Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let N be a prime integer. We know that the element $c=(0)-(\infty)$ generates the torsion subgroup of $J_0(N)$ and it has order Num( (N-1)/12). Now, there is a natural map $\pi^*:J_0(N) \rightarrow J_1(N)$, coming from the covering map $\pi:X_1(N) \rightarrow X_0(N)$. My question is what is the image of c under this map? Specifically, is it possible for $\pi^*(c)=0$?

share|improve this question
6  
This can be made "explicit" in the special case $N=11$, for which both curves have genus 1. By moduli interpretation of cusps, $X_1(p)$ has $p-1$ geometric cusps: $(p-1)/2$ are rational points (the $p$-gon cusps) and $(p-1)/2$ are a single Galois orbit (the $1$-gon cusps, residue field $\mathbf{Q}(\zeta_p)^+$). Making $X_1(11)$ an elliptic curve using a rational cusp and $X_0(11)$ an elliptic curve using its image (the cusp $0$, not $\infty$!!), the deg-5 map $\pi$ has kernel given by the rat'l cusps: $\ker \pi = \mathbf{Z}/(5)$. Hence, dual map has kernel $\mu_5$, and $\mu_5(\mathbf{Q})=1$. –  BCnrd Sep 27 '10 at 5:53
    
Thanks! This example was actually confusing me, and that clarifies the situation. –  Soroosh Sep 27 '10 at 17:41
add comment

2 Answers 2

up vote 4 down vote accepted

The fact you mention about $(0) - (\infty)$ generating the torsion subgroup of $J_0(N)$ is Theorem 1 (Ogg's conjecture) on the first page of Mazur's paper "The Eisenstein Ideal". I recommend you actually read this paper. If you get as far as page 2, you will find a "Theorem 2 (twisted Ogg's conjecture)" which concerns the Shimura subgroup $\Sigma$. The construction of this subgroup essentially identifies it with the kernel of $J_0(N) \rightarrow J_1(N)$, and Proposition 11.6 of ibid. shows that $\Sigma$ is of multiplicative type, so BCnrd's remarks apply to the general case.

share|improve this answer
add comment

I don't think pi^*(c) can be 0. Suppose pi^*(c) = div(f); then f would be a map from X_1(N) to P^1 of degree about N. But in fact any such map is of degree at least ~N^2, i.e. the gonality of X_1(N) is bounded below by a constant multiple of N^2. This was proved independently by Zograf ("Small eigenvalues of automorphic Laplacians in spaces of cusp forms") and Abramovich ("A linear lower bound on the gonality of modular curves.")

Update: As Kevin points out in comments I should say "for N large enough." But "large enough" is effective here since the constants in the gonality bounds are effective (albeit small.)

share|improve this answer
    
It's zero when $N=2$, right? So maybe you mean "...can be zero for $N$ suff large" or something. –  Kevin Buzzard Sep 27 '10 at 12:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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