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Expanding the two covariant derivatives of a vector, the second partial derivatives cancel in the commutator. The remaining derivatives and products of Christoffel symbols combine into the coordinate definition of the Riemann curvature tensor, giving the Ricci identityFor a type-$(2,0)$ tensor, the connection acts on both upper indices, so the curvature commutator on a contravariant tensor is

The coordinate vector fields on the parameter surface commute, so their pushforwards satisfyBecause the Levi-Civita connection is torsion free, part (i) then givesEvery $t$-curve is an affinely parametrized geodesic, hence ${\nabla }_{T}T=0$. Applying the definition of the Riemann curvature tensor,Therefore the connecting vector is a Jacobi field and obeys the geodesic deviation equation

Differentiate $B_{\alpha \beta }={\nabla }_{\beta }{X}_{\alpha }$ along $X$ and commute the two covariant derivatives:where the sign is the covector form of the Ricci identity. Apply the product rule to the first term:The first term vanishes by the geodesic equation, and the second isFinally the pair symmetries and antisymmetries of the Riemann curvature tensor giveConsequentlyThis is the Riccati equation for a timelike geodesic congruence; taking its spatial trace is the starting point for the timelike Raychaudhuri equation.

Cartan's second structure equation givesand, for $i\ne j$,With all indices lowered and the curvature convention in the question, the independent nonzero orthonormal Riemann curvature tensor components aretogether with those obtained from the Riemann symmetries.

Contract the defining formula in its first and third indices. In four dimensions,Thus

The Weyl tensor has the same algebraic symmetries as the Riemann curvature tensor: it is antisymmetric within each index pair, symmetric under exchanging the pairs, and obeys the first Bianchi identity. A contraction within an antisymmetric pair vanishes against the symmetric metric. Every contraction between the two pairs can be converted by the pair symmetries to the one just calculated, possibly with a sign. Hence every contraction vanishes, proving the Trace-free property of the Weyl tensor.

This curvature component produces radial tidal acceleration: nearby freely falling particles separated radially are stretched relative to one another, while transverse separations are compressed. At the Earth, after removing the Earth's own field, the Sun and especially the Moon produce the familiar ocean tides known since antiquity.

There is no conflict with the equivalence principle. A freely falling local frame can remove the uniform gravitational acceleration at one event by setting the connection to zero there. It cannot remove the Riemann curvature tensor, which controls relative acceleration through geodesic deviation and is detectable across an extended body such as the Earth. Tides measure this spatial variation of gravity rather than gravity at a single point.

Using Cartan's second structure equation givesSince ${\Theta }_{\mu \nu }=R_{\mu \nu \rho \sigma }{e}^{\rho }\wedge e^{\sigma }/2$, the independent nonzero lowered Riemann curvature tensor components arewith all others determined by the symmetries of the Riemann curvature tensor.

At any point choose orthonormal coordinates. The antisymmetries and pair symmetry of the Riemann curvature tensor leave only $R_{1212}$ independent. In these coordinates,while $R=2R_{1212}$. Since this is a tensor identity, it follows in every coordinate system that the Ricci tensor in two dimensions satisfiesThe Einstein tensor consequently vanishes identically. Hence the two-dimensional vacuum field equation contains no local gravitational dynamics, while coupling the unmodified equation to matter would require $T_{\alpha \beta }=0$. This is the basic degeneracy of two-dimensional general relativity.

Fix the conventionEquivalently,This defines the Riemann curvature tensor because it is $C^{\infty }(M)$-linear in $U,V,X$: derivatives of a multiplying function cancel between the three terms. Thus its value at a point depends only on the three tangent vectors there, rather than their extensions.

Using the stated convention for the Riemann curvature tensor, the first independent component isThe other isusing the hyperbolic Pythagorean identity ${\coth }^{2}r-{\mathrm{csch}}^{2}r=1$. Antisymmetry in the final two indices supplies the components with those indices reversed. These are the curvature of the two-dimensional hyperbolic polar metric components.

Contracting the Riemann curvature tensor gives the nonzero components of the Ricci tensor:The Ricci scalar is the contraction with the inverse metric. ThereforeIt is independent of both coordinates, so the hyperbolic plane has constant negative scalar curvature.

The phase isBecause the transverse-traceless gauge perturbation has no component alongonly the third term of the linearized Riemann curvature operator contributes to the required contraction:The derivatives and transverse polarization contraction areTherefore the Newman--Penrose scalar Psi4 isIts real and imaginary parts encode the plus and cross gravitational wave polarizations, up to the stated Riemann curvature tensor and complex null tetrad conventions.

Using the stated Christoffel symbols in the definition of the Riemann curvature tensor, the two nontrivial contractions areHenceand the vacuum Einstein field equations reduce toThis is the diagonal Rosen coordinates for a plane gravitational wave equation.

For the plus-polarized wave, setComparison of the transverse metric components gives, to linear order,orThereforeThe Vacuum Einstein equations are consequently satisfied at linear order. A real gravitational wave is obtained by taking the real part of the complex plane-wave notation.

Fix the conventionAlthough a single covariant derivative of a vector is tensorial, the second derivative contains connection-dependent terms. Their antisymmetric difference cancels every second derivative of a coordinate change. More intrinsically, the mapis $C^{\infty }(M)$-linear in each of $U,V,X$, so it defines the Riemann curvature tensor.

In the Levi-Civita connection, the background metric is constant and replacing $g^{\alpha \rho }$ by its $O(ϵ)$ correction would multiply a derivative of $h$ and produce an $O({ϵ}^2)$ term. Hence the linearized Levi-Civita connection isThe products of two connection coefficients in the Riemann curvature tensor are also quadratic and may be discarded. Lowering its first index with ${\eta }_{\mu \nu }$ gives

Because the coordinate change is ${\tilde{x}}^{\alpha }=x^{\alpha }-{\xi }^{\alpha }$, the perturbation changes at linear order bySubstitution into $R_{\mu \rho \alpha \beta }^{(1)}$ produces terms containing three partial derivatives of $\xi$. Since partial derivatives commute, every term cancels another with the opposite sign. ThereforeThis is the gauge invariance of the linearized Riemann tensor.

Let $\gamma :[0,\ell ]\to M$ be a unit-speed geodesic, let $F(s,t)$ be a smooth variation with fixed endpoints, and letbe its variation vector field. Then $V(0)=V(\ell )=0$. If $V^{\perp }=V-⟨V,T⟩T$ is its component normal to $\gamma$, the second variation of Riemannian arc length isHere $D_t={\nabla }_T$ is the covariant derivative along $\gamma$, $R$ is the Riemann curvature tensor, and $I$ is the Riemannian index form. Fixed endpoints remove the boundary term. The normal projection removes a tangential change of parametrization, which does not change length to second order.

The connection 1-forms are defined by $\nabla e_b={\omega }^a{}_b\otimes e_a$. Metric compatibility gives ${\omega }_{ab}=-{\omega }_{ba}$, while vanishing torsion gives Cartan's first structure equationThe curvature 2-forms arewhich displays their relation to the Riemann curvature tensor.

A Ricci-flat manifold has vanishing Ricci tensor. This is weaker than vanishing Riemann curvature tensor: the latter describes complete local flatness. A Kerr black hole supplies a Lorentzian manifold that is Ricci flat in its vacuum region but has nonzero Riemann curvature tensor.