Summary : Still More Tensor Identities Three more identities with gradients, divergences, Laplacians, and cross products. Later today, the fun stuff, special relativity begins! $\vec{A} \cdot \left(\vec{B} \times \vec{C}\right) = \vec{B} \cdot \left(\vec{C} \times \vec{A}\right)$ $= A_i \epsilon_{ijk} B_j C_k$ As long as we only cycle the indices in the Levi-Civita symbol, we won't cause a sign change, so the above is also equal to $= A_k \epsilon_{ijk} B_i C_j$ Which we can commute to get $= B_i \epsilon_{ijk} C_j A_k = \vec{B} \cdot \left(\vec{C} \times \vec{A}\right)$ Done! $\vec{\nabla} \cdot \left(\vec{\nabla} \times \vec{A} \right) = 0$ $=\partial_i \epsilon_{ijk} \partial_j A_k$ $= 0$ If $i$ and $j$ are equal, then the Levi-Civita evaluates to zero. If they are not equal, then swapping the two indices produces the same mixed partial derivative result, but with a negative sign inserted by swapping indices in the Levi-Civita symbol. ...