# Eisenstein series associated with the cusps of a congruence subgroup

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 $SL_{2}(\mathbb{Z})$?

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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

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.