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The first Bianchi identity follows by applying cyclic curvature commutators to the gradient of an arbitrary scalar field: its symmetric Hessian makes the third-derivative terms cancel, leaving $R_{abc}{}^d+R_{bca}{}^d+R_{cab}{}^d=0$. For the second Bianchi identity, write the vector-bundle derivative as $D_a={\partial }_a+{\Gamma }_a$, so $[D_b,D_c]={\mathcal{R}}_{bc}$. The operator Jacobi identity gives the cyclic sum of ${\partial }_a{\mathcal{R}}_{bc}+[{\Gamma }_a,{\mathcal{R}}_{bc}]$ equal to zero. Restoring the connection terms on the two curvature-form indices gives precisely ${\nabla }_a{R}_{bcd}{}^e+{\nabla }_b{R}_{cad}{}^e+{\nabla }_c{R}_{abd}{}^e=0$: the additional terms cancel because the connection is torsion-free and curvature is antisymmetric in its first pair.

For the Levi-Civita connection, the second Bianchi identity gives ${\nabla }_i{R}^{ik}{}_{mn}={\nabla }_m{R}^k{}_n-{\nabla }_n{R}^k{}_m$. Applying another covariant derivative and the contracted Bianchi identity reduces its divergence to a commuting scalar Hessian plus curvature commutators on the Ricci tensor. The latter are the difference of symmetric tensors: a square of the Ricci tensor and $R^{ab}{R}_{anbm}$, which is symmetric in $m,n$ by curvature pair exchange and Ricci symmetry. Both differences vanish. This identity does not require vacuum field equations and is independent of the overall curvature sign convention.

Use the Levi-Civita connection $\nabla$ of the Riemannian metric $g$ and fix the conventionThe Leibniz rule shows that this expression is linear over smooth functions in all three arguments: the differentiated coefficient terms in the first two derivatives cancel those in the bracket term. Thus $R$ is a section of ${\Lambda }^2{T}^{\ast }M\otimes \mathrm{End}(TM)$, the Riemann curvature tensor, and $R_4$ is a covariant tensor field. This convention gives positive sectional curvature on the round sphere.

Its algebraic symmetries areThe first identity follows immediately from the definition. For the second, apply $XY-YX-[X,Y]$ to $g(Z,W)$ and use compatibility of the Levi-Civita connection with the Riemannian metric. The left side is zero and the differentiated terms combine toFor the first Bianchi identity, use normal coordinates at a point $p$, so the Christoffel symbols vanish there. Torsion-freeness gives ${\Gamma }_{ij}^{\ell }={\Gamma }_{ji}^{\ell }$, andThe cyclic sum over $i,j,k$ cancels term by term. Since this is a tensor identity, it holds for arbitrary tangent vectors everywhere.

To prove pair interchange rather than assume it, abbreviate $R_4(a,b,c,d)$ to $R_{abcd}$ and put $A=R_{abcd}$, $B=R_{cdab}$. The first Bianchi identity and the two antisymmetries giveThe first Bianchi identity applied to $(d,b,a)$ with final argument $c$ gives $R_{dbca}+R_{adcb}=R_{badc}=A$. Hence $A=2B-A$, proving $A=B$.

There is also the differential second Bianchi identity,For its proof take commuting coordinate fields, normal at $p$. The Jacobi identity for operator commutators says $[{\nabla }_X,[{\nabla }_Y,{\nabla }_Z]]+\text{cyclic}=0$. On these commuting fields, $[{\nabla }_Y,{\nabla }_Z]=R(Y,Z)$; at $p$, the covariant derivatives of the coordinate arguments vanish. The commutator identity is therefore the displayed covariant derivative identity at $p$, and tensoriality establishes it everywhere.

For a two-dimensional plane $\sigma =\mathrm{span}\{X,Y\}\subset T_{p}M$, define the sectional curvature byThe denominator is the squared area of $X,Y$. Under an invertible change of plane basis, the alternating pairs in the numerator and the area in the denominator both acquire the square of the change-of-basis determinant. Thus the quotient depends only on $\sigma$. For an orthonormal basis of $\sigma$, it is simply $g(R(X,Y)Y,X)$.

The Ricci curvature is the tracewhere $(e_i)$ is any orthonormal basis of $T_{p}M$. This is independent of the basis because it is a trace. Applying the first Bianchi identity to $(e_i,X,Y)$ and pairing with $e_i$ proves that the summand difference for $\mathrm{Ric}(X,Y)-\mathrm{Ric}(Y,X)$ vanishes; the middle term $g(R(X,Y)e_i,e_i)$ is zero by the last-pair antisymmetry. Hence Ricci curvature is a symmetric bilinear form. For a unit vector $v=e_1$ extended to an orthonormal basis,For nonzero $X$, take $v=X/|X|$ and multiply the right side by $|X{|}^2$. This determines the entire Ricci curvature by the polarization identity:Finally, the scalar curvature is the metric trace of Ricci curvature. Taking its diagonal entries in an orthonormal basis and pairing the equal contributions from $(i,j)$ and $(j,i)$ yieldsIn particular, constant sectional curvature $K_0$ gives $\mathrm{Ric}=(n-1)K_{0}g$ and $\mathrm{Scal}=n(n-1)K_0$. In dimension one the traces are zero; there are no two-dimensional tangent planes and the sums are empty.

A linear connection on a manifold is a rule $(X,Y)\mapsto {\nabla }_{X}Y$ for differentiating vector fields which is additive, is $C^{\infty }$-linear in $X$, is real-linear in $Y$, and obeys ${\nabla }_X(fY)=X(f)Y+f{\nabla }_{X}Y$. It extends to covector fields and arbitrary tensor fields by the Leibniz rule and compatibility with contraction. In coordinates,This is the covariant derivative; the affine connection terms correct the changes of the coordinate basis. A symmetric affine connection has zero torsion tensor, $T(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]=0$, or ${\Gamma }^c{}_{ab}={\Gamma }^c{}_{ba}$ in a coordinate basis. Symmetry does not imply metric compatibility; no metric is required for this question.

Additivity of each covariant derivative immediately gives additivity of ${\Delta }_{ab}$. Expanding the second derivative of a tensor product givesSubtract the expression with $a,b$ exchanged. The two mixed terms cancel, proving that the curvature commutator is a tensor derivation:For a scalar, ${\nabla }_a{\nabla }_{b}f={\partial }_a{\partial }_{b}f-{\Gamma }^c{}_{ab}{\partial }_{c}f$. Commuting partial derivatives and using the symmetric affine connection gives ${\Delta }_{ab}f=0$. Consequently ${\Delta }_{ab}(fU)=f{\Delta }_{ab}U$: the commutator is linear over smooth functions, not merely over constant scalars. Write $U=U^c{e}_c$ in a local basis. The tensor-product rule gives ${\Delta }_{ab}U=U^c{\Delta }_{ab}{e}_c$, showing that its value at a point depends only on $U$ at that point. Thus it defines the Riemann curvature tensor byThe antisymmetrized second covariant derivative is itself tensorial in the derivative indices, so this pointwise map transforms as a tensor. The formula or the defining commutator gives $R_{abc}{}^d=-R_{bac}{}^d$, hence $R_{abc}{}^d=R_{[ab]c}{}^d$ with normalized antisymmetrization.

Apply the commutator to the scalar contraction ${\omega }_c{U}^c$. Its action vanishes, soSince $U$ is arbitrary, ${\Delta }_{ab}{\omega }_c=-R_{abc}{}^d{\omega }_d$. Repeated use of the tensor-product rule gives one plus curvature action for an upper index and one minus action for each lower index. In particular,The type notation in the PDF means one contravariant and two covariant indices, not a fractional tensor order.

For the first Bianchi identity, put ${\omega }_c={\nabla }_{c}f$ and $H_{bc}={\nabla }_b{\nabla }_{c}f$. The symmetric Hessian of a scalar field satisfies $H_{bc}=H_{cb}$. Therefore the six third-derivative terms cancel inSubstituting the covector curvature action gives $(R_{abc}{}^d+R_{bca}{}^d+R_{cab}{}^d){\nabla }_{d}f=0$. At a chosen point the gradient of a smooth scalar can be any covector, soThe second form follows from the already proved first-pair antisymmetry.

For the second Bianchi identity, the same third-derivative cancellation can be organized as an operator identity. On vector-valued components set $D_a={\partial }_a+{\Gamma }_a$, with $({\Gamma }_a{)}^e{}_d={\Gamma }^e{}_{ad}$. Although a full second covariant derivative includes a term $-{\Gamma }^f{}_{ab}{D}_f$ for its lower derivative index, that term cancels on antisymmetrization because the affine connection is symmetric. Thus $[D_b,D_c]={\mathcal{R}}_{bc}$, where $({\mathcal{R}}_{bc}{)}^e{}_d=R_{bcd}{}^e$ acts by multiplication. Expanding the three nested operator commutators makes all six triple products cancel:Equivalently, the cyclic sum of ${\partial }_a{\mathcal{R}}_{bc}+[{\Gamma }_a,{\mathcal{R}}_{bc}]$ vanishes. These are the covariant derivative terms on the upper and last lower curvature indices. The two remaining lower indices add $-{\Gamma }^f{}_{ab}{R}_{fcd}{}^e-{\Gamma }^f{}_{ac}{R}_{bfd}{}^e$. In the cyclic sum these additions cancel in pairs, using ${\Gamma }^f{}_{ab}={\Gamma }^f{}_{ba}$ and $R_{fcd}{}^e=-R_{cfd}{}^e$. Hence the Bianchi identities from covariant derivative commutators giveBoth identities here apply to any torsion-free linear affine connection; the second derivation does not require choosing a metric affine connection.

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.

For the torsion-free Levi-Civita connection, cyclically summing the displayed coordinate expression and using ${\Gamma }^a{}_{bc}={\Gamma }^a{}_{cb}$ gives the algebraic first Bianchi identityApply the Jacobi identity to three covariant-derivative commutators, or differentiate the coordinate formula in normal coordinates. The third-derivative terms and the derivatives of quadratic connection terms cancel cyclically, giving the differential second Bianchi identityContracting the first and third curvature indices and then contracting once more yieldsor the contracted Bianchi identity

For the metric variation $h_{ab}=\delta g_{ab}$,while the two double-divergence terms in the supplied metric variation of scalar curvature become a boundary term after integration by parts for tensor fields. The Einstein-Hilbert action therefore contributes $G_{ab}+\Lambda g_{ab}$. Varying the Proca action with respect to the metric gives the Proca stress-energy tensor, and the full Einstein-Proca theory equation is

Since $\delta F_{ab}=2{\nabla }_{[a}\delta A_{b]}$, variation with respect to $A_b$ and one integration by parts give the covariant Proca equationTaking its covariant divergence givesbecause the resulting contractions pair the symmetric Ricci tensor with the antisymmetric electromagnetic field tensor. Since $m\ne 0$, the Lorenz constraint in Proca theory follows:Moreover, $F=dA$ and $d^2=0$, so the Electromagnetic Bianchi identity is

Finally, use that identity to differentiate the matter tensor. The result groups naturally asBoth terms vanish by the Proca equation and its Lorenz constraint in Proca theory. Thus ${\nabla }^a{T}_{ab}=0$, exactly as required by the contracted Bianchi identity applied to the Einstein field equations. The vector equation is therefore consistent with the gravitational equation.

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 identityDifferentiate the coordinate curvature expression and cyclically antisymmetrize. Third derivatives cancel, giving the second Bianchi identityBoth statements are tensorial and hence hold in every coordinate system. Contracting the differential identity, using the curvature symmetries and metric compatibility, yieldsthe contracted Bianchi identity.