Use the paper's curvature convention throughout, and define the Ricci tensor and scalar curvature by $R_{kn}=R^i{}_{kin}$ and $R=g^{kn}{R}_{kn}$. In particular, if semicolons are read from left to right, $X^i{}_{;km}={\nabla }_m({\nabla }_{k}X{)}^i$, including the connection on the derivative index. This ordering matters for the Ricci identity sign.

Choose Riemann normal coordinates at an arbitrary point $P$. The Christoffel symbols vanish there, and the Levi-Civita connection formula givesMetric symmetry and commuting partial derivatives show directly that exchanging $m,n$ or $i,k$ changes the sign, while interchanging the pairs $(i,k)$ and $(m,n)$ leaves the expression unchanged. Adding its three cyclic versions over $k,m,n$ cancels every second derivative. These are tensorial statements and $P$ was arbitrary, soThe cyclic relation is the first Bianchi identity; its reduction to the total antisymmetrization uses last-pair antisymmetry. These arguments use zero torsion tensor and metric compatibility, not a field equation.

At $P$, the covariant derivative of curvature equals its partial derivative. Differentiating the connection expression for curvature gives derivatives of ${\partial }_n{\Gamma }^i{}_{km}-{\partial }_m{\Gamma }^i{}_{kn}$; derivatives of the quadratic connection terms vanish because $\Gamma (P)=0$. In the cyclic sum over $m,n,p$, every second derivative of a connection coefficient occurs twice with opposite signs. ThusThis is the second Bianchi identity, valid everywhere by tensoriality.

Lower the first curvature index and contract the differential identity with $g^{im}$. Metric compatibility allows the metric to pass through the covariant derivatives. Using the pair antisymmetries yieldsContract again with $g^{km}$. The left side is $-{\nabla }^i{R}_{in}$, while the right side is ${\nabla }^k{R}_{kn}-{\nabla }_{n}R$. Hence the contracted Bianchi identity isThe Ricci tensor is symmetric, as follows by contracting the pair-exchange symmetry. The final divergence is that of the Einstein tensor.

For the vector commutator, at the same normal-coordinate point,The terms involving undifferentiated connection coefficients vanish at $P$. Subtracting the reversed expression leaves $({\Gamma }^i{}_{jk,m}-{\Gamma }^i{}_{jm,k})X^j$, exactly the stated curvature convention at $P$. ThereforeFor a covector field, its connection term has the opposite sign: ${\omega }_{i;k}={\partial }_k{\omega }_i-{\Gamma }^j{}_{ik}{\omega }_j$. The same calculation givesEquivalently, apply the commutator to the scalar ${\omega }_i{X}^i$, whose two covariant derivatives commute, and cancel the vector contribution. Each covariant index contributes a negative curvature action and each contravariant index a positive one in this semicolon ordering.

To prove the requested double divergence of the Riemann tensor, letThe once-contracted identity givesCommuting the outer derivative past ${\nabla }_m$ and ${\nabla }_n$ leaves a difference of scalar Hessians, $\frac{1}{2}({\nabla }_m{\nabla }_n-{\nabla }_n{\nabla }_m)R=0$, plus curvature terms. The operator commutator $[{\nabla }_k,{\nabla }_m]$ has the negative of the paper's curvature sign, since it reverses the semicolon order. On the mixed Ricci tensor it givesThe first term is symmetric in $m,n$: it is the metric contraction of two symmetric Ricci tensors. The second is $S_{mn}=R^{ak}{R}_{ankm}$ and is also symmetric, sincewhere the first equality uses curvature pair exchange and the second swaps the dummy indices $a,k$. Thus the two commutator terms cancel on antisymmetrization in $m,n$, provingNo assumption such as vacuum, constant curvature or vanishing Ricci tensor was used.