In some calculations, I saw the following formula


$$\int\_{\mathrm{SU}(2)}\,\mathrm{d}g\,D^{j\_{1}}\_{m\_{1}n\_{1}}(g)D^{j\_{2}}\_{m\_{2}n\_{2}}(g)D^{j\_{3}}\_{m\_{3}n\_{3}}(g)=(-1)^{j\_{1}+j\_{2}+j\_{3}}\begin{pmatrix}j\_{1} & j\_{2} & j\_{3}\\m\_{1} & m\_{2} & m\_{3}\end{pmatrix}\begin{pmatrix}j\_{1} & j\_{2} & j\_{3}\\n\_{1} & n\_{2} & n\_{3}\end{pmatrix}$$


where $\mathrm{d}g$ denotes the (normalized) Haar measure on $SU(2)$and where $D^{j}$ denote Wigner's D-matrices, i.e. the unitary irreducible representation of $SU(2)$ of dimension $2j+1$. The matrix-looking objects on the right hand side denote [Wigner's 3j symbols](https://en.wikipedia.org/wiki/3-j_symbol), which are related to the Clebsch-Gordon coefficients.


The formula appears in the mathematical derivation of the so-called "Ponzano-Regge model" of 3d Euclidean Quantum Gravity. I am wondering how this formula is derived. Furthermore, I have seen in different sources that sometimes the sign factor in front is missing. Another source is mentioning that this formula is only true if $\vert j\_{2}-j\_{3}\vert\leq j\_{1}\leq j\_{2}+j\_{3}$. So all in all I become very confused and would like to know whether this formula is correct and which assumptions are needed. Does maybe someone also have some reference on it?



Formula involving Wigner's 3j symbols and integration over irreducible representations of SU(2)
You can find a fully worked-out derivation in these [lecture notes.](http://www.hep.caltech.edu/~fcp/physics/quantumMechanics/angularMomentum/angularMomentum.pdf) The formula you are looking for is equation (404), written in terms of the Wigner (small)-$d$ matrix. The relationship to the (large)-$D$ matrix goes via the Euler angle parameterization,
$$D^{j}\_{mm'}(\psi,\theta,\phi)=e^{-im\psi-im'\phi}d^{j}\_{mm'}(\theta).$$
The integration over SU(2) with the Haar measure is given in terms of Euler angles by
$$U(\psi,\theta,\phi)=\exp(-i(\psi/2)\sigma\_1)\exp(-i(\theta/2)\sigma\_2)\exp(-i(\phi/2)\sigma\_3,$$
$$\int\_{\rm{SU}(2)}f(U)\,dU=\frac{1}{16\pi^2}\int\_0^{2\pi}d\psi\int\_0^\pi\sin\theta d\theta\int\_0^{4\pi}d\phi\, f[U(\psi,\theta,\phi)].$$


So the desired integral over a product of three $D$-matrices vanishes unless $m\_1+m\_2+m\_3=0=n\_1+n\_2+n\_3$. In that case the integrations over $\psi$ and $\phi$ give a factor $8\pi$, what remains is the integration over $\theta$. Eq. (404) in the lecture notes shows how that is related to the product of $3j$-symbols,
$$\frac{1}{2}\int\_0^\pi d\theta\,d^{j\_1}\_{m\_1n\_1}d^{j\_2}\_{m\_2n\_2}d^{j\_3}\_{m\_3n\_3}=\begin{pmatrix}j\_{1} & j\_{2} & j\_{3}\\m\_{1} & m\_{2} & m\_{3}\end{pmatrix}\begin{pmatrix}j\_{1} & j\_{2} & j\_{3}\\n\_{1} & n\_{2} & n\_{3}\end{pmatrix}.$$
Both sides of the equation are equal to zero unless the $j$'s satisfy the triangle inequality, $\vert j\_{2}-j\_{3}\vert\leq j\_{1}\leq j\_{2}+j\_{3}$.


This formula differs from the one in the OP by a factor $(-1)^{j\_1+j\_2+j\_3}$. As a test, I evaluated both sides of the equation with Mathematica, for $j\_1=2$, $j\_3=3$, $j\_4=4$, $m\_1=n\_1=1$, $m\_2=n\_2=-1$, $m\_3=n\_3=0$, and find $5/126$, without a minus sign.


(1/2)\*Integrate[
 WignerD[{2, 1, 1}, theta]\*WignerD[{3, -1, -1}, theta]\*
 WignerD[{4, 0, 0}, theta]\*Sin[theta], {theta, 0, Pi}] 
ThreeJSymbol[{2, 1}, {3, -1}, {4, 0}]\*
ThreeJSymbol[{2, 1}, {3, -1}, {4, 0}]
