# Vector Calculus Theorems - Duality Between Differentiation & Integration?

I'm trying to go over some vector analysis using forms & kind of noticed what look like random vector identities are more appropriately thought of, to me at least, as differential analogues of the classical integral theorems in the way Maxwell's equations can be cast in differential & integral form. However I'm missing a theorem:

Integral Gradient Theorem: $\smallint_{\vec{a}}^{\vec{b}} \nabla f \cdot d \vec{r} = f(\vec{b}) - f(\vec{a})$

Integral Curl Theorem: $\smallint_S (\nabla \times \vec{F}) \cdot \hat{n} dS = \smallint_{\partial S} \vec{F} \cdot d \vec{r}$

Integral Divergence Theorem: $\smallint_V \nabla \cdot \vec{F} dV = \smallint_{\partial V} \vec{F} \cdot \hat{n} dS$

Derivative Gradient Theorem:

Derivative Curl Theorem: $\nabla \times (\nabla \phi) = 0$

Derivative Divergence Theorem: $\nabla \cdot (\nabla \times \vec{F}) = 0$

Any ideas as to what I should put in there?

Also, could one bluff an appropriate differential & integral theorem for the scalar or vector laplacians?

## 1 Answer

The derivative gradient theorem is extremely trivial: it says that the gradient of the zero function is zero.

To justify this, note that the integral theorems you write down are all specializations of Stokes' theorem, of which there should be $n$ in $n$ dimensions. The derivative theorems you write down are all specializations of the fact that the de Rham differential squares to zero, of which there should be $n$ in $n$ dimensions, except one of them just says that the composition $0 \to \Omega^0(M) \to \Omega^1(M)$ is zero and this is the least interesting one.

It's not clear to me that it's meaningful to consider these theorems dual to each other, though. Note that the integral theorems all involve one specialization of the de Rham differential but the derivative theorems all involve two.

• If you really want, I suppose the derivative gradient theorem could also say that the gradient of a constant function is zero. This corresponds to modifying the de Rham complex slightly so that it computes reduced cohomology rather than cohomology. – Qiaochu Yuan Aug 15 '13 at 8:54
• As regards the double specialization of the de Rham differential, one could say the same thing about Maxwell's equations no? One moves between the differential & integral form of Maxwell's equations via Stokes & Gauss, so on face value if you can't think of the theorems in my post as dual to each other you can't think of the differential & integral form of Maxwell's equations as dual either? Thanks for linking this to de Rham cohomology, it definitely links in with the rest of what I've done. – bolbteppa Aug 15 '13 at 14:36