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Contract the printed Riemann curvature tensor using $R_{ab}=R^{\mu }{}_{a\mu b}$. Only the derivative terms can contain second derivatives of the metric tensor:Substituting the Christoffel symbol formula, its second derivative part isSet $H_b=g_{b\lambda }{H}^{\lambda }=g^{\mu \nu }{\partial }_{\mu }{g}_{\nu b}-\frac{1}{2}{g}^{\mu \nu }{\partial }_b{g}_{\mu \nu }$. The first, second and fourth terms are the second derivative terms of $\frac{1}{2}({\partial }_a{H}_b+{\partial }_b{H}_a)$. Derivatives of the inverse metric tensor produce only products of first derivatives. Thuswhere $Q$ contains at most first metric derivatives. In harmonic coordinates, $H^{\lambda }$ vanishes throughout the chart, so $H_b$ and its derivatives vanish. The vacuum equations have the wave principal partThese are quasilinear partial differential equations with the principal part of a wave equation, because $g^{\mu \nu }$ has Lorentzian signature. This is the harmonic reduction of the vacuum Einstein equations.

At any chosen point take normal coordinates for $g$, so $\Gamma =0$ there. Substitute $\tilde{\Gamma }=\Gamma +T$ into the stated Riemann curvature tensor formula and subtract the original curvature. At that point the mixed products with $\Gamma$ vanish and ordinary derivatives of $T$ equal its covariant derivatives. The result isIn unit-weight antisymmetrization notation this is $\boxed{2T^a{}_{b[d;c]}-2T^a{}_{e[d}{T}^e{}_{c]b}}$. Every term is a tensor, so the identity established in normal coordinates holds in every coordinate system.

Use the Riemann curvature tensor convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. A Jacobi field is a smooth vector field $J$ along $\gamma$ satisfyingwhere $D_t$ is the covariant derivative along a curve for the Levi-Civita connection. Differentiating a geodesic variation gives this equation, and conversely every Jacobi field arises from a geodesic variation by varying its initial point and velocity.

Choose a parallel frame $E_1(t),\dots ,E_n(t)$ along $\gamma$ and write $J={\sum }_a{y}_a{E}_a$. The Jacobi equation becomes the linear systemThe existence and uniqueness theorem for linear ordinary differential equations says that each pair $(y(0),y^{\prime }(0))\in {\mathbb{R}}^n\oplus {\mathbb{R}}^n$ determines exactly one solution on $[0,1]$. Addition and scalar multiplication preserve the equation. Thus evaluation of initial position and covariant velocity is a linear isomorphismand the dimension is

For the convexity assertion, use the convex-normal meaning of a geodesically convex open set: its points are joined by a unique geodesic within the set, and the joining geodesic depends smoothly on its endpoints. Equivalently, the appropriate star-shaped restriction of ${\exp }_p$ is a diffeomorphism onto the set. The convex-normal-neighbourhood theorem supplies such sets around every point. Mere existence of some minimizing geodesic, without this uniqueness and normality, would not imply the assertion.

Let $p=\gamma (0)$ and $v=\dot{\gamma }(0)$. For a Jacobi field with $J(0)=0$ and $D_{t}J(0)=w$, differentiating ${\exp }_p(t(v+sw))$ gives the standard differential-of-the-exponential map identityIn a convex normal neighbourhood, $(d{\exp }_p{)}_v$ is invertible. Therefore a Jacobi field vanishing at both endpoints has $w=0$ and hence is identically zero. The difference of two fields with the same endpoint values consequently vanishes. More precisely, the mapis an isomorphism: it is injective and both spaces have dimension $2n$. The constant-geodesic case follows directly from $D_t^{2}J=0$. This uses the convex-normal-neighbourhood theorem, the differential-of-the-exponential map identity, and linear ordinary differential equation uniqueness.

For the special unitary group example, use Jacobi fields from conjugation at a central endpoint. WriteThe endpoints of $\gamma (t)=\exp (tA)$ are $I$ and $zI$. The latter is central. For every $X\in \mathfrak{su}(3)$ considerConjugation is an isometry for a bi-invariant Riemannian metric, and geodesics of a bi-invariant metric are one-parameter subgroups. Thus $F_X$ is a geodesic variation. Its Jacobi field iswhich vanishes at $t=0$ and $t=1$ because both endpoints are central. Its initial covariant derivative is $[X,A]$.

The linear map $X\mapsto J_X$ has kernel equal to the centralizer of an element of a Lie algebra of $A$. Indeed, $J_X\equiv 0$ implies $[X,A]=0$ on differentiating at $t=0$, and that commutation conversely implies $J_X\equiv 0$. The repeated first two eigenvalues giveAs ${\dim }_{\mathbb{R}}\mathfrak{su}(3)=8$, rank-nullity theorem givesExplicit independent generators are the fields belonging to $E_{13}-E_{31}$, $i(E_{13}+E_{31})$, $E_{23}-E_{32}$ and $i(E_{23}+E_{32})$. Their initial derivatives $[X,A]$ are independent because the eigenvalue differences in the $13$ and $23$ entries are $2\pi i$. This also demonstrates directly why the endpoint-value conclusion fails along this geodesic.

For the Vacuum Einstein equations with zero cosmological constant, $R_{\alpha \beta }=0$, and its tensor contraction gives $R=0$. Every Ricci tensor and scalar curvature term in the Weyl tensor consequently disappears:Thus the Weyl tensor contains all vacuum curvature. Here vacuum means Ricci-flat; a nonzero cosmological constant would retain a constant-curvature contribution to the Riemann curvature tensor.

The electric part of the Weyl tensor has one normal in each antisymmetric pair of the Weyl tensor. ThereforeFor the magnetic part of the Weyl tensor, contracting the first slot gives $n^{\alpha }{n}^{\lambda }{ϵ}_{\alpha \lambda \mu \nu }=0$ by antisymmetry of the metric volume tensor; contracting the second gives $n^{\beta }{n}^{\rho }{C}^{\mu \nu }{}_{\beta \rho }=0$ by antisymmetry of the last curvature pair. Consequently both tensors are entirely spatial:These spatial projection tensor identities actually hold for any Weyl tensor; the vacuum assumption is needed subsequently to replace it by the Riemann curvature tensor.

Put $A,B\in \{1,2\}$ and write $h_{AB}=H_{AB}{e}^{ik\cdot x}$, where $H_{11}=H_{+}$, $H_{22}=-H_{+}$ and $H_{12}=H_{21}=H_{\times }$. Physical fields are the real parts. Use $c=1$ here, and take the initial separation at equal time, so $L^0=0$. For propagation in the positive $x^3$ direction one may set $k_{\mu }=(-\omega ,0,0,\omega )$; only $k_0^2={\omega }^2$ enters the answer.

The linearized Levi-Civita connection in this transverse-traceless gauge has the potentially nonzero componentswith all others zero to this order. In particular ${\Gamma }^{\mu }{}_{00}=0$, so a particle initially at rest has fixed spatial coordinates in transverse-traceless gauge, and $t$ equals its proper time. Expanding the Riemann curvature tensor givesThe plus sign in the last expression agrees with $R(T,S)T$ in the geodesic deviation convention above.

For the separation measured in a parallel-propagated orthonormal frame, write $S^{\hat{\mu }}=L^{\hat{\mu }}+{\xi }^{\hat{\mu }}$. The frame removes the connection terms from the time derivative, so the geodesic deviation equation reduces to ${\ddot{\xi }}^{\hat{A}}=R^{\hat{A}}{}_{\hat{0}\hat{0}\hat{B}}{L}^{\hat{B}}={\ddot{h}}_{AB}{L}^{\hat{B}}/2$ at first order. The observable transverse relative acceleration isThese are the plus polarization and cross polarization tidal patterns; there is no longitudinal acceleration. The curvature is evaluated along the reference worldline, and replacing $S$ by $L$ on the right changes only second-order terms. This is the infinitesimal-separation approximation intrinsic to geodesic deviation.

There is an important component distinction if the symbols $L^{\mu }+{\xi }^{\mu }$ are instead interpreted literally in the transverse-traceless gauge coordinate basis. Expanding the left side, rather than replacing a covariant derivative by an ordinary derivative, givesThus in TT coordinatesWith zero initial coordinate displacement and velocity, ${\xi }_{\mathrm{coord}}=0$ throughout. This does not remove the physical gravitational wave strain. To first order an orthonormal coframe in spacetime along the reference worldline is ${\vartheta }^{\hat{A}}=({\delta }^A{}_B+h_{AB}/2)d{x}^B$; its dual frame is parallel propagated because ${\partial }_0({\delta }^A{}_B-h_{AB}/2)+{\Gamma }^A{}_{0B}=0$. Henceafter choosing the initial flat-frame reference. The TT coordinate separation and measured separation distinction accounts exactly for the apparently different accelerations. In a region initially free of the wave, the measured displacement is ${\xi }^{\hat{A}}=h_{AB}{L}^{\hat{B}}/2$ with the corresponding initial conditions.

Let $\delta h_{\mu \nu }={\partial }_{\mu }{V}_{\nu }+{\partial }_{\nu }{V}_{\mu }$. Substitution into the linearized Levi-Civita connection cancels four terms and leavesTo first order in the metric perturbation, the products of Christoffel symbols are quadratic and may be discarded. Therefore the change in the Riemann curvature tensor isThis is the gauge invariance of the linearized Riemann tensor. Lowering the first index with the Minkowski metric does not change the conclusion; the change from using the full metric tensor instead is of second order because the background curvature is zero. HenceThe result is invariance at linear order about flat spacetime, not componentwise invariance of arbitrary nonlinear curvature under finite coordinate changes. Directly in the four-term formula, the eight third derivatives of $V$ cancel in pairs by commuting partial derivatives.

Use $t$ as an affine parameter on each member of the geodesic variation, so ${\nabla }_{T}T=0$. The parameter directions commute, giving $[T,S]=0$ along the variation. The torsion-free connection then givesConsequently the covariant derivative and curvature definition giveThis is the geodesic-deviation equation with the stated curvature sign:The component expression defines the slot order: $\beta$ is the vector acted on, and $\gamma ,\delta$ are the two derivative directions of the Riemann curvature tensor. Both the commuting parameter directions and affine parametrization are essential; a general connecting vector field or a nonaffine parameter would add terms.

For a closed universe, the spatial Cauchy hypersurface is compact without boundary. There is consequently no asymptotic surface term in the gravitational Hamiltonian. On solutions, the Hamiltonian constraint and momentum constraint giveThis is the vanishing canonical Hamiltonian of a closed universe: evolution generated by lapse and shift is a gauge transformation, and there is no asymptotic time-translation charge. It does not mean the local stress-energy tensor or Riemann curvature tensor is zero. In an asymptotically flat universe the same bulk constraint equations in field theory hold, but a boundary Arnowitt-Deser-Misner energy can be nonzero.

An affine connection on $M$ is an operation $(Y,X)\mapsto {\nabla }_{Y}X$ on smooth vector fields that is real-bilinear, linear over $C^{\infty }(M)$ in the differentiating field, and satisfiesIt is a connection on a vector bundle for the bundle $TM$, also called a linear connection on the manifold. Its curvature form of a connection has the operator formThis convention fixes its sign. Expanding the two connection product rules shows that the extra derivatives of a scalar cancel, so $R$ is linear over smooth functions in all three fields. For a Levi-Civita connection, this is the Riemann curvature tensor.

The torsion form is $T(Y,Z)={\nabla }_{Y}Z-{\nabla }_{Z}Y-[Y,Z]$. For a Riemannian metric $g$, its Levi-Civita connection is the unique torsion-free connection satisfying metric compatibilityExistence and uniqueness follow from the Koszul formula. In coordinates its coefficients areSymmetry in $b,c$ gives zero torsion, and substitution verifies metric compatibility.

At $(p,q)$, choose neighborhoods on which the two factors admit local isometries to open subsets of Euclidean space. Their product is a local isometry of the product Riemannian metric to the standard Euclidean metric on ${\mathbb{R}}^{2n}$. In these product coordinates the metric matrix is constant, so every Christoffel symbol of the Levi-Civita connection is zero. Its Riemann curvature tensor therefore vanishes at every point of that neighborhood. Curvature has tensoriality, so this is independent of the chosen coordinates; the neighborhoods cover the product.

The splitting provides another useful direct check. Expanding the definition on lifted fields, using zero mixed derivatives and commuting opposite-factor fields, gives the curvature splitting for a product connectionTensoriality extends this identity from lifts to arbitrary tangent vectors at a point. Each factor curvature is zero under the stated local-isometry hypothesis. ConsequentlyThis is a local conclusion about the connection; no global identification of either factor or their product with Euclidean space is required.

Fix the Riemann curvature tensor conventionFor the limiting affine connection, every product of two nonzero connection coefficients is zero, because its only nonzero coefficients have spatial upper index and both lower indices $t$. Its only possibly nonzero curvature components, apart from antisymmetry in the last two indices, areThe Ricci tensor is thereforeAll other components vanish, so the limiting connection is a connection with vanishing Ricci tensor exactly when $\varphi$ satisfies the Laplace equation on spatial Euclidean space. This is Ricci-flatness, not necessarily vanishing curvature: $\varphi =x^2-y^2$ is a harmonic function but has nonzero Hessian matrix and hence a nonflat limiting connection. Reversing the curvature sign convention reverses the displayed Ricci sign but leaves the equivalence unchanged.

Let $E_a$ denote the frame dual to the coframe. For the convention just stated, the Ricci tensor is obtained from the curvature 2-forms through the Ricci one-forms ${\mathcal{R}}_b={\sum }_a{\iota }_{E_a}{\Theta }^a{}_b=R_{bd}{e}^d$. In particular ${\iota }_{E_1}(\ell \wedge e^1)=-\ell$ and likewise for index two. It follows thatThusThe scalar curvature vanishes even before imposing vacuum because $du$ is null, but the full tensor vanishes exactly whenThis is the Vacuum Einstein equations in this family. The nonzero-coframe hypothesis makes division by $A$ and $B$ legitimate. A nonzero $P=-Q$ can leave the Riemann curvature tensor nonzero even in vacuum.

In Cartesian coordinates on Euclidean space, the Levi-Civita connection and the Riemann curvature tensor vanish. The second covariant derivative of a Killing vector therefore givesEvery first partial derivative is constant on the connected space ${\mathbb{R}}^n$. Integrating once more gives $K^{\nu }=A^{\nu }+M_{\mu }{}^{\nu }{x}^{\mu }$, with constant coefficients. The Killing equation now becomesThus the coefficient is an antisymmetric matrix. Conversely, any such constant coefficients satisfy the Killing equation. The complete Euclidean Killing vector field is thereforeThe $n$ translation parameters and $n(n-1)/2$ rotation parameters give $n(n+1)/2$ independent Killing vector fields.

Let $D_{abc}={\nabla }_a{\nabla }_b{K}_c$. Differentiating the Killing equation gives $D_{abc}=-D_{acb}$. For a covector, the Ricci identity isCall this difference $C_{abc}$. The differentiated Killing equation then givesTaking the first minus the second plus the third yieldsThe pair symmetries of the Riemann curvature tensor and the first Bianchi identity reduce the coefficient to $2R_{cbad}$. Therefore the second covariant derivative of a Killing vector is

For an affine parameter $\lambda$ on a geodesic, its tangent obeys $V^b{\nabla }_b{V}^a=0$. ThusThe second term vanishes by the geodesic equation; the first contracts a symmetric product $V^a{V}^b$ with the antisymmetric derivative from the Killing equation. This establishes the geodesic conserved quantity from a Killing vector:

Use the curvature convention $[{\nabla }_b,{\nabla }_c]V^d=R^d{}_{abc}{V}^a$, consistent with the coordinate formula later in the paper. Choose normal coordinates at an arbitrary point $p$. The Levi-Civita connection has ${\Gamma }^d{}_{ab}(p)=0$ and is torsion free, so ${\Gamma }^d{}_{ab}={\Gamma }^d{}_{ba}$. At $p$,Hence the cyclic sum isEach derivative cancels using symmetry of the two lower connection indices. Because the Riemann curvature tensor is antisymmetric in its last two indices, full antisymmetrization over $a,b,c$ equals one third of this cyclic sum. ThusThe statement is tensorial and $p$ was arbitrary, so it holds in all coordinates everywhere. This is the first Bianchi identity.

Put $h_{ab}=\delta g_{ab}$, $h=g^{ab}{h}_{ab}$, and $h^{ab}=g^{ac}{g}^{bd}{h}_{cd}$. Throughout this part use the Riemann curvature tensor convention printed in the paper. Varying $g^{ac}{g}_{cb}={\delta }^a{}_b$ gives the inverse-metric variation. The determinant identity $\delta \log |\det g|=g^{ab}{h}_{ab}$ gives the volume form variation. Together these areIn Lorentzian coordinates the latter reads $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}h$.

For the connection, vary metric compatibility ${\nabla }_a{g}_{bc}=0$:Add the equations with derivatives $a,b$, subtract that with derivative $c$, and use symmetry of the lower indices of the Levi-Civita connection. This givesThe difference of two connections is a tensor, so this formula is covariant.

Varying the coordinate curvature formula yields the Palatini identityThe contractions of the connection variation areFinally, $\delta R=(\delta g^{ab})R_{ab}+g^{ab}\delta R_{ab}$ gives the metric variation of scalar curvature:The double divergence may equivalently be written ${\nabla }_a{\nabla }_b{h}^{ab}$: relabeling the contracted dummy indices and using symmetry of $h^{ab}$ gives the same expression. The derivative terms are a covariant divergence, useful when varying a curvature-dependent action.

For a local parametrization $\varphi (u,v)$, write the first fundamental form and second fundamental form asThe Gaussian curvature is

To prove the Theorema Egregium, set $g_{ij}=⟨{\varphi }_i,{\varphi }_j⟩$ and decomposeDifferentiating $g_{ij}$ and solving for the tangential coefficients givesso every Christoffel symbol depends only on $E,F,G$ and their first derivatives. Equality of mixed third derivatives yields the Gauss equationThe Riemann curvature tensor is formed from first derivatives and quadratic products of the Christoffel symbols, and hence from the first and second derivatives of $E,F,G$. Consequentlydepends only on the first fundamental form.

Contract the Gauss equation with $h^{ac}{h}^{bd}$. Since $h^{ab}=g^{ab}+n^a{n}^b$ and the antisymmetries of the Riemann curvature tensor eliminate the term containing four normals,The extrinsic terms contract to $K^2$ and $K_{ab}{K}^{ab}$. Hence the intrinsic Ricci scalar is

Let $D$ be the Levi-Civita connection of $h$. Apply $(D_c{D}_d-D_d{D}_c)$ to a tangent vector, replace each $D$ by the tangential projection of $\nabla$, and decompose $\nabla n$ using $K_{ab}$. Comparison with the intrinsic Riemann curvature tensor gives the Gauss equationChanging the sign convention for $K_{ab}$ does not alter these quadratic terms.