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

Hi, I've been thinking about this questions:

Is there a modular form in $M_2 (\Gamma_0(11))$ which only vanishes at the cusp 0? If so how can we find it?

Edit: There was another question which I removed.

share|improve this question
    
Yes, take a suitable linear combination of the Eisenstein form with the cuspform $\eta(q)^2 \eta(q^{11})^2$. –  Noam D. Elkies Jan 12 '12 at 16:36
    
This cuspform vanishes at 0, does the Eisenstein form vanish at 0 as well? Also, how do I make sure the linear combination does not vanish anywhere else on the upper half plane? –  Nadim Rustom Jan 12 '12 at 17:01
3  
Sorry, I didn't notice that you asked for no other zeros at all (though come to think of it what I wrote doesn't make sense anyway for the reason you suggest). OK, let's try again: there is no such form $\phi$, because if it existed then $\phi / (\eta(q)^2 \eta(q^{11}))^2$ would be a function on degree 1 on $X_0(11)$ [pole at the cusp $\infty$, zero at the cusp $0$], which is impossible because $X_0(11)$ is known to have genus 1. –  Noam D. Elkies Jan 12 '12 at 17:12
2  
Alternatively, the space of cusp forms has codimension one, and the Atkin-Lehner involution switches the two cusps. Thus, any form must either vanish at both cusps or vanish at neither. –  Kevin Ventullo Jan 12 '12 at 20:09
    
@Nadim : What do you mean exactly by vanishing ? Do you mean that there is no constant term in the Fourier expansion (at the cusp 0), or do you mean vanishing as a differential form on $X_0(11)$ ? This is not the same, for example the cusp form $\eta(q)^2 \eta(q^11)^2 \frac{dq}{q}$ doesn't vanish anywhere as a differential form. –  François Brunault Jan 12 '12 at 20:57
show 1 more comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.