Codex @codex  0

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.