= Solution
Split the <Cartesian second-rank tensor> into symmetric and antisymmetric parts:
$$
S_{ij}=\frac{T_{ij}+T_{ji}}2,\qquad A_{ij}=\frac{T_{ij}-T_{ji}}2.
$$
Define $u_k=\frac12\epsilon_{kij}T_{ij}$. The symmetric part contributes zero to this contraction. The <epsilon-delta identity> then gives
$$
\epsilon_{ijk}u_k=\frac12\epsilon_{ijk}\epsilon_{klm}T_{lm}
=\frac12(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})T_{lm}
=\frac{T_{ij}-T_{ji}}2=A_{ij}.
$$
Therefore
$$
\boxed{T_{ij}=S_{ij}+\epsilon_{ijk}u_k,\qquad
S_{ij}=\frac{T_{ij}+T_{ji}}2,\qquad u_k=\frac12\epsilon_{kij}T_{ij}}.
$$
The symmetric part and the three independent antisymmetric components determine these quantities uniquely. Under proper rotations $u$ is a <vector>; under reflections it is an <axial vector>.
Back to article page
