Example of an incompressible vector field on $\mathbb{R}^3 \setminus 0$ without a vector potential As is well-known, the vector field $F = (-y,x)/(x^2+y^2)$ on the punctured plane is irrotational yet not conservative, i.e., there is no scalar potential $f$ such that $F = \nabla f$, because for instance $\oint_{S^1} F = 2\pi$. From a more abstract point of view, vector fields on the punctured plane correspond to 1-forms, irrotational fields correspond to closed forms, conservative fields correspond to exact forms, and $H^1_{dR}(\mathbb R^2 \setminus 0) \cong H^1_{dR}(S^1)$ is 1-dimensional and spanned by $d\theta$, so that any irrotational and non-conservative field on the punctured plane is a scalar multiple of $F$ plus a conservative field.
I am looking for the 3D analogue of the above vector field. Since $H^2_{dR}(\mathbb R^3 \setminus 0) \cong H^2_{dR}(S^2)$ is 1-dimensional, there must exist an incompressible vector field $F$ (meaning that $\nabla \cdot F = 0$, or that the corresponding 2-form is closed) on $\mathbb R^3 \setminus 0$ that does not admit a vector potential (meaning that there is no field $G$ such that $F = \nabla \times G$, or that the corresponding 2-form is not exact). How could one come up with a formula for such a field? I tried to understand how one could get the formula for $F$ in the 2D case from the 1-form $d\theta$ on the circle by pulling it back through the deformation retract $\mathbb R^2 \setminus 0 \to S^1$ given by $(x,y) \mapsto (x,y)/\sqrt{x^2 + y^2}$, but I was not able to fill in the details.
 A: As suggested in the comments, the "point mass field" (better known to PDE people as the gradient of the fundamental solution of Laplace's equation) $$\def\x{\mathbf{x}}F(\x)=|\x|^{-3}\x$$ is such an example. A routine calculation shows that $\nabla \cdot F = 0.$ In order to show $F$ is not a curl, we can use the same idea as in your 2-dimensional example: test its integral over a closed chain. In this case the unit sphere $S^2 \subset \mathbb R^3 \setminus \{0\}$ is the obvious choice, since $F$ coincides with the outwards normal and thus $$\int_{S^2}F\cdot\nu\;dA=4 \pi.$$
Since $S^2$ is closed (i.e. has no boundary), Stokes' theorem tells us that $$\int_{S^2} (\nabla \times G) \cdot \nu \, dA = 0$$ for any $G;$ so we conclude that $F$ is not a curl.
As you seem to have guessed, this is really a very direct generalization of the two-dimensional version: in terms of differential forms, they're both constructed from the volume form of the sphere, which is a natural choice because we can view $\mathbb R^{n+1}\setminus \{0\}$ as the cylinder $S^n \times \mathbb R.$
In both cases (and in higher dimensions too), $F$ is a vector field representation of the differential $n$-form $\omega = \phi^* dA$ where $\phi : \mathbb R^{n+1} \setminus\{0\}\to S^n$ is the radial projection map and $dA$ is the $n$-volume form of $S^n$. By pullback invariance we can conclude $d\omega = \phi^*d(dA) = 0,$ but integration over the sphere gives $\int_{S^n} \omega = \int_{S^n} dA > 0.$
The correspondence between the vector field and form perspectives comes from the Hodge dual and the musical isomorphism. In spherical coordinates $(r,\theta,\varphi)$ on $\mathbb R^3\setminus \{0\}$ and $(\theta,\varphi)$ on $S^2,$ the projection map is simply $\phi(r,\theta,\varphi)=(\theta,\varphi).$  Thus $$\omega = \phi^*(\sin \theta\, d \varphi\wedge d \theta) = \sin \theta \, d \varphi \wedge d \theta.$$ If we let $\hat\cdot$ denote metric normalization, then since $|d\varphi|=1/(r \sin \theta), |d \theta| = 1/r$ we have $$\omega = \frac{1}{r^2} \widehat{d \varphi} \wedge \widehat{d \theta}.$$ Thus $$\star \omega = \frac{1}{r^2}\widehat{dr} = \frac1{r^2}dr,$$ which has corresponding vector field $$(\star \omega)^\sharp = \frac1{r^2}\partial_r = \frac\x{|\x|^3}.$$
