At an arbitrary point choose <normal coordinates>, so the <Christoffel symbols> vanish there. Torsion freedom and commutation of partial derivatives then give the algebraic <first Bianchi identity>
$$
\boxed{R^a{}_{[bcd]}=0}.
$$
Differentiate the coordinate curvature expression and cyclically antisymmetrize. Third derivatives cancel, giving the <second Bianchi identity>
$$
\boxed{\nabla_{[e}R^a{}_{|b|cd]}=0}.
$$
Both statements are tensorial and hence hold in every coordinate system. Contracting the differential identity, using the curvature symmetries and <metric compatibility>, yields
$$
\boxed{\nabla_aR^a{}_b-\frac12\nabla_bR=0},
$$
the <contracted Bianchi identity>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 / 1 / a / iii / Solution

{wiki=Bianchi_identity}

For the <Levi-Civita connection>, the cyclic covariant derivative of curvature vanishes: $\nabla_{[e}R^a{}_{|b|cd]}=0$. Contracting this identity gives the <contracted Bianchi identity>.

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