Finding a basis for the Eisenstein space $\mathcal{E}_1(12,\chi)$ I am trying to find a basis of the Eisenstein space $\mathcal{E}_1(12,\chi)$ of modular forms of weight 1, level 12 and Dirichlet character
\begin{equation*}
    \chi(m)=\genfrac{(}{)}{}{}{-12}{m}=\begin{cases}
    1 ~~&\text{  if  } m\equiv 1,\,7\pmod{12}\\
    -1 ~~&\text{  if  } m\equiv 5,\,11\pmod{12}\\
    0 ~~&\text{  if  } m\equiv 0,\,2,\,3,\,4,\,6,\,8,\,9,\,10 \pmod{12}.
    \end{cases}
\end{equation*}
For this I am following the approach described in Section 4.8 of "A first course in modular forms" by Diamond and Shurman. 
Since $\chi$ is induced by the primitive character modulo 3, it has conductor 3. Therefore I am led to believe that a basis of $\mathcal{E}_1(12,\chi)$ is given by $E_1^{\mathbf{1}_1,\chi,1}(z)=E_1^{\mathbf{1}_1,\chi}(z)$, $E_1^{\mathbf{1}_1,\chi,2}(z)=E_1^{\mathbf{1}_1,\chi}(2z)$ and $E_1^{\mathbf{1}_1,\chi,4}(z)=E_1^{\mathbf{1}_1,\chi}(4z)$. Here $E_1^{\mathbf{1}_1,\chi}(z)$ has $q$-expansion
$$ E_1^{\mathbf{1}_1,\chi}(z)=L(0,\chi) + 2 \sum_{m=1}^{\infty}\sigma_0^{\mathbf{1}_1,\chi}(m)q^m, $$
where 
$$\sigma_0^{\mathbf{1}_1,\chi}(m) = \sum_{\substack{d|m\\d>0}}\chi(d).$$
Writing out the first few coefficients in the $q$-expansions I obtain
\begin{align*}
    E_1^{\mathbf{1}_1,\chi,1}(z) &= \frac{2}{3} + 2q + 2q^2 + 2q^3 + 2q^4 + 2q^6 + 4q^7 + 2q^8 + \dots \\
    E_1^{\mathbf{1}_1,\chi,2}(z) &= \frac{2}{3} + 2q^2 + 2q^4 + 2q^6 + 2q^8 + \dots \\
    E_1^{\mathbf{1}_1,\chi,4}(z) &= \frac{2}{3} + 2q^4 + 2q^8 + \dots .
\end{align*}
However, the span of these forms does not agree with the span of the forms that Sage outputs as a basis for $\mathcal{E}_1(12,\chi)$ and I can not manage to express the theta function of the quadratic form $Q(X)=X_1^2 + 3X_2^2$ in terms of the basis I obtained. 
Something is going wrong here, but I can not seem to put my finger on where I am making a mistake. Any help will be greatly appreciated! 

I don't know about how to generate that Eisenstein space, but from quadratic reciprocity for $O_K = Z[\frac{\sqrt{-3}+1}{2}]$ which  is a PID we have 
$$\sum_{a,b \ne (0,0)} |a+b \frac{\sqrt{-3}+1}{2}|^{-2s} = |O_K^\times| \zeta_K(s) =6 \zeta(s) L(s,(\frac{-3}{.}))=6 \zeta(s) L(s,(\frac{.}{12}))\\ =6 \sum_{n=1}^\infty n^{-s} \sum_{d | n, d \ odd} (\frac{d}{3})(-1)^{(d-1)/2}$$
So that with the quadratic form $f(a,b) = |a+b \frac{\sqrt{-3}+1}{2}|^2$ we have 
$$\sum_{a,b} q^{f(a,b)} = 1 +6 \sum_{n=1}^\infty q^n \sum_{d | n, d \ odd} (\frac{d}{3})(-1)^{(d-1)/2} \in M_1(\Gamma_1(12))$$
Then with the quadratic form $Q(a,b) = |a+b\sqrt{-3}|^2$ I think it is not of level $12$ but of level $12 [O_K : Z[\sqrt{-3}]] = 24$, that's why you won't find it in your Eisenstein space.
