Cartan homotopy formula and curl In Topological Methods in Hydrodynamics, V. I. Arnol'd writes that the following expression
$$curl(\mathbf a \times \mathbf b)=[\mathbf a, \mathbf b]+ \mathbf a \ div \ \mathbf b - \mathbf b \ div \ \mathbf a$$
could be obtained "repeatedly applying" the Cartan homotopy formula
$$L_v = i_vd+di_v$$
And (in another book) adds some hints:


*

*$i_{curl(\mathbf a \times \mathbf b)}\tau = di_{\mathbf a} i_{\mathbf
   b}\tau$

*$div \ \mathbf a = di_{\mathbf a} \tau$

*$[\mathbf a, \mathbf b] = L_{\mathbf a} \mathbf b$
(where $\mathbf a, \mathbf b$ are two vectors in $R^3$, $i$ is the interior product, and $\tau$ the volume element).
I really could not figure how to proceed. Any suggestions?

Apply (plug in by means of the operator $i$) both sides to a volume 3-form $\tau$.
The equality of vector fields on both sides is equivalent to the equality of the obtained 2-forms.
On the left we have 
$i_{curl(a\times b)}\tau= di_b i_a \tau$ (note the order: vector field $a$ is plugged in first).
On the right first use the Cartan calculus formula 
$$
i_{[a,b]}\tau=[i_b, L_a]\tau = i_b L_a \tau - L_a i_b\tau 
= i_b d i_a\tau - di_a i_b\tau -i_a d i_b\tau 
$$
(note that $d\tau = 0$).
Since $L_b\tau = ({div}\,b)\tau$, we have 
$i_{a\, {div}\, b}\tau = i_a(L_b\tau)= i_a d i_b\tau$
and similarly for $i_{b\, {div}\, a}\tau =  i_b d i_a\tau$.
Hence the inner multiplication of $[a,b]+ a\, {div}\, b - b\, {div}\, a$ to the 3-form $\tau$ is
$$ 
i_b d i_a\tau - di_a i_b\tau -i_a d i_b\tau +  i_a d i_b\tau -  i_b d i_a\tau = - di_a i_b\tau,
$$ 
which proves the formula.
