I'm not sure if this is research level, so if this result is known, please excuse the intrusion. I am trying to find a relation between solutions of the Laplacian equation in $4$ dimensions and those in $2$ dimensions, given the following:
Let $n,l$ be integers with $n-l\geq 0$. Set $\alpha =\frac{n-l}{2}$ and $\beta =\frac{n+l}{2}.\ $Then the radial part of the Zernike polynomials in dimension 4 is given by
$\tag 1R_{n}^{(l)}(\rho )=(-1)^{\alpha }\sqrt{2n+4}\binom{\beta +1}{\alpha }\rho ^l\ _{2}F_{1}(-\alpha ,\beta +2;l+2;\rho ^{2})$
whenever $n-l$ is even; and zero otherwise.
We start with the following coordinates:
$x_0=\rho _1\cos \theta _1$
$x_1=\rho _1\sin \theta _1$
$x_2=\rho _2\cos \theta _2$
$x_3=\rho _2\sin \theta _2$
and introduce $\rho \ $ and $\gamma \ $ such that
$\rho _1=\rho \cos \gamma $
$\rho _2=\rho \sin \gamma ,\ $
so that $\rho ^{2}=\rho_{1} ^{2}+\rho_{2} ^{2}$ and therefore $0\leq \gamma \leq \pi /2\ $and $0\leq \rho_1,\rho _2\leq 1\ $.
It turns out that $(1)$ are the radial part of the solutions to the Laplacian equation in the coordinates $(\rho ,\theta _1,\theta _2, \gamma )$.
Now I want to find a relation between $(1)$ and the radial solutions to the Laplacian equation in $2$ dimensions, the radial part of which is
$\tag 2R_{n}^{(l)}(\rho _1)=(-1)^{\alpha }\sqrt{2n+2}\binom{\beta }{\alpha }\rho _1^l\ _{2}F_{1}(-\alpha ,\beta +1;l+1;\rho _1^{2})$
and of course, for $\rho _2$:
$\tag 3R_{n}^{(l)}(\rho _2)=(-1)^{\alpha }\sqrt{2n+2}\binom{\beta }{\alpha }\rho _2^l\ _{2}F_{1}(-\alpha ,\beta +1;l+1;\rho _1^{2})$.
Of course, it is tempting to propose that $R_{n}^{(l)}(\rho )=R_{n}^{(l)}(\rho _1)R_{n}^{(l)}(\rho _2)$, but after slogging through the many recurrence relations for the hypergeometric series, and the gamma function formulation, which fails in this case, since $\Gamma $ has a pole at $-\alpha $, I am fairly convinced that this is not true. So my question is: is there a reasonably simple relation between $(1), (2)\ $ and $(3)$?
