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

I've seen stated offhand in many sources that the cuspidal subgroup of the Jacobian of $X_0(N)$ is finite.

Do they mean that the subgroup of the jacobian generated by $\mathbb{Q}$-rational cusps is finite, or do they mean the subgroup generated by all cusps is finite?

I know that Ogg (and also Mazur) does the case $N = p$ prime in a rather explicit manner (in this case all cusps are rational), but says that the case for $N$ composite is more complicated and doesn't state any results to that end.

Ideally, it would be nice to see a moduli-theoretic explanation of this, if it exists.

thanks,

  • will
share|improve this question
    
I have put the number theory tag as this question is of interest to number theorists as well. –  Arijit Apr 23 '13 at 10:01
1  
The statement is one about all cusps (not necessarily $\mathbb{Q}$-rational ones). The original paper by Drinfeld, "Two theorems on modular curves", is very readable. –  David Loeffler Apr 24 '13 at 7:54
    
@David Loeffler - I've managed to find that paper, though I'm having trouble understanding his "proof of assertion 1". Given any $g\in M$ and $\gamma\in\text{PSL}_2(\mathbb{Z})$, how does he conclude that $g\gamma(i\infty) = \gamma h(i\infty)$ for some $h\in\Gamma(N)$? –  oxeimon Apr 24 '13 at 11:16
    
@David Loeffler - Also, I don't understand how he deduces that $\omega|_g = \omega$ in his proof of assertion 2. Any chance you could help me with that? –  oxeimon Apr 24 '13 at 11:38
    
See also R. Elkik, "Le théorème de Manin-Drinfeld." Astérisque, pages 59-67, 1990. Séminaire sur les Pinceaux de Courbes Elliptiques (Paris, 1988). The proof given there follows an idea of Deligne and uses mixed Hodge structures. –  Damian Rössler May 2 '13 at 10:30
add comment

1 Answer

up vote 2 down vote accepted

This is the Manin-Drinfeld theorem.See related MO question

share|improve this answer
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.