Curl of a vector field cross itself? 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?
 A: $$\begin{align} (\delta_{im}\delta_{kl} - \delta_{il}\delta_{km})\frac{\partial f_m}{\partial x_l}f_k&=\frac{\partial f_i}{\partial x_k}f_k-\frac{\partial f_k}{\partial x_i}f_k \\
&\ne \frac{\partial}{\partial x_k}f_if_k - \frac{\partial}{\partial x_i}f_k^2
\end{align}$$
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
$$\frac{\partial f_k}{\partial x_i}f_k =\frac12 \frac{\partial f_k^2}{\partial x_i}=\frac12 \nabla(|\vec f|^2)$$
A: It's a bit confusing keeping the two $f$'s together. But your work looks correct up till the end. I think the correct expression should be
$$ (\nabla \times f)\times f=(f \cdot\nabla)f-\frac12 \nabla(f\cdot f) $$
A: For some reason you made the derivative operator act on both $f$'s. The calculation should go as follows,
$$
\left((\nabla \times \vec{f} ) \times \vec{f} \right)_k = \epsilon_{ijk} ( \nabla \times \vec{f} )_i f_j  
$$
$$
 = \epsilon_{ijk} \epsilon_{\alpha \beta i} \left( \partial_\alpha f_\beta \right) f_j  
$$
$$
 = \epsilon_{ijk} \epsilon_{ i \alpha \beta }  f_j \partial_\alpha f_\beta  
$$
$$
 = \left( \delta_{j \alpha} \delta_{ k \beta} - \delta_{ j \beta} \delta_{k \alpha} \right)  f_j \partial_\alpha f_\beta  
$$
$$
 =  f_\alpha \partial_\alpha f_k -  f_\beta \partial_k f_\beta    
$$
$$
 = \left( (\vec{f} \cdot \nabla  ) \vec{f} \right)_k - \frac12 \partial_k f_\beta^2    
$$
$$
 = \left( (\vec{f} \cdot \nabla  ) \vec{f} \right)_k - \left( \frac12 \nabla (\vec{f}^2) \right)_k    
$$
So we would conclude that,
$$ (\nabla \times \vec{f} ) \times \vec{f} = ( \vec{f} \cdot \nabla ) \vec{f} - \frac12 \nabla (\vec{f}^2) $$
