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

Under a proper rotation matrix $R$, invariance of a Cartesian second-rank tensor reads $T=RT{R}^T$, equivalently $TR=RT$. The half-turn $R_z(\pi )=\mathrm{diag}(-1,-1,1)$ forces all entries mixing the $z$ direction with the $xy$ plane to vanish. Thus $T$ consists of a planar $2\times 2$ block $B$ and the entry $T_{33}=c$.

Commutation with the planar quarter-turn $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ gives $B=aI+bJ$. This block already commutes with every planar rotation. Now use the half-turn $R_x(\pi )=\mathrm{diag}(1,-1,-1)$: on the planar block its conjugation fixes $I$ and sends $J$ to $-J$, so invariance forces $b=0$. ThereforeThe formula uses the Kronecker delta and is sufficient as well as necessary: rotations preserving the unoriented $z$-axis fix both $I$ and $e_3{e}_3^T$. This is an axially invariant second-rank tensor with the planar antisymmetric part removed by the horizontal half-turn.

Let $R$ be rotation through $2\pi /3$ about $z$, and let $S$ be rotation through $\pi$ about $x$. These generate the order-six dihedral group with relationsnamely $\{I,R,R^2,S,RS,R^{2}S\}$. We specify its order explicitly because dihedral notation has two conventions. It is a proper finite subgroup of the full axial rotation group.

To see that it is sufficient, write a general matrix in blocks $T=\left(\begin{array}{cc}B& u\\ v^T& c\end{array}\right)$. Commutation with $R$ forces $u=v=0$, since a nontrivial planar $120$-degree rotation fixes no nonzero planar vector. Its planar part is $\cos (2\pi /3)I+\sin (2\pi /3)J$, with nonzero sine, so $B$ must commute with $J$ and has the form $aI+bJ$. Commutation with $S$ again kills $b$. Hence this six-element subgroup has precisely the required invariant tensors.

It is also smallest by order. Groups of orders $2,3,5$ are cyclic: a nonidentity element has order dividing the prime group order by Lagrange theorem. For a cyclic rotation subgroup, the nonzero antisymmetric tensor representing cross product with its rotation-axis direction is invariant, so it cannot force the symmetric form above. The trivial subgroup plainly imposes no constraint. A group of order four is either cyclic or has three nonidentity elements of order two; in the latter case they commute, since $(AB{)}^{-1}=AB$ and also equals $BA$. Commuting distinct half-turns in three dimensions have perpendicular axes: conjugation by one half-turn must preserve the other's axis, and a distinct preserved axis lies in its perpendicular plane. In the specified axial group, their three axes are consequently $z$ and two perpendicular horizontal directions. In coordinates along them every diagonal matrix is invariant, including one with unequal horizontal entries. Such a group again does not force the required form.

Thus the smallest subgroup has six elements, generated by the axial $120$-degree rotation and one horizontal half-turn. It is the smallest axial rotation group forcing second-rank transverse isotropy; conjugating the horizontal axis gives equally valid choices.