Exercise 3.6.4 of A First Course in Modular Forms asks readers to prove the final part of Theorem 3.6.1, which provides formulas for the dimension of modular forms and cusp forms of weight $1$, $\mathcal{M}_1(\Gamma)$ and $\mathcal{S}_1(\Gamma)$. Here $\Gamma$ is a congruence subgroup of $\mathrm{SL}_2(\mathbb{Z})$ such that the negative identity matrix $-I \notin \Gamma$.
(Some parts of) Theorem 3.6.1. ...let $g$ the genus of $X(\Gamma)$, $\varepsilon_3$ the number of elliptic points with period $3$, $\varepsilon^{\mathrm{reg}}_{\infty}$ the number of regular cusps, and $\varepsilon^{\mathrm{irr}}_{\infty}$ the number of irregular cusps. If $\varepsilon^{\mathrm{reg}}_{\infty} > 2g-2$ then $\dim(\mathcal{M}_1(\Gamma)) = \frac{\varepsilon^{\mathrm{reg}}_{\infty}}{2}$ and $\dim(\mathcal{S}_1(\Gamma)) = 0$. If $\varepsilon^{\mathrm{reg}}_{\infty} \leq 2g-2$ then $\dim(\mathcal{M}_1(\Gamma)) \geq \frac{\varepsilon^{\mathrm{reg}}_{\infty}}{2}$ and $\dim(\mathcal{S}_1(\Gamma)) = \dim(\mathcal{M}_1(\Gamma)) - \frac{\varepsilon^{\mathrm{reg}}_{\infty}}{2}$.
The former parts can be proved using Riemann-Roch Theorem. Let $f \in \mathcal{A}_1(\Gamma)$ and $\omega \in \Omega^{\otimes k}(X(\Gamma))$ be the differential of $X(\Gamma)$ (the compactified modular curve) pulling back to $f(\tau)^2 (d \tau)^k$ on $\mathcal{H}$. Let $g$ be the genus of $X(\Gamma)$, $\{x_{3, i}\}$ be the period $3$ elliptic points, $\{x_i\}$ the regular cusps, and $\{x'_i\}$ the irregular cusps. Then the divisors \begin{align*} D_1 := \lfloor f \rfloor = \frac{1}{2} \mathrm{div}(\omega) + \sum\limits_i \frac{1}{2} x_i \\ D_2 := \left\lfloor f - \sum\limits_{i} x_i - \sum\limits_{i} \frac{1}{2} x'_i \right\rfloor = \frac{1}{2} \mathrm{div}(\omega) - \sum\limits_i \frac{1}{2} x_i \end{align*} have linear spaces isomorphic to $\mathcal{M}_1(\Gamma)$ and $\mathcal{S}_1(\Gamma)$, respectively. The degree of the divisors are \begin{align*} \deg(D_1) = g-1+\frac{1}{2} \varepsilon^{\mathrm{reg}}_{\infty}, \quad \deg(D_2) = g-1-\frac{1}{2} \varepsilon^{\mathrm{reg}}_{\infty}. \end{align*}
The book describes Riemann-Roch Theorem and a corollary as follows:
Theorem 3.4.1 (Riemann-Roch). Let $X$ be a compact Riemann surface of genus $g$. Let $\mathrm{div}(\lambda)$ be a canonical divisor on $X$. Then for any divisor $D$ on $X$, \begin{align*} l(D) = \deg(D) - g + 1 + l(\mathrm{div}(\lambda) - D). \end{align*}
Corollary 3.4.2 Let $X, g, \mathrm{div}(\lambda)$, and $D$ be as above. Then
- $l(\mathrm{div}(\lambda)) = g$.
- $\deg(\mathrm{div}(\lambda)) = 2g-2$.
- If $\deg(D) < 0$ then $l(D) = 0$.
- If $\deg(D) > 2g-2$ then $l(D) = \deg(D) - g + 1$.
If $\varepsilon^{\mathrm{reg}}_{\infty} > 2g-2$, then the results follow from (c) and (d) of the corollary. If $\varepsilon^{\mathrm{reg}}_{\infty} \leq 2g-2$, then by Riemann-Roch \begin{align*} l(D_1) = \frac{1}{2}\varepsilon^{\mathrm{reg}}_{\infty} + l(\mathrm{div}(\lambda) - D_1) \geq \frac{1}{2}\varepsilon^{\mathrm{reg}}_{\infty}. \end{align*}
The problem is that I don't know how to show $\dim(\mathcal{S}_1(\Gamma)) = \dim(\mathcal{M}_1(\Gamma)) - \frac{\varepsilon^{\mathrm{reg}}_{\infty}}{2}$. It seems that \begin{align*} l(\mathrm{div}(\lambda) - D_2) = l(D_1) \end{align*} but I have no idea how to explain this.
Since $k=1$, we can take $\lambda=\omega$, then
$$ \begin{aligned} \dim(\mathcal M_1(\Gamma))&=l\left(\frac{1}{2} \mathrm{div}(\omega) + \sum\limits_i \frac{1}{2} x_i\right)\\ &=\frac{\varepsilon^{\mathrm{reg}}_{\infty}}{2}+l\left(\frac{1}{2} \mathrm{div}(\omega) - \sum\limits_i \frac{1}{2} x_i\right)&(\text{Riemann-Roch})\\ &=\frac{\varepsilon^{\mathrm{reg}}_{\infty}}{2}+\dim(\mathcal S_1(\Gamma)). \end{aligned} $$