How to deduce $\nabla\cdot (A\times B) = (\nabla\times A)\cdot B - (\nabla\times B)\cdot A $ using differential forms?

Question

I am trying to prove the vector identity:

$$\nabla\cdot (A\times B) = (\nabla\times A)\cdot B - (\nabla\times B)\cdot A $$

Let $A = grad(\Phi)$ is a vector field on a given riemannian manifold, $M$ embedded in $\mathbf{R}^n$.

Using the following correspondence properties:

• 0-forms correspond scalar functions $\Phi$
• 1-forms correspond gradient operations $\omega^1_{grad(\Phi)}$
• 2-forms correspond curl operations $\omega^2_{curl(A)}$
• 3-forms correspond divergence operations $\omega^3_{div(A)}$

I am stuck because if the following is correct:

$$ \omega^3 _{div(A \times B)} = d(\omega_{A \times B}^2) = d(\omega_{A}^1 \wedge \omega_{B}^1)$$

The next step is $$d(\omega_{A}^1 \wedge \omega_{B}^1) = d\omega_{A}^1 \wedge \omega_{B}^1 - \omega_{A}^1 \wedge d\omega_{B}^1= curl(A) \times B - A \times curl(B)$$

But I would expect it to be $curl(A) \cdot B - A \cdot curl(B)$ and not $curl(A) \times B - A \times curl(B)$

The question was originally published at Arnold's book page 194:

Answer 1

Observe that $d\omega^1_A$ is a 2-form, not a 1-form. So the wedge product of this with the 1-form $\omega_B^1$ gives a 3-form, not a 2-form. More precisely, one has $d\omega_A^1\wedge \omega_B^1=\omega_{\nabla\times A}^2\wedge \omega_B^1=\omega^3_{(\nabla \times A)\cdot B}$.

Answer 2

The product rule you are using is wrong.

The correct rule is:

If $\alpha$ is a $p$-form, and $\beta$ is a $q$-form, then $$d(\alpha\wedge\beta) = d\alpha\wedge\beta + (-1)^p \alpha\wedge d\beta$$

In particular, if both $\alpha$ and $\beta$ are $1$-forms, then $$d(\alpha\wedge\beta) = d\alpha\wedge\beta + (-1)^1 \alpha\wedge d\beta = d\alpha\wedge\beta - \alpha\wedge d\beta$$

Note that, as seen in the image, this equation is also found in the hint to the problem.