Starting from this question How to express multiplication of two spherical harmonics expansions in terms of their coefficients?, I wanted to find the formula for the point-wise multiplication of spherical harmonic coefficients back in real space.
Formally, using the usual spherical harmonic expansion: $$\begin{align} f(\theta,\phi) &= \sum_{l=0}^\infty\sum_{m=-l}^l Y_l^m(\theta,\phi) f_l^m \\ f_l^m &= \int_0^{2\pi}\int_0^\pi Y_l^m(\theta,\phi)^*f(\theta,\phi)\sin\theta d\theta d\phi \end{align}$$ I want to calculate the $h$ in real space in terms of the real space functions $f,g$ when they are related by: $$ h_l^m=f_l^mg_l^m $$
I arrived at the point where I need to calculate with $\vec n,\vec n_1,\vec n_2$ unit vectors: $$ K(n,n_1,n_2) = \sum_{l=0}^\infty\sum_{m=-l}^l Y_l^m(\vec n)Y_l^m(\vec n_1)^*Y_l^m(\vec n_2)^* $$ Is there any way to simplify this? By this, I mean is there a formula without the sum allowing to easily look at asymptotics ? Perhaps that there are not, but since there is for "dual" problem using Clebsch Gordon, I have high hopes that a standard formula exists. I'm sure that I am not the first to stumble on this problem, but I lack the keywords.
I naturally thought of the addition theorem: $$ \sum_{m=-l}^lY_l^m(\vec n_1)Y_l^m(\vec n_2) = \frac{2l+1}{4\pi}P_l(\vec n_1\cdot \vec n_2) $$ However, it is not directly applicable in this case due to the $m$ dependence in the third factor.
Furthermore, by rotational symmetry, by acting via a rotation $R$: $$ K(R\vec n,R\vec n_1,R\vec n_2)=K(\vec n,\vec n_1,\vec n_2) $$ so it is a function of $(\vec n\cdot \vec n_1,\vec n\cdot \vec n_2,\vec n\cdot \vec n_3)$, ie the relative angles, which is not obvious from the series expansion. There is a well defined function taking these 3 arguments rather than the 6 arguments when using $\vec n,\vec n_1,\vec n_2$.
Question
Is there a simplification for: $$ K(\vec n,\vec n_1,\vec n_2) = \sum_{l=0}^\infty\sum_{m=-l}^l Y_l^m(\vec n)Y_l^m(\vec n_1)^*Y_l^m(\vec n_2)^* $$