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I am trying to understand a point made by Zagier in the paper Elliptic Modular Forms and Their Applications. Proposition 22 is the statement that if $\tau$ in the upper half plane $\mathcal{H}$ is a CM-point, then $j(\tau)$ is an algebraic number. Here $j(\tau)$ is the $j$-function.

In the proof, he considers a matrix $$M = \left(\matrix{B &C\\-A & 0}\right)$$ containing the coefficients for the quadratic relation $A\tau^2+B\tau+C = 0$ satisfied by $\tau$, where $A,B,C\in\mathbb{Z}$ (since $\tau$ is a CM point). As $\tau\in\mathcal{H}$ we know that $\det(M)>0$, although $M$ may not be in $SL_2(\mathbb{Z})$. We consider the finite-index subgroup

$$H:=SL_2(\mathbb{Z})\cap M^{-1}SL_2(\mathbb{Z})M.$$

Zagier then says that since the functions $j(z)$ and $j(Mz)$ are modular functions for the subgroup $H$ they are algebraically dependent (as functions). Note that $\tau$ is fixed by the action of $M$ on $\mathcal{H}$ by definition of $M$.

I don't understand why this is the case. I do know that if $\Gamma\subseteq SL_2(\mathbb{Z})$ is a finite-index subgroup then the dimension of the vector space $M_k(\Gamma)$ of modular forms of weight $k$ for $\Gamma$ is finite. In this case, we are considering modular functions i.e. modular forms of weight $k=0$ for the subgroup $H$. However, there are two important differences: firstly, we are not considering the linear structure but rather the multiplicative structure, of which I know nothing about. Secondly, $j(\tau)$ is not a modular form in the classical sense but rather a meromorphic modular form with a pole at infinity.

The only idea I have is that successive powers of $j(M\tau)$ generate copies of the functions $$j(A_1\tau), \dots, j(A_n\tau)$$ for a complete set of coset representatives of $H$ in $SL_2(\mathbb{Z})$. However, I'm not really sure how to prove this.

Is there a way to understand why $j(z)$ and $j(M z)$ must be algebraically dependent in this case?

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  • $\begingroup$ First, let me remark that for most authors a modular function is not the same thing as a modular form of degree 0. Modular forms are required to be holomorphic at infinity, whereas modular functions are just meromorphic. $\endgroup$ – user316327 Oct 30 '17 at 15:11
  • $\begingroup$ @DocteurCottard oops, good point! I will edit the question to point out that $j(z)$ is not a modular form in the usual sense. $\endgroup$ – Alex Saad Oct 30 '17 at 15:15
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Here is a geometric picture which is useful to have in mind.

The quotient of the Poincaré upper half-plane by the group $H$ is a Riemann surface, which you can compactify to obtain a compact Riemann surface $X$ (which is a fortiori an algebraic curve). Modular functions for $H$ may be identified with meromorphic functions on $X$. As $X$ is of dimension one, its function field $\mathbf{C}(X)$ has transcendence degree 1 over $\mathbf{C}$, so any two meromorphic functions on $X$ are algebraically dependent.

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  • $\begingroup$ $X$ of dimension $1$ ? You meant genus $0$, which can be shown from how the edges of the fundamental domain are identified (making it homeomorphic to the Riemann sphere) or from the fact $j(\tau)$ has a unique simple pole $\endgroup$ – reuns Oct 30 '17 at 15:19
  • $\begingroup$ I mean that $X$ is a curve. $\endgroup$ – user316327 Oct 30 '17 at 15:20
  • $\begingroup$ Ok, but the OP will need that it is of genus $0$ to show $j(\tau)$ is algebraic whenever $\tau \in \mathbb{Q}(\sqrt{-d})$ $\endgroup$ – reuns Oct 30 '17 at 15:22
  • $\begingroup$ @reuns why do you need to know the genus is $0$? Surely the fact that the function field has tr-degree $1$ over $\mathbb{C}$ suffices? $\endgroup$ – Alex Saad Oct 30 '17 at 15:28
  • $\begingroup$ @reuns also it is not immediately clear how you can show the genus is $0$ from the definition of $H$. Doesn't this depend on $M$ in some way? $\endgroup$ – Alex Saad Oct 30 '17 at 15:29

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