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Although ${\nabla }_{\beta }f={\partial }_{\beta }f$ for a scalar field, its first derivative is a covector field. ThereforeCommuting partial derivatives and using the torsion-free connection gives $\boxed{[{\nabla }_{\alpha },{\nabla }_{\beta }]f=0}$. For a vector field, differentiate the $(1,1)$ tensor ${\nabla }_{\beta }{V}^{\mu }$ with both its indices included:Subtract the expression with $\alpha ,\beta$ exchanged. Second derivatives of $V$ cancel; the terms multiplying first derivatives cancel pairwise, and the final lower-index connection terms cancel by symmetry. The remaining coefficient of $V^{\nu }$ isThis is the curvature commutator. On a covariant index the sign reverses, giving the curvature commutator on a covariant tensor used in the next part.

For a torsion-free connection, the connection terms in the antisymmetrized covariant derivative cancel: $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$. Thus the electromagnetic field tensor is the exterior derivative of the four-potential. It is a tensor, is invariant under $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$, and obeys the homogeneous Maxwell equations ${\nabla }_{[\alpha }{F}_{\beta \gamma ]}=0$. The covariant divergence equation is the generally covariant form of the local source equation; use its minus sign as specified in the paper.

Expand that equation before imposing the Lorenz gauge condition:The curvature commutator on a covariant tensor givesHere the contraction $g^{\mu \rho }{R}^{\alpha }{}_{\mu \rho \nu }=-R^{\alpha }{}_{\nu }$ accounts for the second sign reversal. Hence in Lorenz gaugeThis curved-spacetime electromagnetic potential wave equation contains a Ricci tensor term even in a locally inertial coordinate system: a connection can vanish at one point while its derivatives and curvature remain nonzero.

The Killing equation gives ${\nabla }_{\mu }{K}^{\mu }=0$ and ${\nabla }_{[\mu }{K}_{\lambda ]}={\nabla }_{\mu }{K}_{\lambda }$. Commuting derivatives once and contracting the Ricci identity givesTherefore, using the contracted Bianchi identity and symmetry of $R_{\mu \beta }$,Since ${\nabla }_{[\mu }{K}_{\lambda ]}$ is antisymmetric, swapping the two contracted index names shows that the left side equals its derivative-antisymmetrized form:Now set $A_{\mu \lambda }={\nabla }_{[\mu }{K}_{\lambda ]}$. The right side is the contracted commutator $[{\nabla }^{\mu },{\nabla }^{\lambda }]A_{\mu \lambda }$. By the curvature commutator on a covariant tensor, it is a sum of contractions of the symmetric Ricci tensor with the antisymmetric tensor $A$, and hence vanishes. Thusso the Killing vector preserves scalar curvature.

Use the sequential convention ${\phi }_{;lmn}={\nabla }_n{\nabla }_m{\nabla }_l\phi$ and define $R$ by $X^a{}_{;mn}-X^a{}_{;nm}=R^a{}_{bmn}{X}^b$. The symmetric Hessian of a scalar field gives symmetry in $l,m$, but the curvature commutator on a covariant tensor gives ${\phi }_{;l[mn]}=-R^p{}_{lmn}{\phi }_{;p}/2$. Averaging the three distinct permutations yields ${\phi }_{;(lm)n}={\phi }_{;(lmn)}-(R^p{}_{lmn}+R^p{}_{mln}){\phi }_{;p}/3$. Thus a scalar's third covariant derivative is not generally totally symmetric.