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.

  • $\begingroup$ @JyrkiLahtonen ok, took into consideration $\endgroup$ – janmarqz Jul 1 '18 at 18:00

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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  • $\begingroup$ 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. $\endgroup$ – k.stm Sep 30 '12 at 17:28

Although this is an old question, I fell into the same problem, and will explain what I think was Miyake's original line of thought with this.

Looking back at the proof of Theorem 1.9.1, there was no point in considering $z$ a vertex. Rather, we can pick it on any point of the border of $F$ where we can think of it as being a vertex, after cutting the line containing it at it.

But then, we would have, by the very same proof (Unless he used the fact that it's a vertex and I didn't see it.), that $\omega_1+\ldots+\omega_v=\frac{2\pi}{e}$.

Now, since $\tau$ and $\omega$ are in $\Gamma$, then for any point $z$ on $\partial F$ except $\zeta,\zeta',i$, there is a different point on $\partial F$($\tau z$ if it is on the straight lines, $\omega z$ if it is on the circular arc) equivalent to it. But then, for all these points, we have $\omega_v=\pi$, and thus, $e=\frac{2\pi}{\omega_1+\ldots+\omega_v}\leq\frac{2\pi}{\pi+\pi}=1$, and therefore, these points are not elliptic.

The difference is that $i$ is taken by $S$ to itself, and that the interior angles at $\zeta$ and $\zeta'$ are less than $\pi$, which is what makes these points special.

The more reason to think my argument is correct is because he uses the formula for $i$, which is not a vertex as he defined it before, but I think it doesn't matter as he doesn't use that back in the proof.

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  • $\begingroup$ Unfortunately, I don’t have Miyake’s book at hand anymore to check what you claim. $\endgroup$ – k.stm Jul 4 '19 at 19:53

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