Varying the inverse metric in $H_{abc}H^{abc}$ produces three identical terms. With $T_{ab}=-(2/\sqrt{-g})\delta S_H/\delta g^{ab}$,
$$
\boxed{T_{ab}=H_{acd}H_b{}^{cd}-\frac16g_{ab}H_{cde}H^{cde}}.
$$
In four dimensions use the <Hodge star operator> to define the dual covector $v^a=(1/3!)\epsilon^{abcd}H_{bcd}$. Then $H^2=-6v^2$ and $H_{acd}H_b{}^{cd}=2(v_av_b-g_{ab}v^2)$, so equivalently
$$
\boxed{T_{ab}=2v_av_b-g_{ab}v^2}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-309/2/a/i/solution