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For smooth spatial fields $E^i,C^i{}_j$, define a torsion-free connection on $\mathbb{R}\times {\mathbb{R}}^n$ by ${\Gamma }^i{}_{00}=-E^i$, ${\Gamma }^i{}_{0j}={\Gamma }^i{}_{j0}=-C^i{}_j/2$ and all other coefficients zero. The time geodesic equation is $\ddot{t}=0$. For $\dot{t}\ne 0$, use $t$ as affine parameter; the spatial geodesic equation becomes $d^2{x}^i/d{t}^2=E^i+C^i{}_{j}d{x}^j/dt$. In dimension three, $C^i{}_j=2{ϵ}^i{}_{kj}{B}^k$ realizes $\ddot{\mathbf{x}}=\mathbf{E}+2\mathbf{B}\times \dot{\mathbf{x}}$. Such a connection need not preserve a metric; a spacetime description of these trajectories only needs affine geometry.

For a smooth one-form $A$, the components ${\partial }_a{A}_b-{\partial }_b{A}_a$ transform as a covariant rank-two tensor. The non-tensorial second derivatives in the ordinary covector derivative under nonlinear coordinate changes are symmetric in $a,b$ and cancel. For a torsion-free connection the same components are ${\nabla }_a{A}_b-{\nabla }_b{A}_a$. This is the coordinate expression of $dA$, without choosing a metric.

The average of the two flat translation connections on a Lie group is a torsion-free connection. For left-invariant vector fields, the Jacobi identity gives $R(X,Y)Z=-\frac{1}{4}[[X,Y],Z]$ with $R(X,Y)=[{\nabla }_X,{\nabla }_Y]-{\nabla }_{[X,Y]}$. Contracting gives $\mathrm{Ric}(X,Y)=-\frac{1}{4}K(X,Y)$ for the Killing form $K$. It is flat exactly when all double Lie brackets vanish, including the case of a two-step nilpotent group. For a semisimple Lie algebra, $K$ is nondegenerate and supplies a bi-invariant pseudo-Riemannian metric whose Levi-Civita connection is ${\nabla }^0$.

The canonical vorticity two-form is ${\omega }_{ab}={\nabla }_a(ϵu_b)-{\nabla }_b(ϵu_a)$, with specific enthalpy per particle $ϵ$. Potential flow $ϵu_a={\nabla }_a\chi$ has $\omega =0$ for a torsion-free connection. Contracting this identity with $u^a$ yields $ϵu^a{\nabla }_a{u}_b={\nabla }_bϵ-u_b{u}^a{\nabla }_aϵ$, the isentropic relativistic Euler equation in signature $(+---)$. This canonical condition does not mean the unweighted four-velocity one-form is closed.

The spinor covariant derivative commutator is a differential two-form in its derivative indices, so the two-component spinor calculus decomposition defines symmetric curvature operators ${\Delta }_{AB}$ and ${\Delta }_{A^{\prime }{B}^{\prime }}$. With ${ϵ}^{AB}{ϵ}_{AB}=2$, the unprimed projection is ${\Delta }_{AB}={ϵ}^{A^{\prime }{B}^{\prime }}[{\nabla }_{A{A}^{\prime }},{\nabla }_{B{B}^{\prime }}]/2$, and the primed projection is analogous. For a torsion-free connection these are algebraic curvature derivations: they annihilate scalar functions and obey the Leibniz rule on tensor products.

For a torsion-free connection, the commutator of coordinate covariant derivatives vanishes on a scalar field and acts on a vector field through the displayed Riemann curvature tensor. On a covector field it instead gives $[{\nabla }_{\alpha },{\nabla }_{\beta }]A_{\mu }=-R^{\nu }{}_{\mu \alpha \beta }{A}_{\nu }$. The curvature commutator is a tensor derivation, so each index of a general tensor contributes independently with the corresponding sign. Including the lower derivative index in the second covariant derivative is essential for this identity.

For a local basis $e_a$ of the tangent bundle, an affine connection has components defined by ${\nabla }_{e_c}{e}_b={\Gamma }^a{}_{bc}{e}_a$. If $V=V^b{e}_b$, then $({\nabla }_{e_c}V{)}^a=e_c(V^a)+{\Gamma }^a{}_{bc}{V}^b$. The connection components depend on the chosen basis and have an inhomogeneous transformation law; they are not themselves a tensor field. In a coordinate basis, a torsion-free connection has symmetry in the two lower indices.

For a geodesic variation $F(s,\tau )$, the deviation vector $S=F_{\ast }{\partial }_s$ describes infinitesimal separation at equal parameter values. With $T=F_{\ast }{\partial }_{\tau }$, commuting parameter derivatives give $[S,T]=0$. A torsion-free connection then gives ${\nabla }_{T}S={\nabla }_{S}T$. For an affinely parametrized variation, geodesic deviation states ${\nabla }_T^{2}S=R(T,S)T$ with the convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$.

A Riemannian metric determines a unique torsion-free connection that is a metric connection. Metric compatibility and the torsion identity imply the Koszul formula, which proves uniqueness. Conversely the right-hand side of the Koszul formula is a smooth one-form in its last vector-field argument. The musical isomorphism defines $D_{X}Y$ from it; expanding the Lie bracket of vector fields verifies the connection rules, metric compatibility and zero torsion. Thus it constructs the Levi-Civita connection intrinsically.

For smooth stationary fields on ${\mathbb{R}}^3$, set ${\Gamma }^i{}_{tt}=-E^i$ and ${\Gamma }^i{}_{tj}={\Gamma }^i{}_{jt}={ϵ}^i{}_{jk}{B}^k$, with all other Christoffel symbols zero. This torsion-free connection makes the equation $\ddot{\mathbf{x}}=\mathbf{E}+2\mathbf{B}\times \dot{\mathbf{x}}$ a geodesic equation with time as affine parameter. Arbitrary smooth fields suffice; a metric realization imposes additional potential conditions.

For $\tilde{g}=e^{2\sigma }g$, the displayed correction is a symmetric tensor, so adding it to $\nabla$ gives a torsion-free connection. The two extra metric-compatibility pairings sum to $2d\sigma (X)g(Y,Z)$; this is exactly the derivative of the conformal factor. Uniqueness of the Levi-Civita connection proves the formula. It also yields the critical-point criterion for equal conformal connections.

For any torsion-free connection, expand the coordinate Lie derivative of a tensor field and cancel the symmetric connection terms to obtain the displayed identity. If the connection is metric-compatible, lowering the vector index commutes with the covariant derivative and it becomes ${\mathcal{L}}_{\xi }{g}_{ab}=2{\nabla }_{(a}{\xi }_{b)}$. Symmetry of the connection alone does not suffice: with $g=e^{2x}d{x}^2$, zero connection coefficients and $\xi ={\partial }_x$, the two expressions ${\mathcal{L}}_{\xi }g$ and $2{\nabla }_x{\xi }_x$ are $2e^{2x}$ and $4e^{2x}$, respectively.

For a torsion-free connection, the Lie derivative of a metric tensor obeysThus the abbreviated formula ${\mathcal{L}}_{K}g=2{\nabla }_{(a}{K}_{b)}$ requires metric compatibility. The familiar Killing equation and second covariant derivative of a Killing vector use the Levi-Civita connection, which has both vanishing torsion tensor and metric compatibility. On Minkowski spacetime with metric signature $(-+++)$, a torsion-free connection with only ${\Gamma }^0{}_{00}=1$ supplies a counterexample: $K={\partial }_0$ is a Killing vector field for the constant Minkowski metric, but ${\nabla }_0{K}_0=1$.

For smooth spatial $\varphi$ and $g(c)=-c^2{e}^{2\varphi /c^2}d{t}^2+{\delta }_{ij}d{x}^{i}d{x}^j$, the only nonzero Christoffel symbols are ${\Gamma }^t{}_{ti}={\Gamma }^t{}_{it}=c^{-2}{\partial }_i\varphi$ and ${\Gamma }^i{}_{tt}=e^{2\varphi /c^2}{\partial }_i\varphi$. Consequently the Levi-Civita connection converges smoothly on compact sets as $c\to \infty$ although the metric itself diverges in these coordinates. The limiting torsion-free connection has only ${\Gamma }^i{}_{tt}={\partial }_i\varphi$ and gives $d^2{x}^i/d{t}^2=-{\partial }_i\varphi$ when $t$ is an affine parameter. With $R^a{}_{bcd}={\partial }_c{\Gamma }^a{}_{db}-{\partial }_d{\Gamma }^a{}_{cb}+\cdots$, its Ricci tensor is $(\Delta \varphi )dt\otimes dt$: quadratic connection terms vanish and $R^i{}_{tjt}={\partial }_j{\partial }_i\varphi$. Thus zero Ricci tensor is the Laplace equation, while full flatness would require the entire Hessian to vanish.

Use metric signature $(-,+,+,+)$, $x^0=ct$, and retain terms linear in $\Phi /c^2$ or $\Psi /c^2$. The inverse metric is $g^{00}=-(1-2\Phi /c^2)$ and $g^{ij}=(1+2\Psi /c^2){\delta }^{ij}$. In the Christoffel symbol formulathe differentiated metric perturbation is already first order, so the prefactor can be replaced by the flat inverse metric. For example ${\Gamma }^0{}_{ij}=\frac{1}{2}{\partial }_0{g}_{ij}=-{\delta }_{ij}{\partial }_0\Psi /c^2$ and ${\Gamma }^k{}_{00}=-\frac{1}{2}{\partial }_k{g}_{00}={\partial }_k\Phi /c^2$. Evaluating the remaining components givesThe lower-index symmetry follows from the torsion-free connection. Products of these first-order Christoffel symbols are beyond the retained weak-field approximation.

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 paper uses $T_{\mu \nu }{}^{\lambda }={\Gamma }_{\nu \mu }{}^{\lambda }-{\Gamma }_{\mu \nu }{}^{\lambda }$, the negative of a common torsion tensor convention. Keep that sign throughout. Also, ${\nabla }_{\mu }{\nabla }_{\nu }\phi$ denotes the covariant Hessian, so the second derivative acts on the lower index $\nu$ of the covector ${\nabla }_{\nu }\phi$; it is not just a composition of two directional derivatives on a scalar.

Since the covariant derivative of a scalar field is its ordinary derivative,Commuting the ordinary partial derivatives therefore givesFor a vector field, write $D_{\nu }{X}^{\lambda }={\partial }_{\nu }{X}^{\lambda }+{\Gamma }_{\nu \rho }{}^{\lambda }{X}^{\rho }$. This is a tensor with one covariant and one contravariant index. HenceSubtracting the expression with $\mu ,\nu$ exchanged, the first two terms combine toThe coefficients of first derivatives of $X$ in those two terms cancel. The last terms give $({\Gamma }_{\nu \mu }{}^{\sigma }-{\Gamma }_{\mu \nu }{}^{\sigma })D_{\sigma }{X}^{\lambda }$. Identifying the curvature coefficient yieldsHere the bracket denotes antisymmetrization of the tensorial second covariant derivative, consistently with the preceding calculation.

To check tensoriality directly, antisymmetrize the supplied transformation law of the connection coefficients in the two lower indices. The inhomogeneous term vanishes because mixed coordinate second derivatives are symmetric. The remaining transformation law iswhich is precisely that of a $(1,2)$ tensor. Intrinsically, with the paper's sign,The derivative terms involving the components of $X,Y$ cancel against their Lie bracket, leaving $({\Gamma }_{\nu \mu }{}^{\lambda }-{\Gamma }_{\mu \nu }{}^{\lambda })X^{\mu }{Y}^{\nu }$. This also shows that the expression is linear over smooth functions in each argument.

Finally take the torsion-free connection and the commuting parameter fields $Y,Z$ along the geodesic family. Vanishing torsion gives ${\nabla }_{Y}Z={\nabla }_{Z}Y$, while affine geodesic parametrization gives ${\nabla }_{Y}Y=0$. The curvature operator in the specified convention isThereforeIn components this is the geodesic deviation equationThe connecting field represents infinitesimal separation within the family, so this is a linear equation for relative free-fall acceleration, rather than an exact coordinate difference for widely separated worldlines.

For a contravariant vector field $A^{\lambda }$, the covariant derivative isThe Levi-Civita connection is the unique torsion-free connection satisfying metric compatibility. Its Christoffel symbols in a coordinate basis areThere is generally no symmetry interchanging the upper index with a lower one. The Christoffel symbols themselves are not the components of a tensor.

An ordinary partial derivative of the components of a vector field includes the apparent change caused by the varying coordinate basis. The connection term corrects for that change, so ${\nabla }_{\nu }{A}^{\lambda }$ obeys the tensor component transformation law. It gives coordinate-independent ways to express parallel transport, geodesics and stress-energy conservation in curved spacetime.

An affine connection on the tangent bundle of a smooth manifold assigns a vector field ${\nabla }_{X}Y$ to two vector fields, is $\mathbb{R}$-bilinear, and satisfiesThus its first slot is tensorial and its second obeys the Leibniz rule. Its action on covectors is defined by differentiating the pairing, and this extends it to tensor fields. A torsion-free connection has ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$.

Assume that $g$ is a smooth nondegenerate metric tensor. Metric compatibility and a vanishing torsion tensor imply the Koszul formula,The right-hand side is determined by $g$ and the Lie bracket of vector fields. Nondegeneracy determines ${\nabla }_{X}Y$ uniquely; conversely this formula defines an affine connection with both required properties, proving existence as well as uniqueness. In a coordinate frame the brackets vanish. Combining the three differentiated metric tensor identities givesThis is the Levi-Civita connection. Positive definiteness is unnecessary: the argument also works for a pseudo-Riemannian metric.

For the other affine connection define $S(X,Y)={\bar{\nabla }}_{X}Y-{\nabla }_{X}Y$. The derivative of $f$ cancels between the two Leibniz rules, so $S$ is $C^{\infty }$-linear in both slots. The difference of affine connections is a tensor: here $S$ is a section of $TM\otimes T^{\ast }M\otimes T^{\ast }M$ with components ${\bar{\Gamma }}^a{}_{bc}-{\Gamma }^a{}_{bc}$. Both affine connections are torsion-free, so $S(X,Y)=S(Y,X)$.

Agreement of geodesics means agreement of their unparametrized curves; different affine parameters are allowed. For every nonzero tangent vector $v$, existence of a geodesic with that initial velocity implies that $S(v,v)$ is parallel to $v$. The symmetric bilinear diagonal-parallel lemma now determines $S$. Choose a basis $e_i$ and write $S(e_i,e_i)={\alpha }_i{e}_i$. Applying the parallelism condition to $e_i+e_j$ and $e_i-e_j$ givesWith $V(e_i)={\alpha }_i/2$, bilinearity yieldsThis is projective equivalence of affine connections. It also covers dimension one. For two Levi-Civita connections, the projective covector from metric volume densities gives the explicit answerIndeed ${\Gamma }^a{}_{ac}={\partial }_c\log \sqrt{|\det g|}$. The determinant ratio is a scalar, so this expression is a genuine covector, although either determinant alone is coordinate-dependent. This determinant expression requires nondegeneracy of $\bar{g}$; the difference-tensor and earlier trace formula require only the two torsion-free affine connections.

Conversely, the displayed tensor gives ${\bar{\nabla }}_{\dot{\gamma }}\dot{\gamma }=2V(\dot{\gamma })\dot{\gamma }$ along an affinely parametrized $\nabla$-geodesic. Choosing a new parameter $t(\tau )$ satisfying $t^{\prime \prime }/t^{\prime }=2V(\dot{\gamma })$ removes that tangential acceleration. This geodesic reparametrization under projective equivalence proves sufficiency. If agreement were required with the very same affine parameter, polarization would instead force $S=0$ and $V=0$.