Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

While reading a paper I came across the phrase "the Eisenstein series of weight 2 associated with the cusps of $\Gamma_{0}(6)$". Can anyone give me the definition of Eisenstein series of weight $k$ associated with the cusps of $\Gamma$ where $\Gamma$ is a congruence subgroup of $SL_{2}(\mathbb{Z})$?

share|improve this question
4  
Dear Shayla, How much do you know about modular forms and modular curves? (The answer to this will be very helpful in formulating an answer to your question.) Regards, –  Matt E Jul 14 '11 at 6:55
    
Aside from the basic definitions related to modular forms, not very much. –  Shayla Jul 14 '11 at 6:57
add comment

1 Answer

In general, for congruence subgroups $\Gamma$ of $SL(2,\mathbb Z)$, the collection of weight $k$ Eisenstein series is spanned by functions formed by $$ E(z)=E(z,c_o,d_o,N,k)=\sum''_{c=c_o,d=d_o(N)}\;\;\frac{1}{(cz+d)^k}$$ where $c,d$ run over _relatively_prime_ integers congruence to $c_o,d_o$ mod $N$. Especially for composite $N$, there are (elementary) relations among these. Eisenstein series are far more elementary objects than most other modular forms.

However, those sums only converge absolutely for $k>2$. Many decades ago, E. Hecke had already addressed this "problem", by what is now called "Hecke summation" (although it can be understood more systematically, too): throw in a "summation factor" to make the series converge when $k=2$: $$ E(z,s) = \sum''_{c=c_o,d=d_o(N)}\;\frac{1}{|cz+d|^{2s}\cdot (cz+d)^2}$$ with $\Re(s)>0$. The plan is to analytically continue in $s$ to a neighborhood of $s=0$, and then set $s=0$.

The latter plan does largely succeed, except that it does not always produce holomorphic modular forms at $s=0$. This "disappointment" already occurs for $\Gamma=SL(2,\mathbb Z)$, where there is an extra summand of $1/y$ as part of the $0$th Fourier coefficient of the Hecke-summed Eisenstein series.

For Hilbert modular forms, that is, the analogue of elliptic modular forms but over larger (totally real) number fields in place of $\mathbb Q$, the "disappointment" never occurs. Although the computation to see the outcome is essentially elementary, the reason for it is not so elementary.

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.