Simplifying a sum of products of Clebsch-Gordan Coefficients I'm trying to simplify this sum involving the products of four Clebsch Gordan coefficients:
\begin{align}
&\sum_{m_1 m_2 m_3 K K' M} (-1)^{j_1+j_2+j_3+G+m_1+m_2+m_3-M}
\langle j_1\ m_1; j_2\ m_2|J\ K\rangle \\
&\times 
\langle j_2\ -m_2;j_3\ m_3|J'\ K'\rangle \times 
\langle j_3\ -m_3;j_1\ -m_1|G\ -M\rangle \times 
\langle J\ K;J'\ K'|G\ M\rangle  
\end{align}
where each of $j_1$, $j_2$, and $j_3$ are half-integers, and the sums are taken over all allowed values for each parameter ($-j_1$ to $j_1$, $-j_2$ to $j_2$, etc.). I looked through several identities in Wolfram's list here, but I couldn't find any among the two-term identities that I could apply, and I'm not familiar (and unfortunately don't have time to learn at the moment) with the 6-j and 9-j symbols used in the more complicated ones. I expect the sum to simplify nicely (as all of the others that I've encountered in this context so far have), but I can't see how.
 A: You should double-check but the summation appears to be given by
$$
(2G+1)\sqrt{(2J'+1)(2J+1)}
\left\{\begin{array}{ccc}
j_1&j_2&J\\
J'&G&j_3
\end{array}\right\}\tag{1}
$$
To get there the simplest way is to start from the definition of the $6j$ symbol:
\begin{align}
&\sum_{\bar{m}_1\bar{m}_2\bar{m}_3\bar{m}_{12}\bar{m}_{23}}
\langle \bar{j}_{12}\bar{m}_{12};\bar{j}_3\bar{m}_3\vert \bar{j}\bar{m}\rangle
\langle \bar{j}_{1}\bar{m}_{1};\bar{j}_2\bar{m}_2\vert \bar{j}_{12}\bar{m}_{12}\rangle\, ,\\
&\qquad\qquad \times \langle \bar{j}_{1}\bar{m}_{1};\bar{j}_{23}\bar{m}_{23}\vert \bar{j'}\bar{m'}\rangle
\langle \bar{j}_{2}\bar{m}_{2};\bar{j}_3\bar{m}_3\vert \bar{j}_{23}\bar{m}_{23}\rangle\\
&=\delta_{\bar{j}\bar{j'}} (-1)^{\bar{j}_1+\bar{j_2}+\bar{j}_3+\bar{j}}
\sqrt{(2\bar{j}_{12}+1)(2\bar{j}_{23}+1)}
\left\{\begin{array}{ccc}
\bar{j}_1&\bar{j}_2&\bar{j}_{12}\\
\bar{j}_3&\bar{j}&\bar{j}_{23}
\end{array}\right\}
\end{align}
This is equation (9.1.8) from D.A. Varshalovich et al, Quantum Theory of angular momentum (1988 English edition by WorldScientific; in the Russian edition some of the material is in different places).
There are some CG's to manipulate to the right form but basically the identification is
$$
\bar{j}_1\to j_1\, ,\quad \bar{j}_2\to j_2\, ,\quad \bar{j}_3\to J'
\, ,\quad \bar{j}_{12}\to J\, ,\quad \bar{j}_{23}\to j_3\, ,\quad \bar{j}=\bar{j'}\to G\, ,
$$
Your expression has a final sum on $M$ which provides an additional $(2G+1)$ factor, giving (1) as final expression.
I've checked it with about half-a-dozen values and it seems to work, but please double-check this as I could have made a typesetting error.

Edit: after comments I double checked and found that my original expression had the incorrect overall phase.  I believe the current Eq.(1) is correct, i.e. the overall phase is $+1$.  I've checked the result for various half-integer and integer values of $j_1,j_2$ and $j_3$ using the built-in ClebschGordan and SixJSymbol routines of Mathematica.
