Varying the gauge field and integrating by parts gives
$$
\boxed{\nabla_aF^{ab}=0},
\qquad
\boxed{\nabla_{[a}F_{bc]}=0},
$$
the second identity following from $F=dA$. Varying the inverse metric, using
$$
\delta F^2=2F_{ac}F_b{}^c\delta g^{ab},
$$
and discarding the Einstein--Hilbert boundary term gives
$$
\boxed{R_{ab}-\frac12Rg_{ab}+\Lambda g_{ab}
=2\left(F_{ac}F_b{}^c-\frac14g_{ab}F_{cd}F^{cd}\right)}.
$$
For AdS$_4$, $\Lambda=-3/L^2$.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 354 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-354/2/a/solution