Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 3 / 12A (Vector Calculus)

DefineThey sum to $t_{ij}$, with $s_{ji}=s_{ij}$ and $a_{ji}=-a_{ij}$. Interchanging two slots of a tensor still produces a tensor, so these are a symmetric second-rank tensor and an antisymmetric second-rank tensor, not merely symmetric and antisymmetric arrays in a selected frame.

For uniqueness, if $t=s+a=s^{\prime }+a^{\prime }$, then $s-s^{\prime }=a^{\prime }-a$. The common difference is both symmetric and antisymmetric, so each of its components equals its own negative and vanishes over $\mathbb{R}$. Thus the decomposition is unique. These are the symmetric and antisymmetric parts of a matrix, with their tensor transformation law preserved.

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.

Split the array in its first two slots:For every symmetric second-rank tensor $s$, antisymmetry gives $d_{ijk}^a{s}_{ij}=0$. Hence the stipulated vector equals $d_{ijk}^s{s}_{ij}$.

Let $R$ change the orthonormal frame, so $s_{ab}^{\prime }=R_{ai}{R}_{bj}{s}_{ij}$. The vector transformation law givesfor every symmetric test $s$. Both coefficient arrays in $i,j$ are symmetric, so equality against every symmetric matrix forces equality of the coefficients; one may test the individual diagonal entries and the symmetric off-diagonal basis matrices. Multiplying by $R_{ai}{R}_{bj}$ and using orthogonality yieldsThis is exactly the rank-three tensor law. It is the symmetric-test criterion for a third-rank Cartesian tensor, a form of the quotient theorem for Cartesian tensors. Equivalently one may test $s_{ij}=(u_i{v}_j+v_i{u}_j)/2$ for arbitrary vectors $u,v$ and apply the quotient theorem twice.

The antisymmetric part need not be a tensor. Its contraction with every symmetric test is zero, so the premise imposes no transformation law on it. This is the blindness of symmetric contraction tests to antisymmetric arrays, now with a third free index.

For a concrete example, prescribe the same numerical array in every orthonormal frame, with onlynonzero, and take $d^s=0$. Every contraction $d_{ijk}^a{s}_{ij}$ vanishes, and the zero result is a genuine vector in every frame. But under $R=\mathrm{diag}(-1,-1,1)$ the rank-three tensor transformation law would send the $121$ component to $-1$, whereas our framewise prescription leaves it equal to $1$. Thus the array is not a tensor despite satisfying every stipulated contraction test. An antisymmetric part could be a tensor if an additional transformation law were supplied; it is simply not forced to be one.