TITLE: Curl of a vector field cross itself?
QUESTION [5 upvotes]: Is there a neat expression for $(\nabla \times f ) \times f$ for some vector field $f$? Here is my attempt at a solution:
$$((\nabla \times f ) \times f)_i = \epsilon_{ijk}(\nabla \times f )_jf_k$$
$$ = \epsilon_{ijk}\epsilon_{jlm} \frac{d}{dx_l}f_mf_k$$
$$ = (\delta_{im}\delta_{kl} - \delta_{il}\delta_{km})\frac{d}{dx_l}f_mf_k$$
$$ = \frac{d}{dx_k}f_if_k - \frac{d}{dx_i}f_k^2$$
I had interpreted this as being $f  (\nabla \cdot f) - \nabla (f \cdot f)$ but I don't believe this is correct. Can anyone tell me where i went wrong?

REPLY [2 votes]: $$\begin{align} (\delta_{im}\delta_{kl} - \delta_{il}\delta_{km})\frac{\partial f_m}{\partial x_l}f_k&=\frac{\partial f_i}{\partial x_k}f_k-\frac{\partial f_k}{\partial x_i}f_k \\
&\ne \frac{\partial}{\partial x_k}f_if_k - \frac{\partial}{\partial x_i}f_k^2
\end{align}$$
which was the incorrect result.
It is straightforward to see that 
$$\frac{\partial f_i}{\partial x_k}f_k=(\vec f \cdot \nabla) \vec f$$
while using the product rule reveals that
$$\frac{\partial f_k}{\partial x_i}f_k =\frac12 \frac{\partial f_k^2}{\partial x_i}=\frac12 \nabla(|\vec f|^2)$$