Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 6 / a

Under the specified passive change of orthonormal basis, each Cartesian tensor index acquires one copy of $L$. Thus a tensor of order $n$ transforms asRepeated old-basis indices are summed from one to three. Orthogonality of the new basis gives $L{L}^T=I$. Taking determinants yields $(\det L{)}^2=1$, hence $\boxed{\det L=+1\text{ or }-1}$. The positive case preserves orientation and the negative case reverses it.

A pseudotensor has an additional determinant factor under an orthogonal transformation. Thus its rank-three transformation law isThe contracted expression $L_{ia}{L}_{jb}{L}_{kc}{ϵ}_{abc}$ is fully antisymmetric in $i,j,k$, so it equals a scalar multiple of ${ϵ}_{ijk}$. For $(i,j,k)=(1,2,3)$ it is precisely the determinant expansion of $L$, fixing that scalar to $\det L$. ThereforeThe Levi-Civita symbol consequently has the same numerical entries in every orthonormal frame when interpreted as a pseudotensor. With an ordinary rank-three tensor transformation law alone it would instead acquire a minus sign under an orientation-reversing transformation.