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On Euclidean space, every Killing vector field is a constant translation generator plus a constant infinitesimal rotation generator:Here $A$ is a constant vector and $B$ is an antisymmetric matrix. The second covariant derivative of a Killing vector vanishes for the Euclidean metric, so the components are affine functions of Cartesian coordinates; the Killing equation makes the linear coefficient antisymmetric.

In the Euclidean plane with metric tensor $g=d{x}^2+d{y}^2$, two useful examples areThe notation ${\partial }_x$ denotes a coordinate basis vector; $dx$ denotes a differential one-form. The corresponding flow of a vector field maps areFor translation, all arrows have the same direction and length. For rotation, arrows are tangent to circles, have length $\sqrt{x^2+y^2}$, and vanish at the origin. The rotation flow keeps $x^2+y^2$ fixed, and its nonstationary integral curves of a vector field are circles. Both flows preserve the Euclidean metric. See Carroll's original notes, section 5, equations (5.45)-(5.46).

The circular integral curves of a vector field are symmetry orbits, whereas geodesics of the Euclidean plane are straight lines. A Killing vector field does not require its own orbits to be geodesics. The geodesic conserved quantity from a Killing vector pairs the field with the velocity of a separate geodesic; it does not require the velocity to equal the Killing vector field.

On a connected Riemannian manifold, these data determine a Killing field uniquely, without a completeness assumption. The second covariant derivative of a Killing vector makes its restriction to every geodesic a Jacobi field. The difference of two fields with identical data has zero Jacobi initial values and vanishes in a convex normal neighborhood. The set on which both its value and covariant derivative vanish is closed by continuity and open by the same local argument, so connectedness makes it the whole manifold.

The Killing equation and the second covariant derivative of a Killing vector transport a Killing field and its antisymmetric derivative along any curve. Here $L_{ab}={\nabla }_b{\xi }_a$ and the curvature convention is the one in curvature-sign convention in Killing derivative identities. In a parallel frame the equations form a homogeneous linear system of ordinary differential equations. Initial value zero has unique zero continuation. Thus on a connected pseudo-Riemannian manifold, not just on a Riemannian manifold, the field is determined by its value and first derivative at one point.

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, 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 a Killing vector field $K$, its metric-dual differential one-form produces the electromagnetic field $F_{ab}=2{\nabla }_{[a}{K}_{b]}=2{\nabla }_a{K}_b$. The exterior derivative gives $dF=0$, while the second covariant derivative of a Killing vector gives ${\nabla }^a{F}_{ab}=-2R_{bd}{K}^d$. Thus this construction solves the source-free Maxwell equations in a vacuum spacetime with zero cosmological constant. The field is a test field: adding its stress-energy tensor would generally change the original metric tensor.

A Killing vector field $K$ generates local isometries: its flow of a vector field ${\Phi }_s$ satisfies ${\Phi }_s^{\ast }g=g$. Equivalently, the Lie derivative of the metric tensor vanishes. For the Levi-Civita connection, metric compatibility converts the coordinate expression into the Killing equation:The Levi-Civita connection is an implicit extra assumption in this question. The printed assumption of vanishing torsion tensor alone does not imply metric compatibility. For example, on Minkowski spacetime, take the affine connection whose only nonzero Christoffel symbol is ${\Gamma }^0{}_{00}=1$. The vector field $K={\partial }_0$ preserves the Minkowski metric, but $K_0=-1$ gives ${\nabla }_0{K}_0=1$, contradicting the Killing equation. The following calculation, and the covariant formulas in part (b), use the Levi-Civita connection; this is the role of metric compatibility in Killing identities.

For the stated curvature sign convention, the Ricci identity on a covector has a minus sign:Write $A_{\mu \nu \rho }={\nabla }_{\mu }{\nabla }_{\nu }{K}_{\rho }$. Differentiating the Killing equation gives $A_{\mu \nu \rho }=-A_{\mu \rho \nu }$. Alternately using this relation and the Ricci identity givesThe first Bianchi identity and antisymmetry of the last two indices of the Riemann curvature tensor reduce the three curvature terms to $2R^{\sigma }{}_{\mu \nu \rho }{K}_{\sigma }$. Consequently the second covariant derivative of a Killing vector is

For a scalar, the Lie derivative of a function is its directional derivative:The Lie derivative of a vector field is their Lie bracket of vector fields:For a torsion-free connection,The last term vanishes because the connection coefficients are symmetric in $b,c$. This proves ${\mathcal{L}}_{\xi }X={\nabla }_{\xi }X-{\nabla }_X\xi$.

For the metric tensor statements use the Levi-Civita connection, which also has metric compatibility. The tensor derivation property of the Lie derivative of a tensor field givesExpanding the first term using $\nabla g=0$ and using the vector identity cancels the derivatives of $X$ and $Y$. What remains isSince $X,Y$ are arbitrary, the Killing equation isMetric compatibility is needed here, even though symmetry alone sufficed for the vector identity. In general the Lie derivative of a metric with nonmetricity contains a term ${\xi }^c{\nabla }_c{g}_{ab}$. For example, with the metric tensor $g=e^{2x}d{x}^2$, zero connection coefficients and $\xi ={\partial }_x$, the Lie derivative of a tensor field is $2e^{2x}$, whereas $2{\nabla }_x{\xi }_x=4e^{2x}$. The usual metric connection interpretation of the question supplies the required compatibility.

For the Riemann curvature tensor calculation, keep the original PDF's convention. With $R_{abcd}=g_{ae}{R}^e{}_{bcd}$ and $R_{abc}{}^d=g^{de}{R}_{abce}$, it meansThus the commutator on a covector has the opposite sign. Write $K_{abc}={\xi }_{a;bc}={\nabla }_c{\nabla }_b{\xi }_a$. Taking a covariant derivative of the Killing equation makes $K$ antisymmetric in its first two indices, while the Ricci identity givesCombining this with the first-two-index antisymmetry gives the three relationsAdd the first and third and subtract the second. Pair symmetry and the first Bianchi identity imply $R_{adbc}+R_{bdca}-R_{cdab}=-2R_{abcd}$. ConsequentlyThis is the second covariant derivative of a Killing vector in the paper's convention; its sign is tied to the declared curvature-sign convention in Killing derivative identities, not to raising a particular index.

Set $L_{ab}={\xi }_{a;b}$. Along any smooth curve with tangent vector $V$, the chain rule and the just-proved identity give the Killing transport equationsChoose a parallel-propagated orthonormal frame along the curve. These become a homogeneous linear system of ordinary differential equations for the components of $(\xi ,L)$. Uniqueness of its initial value problem implies that zero initial data remain zero. Any point in the same connected component as $P$ can be joined to $P$ by a piecewise smooth curve. Therefore zero value and zero first derivative at $P$ force the Killing field and its derivative to vanish throughout that component, with no completeness assumption. On a manifold with one connected component this is the requested global result. Connectedness cannot be dropped: on two disjoint flat spacetimes, take the zero field on one and a nonzero translation on the other.

On a pseudo-Riemannian manifold of dimension $n$ with one connected component, the map $\xi \mapsto ({\xi }_a(P),L_{ab}(P))$ is therefore injective. There are $n$ independent components of the value and $n(n-1)/2$ of its antisymmetric derivative. The dimension bound for Killing vector fields isIt is attained in flat space by ${\xi }_a=A_a+B_{ab}{x}^b$ with constant antisymmetric $B$, giving translations and infinitesimal rotations or Lorentz transformations. The four-dimensional maximum is 10. For a disconnected manifold the bound applies componentwise; there is no uniform global bound depending only on $n$ if the number of components is unrestricted.

A Riemannian isometry $F:(M,g)\to (N,h)$ is a diffeomorphism satisfying $F^{\ast }h=g$. A local isometry is a local diffeomorphism with the same pullback identity; equivalently $d{F}_p$ is an isometric linear isomorphism for every $p$. An isometric immersion into a higher-dimensional manifold does not satisfy this local-diffeomorphism requirement.

Let ${\varphi }_t$ be the local flow of the generating vector field $K$. The identity ${\varphi }_t^{\ast }g=g$ differentiates to ${\mathcal{L}}_{K}g=0$, where $\mathcal{L}$ is the Lie derivative of a tensor field. For arbitrary vector fields $X,Y$, metric compatibility and the torsion-free connection giveThis proves the Killing equation. Conversely, a smooth Killing field has a local flow, andSince ${\varphi }_0$ is the identity, ${\varphi }_t^{\ast }g=g$ wherever the flow is defined. Its inverse is ${\varphi }_{-t}$ locally, so the local flow consists of local isometries. Neither direction assumes that $K$ is complete.

Put $H(X,Y)={\nabla }_X({\nabla }_{Y}K)-{\nabla }_{{\nabla }_{X}Y}K$ and $B(X,Y,Z)=⟨H(X,Y),Z⟩$. These expressions are tensorial in $X,Y,Z$. Take the covariant derivative in direction $X$ of the Killing equation, subtracting the terms from differentiating $Y$ and $Z$. It follows thatWith the curvature convention $\mathcal{R}$ of Solution 1, the torsion-free connection also givesDefine $C(X,Y,Z)=⟨\mathcal{R}(X,K)Y,Z⟩$. It is skew in $Y,Z$ by skew-adjointness of the Riemann curvature tensor. The first Bianchi identity givesso $C$ has the same antisymmetrization in $X,Y$ as $B$. Consequently $D=B-C$ is symmetric in its first two slots and skew in its last two. These symmetries force it to vanish:We have proved the second covariant derivative of a Killing vector identity, including its sign:The printed identity uses the opposite curvature-sign convention in Killing derivative identities. If $R=-\mathcal{R}$, namely $R(X,Y)Z={\nabla }_Y{\nabla }_{X}Z-{\nabla }_X{\nabla }_{Y}Z+{\nabla }_{[X,Y]}Z$, the same result reads ${\nabla }_{X,Y}^{2}K=R(K,X)Y$, exactly as printed. The convention must change with the sign; under $\mathcal{R}$, writing $\mathcal{R}(K,X)Y$ would be incorrect.

Finally let $H_0=K-\tilde{K}$ have zero value and zero first covariant derivative at $p$. Along any geodesic issuing from $p$, the identity just proved saysThus $H_0$ is a Jacobi field with zero initial value and derivative. Uniqueness for this linear ordinary differential equation implies $H_0=0$ on each such geodesic, hence on a convex normal neighborhood of $p$. Consider the set of points where both $H_0$ and $\nabla H_0$ vanish. It is closed by continuity and open by this same local argument; on the resulting neighborhood $H_0$ vanishes identically, so its derivative does too. It is nonempty, and $M$ is connected, therefore it is all of $M$. This proves $K=\tilde{K}$ everywhere and the general statement that a Killing field is determined by its value and first covariant derivative.

Work on a regular level surface, so $d\beta \ne 0$. A nonzero normal covector has the local form $n_a=f{\nabla }_a\beta$. Extend it as a normal to neighbouring level surfaces. Since the Levi-Civita connection is torsion-free,Antisymmetrizing the product with $n_c=f{\beta }_{;c}$ over three indices kills the symmetric second-derivative term and the term containing two copies of $d\beta$. HenceEquivalently $n\wedge dn=0$, the Frobenius theorem condition for hypersurface orthogonality.

For the second covariant derivative of a Killing vector, the curvature sign needs explicit care. Denote the curvature defined on the cover by $C^a{}_{bcd}$, and obtain its other index positions by ordinary raising and lowering with the metric tensor. With $D_{abc}=k_{a;bc}$, the Killing equation gives $D_{abc}=-D_{bac}$, while the supplied Ricci identity givesSuccessive use of these two facts givesSubstitute $D_{acb}=D_{abc}-C_a{}^d{}_{bc}{k}_d$. The curvature symmetries and first Bianchi identity then implyThus the identity consistent with the cover convention isThe plus-sign version is valid for the opposite curvature tensor $\tilde{R}=-C$. With ordinary metric index raising, the printed plus sign and the cover's curvature convention cannot both be used. This is the curvature-sign convention in Killing derivative identities, not a failure of the Killing equation.

An explicit sign check uses $d{s}^2=e^{2x}d{t}^2-d{x}^2-d{y}^2-d{z}^2$ and the timelike Killing vector $k={\partial }_t$. Here $k_t=e^{2x}$, ${\Gamma }^t{}_{tx}=1$ and ${\Gamma }^x{}_{tt}=e^{2x}$. Direct differentiation gives $k_{t;xx}=e^{2x}$, whereas the cover definition gives $C^t{}_{xxt}=-1$ and $C_{txx}{}^d{k}_d=-e^{2x}$. This also verifies the corrected sign without relying on a remembered curvature convention.

To prove the static Killing field is a Ricci eigenvector, use the permitted alternative of adapted coordinates. Hypersurface orthogonality supplies hypersurfaces orthogonal to $k$. Choose spatial coordinates on one such hypersurface and carry them along the flow of the Killing vector, using its flow parameter as $t$. The flow preserves both the metric tensor and $k$, so it preserves orthogonality to $k$. Consequently the mixed metric components vanish, and the Killing equation makes all coefficients independent of $t$:The only nonzero Christoffel symbols involving time can bein particular ${\Gamma }^0{}_{ij}={\Gamma }^i{}_{0j}={\Gamma }^0{}_{00}=0$. In $C_{0i}$, the derivative of the connection trace vanishes because ${\Gamma }^a{}_{0a}=0$; the other derivative is either a time derivative of a static coefficient or a derivative of ${\Gamma }^j{}_{0i}=0$. Every connection-product term contains a vanishing coefficient of these same types. Thus $C_{0i}=0$. The one-form $C_{ab}{k}^b=C_{a0}$ has only a time component, and is proportional to $k_a$. ThereforeChanging the overall curvature sign leaves this zero identity unchanged.

Finally a perfect fluid in signature $(+---)$ has $T_{ab}=(\rho +p)u_a{u}_b-p{g}_{ab}$ and $u^a{u}_a=1$. Contract the Einstein field equations with $k^b$, and antisymmetrize with $k_a$. Terms proportional to $g_{ab}$ disappear, as does the Ricci term just proved. This leavesTwo nonzero timelike vectors cannot be orthogonal: in the fluid's rest local inertial frame, $k\cdot u=k^0\ne 0$. The hypothesis $\rho +p>0$ therefore forces $u_{[a}{k}_{b]}=0$. Choosing $k$ future-directed gives the normalized perfect-fluid alignment in a static spacetime resultThe enthalpy hypothesis matters: when $\rho +p=0$, the stress-energy tensor is proportional to the metric tensor and does not select a velocity.

Use metric compatibility and a torsion-free connection. The coordinate formula for the Lie derivative of the metric tensor isSubstituting ${\partial }_a{g}_{bc}={\Gamma }^d{}_{ab}{g}_{dc}+{\Gamma }^d{}_{ac}{g}_{bd}$ and collecting connection terms gives $({\mathcal{L}}_{\xi }g{)}_{ab}={\nabla }_a{\xi }_b+{\nabla }_b{\xi }_a$. Thus a Killing vector obeys ${\nabla }_{(a}{\xi }_{b)}=0$, and conversely this Killing equation makes its flow preserve the metric. Its contraction gives ${\nabla }_a{\xi }^a=0$.

With the curvature convention fixed in Q1, contract the vector Ricci identity to obtainThe second term vanishes because the Killing divergence is zero. Metric compatibility converts the first term to ${\nabla }^b{\nabla }_a{\xi }_b$, and the Killing equation changes this to $-{\nabla }^b{\nabla }_b{\xi }_a$. Therefore the second covariant derivative of a Killing vector satisfiesThis derivation fixes the sign by the stated Ricci identity rather than by an unspecified convention for curvature.

For the axial Killing vector set $F^{ab}={\nabla }^b{\xi }^a$. It is antisymmetric and its divergence is ${\nabla }_b{F}^{ab}=-R^a{}_b{\xi }^b$. Work in geometrized units $G=c=1$ and with zero cosmological constant, as required by the asymptotically flat setting. If a spatial shell $\Omega$ has oriented boundary $S_2-S_1$, the given form of Stokes theorem applied to the Komar angular momentum givesThe vacuum Einstein field equations give $R_{ab}=0$ everywhere in the shell outside the matter source. The shell can still contain gravitational waves, since those have nonzero Weyl curvature rather than a nonzero Ricci tensor. HenceThis is angular momentum conservation in an axisymmetric vacuum region. Moving the outer surface through the radiative region adds no angular momentum. In an exactly axisymmetric spacetime there is no outward transport of the axial angular momentum by the waves. Axisymmetry does not imply stationarity, so this conclusion does not exclude radiation of energy.

For completeness the Einstein equations express the bulk current asOn an axisymmetric spacelike slice, $\xi$ is tangent to the slice and its contraction with $d^3{\Sigma }_a$ vanishes; there the trace contribution to the angular-momentum volume integral drops out. One must retain any inner surface charges if the integration region has inner boundaries. The vacuum shell argument itself does not require a matter-only representation of the total charge.

The current $Y$ is conserved on arbitrary slices, not just those to which $\xi$ is tangent. Indeed symmetry of the Ricci tensor and antisymmetry of ${\nabla }_a{\xi }_b$ giveThe last equality follows because a metric-preserving Killing flow also preserves the scalar curvature. Equivalently, the stress-energy conservation law and invariance of the stress trace give the same conserved current.

Let two spacelike hypersurfaces terminate on large enclosing spheres $S_1,S_2$ at the same asymptotic end. Join the spheres by a timelike tube lying wholly in the vacuum exterior. On this tube ${\nabla }_b{F}^{ab}=0$, so Stokes theorem gives zero difference of their Komar surface integrals, independently of the tube's shape. Alternatively, use the given Gauss theorem on a spacetime volume between the slices: ${\nabla }_a{Y}^a=0$, and the lateral flux is zero because $Y=0$ in the exterior vacuum tube. Taking the enclosing spheres to infinity with the usual asymptotic falloff therefore provesAll surface orientations in these equations use the same convention as the stated Komar integral. The argument concerns enclosing surfaces at the same asymptotic end, so it does not discard additional charges from other ends or inner boundaries.

The Levi-Civita connection has symmetric Christoffel symbols, so antisymmetrization removes its connection terms. Hence the Papapetrou electromagnetic field iswhere the last equality uses the Killing equation. As a differential two-form, $F=d{V}^{♭}$, so the identity $d^2=0$ for the exterior derivative proves ${\nabla }_{[a}{F}_{bc]}=0$. Contracting the preceding second covariant derivative of a Killing vector givesA vacuum spacetime here has $R_{ab}=0$, with no cosmological constant. Therefore both source-free Maxwell equations hold:This is a test electromagnetic field on the given vacuum spacetime; its own stress-energy tensor is not included in that background's Einstein field equations.

Use signature $(-,+,+,+)$ and the Riemann curvature tensor convention $[{\nabla }_a,{\nabla }_b]X^c=R^c{}_{dab}{X}^d$. Lower the index of the Killing vector field and set $T_{abc}={\nabla }_a{\nabla }_b{V}_c$. Differentiating the Killing equation gives $T_{abc}=-T_{acb}$, while the Ricci identity givesDefine $S_{abc}=R_{cbad}{V}^d$. The first-pair antisymmetry of the Riemann curvature tensor gives $S_{abc}=-S_{acb}$. The pair symmetry and the first Bianchi identity giveThus $D=T-S$ is symmetric in its first two indices and antisymmetric in its last two. These two symmetries force it to vanish:Consequently $T_{abc}=R_{cbad}{V}^d$. Raising $c$ using metric compatibility proves the required second covariant derivative of a Killing vector:

Use $G=c=1$, the Einstein field equations with zero cosmological constant, and metric signature $(-,+,+,+)$. Write $m=m_{a}d{x}^a$ for the differential one-form dual to the axial Killing vector field. To fix the sign of the volume formula, use the component Hodge star operator convention $(\star j{)}_{abc}={ϵ}_{abcd}{j}^d$ and $(\star F{)}_{ab}=\frac{1}{2}{ϵ}_{abcd}{F}^{cd}$. The Killing equation gives $(dm{)}_{ab}=2{\nabla }_a{m}_b$, and its contracted curvature identity givesTo see the curvature step, tracing the Killing equation gives ${\nabla }_b{m}^b=0$. The second covariant derivative of a Killing vector then gives ${\nabla }^b{\nabla }_b{m}_a=-R_{ab}{m}^b$, while commuting the covariant derivatives gives ${\nabla }^b{\nabla }_a{m}_b=R_{ab}{m}^b$. Subtracting these two terms is precisely ${\nabla }^b(dm{)}_{ab}$ above. The component Hodge star operator converts this divergence to $d\star dm$ with the displayed minus sign.

In a vacuum spacetime region $R_{ab}=0$, so $d\star dm=0$. If $S_1$ and $S_2$ are enclosing spacelike submanifolds in the same homology class bounding a vacuum three-dimensional region $W$, Stokes theorem yieldsThus the Komar angular momentum is independent of the enclosing vacuum spacelike two-manifold, provided the surfaces have the same orientation and enclose the same sources and inner boundaries. The vacuum region need not be stationary: an axial Killing vector field suffices.

For a regular filling hypersurface $\Sigma$ with $\partial \Sigma =S$, the Einstein field equations giveThe pullback of $\star m$ to $\Sigma$ vanishes because $m^a$ is tangent to $\Sigma$: the dual three-form measures the normal component, which is zero. ThereforeThis is the stress-energy current from a Killing vector integrated over the slice, with the signs fixed by the stated Hodge star operator convention.

There is a necessary boundary qualification omitted from the printed formulation. If the slice has inner boundaries $S_i$, orient them so $\partial \Sigma =S-{\bigcup }_i{S}_i$. The actual Komar angular momentum with inner boundaries identity isFor a vacuum Kerr black hole, $T_{ab}=0$ on the exterior slice but $J\ne 0$; its horizon supplies precisely the inner boundary contribution. Thus the matter-only formula is false for arbitrary exterior slices. It is valid when a nonsingular filling with no inner boundaries exists, or when all omitted inner charges vanish. The Komar angular momentum independence likewise concerns homologous surfaces in the same vacuum region, not an unrestricted comparison of differently enclosed objects.

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:

Lower the free index of $K$. The Killing equation makes the last two indices of ${\nabla }_{\nu }{\nabla }_{\mu }{K}_{\alpha }$ antisymmetric after the appropriate derivative is moved. Combining the three resulting Ricci identities cyclically givesApplying the covector form of the Ricci identity and then the first Bianchi identity reduces this toRaising $\alpha$ proves the second covariant derivative of a Killing vector identityContracting $\mu$ and $\nu$ and using the definition and symmetries of the Ricci tensor gives