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Spherical symmetry makes moments over a ball isotropic tensors. Odd moments vanish under $x\mapsto -x$, while the second and fourth moments areFor the second moment, a tensor contraction gives $3\alpha =\int r^{2}dV=4\pi a^5/5$. For the fourth, complete index symmetry makes the three coefficients equal and a double tensor contraction gives $15\beta =\int r^{4}dV=4\pi a^7/7$. These identities turn polynomial volume integrals into scalar radial integrals and are useful for inertia and rotational averages.

In Cartesian coordinates on four-dimensional Minkowski spacetime, all Conformal Killing vector fields have the formTheir conformal Killing factor is $c+2b\cdot x$. There are four translation generators, six Lorentz transformation generators, one dilation generator, and four special conformal transformation generators. These form a fifteen-dimensional Lie algebra under the Lie bracket of vector fields.

For completeness, differentiate the Conformal Killing equation in three index orders to obtainIts divergence gives $2{\partial }_{\mu }{\partial }_{\nu }\lambda =-{\eta }_{\mu \nu }\square \lambda$; its tensor contraction gives $\square \lambda =0$. Hence $\lambda$ is an affine function. Subtracting the displayed quadratic and dilation terms leaves a Killing vector field whose second partial derivatives vanish, proving completeness of the list. The vector fields are globally smooth, but some special conformal transformations have incomplete flows on affine Minkowski spacetime; the dilation flow $x\mapsto e^{s}x$ is global.

For a Conformal Killing vector field $X$ on an $n$-dimensional pseudo-Riemannian manifold, the conformal Killing factor is the scalar field $\lambda$ in ${\mathcal{L}}_{X}g=2\lambda g$. For the Levi-Civita connection, taking the tensor contraction of the Conformal Killing equation gives $\lambda ={\nabla }_{\mu }{X}^{\mu }/n$. A Killing vector field has $\lambda =0$, while the dilation generator $x^{\mu }{\partial }_{\mu }$ on Minkowski spacetime has $\lambda =1$.

A tensor contraction over two paired indices. For real second-order tensors, it gives the displayed Frobenius inner product. For a fourth-order elasticity tensor and a second-order infinitesimal strain tensor, $(C:ϵ{)}_{ij}={\sum }_{kl}{C}_{ijkl}{ϵ}_{kl}$ remains a second-order tensor; double contraction need not always give a scalar.

A dummy index is summed over its full specified range. Its name can be changed consistently without changing the expression, as in $A_{ij}{B}_{ij}=A_{ji}{B}_{ji}$. A free index specifies a component and cannot be renamed in only one term of an equation. Relabelling dummy indices is useful for tensor contraction arguments involving a symmetric tensor and an antisymmetric tensor.

A smooth endomorphism $A$ of the tangent bundle defines the tensor derivation $D_A$ by $D_{A}f=0$ and $D_{A}Y=A(Y)$. On a differential one-form, $D_A\omega =-\omega \circ A$. On a general tensor field, it acts by $A$ in each vector factor and by the negative dual action in each covector factor. The two actions cancel in each contracted pairing, proving compatibility with tensor contraction.

For the local flow ${\varphi }_t$ of a smooth vector field $X$, pull a tensor field $T$ back to the fixed base point before differentiating. Pullback uses $d{\varphi }_t^{-1}$ on vector factors and the dual of $d{\varphi }_t$ on covector factors. The chain rule gives ${\mathcal{L}}_{X}f=Xf$. In coordinates, $d{\varphi }_t^{-1}Y({\varphi }_t(x))=Y+t(dYX-dXY)+O(t^2)$, so ${\mathcal{L}}_{X}Y=[X,Y]$. Differentiation of pullback preserves tensor contractions and obeys the Leibniz rule for tensor products; hence this is the tensor derivation defining the Lie derivative of a tensor field.

Each Killing vector field supplies a conserved quantity for freely moving test particles: it turns a geometric symmetry into a first integral of the geodesic equation. Let $\gamma (\lambda )$ be a geodesic with affine parameter $\lambda$, tangent vector $U^{\mu }=d{x}^{\mu }/d\lambda$, and ${\nabla }_{U}U=0$. Wherever a Killing vector field $K$ is defined along the geodesic,The statement applies to Riemannian manifolds and to timelike geodesics, null geodesics and spacelike geodesics in a Lorentzian manifold. It requires an affine parameter; for a massive test particle, proper time is one choice. See Tong, section 4.3.2, equation (4.32).

For a worked example in the Euclidean plane, take $\gamma (t)=(1+t,2-t)$ with affine parameter $t$ and velocity $U=(1,-1)$. The Euclidean Killing vector fields for translation and rotation giveThe rotational field along the geodesic is $K_{\mathrm{rot}}(\gamma (t))=(t-2,1+t)$: it changes from $(-2,1)$ at $t=0$ to $(-1,2)$ at $t=1$, yet its pairing with $U$ stays $-3$. With toy mass $m=1$ and $t$ interpreted as time, the first constant is the $x$ component of linear momentum, and the second is the signed $z$ component of angular momentum about the origin. Their numerical values use toy units. The geodesic is straight; the rotation flow orbits are circles.

This conservation law follows from metric compatibility and the Killing equation:The geodesic equation kills the acceleration term, and the symmetric product $U^{\mu }{U}^{\nu }$ selects the vanishing symmetric derivative. This tensor contraction is the whole proof outline; see Carroll, equation (5.43).

In a stationary spacetime, a chosen time-translation Killing vector field $K$ gives Killing energy. With metric signature $(-+++)$, $c=1$, and a massive test particle's four-momentum $p^{\mu }=m{U}^{\mu }$, the physical normalization isHere $\Phi$ generates the axial rotational symmetry when one exists. In coordinates with $K={\partial }_t$ and $\Phi ={\partial }_{\varphi }$, these are $E=-p_t$ and $L_z=p_{\varphi }$. Both are conserved quantities along the geodesic, even if the metric has a time-angle cross term. A future timelike $K$ makes $E$ positive for a future massive particle; the scale of $E$ depends on the normalization of $K$. Not every spacetime has a global timelike Killing vector field, and a stationary field can become spacelike in an ergoregion of a stationary spacetime. Conservation on the field's domain does not imply a universal global positive energy; see Tong, sections 4.3.1-4.3.2 on the timelike-domain qualification and conserved charges.

For a Newtonian Kepler orbit, let $M=M_1+M_2$ be the total mass and $\mu =M_1{M}_2/M$ the reduced mass; put $k=GM$, $\mathbf{n}=\mathbf{r}/r$, $u=\dot{r}$ and ${\mathbf{v}}_{\perp }=\dot{\mathbf{r}}-u\mathbf{n}$, with $v^2=\dot{\mathbf{r}}\cdot \dot{\mathbf{r}}$. In the convention $q_{ij}=\mu (3r_i{r}_j-r^2{\delta }_{ij})/2$, differentiating the acceleration $\ddot{\mathbf{r}}=-k\mathbf{r}/r^3$ givesThe two terms have zero cross tensor contraction, with squared norms $6u^2$ and $72v_{\perp }^2$. Therefore ${\stackrel{...}{q}}_{ij}{\stackrel{...}{q}}_{ij}=72{\mu }^2{k}^2(v^2-11u^2/12)/r^4$. Substitution into the quadrupole formula in this convention, $L=4G{\stackrel{...}{q}}_{ij}{\stackrel{...}{q}}_{ij}/(45c^5)$, proves the luminosity. It is nonnegative since $v^2-11u^2/12=v_{\perp }^2+u^2/12$. The formula uses a weak gravitational field, slow motion and leading radiation order; orbital averaging gives a secular luminosity.

An invariant tensor is fixed by every element of the represented group, so it spans or belongs to a trivial group representation. The special unitary group preserves its identity tensor and the upper and lower volume tensors. Such tensors make tensor contraction equivariant and permit invariant inner products and real structures.

The Kretschmann scalar is the scalar tensor contraction of two copies of the Riemann curvature tensor. In Schwarzschild spacetime, with metric coefficient $1-2M/r$ and $G=c=1$, it is $48M^2/r^6$. Divergence of this scalar along a curve approaching a proposed regular point proves that there is no twice continuously differentiable, nondegenerate metric extension through that point: such an extension would have continuous finite curvature components and hence a finite scalar contraction. This gives a coordinate-independent test of a curvature singularity. Boundedness of this single scalar does not prove regularity.

A semicolon denotes the covariant derivative for the Levi-Civita connection. For example, the four lower indices of the Riemann curvature tensor contribute four connection terms:Here ${\partial }_{\eta }=\partial /\partial x^{\eta }$ is a partial derivative, ${\Gamma }_{b\eta }^a$ are Christoffel symbols, and repeated indices are summed from $0$ to $3$. Write $g_{\mu \nu }$ for the metric tensor, $g^{\mu \nu }$ for its inverse metric, $R_{\mu \nu }=g^{\lambda {\nu }^{\prime }}{R}_{\lambda \mu {\nu }^{\prime }\nu }$ for the Ricci tensor, and $R=g^{\mu \nu }{R}_{\mu \nu }$ for the Ricci scalar. This tensor contraction agrees with the paper's $R_{\mu \nu }=-R^a{}_{\mu \nu a}$.

By metric compatibility, the inverse metric may pass through each covariant derivative. Contracting the second Bianchi identity with $g^{\lambda \nu }$ giveswhere ${\nabla }_{\eta }$ is the same covariant derivative as the semicolon. Next contract with $g^{\mu \kappa }$. Antisymmetry of the first pair gives $g^{\mu \kappa }{R}_{\lambda \mu \kappa \eta }=-R_{\lambda \eta }$, soThus the contracted Bianchi identity is ${\nabla }^{\mu }{R}_{\mu \eta }=\frac{1}{2}{\partial }_{\eta }R$. Raising the remaining index and using metric compatibility provesThe Einstein tensor $G^{\mu \nu }$ therefore has identically zero covariant divergence, without imposing any field equation. This makes the Einstein field equations compatible with stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, where $T^{\mu \nu }$ is the stress-energy tensor. The connection terms describe how energy and momentum are compared at neighboring points; ordinary componentwise partial derivatives would not give a coordinate-independent conservation law in curved spacetime.

Take the tensor contraction of the Einstein field equations with the inverse metric $g^{\mu \nu }$. In four-dimensional spacetime, $g^{\mu \nu }{g}_{\mu \nu }=4$, henceHere $T$ is the tensor contraction of the stress-energy tensor. Substituting this result back into the Einstein field equations, while keeping both free indices lowered, gives the trace-reversed stress-energy tensor:The factor $1/2$ uses the dimension four: in dimension $n>2$, the Trace-reversed Einstein field equations in D dimensions have $1/(n-2)$ in place of $1/2$.

Use $x^0=ct$ and metric signature $(-,+,+,+)$. With the source's curvature sign convention, the Ricci tensor isIts time-time component is thereforeFor a slowly moving, pressure-negligible source, the stress-energy tensor is approximated by relativistic dust: $T^{\mu \nu }=\rho U^{\mu }{U}^{\nu }$, where $\rho$ is mass density and $U^{\mu }$ is the four-velocity. In its approximate rest local inertial frame, $U^{\mu }\simeq (c,0,0,0)$ and $U_0\simeq -c$. Thusand the trace-reversed stress-energy tensor satisfies

Independently, introduce the metric perturbation $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$, with $|h_{\mu \nu }|\ll 1$ and constant Minkowski metric ${\eta }_{\mu \nu }$. The Christoffel symbols are first order in $h$, so their products are discarded in the weak-field approximation. With a comma denoting a partial derivative, the linearized Riemann curvature tensor isTaking the tensor contraction gives, to this order,In the static Newtonian limit, all time partial derivatives vanish. The constant ${\eta }_{00}$ also has zero partial derivatives, leavingHere ${\nabla }^2={\sum }_i{\partial }_i^2$ is the spatial Laplace operator in the approximately Cartesian background coordinates. This sign follows the paper's curvature sign convention.

In Cartesian coordinates on Minkowski spacetime, the Conformal Killing equation becomesDifferentiate the equations with index pairs $(\nu ,\rho )$, $(\mu ,\rho )$ and $(\mu ,\nu )$ by ${\partial }_{\mu }$, ${\partial }_{\nu }$ and ${\partial }_{\rho }$, respectively. Adding the first two and subtracting the third, Clairaut's theorem yields the flat conformal Killing integrability formulaApply ${\partial }^{\rho }$ and use ${\partial }^{\rho }{X}_{\rho }=4\lambda$. With the d'Alembert operator $\square ={\eta }^{\alpha \beta }{\partial }_{\alpha }{\partial }_{\beta }$, this givesA tensor contraction with ${\eta }^{\mu \nu }$ gives $2\square \lambda =-4\square \lambda$, so $\square \lambda =0$. Returning to the preceding equation proves the stronger statementThus the conformal Killing factor is an affine function, not an arbitrary solution of the wave equation.

For a Conformal Killing vector field, the same metric compatibility calculation givesThus its Conformal Killing equation is ${\nabla }_{(\mu }{X}_{\nu )}=\lambda g_{\mu \nu }$. Taking the tensor contraction with the inverse metric, and using $g^{\mu \nu }{g}_{\mu \nu }=4$, determines the conformal Killing factor:Parentheses include the factor $1/2$ in tensor index symmetrization.

Write $D=u^a{\nabla }_a$, $\theta ={\nabla }_a{u}^a$ for the expansion scalar, $a_b=u^a{\nabla }_a{u}_b$ for the four-acceleration, and $P_b{}^c={\delta }_b{}^c+u_b{u}^c$ for the spatial projector of a timelike congruence. The normalization of the four-velocity gives $u^b{a}_b=0$ and $u^b{\nabla }_a{u}_b=0$. The product rule and metric compatibility give the covariant derivative of the perfect fluid in general relativity stress-energy tensor:Its tensor contraction with the four-velocity isand its projection with the spatial projector of a timelike congruence isThus stress-energy conservation implies the relativistic perfect-fluid energy equation and the relativistic Euler equation:Conversely, the two displayed equations set both projections of $F_b$ to zero. Since $F_b=P_b{}^c{F}_c-u_b{u}^c{F}_c$, they imply $F_b=0$. This establishes equivalence, including points where $\rho +p=0$, without dividing by that quantity.

Use the Levi-Civita symbol and Einstein summation convention. The tensor contraction ${ϵ}_{ijk}{ϵ}_{klm}={\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}$, expressed through the Kronecker delta, givesThe last two contractions are dot products, proving the vector triple product identity $\boxed{a\times (b\times c)=(a\cdot c)b-(a\cdot b)c}$.

Choose the linear relationThen the displayed tensor contraction is $(a_i{c}_i)b_k-(a_i{b}_i)c_k=(a\cdot c)b_k-(a\cdot b)c_k$, precisely the $k$th vector component of $a\times (b\times c)$ by the vector triple product identity. The matrix $M$ need not be a symmetric matrix.

A Levi-Civita connection is a connection on the tangent bundle which is torsion-free and compatible with the Riemannian metric. Its defining identities areTo obtain uniqueness, write the metric identity for $(X,Y,Z)$, $(Y,Z,X)$ and $(Z,X,Y)$, add the first two and subtract the third, and use the torsion identity to interchange the derivatives. This gives the Koszul formulaThe metric is nondegenerate, so these inner products determine ${\nabla }_{X}Y$ uniquely.

For existence, let $K(X,Y,Z)$ denote the right-hand side. The bracket rules $[X,fY]=f[X,Y]+X(f)Y$ and $[fX,Y]=f[X,Y]-Y(f)X$ give, by cancellation of all extra derivatives,Thus the nondegenerate metric defines a unique smooth vector field ${\nabla }_{X}Y$ from $2g({\nabla }_{X}Y,Z)=K(X,Y,Z)$, and these identities give the connection rules. In coordinates the construction readsThe coefficients are smooth and symmetric in $i,j$, proving zero torsion. Substituting them giveswhich is exactly metric compatibility. Since the construction was given intrinsically by $K$, the local formulas agree on overlaps. This proves existence and uniqueness of the Levi-Civita connection on every Riemannian manifold.

Define the Riemann curvature tensor byThe connection rules make this expression tensorial in all three arguments. To fix the component convention, throughout this solution useWith this convention, the unit sphere has $R_{ijkl}=g_{ik}{g}_{jl}-g_{il}{g}_{jk}$ and positive Ricci curvature. In a coordinate frame, the components are obtained fromThe algebraic curvature symmetries aretogether with the first Bianchi identityThe first antisymmetry follows from the definition. Metric compatibility gives $g(R(X,Y)Z,W)+g(Z,R(X,Y)W)=0$, giving the second. Torsion-freeness reduces the cyclic sum to the Jacobi identity for Lie brackets of vector fields; pair interchange follows algebraically from these antisymmetries and the cyclic identity. These explain the identities rather than choosing unrelated sign conventions.

Define the Ricci curvature by the tensor contractionThis is a bilinear form, since $R$ is tensorial and trace is linear. At a point choose an orthonormal basis. Pair interchange givesso Ricci curvature is symmetric.

For the final determination in dimension three, an algebraic curvature tensor is determined by its entries in the three unordered index pairs $12,13,23$: antisymmetry removes repeated indices, and pair interchange makes the resulting $3\times 3$ array symmetric. It therefore has at most six independent entries. WriteThe diagonal Ricci components giveso, for example, $K_{12}=({\mathrm{Ric}}_{11}+{\mathrm{Ric}}_{22}-{\mathrm{Ric}}_{33})/2$, and the other two follow cyclically. The three off-diagonal entries are also recovered explicitly:Thus every independent curvature entry is determined by Ricci; this proves injectivity, not merely a dimension count.

An invariant formula making the reconstruction explicit is, with $S={\mathrm{tr}}_g\mathrm{Ric}$,To verify it, the right-hand side has the two antisymmetries, pair interchange and the cyclic identity by symmetry of $g$ and Ricci. Contracting its first and third indices in dimension three givesSubtract it from the original tensor. The difference has zero Ricci contraction, and the six recovered-entry formulas above force it to vanish. This completes the proof of three-dimensional curvature from the Ricci tensor in arbitrary coordinates.