I am beginning to understand the very basics of modular forms, in that I understand the concept of a weakly modular function, I have seen the examples of $G_k(z)$ and $E_k(z)$ as weakly modular functions, and modular forms resp. I have also seen the importance of $D:=\{z\in \mathbb{H} : -\frac 12 \leq \Re(z) \leq \frac 12 , |z|\geq 1 \}$, where $\mathbb{H}$ is the upper half plane, and the proof that $\gamma \in\Gamma$ sends $z\in\mathbb{H}$ to $\gamma z \in D$, and also that the points on the boundary of $D$ are the unique points where $\gamma z=z$, so that there are three points at which the stablilizer subgroup of $SL_2(\mathbb{Z})$ is not the identity matrix.

My question is regarding the theorem that seems to follow on from these observations.

We are given a function, $f$, that is meromorphic at $p$, and we define $v_p(f)$ to be $k$, where $f(z)=(z-p)^k+\sum_{n=k+1}^\infty a_n(z-p)^n$. We also define $e_p$ to be $\frac{\#Stab(p)\subset SL_2(\mathbb{Z})}{2}$, which can take values of $1,2$ or $3$.

Finally, if $f$ is weakly modular of weight $k$, define $v_{\infty}(f):=k$ s.t. $f=a_kq^k+\sum_{n=k+1}^\infty a_nq^n, a_k\neq 0$

The theorem states:

$$v_{\infty}(f)+\sum_{p\:\in \:\mathbb{H}\:/\:SL_2(\mathbb{Z})} \frac{v_p(f)}{e_p}=\frac{k}{12}$$

My main problem is that I don't understand the definition of $\mathbb{H}/SL_2(\mathbb{Z})$. Is it the set given by the action of $SL_2(\mathbb{Z})$ on $\mathbb{H}$? Is it merely all the points at which $f$ is meromorphic?

I ask because it is used repeatedly later on and although I have a sense of what it is I don't quite get it.

  • $\begingroup$ The set of orbits $SL_2(\mathbb{Z})\setminus \mathbb{H}$ is a Riemann surface, whose construction is similar to $(\mathbb{Z}+i\mathbb{Z})\setminus \mathbb{C}$. With $\mathbb{H}^* = \mathbb{H} \cup \mathbb{Q} \cup \{i\infty\}$ we obtain that $SL_2(\mathbb{Z})\setminus \mathbb{H}^*=SL_2(\mathbb{Z})\setminus \mathbb{H} \ \cup \{i \infty\}$ is a compact Riemann surface. The meromorphic functions $SL_2(\mathbb{Z})\setminus \mathbb{H}^* \to \mathbb{C}$ are precisely the meromorphic weight $0$ modular forms. $\endgroup$
    – reuns
    Jun 11 '17 at 5:19

Typically, it's actually written $\text{SL}_2(\mathbb{Z}) \backslash \mathbb{H}$ since we usually write the group action on the left. I will do so here.

Then $\text{SL}_2(\mathbb Z) \backslash \mathbb{H}$ is the set of elements in $\mathbb{H}$ quotiented out be the relation that $z \sim \gamma z$ for all $\gamma \in \text{SL}_2(\mathbb Z)$. Some more naturally identify this with the "cannonical" fundamental domain $$ \mathcal{D} = \left\{ z | -\frac{1}{2} \leq \text{Re} z \leq \frac{1}{2}, \lvert z \rvert \geq 1 \right\}. $$

There is a more sophisticated point of view that makes this a bit better, too. We can identify $\mathbb{H}$ with the matrix $\begin{pmatrix} y&x \\ 0&1 \end{pmatrix}$ appearing in the Iwasawa decomposition of matrices in $\text{GL}_2(\mathbb{R})$. That is, every matrix $g \in \text{GL}_2(\mathbb{R})$ can be written uniquely as $$ g = \begin{pmatrix} y&x \\ 0&1 \end{pmatrix} \Theta R $$ where $\Theta \in \text{O}_2(\mathbb{R})$ is a (uniquely determined) orthogonal matrix and $R = \begin{pmatrix} r&0 \\ 0&r \end{pmatrix}$ is a diagonal matrix with $r \neq 0$.

In this view, $$ \mathbb{H} \simeq \text{GL}_2(\mathbb{R}) /( \text{O}_2(\mathbb{R}) \cdot\mathbb{R}^\times). $$ We may now quotient on the left by $\text{SL}_2(\mathbb{Z})$ as an actual group quotient. It is remarkable (and not immediately obvious) that this quotient is the exact same as identification under the standard action, but it is.

  • $\begingroup$ Thanks for the detail! For all intents and purposes, then, (That of a fourth-year modular forms course), can I think of $ SL_2(\mathbb{Z})\backslash\mathbb{H}$ as the canonical domain that you speak of? Merely because I need to use it to determine how many zeroes are in the set, so it is easier for me to think of it as an actual set in the complex plane in which I can 'physically see' the points $p$. $\endgroup$ Apr 19 '15 at 23:26
  • $\begingroup$ Also I think I still have slight trouble getting my head round the concept of 'quotienting out by an equivalence relation', which probably doesn't help my understanding. $\endgroup$ Apr 19 '15 at 23:29

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