TITLE: What is the geometric reason of why is the divergence of the curl of a vector field equal to zero?
QUESTION [7 upvotes]: What is the geometric reason of why is the divergence of the curl of a vector field equal to zero? I know how to prove it but I can't quite get some intuition behind it.    
I have seen some arguments that treat the del operator as a vector function, but I think this is not so correct as in some cases this analogy fails.
This is described in http://www.feynmanlectures.caltech.edu/II_02.html in sections 2-7 and 2-8 but gives poor explanations on why thinking about the del operator as a normal vector works in some cases while does not work in other cases.

REPLY [2 votes]: A very nice point of view on this is given by considering the deRham complex of $\mathbb{R}^3$ (as already mentioned in the comments and in Anthony's answer):
We have of course $\Omega^0(\mathbb{R}^3) = C^\infty(\mathbb{R}^3)$ by definition, and we can also identify $\Omega^3(\mathbb{R}^3)\cong C^\infty(\mathbb{R}^3)$ by sending $f\,dx\wedge dy\wedge dz$ to $f$. Further, we can identify both $\Omega^1(\mathbb{R}^3)$ and $\Omega^2(\mathbb{R}^3)$ with the space $\Gamma(\mathbb{R}^3)$ of vector fields (I will not write these identifications down, but you'll be able to recover them from what follows). Then, under these identifications, the differentials $d$ become
$$0\longrightarrow C^\infty(\mathbb{R}^3)\stackrel{\text{grad}}{\longrightarrow}\Gamma(\mathbb{R}^3)\stackrel{\text{curl}}{\longrightarrow}\Gamma(\mathbb{R}^3)\stackrel{\text{div}}{\longrightarrow}C^\infty(\mathbb{R}^3)\longrightarrow0.$$
As the cohomology ring of $\mathbb{R}^3$ is given by $H^\bullet_{dR}(\mathbb{R}^3)=\mathbb{R}$ (concentrated in degree $0$), we get the usual identities
\begin{align}
\text{curl}\circ\text{grad} = & 0,\\
\text{div}\circ\text{curl} = & 0,
\end{align}
plus the facts that:

If $X\in\Gamma(\mathbb{R}^3)$ is such that $\text{curl}(X) = 0$, then there is $f\in C^\infty(\mathbb{R}^3)$ such that $\text{grad}(f) = X$, and two such $f$ can only differ by a constant.
If $Y\in\Gamma(\mathbb{R}^3)$ is such that $\text{div}(Y) = 0$, then there is $X\in\Gamma(\mathbb{R}^3)$ such that $\text{curl}(X) = Y$, and two such $X$ can only differ by the gradient of a function.