# Elliptic Points of Modular Group in Upper Half Plane

This is a very small question.

Let $\mathbb{\Gamma} = \mathrm{SL_2}(\mathbb{Z})$ be the modular group, $\mathcal{F} = \{z \in \mathbb{C} ;\; \lvert z \rvert \geq 1,\; \lvert \Re (z) \rvert \leq 1/2\}$ its fundamental domain.

I (probably) don't understand the following argument made by Toshitsune Miyake in "Modular Forms", in the proof of Theorem 4.1.3, page 98.

Since $\mathbb{\Gamma}$ contains $\tau = \left[\begin{smallmatrix}1 & 1 \\ 0 & 1\end{smallmatrix}\right]$, and $\omega = \left[\begin{smallmatrix}0 & -1 \\ 1 & 0\end{smallmatrix}\right]$, the boundary points of $\mathcal{F}$, other than $i$, [$\zeta_6 = e^{2\pi i/6}$, $\zeta_6^2$], are [...] ordinary points.

This is what I thought: Let $p$ be an elliptic point on the boundary of $\mathcal{F}$, stabilized by $\gamma \in \mathbb{\Gamma}$. Suppose $p$ isn't $\zeta_6$ nor $\zeta_6^2$. Since $\gamma \mathcal{F} \cap \mathcal{F} \neq \emptyset$, $\gamma$ must now be either $\tau$, $\tau^{-1}$ or $\omega$, but it can't be $\tau$ nor its inverse, for it has to be elliptic. So the only other possible elliptic points have to be stabilized by $\omega$, which leaves one with $i$.

But now I don't know how to prove the step $\gamma \mathcal{F} \cap \mathcal{F} \neq \emptyset\; \Rightarrow\; \gamma = \tau, \tau^{-1}, \omega$, which only is visually clear to me. I feel that the original, intended argument is easier than this and I'm being blind.

-

A bit more checking is needed here than the argument you sketch: there are more elements than just $\tau, \tau^{-1}$ and $\omega$ such that $\gamma D \cap D \ne \varnothing$ -- in fact the set $\{ \gamma \in \overline{\Gamma} : \gamma D \cap D \ne \varnothing\}$ has 10 elements if I remember correctly. Here $\overline{\Gamma} = \Gamma / \{\pm 1\} = PSL_2(\mathbb{Z})$. However, you can check that none of these ten elements (except the identity) fix any point other than $i, e^{\pi i/3}, e^{2\pi i / 3}$. This argument is worked out very carefully in the last chapter of Serre's book "A Course in Arithmetic".
Ok, thanks for that. Though I think, if $\gamma$ doesn't stabilize as $\zeta_6$ or $\zeta_6^2$ as assumed, the only possibilities are indeed $\tau$, $\tau^{-1}$ and $\omega$ as suggested by this picture. – k.stm Sep 30 '12 at 17:28