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

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 $$\pi$$, $$2 \pi$$,...?

My dim memory is that the notation in the exercise changed $$7$$ to $$\infty$$ as the book went through successive printings.
• Apparently changing the last coset representative did break part of the exercise, sorry, but if $\beta_7$ is used (as the book originally had it) rather than $\beta_\infty$ then things work out. Commented Jul 31 at 20:56
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.