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4

Yes, a modular form on $\text{SL}_2(\mathbb{Z})$ is a modular form on any congruence subgroup $\Gamma < \text{SL}_2(\mathbb{Z})$. It's quite clear that the transformation law holds, as you've mentioned. Suppose that $\alpha \neq \infty$ is a cusp of $\Gamma$. Then since $\alpha$ is equivalent to $\infty$ in $\text{SL}_2(\mathbb{Z})$, i.e. there is some ...

2

The short answer is very straightforward. I'll let you fill in the details. As $f$ is cuspidal, you have that $f(iy) \to 0$ super-polynomially fast as $y \to \infty$. Consider the integral function $$I(s) = \int_0^\infty f(iy) y^s \frac{dy}{y}.$$ The super-decay of $f(iy)$ guarantees convergence for every $s$, and this is clearly an analytic function in ...

1

Note that $q^{n}$ corresponds to $nz$. Hence we have \begin{align} F(q) &= \left(\frac{\Delta(6z)\Delta(z)}{\Delta(3z)\Delta(2z)}\right)^{1/2}\notag\\ &= \left(\frac{q^{6}\prod_{n = 1}^{\infty}(1 - q^{6n})^{24}\cdot q\prod_{n = 1}^{\infty}(1 - q^{n})^{24}}{q^{3}\prod_{n = 1}^{\infty}(1 - q^{3n})^{24}\cdot q^{2}\prod_{n = 1}^{\infty}(1 - ...

1

Remember that if you take $n$-th roots of an equation, there are $n$ (complex-valued) solutions corresponding to the $n$ roots of unity. So taking $24$-th roots of unity of $\Delta(\gamma z) = (cz + d)^{12} \Delta(z)$ for $\gamma = \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)$, we must have that \[\eta(\gamma z) = ...

1

This is a little confusing. Let $Y_1(6)$ be the open modular curve $\mathcal{H}/\Gamma_1(6)$. It has a canonical compactification $X_1(6)$ on which the vector space of functions is indeed one-dimensional. The field of modular functions is the function field of $X_1(6)$, which has transcendence degree 1 over the complex numbers (and in particular is certainly ...

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