Characterization of spherically symmetric functions A function $f:\mathbb{R}^3 \times \mathbb{R}^3 \rightarrow \mathbb{R}$ is said to be spherically symmetric if:
$$f(x,y) = f(Ax,Ay) \quad \forall (x,y) \in \mathbb{R}^3\times \mathbb{R}^3 \quad \forall A\in O(3)$$
(where $O(3)$ are isommetries in $\mathbb{R}^3$)
Does it hold that if $f$ is spherically symmetric then there exists $\phi: [0,+\infty[ \times [0,+\infty[ \times [0,\pi[ \rightarrow \mathbb{R}$ verifying $f(x,y) = \phi(|x|, |y|, \sphericalangle(x,y))$ for every $x,y\in \mathbb{R}^3$?
 A: If you replace $[0,\pi[$ with $[0,\pi]$ (to cover antipodal pairs) then yes. 
If $(x_1,y_1)$ and $(x_2,y_2)$ are two pairs of points such that $|x_1|=|x_2|,$ $|y_1|=|y_2|$ and $\angle(x_1,y_1) = \angle(x_2,y_2),$ then we can construct an isometry $A$ such that $Ax_1 = x_2$ and $Ay_1 = y_2:$ first rotate $x_1$ to $x_2$ however you like (possible thanks to their matching lengths), then rotate about the axis determined by $x_2$ to send $y_1$ to $y_2.$ This second step works because the locus of points with the desired length and angle to $x_2$ is a circle about this axis.
Thus in this situation we have $f(x_1,y_1) = f(x_2,y_2),$ so $f$ factors via such a $\phi$. 
To construct $\phi$ explicitly: for each element $(a,b,\theta) \in [0,+\infty[ \times [0,+\infty[ \times [0,\pi],$ pick a pair $(x,y)$ such that $|x|=a,|y|=b,\angle(x,y)=\theta$ and define $\phi(a,b,\theta)=f(x,y).$ The claim I proved above shows that this is independent of the choice of $x,y$ and thus gives a well-defined $\phi.$
