Dot product of curl (curl A * curl A) i want to compute the value of $$curl A \space \space * \space \space  curl A$$, that is, the dot product of the curl of the same vector, also know as the square of the norm of the curl of A.
But, i would like to compute in terms of another vectors operators, and it has been hard to me do it. Particurlally, i opened it in cartesian coordinates and tried, but it got messy.
For example, arised terms like $$(\partial_{i}A_{i})^2$$ where i  could see that it seems a mixture of divergence, with dot product, and something else, which i wasn't able to find.
 A: $
\def\e{\varepsilon}
\def\p{{\partial}}
\def\n{{\nabla}}
\def\l{\left(}
\def\r{\right)}
$The vector-valued curl can be written in index notation using the Levi-Civita tensor
$$\eqalign{
c_k &= (\n\times A)_k &= \l\n_i A_j\r\e_{ijk} &= \e_{kij}\l\n_iA_j\r \\
c &= \n\times A &= \l\n A\r:\e &= \e:\l\n A\r \\
}$$
where the colon denotes the double-dot product.
The matrix-valued gradient can also be written in index notation
$$\eqalign{
G &= \n A \\
G_{ij} &= \n_{i}A_j \\
}$$
Also recall that $\e$ satisfies the delta-epsilon identity
$$\eqalign{
\e_{ijk}\e_{kpq} = \delta_{ip}\delta_{jq} - \delta_{iq}\delta_{jp} \\ 
}$$
Combining these ideas yields
$$\eqalign{
(\n\times A)_k(\n\times A)_k &= \l\n_i A_j\e_{ijk}\r \l\e_{kpq}\n_pA_q\r \\
  &= \l\n_i A_j\r\Big(\e_{ijk}\e_{kpq}\Big)\l\n_pA_q\r \\
  &= \l\n_i A_j\r\Big(\delta_{ip}\delta_{jq} - \delta_{iq}\delta_{jp}\Big)\l\n_pA_q\r \\
  &= \l\n_i A_j\r\l\n_iA_j\r - \l\n_i A_j\r\l\n_jA_i\r \\
(\n\times A)\cdot(\n\times A) &= G:G - G:G^T \\
 &= {\rm Tr}(G^TG) - {\rm Tr}(G^2) \\
}$$
where ${\rm Tr}(M)$ denotes the trace of a matrix.
