TITLE: Expression for derivative of cross product of two vectors in $S^2$
QUESTION [5 upvotes]: Let $a_1,a_2 \in S^2$ where $S^2= \{x\in \mathbb{R}^3 | x \cdot x = 1  \}$. 
From the tangent space structure we have,
$\frac{d}{dt} a_1 = a_1 \times w_1, \quad \frac{d}{dt} a_2 = a_2 \times w_2, \quad w_1,w_2\in \mathbb{R}^3$  
where $\times$ represents the standard vector product in $\mathbb{R}^3$.
I am trying to find the following - if $a_3 = a_1 \times a_2$, can I find $w_3\in \mathbb{R}^3$ such that $\frac{d}{dt} a_3 = a_3 \times w_3$ ?
Attempts so far :
$\frac{d}{dt}(a_1 \times a_2) = (a_1 \times (a_2 \times w_2)) + ((a_1 \times w_1)\times a_2) $
$ \quad = -(w_2 \times (a_1 \times a_2)) - (a_2 \times (w_2 \times a_1))- (a_2\times a_1) \times w_1 - (w_1\times a_2) \times a_1 $
applying the vector triple product property,
$ \quad = (a_1 \times a_2) \times w_2 - (a_2 \times (w_2 \times a_1))+ (a_1\times a_2) \times w_1 - (w_1\times a_2) \times a_1 $
stuck here.

REPLY [3 votes]: The complete answer:

If $a_1 \cdot a_2 \ne 0$, and $a_1$ and $a_2$ are linearly independent, then there is only a solution if $(a_1 \times a_2) \cdot (w_2 - w_1) = 0$. In that case the solution is given by
$$ w_3 = \frac{b_1 \cdot w_2}{1 - (a_1 \cdot a_2)^2}a_1 + \frac{b_2 \cdot w_1}{1 - (a_1 \cdot a_2)^2}a_2 + k (a_1 \times a_2), $$
where $b_1 = a_1 - (a_1 \cdot a_2) a_2$ and $b_2 = a_2 - (a_1 \cdot a_2) a_1$, and again $k$ is an arbitrary number.
If $a_1 \cdot a_2 = 0$, then there is always a solution; it is given by
$$ w_3 = (a_1 \cdot w_2)a_1 + (a_2 \cdot w_1)a_2 + k (a_1 \times a_2), $$
where $k$ is an arbitrary number.
If $a_1$ and $a_2$ are linearly dependent, i.e. $a_2$ is a multiple of $a_1$, then there is only a solution if $w_2 - w_1$ is also a multiple of $a_1$. In this case, $w_3$ is arbitrary, because
$\frac{d}{dt} (a_1 \times a_2) = 0$.

Note that a solution exists only if $a_3 \cdot \frac{d}{dt}a_3 = (a_1 \cdot a_2)(a_1 \times a_2) \cdot (w_2 - w_1) = 0$. Also, if $w_1 = w_2$, then $w_3 = w_1$ is always a solution.

Proof. Suppose $\frac{d}{dt}(a_1 \times a_2) = (a_1 \times a_2) \times w_3$. Using the vector identity $a \times (b \times c) = (a \cdot c)b - (a \cdot b) c$, we can find that
\begin{align}
0&=\frac{d}{dt}(a_1 \times a_2) - (a_1 \times a_2) \times w_3 \\
&= (a_1 \times w_1) \times a_2 + a_1 \times (a_2 \times w_2) - (a_1 \times a_2) \times w_3 \\
&= (a_1 \cdot a_2)(w_2 - w_1) + (a_1 \cdot (w_3 - w_2)) a_2 - (a_2 \cdot (w_3 - w_1)) a_1. \tag{1}
\end{align}
Case 1: $a_1$ and $a_2$ are linearly independent. Then $(a_1 \cdot a_2)^2 < 1$.
Since $\{a_1,a_2,a_1 \times a_2\}$ is a basis of $\mathbb{R}^3$, any solution takes the form
$$w_3 = k_1 a_1 + k_2 a_2 + k_3 (a_1 \times a_2). \tag{2}$$
By dotting $(1)$ with $a_1$, $a_2$, and $a_1 \times a_2$, respectively, we get three equations
\begin{align}
0 &= b_1 \cdot (w_3 - w_2), \tag{3} \\
0 &= b_2 \cdot (w_3 - w_1), \tag{$3'$} \\
0 &= (a_1 \cdot a_2)(a_1 \times a_2) \cdot (w_2 - w_1), \tag{$3''$} \\
\end{align}
where $b_1 = a_1 - (a_1 \cdot a_2) a_2$ and $b_2 = a_2 - (a_1 \cdot a_2) a_1$. Because $\{a_1,a_2,a_1\times a_2\}$ is a basis of $\mathbb{R}^3$, these equations together are equivalent to $(1)$.
Note that if $a_1 \cdot a_2 \ne 0$, then $(3'')$ implies that $(a_1 \times a_2) \cdot (w_2 - w_1) = 0$. That means that if $a_1 \cdot a_2 \ne 0$, then $a_1,a_2,w_2-w_1$ must lie in the same plane; otherwise there is no solution.
Observe that
\begin{align}
b_1 \cdot a_1 = b_2 \cdot a_2 &= 1 - (a_1 \cdot a_2)^2, \\
b_1 \cdot a_2 = b_2 \cdot a_1 &= 0.
\end{align}
Then plugging $(2)$ into $(3),(3')$ gives
\begin{align}
k_1 = \frac{b_1 \cdot w_2}{1 - (a_1 \cdot a_2)^2}, \\
k_2 = \frac{b_2 \cdot w_1}{1 - (a_1 \cdot a_2)^2},
\end{align}
so that the general solution is
$$ w_3 = \frac{b_1 \cdot w_2}{1 - (a_1 \cdot a_2)^2}a_1 + \frac{b_2 \cdot w_1}{1 - (a_1 \cdot a_2)^2}a_2 + k_3 (a_1 \times a_2), $$
where $k_3$ is arbitrary. If $a_1 \cdot a_2 = 0$, then this simplifies to
$$ w_3 = (a_1 \cdot w_2)a_1 + (a_2 \cdot w_1)a_2 + k_3 (a_1 \times a_2). $$
Case 2: $a_1$ and $a_2$ are linearly dependent. Thus $a_1 = \epsilon a_2$ with $\epsilon = \pm 1$. Then $(1)$ reduces to
$$0 = (w_2 - w_1) - (a_1 \cdot(w_2 - w_1))a_1,$$
which implies that $w_2 - w_1$ must also be a multiple of $a_1$.
If $w_2 - w_1$ is not a multiple of $a_1$, then there is no solution.
Note that $w_3$ no longer appears in this equation. That means any $w_3$ works. How can this be? Well, in this case we have $a_1 \times a_2 = 0$, so $\frac{d}{dt}(a_1 \times a_2) = (a_1 \times a_2) \times w_3 = 0$, so the particular value of $w_3$ is irrelevant.