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Write $A_{ij}={\partial }_j{u}_i$. The trace fixes the isotropic part, and splitting the remaining matrix into its symmetric and antisymmetric parts gives the scalar, symmetric-traceless and axial tensor decomposition:Here $s_{ij}=s_{ji}$ and $s_{ii}=0$. The antisymmetric part is $a_{ij}=({\partial }_j{u}_i-{\partial }_i{u}_j)/2$. Its associated axial vector isIndeed the contraction identity ${ϵ}_{ijk}{ϵ}_{k\ell m}={\delta }_{i\ell }{\delta }_{jm}-{\delta }_{im}{\delta }_{j\ell }$ gives ${ϵ}_{ijk}{\omega }_k=({\partial }_j{u}_i-{\partial }_i{u}_j)/2$. Adding the three parts reconstructs $A_{ij}$.

The minus sign is required by the index order $A_{ij}={\partial }_j{u}_i$. The symmetric gradient and curl decomposition written instead for ${\partial }_i{u}_j$ has the opposite axial sign. The divergence supplies $3p$, the trace-free symmetric part describes shape change, and the antisymmetric part describes local rotation.

Apply the symmetric gradient and curl decomposition separately to the electric field and the magnetic field:The corresponding axial parts areThe second equalities use the vacuum Maxwell equations. Their divergence constraints also show that both symmetric gradient tensors are traceless.