Show with index notation that $||\nabla \times \underline{u}||^2=||\nabla \underline{u}||^2 - \mathbf{Tr}[(\nabla \underline{u})^2]$

Question

So far I have: $$||\nabla\times\underline{u}||^2 = \left[(\nabla\times\underline{u})\right]_i\left[(\nabla\times\underline{u})\right]_k =\left[\varepsilon_{ijk}\frac{\partial u_k}{\partial x_j}\right]\left[\varepsilon_{kmn}\frac{\partial u_n}{\partial x_m}\right]$$ $$=\varepsilon_{kij}\varepsilon_{kmn}\frac{\partial u_k}{\partial x_j}\frac{\partial u_n}{\partial x_m} = \left[\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}\right]\frac{\partial u_k}{\partial x_j}\frac{\partial u_n}{\partial x_m}$$ $$= \frac{\partial u_k}{\partial x_j}\frac{\partial u_j}{\partial x_i} - \frac{\partial u_k}{\partial x_j}\frac{\partial u_i}{\partial x_j}$$ Here is where I don't understand how I am to get the final result. I struggle on this a lot so if someone could show me that would be great.

Answer 1

By $\boldsymbol \nabla \mathbf u$, I assume you mean the Jacobian matrix of $\mathbf u$, with components $$J_{ij} = \partial_j u_i = \frac{\partial u_i}{\partial x_j}.$$ Then, \begin{align*} \lVert \boldsymbol \nabla \times \mathbf u \rVert^2 &= [\boldsymbol \nabla \times \mathbf u]_i[\boldsymbol \nabla \times \mathbf u]_i \\ &= \varepsilon_{ijk}\partial_ju_k \varepsilon_{ipq}\partial_pu_q \\ &= (\delta_{jp}\delta_{kq} - \delta_{jq}\delta_{kp})\partial_ju_k\partial_p u_q \\ &= \partial_ju_k \partial_j u_k - \partial_j u_k \partial_k u_j \\ &= J_{kj}J_{kj} - J_{jk} J_{kj} \\ &= \left[\mathbf J \mathbf J^T\right]_{kk} - \left[\mathbf J^2\right]_{jj} \\ &= \operatorname{Tr}\left(\mathbf J \mathbf J^T\right) - \operatorname{Tr} \left(\mathbf J^2\right). \end{align*} The original notation in your question is confusing and ambiguous.