Radial curl of angular gradient fields ($\mathbf{r}\cdot \nabla \times \mathbf{f}=r^{-1}{\Delta }_{S^2}F$)

Let ${\nabla }_{\tan }F=\nabla F-{\mathbf{e}}_r{\partial }_{r}F$ 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 $F$:To prove this, use the radial curl formula $(r\sin \theta {)}^{-1}[{\partial }_{\theta }(\sin \theta f_{\varphi })-{\partial }_{\varphi }{f}_{\theta }]$. The radial component $R$ never enters it. The $G$ terms cancel as $G_{\varphi \theta }-G_{\theta \varphi }=0$ 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.