Why $\nabla \times (\nabla \times a) = -(\nabla \cdot \nabla)a + \nabla(\nabla \cdot a)$?

Question

Why $\nabla \times (\nabla \times a) = - (\nabla \cdot \nabla)a + \nabla(\nabla \cdot a)$ ? I can derive this equation with the identity $\varepsilon_{ijk}\varepsilon_{jkl}=-\delta_{jl}\delta_{im} + \delta_{jm}\delta_{li}$. But it's not obvious compare to the identities like $\nabla \cdot \nabla \cdot (\nabla a)=0$. Thinking scalar vectors $a,b, b\times (b\times a)$ is parallel to a,but when b is replaced by $\nabla, \nabla(\nabla \cdot a)$　appear which is not parallel to a. What does $\nabla(\nabla \cdot a)$ represents for ? Is there any intuitive or easy way to derive the equation above ?

Answer 1

$\vec{\nabla}=\sum \vec{e_i}\frac{\partial}{\partial x_i}$.

So you are only doing "classical" vector products. If it holds true for any vector $\vec{b}$, there is no reason it should not hold true for $\vec{\nabla}$. Can you show us what was you issue in detail?

Answer 2

\begin{align} \nabla \times (\nabla \times a) =& \nabla \times (\partial_i a_j \varepsilon_{ijk} \bf{e_k} )\\ =& \partial_n \partial_i a_j \varepsilon_{ijk} \varepsilon_{nkp} \bf{e_p} \\ [\nabla \times (\nabla \times a)]_p=& \partial_n \partial_i a_j \varepsilon_{ijk} \varepsilon_{pnk}\\ =& \partial_n \partial_i a_j(\delta_{ip} \delta_{jn} - \delta_{in} \delta_{jp})\\ =& \partial_j \partial_p a_j - \partial_i \partial_i a_p\\ [(\nabla \cdot \nabla)a]_p = & (\nabla \cdot \nabla) a_p = \partial_i \partial_i a_p\\ [\nabla(\nabla \cdot a)]_p = & \partial_p \partial_j a_j = \partial_j \partial_p a_j \\ \nabla \times (\nabla \times a) =& -(\nabla \cdot \nabla)a + \nabla(\nabla \cdot a) \end{align}