Regarding expressing $j_p $ as a polynomial in $\Phi $

Question

I am self studying apostol Modular functions and Dirichlet series in Number Theory and I have a doubt in an argument of theorem 4.11 .

I have posted two images and I hope both of them are clear. I posted them because it is very time consuming to write all details involving proof and also due to the fact that I have doubt only in last part of proof .

Doubt is - Function f is proved to be analytic at vertex $\tau $ =0 and each point $\tau $ =0 in H. ( I clearly understood it) .

But now I don't know how to deduce next line which is

Therefore f is bounded in H so f is constant.

How does f becomes bounded in H?

Edit1 ->

I am looking for an answer that doesn't involves Riemann surfaces.

Can someone please give some hint of this

Answer 1

You have to look at the fundamental region of $\Gamma_0(p)$, it is the union of $ST^j(R)$ for $0\le j <p$, where $R$ is the fundamental region of $\Gamma$ and $S$ and $T$ are the usual $S\tau=-1/\tau$ and $\tau = \tau+1$. The only points of concern (outside $\mathbb H$) are easily seen to be $0$ and $\infty$.

As $f$ is automorphic for $\Gamma_0(p)$ and analytic in all $\mathbb H$ and in $\tau = 0$ it is enough to look at $\tau \to \infty$. But $\Phi(\tau)$ has a zero in $\tau=\infty$ looking the expression for $f$ if it is enough to check that $j_p(\tau)$ is bounded as $\tau \to \infty$ but then you can use the relation for $j_p(-1/\tau)$ on top of the first page you have quoted to see that it is also bounded as $\tau \to \infty$.

Answer 2

$f$ is analytic on $\Bbb{H}$, it is $\Gamma_0(p)$ invariant and it is holomorphic at the two cusps $\Gamma_0(p)i\infty,\Gamma_0(p)0$. Covering the Riemann surface $X_0(p)=\Gamma_0(p)\setminus(\Bbb{H}\cup \Bbb{Q}\cup i\infty)$ by finitely many charts from $|s|\le 1$ we obtain that $f$ is bounded uniformly on $X_0(p)$, it attains its maximum modulus at some point $\Gamma_0(p)z$ and taking a chart $\phi(0)=z$ we obtain that $f\circ \phi$ is analytic and it attains its maximum modulus thus it is constant. By analytic continuation $f$ is constant.