Problem about Eisenstein series on $\Gamma_1(N)$ I'm learning about Eisenstein series on $\Gamma_1(N)$ and it seems to me that I have misunderstood something. I imagine the following situation :
Let $\nu$ be a function on $(\mathbb{Z}/ N \mathbb{Z})$ to $\mathbb{C}$. One can define an "Eisenstein form" for $k\geq 3$, $$G_k(\nu,\tau)=\frac{(k-1)!}{2(2i\pi)^k}\sum_{m,n}{\frac{\nu(\bar{m})}{(m\tau +n)^k}}$$
Since 
\begin{align}
G_k\left(\nu,\frac{a\tau+b}{c\tau +d}\right) &= (c\tau+d)^k\frac{(k-1)!}{2(2i\pi)^k}\sum_{m,n}{\frac{\nu(\bar{m})}{((am+cn)\tau +(bm+dn))^k}} \\ &=(c\tau+d)^k\frac{(k-1)!}{2(2i\pi)^k}\sum_{u,v}{\frac{\nu(\overline{du-cv})}{(u\tau +v)^k}} \\ &=(c\tau+d)^k G_k(\nu,\tau)
\end{align}
when $ad-bc=1$, $c\equiv 0 \pmod{N}$, $d\equiv 1 \pmod N$, $G_k(\nu,\cdot)$ is a modular form of weight $k$ on $\Gamma_1(N)$. But we also have the Fourier serie :
$$G_k(\nu,\tau)=-\frac{B_k}{2k}+\sum_{n=1}^{+\infty}{\sigma_{k-1}(\nu,n)q^n} \quad \quad q=e^{2i\pi \tau}$$
where $$\sigma_{k-1}(\nu,n)=\sum_{d|n}{\nu(\bar{d})\left(\frac{n}{d}\right)^{k-1}}$$
So $G_k \in E_k(\Gamma_1(N))$ - the space of Eisenstein series - when $k$ is even. Then for $k\geq 4$ even and $N=4$, since $\dim(E_k(\Gamma_1(4)))=3$, one can find $(a,b,c,d) \in \mathbb{C}^4$ such that
$$aG_k(\alpha,\tau)+bG_k(\beta,\tau)+cG_k(\gamma,\tau)+dG_k(\delta,\tau)=0$$
but it's false in general (for instance, try $(\alpha,\beta,\gamma,\delta)=(\nu_0,\nu_1,\nu_2,\nu_3)$ where $\nu_i(\bar{m})$ is $1$ if $m\equiv i \pmod 4$ and $0$ otherwise).
So where am I wrong ? Thanks a lot for your help !
 A: Let me know if this answer is complete rubbish, I don't think I've seen this construction of Eisenstein series before, but here goes.
To solve
$$aG_k(\alpha,\tau)+bG_k(\beta,\tau)+cG_k(\gamma,\tau)+dG_k(\delta,\tau)=0$$
we may clear coefficients to get
$$a\sum_{m,n}{\frac{\alpha(\bar{m})}{(m\tau +n)^k}} + b\sum_{m,n}{\frac{\beta(\bar{m})}{(m\tau +n)^k}} + c\sum_{m,n}{\frac{\gamma(\bar{m})}{(m\tau +n)^k}} + d\sum_{m,n}{\frac{\delta(\bar{m})}{(m\tau +n)^k}}$$
but this equals
$$a\sum_{m,n}{\frac{\alpha(\bar{m})}{(m\tau +n)^k}} + b\sum_{m,n}{\frac{\beta(\bar{m})}{(m\tau +n)^k}} + c\sum_{m,n}{\frac{\gamma(\overline{-m})}{(-m\tau -n)^k}} + d\sum_{m,n}{\frac{\delta(\overline{-m})}{(-m\tau -n)^k}}$$
$$=a\sum_{m,n}{\frac{\alpha(\bar{m})}{(m\tau +n)^k}} + b\sum_{m,n}{\frac{\beta(\bar{m})}{(m\tau +n)^k}} + c\sum_{m,n}{\frac{\gamma(\overline{-m})}{(m\tau +n)^k}} + d\sum_{m,n}{\frac{\delta(\overline{-m})}{(m\tau +n)^k}}$$
as $k$ is even.
But now as $\gamma(\overline{-m}) = \alpha(m)$ and $\delta(\overline{-m}) = \beta(m)$ we can take $a=b=1$ and $c=d=-1$ and the resulting sum will be $0$.
I'm inclined to believe something like this would work for other examples too, but really don't know much about this area as I said.
