Proving Diamond & Shurman Exercise 3.1.4 (f), identifying boundaries of fundamental domain for $\Gamma_0(13)$.

Question

I am studying modular forms through Diamond and Shurman's book A First Course in Modular Forms. In section 3.1, the author presents the following figure of fundamental domain for $\Gamma_0(13)$.

where according to the texts, the light dots are the two elliptic points of order $2$ and the dark ones are the two elliptic points of order $3$. They explains how this figure was generated and how to identify the boundary arcs in Exercise 3.1.4 (f).

Exercise 3.1.4 (f) Figure 3.1 was generated using the coset representatives \begin{align*} \beta_j = \begin{bmatrix} 1 & 0 \\ j & 1 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \quad j = -6, \dotsc, 6, \end{align*} and $\beta_{\infty} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. Find each translate $\beta_j(\mathcal{D})$ in the figure, where $\mathcal{D}$ is the fundamental domain from Chapter 2 (that is, the fundamental domain of $\mathrm{SL}_2(\mathbb{Z})$). Show that the $13$ points of $\mathrm{SL}_2(\mathbb{Z})(i)$ in the figure are $\beta_j(i)$ for $j = -6, \dotsc, 6$. Since $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ fixes $i$, the elliptic points of order $2$ are $\beta_j(i)$ when $j^2 + 1 \equiv 0 \pmod{13}$. Show that $\gamma \beta_j(i) = \beta_{j'}(i)$ for some $\gamma \in \Gamma_0(13)$ with $j' \neq j$ if and only if $jj' + 1 \equiv 0 \pmod{13}$. Use this to partition the 13 points of $\mathrm{SL}_2(\mathbb{Z})(i)$ in the figure into eight equivalence classes under $\Gamma_0(13)$, five with two points each where the angle is $\pi$, giving a total of $2 \pi$; one with one point where the angle is $2 \pi$; and two with one point where the angle is $\pi$ as it is at $i$ in $\mathcal{D}$, representing the unramified points. Identify the boundary arcs pairwise except for two arcs that fold in on themselves. Note that $\mathrm{SL}_2(\mathbb{Z})(\mu_3) = \mathrm{SL}_2(\mathbb{Z})(\mu_6)$. (Here $\mu_n$ is the $n$-th root of unity, $e^{2 \pi i/n}$.) Since $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ takes $\mu_6$ to $\mu_3$, the elliptic points of order $3$ are $\beta_j(\mu_6)$ when $j^2 - j + 1 \equiv 0 \pmod{13}$. Show that the $14$ points of $\mathrm{SL}_2(\mathbb{Z})(\mu_3)$ in the figure are $\beta_j(\mu_6)$ for $j = -6, \dotsc, 6, \infty$. Show that $\gamma \beta_j(\mu_6) = \beta_{j'}(\mu_6)$ for some $\gamma \in \Gamma_0(13)$ with $j' \neq j$ if and only $jj' - j + 1 \equiv 0 \pmod{13}$ or $jj' - j' + 1 \equiv 0 \pmod{13}$. Use this to partition the $14$ points of $\mathrm{SL}_2(\mathbb{Z})(\mu_3)$ in the figure into six equivalence classes under $\Gamma_0(13)$, one with the four points where two of the angles are $2\pi / 3$ and two of the angles are $\pi/3$, giving a total of $2 \pi$; two with three points where the angle is $2 \pi/3$, again giving $2 \pi$; one with the two points where the angle is $\pi$, giving $2 \pi$; and two classes with one point each where the angle is $2 \pi/3$ as it is at $\mu_3$ in $\mathcal{D}$, representing the unramified points. Determine whether each pair of boundary arcs is identified with orientation preserved or reversed. Show that under identification the figure is topologically a sphere.

I am fine with partitioning the $13$ points of $\mathrm{SL}_2(\mathbb{Z})(i)$, but I get confused when partitioning the $14$ points of $\mathrm{SL}_2(\mathbb{Z})(\mu_3)$. The equivalence classes I found for $j = -6, \dotsc, 6, \infty$ defined by $\gamma \beta_j(\mu_6) = \beta_{j'}(\mu_6)$ for some $\gamma \in \Gamma_0(13)$ are \begin{align*} \{\infty, 0, 1\}, \{2, -6, -1\}, \{3, 5, 6\}, \{4\}, \{-2, -5, -4\}, \{-3\} \end{align*} which is different from the description in the Exercise. I am also confused what the term "angle" refers to. How to see that a point has angle $\pi$, $2 \pi$,...?

Answer 1

My dim memory is that the notation in the exercise changed $7$ to $\infty$ as the book went through successive printings.

Answer 2

The 14 points of $\text{SL}_2(\mathbb{Z})\mu_3=\text{SL}_2(\mathbb{Z})\mu_6$ in this fundamental domain are $\{\beta_j(\mu_6)\}$ for $j\in\{-6,\dots,6,7\}$ and NOT $j\in\{-6,\dots,6,\infty\}$. (I think you incorrectly copied the problem from Diamond and Shurman at the point where this is suggested.)

Note that in your set $\{-6,\dots,6,\infty\}$, $\infty$ is redundant since $\beta_\infty(\mu_6)=\mu_6=\beta_1(\mu_6)$.

Now when you find the partition you should get \begin{align*} \{-6, -1, 2, 7\},\ \{-2, -5, -4\},\ \{3, 5, 6\},\ \{0, 1\},\ \{4\},\ \{-3\}, \end{align*} which matches the description in the exercise.