 Solution

Use the curvature convention $[{\nabla }_b,{\nabla }_c]V^d=R^d{}_{abc}{V}^a$, consistent with the coordinate formula later in the paper. Choose normal coordinates at an arbitrary point $p$. The Levi-Civita connection has ${\Gamma }^d{}_{ab}(p)=0$ and is torsion free, so ${\Gamma }^d{}_{ab}={\Gamma }^d{}_{ba}$. At $p$,Hence the cyclic sum isEach derivative cancels using symmetry of the two lower connection indices. Because the Riemann curvature tensor is antisymmetric in its last two indices, full antisymmetrization over $a,b,c$ equals one third of this cyclic sum. ThusThe statement is tensorial and $p$ was arbitrary, so it holds in all coordinates everywhere. This is the first Bianchi identity.