Show that $\Re(\tau)=\frac12+n$ cannot be inside any fundamental region

Question

I am currently reading Hardy's book Ramanujan. He claimed (in modern notations) that when $D$ denotes the fundamental domain of the full modular group, for each transformation $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in PSL(2,\mathbb Z)$, there exists a unique integer $n$ such that for all $\tau\in D$, we have

$$ \tau'={a\tau+b\over c\tau+d},\quad n-\frac12\le\Re(\tau')\le\frac12+n.\tag1 $$

I understood we can always find a unique $n=n(\tau)$ for each $\tau\in D$, but I could not see why this chosen $n$ will work for all $\tau\in D$. After some considerations, I formulated the following proof-by-contradiction argument:

Suppose $n$ is chosen by specifying a particular $\tau_0\in D$ and for some other $\tau_1\in D$, the image of $\tau_1$ lies outside the region described by (1). Then by the simple connectedness of $D$, there exists some interior point $\tau_2\in D$ that gets mapped to the vertical line with real part of some half odd integer $\frac12+m$, and a contradiction will happen if we can prove the statement that $\Re(\tau)=\frac12+m$ cannot lie inside any fundamental domains.

This statement is quite obvious from the visualizations, but I do not know how to prove it:

Answer 1

We are going to need some facts of hyperbolic geometry

• angle sum at a point is still $2\pi$, angle of a geodesic is $\pi$ at any point.
• geodesics in $\mathbb{H}$ are portions of Euclidean circles or lines that intersect the real axis orthogonally.
• Given any geodesic $\gamma$, there is an (orientation-reversing) isometry "reflection in $\gamma$" that fixes only every point on the geodesic $\gamma$. Reflections map geodesics to geodesics and preserves angles (up to orientation if you care about the sign).
• the hyperbolic metric is conformal to the Euclidean one on $\mathbb{H}$ (i.e., $\lvert ds\rvert=e^{f(\tau)}\lvert d\tau\rvert$ some $f\colon\mathbb{H}\to\mathbb{R}$), so we can calculate angles using Euclidean geometry.
• (To prove the full tiling, we will need uniqueness of geodesic and the area of polygons, but we don't actually need that for this question)

We easily calculate the angles of region $D$ at the corners $\pm\frac12+\frac{\sqrt{3}}2i$ are $\pi/3$, and the two boundary curves are hyperbolic geodesics. So we can reflect repeatedly and get 6 copies of $D$ meeting at the point. In particular, $3\times(\pi/3)=\pi$ so the geodesic $\operatorname{Re}z=\pm\frac12$ continues to be a boundary of a copy of $D$ for a little more (down to $\operatorname{Im}z=\frac1{2\sqrt3}$). Again same argument, reflecting and you get down to the real axis following a boundary of a copy of $D$.

Finally, note that there is a copy of $D$ sharing the $\operatorname{Re}z=\pm\frac12$ is going to have the same orientation (in fact it is the one on the "same side"), so you can find an element of $PSL(2,\mathbb{Z})$ that maps $D$ to this copy. So if $\operatorname{Re}\tau=\pm\frac12$ then it is on the boundary of our tiling. Now we translate using $(\tau\mapsto \tau+m)\in PSL(2,\mathbb{Z})$ to get other $\Re(\tau)=m+\frac12$.