TITLE: Can a vector field always be written as a cross-product of two other vector fields?
QUESTION [11 upvotes]: As the title suggests, can any arbitrary conservative vector field, $\bf{F}$ $= \langle P,Q,R \rangle$, where the component functions are functions of $(x,y,z)$, always be written as a cross-product of two other vector fields, namely $\bf{A} \times \bf{B}$? Going further, let $\bf{A}$ $= \nabla a$, where $a = a(x,y,z)$ and has continuous first partial derivatives. Also, let $\bf{B}$ $=\langle b_1,b_2,b_3\rangle$ where the component functions are functions of $(x,y,z)$.
What I have so far is the following:
$\cases{
P=\frac{\partial a}{\partial y}b_3-\frac{\partial a}{\partial z}b_2\\
\\
Q=\frac{\partial a}{\partial z}b_1-\frac{\partial a}{\partial x}b_3\\
\\
R=\frac{\partial a}{\partial x}b_2-\frac{\partial a}{\partial y}b_1
}$
The reason I am asking is that I recently solved the following question:
Let $f$ be a smooth function defined on $\mathbb{R}^3$. Let $S$ be the level surface, $\{(x,y,z):f(x,y,z)=c\}$ for some $c \in \mathbb{R}$. Assume $\nabla f$ is never the zero vector on $S$ and let $\bf{F}$ $=\nabla f$. Show that the surface integral over $S$ of $\bf{F} \times \bf{G}$ is $0$ for any $\bf{G}$. ($\star$)
And I thought I could use this property if I can take a complicated surface integral and attempt to break it down to this case to easily see answers that are zero. So I suppose I am asking for the conditions that arbitrary vector fields need to satisfy to use the result of the property in the above problem.
Thanks!
$\bf{EDIT}$
Summary of questions:
Q1. Can a vector field always be written as a cross-product of two other vector fields? $\bf{Answered.}$
Q2. Given that $\bf{F}$ $=\nabla f$ can be written as a cross-product of two vector fields, what other conditions are necessary to use the result of $(\star)$?    $\bf{Answered.}$

REPLY [6 votes]: Globally, the answer is in general no, as has been shown by Chris Culter in his answer and achille hui in his comment.  But locally, near any point $p \in R^3$ where $Y(p) \ne 0$, the answer is always yes.
To see this, suppose $Y(p) \ne 0$.  I assume that by vector field you are allowing a certain amount of smoothness, so let's say $Y$ is at least $C^1$.  Then we can easily pick, near $p$, a vector field $W$ such that $Y$ and $W$ are linearly independent (again, sufficiently near $p$).  Let $Z = W \times Y$.  Then $\langle Z, Y \rangle = \langle Z, W \rangle = 0$, and furthermore $Z \ne 0$.  Let $e_Y = \frac{Y}{\Vert Y \Vert}$; we have $\Vert e_Y \Vert = 1$.  Let $X = W - \langle W, e_Y \rangle e_Y$; then $X \ne 0$ since $W$ and $Y$, hence $W$ and $e_Y$, are linearly independent.  Furthermore,
$\langle X, Y \rangle = \langle W, Y \rangle - \langle W, e_Y \rangle \langle Y, e_Y \rangle = \langle W, Y \rangle - \langle W, e_Y \rangle \Vert Y \Vert = \langle W, Y \rangle - \langle W, \Vert Y \Vert e_Y \rangle   = 0$,
and $\langle X, Z \rangle = \langle  W - \langle W, e_Y \rangle e_Y, W \times Y \rangle = 0$.  These calculations show that the three vectors fields $X$, $Y$, $Z$ are orthogonal and non-vanishing at least on some small neighborhood of $p$.  Taking the unit vector fields $e_X = \frac{X}{\Vert X \Vert}$, $e_Y = \frac{Y}{\Vert Y \Vert}$,  $e_Z = \frac{Z}{\Vert Z \Vert}$, we must have $e_Y = \pm e_X \times e_Z$.  So $Y = Ye_Y = \pm \Vert Y \Vert e_X \times e_Z$.  Reversing the sign of $e_X$ or $e_X$ if necessary we obtain $Y = (\Vert Y \Vert e_X) \times e_Z$, showing that $Y$ is in fact expressible as the cross product of two vectors, at least locally, near $p \in R^3$.
Finally, not to put too fine a point on it, but I believe that Chris Culter was correct, in his comment, in observing that $\nabla$ is not a vector field.  The result given here, being basically algebraic, applies whether $\nabla \cdot Y = 0$ or not; the Helmholtz theorem does not, in my reckoning, apply.