Proof of $u\times\omega = \nabla\ (\frac{ u\cdot\ u}{2}) - u\cdot\nabla\ u$ I'm having difficulty proving the formula: $$u\times\omega = \nabla\ (\frac{ u\cdot\ u}{2}) - u\cdot\nabla\ u$$ I should be using tensor notation. Given is that: $$\omega\ = \nabla\times\ u$$ and $$\nabla\cdot\ u\ = 0$$
I've done this so far: $$ (u\times\omega)_i = (u \times\ (\nabla\times\ u))_i = \epsilon_{ijk} u_j(\epsilon_{klm}\frac{\partial}{\partial\ x_l}u_m)=\epsilon_{ijk}\epsilon_{klm}\ u_j\frac{\partial}{\partial\ x_l}u_m=\epsilon_{kij}\epsilon_{klm}\ u_j\frac{\partial}{\partial\ x_l}u_m=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})u_j\frac{\partial}{\partial\ x_l}u_m=u_j\frac{\partial}{\partial\ x_i}u_j-u_j\frac{\partial}{\partial\ x_j}u_i $$
But that is as far as I come. I could really need some help+input, thanks on beforehand.
 A: If $\vec w=\nabla \times \vec u$, then using implied summation notation reveals
$$\begin{align}
\vec u\times \vec w&=\vec u\times \nabla \times \vec u\\\\
&=u_i\hat x_i\times \partial_j(\hat x_j\times \hat x_ku_k)\\\\
&=(\delta_{ik}\hat x_j-\delta_{ij}\hat x_k)u_i\partial_j(u_k)\\\\
&=\hat x_ju_i\partial_j(u_i)-\hat x_ku_i\partial_i(u_k)\\\\
&=\frac12\nabla (|\vec u|^2)-(\vec u\cdot \nabla)\vec u
\end{align}$$
as was to be shown!


Alternativley, using the Levi-Civita notation, we can write
$$\begin{align}
(\vec u\times \vec w)_i&=(\vec u\times \nabla \times \vec u)_i\\\\
&=\epsilon_{ijk}u_j(\nabla \times \vec u)_k\\\\
&=\epsilon_{ijk}u_j\epsilon_{k\ell m}\partial_\ell (u_m)\\\\
&=(\delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell})u_j\partial_\ell (u_m)\\\\
&=u_j\partial_i(u_j)-u_j\partial_j(u_i)\\\\
&=\frac12\partial_i(u_j u_j)-(u_j\partial_j)(u_i)\\\\
&=\left(\frac12 \nabla(\vec u\cdot \vec u)-(\vec u\cdot \nabla)\vec u \right)_i
\end{align}$$
Hence, we conclude that
$$\vec u\times \vec w=\frac12 \nabla(\vec u\cdot \vec u)-(\vec u\cdot \nabla)\vec u$$
as expected!
