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A Killing horizon is a null hypersurface with a Killing vector field $\xi$ as its normal and generator. Write $\xi =d/d\alpha$ along its orbits. These are null geodesics, but their Killing parameter need not be affine:The coefficient is the surface gravity for the chosen normalization; nondegeneracy means $\kappa \ne 0$. If $\lambda$ is affine and $q=d\lambda /d\alpha$, then $\xi =qd/d\lambda$ and $d(\log q)/d\alpha =\kappa$. For constant $\kappa$, $\lambda ={\lambda }_0+C{e}^{\kappa \alpha }$.

The Killing equation yields ${\partial }_{\mu }({\xi }^2)=2{\xi }^{\nu }{\nabla }_{\mu }{\xi }_{\nu }=-2{\xi }^{\nu }{\nabla }_{\nu }{\xi }_{\mu }$. On the horizon, $\boxed{{\partial }_{\mu }{\xi }^2=-2\kappa {\xi }_{\mu }}$. Multiplying $\xi$ by a positive constant multiplies $\kappa$ by that constant; physical temperature therefore needs a stated time normalization.

For a Killing vector field $\xi$, set $K^{\mu \nu }={\nabla }^{\mu }{\xi }^{\nu }=-K^{\nu \mu }$. Choose surface orientation conventions so that the Komar charge, up to overall normalization, isThis is the Komar current from trace-reversed stress-energy construction. The equality is the spacetime Stokes theorem. With $[{\nabla }_{\mu },{\nabla }_{\nu }]V^{\rho }=R^{\rho }{}_{\sigma \mu \nu }{V}^{\sigma }$, the Killing equation and ${\nabla }_{\nu }{\xi }^{\nu }=0$ give $J_{\xi }^{\mu }=R^{\mu }{}_{\nu }{\xi }^{\nu }$. In four dimensions, at zero cosmological constant, the Einstein field equations give the currentReversing orientation changes the common overall sign, not conservation. The contracted Bianchi identity givesAn isometry preserves scalar curvature, and the second term contracts symmetric and antisymmetric tensors. Equivalently stress-energy conservation, the Killing equation, and ${\mathcal{L}}_{\xi }T=0$ cancel the matter expression; its trace derivative needs this last observation. Hence $\boxed{{\nabla }_{\mu }{J}_{\xi }^{\mu }=0}$. Nonzero cosmological constant adds $\Lambda {\xi }^{\mu }$ to the geometric current, also divergence-free.

The two identities. Set $F_{\beta \gamma }={\nabla }_{\beta }{K}_{\gamma }$. The Killing equation says $F_{\beta \gamma }=-F_{\gamma \beta }$. The Levi-Civita connection has symmetric lower connection indices, so they cancel in the antisymmetric difference:Use normalized antisymmetrization, soPut $a^{\sigma }=K^{\tau }{\nabla }_{\tau }{K}^{\sigma }$. Contracting with $F^{\beta \gamma }$, the second term is $a^{\gamma }{F}_{\gamma \alpha }$ and the third is the same after using antisymmetry. ThusRearranging proves the Killing derivative contraction identity

Restriction to a Killing horizon. On a Killing horizon, the generator $K$ is normal to the horizon as well as tangent to its null generators. Consequently $K_{[\alpha }{\nabla }_{\beta }{K}_{\gamma ]}=0$ there. One can see this without assuming hypersurface orthogonality away from the horizon: if the horizon is locally $\Phi =0$, write its dual one-form as $K=fd\Phi +\Phi q$ near it; then $K\wedge dK=0$ on $\Phi =0$. By the definition of surface gravity, $a^{\sigma }=\kappa K^{\sigma }$ on the horizon, and lowering that relation gives $K^{\sigma }{\nabla }_{\sigma }{K}_{\alpha }=\kappa K_{\alpha }$. The contraction identity becomesAway from points where $K$ vanishes, cancel a nonzero component of $K_{\alpha }$; at a regular bifurcation surface extend by continuity. ThereforeThis is surface gravity from the Killing derivative; it concerns the horizon, not every point in the exterior.

Static spherical metric. Choose the Killing vector $K={\partial }_t$. Its dual one-form is $-Adt$, and the first identity immediately giveswith all other components zero. HenceUsing the horizon identity and the simultaneous simple zeros,Here the positive value is chosen for a regular outer Killing horizon with $A,B>0$ on the exterior side. This is surface gravity of a static spherical horizon. A different normalization $K\mapsto cK$ rescales surface gravity by $c$; when an asymptotically flat normalization is wanted, the time coordinate is chosen so $A\to 1$ at infinity.

Euclidean regularity and temperature. Let $\delta =r-r_{+}$, $a=A^{\prime }(r_{+})$ and $b=B^{\prime }(r_{+})$. On the exterior side, define the proper radial coordinate $R=2\sqrt{\delta /b}$. After Wick rotation, the near-horizon metric isThe radial-time plane is a polar plane with angular coordinate $\vartheta =\kappa \tau$. Its circumference-to-radius ratio is $\kappa \Delta \tau$; the conical singularity disappears precisely whenThis is the Euclidean black-hole regularity condition. If a signed surface gravity convention is used, the period uses $|\kappa |$.

In a thermal quantum field theory, imaginary-time periodicity is the inverse-temperature condition expressed by the KMS condition. Therefore smoothness identifies the Hawking temperaturein units $\hslash =k_B=1$, or $T_H=\hslash \kappa /(2\pi k_B)$ when $\kappa$ is measured as an inverse time. It provides the thermal interpretation of surface gravity in black-hole thermodynamics. The regular Euclidean construction describes an equilibrium thermal state; the collapse calculation of Hawking radiation in the last solution supplies the outgoing spectrum. The simple-zero hypothesis excludes an extremal black hole, for which this polar-plane argument changes.

Write $F=V_{\rho }{V}^{\rho }<0$ and $A_{\mu \nu }={\nabla }_{\mu }{V}_{\nu }$. The Killing equation makes $A_{\mu \nu }$ antisymmetric. Expanding hypersurface orthogonality and contracting with $V^{\alpha }$ givesHere metric compatibility gives ${\partial }_{\mu }F=2V^{\rho }{A}_{\mu \rho }$. Thereforeand the normalized Killing one-form satisfiesThus the required exponent is $\boxed{n=-2}$. The notation $|V{|}^2$ in this paper denotes the negative scalar $F$, rather than a positive squared norm; the answer uses only its real reciprocal. More generally, for a multiplier $h(F)$ the antisymmetric derivative is $(h/F+h^{\prime })(V_{\nu }{\partial }_{\mu }F-V_{\mu }{\partial }_{\nu }F)$, so $h=C/F$ is the universally valid choice. If $F$ is constant, other exponents can work in that particular metric.

In a stationary spacetime chart, take $V={\partial }_0$. Its squared length is $g_{00}<0$, so $V$ is a timelike vector. Since its components are constant, the Lie derivative of a tensor field obeys $({\mathcal{L}}_{V}g{)}_{\mu \nu }={\partial }_0{g}_{\mu \nu }=0$. For the Levi-Civita connection,so $V$ satisfies the Killing equation.

Conversely, a timelike Killing vector field is nowhere zero. Apply the flow-box theorem to choose coordinates with $V={\partial }_0$. The Killing equation then gives ${\partial }_0{g}_{\mu \nu }=0$, and $g_{00}=g(V,V)<0$. Thus local stationarity is equivalent to the existence of a timelike Killing vector field.

The normalized Killing one-form $\eta =F^{-1}{V}_{\mu }d{x}^{\mu }$ is a closed differential form. The Poincaré lemma gives a local scalar potential $\varphi$ with $\eta =d\varphi$, hence $V_{\mu }=F{\partial }_{\mu }\varphi$. Choose the adapted chart of part (i), so $V={\partial }_0$. ThenThe closed differential form condition supplies the integrability of these spatial derivatives. For a fixed $g$ and $V$, $f$ is determined locally up to an additive constant; its derivatives cannot be chosen freely despite the wording in the paper.

Use ${\bar{x}}^0=\varphi$, ${\bar{x}}^i=x^i$. The Jacobian determinant is one, so this is a local coordinate change. Holding ${\bar{x}}^i$ fixed gives $V={\partial }_{\bar{0}}$, while its metric-dual one-form is $v=Fd{\bar{x}}^0$. Thus ${\bar{g}}_{00}=F<0$ and ${\bar{g}}_{0i}=0$. The Killing equation still holds, so ${\partial }_{\bar{0}}{\bar{g}}_{\mu \nu }=0$. The metric is locally static:A closed differential form need not be globally exact; neither a global potential nor a global static chart follows without additional topological assumptions.

An electromagnetic field tensor obeys the homogeneous Maxwell equations $dF=0$. Together with its invariance under the Killing vector field $V$, Cartan's magic formula gives $d({\iota }_{V}F)={\mathcal{L}}_{V}F-{\iota }_{V}dF=0$. The local potential of a closed differential one-form construction, the one-form Poincaré lemma, therefore supplies a scalar field $\Phi$ on a sufficiently small contractible neighborhood withFor an explicit local proof, write ${\iota }_{V}F={\alpha }_i(x)d{x}^i$ on a coordinate ball centered at zero and take $\Phi (x)={\int }_0^1{x}^i{\alpha }_i(sx)ds$. Since $d\alpha =0$, differentiating under the integral gives ${\partial }_j\Phi ={\int }_0^1[{\alpha }_j(sx)+s{x}^i{\partial }_i{\alpha }_j(sx)]ds={\alpha }_j(x)$. There need not be a global potential when this closed one-form has nonzero periods. The homogeneous Maxwell equation is essential; an arbitrary invariant two-form would not suffice.

Use proper time for the massive particle, let $u$ be its tangent, and assume $m\ne 0$. The Killing equation makes $u^a{u}^b{\nabla }_a{V}_b=0$. The Lorentz force equation then impliesConsequently the local conserved quantity isThe sign follows from contracting $F$ in its first argument, exactly as in the source. Adding a constant to $\Phi$ merely shifts the conserved quantity. This is a charged-particle conserved quantity from a Killing symmetry.

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:

Let $k$ generate the Killing horizon and let $\ell =fk$ be an affinely parametrized generator wherever $k\ne 0$. Choose an auxiliary null vector $n$ with $n\cdot \ell =-1$ and the screen-space projector $q_{ab}=g_{ab}+{\ell }_a{n}_b+n_a{\ell }_b$. The optical tensor is the screen projection $B_{ab}=q_a{}^c{q}_b{}^d{\nabla }_d{\ell }_c$.

Since the projector annihilates $k$, the symmetric part isby the Killing equation. Its trace is the null expansion and its trace-free symmetric part is the null shear, so both vanish. Also $\ell$ is normal to the null hypersurface. Frobenius theorem gives ${\ell }_{[a}{\nabla }_b{\ell }_{c]}=0$, which implies $q_a{}^c{q}_b{}^d{\nabla }_{[d}{\ell }_{c]}=0$: the null twist vanishes. These arguments use derivatives tangent to the horizon and hence apply to any smooth null geodesic congruence containing its generators, independently of its extension off the horizon. At a regular bifurcation surface, use the smooth affine generator and continuity. ThusThis is the vanishing of the optical scalars on a Killing horizon.

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

The Killing equation gives ${\nabla }_{\mu }{K}^{\mu }=0$ and ${\nabla }_{[\mu }{K}_{\lambda ]}={\nabla }_{\mu }{K}_{\lambda }$. Commuting derivatives once and contracting the Ricci identity givesTherefore, using the contracted Bianchi identity and symmetry of $R_{\mu \beta }$,Since ${\nabla }_{[\mu }{K}_{\lambda ]}$ is antisymmetric, swapping the two contracted index names shows that the left side equals its derivative-antisymmetrized form:Now set $A_{\mu \lambda }={\nabla }_{[\mu }{K}_{\lambda ]}$. The right side is the contracted commutator $[{\nabla }^{\mu },{\nabla }^{\lambda }]A_{\mu \lambda }$. By the curvature commutator on a covariant tensor, it is a sum of contractions of the symmetric Ricci tensor with the antisymmetric tensor $A$, and hence vanishes. Thusso the Killing vector preserves scalar curvature.

Every metric coefficient is independent of $t$ and $\varphi$. The corresponding coordinate flows therefore preserve the metric, sosatisfy the Killing equation and are Killing vector fields. They generate stationarity and axial symmetry respectively.