Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 6 / a

Under an orthogonal matrix $R$, the Levi-Civita symbol satisfiesUse this twice, and use $R_{mc}{R}_{mf}={\delta }_{cf}$ for the contracted index. It follows thatTherefore $\boxed{T_{ijkl}^{\prime }=R_{ia}{R}_{jb}{R}_{kc}{R}_{ld}{T}_{abcd}=T_{ijkl}}$: it is a fourth-rank isotropic tensor. Although a single alternating symbol is a pseudotensor under reflections, the two determinant factors multiply to one, so this product is an ordinary tensor even for improper orthogonal transformations.

Use three components of the proposed isotropic tensor form. Since $T_{1122}=0$, one gets $\alpha =0$. Since $T_{1212}={ϵ}_{123}^2=1$, one gets $\beta =1$. Finally $T_{1221}={ϵ}_{123}{ϵ}_{213}=-1$ gives $\gamma =-1$. Hence contraction of two Levi-Civita symbols gives