Some constructions related to Helmholtz decomposition
A general question is as follows:
Claim 1 Can a vector field always be written as a crossproduct of two other vector fields?
The answer is apparently “No”. Since if the vector field $\mathbf F(\mathbf x)=\mathbf x$ could be written as $\mathbf G\times\mathbf H$, then $\mathbf G$ restricted to the sphere $S^2$ would be a nonvanishing tangent vector field, contradicting the hairy ball theorem.
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 divergencefree, then the answer can possible be “yes”.
My claim:
Claim 2 there exists a vector field that is in the form of cross product not differing with the original one under $L^2$norm in a compact set that is a subset of $\mathrm{supp}(\mathbf{F})$.
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).
The proof is to construct an $\mathbf{F}’$ first, for $\mathbf{p} = (x,y,z)\in \mathbb{R}^3$, let
\[\displaystyle\mathbf{F}'(x,y,z) = \mathbf{p}\times \int^1_0 t\nabla \times \mathbf{F}(t\mathbf{p})\,dt.\tag{$\star$}\]Applying certain mollification technique (this needs to be written out in details) 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
\[\displaystyle \int_{\mathbb{R}^3} \nabla (\mathbf{F}  \mathbf{F}')^2 = \int_{\mathbb{R}^3} \nabla\times (\mathbf{F}  \mathbf{F}')^2 + \int_{\mathbb{R}^3} \nabla\cdot (\mathbf{F}  \mathbf{F}')^2 .\]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.
Another interest claim is that:
Claim 3: if $f$ is 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 $\bf{F}$ $=\nabla f$. Then surface integral over $S$ of $\bf{F} \times \bf{G}$ is $0$ for any $\bf{G}$.
The surface integral can rewritten as follows:
\[\displaystyle\int_S \nabla f \times \mathbf{G} \cdot {\mathbf{n}} \,dS = \int_S {\mathbf{n}}\times \nabla f \cdot \mathbf{G} \,dS = 0. \tag{$\dagger$}\]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 $\pm \nabla f/\vert\nabla f\vert$, this is where $\nabla f \neq 0$ comes into play).
A necessary condition to for this claim to hold 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 for a level surface ${f= c}$, 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:
The potential $f$ 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 $\mathbf{F} = \nabla e^{(x^2+y^2+z^2)}$.
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