# 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.

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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".

As for what Miyake had in mind at that point in his book, I'm not sure!

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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