On the unit 3-sphere in 4-dimensional Cartesian coordinate space, one can find a set of three orthonormal vector fields that parallelize the 3-sphere. (Utilize the correspondence between the unit 3-sphere and the coordinates of a geometric algebra rotor, and then infinitesimal displacements on the 3-sphere map to/from infinitesimal rotations in 3 orthogonal directions. See below for explicit expressions of the vector fields.)
It turns out it is easiest to find the vector fields in terms of the extrinsic 4d coordinates on the 3-sphere - they are just linear functions (and very simple ones at that) of the 4d coordinates. But there are some properties of these vector fields that I want to prove - namely, that they satisfy certain commutation relations (analogous to the commutation relations of the angular momentum operators in quantum mechanics) and that the vector fields are divergence-free - that I only know how to prove by introducing an intrinsic 3d coordinate system. But doing so is laborious.
My question is, is there a simple/elegant way to prove the commutation and divergence properties of these vector fields without introducing an intrinsic coordinate system, e.g. by using their symmetry properties or the fact that they generate infinitesimal rotations in orthogonal directions?
For further info: on the 3-sphere characterized by the set of points $(w,x,y,z)$ such that $w^2+x^2+y^2+z^2=1$, the three vector fields are: $$e_1=(-x,w,z,-y)$$ $$e_2=(-y,-z,w,x)$$ $$e_3=(-z,y,-x,w)$$
And the Lie brackets that I want to prove are: $$[e_i,e_j]=2\epsilon_{ijk}e_k$$
Clearly, $e_1$, $e_2$ and $e_3$ correspond to multiplications of unit quaternions by ${\rm i}$, ${\rm j}$ and ${\rm k}$. This means that one can compute their flows with exponentials. For instance:
$$\Phi_{e_1}(t, (w,x,y,z)) = {\rm e}^{{\rm i}t}(w+{\rm i}x+{\rm j}y+{\rm k}z)$$reads only $$\Phi_{e_1}(t,(w,x,y,z)) = (w\cos t-x\sin t, x\cos t+w\sin t, y\cos t-z\sin t, z\cos t+y\sin t).$$Clearly $\Phi_{e_1}(t,(w,x,y,z)) \in S^3$ whenever $(w,x,y,z) \in S^3$. In fact, $\Phi_{e_1}(t,\cdot)\colon S^3 \to S^3$ is an orientation-preserving isometry, which in particular preserves the volume form. This means that $\mathcal{L}_{e_1}{\rm vol} = 0$. But in general, $\mathcal{L}_X{\rm vol} = ({\rm div}(X)){\rm vol}$, so we conclude that ${\rm div}(e_1)= 0$. Now, the three quaternionic imaginary units are in equal standing, so we also conclude that ${\rm div}(e_2) = {\rm div}(e_3) = 0$.
As for the Lie bracket, let me elaborate on why there is no problem in regarding them as fields in $\Bbb R^4$. We have fields $E_1$, $E_2$ and $E_3$ in $\Bbb R^4$, given by the same formulas as $e_1$, $e_2$ and $e_3$, so that $E_1|_{S^3} = e_1$, $E_2|_{S^3} = e_2$ and $E_3|_{S^3} = e_3$. This implies that $[E_i,E_j]|_{S^3} = [e_i,e_j]$, where the former Lie bracket is computed in $\Bbb R^4$ and the latter in $S^3$. Now, $E_1$, $E_2$ and $E_3$ correspond to multiplications of arbitrary quaternions by ${\rm i}$, ${\rm j}$ and ${\rm k}$, and the Lie bracket in $\Bbb R^4$ corresponds to the commutator of such $\Bbb R$-linear maps. So, for example, we have that $$[E_1,E_2] \cong {\rm ij} - {\rm ji} = 2{\rm k} \cong 2E_3.$$Similarly for the other relations.