Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 3 / 12A / c

Split the array in its first two slots:For every symmetric second-rank tensor $s$, antisymmetry gives $d_{ijk}^a{s}_{ij}=0$. Hence the stipulated vector equals $d_{ijk}^s{s}_{ij}$.

Let $R$ change the orthonormal frame, so $s_{ab}^{\prime }=R_{ai}{R}_{bj}{s}_{ij}$. The vector transformation law givesfor every symmetric test $s$. Both coefficient arrays in $i,j$ are symmetric, so equality against every symmetric matrix forces equality of the coefficients; one may test the individual diagonal entries and the symmetric off-diagonal basis matrices. Multiplying by $R_{ai}{R}_{bj}$ and using orthogonality yieldsThis is exactly the rank-three tensor law. It is the symmetric-test criterion for a third-rank Cartesian tensor, a form of the quotient theorem for Cartesian tensors. Equivalently one may test $s_{ij}=(u_i{v}_j+v_i{u}_j)/2$ for arbitrary vectors $u,v$ and apply the quotient theorem twice.

The antisymmetric part need not be a tensor. Its contraction with every symmetric test is zero, so the premise imposes no transformation law on it. This is the blindness of symmetric contraction tests to antisymmetric arrays, now with a third free index.

For a concrete example, prescribe the same numerical array in every orthonormal frame, with onlynonzero, and take $d^s=0$. Every contraction $d_{ijk}^a{s}_{ij}$ vanishes, and the zero result is a genuine vector in every frame. But under $R=\mathrm{diag}(-1,-1,1)$ the rank-three tensor transformation law would send the $121$ component to $-1$, whereas our framewise prescription leaves it equal to $1$. Thus the array is not a tensor despite satisfying every stipulated contraction test. An antisymmetric part could be a tensor if an additional transformation law were supplied; it is simply not forced to be one.