Benutzer:Jml314/Spielwiese

Maaßsche Wellenformen

Allgemeines

Die Gruppe $\Gamma :=SL_{2}(\mathbb {R} )$ operiert auf der oberen Halbebene $\mathbb {H} =\{z\in \mathbb {Z} :Im(z)>0\}$ durch $\gamma .z:={\frac {az+b}{cz+d}}$ wobei $\gamma ={\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}\in \Gamma .$ Für festes $\gamma \in \Gamma$ ist die Abbildung $\phi :\mathbb {H} \to \mathbb {H} \,,\,z\mapsto \gamma .z$ ein Diffeomorphismus. Damit operiert $\Gamma$ auch auf $C^{\infty }(\mathbb {H} )$ durch $\gamma .f(z):=L_{\gamma }(f)(z):=f(\gamma ^{-1}z)\,,\,z\in \mathbb {H}$.

Definition des hyperbolischen Laplace Operators

Der hyperbolische Laplace Operator auf $\mathbb {H}$ wird definiert durch

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

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta (f)=-y^{2}({\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y^{2}}})$

Definition einer Maaßschen Wellenform

Eine Maaßsche Wellenform zur Gruppe $\Gamma (1):=SL_{2}(\mathbb {Z} )$ ist eine glatte Funktion $f$ auf $\mathbb {H}$ so dass

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1)\,f(\gamma z)=f(z)\,\,\,\forall \,\gamma \,\in \,\Gamma (1)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2)\,\Delta (f)=\lambda f$ für ein $\lambda \in \mathbb {C}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)$ es existiert ein $N\in \mathbb {N}$ mit $f(x+iy)={\mathcal {O}}(y^{N})$ für $y\geq 1$

Gilt außerdem

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

dann nennen wir $f$ Maaß-Spitzenform.

Zusammenhang von Maaßschen Wellenformen und Dirichletreihen

Sei nun $f$ eine Maaßsche Wellenform. Dann gilt wegen $\gamma :={\begin{pmatrix}1&1\\0&1\\\end{pmatrix}}\in \Gamma (1)\,f(z)=f(\gamma .z)=f(z+1)\,\forall \,z\in \mathbb {H}$ . Damit hat $f$ eine Fourier-Entwicklung der Gestalt $f(x+iy)=\sum _{n=-\infty }^{\infty }a_{n}(y)e^{2\pi inx}$ , wobei die Koeffizientenfunktionen glatt sind.

Wir beobachten außerdem : $f$ ist eine Maaß-Spitzenform genau dann wenn $a_{0}(y)=0\,\,\,\forall y>0$, denn

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=\int _{0}^{1}f(z+t)dt$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int _{0}^{1}\sum _{n=-\infty }^{\infty }a_{n}(y)e^{2\pi in(x+t)}dt$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overset {(1)}{=}}\sum _{n=-\infty }^{\infty }a_{n}(y)e^{2\pi inx}\int _{0}^{1}e^{2\pi int}dt$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=a_{0}(y)$

wobei in (1) benutzt wurde, dass die Reihe $\sum _{n=-\infty }^{\infty }a_{n}(y)e^{2\pi in(x+t)}dt$ für festes $y>0$ lokal gleichmäßig konvergiert.

Lemma : Fourierkoeffizienten einer Maaßschen Wellenform

Sei $\lambda \in \mathbb {C}$ der Eigenwert der Maaßschen Wellenform f bezüglich $\Delta$. Sei $\nu \in \mathbb {C}$ die bis aufs Vorzeichen eindeutige komplexe Zahl mit $\lambda ={\frac {1}{4}}-\nu ^{2}$. Sei $K_{s}(y):={\frac {1}{2}}\int _{0}^{\infty }e^{-y(t+t^{-1})/2}t^{s}{\frac {dt}{t}}\,\,,s\in \mathbb {C} \,,\,y>0$ die K-Besselfunktion. Dann gilt für die Fourierkoeffizientenfunktionen von $f$

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

falls $n\neq 0$. Ist $n=0$ so gilt

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

Beweis : Es gilt $\Delta (f)=({\frac {1}{4}}-\nu ^{2})f$. Nach der Definition von Fourierkoeffizienten gilt für $n\in \mathbb {Z}$

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

Zusammen folgt für $n\in \mathbb {Z}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\frac {1}{4}}-\nu ^{2})a_{n}$ $=\int _{0}^{1}({\frac {1}{4}}-\nu ^{2})f(x+iy)e^{-2\pi inx}dx$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int _{0}^{1}(\Delta f)(x+iy)e^{-2\pi inx}dx$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-y^{2}\left(\int _{0}^{1}{\frac {\partial ^{2}f}{\partial x^{2}}}(x+iy)e^{-2\pi inx}dx+\int _{0}^{1}{\frac {\partial ^{2}f}{\partial y^{2}}}(x+iy)e^{-2\pi inx}dx\right)$

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

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-y^{2}(2\pi in)^{2}a_{n}(y)-y^{2}{\frac {\partial ^{2}}{\partial y^{2}}}a_{n}(y)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=4\pi ^{2}n^{2}y^{2}a_{n}(y)-y^{2}{\frac {\partial ^{2}}{\partial y^{2}}}a_{n}(y)$

In (1) wurde für den ersten Summanden benutzt, dass der n-te Fourierkoeffizient von ${\frac {\partial ^{2}f}{\partial x^{2}}}$ genau $(2\pi in)^{2}a_{n}(y)$ ist, da wir Fourierreihen gliedweise differenzieren dürfen. Im zweiten Summanden wurde die Reihenfolge von Integration und Differentiation geändert, was erlaubt ist, da f beliebig oft stetig differenzierbar nach y ist und man über ein Kompaktum integriert. Es ergibt sich folgende lineare Differentialgleichung zweiter Ordnung

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

Für $n=0$ kann man zeigen, dass für jede Lösung $f$ dieser Differentialgleichung eindeutige Koeffizienten $c_{0},d_{0}\in \mathbb {C}$ existieren so dass gilt $f(y)=c_{0}y^{{\frac {1}{4}}-\nu }+d_{0}y^{{\frac {1}{4}}+\nu }$ .

Für $n\neq 0$ ist jede Lösung $f$ der obigen Differentialgleichung von der Form $f(y)=c_{n}{\sqrt {y}}K_{v}(2\pi |n|y)+d_{n}{\sqrt {y}}I_{v}(2\pi |n|y)$ für eindeutige $c_{n},d_{n}\in \mathbb {C}$, wobei $K_{v}(s)$ die K-Besselfunktion und $I_{v}(s)$ die I-Besselfunktionen ist (Siehe dazu zum Beispiel O.Forster : Analysis 2). Da die I-Besselfunktion exponentiell wächst und die K-Besselfunktion exponentiell fällt, folgt mit der Forderung 3) des höchstens polynomialen Wachstums von $f\,\,\,$ $a_{n}(y)=c_{n}{\sqrt {y}}K_{v}(2\pi |n|y)$ (also $d_{n}=0$) für ein eindeutiges $c_{n}\in \mathbb {C} .\,\,\,\square$

Satz : L-Funktion einer Maaßschen Wellenform

Sei $f(x+iy)=\sum _{n\neq 0}c_{n}{\sqrt {y}}K_{\nu }(2\pi |n|y)e^{2\pi inx}$ eine Maaß-Spitzenform. Wir definieren die sogenannte L-Funktion von $f$ als

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

Dann konvergiert die Reihe $L(s,f)$ für $Re(s)>{\frac {3}{2}}$ und man kann sie zu einer ganzen Funktion auf $\mathbb {C}$ fortsetzen.

Ist f gerade oder ungerade so definiert man

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

wobei $\epsilon =0$ falls $f$ gerade und $\epsilon =-1$ falls $f$ ungerade ist. Dann erfüllt $\Lambda$ die Funktionalgleichung

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

Beweis:

Sei f eine Maaß-Spitzenform. Zuerst machen wir uns klar wie schnell die Fourierkoeffizienten von f wachsen.

Behauptung: Es gilt $c_{n}={\mathcal {O}}(n^{\frac {1}{2}})$

Beweis: Da f eine Maas-Spitzenform ist, existieren $C,N>0$ so dass für $y>1$ die Ungleichung $|f(x+iy)|\leq Cy^{N}$ gilt. Ist $y<{\frac {1}{2}}$ , und ist $\omega \in D$ konjugiert zu $z=x+iy$ modulo $\Gamma (1)$ so rechnet man leicht nach, dass $Im(\omega )\leq {\frac {1}{y}}$ gilt. Da f invariant unter $\Gamma (1)$ ist, gilt für $y<{\frac {1}{2}}$ : $|f(x+iy)|\leq Cy^{-N}$. Also gilt für $y<{\frac {1}{2}}$ die Abschätzung

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

Für $n\in \mathbb {Z} \,,|n|>2$ und $y={\frac {1}{|n|}}$ gilt damit

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

Damit finden wir eine Konstante $D\geq C$ so dass für jedes $n\neq 0$ gilt

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

Nun fällt die K-Besselfunktion aber exponentiell schnell und f ist eine Maas-Spitzenform. Zusammen folgt, dass f auf $D$ beschränkt ist und damit auf $\mathbb {H}$. Damit können wir den obigen Beweis mit $N=0$ wiederholen und erhalten $|c_{n}|\leq Kn^{\frac {1}{2}}$ für ein $K<\infty$ also $c_{n}={\mathcal {O}}(n^{\frac {1}{2}})\,.\,\,\square$.

Um den Satz zu beweisen brauchen wir noch die Mellin-Transformierte von $K_{\nu }$.

Behauptung: Für $Re(s)>|Re(\nu )|$ konvergiert das Integral

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int _{0}^{\infty }K_{\nu }(y)y^{s}{\frac {dy}{y}}$

absolut und es gilt

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

Beweis: Nach Definition gilt

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int _{0}^{\infty }K_{\nu }(y)y^{s}{\frac {dy}{y}}={\frac {1}{2}}\int _{0}^{\infty }\int _{0}^{\infty }e^{-y(t+t^{-1})/2}t^{s}{\frac {dt}{t}}{\frac {dy}{y}}$

Wir wenden nun die Transformationsformel auf den Diffeomorphismus

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Phi :(0,\infty )\times (0,\infty )\to (0,\infty )\times (0,\infty )$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Phi (t,y)({\frac {1}{2}}t^{-1}y,{\frac {1}{2}}ty)=(u,v)$

an. Wir erhalten $y=2{\sqrt {uv}}$ und $t={\sqrt {\frac {v}{u}}}$. Das Jacobi-Matrix ergibt sich als

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,D\Phi (t,y)={\frac {1}{2}}{\begin{pmatrix}-{\frac {y}{t^{2}}}&{\frac {1}{t}}\\y&t\\\end{pmatrix}}$

mit Determinante $det(D\Phi )(t,y)=-{\frac {y}{2t}}$. Benutzt man nun die Transformationsformel vereinfacht sich obiges Integral zu

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\frac {1}{2}}2^{s-1}\int _{0}^{\infty }\int _{0}^{\infty }e^{-u-v}v^{(s+\nu )/2}u^{(s-v)/2}{\frac {dv}{v}}{\frac {du}{u}}=2^{s-2}\Gamma ({\frac {s+\nu }{2}})\Gamma ({\frac {s-\nu }{2}})$

und dieses konvergiert absolut für $Re(s)>|Re(\nu )|\,\,\,\square$.

Nun zum Beweis des Satzes. Ist f gerade oder ungerade folgt aus der Eindeutigkeit der Fourierkoeffizienten $a_{n}=(-1)^{\epsilon }$ für alle $n\in \mathbb {N}$. Sei f zuerst gerade. Dann gilt

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int _{0}^{\infty }f(iy)y^{s-{\frac {1}{2}}}{\frac {dy}{y}}$ $=2\int _{0}^{\infty }\sum _{n=1}^{\infty }c_{n}y^{s-1}K_{\nu }(2\pi ny)\,dy$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sum _{n=1}^{\infty }c_{n}\int _{0}^{\infty }y^{s-1}K_{\nu }(2\pi ny)dy$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sum _{n=1}^{\infty }c_{n}(2\pi n)^{-s}\int _{0}^{\infty }y^{s-1}K_{\nu }(y)dy$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2(2\pi )^{-s}\left(\sum _{n=1}^{\infty }c_{n}n^{-s}\right)2^{s-2}\Gamma \left({\frac {s+\nu }{2}}\right)\Gamma \left({\frac {s-\nu }{2}}\right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,={\frac {1}{2}}\pi ^{s}\Gamma \left({\frac {s+\nu }{2}}\right)\Gamma \left({\frac {s-\nu }{2}}\right)L(s,f)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,={\frac {1}{2}}\Lambda (s,f)$

Das vertauschen der Reihenfolge von Integral und Summe zeigt man zum Beispiel mit majorisierter Konvergenz, wobei man ausnutzt, dass für die K-Besselfunktion für $y>4$ gilt : $|K_{s}|\leq e^{-{\frac {y}{2}}}K_{Re(s)}(2)$. Ebenso zeigt man dass $f(iy)$ für $y\to \infty$ exponentiell fällt.

Wir definieren nun

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Lambda _{1}(s,f):=\int _{1}^{\infty }f(iy)y^{s-{\frac {1}{2}}}{\frac {dy}{y}}$ ,

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Lambda _{2}(s,f):=\int _{0}^{1}f(iy)y^{s-{\frac {1}{2}}}{\frac {dy}{y}}$.

Damit gilt $\Lambda =2(\Lambda _{1}+\Lambda _{2})$. Da $f(iy)$ exponentiell fällt für $y\to \infty$, konvergiert $\Lambda _{1}(s,f)$ für jedes $s\in \mathbb {C}$ und damit ist $\Lambda _{1}$ eine ganze Funktion (Komplexe Analysis). Nun ist $f$ aber invariant unter $\Gamma (1)$ womit insbesondere $f(iy)=f\left({\begin{pmatrix}0&1\\-1&0\\\end{pmatrix}}.iy\right)=f\left({\frac {1}{-iy}}\right)=f\left(i{\frac {1}{y}}\right)$ folgt.

Wir erhalten nun

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Lambda _{2}(s,f):=\int _{0}^{1}f(iy)y^{s-{\frac {1}{2}}}{\frac {dy}{y}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int _{1}^{\infty }f(i{\frac {1}{y}})y^{-s+{\frac {1}{2}}}{\frac {dy}{y}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int _{1}^{\infty }f(iy)y^{(1-s)-{\frac {1}{2}}}{\frac {dy}{y}}$

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

Damit ist auch $\Lambda _{2}$ eine ganze Funktion und damit ist $\Lambda$ ganz. Insbesondere kann man damit $L\left(s,f\right)$ zu einer ganzen Funktion auf $\mathbb {C}$ fortsetzen. Weiterhin gilt für $\Lambda$ die Funktionalgleichung

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

Wenn f ungerade ist, definiert man

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,g(z)={\frac {1}{4\pi i}}{\frac {\partial f}{\partial x}}(z)=\sum _{1}^{\infty }c_{n}n{\sqrt {y}}K_{\nu }(2\pi ny)cos(2\pi nx)$ .

Dann rechnet man analog zu oben

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int _{0}^{\infty }g(iy)y^{s+{\frac {1}{2}}}{\frac {dy}{y}}=\Lambda (s,f)$

indem man wieder benutzt, dass die K-Besselfunktion exponential fällt. Wir definieren wieder

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Lambda _{1}(s,f):=\int _{1}^{\infty }g(iy)y^{s+{\frac {1}{2}}}{\frac {dy}{y}}$ ,

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Lambda _{2}(s,f):=\int _{0}^{1}g(iy)y^{s+{\frac {1}{2}}}{\frac {dy}{y}}$ .

Auch $g(iy)$ fällt exponentiell für $y\to \infty$. Damit ist auch $\Lambda _{1}$ wieder eine ganze Funktion. Man rechnet leicht nach, dass gilt $g(iy)=-{\frac {1}{y^{2}}}g\left({\frac {i}{y}}\right)$. Damit folgt mit einer analogen Rechnung $\Lambda _{2}(s,f)=-\Lambda _{1}(1-s,f)$. Damit ist $\Lambda$ auch im ungeraden Fall ganz und der Satz ist bewiesen.$\,\,\,\square$.

Beispiel : Die nichtholomorphe Eisensteinreihe

Die nichtholomorphe Eisenstein-Reihe wird für $z\in \mathbb {H}$ und $s\in \mathbb {C}$ definiert durch

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

wobei $\Gamma (s)$ die Gammafunktion ist.

Dann ist E(z,s) eine Maaßsche Wellenform. Einen Beweis dazu findet man zum Beispiel im Buch Automorphe Formen von Anton Deitmar.

Quellen

1) Anton Deitmar, Automorphe Formen S.52-74