Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 6

For a rotation matrix $R\in SO(3)$, the Cartesian tensor component transformation law isAn isotropic tensor is unchanged under every such rotation, so its components satisfy $T_{ijkl}=R_{ia}{R}_{jb}{R}_{kc}{R}_{ld}{T}_{abcd}$ for every $R\in SO(3)$. The repeated indices are summed by the Einstein summation convention.

Take the trace first: the antisymmetric tensor has zero diagonal and $X$ is traceless, so $T_{ii}=3t$. Taking symmetric and antisymmetric parts then gives the scalar, symmetric-traceless and axial tensor decomposition:The Kronecker delta satisfies ${\delta }_{ii}=3$, which verifies $X_{ii}=0$. Substitution verifies the original decomposition and also proves its uniqueness.

Relabel the two dummy indices and use the respective symmetries:Thus the symmetric tensor and antisymmetric tensor subspaces are orthogonal in the Frobenius inner product.

The contraction is $\mathrm{tr}(A^3)$. Since $A^T=-A$, transpose and invariance of the trace giveThis is the odd-power case of vanishing odd complete contractions of an antisymmetric matrix.

A complete pairwise tensor contraction of copies of $A$ is represented by closed index cycles. Each cycle is a trace $\mathrm{tr}(A^r)$, possibly with a sign from reversing an index pair. If a contraction is internal to one copy, it contains the already vanishing $\mathrm{tr}A$. Otherwise an odd total number of copies forces at least one cycle of odd length, since a sum of even lengths is even. That trace vanishes by the transposition argument in part (ii). Therefore every such complete contraction is zero. This proves vanishing odd complete contractions of an antisymmetric matrix without assuming that all contractions form a single trace.

Let $J_{ij}={\int }_B{x}_i{x}_j{d}^{3}x$. Rotational symmetry makes $J$ an isotropic second-rank tensor, so $J_{ij}=C{\delta }_{ij}$. Contracting the indices and integrating in spherical coordinates givesThe isotropic second moment of a ball therefore givesOnly the scalar part of $M$ contributes; its antisymmetric tensor and symmetric-traceless parts give zero.