If $F:S^3 \rightarrow \mathbb{R}^3$ is a nonvanishing smooth function then exists a "smooth basis"

Question

I think this is easy, but I'm stuck on this problem.

I am trying (and failing miserably) to demonstrate the following theorem

Theorem: Let $F: S^3 \rightarrow \mathbb{R}^3$ be a nonvanishing smooth function. Then exists smooth functions $G, H: S^3 \rightarrow \mathbb{R}^3$, such that $\{ F(p), G(p), H(p) \}$ is a basis of $\mathbb{R}^3$ for every $p$ $\in$ $S^3$.

If $F: S^3 \rightarrow \mathbb{R}^2$ $(F(p) = (F_1(p) , F_2(p) )$ the demonstration would be trivial, we would define $G(p) = (-F_2(p) , F_1(p))$ and and the proof would be ready. For any reason, I can't do this kind of trick in $\mathbb{R}^3$ (I'm just able to do this locally and not globally).

Does anyone know some magic trick that demonstrates the existence of the functions $H$ and $G$?

EDIT: Couple days ago I figured out the solution

---

Solution

Using that $S^3$ is a Lie group ($S^3$ is diffeomorphic to $SU (2) $), there is a trivialization of $TS^3$, i.e. there is a smooth function

$$H: S^3 \times \mathbb {R}^3\rightarrow TS^3$$ $$(p,v) \mapsto (p,A_p v),$$ where $A_p$ is a isomorphism of $\mathbb {R}^3$ to $T_p S^3$.

So we can define the following vector field in $S^3$, $\tilde{F}: S^3 \rightarrow TS^3$, $\tilde {F}(p) = (p,A_p F (p))$.

Let $F_t (p)$ be the flow of $\tilde {F} $, i. e;

1. $F_0 (p) = p$, for all $p$ $\in $ $S^3$.
2. $\frac {d}{dt} F_t (p) = A_{F_t (p)} F (F_t (p))$, for all $(t,p)$ $\in$ $\mathbb {R} \times S^3$. (Since $S^3$ is compact $\tilde {F} $ has complete flow).

Define $T (p,t) = \frac {F (F_t (p))}{\|F (F_t (p)) \|}$ and $G (p) = \left.\frac {d}{dt}T (p,t)\right|_{t=0}$

Finally, $G $ is the required function because

\begin{align*} \left\langle G (p), \frac {F (p)}{\|F (p)\|}\right\rangle &= \left\langle \left.\frac {d}{dt}T (p,t)\right|_{t=0} , T (p,0)\right\rangle\\ &=\frac {1}{2}\left.\frac {d}{dt} \left\langle T (p,t), T (p,t)\right\rangle\right|_{t=0}\\ &=\frac {1}{2}\frac {d}{dt} 1\\ &=0. \end{align*}

$H $ can be taken as the cross product between $F $ and $G $.

---

Bonus Information:

If $F$ is just a continuous function we can find a sequence of smooth functions $\{F_n\}_{n\in\mathbb{N}}$ ($F_n: S^3 \rightarrow \mathbb{R}^3$), such that $F_n \rightarrow \frac{F}{\|F\|}$ uniformly.

Note that if $F_n \rightarrow \frac{F}{\|F\|}$ uniformly, then $\frac{F_n}{\|F_n\|} \rightarrow \frac{F}{\|F\|}$ uniformly. For each $n$ $\in$ $\mathbb{N}$, there is (by the demonstration of the theorem above) a smooth function $G_n:S^3 \rightarrow \mathbb{R}^3$, satisfying $\left\langle G_n, F_n \right\rangle = 0 $.

Let $n_0$ be a natural number such that $\left\|\frac{F}{\|F\|} - \frac{F_{n_0}}{\|F_{n_0}\|} \right\|<\frac{1}{2}$. Realize that

\begin{align*} \left|\left\langle \frac{G_{n_0}(p)}{\|G_{n_0}(p)\|} , \frac{F(p)}{\|F(p)\|}\right\rangle\right|&= \left|\left\langle \frac{G_{n_0}(p)}{\|G_{n_0}(p)\|} , \frac{F(p)}{\|F(p)\|} - \frac{F_{n_0}(p)}{\|F_{n_0}(p)\|}\right\rangle \right|\\ &\leq \left\|\frac{G_{n_0}(p)}{\|G_{n_0}(p)\|}\right\|\cdot \left\| \frac{F(p)}{\|F(p)\|} - \frac{F_{n_0}(p)}{\|F_{n_0}(p)\|}\right\|\\ &\leq \frac{1}{2} \end{align*}

Then $G_{n_0}(p)$ is l.i with $F(p)$ for every $p$ $\in$ $S^{3}$ (otherwise we would have the existence of $p$ satisfying $\left|\left\langle \frac{G_{n_0}(p)}{\|G_{n_0}(p)\|} , \frac{F(p)}{\|F(p)\|}\right\rangle\right|=1$).

One more time we can take $H$ as the cross product between $G_{n_0}$ and $F$.

Answer 1

Given a point $p\in S^3$ the orthogonal complement of $F(p)$ gives you a plane in $R^3$. This patches together to form a two plane bundle over $S^3$. You are looking for two linearly independent sections of this bundle. But this bundle is trivial as all vector bundles over the three sphere are (this is a consequence of the fact that $\pi_2(G)=0$ for any Lie group $G$). It follows that we can find two linearly independent sections of this bundle. By the Gram-Schmidt process they can be made orthogonal.

This is probably a bit too advanced for such a problem, but I couldn't get an explicit solution via some trick. Maybe the fact that $S^3$ is the space of unit quaternions can help you.