In three dimensions, it seems pretty straightforward to prove the identity that for any vector field $\mathbf{A}$,
$$\nabla \cdot (\nabla \times \mathbf{A}) = 0$$
Does this identity still hold true when $\mathbf{A}$ is a 2-dimensional vector field? For instance, if
$$\mathbf{A} = \langle xy^2, xy^2 \rangle$$
Then the curl of that field would end up being a scalar field defined by
$$y^2 – x2y$$
Which, if reinterpreted as a vector field $\langle y^2-x2y, y^2-x2y \rangle$, does not have zero-divergence.
Does this interpretation make sense, and therefore the original identity doesn't hold in $2$-dimensions, or is taking the divergence of the curl of a vector field simply not applicable in two dimensions?
Best Answer
No, it doesn't make sense, it is fundamental that it is in three dimensions, normally you would write:
$$\mathbf{A}=(0,0,f(x,y))$$
Then the identity holds true, for your $\mathbf{A}$, you have:
$$\mathbf{A}=(f(a,y),g(x,y),0)$$ and then your curl will have components in the $z$ direction which you need to take into account.