An automorphic function with no poles is constant.

Question

Daniel Bump calls $f$ an automorphic function if it satisfies the formula $$f\left(\frac{az+b}{cz+d}\right)=f(z)$$ where $\begin{pmatrix} a&b\\ c&d \end{pmatrix}\in\mathrm{SL}(2,\mathbb{Z})$ in his Automorphic Forms and Representations. Now I want to verify that an automorphic function with no poles is constant. Bump claims it follows from the maximum modulus principle, but how to explain this in more detail? Should I try to show that $|f(x)|$ has a maximum in $U$ and how to demonstrate it?

Answer 1

First, note that the notion of “no pole” needs to be made very specific: the $j$-invariant map has no pole on $\mathbb{H}$ and is automorphic, yet is not constant. Of course, it has a “pole at infinity”. If we want to split hairs, we could then consider $\exp{j}$, which does not have a pole at infinity, rather an essential singularity.

We actually want not to consider any of these somewhat pathological situations. Here’s how the classical approach goes.

Because $f(z)=f(z+1)$, you can write $f(\tau)=g(e^{2i\pi \tau})$ for some holomorphic function on the punctured open unit disk $g$.

In particular, $g$ has a Laurent series $g(q)=\sum_{n \in \mathbb{Z}}{a_nq^n}$.

We assume now that $g$ is holomorphic at zero ($g(0)$ corresponds to $f(i\infty)$ – if such a thing could be given sense – so we usually say that $f$ is holomorphic at infinity or that $f$ has no pole at infinity. In other words, the “no pole” assumption has to include infinity).

So $g(q)=\sum_{n \geq 0}{a_nq^n}$ is holomorphic on the full open unit disk. This implies in particular that $f(z)$ is bounded on the subspace $S$ defined by $Im(z) \geq 1/2$ (ie $|e^{2i\pi z}| \leq e^{-\pi}$).

It’s an elementary exercise that every point of $\mathbb{H}$ has a $SL_2(\mathbb{Z})$-translate in $S$: in other words, every value taken by $g$ is in $g(\overline{D}(0,e^{-\pi}))$.

So the local maximum of $g$ on that disk is a global maximum on the open disk $D(0,1)$. By the strong maximum principle, $g$ is constant, hence so is $f$.