Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 309 / 2 / a / i / Solution

Varying the inverse metric in $H_{abc}{H}^{abc}$ produces three identical terms. With $T_{ab}=-(2/\sqrt{-g})\delta S_H/\delta g^{ab}$,In four dimensions use the Hodge star operator to define the dual covector $v^a=(1/3!){ϵ}^{abcd}{H}_{bcd}$. Then $H^2=-6v^2$ and $H_{acd}{H}_b{}^{cd}=2(v_a{v}_b-g_{ab}{v}^2)$, so equivalently