Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 3 / 12C (Vector Calculus)

Taking the symmetric part givesContracting the antisymmetric part with the Levi-Civita symbol givesIndeed, $\frac{1}{2}{\epsilon }_{ijk}{\epsilon }_{k\ell m}{T}_{\ell m}=\frac{1}{2}(T_{ij}-T_{ji})$, so $S_{ij}+{\epsilon }_{ijk}{V}_k=T_{ij}$.

In three dimensions, the most general isotropic tensors of ranks zero through three areThe scalar and zero vector are unchanged by rotations. For every $R\in SO(3)$, $R_{ia}{R}_{jb}{\delta }_{ab}={\delta }_{ij}$ and $R_{ia}{R}_{jb}{R}_{kc}{\epsilon }_{abc}=(\det R){\epsilon }_{ijk}={\epsilon }_{ijk}$, proving isotropy.

The relation isTaking its trace, symmetric traceless part, and antisymmetric part separately givesEquivalently,

If $\beta =-\gamma \ne 0$, $\alpha \ne 0$, and $A_{ij}=0$, its symmetric part forces $B_{kk}=0$ and its antisymmetric part forces $B_{[ij]}=0$. Thus the solutions are exactly the symmetric traceless tensors $B_{ij}$.

Dualize the antisymmetric tensor to a vector,An isotropic symmetric rank-two tensor depending quadratically on one vector can only be a linear combination of ${\delta }_{ij}{d}_k{d}_k$ and $d_i{d}_j$: after rotating $d$ to the third axis, invariance under all rotations about that axis forces the matrix to be $\mathrm{diag}(a,a,b)$.

Since $D_{ik}{D}_{jk}={\delta }_{ij}{d}^2-d_i{d}_j$ and $D_{kl}{D}_{kl}=2d^2$, the most general answer may be writtenThe two displayed tensors are independent, for example by taking $d$ along a coordinate axis and comparing the components parallel and perpendicular to it. Hence there are exactly two independent terms.