= Radial curl of angular gradient fields
{title2=$\mathbf r\cdot\nabla\times\mathbf f=r^{-1}\Delta_{S^2}F$}
Let $\nabla_{\!\tan}F=\nabla F-\mathbf e_r\partial_rF$ be the tangential <surface gradient> on a <sphere> of <radius> $r>0$. For smooth scalar fields, set $\mathbf f=R\mathbf e_r+\nabla_{\!\tan}G+\mathbf e_r\times\nabla_{\!\tan}F$. Its radial <curl> detects only $F$:
$$
(\nabla\times\mathbf f)\cdot\mathbf e_r=\frac1{r^2}\Delta_{S^2}F,\qquad\Delta_{S^2}=\frac1{\sin\theta}\partial_\theta(\sin\theta\,\partial_\theta)+\frac1{\sin^2\theta}\partial_\phi^2.
$$
To prove this, use the radial <curl> formula $(r\sin\theta)^{-1}[\partial_\theta(\sin\theta f_\phi)-\partial_\phi f_\theta]$. The radial component $R$ never enters it. The $G$ terms cancel as $G_{\phi\theta}-G_{\theta\phi}=0$ by <symmetry of second derivatives>, while the $F$ terms give the unit-sphere <Laplace-Beltrami operator>. Taking the <dot product> with the position <vector> $\mathbf r=r\mathbf e_r$ gives the displayed $r^{-1}$ version. This tangential gradient decomposition does not assert that the whole <vector field> is <solenoidal>.
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