Consider $\mathbb R^3$ and the following equation \begin{equation} \star d\omega=df \end{equation} where $f$ is a 0-form and $\omega$ is a 1-form. My question is whether this equation is satisfied by \begin{align} \omega&=\hat z \delta(x)\delta(y) \\ f&=\phi \end{align} where $(x,y,z)$ are the Cartesian coordinates, $\hat z$ is the unit vector in the $z$ direction, $(r,\theta,\phi)$ are the spherical polar coordinates and $\delta(\cdot)$ is the Dirac delta function. I find this difficult to compute because of the delta functions.
Edit: It turns out that this is false. The formula I was looking for is actually $\epsilon_{ijk}[\partial_{j},\partial_{k}]\phi=\hat z_i \delta(x)\delta(y)$.
Part I.
I am assuming that the polar coordinates are related to the Cartesian coordinates by $$ x=r\sin\theta\cos\phi\,,\quad y=r\sin\theta\sin\phi\,,\quad z=r\cos\theta $$ so that $$ f(r,\theta,\phi)=\phi=\arccos\Big(\frac{x}{\sqrt{x^2+y^2}}\Big)\,. $$ From this answer we know that $$ df=d\phi=\frac{x\,dy-y\,dx}{x^2+y^2}\,. $$ The Hodge dual of this is $$ \star df=\frac{x\,dz\wedge dx-y\,dy\wedge dz}{x^2+y^2}\,. $$ The right hand side of this is $d\omega$ which can be written as $$ d\omega=dz\wedge\frac{x\,dx+y\,dy}{x^2+y^2}\,. $$ This shows that $$ \omega=-\frac{\log(x^2+y^2)}{2}\,dz\,. $$ The function $-\frac{\log(x^2+y^2)}{2}$ is not the product of two delta functions $\delta(x)\delta(y)$ but it has a pole at the origin and its integral over the unit disc is $$ -\pi\int_0^1\log (r^2)\,r\,dr=-2\pi\int_0^1(\log r)\,r\,dr =2\pi\int_0^1 \frac{r^2}{2r}\,dr-2\pi(\log r)r^2\Big|_0^1=\frac{\pi}{2}\,. $$
Part II.
You want to know if $$\tag{1} \epsilon_{ijk}\,[\partial_{j},\partial_{k}]\,\phi=\hat z_i\,\delta(x)\delta(y) $$ is true. The left hand side of this contains the gradient of $\phi$ which we know from above is $$ \nabla\phi=\partial_k\,\phi=\frac{1}{x^2+y^2}\begin{pmatrix}-y\\x\\0 \end{pmatrix}\,. $$ Since $\epsilon_{ijk}\,a_j\,b_k=(\mathbf{a}\times\mathbf{b})_i$ we have $$ \epsilon_{ijk}\,\partial_j\,\partial_k\,\phi=(\nabla\times(\nabla\phi))_i $$ which is zero for all $x,y\not=0\,.$ Therefore, $$ \epsilon_{ijk}\,[\partial_j,\partial_k]\,\phi=0\,\quad\text{ for all }x,y\not=0\,. $$ Let $S^1$ be the unit circle around the origin in the $xy$-plane. Since $$ \int_{S^1}d\phi=\int_0^{2\pi}\,d\phi=2\pi $$ there seems a contradiction to the Stokes theorem because from $dd\phi=0$ it should follow that this integral must be zero. The problem is however that $\phi$ is undefined in the origin and the Stokes theorem can be "preserved" when in the distributional sense the rotation of $d\phi$ is given by those delta functions in (1).