# Is there a vector identity for $\left( \nabla \times \mathbf{\vec F}\right) \times \mathbf{\vec F}$?

I have looked through several lists of curl identities but cannot find anything for this.

$$\nabla \times$$ denotes the curl, and $$\mathbf{\vec F}$$ is just some arbitrary vector field in $$\Bbb R^3$$

We can use Einstein summation and the Levi-Civita tensor $$\varepsilon_{ijk}$$ to write the $$i$$-th component of a more general expression as $$[(\vec{\nabla}\times\vec{a})\times\vec{b}]_i=\varepsilon_{ijk}(\varepsilon_{j\ell m}\partial_\ell a_m)b_k,$$ now use the relation $$\varepsilon_{ijk}\varepsilon_{j\ell m}=-\varepsilon_{jik}\varepsilon_{j\ell m}=\delta_{im}\delta_{k\ell}-\delta_{i\ell}\delta_{km}$$ and write (note that $$\partial_i$$ only acts on $$a_k$$ and not $$b_k$$) $$\varepsilon_{ijk}\varepsilon_{j\ell m}(\partial_\ell a_m)b_k=(b_k\partial_k)a_i-(\partial_i a_k)b_k.$$ Letting the sum run over all $$i$$ gives the result: $$(\vec{\nabla}\times\vec{a})\times\vec{b}=(\vec{b}\cdot\vec{\nabla})\vec{a}-\sum_i\sum_k(\partial_i a_k)b_k\hat{e}_i.$$ Looking at the last part, this is equivalent to $$\sum_i\sum_k(\partial_i a_k)b_k\hat{e}_i=\sum_n\delta_{nk}b_n\sum_i\sum_k(\partial_i a_k)\hat{e}_i\hat{e}_k$$ but this is just a "dot product" between a vector and the Jacobian of $$\vec{a}$$, so we have $$(\vec{\nabla}\times\vec{a})\times\vec{b}=(\vec{b}\cdot\vec{\nabla})\vec{a}-\vec{b}\cdot\boldsymbol{\mathrm{J}}_{\vec{a}},$$ where the last part is in the sense of performing the dot product column wise and use the result as new row entry.
For the special case of $$\vec{a}=\vec{b}=\vec{F}$$ note that $$\vec{F}\cdot\boldsymbol{\mathrm{J}}_{\vec{F}}$$ equals half of the gradient of the squared norm (i.e. of $$F_1^2+F_2^2+\dotsm$$), so this becomes: $$\boxed{(\vec{\nabla}\times\vec{F})\times\vec{F}=(\vec{F}\cdot\vec{\nabla})\vec{F}-\frac{1}{2}\vec{\nabla}\lVert\vec{F}\rVert^2}$$