What is the Laplace operator's representation in 3-sphere-coordinates?

Solution 1:

Let's use constructive induction to show that the $d$-dimensional Laplacian $\Delta_d$ is given by

$$\Delta_d = \frac1{r_d^{d-1}}\partial_{r_d}(r_d^{d-1}\partial_{r_d}) + \frac1{r_d^2}L_d^2$$

where $r_d$ is the $d$-dimensional radius and $L_d$ the $d$-dimensional angular momentum operator, the description of which will be constructed.

• For $d=1$, there is no angle, i.e. the radius is the $x$-coordinate and $L_1=0$. Therefore $$\Delta_1=\partial_x^2=\frac{d^2}{dx^2}$$

• For $d=2$, we have $(x_1,x_2) = \rho(\cos\phi, \sin\phi)$ and from the well-known polar coordinates one obtains $L_2^2=\partial_\phi^2$. This case is treated separate due to the inconsistency with the anon angles

• For $d-1\to d\ge3$, let $r\equiv r_d$ and $z\equiv x_d \equiv r\cos\theta$ (where the additional angle is named $\theta$, while the other ones may remain anonymous) such that $\rho\equiv r_{d-1} = r\sin\theta$. Note by comparing a total differential that $$\text d\rho\cdot\partial_\rho + \text dz\cdot\partial_z = \text dr\cdot\partial_r + \text d\theta\cdot\partial_\theta$$ and together with $\text d\rho = \text dr\cdot\sin\theta + \text d\theta\cdot r\cos\theta$ and $\text dz=\text dr\cdot\cos\theta-\text d\theta\cdot r\sin\theta$ one obtains $$\begin{pmatrix}\partial_\rho\\\partial_z\end{pmatrix} = \begin{pmatrix}\sin\theta & \frac1r\cos\theta\\ \cos\theta&-\frac1r\sin\theta\end{pmatrix} \begin{pmatrix}\partial_r\\\partial_\theta\end{pmatrix}.$$ Since $\Delta_d = \Delta_{d-1}+\partial_z^2$ and $\frac1{\rho^{d-2}}\partial_\rho(\rho^{d-2}\partial_\rho) = \partial_\rho^2 + \frac{d-2}\rho\partial_\rho$, one obtains together with $$\partial_\rho^2 + \partial_z^2 = (\partial_\rho,\partial_z)\begin{pmatrix}\partial_\rho\\\partial_z\end{pmatrix} = \partial_r^2 + \frac1r\partial_r + \frac1{r^2}\partial_\theta^2 $$ and $$\frac{d-2}\rho\partial_\rho = \frac{d-2}{r\sin\theta}(\sin\theta\partial_r+\frac1r\cos\theta\partial_\theta)$$ $$\boxed{\Delta_d = \underbrace{\partial_r^2 + \frac{d-1}r\partial_r}_{\frac1{r^{d-1}}\partial_r(r^{d-1}\partial_r)} + \frac1{r^2}\Big(\underbrace{\partial_\theta^2+(d-2)\cot\theta\partial_\theta}_{=\frac1{\sin^{d-2}\theta}\partial_\theta(\sin^{d-2}\theta\partial_\theta)}+\frac1{\sin^2\theta}L_{d-1}^2\Big).}$$

So in a recursive form, when $\theta$ denotes the new angle added, one obtains

$$L_d^2 = \frac1{\sin^{d-2}\theta}\partial_\theta(\sin^{d-2}\theta\partial_\theta) + \frac1{\sin^2\theta} L_{d-1}^2.$$

This recursion can be solved directly to finally obtain

$$\boxed{L_d^2 = \sum_{i=2}^d \left(\prod_{j=i+1}^d\frac1{\sin^2\theta_j}\right)\frac1{\sin^{i-2}\theta_i}\partial_{\theta_i}(\sin^{i-2}\theta_i\partial_{\theta_i})}$$

where $\theta_j$ denotes the $j-1$-th angle (i.e. there is no $\theta_1$). Note that the $d=2$ inconsistency is compensated by the $i-2$ exponent.

So for the 3-sphere, let the angles be, $(\phi,\theta,\omega)$ then $$L_4^2 = \frac1{\sin^2\omega}\left(\partial_\omega(\sin^2\omega\partial_\omega) + \frac1{\sin^2\theta}\left((\sin\theta\partial_\theta)^2 + \partial_\phi^2\right)\right)$$