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Under an orthogonal coordinate transformation $R$, a Cartesian pseudotensor has the usual tensor transformation law multiplied by $\det R$. Thus proper rotation matrices act as on ordinary tensors, while an orientation-reversing change contributes an extra minus sign. Interpreted as a rank-three pseudotensor, the Levi-Civita symbol has the same numerical components in every orthonormal frame, because its ordinary three-index transformation already contributes a factor $\det R$.

For constant $K$, this quadratic combination of two-by-two gradient minors is a divergence. Indeed,because the term containing $u_{j,\beta \alpha }$ vanishes by antisymmetry of the Levi-Civita symbol. Its volume integral vanishes for homogeneous displacement boundary data. It also vanishes pointwise on rank-one matrices, since ${\nu }_{\alpha }{\nu }_{\beta }$ is symmetric. Consequently one can strengthen an elastic energy density by such a term without changing fixed-boundary incremental energies or acoustic tensors.

The determinant transformation formula for the Levi-Civita symbol shows that it is invariant under every proper rotation matrix, for which $\det R=1$. Under an orthogonal reflection its sign reverses. This distinguishes proper-rotation isotropy from the pseudotensor transformation convention when both orientations are included.

For a differentiable three-dimensional vector field $F$, split its gradient tensor into symmetric and antisymmetric parts. The symmetric part is $S_{ij}=({\partial }_i{F}_j+{\partial }_j{F}_i)/2$. Contracting the antisymmetric part with the Levi-Civita symbol gives the axial vector $u_k={ϵ}_{kij}{\partial }_i{F}_j/2=(\nabla \times F{)}_k/2$, proving the decomposition. If $\nabla \cdot F=0$, the divergence of the symmetric part is ${\partial }_j{S}_{ij}=\Delta F_i/2$. This lets a vector Laplacian be converted into a boundary integral of a symmetric tensor by the divergence theorem.

The first coefficient of the alpha tensor is ${\alpha }_{ij}^{(1)}={ϵ}_{ikl}⟨u_k{\partial }_j{u}_l⟩$. Its contraction with the Levi-Civita symbol isby incompressibility and vanishing averages of total derivatives. Every antisymmetric second-rank tensor in three dimensions is detected by this contraction, so the coefficient is a symmetric matrix. The proof needs no Fourier analysis.

// Target: fluid-mechanics.bigb

Split the two dual factors into antisymmetric and symmetric tensors. The invariant Levi-Civita symbol identifies the antisymmetric pair with $3$, giving $6\oplus \overline{3}$ after multiplying by the remaining triplet. On the symmetric part, contraction $T^a{}_{ba}$ maps onto $\overline{3}$; the equivariant splitting is $T^a{}_{bc}=({\delta }_b^a{v}_c+{\delta }_c^a{v}_b)/4$. Its trace-free kernel has dimension $18-3=15$ and highest weight with Dynkin labels $(1,2)$. The Weyl dimension formula gives dimension $15$, so this is exactly the irreducible kernel. It is conjugate to the other fifteen-dimensional module with labels $(2,1)$; dimension alone does not distinguish the two.