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Let $r>2$ and let $b_1,b_2,\ldots,b_r$ be in $\mathbf{P}^1(\mathbf{Q})$. Let $B$ be the divisor $$B:= \sum [b_i].$$ We consider this data to be fixed. For $d>1$, we define $\textrm{Ell}(b_1,b_2,\ldots,b_r,d)$ as the set of (isomorphism classes of) elliptic curves $E$ over $\mathbf{Q}$ that admit a finite morphism $f:E\longrightarrow \mathbf{P}^1_\mathbf{Q}$ of degree $d$ which is etale outside $\{b_1,b_2,\ldots,b_r\} \subset \mathbf{P}^1(\mathbf{Q})$.

Question 1. Let $E$ be in $\textrm{Ell}(b_1,b_2,\ldots,b_r,d)$ and choose a finite morphism $f:E\longrightarrow \mathbf{P}^1_\mathbf{Q}$ of degree $d$ which is etale outside $\{b_1,b_2,\ldots,b_r\} \subset \mathbf{P}^1(\mathbf{Q})$. Let $X$ be the analytification of $E_\mathbf{C}$. There exists a $\tau$ in the complex upper half plane such that $X = \mathbf{C}/\mathbf{Z}+\tau\mathbf{Z}$. Can we choose $\tau$ (or $q=e^{2\pi i \tau}$) using the data $(b_1,b_2,\ldots,b_r,d,f)$?

Question 2. It follows from Faltings's theorem that the set $\textrm{Ell}(b_1,b_2,\ldots,b_r,d)$ is finite. Is there a more elementary proof of this?

EDIT: Let me describe how the elliptic curve is given (in the set-up I have in mind).

Let $U$ be an open subscheme of $\mathbf{P}^1_\mathbf{Z}$ with complement $D$. We suppose that the closed subscheme $D$ is a horizontal divisor on $\mathbf{P}^1_\mathbf{Z}$ such that the base change $D_\mathbf{Q}$ equals $B$ defined above. Let $V\longrightarrow U$ be a finite etale morphism, with $V$ connected. Let $g:Y\longrightarrow \mathbf{P}^1_\mathbf{Q}$ be the normalization of $\mathbf{P}^1_\mathbf{Q}$ in the function field of $V$. We make the following extra assumptions:

1. $Y$ has a $\mathbf{Q}$-rational point.

2. The genus of $Y$ equals 1.

So the morphism $f$ arises like this.

I'm actually more interested in the set-up described above without assumptions 1 and 2. I just figured it would be an easy case to start with because it could/should be handled more directly.

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1 Answer 1

I am having difficulty making sense of question 1. If you already know $f$, then you have $E$, which gives you an $SL_2(\mathbb{Z})$-orbit of values of $\tau$. This seems to be the best you can do.

For question 2, I think you can bound the number of degree $d$ field extensions of $\mathbb{C}(z)$ whose discriminant divides a certain polynomial (e.g., $\prod_{i=1}^r (z-b_i)^d$). This puts a bound on the number of curves of any genus with a degree $d$ map to the line ramified at the chosen points, and hence on the genus one curves.

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I want to know how this orbit you mention looks like and choose a "small" representative in a way. This could mean something like finding a tau such that its absolute value is bounded by d times r, say. I really hope I'm making sense here. –  Ari Jan 3 '11 at 14:58
    
For any orbit, there are infinitely many points within any ball of positive radius around zero. If instead you want $e^{2 \pi i \tau}$ to be small, then you should get a well-defined answer in most cases (namely, inside the injectivity radius). Given the $j$-invariant of $E$, you should be able to approximate $q$ in various absolute values by power series methods (but I haven't tried this). –  S. Carnahan Jan 3 '11 at 15:24
    
What is the injectivity radius? What does it depend on? I really don't know anything about X as a compact Riemann surface besides its genus and the given morphism f. So for example I "dont know" the j-invariant. So I ask, can one "explicitly" write down the j-invariant in terms of the data given (branch points and degree of f)? –  Ari Jan 3 '11 at 16:43
    
Sorry, I shouldn't have called it the injectivity radius, but I meant the maximum radius in the $q$-disk where $j$ is injective, which is $e^{-2\pi}$. Again, when you say that you know the morphism $f$, it sounds like you have some datum that gives you the isomorphism type of the source $E$. How is the map $f$ described to you otherwise? A set of branch points in the line is not sufficient to characterize $E$. –  S. Carnahan Jan 4 '11 at 5:55
    
For the sake of brevity, I left out the exact set-up of my problem. I now see that this wasn't a good idea. I edited the question. Hopefully it'll be a bit clearer now. –  Ari Jan 4 '11 at 11:07

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