Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 3 / 12A / c

True. In all four tests, an array is understood to have components specified in each rotated orthonormal basis; being a scalar means that the contraction is invariant under those changes. Write the Frobenius inner product as $A:B={\sum }_{ij}{A}_{ij}{B}_{ij}$. Every test Cartesian second-rank tensor obeys $B^{\prime }=RB{R}^T$. The assumption saysChoosing the elementary matrices as $B$, or choosing $B=R^T{A}^{\prime }R-A$, gives $R^T{A}^{\prime }R=A$. Thereforewhich is the required Cartesian second-rank tensor transformation law. This is the scalar contraction test for a Cartesian tensor.

False. Tests against symmetric second-rank tensors see only the symmetric part $A_s=(A+A^T)/2$, because the Frobenius inner product of a symmetric matrix and an antisymmetric matrix is zero:An arbitrary basis-dependent antisymmetric part is invisible to these tests. For example, prescribein one basis, and $A^{\prime }=0$ in another rotated basis, choosing antisymmetric arrays in every remaining basis. Every contraction with any symmetric second-rank tensor is the invariant scalar zero. But an invertible rotation cannot transform this nonzero matrix to zero, so the array is not a Cartesian second-rank tensor. This is the blindness of symmetric contraction tests to antisymmetric arrays.

False. The counterexample just given is antisymmetric in every basis and has zero contraction with every symmetric second-rank tensor. It still fails the Cartesian second-rank tensor transformation law, because its value is nonzero in one basis and zero in another. Thus antisymmetry alone, together with symmetric tests, proves no tensor transformation law. The same blindness of symmetric contraction tests to antisymmetric arrays applies even when the array is assumed antisymmetric.

True. Since $A$ and $A^{\prime }$ are antisymmetric, the matrix $D=R^T{A}^{\prime }R-A$ is antisymmetric. Invariance of contractions with every antisymmetric second-rank tensor gives $D:B=0$ for every antisymmetric test matrix $B$. Such a matrix can be freely prescribed in one basis and then transformed to define a valid test tensor. Choose $B=D$ in that basis. Thenso $D=0$ and $A^{\prime }=RA{R}^T$. Together with the given antisymmetry, this proves that $A$ is an antisymmetric second-rank tensor. The antisymmetric contraction test for an antisymmetric tensor works because the Frobenius inner product is nondegenerate on the antisymmetric subspace.