A special modular function: $ j $-invariant.

Question

It is known that j invariant $$j(\tau)= 1728 \frac{g_2^3(\tau)}{\Delta(\tau)} $$ $\tau \in \mathbb{H}$

attains every complex value. Can someone guide me its proof?

where $L(\tau ) = \{\tau m + n | \quad \tau \in \mathbb{C} ; m,n \in \mathbb{Z}\} $ $G_k=\sum_{\omega \in L , \omega \neq 0}\frac{1}{\omega^k} $

$g_2(\tau)=60G_4(\tau);g_3(\tau)=140G_6(\tau);\Delta = g_2^3 -27g_3^2 $

There is a Theorem given in Advance topics in arithmetic in elliptic curves. It states that :$$j:X(1)\rightarrow \mathbb{P}^1 (\mathbb{C}) $$ So do is this question linked to mine??

Also what is cusp $\infty$?

There is an argument written "The $j$ has a simple pole at the cusp $\infty \in X(1)$, so the map is analytic with degree $1$ between compact Riemann surface. It is therefore isomorphism". Can someone explain this ?

Answer 1

So we have to prove that $j:\mathbb{H} \longrightarrow \mathbb{C}$ is surjective. By the given q series of $j(\tau) $it is clear that $j(\tau)$ $$ j(\tau ) = \frac{1}{q} +744+ 196884q+ 21493760q^2.+ c_2q^3......\qquad where \quad q = e^{2\pi i \tau}$$only goes to infinty when q=0 or the imaginary part of $\tau$ goes to $\infty$. and the pole is a order of pole is 1 and it is a ratio of two modular forms. So $j(\tau)$is holomorphic on $\mathbb{H}$.Hence by the open mapping theorem , the image of $j(\tau)$ must be a open set in $\mathbb{C}$.Now we will claim that $j(\mathbb{H})$ is also closed. Let $j (\tau_{k})$ be a sequence in $j(\mathbb{H})$ converging to some $\omega$ in $\mathbb{C}$ (where $\tau_{k} \in \mathbb{H}$).Since $j(\tau)$ is invariant under $SL_{2}(\mathbb{Z})$, then by only considering$ \tau_k$ such that $$ |RE(\tau_k)|\leq 1/2 \quad \text{and} \quad |Im (\tau_k)|\geq \frac{\sqrt{3}}{2}$$ We can consider because $ S= \left( \begin{array}{ccc} 0 & -1 \\ 1 & 0 \end{array} \right) \quad T= \left( \begin{array}{ccc} 1 & 1 \\ 0 & 1 \end{array} \right)$ also belong to $SL_{2}(\mathbb{Z})$ and thus $j(\tau)$ will be just repeat its value as $$ S(\tau)= -\frac{1}{\tau} \mbox{ and } T(\tau)= \tau +1$$ So we can restrict our attention to the specified region. \

Suppose the imaginary parts of the $\tau_k$'s are unbounded. But,then $j(\tau_k)$ contains a subsequence converging to $\infty$. Since $j(\tau_k)$converges to $w$, this cannot happen. So, the imaginary parts of the $\tau_k$'s are bounded, say by some$M\in \mathbb{R}$. Hence, each$\tau_k$ lies in the region. \begin{equation} R = \left\{\tau\in\mathbb{H}:|Re(\tau)|\leq\frac{1}{2},\frac{\sqrt{3}}{2}\leq |Im(\tau)|\leq M\right\}, \end{equation} a compact subspace of $\mathbb{H}$. But, this implies that $\tau_k$ has a subsequence converging to some $\tau_0\in \mathbb{H}$. Since $j(\tau)$ is continuous and $j(\tau_k)$ converges to $w$, then $j(\tau_0)=w$, and hence $w\in j(\mathbb{H)}$. Thus $j(\mathbb{H})$ is closed, and so $j(\mathbb{H})=C$.

Answer 2

This is not true. For instance $j(\tau)=j(\tau+1)=j(-1/\tau)$ for every $\tau\in\mathbb{H}$. More generally, $$j(\tau) = j\left(\frac{a\tau+b}{c\tau+d}\right)$$ for any matrix $\left(\begin{array}[cc] aa & b\\ c& d\end{array}\right)$ in $\Gamma(1)=\operatorname{SL}(2,\mathbb{Z})$. It is true, however, that $j:\mathbb{H}/\Gamma(1)\to \mathbb{C}$ is a bijection. See Chapter 1, Section 3, of Silverman's "Advanced Topics in the Theory of Elliptic Curves".