to \infty }$

We define now

${\displaystyle {\begin{aligned}\Lambda _{1}(s,f)&:=\int _{1}^{\infty }f(iy)y^{s-{\frac {1}{2}}}{\frac {\mathrm {d} y}{y}}\\\Lambda _{2}(s,f)&:=\int _{0}^{1}f(iy)y^{s-{\frac {1}{2}}}{\frac {\mathrm {d} y}{y}}\end{aligned}}}$

This applies . Since it falls exponentially for , it converges for each and thus is a whole function (complex analysis). But now is invariant under , which in particular follows. ${\displaystyle \Lambda =2(\Lambda _{1}+\Lambda _{2})}$ ${\displaystyle f(iy)}$ ${\displaystyle y\to \infty }$ ${\displaystyle \Lambda _{1}(s,f)}$ ${\displaystyle s\in \mathbb {C} }$ ${\displaystyle \Lambda _{1}}$ ${\displaystyle f}$ ${\displaystyle \Gamma (1)}$ ${\displaystyle f(iy)=f\left({\begin{pmatrix}0&1\\-1&0\\\end{pmatrix}}.iy\right)=f\left({\frac {1}{-iy}}\right)=f\left(i{\frac {1}{y}}\right)}$

We now get:

${\displaystyle {\begin{aligned}\Lambda _{2}(s,f)&:=\int _{0}^{1}f(iy)y^{s-{\frac {1}{2}}}{\frac {\mathrm {d} y}{y}}\\&=\int _{1}^{\infty }f(i{\frac {1}{y}})y^{-s+{\frac {1}{2}}}{\frac {\mathrm {d} y}{y}}\\&=\int _{1}^{\infty }f(iy)y^{(1-s)-{\frac {1}{2}}}{\frac {\mathrm {d} y}{y}}\\&=\Lambda _{1}(1-s,f)\end{aligned}}}$

So that is also a whole function and so is whole. In particular, it can be used to continue an entire function . Furthermore applies to the functional equation ${\displaystyle \Lambda _{2}}$ ${\displaystyle \Lambda }$ ${\displaystyle L\left(s,f\right)}$ ${\displaystyle \mathbb {C} }$ ${\displaystyle \Lambda }$

${\displaystyle \Lambda (s,f)=2(\Lambda _{1}(s,f)+\Lambda _{2}(s,f)=2(\Lambda _{1}(s,f)+\Lambda _{1}(1-s,f))=\Lambda (1-s,f)}$.

In particular, it can be continued to completely holomorphic and the theorem is proven. ${\displaystyle L}$ ${\displaystyle \mathbb {C} }$ ${\displaystyle \,\,\,\,\,\,\,\,\square }$

Example: The non-holomorphic Eisenstein series E

The non-holomorphic Eisenstein series is defined for and by ${\displaystyle z\in \mathbb {H} }$ ${\displaystyle s\in \mathbb {C} }$

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

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

The above series converges absolutely in for and locally uniformly in , because one can show that the series converges absolutely in if . More precisely, the sum even converges evenly on every quantity , for every compact and each . ${\displaystyle z\in \mathbb {H} }$ ${\displaystyle \operatorname {Re} (s)>1}$ ${\displaystyle \mathbb {H} \times \{s\mid \operatorname {Re} (s)>1\}}$ ${\displaystyle S(z,s):=\sum _{(m,n)\neq (0,0)}{\frac {1}{|mz+n|^{s}}}}$ ${\displaystyle z\in \mathbb {H} }$ ${\displaystyle \operatorname {Re} (s)>2}$ ${\displaystyle K\times \{s\mid \operatorname {Re} (s)\geq \alpha \}}$ ${\displaystyle K\subset \mathbb {H} }$ ${\displaystyle \alpha >2}$

In particular, the limit of continuous functions is continuous in . For solid is even holomorphic in , since according to Weierstrass the locally uniform limit of holomorphic functions is again holomorphic. ${\displaystyle E}$ ${\displaystyle \mathbb {H} \times \{s\mid \operatorname {Re} (s)>1\}}$ ${\displaystyle z\in \mathbb {H} }$ ${\displaystyle E}$ ${\displaystyle \{s\mid \operatorname {Re} (s)>1\}}$

Theorem: E is a Maaß waveform

We only show the -invariance and the eigen-equation here. Proof of smoothness can be found in Deitmar or Bump. The growth condition follows from the theorem of the Fourier expansion of E. ${\displaystyle SL_{2}(\mathbb {Z} )}$

First about -invariance. Be ${\displaystyle SL_{2}(\mathbb {Z} )}$

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

the stabilizer group from regarding the operation from on . Then the following applies. ${\displaystyle \infty }$ ${\displaystyle SL_{2}(\mathbb {Z} )}$ ${\displaystyle \mathbb {H} \cup \{\infty \}}$

Lemma: The figure

${\displaystyle {\begin{aligned}\Gamma _{\infty }\backslash \Gamma &\to \{\pm (x,y)\in \mathbb {Z} ^{2}/\{\pm 1\}\mid \operatorname {ggT} (x,y)=1\}\\\Gamma _{\infty }{\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}&\mapsto \pm (c,d)\end{aligned}}}$

is a bijection.

Proposition: E is Γ (1) -invariant

(a) Be . Then absolutely converges in for and we have: ${\displaystyle {\tilde {E}}(z,s):=\sum _{\gamma \in \Gamma _{\infty }\backslash \Gamma }\operatorname {Im} (\gamma z)^{s}}$ ${\displaystyle {\tilde {E}}}$ ${\displaystyle z\in \mathbb {H} }$ ${\displaystyle \operatorname {Re} (s)>1}$

${\displaystyle E(z,s)=\pi ^{-s}\Gamma (s)\zeta (2s){\tilde {E}}(z,s)}$

(b) It applies to everyone . ${\displaystyle E(\gamma z,s)=E(z,s)}$ ${\displaystyle \gamma \in \Gamma (1)}$

Proof:

To (a): For true . This follows from the above lemma ${\displaystyle \gamma ={\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}\in \Gamma (1)}$ ${\displaystyle \operatorname {Im} (\gamma z)={\frac {\operatorname {Im} (z)}{|cz+d|^{2}}}}$

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

This leads to the absolute convergence in for . ${\displaystyle z\in \mathbb {H} }$ ${\displaystyle \operatorname {Re} (s)>1}$

It also follows

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

because the image is a bijection. ${\displaystyle \mathbb {N} \times \{(x,y)\in \mathbb {Z} ^{2}-\{(0,0)\}\mid \operatorname {ggT} (x,y)=1\}\to \mathbb {Z} ^{2}-\{(0,0)\},(n,(x,y))\mapsto (nx,ny)}$

This implies (a).

For to (b) applies ${\displaystyle {\tilde {\gamma }}\in \Gamma (1)}$

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

According to (a) is therefore also invariant under . ${\displaystyle E}$ ${\displaystyle \Gamma (1)}$ ${\displaystyle \,\,\,\,\square }$

Proposition: E is an eigenmode of the hyperbolic Laplace operator

We need the following.

Lemma: interchanged with the operation of on . More precisely applies to each : ${\displaystyle \Delta }$ ${\displaystyle G}$ ${\displaystyle C^{\infty }(\mathbb {H} )}$ ${\displaystyle g\in G}$

${\displaystyle L_{g}\Delta =\Delta L_{g}}$

Proof: The group is created by the elements of the form with , with and . One recalculates the assertion on these generators and thus obtains the assertion for each one . ${\displaystyle SL_{2}(\mathbb {R} )}$ ${\displaystyle {\begin{pmatrix}a&0\\0&{\frac {1}{a}}\\\end{pmatrix}}}$ ${\displaystyle a\in \mathbb {R} ^{\times }}$ ${\displaystyle {\begin{pmatrix}1&x\\0&1\\\end{pmatrix}}}$ ${\displaystyle x\in \mathbb {R} }$ ${\displaystyle S={\begin{pmatrix}0&-1\\1&0\\\end{pmatrix}}}$ ${\displaystyle g\in SL_{2}(\mathbb {R} )}$ ${\displaystyle \,\,\,\square }$

Because of (compare above) it is sufficient to show the eigen equation for . The following applies: ${\displaystyle E(z,s)=\pi ^{-s}\Gamma (s)\zeta (2s){\tilde {E}}(z,s)}$ ${\displaystyle {\tilde {E}}}$

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

Also applies

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

Since the Laplace operator interchanges with the operation of , it follows for each ${\displaystyle \Gamma (1)}$ ${\displaystyle \gamma \in \Gamma (1)}$

${\displaystyle \Delta (\operatorname {Im} (\gamma z)^{s})=s(1-s)\operatorname {Im} (\gamma z)^{s}}$and with it . ${\displaystyle \Delta {\tilde {E}}(z,s)=s(1-s){\tilde {E}}(z,s)}$

It follows for the eigen equation also for . To get the claim for each , consider the function . One writes this function out explicitly with the help of the Fourier expansion of and recognizes that it is meromorphic. But now it disappears for , so according to the identity theorem it is identical to zero and the eigen-equation holds for each . ${\displaystyle \operatorname {Re} (s)>3}$ ${\displaystyle E}$ ${\displaystyle s\in \mathbb {C} }$ ${\displaystyle \Delta E(z,s)-s(1-s)E(z,s)}$ ${\displaystyle E}$ ${\displaystyle \operatorname {Re} (s)>3}$ ${\displaystyle s\in \mathbb {C} }$ ${\displaystyle \,\,\,\,\square }$

Theorem for the Fourier expansion of E

The non-holomorphic Eisenstein series has a Fourier expansion

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

where the Fourier coefficients are given by:

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

For has a meromorphic sequel in on whole . This is holomorphic except for simple poles in . ${\displaystyle z\in \mathbb {H} }$ ${\displaystyle E(z,s)}$ ${\displaystyle s}$ ${\displaystyle \mathbb {C} }$ ${\displaystyle s=0,1}$

The Eisenstein series satisfies the functional equation for each ${\displaystyle z\in \mathbb {H} }$

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

and it applies locally uniformly in the growth condition ${\displaystyle x\in \mathbb {R} }$

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

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

The meromorphic continuation of E is of great importance in the spectral theory of the hyperbolic Laplace operator.

Maass waveforms from weight k

Congruence subsets

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

${\displaystyle SL_{2}(\mathbb {Z} )\to SL_{2}(\mathbb {Z} \,/\,N\mathbb {Z} )}$.

One calls the main congruence group of the level . A subgroup is called a congruence subgroup if one exists such that . All congruence subsets are discrete. ${\displaystyle \Gamma (N)}$ ${\displaystyle N}$ ${\displaystyle \Gamma \subseteq Sl_{2}(\mathbb {Z} )}$ ${\displaystyle N\in \mathbb {N} }$ ${\displaystyle \Gamma (N)\subseteq \Gamma }$

Be it . For a congruence subgroup, let the image of in . Let S be a system of representatives of , then is ${\displaystyle {\overline {\Gamma (1)}}:=\Gamma (1)\,/\,\pm \,1}$ ${\displaystyle \Gamma }$ ${\displaystyle {\overline {\Gamma }}}$ ${\displaystyle \Gamma }$ ${\displaystyle {\overline {\Gamma (1)}}}$ ${\displaystyle {\overline {\Gamma }}\setminus {\overline {\Gamma (1)}}}$

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

a fundamental domain for . The amount is clearly defined by the fundamental area . In addition, it is finite. ${\displaystyle \Gamma }$ ${\displaystyle S}$ ${\displaystyle SD}$ ${\displaystyle S}$

It's called the points of tips of the fundamental domain . They are completely in . ${\displaystyle \gamma \infty }$ ${\displaystyle \gamma \in S}$ ${\displaystyle SD}$ ${\displaystyle \mathbb {Q} \cup \{\infty \}}$

For each tip there is a with . ${\displaystyle c}$ ${\displaystyle \sigma \in \Gamma (1)}$ ${\displaystyle \sigma \infty =c}$

Definition of Maaß waveforms from weight k

Let be a congruence subgroup of and . ${\displaystyle \Gamma }$ ${\displaystyle SL_{2}(\mathbb {R} )}$ ${\displaystyle k\in \mathbb {Z} }$

We generalize the hyperbolic Laplace operator to the hyperbolic Laplace operator of weight , where: ${\displaystyle \Delta }$ ${\displaystyle \Delta _{k}}$ ${\displaystyle k}$

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

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

For we define a right operation from on through ${\displaystyle k\in \mathbb {Z} }$ ${\displaystyle Sl_{2}(\mathbb {R} )}$ ${\displaystyle C^{\infty }(\mathbb {H} )}$

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

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

One can show that for each , and each applies: ${\displaystyle f\in C^{\infty }(\mathbb {H} )}$ ${\displaystyle k\in \mathbb {Z} }$ ${\displaystyle g\in Sl_{2}(\mathbb {R} )}$

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

So it operates on vector space ${\displaystyle \Delta _{k}}$

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

Definition: A weight to group Maaß waveform is a function that is an eigenmode of and is of moderate growth at the tips. ${\displaystyle k\in \mathbb {Z} }$ ${\displaystyle \Gamma }$ ${\displaystyle f\in C^{\infty }(\Gamma \setminus \mathbb {H} ,k)}$ ${\displaystyle \Delta _{k}}$

On the concept of moderate growth at the tips:

If there is a congruence subgroup, then there is a peak and one calls a function from of moderate growth at , if can be restricted by a polynomial, if . Be now another tip. Then there is a with . Then be . Are expected after that then in located, wherein the congruence subgroup is. It is now said that there is moderate growth at the top if there is moderate growth at the top . ${\displaystyle \Gamma }$ ${\displaystyle \infty }$ ${\displaystyle f}$ ${\displaystyle C^{\infty }(\Gamma \setminus \mathbb {H} ,k)}$ ${\displaystyle \infty }$ ${\displaystyle f(x+iy)}$ ${\displaystyle y\to \infty }$ ${\displaystyle c\in \mathbb {Q} }$ ${\displaystyle \theta \in Sl_{2}(\mathbb {Z} )}$ ${\displaystyle \theta (\infty )=c}$ ${\displaystyle f':=f_{||k}\theta }$ ${\displaystyle f'}$ ${\displaystyle C^{\infty }(\Gamma '\setminus \mathbb {H} ,k)}$ ${\displaystyle \Gamma '}$ ${\displaystyle \theta ^{-1}\Gamma \theta }$ ${\displaystyle f}$ ${\displaystyle c}$ ${\displaystyle f'}$ ${\displaystyle \infty }$

If the main congruence group contains the degree , it is called cuspid at infinity, if ${\displaystyle \Gamma }$ ${\displaystyle N}$ ${\displaystyle f}$

${\displaystyle \int _{0}^{N}\,f(z+u)\,du=0}$ for each ${\displaystyle z\in \mathbb {H} }$

applies. It is called cuspid at a point if cusp is at infinity. ${\displaystyle f}$ ${\displaystyle c}$ ${\displaystyle f'}$

If each tip is cuspid, it is called a tip shape . ${\displaystyle f}$ ${\displaystyle f}$

We call Maaß's waveforms that are cuspidous, Maaß's peak shapes.

We give an example of a Maaß waveform from weight to module group: ${\displaystyle k>1}$

Example: Be a modular form of weight to group . Then there is a Maass waveform from weight to group . ${\displaystyle g\colon \mathbb {H} \to \mathbb {C} }$ ${\displaystyle k\in 2\mathbb {N} }$ ${\displaystyle \Gamma (1)}$ ${\displaystyle f(z):=y^{\frac {k}{2}}g(z)}$ ${\displaystyle k}$ ${\displaystyle \Gamma (1)}$

Proof: Since is a modular form, is holomorphic, i.e. especially smooth in . So that's smooth. Be now . Then applies ${\displaystyle g}$ ${\displaystyle g}$ ${\displaystyle \mathbb {H} }$ ${\displaystyle f}$ ${\displaystyle \gamma \in \Gamma (1)}$

${\displaystyle f(\gamma z)=\operatorname {Im} (\gamma z)^{\frac {k}{2}}g(\gamma z)=\left({\frac {\operatorname {Im} (z)}{|cz+d|^{2}}}\right)^{\frac {k}{2}}(cz+d)^{k}g(z)=\left({\frac {cz+d}{|cz+d|}}\right)^{k}f(z)}$.

In particular, since is a modular shape, in is holomorphic ; H. for . But there is a so that for . ${\displaystyle g}$ ${\displaystyle g}$ ${\displaystyle \infty }$ ${\displaystyle g(x+iy)\to c\in \mathbb {C} }$ ${\displaystyle y\to \infty }$ ${\displaystyle N\in \mathbb {N} }$ ${\displaystyle f(z):=y^{\frac {k}{2}}g(z)={\mathcal {O}}(y^{N})}$ ${\displaystyle y\geq 1}$

We now show the eigen equation for . Since is holomorphic, the Riemann differential equations apply, i.e. ${\displaystyle f}$ ${\displaystyle g}$

${\displaystyle {\frac {\partial g}{\partial x}}=-i{\frac {\partial g}{\partial y}}}$

and thus follows with Black's theorem

${\displaystyle {\frac {\partial ^{2}g}{\partial x^{2}}}=-{\frac {\partial ^{2}g}{\partial y^{2}}}}$.

The following then applies:

${\displaystyle {\begin{aligned}\Delta _{k}(y^{\frac {k}{2}}g)&=-y^{2}\left({\frac {\partial ^{2}y^{\frac {k}{2}}g}{\partial x^{2}}}+{\frac {\partial ^{2}y^{\frac {k}{2}}g}{\partial y^{2}}}\right)+iky{\frac {\partial y^{\frac {k}{2}}g}{\partial x}}\\&=-y^{2}\left(-y^{\frac {k}{2}}{\frac {\partial ^{2}g}{\partial y^{2}}}+{\frac {k}{2}}\left({\frac {k}{2}}-1\right)y^{{\frac {k}{2}}-2}g+ky^{{\frac {k}{2}}-1}{\frac {\partial g}{\partial y}}+y^{\frac {k}{2}}{\frac {\partial ^{2}g}{\partial y^{2}}}\right)+ky^{{\frac {k}{2}}+1}{\frac {\partial g}{\partial y}}\\&={\frac {k}{2}}\left(1-{\frac {k}{2}}\right)y^{\frac {k}{2}}g\end{aligned}}}$

This is a Maaß form of weight to group . ${\displaystyle f}$ ${\displaystyle k}$ ${\displaystyle \Gamma (1)}$ ${\displaystyle \,\,\,\square }$

The spectral problem

Let be a congruence subgroup of . Let it be the vector space of all measurable functions with for each and ${\displaystyle \Gamma }$ ${\displaystyle Sl_{2}(\mathbb {R} )}$ ${\displaystyle L^{2}(\Gamma \setminus \mathbb {H} ,k)}$ ${\displaystyle f\colon \mathbb {H} \to \mathbb {C} }$ ${\displaystyle f_{||k}\gamma =f}$ ${\displaystyle \gamma \in \Gamma }$

${\displaystyle ||f||^{2}:=\int _{\Gamma \setminus \mathbb {H} }|f(z)|^{2}\,{\frac {dx\,dy}{y^{2}}}<\infty }$

modulo functions with . The integral is well-defined because the function is -invariant. The space is a Hilbert space with the scalar product ${\displaystyle ||f||=0}$ ${\displaystyle |f(z)|^{2}}$ ${\displaystyle \Gamma }$ ${\displaystyle L^{2}(\Gamma \setminus \mathbb {H} ,k)}$

${\displaystyle \langle f,g\rangle =\int _{\Gamma \setminus \mathbb {H} }f(z){\overline {g(z)}}\,{\frac {dx\,dy}{y^{2}}}}$.

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

We denote the space of all tip shapes in . Then operates on and has a pure spectrum of eigenvalues ​​there. The spectrum on the orthogonal complement has a continuous part and is described with the help of (modified) non-holomorphic Eisenstein series, their meromorphic continuations and their residuals. For a detailed analysis, see Bump or Iwaniec . If the subgroup is discrete (torsion-free) so that the quotient is compact , the spectral problem is simplified. This is mainly due to the fact that a discrete, co-compact subgroup has no peaks. Here the complete space is a sum of the operator's eigenspaces . ${\displaystyle C(\Gamma \setminus \mathbb {H} ,k)}$ ${\displaystyle L^{2}(\Gamma \setminus \mathbb {H} ,k)\cap C^{\infty }(\Gamma \setminus \mathbb {H} ,k)}$ ${\displaystyle \Delta _{k}}$ ${\displaystyle C(\Gamma \setminus \mathbb {H} ,k)}$ ${\displaystyle \Gamma }$ ${\displaystyle \Gamma \setminus \mathbb {H} }$ ${\displaystyle L^{2}(\Gamma \setminus \mathbb {H} ,k)}$ ${\displaystyle \Delta _{k}}$

Embedding in the space L ² (Γ \ G)

${\displaystyle G=Sl_{2}(\mathbb {R} )}$is a unimodular, locally compact group with the subspace topology of . Be back a congruence subgroup. Since in is discreet, in is locked in . The group is unimodular , and since the counting measure is a hair measure on the discrete group , it is also unimodular. According to the quotient integral formula, there is a right-invariant Radon measure on the locally compact space . We now consider the space belonging to the measure . ${\displaystyle \mathbb {R} ^{4}}$ ${\displaystyle \Gamma }$ ${\displaystyle \Gamma }$ ${\displaystyle G}$ ${\displaystyle \Gamma }$ ${\displaystyle G}$ ${\displaystyle G}$ ${\displaystyle \Gamma }$ ${\displaystyle \Gamma }$ ${\displaystyle G}$ ${\displaystyle \mathrm {d} x}$ ${\displaystyle \Gamma \setminus G}$ ${\displaystyle \mathrm {d} x}$ ${\displaystyle L^{2}}$ ${\displaystyle L^{2}(\Gamma \setminus G)}$

The space breaks down into a direct Hilbert sum ${\displaystyle L^{2}(\Gamma \setminus G)}$

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

where and for . ${\displaystyle L^{2}(\Gamma \setminus G\,,\,k):=\{\phi \in L^{2}(\Gamma \setminus G)\,\mid \,\forall \,x\in \Gamma \setminus G\,\forall \,\theta \in \mathbb {R} :\,\phi (xk_{\theta })=e^{ik\theta }F(x)\,\,\,\}}$ ${\displaystyle k_{\theta }={\begin{pmatrix}\cos(\theta )&-\sin(\theta )\\\sin(\theta )&\cos(\theta )\\\end{pmatrix}}\,\,\in SO(2)}$ ${\displaystyle \theta \in \mathbb {R} }$

The Hilbert space can be isometrically embedded in the Hilbert space . The isometry is given by the illustration ${\displaystyle L^{2}(\Gamma \setminus \mathbb {H} \,,\,k)}$ ${\displaystyle L^{2}(\Gamma \setminus G\,,\,k)}$

${\displaystyle \psi _{k}\colon L^{2}(\Gamma \setminus \mathbb {H} \,,\,k)\to L^{2}(\Gamma \setminus G\,,\,k)\,,\,\psi _{k}(f)(g):=f_{||k}\gamma (i).}$

With this we can understand all Maass tip shapes to the congruence group as elements of . ${\displaystyle \Gamma }$ ${\displaystyle L^{2}(\Gamma \setminus G)}$

${\displaystyle L^{2}(\Gamma \setminus G)}$is a Hilbert space on which operates via legal translation: ${\displaystyle G}$

${\displaystyle R_{g}\phi :=\phi (xg)}$,

where and . ${\displaystyle x\in \Gamma \setminus G}$ ${\displaystyle \phi \in L^{2}(\Gamma \setminus G)}$

One easily calculates that there is a unitary representation of on the Hilbert space . One now wants to break down the representation into a sum of irreducible sub-representations. It turns out that this is only possible if it is co-compact. Otherwise there is also a continuous Hilbert integral. The interesting thing is that solving this problem also solves the spectral problem of Maass shapes. For a detailed analysis of this connection see also Bump . ${\displaystyle R}$ ${\displaystyle G}$ ${\displaystyle L^{2}(\Gamma \setminus G)}$ ${\displaystyle (R,L^{2}(\Gamma \setminus G))}$ ${\displaystyle \Gamma }$

Automorphic representations of the nobility group

The group Eq ₂ (A)

For a ring with one, let the group of matrices with entries in and in an invertible determinant. Let the ring of the (rational) Adele, the ring of the finite (rational) Adele and for a prime number be the field of the p-adic numbers and the ring of the whole p-adic numbers. Be it . Both and are with the respective subspace topologies of or locally compact unimodular groups. The group is isomorphic to the group , which means the restricted direct product (see Adelring) of the groups with respect to the compact, open subgroups of . Then there is a locally compact group with the restricted product topology. ${\displaystyle R}$ ${\displaystyle G_{R}:=Gl_{2}(R)}$ ${\displaystyle 2\times 2}$ ${\displaystyle R}$ ${\displaystyle R}$ ${\displaystyle \mathbb {A} =\mathbb {A} _{\mathbb {Q} }}$ ${\displaystyle \mathbb {A} _{fin}}$ ${\displaystyle p\in \mathbb {N} }$ ${\displaystyle \mathbb {Q} _{p}}$ ${\displaystyle \mathbb {Z} _{p}}$ ${\displaystyle G_{p}:=G_{\mathbb {Q} _{p}}}$ ${\displaystyle G_{p}}$ ${\displaystyle G_{\mathbb {R} }}$ ${\displaystyle \mathbb {Q} _{p}^{4}}$ ${\displaystyle \mathbb {R} ^{4}}$ ${\displaystyle G_{fin}:=G_{\mathbb {A} _{fin}}}$ ${\displaystyle {\widehat {\prod _{p<\infty }}}^{K_{p}}G_{p}}$ ${\displaystyle G_{p}}$ ${\displaystyle K_{p}:=G_{\mathbb {Z} _{p}}}$ ${\displaystyle G_{p}}$ ${\displaystyle G_{fin}}$

The group is isomorphic to the group ${\displaystyle G_{\mathbb {A} }}$

${\displaystyle G_{fin}\times G_{\mathbb {R} }}$

and is a locally compact group with the product topology da and are locally compact groups. ${\displaystyle G_{fin}}$ ${\displaystyle G_{\mathbb {R} }}$

By we denote the ring . The subgroup ${\displaystyle {\widehat {\mathbb {Z} }}}$ ${\displaystyle \prod _{p<\infty }\mathbb {Z} _{p}}$

${\displaystyle G_{\widehat {\mathbb {Z} }}:=\prod _{p<\infty }K_{p}}$

is a maximally compact, open subgroup of and is also understood as a subgroup of through the figure . ${\displaystyle G_{fin}}$ ${\displaystyle x_{fin}\mapsto (x_{fin},1_{\infty })}$ ${\displaystyle G_{\mathbb {A} }}$

With we denote the center of , i.e. diagonal matrices of the form , where . We consider embedding as a subgroup of via . ${\displaystyle Z_{\mathbb {R} }}$ ${\displaystyle G_{\infty }}$ ${\displaystyle {\begin{pmatrix}\lambda &\\&\lambda \\\end{pmatrix}}}$ ${\displaystyle \lambda \in \mathbb {R} ^{\times }}$ ${\displaystyle Z_{\mathbb {R} }}$ ${\displaystyle G_{\mathbb {A} }}$ ${\displaystyle z\mapsto (1_{G_{fin}},z)}$

The group is embedded diagonally in , which is possible, since the four entries have only a finite number of prime divisors and thus lies in for all but a finite number of prime numbers . ${\displaystyle G_{\mathbb {Q} }}$ ${\displaystyle G_{\mathbb {A} }}$ ${\displaystyle x\in G_{\mathbb {Q} }}$ ${\displaystyle x}$ ${\displaystyle K_{p}}$ ${\displaystyle p\in \mathbb {N} }$

Be the group of everyone with , where the amount of the id is meant. You calculate immediately that it is even in (product formula). ${\displaystyle G_{\mathbb {A} }^{1}}$ ${\displaystyle x\in G_{\mathbb {A} }}$ ${\displaystyle |\det(x)|=1}$ ${\displaystyle \det(x)}$ ${\displaystyle G_{\mathbb {Q} }}$ ${\displaystyle G_{\mathbb {A} }^{1}}$

The injection can be used to identify the groups and each other. ${\displaystyle G_{\mathbb {A} }^{1}\hookrightarrow G_{\mathbb {A} }}$ ${\displaystyle G_{\mathbb {Q} }\setminus G_{\mathbb {A} }^{1}}$ ${\displaystyle G_{\mathbb {Q} }Z_{\mathbb {R} }\setminus G_{\mathbb {A} }}$

The following sentence applies to : ${\displaystyle G_{\mathbb {A} }}$

The group lies close in and discreetly in . The quotient is not compact, but has a finite hair measure. ${\displaystyle G_{\mathbb {Q} }}$ ${\displaystyle G_{fin}}$ ${\displaystyle G_{\mathbb {A} }}$ ${\displaystyle G_{\mathbb {Q} }Z_{\mathbb {R} }\setminus G_{\mathbb {A} }=G_{\mathbb {Q} }\setminus G_{\mathbb {A} }^{1}}$

This makes a grid of in particular , as was the module group of in the classic case . It also follows that is unimodular. ${\displaystyle G_{\mathbb {Q} }}$ ${\displaystyle G_{\mathbb {A} }^{1}}$ ${\displaystyle Sl_{2}(\mathbb {R} )}$ ${\displaystyle G_{\mathbb {A} }^{1}}$

Ennobling of lace forms

We now want to understand the classic Maaß tip shapes from weight 0 to the module group as functions . This works with the strong approximation theorem, which says that the mapping ${\displaystyle Z_{\mathbb {R} }\,\,G_{\mathbb {Q} }\setminus G_{\mathbb {A} }}$

${\displaystyle \psi \colon G_{\mathbb {Z} }x_{\infty }\mapsto G_{\mathbb {Q} }(1,x_{\infty })G_{\widehat {\mathbb {Z} }}}$

is an -equivariant homeomorphism. It then applies ${\displaystyle G_{\mathbb {R} }}$

${\displaystyle G_{\mathbb {Z} }\setminus G_{\mathbb {R} }\,\,{\tilde {\to }}\,\,G_{\mathbb {Q} }\setminus G_{\mathbb {A} }\,\,/\,\,G_{\widehat {\mathbb {Z} }}}$

and so too

${\displaystyle G_{\mathbb {Z} }\,Z_{\mathbb {R} }\setminus G_{\mathbb {R} }\,\,{\tilde {\to }}\,\,G_{\mathbb {Q} }\,Z_{\mathbb {R} }\setminus G_{\mathbb {A} }\,\,/\,\,G_{\widehat {\mathbb {Z} }}.}$

Now there are Maaß tip shapes from weight zero to the module group in ${\displaystyle \Gamma (1)}$

${\displaystyle L^{2}(Sl_{2}(\mathbb {Z} )\setminus Sl_{2}(\mathbb {R} ))\,\,{\tilde {=}}\,\,L^{2}(Gl_{2}(\mathbb {Z} )\,Z_{\mathbb {R} }\setminus Gl_{2}(\mathbb {R} ))}$.

However, according to the strong approximation theorem, this space is unitarily isomorphic to

${\displaystyle L^{2}(G_{\mathbb {Q} }\,Z_{\mathbb {R} }\setminus G_{\mathbb {A} }\,\,/\,\,G_{\widehat {\mathbb {Z} }})\,\,{\tilde {=}}\,\,L^{2}(G_{\mathbb {Q} }\,Z_{\mathbb {R} }\setminus G_{\mathbb {A} })^{G_{\widehat {\mathbb {Z} }}},}$

what is a subspace of . ${\displaystyle L^{2}(G_{\mathbb {Q} }\,Z_{\mathbb {R} }\setminus G_{\mathbb {A} })}$

With the same argument one can understand the classical holomorphic tip shapes as elements of . With a small generalization of strong Approximationstheorems one realizes that all the classic Maass's tip shapes (and holomorphic cusp forms) of any weight at any congruence subgroup in may embed. ${\displaystyle L^{2}(G_{\mathbb {Q} }\,Z_{\mathbb {R} }\setminus G_{\mathbb {A} })}$ ${\displaystyle \Gamma }$ ${\displaystyle L^{2}(G_{\mathbb {Q} }\,Z_{\mathbb {R} }\setminus G_{\mathbb {A} })}$

In the literature it is often referred to as the set of automorphic forms (the nobility group). If the condition is replaced by suitable growth conditions, the embedded non-holomorphic Eisenstein series also belong to the automorphic forms, which themselves cannot be integrated. ${\displaystyle L^{2}(G_{\mathbb {Q} }\,Z_{\mathbb {R} }\setminus G_{\mathbb {A} })}$ ${\displaystyle L^{2}}$ ${\displaystyle L^{2}}$

Top forms of the nobility group

For a ring, let the set of all be where . This group is isomorphic to the additive group of . ${\displaystyle R}$ ${\displaystyle N_{R}}$ ${\displaystyle {\begin{pmatrix}1&r\\&1\\\end{pmatrix}}}$ ${\displaystyle \,r\,\in \,R}$ ${\displaystyle R}$

A function is called a tip shape, if ${\displaystyle f\in L^{2}(G_{\mathbb {Q} }\setminus G_{\mathbb {A} }^{1})}$

${\displaystyle \int _{N_{\mathbb {Q} }\setminus N_{\mathbb {A} }}\,f(nx)\,dn=0}$

applies to almost everyone . The vector space of all tip shapes is called or for short with . is closed and invariant under the right-regular representation of . ${\displaystyle x\in G_{\mathbb {Q} }\setminus G_{\mathbb {A} }^{1}}$ ${\displaystyle L_{cusp}^{2}(G_{\mathbb {Q} }\setminus G_{\mathbb {A} }^{1})}$ ${\displaystyle L_{cusp}^{2}}$ ${\displaystyle L_{cusp}^{2}}$ ${\displaystyle G_{\mathbb {A} }^{1}}$

One is now interested in decomposing into irreducible closed subspaces under the right-regular representation. ${\displaystyle L_{cusp}^{2}}$

The following sentence applies more precisely :

The space breaks down into a direct sum of irreducible Hilbert spaces with finite multiples: ${\displaystyle L_{cusp}^{2}}$

${\displaystyle L_{cusp}^{2}={\widehat {\bigoplus }}_{\pi \in {\widehat {G_{\mathbb {A} }}}}N_{cusp}(\pi )\pi }$

The determination of the multiplicities is one of the most difficult and important problems of the theory of automorphic forms. ${\displaystyle N_{cusp}(\pi )}$

Cuspid representations of the aristocratic group

An irreducible representation of the group is called cuspid if it is isomorphic to a sub-representation of . ${\displaystyle \pi }$ ${\displaystyle G_{\mathbb {A} }}$ ${\displaystyle L_{cusp}^{2}}$

An irreducible representation of the group is called admissible if there is a compact set such that for each . ${\displaystyle \pi }$ ${\displaystyle G_{\mathbb {A} }}$ ${\displaystyle K\subset G_{\mathbb {A} }}$ ${\displaystyle \dim _{K}(V_{\pi },V_{\tau })<\infty }$ ${\displaystyle \tau \in \in {\widehat {G_{\mathbb {A} }}}}$

One can show that every cuspid representation is permissible.

The admissibility is used to apply the so-called tensor product theorem, which says that every irreducible unitary representation of the group is isomorphic to an infinite tensor product , the irreducible representations of the group being almost all unbranched. ${\displaystyle G_{\mathbb {A} }}$ ${\displaystyle \bigotimes _{p\leq \infty }\,\pi _{p}}$ ${\displaystyle \pi _{p}}$ ${\displaystyle G_{p}}$

(A representation of the group is called unbranched if the vector space : is not the null space.) ${\displaystyle \pi _{p}}$ ${\displaystyle G_{p}}$ ${\displaystyle (p<\infty )}$
${\displaystyle V_{\pi _{p}}^{K_{p}}=\{v\in V_{\pi _{p}}\,\mid \,\forall \,k\in K_{p}:\,\pi _{p}(k)v=v\}}$

For the construction of an infinite tensor product see for example Deitmar , Chap. 7.

Automorphic L-functions

Let be an irreducible feasible unitary representation of . According to the tensor product theorem, is of the form , where these are irreducible representations of the groups , almost all of which are unbranched. ${\displaystyle \pi }$ ${\displaystyle G_{\mathbb {A} }}$ ${\displaystyle \pi }$ ${\displaystyle \pi =\bigotimes _{p\leq \infty }\,\pi _{p}}$ ${\displaystyle \pi _{p}}$ ${\displaystyle G_{p}}$

Let it be a finite set of places such that and contains all branched places. The global L-function of is defined as ${\displaystyle F}$ ${\displaystyle \infty \in F}$ ${\displaystyle F}$ ${\displaystyle \pi }$

${\displaystyle L^{F}(\pi ,s):=\prod _{p\notin F}L(\pi _{p},s),}$

where is a so-called local L-function of the local representation . For a detailed construction of a local L-function see, for example, Anton Deitmar: Automorphic forms, Chapter 8.2. ${\displaystyle L(\pi _{p},s)}$ ${\displaystyle \pi _{p}}$

If it is a cuspid representation, the L function continues to become a meromorphic function . This is possible because , like the classic L functions, certain functional equations are fulfilled. ${\displaystyle \pi }$ ${\displaystyle L^{F}(\pi ,s)}$ ${\displaystyle \mathbb {C} }$ ${\displaystyle L^{F}(\pi ,s)}$

An ennobled Maass tip shape (or a holomorphic tip shape) can be assigned a cuspidic representation so that the L function corresponds to the classic L function . In this sense the automorphic L-functions are a generalization of the classical L-functions. They were first examined by Robert Langlands in 1969 . ${\displaystyle f}$ ${\displaystyle \pi _{f}}$ ${\displaystyle L^{F}(\pi _{f},s)}$ ${\displaystyle L(f,s)}$

literature

• Anton Deitmar : Automorphic forms. Springer, Berlin / Heidelberg et al. 2010, ISBN 978-3-642-12389-4 .
• Henryk Iwaniec : Spectral Methods of Automorphic Forms (Graduate Studies in Mathematics). American Mathematical Society, 2nd edition, 2002, ISBN 978-0-8218-3160-1 .
• Daniel Bump : Automorphic Forms and Representations (Cambridge Studies in Advanced Mathematics). Cambridge University Press, 1997, ISBN 978-0-521-55098-7 .

supporting documents

1. ↑ See for example Gelbart: Automorphic forms of the adele group.