Can a vector field always be written as a cross-product of two other vector fields?

Question

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 $(\star)$? $\bf{Answered.}$

Answer 1

Can a vector field always be written as a cross-product of two other vector fields?

No. But if we are considering some special vector space in some special space, for a vector field $\mathbf{F}\in C^{\infty}_c(\mathbb{R}^3)$, and divergence-free, then locally yes. There exists a vector field that is in the form of cross product not differing with the original one under $L^2$-norm.

The explicit construction is known as Poincare's lemma or Helmholtz decomposition. For reference you could refer to either Spivak, or Wade's Intro to analysis last chapter, last theorem (the presentation is given for manifold and $k$-form, and this question is relevant.

The proof is to construct an $\mathbf{F}'$ first, for $\mathbf{p} = (x,y,z)\in \mathbb{R}^3$, let $$ \mathbf{F}'(x,y,z) = -\mathbf{p}\times \int^1_0 t\nabla \times \mathbf{F}(t\mathbf{p})\,dt.\tag{$\star$} $$ Applying certain mollification technique to make $\mathbf{F}'$ have a compact support that is a subset of the support of $\mathbf{F}$ (with slightly abuse of notation just denote the compact approximation to identity of $\mathbf{F}'$ as $\mathbf{F}'$), and in this small neighborhood, we can verify that $$ \nabla \times \mathbf{F}' = \nabla \times \mathbf{F},\quad \text{and }\; \nabla \cdot \mathbf{F}'= \nabla \cdot \mathbf{F}=0. $$ Now use the Poincare inequality: $$ \|\mathbf{F} - \mathbf{F}'\|_{L^2(\mathbb{R}^3)} \leq C\|\nabla (\mathbf{F} - \mathbf{F}')\|_{L^2(\mathbb{R}^3)}.\tag{1} $$ Also for smooth vector field: $$ -\Delta = \nabla\times\nabla \times - \nabla \nabla \cdot, $$ integration by parts twice we have $$ \int_{\mathbb{R}^3} |\nabla (\mathbf{F} - \mathbf{F}')|^2 = \int_{\mathbb{R}^3} |\nabla\times (\mathbf{F} - \mathbf{F}')|^2 + \int_{\mathbb{R}^3} |\nabla\cdot (\mathbf{F} - \mathbf{F}')|^2 . $$ Back to (1) we have: $$\|\mathbf{F} - \mathbf{F}'\|_{L^2(\mathbb{R}^3)} =0.$$ In other words, $\mathbf{F}$ and $\mathbf{F}'$ only differ from the other on a measure zero set. Now check $\mathbf{F}'$'s form in $(\star)$.

Caveat: globally it is not possible, ($\star$)'s line integral forbids the direct usage of Poincare inequality globally.

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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 $\bf{G}$.

The surface integral is as follows: $$ \int_S \nabla f \times \mathbf{G} \cdot {\mathbf{n}} \,dS = \int_S {\mathbf{n}}\times \nabla f \cdot \mathbf{G} \,dS = 0.\tag{$\dagger$} $$ If everything is smooth, then this is true for $\nabla f$ is parallel to the unit vector normal to this level surface (actually the unit vector normal to $S$ is defined as $\pm \nabla f/|\nabla f|$, this is where you need $\nabla f \neq 0$).

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What conditions are necessary so that I can use the result of the second quoted problem?

A necessary condition is that $\mathbf{F}\times \mathbf{G} \cdot \mathbf{n}$ has zero average on this surface $S$.

• If $S$ is a closed surface, then $\nabla \cdot(\mathbf{F}\times \mathbf{G}) = 0$ within the domain this surface encloses suffices. This is by divergence theorem.

• If $S$ is not closed, just a level surface $\{f= c\}$ you gave. Then $\mathbf{F} = \nabla f$ suffices. For $\mathbf{F}$ is parallel to the surface normal $\mathbf{n}$, and hence $\mathbf{F}\times \mathbf{G}$ is perpendicular to $\mathbf{n}$ pointwisely, which makes the integral vanish. Because of $(\dagger)$, and arbitrariness of $\mathbf{G}$, a necessary condition, given $\mathbf{F}$ has the form $\nabla f$, is that $\nabla f\times \mathbf{n} = 0$ a.e.. What this condition implies is that: $$ \text{The potential } f \text{ doesn't vary tangentially on this surface.} $$ This is to say, $f$ does not have to be the equation of the surface, it can be any smooth function, as long it is a constant on this level surface $S$, $(\dagger)$ is 0. For example, the level surface is $x^2+y^2+z^2 = c$, and $\mathbf{F} = \nabla e^{-(x^2+y^2+z^2)}$.

Answer 2

Not continuously, no. If the vector field $\mathbf F(\mathbf x)=\mathbf x$ could be written as $\mathbf A\times\mathbf B$, then $\mathbf A$ restricted to the sphere $S^2$ would be a nonvanishing tangent vector field, contradicting the hairy ball theorem.

Answer 3

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.