The Dimension of $S_2(\Gamma _0(N))$

The dimension of $S_2(\Gamma _0(N))$ is

$$\dim_\mathbb{C}S_2(\Gamma_0(N)) = 1+ \frac{\mu}{12} - \frac{\nu_2}{4} - \frac{\nu_3}{3} - \frac{\nu_\infty}{2},$$

where $\mu = N \prod_{p\mid N} (1+1/p)$, and $\nu_2 = \prod_{p\mid N} \left(1+\left(\frac{-4}{p}\right)\right)$ unless $4\mid N$ in which case $\nu_2=0$, and $\nu_3 = \prod_{p\mid N} \left(1+\left(\frac{-3}{p}\right)\right)$ unless $2\mid N$ or $9\mid N$ in which case $\nu_3=0$, and $\nu_\infty = \sum_{d\mid N} \varphi (\gcd(d,N/d))$. For example,

$$\dim_\mathbb{C}S_2(\Gamma_0(2)) = 1+ \frac{3}{12} - \frac{1}{4} - \frac{0}{3} - \frac{2}{2} = 0,$$

and

$$\dim_\mathbb{C}S_2(\Gamma_0(11)) = 1+ \frac{12}{12} - \frac{0}{4} - \frac{0}{3} - \frac{2}{2} = 1.$$

One can prove that the vector space $S_2(\Gamma_0(11))$ has basis

$$f = q\prod_{n=1}^{\infty}(1-q^n)^2(1-q^{11n})^2 = q - 2q^2 - q^3 + 2q^4 + q^5 + 2q^6 - 2q^7 + \cdots.$$

Exercise 3.6   Compute the dimension of $S_2(\Gamma_0(25))$.