Expression for derivative of cross product of two vectors in $S^2$

Question

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.

Answer 1

No, in general such a $w_3$ does not exist, because $a_3$ and $\frac{d}{dt}a_3$ are not orthogonal.

Here's an example. Define the shorthand $s=\frac{1}{\sqrt2}$. Let $$a_1=(1,0,0)\quad\frac{d}{dt}a_1=(0,0,0)$$ $$a_2=(s,s,0)\quad\frac{d}{dt}a_2=(-1,1,0)$$ Then $$a_1\times a_2=(1,0,0)\times(s,s,0) = (0,0,s)$$ and $$\frac{d}{dt}(a_1\times a_2)=(1,0,0)\times(-1,1,0)+(0,0,0)\times(s,s,0) = (0,0,1).$$

Answer 2

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.

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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.