5
$\begingroup$

In Silverman's Advanced Topics in the Arithmetic of Elliptic Curves there are two demonstrations of the Uniformization Theorem for the Elliptic Curves (It says that, given an Elliptic Curve $E$, there's a unique lattice such that $E$ is its associated Elliptic Curve).

The first proof uses a lot of advanced topics in modular forms, but I'm interested in the second proof. It is left as exercise at the end of the first chapter and it seems to use only elementary instruments.

There are four steps in this proof, I need help to solve the following:

Define $$j:\mathbb{H}/SL_2(\mathbb{Z}) \longrightarrow \mathbb{C}$$ as the function that sends $\tau \in \mathbb{H}/SL_2(\mathbb{Z})$ in the $j$-invariant of the torus $\mathbb{C}/\Lambda_{\tau}$, where $\Lambda_{\tau}$ is the lattice $\mathbb{Z}+\tau\mathbb{Z} \subset \mathbb{C}$.

Let $j_0 \in \mathbb{C}$, $H$ be a positive real number, $F_H \subset \mathbb{C}$ the set $\{z \in \mathbb{C} | -\frac{1}{2}<Re(z)<\frac{1}{2}, \, |z|>1, \, Im(z)< H\}$ and $\partial F_H$ its boundary with counterclockwise orientation. Suppose furthermore $j(\tau) \neq j_0$ on $\partial F_H$.

Prove that

$$\lim_{H \rightarrow +\infty}\frac{1}{2\pi i}\int_{\partial F_H} \frac{j'(\tau)}{j(\tau)-j_0}d\tau=1$$

The autor suggest to use the following property of the function $j$ $$j(\tau)=j(\tau+1)=j(-1/\tau)$$ and its series expansion to evaluate the integral, but I don't see how this can help me to evaluate the limit.

Could someone help me? Any hints or references?

I apologize if this question could be very simple for someone: I'm sure I'm missing something trivial!

$\endgroup$
  • $\begingroup$ Are you asking why the limit is $1$, or are you asking why the limit being 1 helps in proving the uniformization theorem, or both? $\endgroup$ – Álvaro Lozano-Robledo Apr 24 '14 at 19:10
  • $\begingroup$ I need help to prove the limit is 1. $\endgroup$ – Sabino Di Trani Apr 25 '14 at 1:30
  • $\begingroup$ Did my hints work then? $\endgroup$ – Álvaro Lozano-Robledo May 10 '14 at 1:47
  • $\begingroup$ Yes, it works.Furthermore I've found athe proof of a more general result for the $k$-modular forms in Lang's book: He uses the same idea! $\endgroup$ – Sabino Di Trani May 10 '14 at 21:05
2
$\begingroup$

I haven't checked the details, but here is how it should work out: the boundary $\partial F_H$ can be divided in 5 segments, namely the points of $\partial F_H$ that satisfy:

  • (A) $\Re(z)=1/2$, or

  • (B) $\Re(z)=-1/2$, or

  • (C) $|z|=1$ and $\Re(z)>0$, or

  • (D) $|z|=1$ and $\Re(z)\leq 0$, or

  • (E) $\Im(z)=H$.

Now you can calculate the integral on each piece:

  • The integrals on (A) and (B) will cancel out, because they are identical (because $j(\tau)=j(\tau+1)$), but in opposite directions.

  • The integrals on (C) and (D) will cancel out, because if you change variables in (D) sending $\tau$ to $-1/\tau$ (and use the fact that $j(\tau)=j(-1/\tau)$), then you get (C) but going in the opposite direction.

So it remains to calculate the integral on (E), and this one can be done using the series expansion of $j(\tau)$, to get a result that depends on $H$, and then take the limit as $H\to \infty$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.