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.