Equivalence of two definitions of modular forms

Question

I am a theoretical computer science student, so have only a limited background in complex geometry. As part of my study on Ramanujan graphs, I came across modular forms.

As I understand it, there are different ways to approach and define a modular form $f$ of weight $k$: 1) A holomorphic function $f:\mathbf{H} \to \mathbb{C}$ that is also holomorphic at the limit $i \infty$ and satisfies a function equation $$f\left( \frac{az+b}{cz+d} \right)= (cz+d)^k f(z)$$ for every $$\begin{pmatrix} a & b\\c & d \end{pmatrix} \in SL(2,\mathbb{Z})$$

2) A complex-valued function $f$ on the set of lattices in $\mathbb{C}$ defined as follows: for a lattice $\Lambda=<1,z>$, $f(\Lambda)=f(z)$ is an analytic function of $z$, and for an equivalent lattice $\alpha \Lambda$, $$f(\alpha \Lambda)=\alpha^{-k}f(\Lambda)$$

So on the one hand, a modular form is a function defined on the upper half plane $\mathbf{H}$ that is "well-behaved" with respect to the action of the modular group $\Gamma$. On the other hand, a modular form is a function on the set of lattices in $\mathbb{C}$ whose "growth" is prescribed. I am trying to see how to relate the two definitions and intuitively understand their equivalence.

The first definition tells us that given $f(z)$, the values taken by $f$ on the orbit of $z$ under the action of $\Gamma$ is fully known. So if this orbit is a lattice in $\mathbf{H}$ (which can then be extended to a lattice in $\mathbb{C}$), then we would find some common ground between the two definitions. So my questions are: 1) Is the orbit of $z \in \mathbf{H}$ under the action of $\Gamma$ a lattice in $\mathbb{C}$? If so, do we have a simple expression for a basis of this lattice?

2) Conversely, can every lattice in $\mathbb{C}$ be expressed (or at least equivalent to) as the orbit of some element of $\mathbf{H}$ under the action of $\Gamma$ ?

I apologize if these questions are trivial, but most references on these topics seem to require too many prerequisites and are way too rigorous to be of much immediate help. Thanks.

Answer 1

Let $\Lambda$ be the lattice of rank $2$ generated by $[u,v]$ two complex numbers. Let's try to find an invariant up to scaling and orientation. A natural choice is $$\tau_\Lambda = \begin{cases}\frac{u}{v} \text{ if } Im(\frac{u}{v}) > 0 \\ \\ -\frac{u}{v} \text{ otherwise} \end{cases}$$ which is in the upper half-plane.

But, as Andres Mejia said, there are many other possible choices : $$\frac{-1}{\tau_\Lambda}= \begin{cases}\frac{v}{u} \text{ if } Im(\frac{v}{u}) > 0 \\ \\ -\frac{v}{u} \text{ otherwise} \end{cases} \qquad \text{or} \qquad \tau_\Lambda+1= \frac{u+v}{v}$$ both being not less acceptable, since $[v,u]$ and $[u+v,v]$ generate the same lattice $\Lambda$.

So for $\tau_\Lambda$ really deserving to be called an invariant, it must depend only on the lattice and not on the particular choice of generators, and hence we need to look at the equivalence class $$\left\{ \frac{a\tau_\Lambda+b}{c\tau_\Lambda+d} \ | \ (a,b,c,d) \in \mathbb{Z}^4, ad-bc = 1 \right\}$$ i.e. the modular curve

Answer 2

Consider the two elements $$T:= \begin{pmatrix} 1 & 1\\0 & 1\end{pmatrix}$$ and $$S:=\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$$

in $SL_2(\mathbb Z)$.

From these, we obtain the identities $f(z+1)=f(z)$ and $f(-1/z)=(-z)^kf(z)$. The first relation tells you why the lattice is natural at all and the second tells us that in particular, let $x:=-1/z$ in the second identity to see that $f(x)=(x^{-k}f(-1/x)$. The action of $T$ and $S$ generate the same lattice, since $z \mapsto z+1$ is equivalent to "picking a third point" in the same grid, and $z \mapsto -1/z$ is equivalent up to scaling and orientation (reference point.) So, these two transformations tell us that we get the same lattice when $f$ is applied, but also with a $x^{-k}$ factored out.