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:

enter image description here


2 Answers 2


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 $d\omega_A^1\wedge \omega_B^1=\omega_{\nabla\times A}^2\wedge \omega_B^1=\omega^3_{(\nabla \times A)\cdot B}$.

  • $\begingroup$ Why $\omega_{\nabla\times A}^2\wedge \omega_B^1=\omega^3_{(\nabla \times A)\cdot B}$ $\endgroup$
    – 0x90
    Dec 6, 2016 at 19:56
  • 1
    $\begingroup$ @0x90 Because $$(A_x dy\wedge dz-A_y dx\wedge dz+A_z dx\wedge dy)\wedge (B_x dx+B_y dy+B_z dz)=(A_x B_x+A_y B_y+A_z B_z)\,dx\wedge dy\wedge dz$$ $\endgroup$ Dec 6, 2016 at 19:58
  • $\begingroup$ Thanks, is there a fast way of translating $\wedge$ with $\cdot$ or $\times$ according the context? $\endgroup$
    – 0x90
    Dec 6, 2016 at 20:00
  • 1
    $\begingroup$ @0x90 Note that 1-forms and 2-forms are both dual to vectors, whereas 0-forms and 3-forms are dual to scalars. The wedge product takes two 1-forms to a 2-form; hence it should be dual to the cross product, which takes two vectors and spits out another vector. On the other hand, the wedge product takes a 1-form & a 2-form and spits out a 3-form; this should be dual to the dot product, which takes two vectors and spits out a scalar. $\endgroup$ Dec 6, 2016 at 20:06
  • 1
    $\begingroup$ There's only one 3-form in $\Bbb R^3$, and that's $dx\wedge dy\wedge dz$. So a 3-form can only be dual to a scalar. (Things would be different in other dimensions---but then, there's also no such thing as a cross product in a generic $\mathbb{R}^n$.) PS: The remaining parts of the problem amount to considering a 0-form paired with a 1-form or a 2-form. $\endgroup$ Dec 6, 2016 at 20:07

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.

  • $\begingroup$ You are right, it was a typo. But my main question still remains how $$d\omega_{A}^1 \wedge \omega_{B}^1 - \omega_{A}^1 \wedge d\omega_{B}^1$$ ends up to be $$curl(A) \cdot B - A \cdot curl(B)$$ $\endgroup$
    – 0x90
    Dec 6, 2016 at 7:41
  • $\begingroup$ For me it looks like $$curl(A) \times B - A \times curl(B)$$ and not $$curl(A) \cdot B - A \cdot curl(B)$$ $\endgroup$
    – 0x90
    Dec 6, 2016 at 7:48
  • $\begingroup$ @0x90: I don't know what the [] and () notations mean, but I'm pretty sure there should be a Hodge dual involved in the definition of the cross product. $\endgroup$
    – celtschk
    Dec 6, 2016 at 7:53
  • $\begingroup$ [] is a vector product $\times$ and () is an inner product $\cdot$ $\endgroup$
    – 0x90
    Dec 6, 2016 at 7:55

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