---
tag: SZEYSY99AM
kind: example
name: Fundamental domain of the modular group
msc: [11F06, 20H05, 30F35]
instance_of: [KNJA2NEZ9F]
uses: [VEKV1EM77B, X6XGGH5FPW, 0SG74WBFA2, 0YQPZB5Q1G]
sources:
  - url: https://en.wikipedia.org/wiki/Fundamental_domain
---

## Example

Let the [[modular group|VEKV1EM77B]] $\Gamma = \mathrm{SL}_2(\Z)$ act on the [[upper half-plane|X6XGGH5FPW]] $\mathbb H$ by [[Möbius transformations|0SG74WBFA2]], generated by $T \tau = \tau + 1$ and $S \tau = -1/\tau$.
Let
$$\mathcal F = \set{\tau \in \mathbb H : \abs{\operatorname{Re} \tau} \le \tfrac12, \ \abs{\tau} \ge 1}.$$
Then $\mathcal F$ is a [[fundamental domain|KNJA2NEZ9F]] for $\Gamma$:

1. every [[orbit|0YQPZB5Q1G]] of $\Gamma$ in $\mathbb H$ meets $\mathcal F$;
2. two distinct points of $\mathcal F$ lie in the same orbit only if they lie on the boundary of $\mathcal F$.

Check of 1: for $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, one has $\operatorname{Im}(\gamma \tau) = \operatorname{Im} \tau / \abs{c\tau + d}^2$.
Only finitely many pairs $(c, d)$ of integers have $\abs{c\tau + d} \le 1$, so some $\gamma$ makes $\operatorname{Im}(\gamma \tau)$ maximal.
Apply a power of $T$ to move $\gamma \tau$ into the strip $\abs{\operatorname{Re} \tau} \le \tfrac12$; this does not change the imaginary part.
If the result $\tau'$ had $\abs{\tau'} < 1$, then $\operatorname{Im}(S \tau') = \operatorname{Im} \tau' / \abs{\tau'}^2 > \operatorname{Im} \tau'$, against maximality, so $\tau' \in \mathcal F$.

Check of 2: there are two boundary identifications.
The map $T$ sends the line $\operatorname{Re} \tau = -\tfrac12$ to the line $\operatorname{Re} \tau = \tfrac12$.
The map $S$ sends the arc $\abs{\tau} = 1$ to itself by $\tau \mapsto -\bar\tau$.
A short computation with $\abs{c\tau + d} \ge 1$ on $\mathcal F$ shows there are no other identifications.

## Remarks

The hyperbolic area of $\mathcal F$ is
$$\int_{-1/2}^{1/2} \int_{\sqrt{1 - x^2}}^{\infty} \frac{dy \, dx}{y^2} = \int_{-1/2}^{1/2} \frac{dx}{\sqrt{1 - x^2}} = 2 \arcsin \tfrac12 = \frac{\pi}{3}.$$
In $\mathcal F$, the points with nontrivial stabilizer in $\mathrm{PSL}_2(\Z)$ are the following.
The point $i$ is fixed by $S$ of order $2$, and the corners $e^{2\pi i/3}$ and $e^{\pi i/3}$ are fixed by elements of order $3$.
Gluing the sides of $\mathcal F$ by $T$ and $S$ gives the modular curve $\Gamma \backslash \mathbb H$, a sphere with one puncture at the cusp $i\infty$.

## Explanations

### L1 human

The modular group moves points of the upper half-plane around in many ways.
The region $\mathcal F$, a strip of width one above the unit circle, contains a copy of every point up to these moves, and only its edges are counted twice.
Its images under the group tile the whole half-plane by curved triangles.
A modular form is determined by its values on this one region.