I am self studying analytic number theory from Tom M Apostol Modular functions and Dirichlet series in number theory and I couldn't think about an argument in proof related to Modular group.

I am adding image of proof highlighting the argument which I don't understand. enter image description here

enter image description here

My doubt is in 7 th line of 2nd image ie

$\phi $ is bounded in $R_\Gamma$ and it has been proved that $\phi $ is invariant under $\Gamma$ , so how does Apostol deduces $\phi $ is bounded in H?

Can someone please give a hint.


Since $R_\Gamma$ is a fundamental region for $\Gamma$, we have that $H = \Gamma \cdot R_\Gamma$. Stated differently, for any point $z \in H$, there is an element $\gamma \in \Gamma$ such that $\gamma z = z' \in R_\Gamma$. This is the defining characteristic of a fundamental domain.

Thus to understand the size of $\phi(z)$, you can use that $\phi(z) = \phi(\gamma z) = \phi(z')$, where we are using that $\phi$ is invariant under $\Gamma$. And thus $\phi$ being bounded on $R_\Gamma$ shows that $\phi$ is bounded on $H$.

  • $\begingroup$ @davilowryduda according to Apostol z' belongs to closure of fundamental region not fundamental region ( See page 30) . But it is given that $\phi $ is bounded in fundamental region, then you need to prove that $\phi $ is also bounded in closure of fundamental region $\endgroup$ – Tim Jan 3 '20 at 16:26
  • $\begingroup$ Very well, then prove that $\phi$ is bounded on the closure of the fundamental domain. As $\phi$ is continuous and $\phi \to 0$ as $\mathrm{Im} z \to \infty$, you can consider only a finite region up to some bounded height. Then merely the fact that $\phi$ is continuous and the remaining region is compact shows that $\phi$ is bounded on the whole fundamental domain (and thus on all of $\mathcal{H}$. $\endgroup$ – davidlowryduda Apr 12 '20 at 17:02
  • 1
    $\begingroup$ thank you very much!! $\endgroup$ – Tim Apr 12 '20 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.