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Let $d \ge 2$, and consider the sphere $S^{d-1}$ embedded in $\mathbb R^d$. Does there exist a family of rotations $\{\mathcal O_v\}_{v \in S^{d-1}}$ which satisfies:

  • $\mathcal O_v e_1 = v$, and

  • $\mathcal O_v w = \mathcal O_w v$

for all $v, w \in S^{d-1}$ ? That is, the map $\mathcal O_v$ rotates the vector $e_1$ to $v$, and the action of rotating $w$ by $v$ is the same as rotating $v$ by $w$. The second property is a weak form of commutativity, as it is equivalent to $\mathcal O_v \mathcal O_w e_1 = \mathcal O_w \mathcal O_v e_1$.

If not, can we do it if we exclude a set of measure zero (such as the point $-e_1$)? That is, there exists a set $X \subseteq S^{d-1}$ of full measure such that above hold for all $v, w \in X$.

The case of $d=2$ is yes, for a very simple reason: $S^1 = \operatorname{SO}(2)$ is an abelian group, so we can define $\mathcal O_v$ as multiplication by $\operatorname{arg}(v)$, and the second property follows immediately by commutativity.

Here's what I am thinking for $d > 2$. Consider $S^{d-1}$ embedded as a subspace of $\mathbb R^d$. Let $v \ne -e_1$, so that there is a unique minimizing geodesic $\gamma_v$ from $e_1$ to $v$. Define $\mathcal O_v e_1 = v$. The vectors $e_2, \dots, e_d$ span the tangent space $T_{e_1}S^{d-1}$ at $e_1$, so define $\mathcal O_v e_i$ to be the result of parallel translation of $e_i$ along $\gamma_v$, for $i \ne 1$. Does this seem reasonable, or is it obviously false and I'm missing something?

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Your first condition $\mathcal O_ve_1 = v$ is equivalent to saying the map $\mathcal O : S^{n-1} \to SO_n$ is a section of the fibre bundle $$SO_{n-1} \to SO_n \to S^{n-1}$$ If you had a section of that bundle and if it were continuous, then the total space would be a product $$SO_n \simeq SO_{n-1} \times S^{n-1}$$ (as a space). Generally this isn't true, in fact, it's true if and only if the tangent bundle to $S^{n-1}$ is trivial, which is when $n=2, 4$ or $8$. Of course, if you were to puncture the sphere $S^{n-1}$ a continuous section would exist, you can write down explicit sections using linear algebra constructions and a little spherical geometry.

The secondary condition amounts to saying that the commutator $[\mathcal O_v,\mathcal O_w]$ fixes $e_1$ for all $v$ and $w$. This is again too much to expect, even over the punctured sphere. Checking that you can't do this in general amounts to looking at the local properties of the conjugation map on the lie group $SO_n$. The map $A \longmapsto A^{-1}BA$ is an automorphism of the Lie group $SO_n$, so it preserves the identity element, to $SO_n$ acts on the tangent space to the Lie group (called the associated Lie algebra), taking the derivative you get the adjoint map, which is a Lie bracket. This Lie bracket is quite non-degenerate so you can't expect what you're asking for. Anyhow, that's just a sketch. Hope it's useful.

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I think you want to include $n=8$ also in your list, for $S^7$ is also parallelisable, though not a Lie group. – José Figueroa-O'Farrill Jun 16 '10 at 2:59
Ah, forgot about that one! – Ryan Budney Jun 16 '10 at 4:02

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