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Consider a signal $U$ defined on the 2-sphere that is expressed as the product of two functions $A,B$, or $$ \begin{aligned} U(\theta,\phi) &= \sum_{n=0}^{\infty}\sum_{m=-n}^{n}u_{nm}Y_n^m(\theta,\phi)\\ &= \left[\sum_{n'=0}^{\infty}\sum_{m'=-n'}^{n'}a_{n'm'}Y_{n'}^{m'}(\theta,\phi)\right]\cdot\left[\sum_{n''=0}^{\infty}\sum_{m''=-n''}^{n''}b_{n''m''}Y_{n''}^{m''}(\theta,\phi)\right], \end{aligned} $$ where the coefficients $a_{nm},b_{nm}$ are known. Now, imagine my signal $A$ is filtered, e.g. $a_{nm}\rightarrow a_{nm} h_{nm}$. Is there a modification of any kind that can be performed on the signal $B$, for instance some kind of compensating filter $b_{nm} \rightarrow b_{nm} k_{nm}$ to retain the same value of $U$? If so, can we derive an expression for the filter coefficients $k_{nm}$? In a strict Fourier basis, this is trivial, but the spherical harmonics add quite a bit of complexity to this problem unfortunately.

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The product of spherical functions can be expanded to series of spherical functions again. It is usually done with Wigner 3-j symbols or Clebsch-Gordan coefficients (formulas there are for complex spherical functions, but one can derive from them real-valued ones). This will give you an infinite linear system on coefficients $b_{mn}$. However, you can trim your expansion at some azimuthal number and solve a finite system

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