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Write $f=1-2M/r$ in the Schwarzschild metric. For the stationary Killing vector field $V={\partial }_t$, its dual is $V^{♭}=-fdt$ and the Papapetrou electromagnetic field isOutside the Schwarzschild event horizon, use the static orthonormal coframe $e^{\hat{0}}=\sqrt{f}dt$, $e^{\hat{r}}=dr/\sqrt{f}$, $e^{\hat{\theta }}=rd\theta$, $e^{\hat{\varphi }}=r\sin \theta d\varphi$. Then $F=(2M/r^2)e^{\hat{0}}\wedge e^{\hat{r}}$. With the electric field convention $E_{\hat{i}}=F_{\hat{i}\hat{0}}$ from the preceding part,Thus the stationary Killing vector field gives a Coulomb electric field, directed inward in this convention. If the charge is normalized by $E_{\hat{r}}=Q/r^2$, its charge is $Q=-2M$. A convention with $F\mapsto -F$ reverses that charge; the invariant content is a radial monopole electric field of magnitude $2M/r^2$.

For the axial Killing vector field $V={\partial }_{\varphi }$, $V^{♭}=r^2{\sin }^2\theta d\varphi$, givingIn the same static orthonormal coframe, this is a purely magnetic field withAt large $r$ these are the spherical components of $2{\mathbf{e}}_z$. This is therefore the regular magnetic test field asymptotic to a uniform axial field, distorted by the Schwarzschild black hole. Its net magnetic monopole flux is zero because the integral of $\cos \theta$ over a sphere vanishes. Both electromagnetic fields are regular at the future event horizon: in Ingoing Eddington-Finkelstein coordinates, the electric field is $(2M/r^2)dv\wedge dr$, and the magnetic two-form already contains only regular spatial coordinates. Static observers themselves cease to exist on the event horizon.

Use the exterior-normalized Killing vector field $K={\partial }_t={\partial }_v$ from the previous part. In Ingoing BTZ coordinates, its only nonzero acceleration component on $r=r_{+}$ isThe defining equation for surface gravity, ${\nabla }_{K}K=\kappa K$, therefore gives the positive future-branch valueThis value uses exactly the prescribed Killing vector field normalization. A constant rescaling of the Killing vector field rescales its surface gravity.

Put $r_{+}=\ell \sqrt{M}$ and $f=(r^2-r_{+}^2)/{\ell }^2$. The apparent failure of the nonrotating BTZ black hole coordinates at $r=r_{+}$ is a coordinate singularity. Define the tortoise coordinate and ingoing time byThe Ingoing BTZ coordinates giveAll coefficients are analytic at $r_{+}>0$, and the determinant is $-r^2$, so the metric tensor extends nondegenerately through that radius. The Killing vector field $K={\partial }_v$ agrees with ${\partial }_t$ in the exterior. On $r=r_{+}$, $K^2=-f=0$ and $K_a=(dr{)}_a$. It is therefore both normal and tangent to this null hypersurface, proving that it is a Killing horizon.

One may include both horizon branches and their bifurcation surface explicitly. Set $\kappa =r_{+}/{\ell }^2$, $u=t-r_{\ast }$ and, initially in the exterior, $U=-e^{-\kappa u}$, $V=e^{\kappa v}$. ThenThis analytic extension of a spacetime is regular near $UV=0$, and $K=\kappa (V{\partial }_V-U{\partial }_U)$ extends analytically. The surfaces $U=0$ and $V=0$ form the extended Killing horizon at $r=r_{+}$; $K$ vanishes only at their regular bifurcation circle.

The stress-energy tensor is already first order, so its energy flux can be evaluated on the unperturbed Schwarzschild event horizon $r=2M_0$, with the infinity-normalized Killing vector field $\xi ={\partial }_v$. On this background $g^{rv}=1$, and henceThe conserved stress-energy current from a Killing vector is $J^a=-T^a{}_b{\xi }^b$. Its radial component is negative, so the positive inward energy flux is $-J^r$. Using the directed null-surface measure $r_H^{2}dvd\Omega$ givesThe initial and final event horizon areas are $A_0=16\pi M_0^2$ and $A_1=16\pi (M_0+\Delta M{)}^2$. Thus $\Delta A=32\pi M_0\Delta M+O((\Delta M{)}^2)$. With Schwarzschild surface gravity ${\kappa }_0=1/(4M_0)$, the Physical-process first law of black-hole mechanics follows:The flux is Killing energy, including gravitational redshift, rather than the energy measured by a sequence of static observers arbitrarily close to the event horizon.

Away from a bifurcation set, the normal to a Killing horizon is its nonzero Killing vector field $K$. Any affine tangent to the same generators has the form $k=hK$. In the optical tensor, derivatives of $h$ contribute a factor of $K_a$ and disappear under the screen-space projector. Consequentlyby the Killing equation. The trace and trace-free symmetric part are therefore zero: $\theta =0$ and ${\sigma }_{ab}=0$.

For the antisymmetric part, locally the horizon normal is proportional to a gradient. The restriction of its dual one-form to every tangent direction on the horizon is zero. Taking its exterior derivative and evaluating on two screen tangents therefore gives zero as well; equivalently this is the hypersurface-orthogonality consequence of the Frobenius theorem. Since both screen indices are tangential, $B_{[ab]}=0$. Thus the optical scalars on a Killing horizon satisfyNo affine normalization of the Killing vector field itself was assumed. At a regular bifurcation surface, $K$ vanishes but the smooth null generators do not; the conclusions for their optical quantities follow by continuity along each horizon branch.

Use metric compatibility and the symmetric Levi-Civita connection. Expanding $k_a=g_{ac}{k}^c$ givesThe Christoffel formula makes the bracket ${\partial }_c{g}_{ab}$. Thus the Killing equation is equivalent to the displayed coordinate formula in the question, also the vanishing Lie derivative of the metric.

Along an affinely parametrized geodesic with tangent $V$,The first term vanishes by the geodesic equation and the second by the Killing equation, because $V^a{V}^b$ is symmetric. This proves the geodesic conserved quantity from a Killing vector.

For $d{s}^2=-d{u}^2+u^{2}d{v}^2$, the three independent coordinate Killing conditions areThe field $(0,1)$ satisfies all three immediately. For $e^{-v}(1,u^{-1})$, the second equation reads $u{e}^{-v}-u{e}^{-v}=0$, and the third reads $e^{-v}-e^{-v}=0$. For $e^v(-1,u^{-1})$, they read $-u{e}^v+u{e}^v=0$ and $e^v-e^v=0$. The first equation holds for both, verifying all three Killing vector fields.

Their conserved contractions with $(\dot{u},\dot{v})$ may be denotedAdding the last two expressions after multiplying by $e^v,e^{-v}$ givesTo identify these loci, set $T=u\cosh v$, $X=u\sinh v$. The metric becomes $-d{T}^2+d{X}^2$, and the equation becomes $(\alpha +\beta )T+(\alpha -\beta )X=2\gamma$, a straight line in Minkowski spacetime. Nontrivial such lines, parametrized affinely, are exactly the geodesics in this coordinate patch. The all-zero choice of constants is a vacuous equation rather than an individual geodesic.

The metric tensor of the planar warped spacetime depends only on $z$. The three translations therefore preserve it, givingThey are linearly independent Killing vector fields everywhere. The associated geodesic conserved quantities from Killing vectors are $-A^2\dot{t}$, $A^2\dot{x}$ and $A^2\dot{y}$, for an affine parameter; the conventional positive energy is $E=A^2\dot{t}$. No restriction on the positive function $A$ is needed for these three symmetries.

Let $u$ be the tangent vector to a geodesic with affine parameter $\lambda$. Its geodesic equation is ${\nabla }_{u}u=0$. For a Killing vector field $\xi$, metric compatibility givesThusThis geodesic conserved quantity from a Killing vector is independent of whether the geodesic is timelike, spacelike or null. For a timelike geodesic one can use proper time; for a null geodesic choose an affine parameter. An arbitrary nonaffine tangent rescales the displayed quantity and need not make it constant.

A Killing vector field $\xi$ generates a local one-parameter family of isometries of the metric tensor. Equivalently, its tensor Lie derivative of the metric vanishes:The latter is the Killing equation; the equivalence uses metric compatibility and the torsion-free connection.

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.

Axisymmetry preserves the horizon, so the axial Killing vector field $m$ is tangent to it, so its null normal $\xi =k+\Omega m$ obeys $\xi \cdot m=0$ and ${\xi }^2=0$. Therefore $k\cdot m=-\Omega m^2$ andThus $\boxed{k^2={\Omega }^2{m}^2}$ there. A generator has $dt/d\alpha =1$, $d\phi /d\alpha =\Omega$, making $\Omega =d\phi /dt$ its angular velocity relative to the nonrotating stationary frame at infinity. For Kerr it is the Kerr horizon angular velocity $a/(r_{+}^2+a^2)$; the argument itself is not Kerr-specific.

An ergoregion of a stationary spacetime is where the stationary Killing vector field, normalized as time translation at infinity, is spacelike; a fixed-coordinate stationary observer cannot remain timelike there. For the Kerr metric, $k^2=-(1-2Mr/\Sigma )$, so the outer stationary-limit surface isFor a subextremal rotating hole $r_{\mathrm{E}}\ge r_{+}=M+\sqrt{M^2-a^2}$, with equality only at the poles. At the equator $r_{\mathrm{E}}=2M>r_{+}$. The ergosphere touches the horizon on the rotation axis and is outside it elsewhere. Its boundary is not the horizon, whose generator is $k+{\Omega }_{H}m$.

The drawing uses oblate Cartesian coordinates to display the intersection and is not an isometric embedding of the spatial geometry.

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 qualification concerns gravitational energy. Local matter energy exists: an observer $u$ measures $T_{\mu \nu }{u}^{\mu }{u}^{\nu }$. What is absent is a generally covariant local gravitational stress-energy density with the required universal conservation properties. The equivalence principle allows the connection to vanish at a point in freely falling coordinates; standard gravitational energy expressions depend on coordinates. Curvature does remain, but supplies no unique gravitational energy tensor of the required kind. A nonstationary geometry also lacks a preferred time-translation Killing vector field. For a trial timelike $t$, stress-energy conservation gives ${\nabla }_{\mu }(T^{\mu \nu }{t}_{\nu })=T^{\mu \nu }{\nabla }_{(\mu }{t}_{\nu )}$, not generally zero.

An internal charge instead has a current ${\nabla }_{\mu }{j}^{\mu }=0$ independently of spacetime time translations. Its hypersurface flux is conserved with suitable boundary conditions. For electric charge, the Gauss law expresses it as a surface flux. Observer-dependent charge density does not prevent a well-defined total charge.

For an asymptotically flat spacetime, take asymptotically Cartesian slice coordinates with ${\gamma }_{ij}={\delta }_{ij}+h_{ij}$ and appropriate falloff. In $c=1$ units, the ADM energy isIt is an asymptotic gravitational energy. The dominant energy condition requires $-T^{\mu }{}_{\nu }{u}^{\nu }$ to be future-directed nonspacelike or zero for every future timelike $u$, in particular $T_{\mu \nu }{u}^{\mu }{u}^{\nu }\ge 0$. Together with the constraint equations, completeness and appropriate asymptotic/boundary hypotheses, it yields $E_{\mathrm{ADM}}\ge |{\mathbf{P}}_{\mathrm{ADM}}|\ge 0$. The energy condition alone is not the complete positive-energy theorem.

Multiplying the matrices gives the Heisenberg group lawDifferentiating at the identity gives all strictly upper-triangular matrices. With $X=E_{12}$, $Y=E_{23}$ and $Z=E_{13}$, the commutator givesThis is the Heisenberg Lie algebra.

The right Maurer-Cartan form isThus the right-invariant coframe of the real Heisenberg group is ${\sigma }^1=dx$, ${\sigma }^2=dy$, ${\sigma }^3=dz-ydx$. Under a right translation,we have $d{x}^{\prime }=dx$, $d{y}^{\prime }=dy$, and $d{z}^{\prime }-y^{\prime }d{x}^{\prime }=dz-ydx$. All three right-invariant differential forms are preserved.

Every right-invariant Riemannian metric is determined by an arbitrary inner product at the identity. In this coframe its most general expression isEquivalently,Products here are symmetric products. If a pseudo-Riemannian metric tensor is intended, replace positive definiteness by nondegeneracy. Since every right translation preserves the coframe, it is an isometry. The action is faithful, and $a\mapsto R_{a^{-1}}$ is a group homomorphism, embedding a copy of $G$ into the isometry group.

The one-parameter right translations produce the Killing frame for a right-invariant Heisenberg metric:Their flows are respectively $(x+t,y,z)$, $(x,y+t,z+xt)$ and $(x,y,z+t)$. Each preserves $h$, and their Lie brackets of vector fields are $[K_X,K_Y]=K_Z$, with the other two zero. This explicitly realizes the Heisenberg Lie algebra. These Killing vector fields are left-invariant vector fields; the vector fields dual to the right-invariant coframe are instead ${\partial }_x+y{\partial }_z,{\partial }_y,{\partial }_z$, and should not be substituted for these generators.

For the diagonal case, putThe Kaluza-Klein decomposition along a Killing field uses $V=h(K_X,K_X)=\alpha +\chi y^2$. Completing the square in $dx$ gives the Heisenberg metric Kaluza-Klein reduction:These depend only on the quotient coordinates $(y,z)$ and satisfy $h=V(dx+A{)}^2+\gamma$. For a diagonal indefinite metric tensor, the same expressions hold wherever $V\ne 0$; a null Killing vector field cannot be treated with this completed-square decomposition.