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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.

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  • $\begingroup$ Do you know what forms are? $\endgroup$ – IAmNoOne Sep 11 '16 at 22:27
  • $\begingroup$ @Nameless Sadly, no.. $\endgroup$ – TheQuantumMan Sep 12 '16 at 0:20
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Remember that in the analogous case $\nabla \times \nabla f = 0$, some intuition for the result can be attained by integration: by Green's theorem this is equivalent to $\int \nabla f \cdot ds = 0$ around every closed loop, which is true because $\int_{\gamma} \nabla f \cdot ds = f(\gamma(1)) - f(\gamma(0)).$ Thus our intuition is that curl measures circulation, and $\nabla f$ cannot circulate because this would introduce a discontinuity in $f$ around a loop.

Let's try the same thing: by the divergence theorem, it suffices to show that $\int_\Sigma (\nabla \times V) \cdot \hat n\ dA = 0$ for every closed surface $\Sigma$. By Stokes' theorem we know $$\int_\Sigma (\nabla \times V) \cdot \hat n\ dA = \int_{\partial \Sigma}V\cdot ds,$$which vanishes because $\Sigma$ is closed (i.e. $\partial \Sigma = \emptyset$).

In more intuitive terms, the divergence measures flux through a small cube; but the flux of a curl through a closed surface must be zero because there is no boundary curve for the circulation to accumulate upon.

As Nameless alluded to in his comment, you can get a more unified understanding of what's going on here by studying differential forms. All these geometric differential operators $\nabla, \nabla \times, \nabla \cdot$ are exterior derivatives $d_0,d_1,d_2$, and the identity $d_{k+1} \circ d_k = 0$ can be seen either by expanding out the partial derivative expression and noting that everything cancels (what I assume you've done in your proof), but also by applying the general Stokes theorem twice and noting that the boundary of a boundary is always empty.

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  • $\begingroup$ +1 for a great answer! I was trying to think about it more locally (with the differentials) rather than collectively (through the integral theorems)..Cool! $\endgroup$ – TheQuantumMan Sep 12 '16 at 0:29
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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:

  1. 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.
  2. 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.
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    $\begingroup$ The fact the compositions are zero is completely independent of the deRham cohomology ring of 3-space, since that is true for any manifold whatsoever. You cannot even begin to compute the cohomology of anything if you don't have a complex to begin with! $\endgroup$ – Pedro Tamaroff Sep 12 '16 at 5:00
  • $\begingroup$ @PedroTamaroff You're right, of course. I modified my answer explaining what the cohomology ring of $\mathbb{R}^3$ being what it is entails in terms of gradients, curls and divergences. $\endgroup$ – Daniel Robert-Nicoud Sep 12 '16 at 10:24

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