Brillouin zone in such a way that no two points in it have values of $\vec{K}$ differing by a reciprocal lattice vector, so that only the term with $\vec{G}=\vec{0}$ contributes. The matrix element of $V$ is then diagonal in $\vec{K}$. From now on we will assume that the Brillouin zone has been chosen in such a way, and drop the reference to $\vec{G}$. Since the total momentum $\vec{K}$ is conserved by ${\cal H}_0$ and $V$, ${\cal G}$ becomes diagonal in it. So, writing (\[Geq\]) in the $\vec{K},\vec{k}$ representation, we have $$\begin{aligned}
\langle\vec{K}\,\vec{k}| {\cal G}(z) |\vec{K}\,\vec{k}'\rangle & = &
\frac{1}{z-E_0(\vec{K},\vec{k})} \left\{ \vphantom{|\sum_{3}|}
\delta_{\vec{k},\vec{k}'}\right.
\nonumber \\
&&+\left.
\frac{2U}{N}\sum_{\vec{k}''} \langle\vec{K}\,\vec{k}''| {\cal G}(z)
|\vec{K}\,\vec{k}'\rangle \right\}.\nonumber\\
\label{Gdiag}\end{aligned}$$ The simplifying feature of (\[Gdiag\]) is that the general matrix element of ${\cal G}$ couples only to the total sum over the first entry of the matrix elements. For the latter we obtain an expression by summing (\[Gdiag\]) over $\vec{k}$ $$\begin{aligned}
\sum_{\vec{k}}\langle\vec{K}\,\vec{k}| {\cal G}(z&)& |
\vec{K}\,\vec{k}'\rangle =
\frac{1}{z-E_0(\vec{K},\vec{k}')}\nonumber \\
& & +2U{\cal R}(z,\vec{K})  \sum_{\vec{k}''}
\langle\vec{K}\,\vec{k}''| {\cal G}(z)
|\vec{K}\,\vec{k}'\rangle,\nonumber\\
\label{sumk}\end{aligned}$$ with ${\cal R}(z,\vec{K}) $ given by $${\cal R}(z,\vec{K}) =\frac{1}{N} \sum_{\vec{k}}
\frac{1}{z-E_0(\vec{K},
\vec{k})}.
\label{rdef}$$ Now (\[sumk\]) is an algebraic equation for the quantity in the left hand side. Using the 