Spherical electrovacuum mass and local rigidity

Fix $p\in S^2$ and put $H_p:=2p{p}^{\mathsf{T}}-I_3$. This matrix fixes $p$ and acts as $-I$ on $p^{\perp }=T_p{S}^2$; its eigenvalues are $1,-1,-1$, so $H_p^{\mathsf{T}}{H}_p=I_3$ and $\det H_p=1$, hence $H_p\in SO(3)$. Its lift fixes $(y,p)$, is the identity on $T_{y}U$, and is $-I$ on $T_p{S}^2$. The decomposition ${\mathrm{\Lambda }}^2(T_y^{\ast }U\oplus T_p^{\ast }{S}^2)={\mathrm{\Lambda }}^2{T}_y^{\ast }U\oplus (T_y^{\ast }U\otimes T_p^{\ast }{S}^2)\oplus {\mathrm{\Lambda }}^2{T}_p^{\ast }{S}^2$ is preserved by the lifted rotation, while the mixed summand changes sign. Invariance therefore forces the mixed component of ${\mathcal{F}}_{(y,p)}$ to equal its negative, so that component is zero. The first and last summands are one-dimensional, with generators ${ϵ}_2$ and ${\omega }_{S^2}$. The base form is preserved because each lift is the identity on $U$, and $(R^{\ast }{\omega }_{S^2}{)}_p(u,v)=\det (Rp,Ru,Rv)=\det (R)\det (p,u,v)={\omega }_{S^2,p}(u,v)$ for $R\in SO(3)$. To prove transitivity, take $p,p^{\mathrm{\prime }}\in S^2$, choose unit vectors $u\perp p$ and $u^{\mathrm{\prime }}\perp p^{\mathrm{\prime }}$, and set $v=p\times u$ and $v^{\mathrm{\prime }}=p^{\mathrm{\prime }}\times u^{\mathrm{\prime }}$. The oriented orthonormal frames $(p,u,v)$ and $(p^{\mathrm{\prime }},u^{\mathrm{\prime }},v^{\mathrm{\prime }})$ are related by a unique orientation-preserving orthogonal map $R$, which lies in $SO(3)$ and sends $p$ to $p^{\mathrm{\prime }}$. Thus invariance makes the coefficients of ${ϵ}_2$ and ${\omega }_{S^2}$ independent of $p$. For a fixed $p_{\ast }\in S^2$, the smooth section $j(y):=(y,p_{\ast })$ gives $j^{\ast }\mathcal{F}=E{ϵ}_2$. In any base chart the coefficient of the nowhere-vanishing smooth form ${ϵ}_2$ is nonzero, so division by it shows that $E$ is smooth locally, hence on $U$. To recover the vertical coefficient instead, take $p_{\ast }=(0,0,1)$, $v=(1,0,0)$ and $w=(0,1,0)$ in $T_{p_{\ast }}{S}^2$. Along $j$, the product tangent vectors $(0,v)$ and $(0,w)$ are smooth sections of $j^{\ast }T\mathcal{M}$: this follows in any fixed sphere chart about $p_{\ast }$, and requires no frame on all of $S^2$. Smoothness of $\mathcal{F}$ therefore makes $y\mapsto {\mathcal{F}}_{(y,p_{\ast })}((0,v),(0,w))$ smooth. Since ${\omega }_{S^2,p_{\ast }}(v,w)=\det (p_{\ast },v,w)=1$, this function is exactly $B(y)$, proving its smoothness. Consequently, uniquely $\mathcal{F}=E(y){ϵ}_2+B(y){\omega }_{S^2}$.

Since ${ϵ}_2$ and ${\omega }_{S^2}$ are top forms on their respective two-dimensional factors, their exterior derivatives vanish, and $d\mathcal{F}=dB\wedge {\omega }_{S^2}$. The form ${\omega }_{S^2}$ never vanishes, so $d\mathcal{F}=0$ implies $dB=0$. A smooth function with zero differential is constant on connected $U$ (differentiate along each smooth path in a coordinate-ball chain); call this constant $q_m$.

In an oriented orthonormal coframe with ${ϵ}_2=-e^0\wedge e^1$ and $r^2{\omega }_{S^2}=e^2\wedge e^3$, the four-volume is $e^0\wedge e^1\wedge e^2\wedge e^3$. The Lorentzian Hodge definition therefore gives ${\star }_g{ϵ}_2=r^2{\omega }_{S^2}$ and ${\star }_g{\omega }_{S^2}=-r^{-2}{ϵ}_2$. Hence ${\star }_g\mathcal{F}=E{r}^2{\omega }_{S^2}-B{r}^{-2}{ϵ}_2$, and $d({\star }_g\mathcal{F})=d(E{r}^2)\wedge {\omega }_{S^2}$ because a base one-form wedged with the base top form ${ϵ}_2$ vanishes. The second Maxwell equation implies $d(E{r}^2)=0$, so $q_e:=E{r}^2$ is constant on connected $U$. Restricting $\mathcal{F}$ and its Hodge dual to $S_y$ gives $q_m{\omega }_{S^2}$ and $q_e{\omega }_{S^2}$; since ${\omega }_{S^2}=\sin \theta d\theta \wedge d\varphi$ and ${\int }_0^{\pi }\sin \theta d\theta {\int }_0^{2\pi }d\varphi =4\pi$, the stated flux normalizations follow. This proves existence and uniqueness of both constants.

Form the four-manifold $\tilde{M}:=U\times {\mathbb{R}}^2$ with product metric $\tilde{g}:=q+(d{x}^2{)}^2+(d{x}^3{)}^2$, product orientation, and the lift of the chosen future timelike field on $U$. This is a smooth time-oriented Lorentzian four-manifold. By [F2], its metric-selected connection has the displayed Christoffel formula. In product coordinates its base Christoffels are exactly $\mathrm{\Gamma }[q{]}^A{}_{BC}$, every Christoffel with a flat-factor index is zero, and its curvature components with all indices in the base are exactly $R[q{]}^A{}_{BCD}$ by [F5]. Thus the base Ricci contraction in the product equals the two-dimensional contraction in the Statement. Moreover, in an oriented base chart, ${ϵ}_{01}=\pm \sqrt{\mathrm{\mid }\det q\mathrm{\mid }}$ and ${\mathrm{\Gamma }}^C{}_{AC}=\frac{1}{2}{q}^{CD}{\mathrm{\partial }}_A{q}_{CD}={\mathrm{\partial }}_A\log \sqrt{\mathrm{\mid }\det q\mathrm{\mid }}$. Substituting these identities in ${\mathrm{\nabla }}_A{ϵ}_{BC}={\mathrm{\partial }}_A{ϵ}_{BC}-{\mathrm{\Gamma }}^D{}_{AB}{ϵ}_{DC}-{\mathrm{\Gamma }}^D{}_{AC}{ϵ}_{BD}$ gives ${\mathrm{\nabla }}^q{ϵ}_2=0$; the only independent component is $BC=01$, for which the connection terms sum to ${\mathrm{\Gamma }}^C{}_{AC}{ϵ}_{01}$ and cancel the partial derivative.

At any point choose a $q$-orthonormal basis $(e_0,e_1)$ with $q(e_0,e_0)=-1$ and $q(e_1,e_1)=1$. In the product four-manifold the algebraic symmetries [F4] and antisymmetry in each pair force every base component of $R_{ABCD}$ to vanish unless both pairs are $(0,1)$ up to order. Write $K:=R_{0101}$. Pair exchange and the two antisymmetries give $R_{1010}=K$ and all components with repeated entries in either pair vanish. Using [F3], ${\mathrm{Ric}}_{00}=q^{11}{R}_{1010}=K$, ${\mathrm{Ric}}_{11}=q^{00}{R}_{0101}=-K$, and ${\mathrm{Ric}}_{01}={\mathrm{Ric}}_{10}=0$. Hence $\mathrm{Ric}(q)=-Kq$ and $S_q=-2K$, so $\mathrm{Ric}(q)=\frac{1}{2}{S}_{q}q$. The point was arbitrary; together with step 1.1 this proves both identities.

Let $A,B,C\in \{0,1\}$ be base indices and $i,j,k\in \{2,3\}$ sphere indices. In base coordinates, write $r_A:={\mathrm{\partial }}_{A}r$ and $r^A:=q^{AB}{r}_B$, the components of the vector field $q^{AB}{\mathrm{\partial }}_{B}r$ (the $q$-dual of $dr$). The inverse metric is $g^{AB}=q^{AB}$ and $g^{ij}=r^{-2}{\gamma }^{ij}$. Substituting the block metric $g_{AB}=q_{AB}$, $g_{ij}=r^2{\gamma }_{ij}$, $g_{Ai}=0$ into [F4] gives the mixed Christoffels ${\mathrm{\Gamma }}^A{}_{ij}=-\frac{1}{2}{q}^{AB}{\mathrm{\partial }}_B(r^2{\gamma }_{ij})=-r{r}^A{\gamma }_{ij}$ and ${\mathrm{\Gamma }}^i{}_{Aj}=\frac{1}{2}{r}^{-2}{\gamma }^{ik}{\mathrm{\partial }}_A(r^2{\gamma }_{kj})=r^{-1}{r}_A{\delta }_j^i$. All other mixed Christoffels vanish because the inverse metric is block diagonal, $q_{AB}$ is independent of sphere coordinates, and ${\gamma }_{ij}$ is independent of base coordinates. The unit round sphere has ${\mathrm{\Gamma }}^{\theta }{}_{\varphi \varphi }=-\sin \theta \cos \theta$ and ${\mathrm{\Gamma }}^{\varphi }{}_{\theta \varphi }={\mathrm{\Gamma }}^{\varphi }{}_{\varphi \theta }=\cot \theta$ on the polar chart of [F6]. Substitution in [F5] and contraction by [F3] give $\mathrm{Ric}(\gamma {)}_{\theta \theta }=-{\mathrm{\partial }}_{\theta }\cot \theta -{\cot }^2\theta =1$ and $\mathrm{Ric}(\gamma {)}_{\varphi \varphi }={\mathrm{\partial }}_{\theta }(-\sin \theta \cos \theta )-{\cos }^2\theta +2{\cos }^2\theta ={\sin }^2\theta$, with mixed component zero. Thus $\mathrm{Ric}(\gamma )=\gamma$ on that chart and everywhere by smoothness.

In an oriented orthonormal coframe for the two-dimensional Lorentzian base and the round sphere, ${ϵ}_{AB}{ϵ}^{AB}=-2$ and the sphere area form has squared norm $2\mathrm{/}{r}^4$. The two summands of $\mathcal{F}$ occupy orthogonal blocks, so ${\mathcal{F}}_{ab}{\mathcal{F}}^{ab}=2(q_m^2-q_e^2)\mathrm{/}{r}^4$. The same two-dimensional contractions give ${\mathcal{F}}_{AC}{\mathcal{F}}_B{}^C=-q_e^2{r}^{-4}{q}_{AB}$ and ${\mathcal{F}}_{ik}{\mathcal{F}}_j{}^k=q_m^2{r}^{-4}{g}_{ij}$; mixed contractions vanish. Substituting these three identities into [F1]'s definition of $Q$ yields the stated base and sphere components.

In the coordinate Ricci formula, the terms in $\mathrm{Ric}(g{)}_{AB}$ with only base indices give $\mathrm{Ric}(q{)}_{AB}$. The remaining sphere-index terms are $-{\mathrm{\partial }}_B(2r_A\mathrm{/}r)+(2r_C\mathrm{/}r){\mathrm{\Gamma }}^C{}_{AB}-2r_A{r}_B\mathrm{/}{r}^2=-2r^{-1}({\mathrm{\partial }}_B{r}_A-{\mathrm{\Gamma }}^C{}_{AB}{r}_C),$ which is $-2r^{-1}({\mathrm{\nabla }}_A^{q}dr{)}_B$. For $\mathrm{Ric}(g{)}_{ij}$, the pure-sphere terms give $\mathrm{Ric}(\gamma {)}_{ij}={\gamma }_{ij}$. The base derivative terms are $-[{\mathrm{\partial }}_A(r{r}^A)+{\mathrm{\Gamma }}^C{}_{CA}r{r}^A]{\gamma }_{ij}=-(r{\mathrm{\Delta }}_{q}r+z){\gamma }_{ij}$. The mixed trace ${\mathrm{\Gamma }}^k{}_{kA}{\mathrm{\Gamma }}^A{}_{ij}=-2z{\gamma }_{ij}$ cancels the two mixed products $-{\mathrm{\Gamma }}^k{}_{jA}{\mathrm{\Gamma }}^A{}_{ik}-{\mathrm{\Gamma }}^A{}_{jk}{\mathrm{\Gamma }}^k{}_{iA}=2z{\gamma }_{ij}$. For $\mathrm{Ric}(g{)}_{Ai}$, the derivative ${\mathrm{\partial }}_i{\mathrm{\Gamma }}^{\alpha }{}_{\alpha A}$ vanishes because ${\mathrm{\Gamma }}^{\alpha }{}_{\alpha A}={\mathrm{\Gamma }}^B{}_{BA}+2r_A\mathrm{/}r$ depends only on the base coordinates, and all derivatives ${\mathrm{\partial }}_{\alpha }{\mathrm{\Gamma }}^{\alpha }{}_{iA}$ vanish because the only nonzero such coefficients are ${\mathrm{\Gamma }}^j{}_{iA}=(r_A\mathrm{/}r){\delta }_i^j$. In the remaining products, the term ${\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{\mathrm{\ell }k}{\mathrm{\Gamma }}^k{}_{iA}=(r_A\mathrm{/}r){\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{\mathrm{\ell }i}$ is canceled by $-{\mathrm{\Gamma }}^j{}_{ik}{\mathrm{\Gamma }}^k{}_{jA}=-(r_A\mathrm{/}r){\mathrm{\Gamma }}^j{}_{ij}=-(r_A\mathrm{/}r){\mathrm{\Gamma }}^j{}_{ji}$; every other product contains a zero mixed Christoffel. Thus $\mathrm{Ric}(g{)}_{Ai}=0$. These identities give the three stated warped Ricci formulas.

The base Einstein equation, steps 2.1 and 1.2, and [F2] yield $\frac{1}{2}{S}_q{q}_{AB}-2r^{-1}({\mathrm{\nabla }}_A^{q}dr{)}_B=(\mathrm{\Lambda }-Q^2\mathrm{/}{r}^4)q_{AB}\mathrm{.}$ Its right-hand side and the first term are scalar multiples of $q$, so ${\mathrm{\nabla }}^{q}dr$ is a scalar multiple of $q$. Taking its trace identifies that multiple as $\frac{1}{2}{\mathrm{\Delta }}_{q}r$. The sphere Einstein equation gives $1-z-r{\mathrm{\Delta }}_{q}r=\mathrm{\Lambda }{r}^2+Q^2\mathrm{/}{r}^2$, which rearranges to the stated formula for ${\mathrm{\Delta }}_{q}r$.

Differentiate $z=q^{AB}{r}_A{r}_B$. Metric compatibility gives $dz=2({\mathrm{\nabla }}^{q}dr)({\mathrm{\nabla }}^{q}r,\cdot )={\mathrm{\Delta }}_{q}rdr$ by step 3.1. Therefore $d\varpi =\frac{1}{2}(1-z)dr-\frac{r}{2}dz-\frac{Q^2}{2r^2}dr-\frac{\mathrm{\Lambda }{r}^2}{2}dr\mathrm{.}$ Substituting ${\mathrm{\Delta }}_{q}r=(1-z)\mathrm{/}r-\mathrm{\Lambda }r-Q^2\mathrm{/}{r}^3$ from step 3.1 makes the coefficient of $dr$ zero term by term. Thus $d\varpi =0$, completing the conditional calculation.

By [F2], $d\varpi =0$. Any two points in a connected smooth manifold can be joined by a piecewise smooth path: the set reachable from a point by such paths and its complement are both unions of coordinate-ball neighborhoods, so connectedness makes the reachable set the whole patch. Along each smooth path segment $\sigma$, $(\varpi \circ \sigma {)}^{\mathrm{\prime }}=d\varpi ({\sigma }^{\mathrm{\prime }})=0$, so the one-variable mean-value theorem makes $\varpi$ constant along the segment. Thus $\varpi =m$ on the patch.

Let $X^A:={ϵ}_2^{AB}{\mathrm{\nabla }}_B^{q}r$, where indices on ${ϵ}_2$ are raised with $q^{-1}$. The volume form is nondegenerate, so $X\mathrm{\ne }0$ because $dr\mathrm{\ne }0$. Antisymmetry gives $X(r)=0$. In an oriented orthonormal frame with ${ϵ}_2=-e^0\wedge e^1$, direct contraction gives $q(X,X)=-(r_1{)}^2+(r_0{)}^2=-z$. By [F2] the Hessian is ${\mathrm{\nabla }}^{q}dr=\frac{1}{2}({\mathrm{\Delta }}_{q}r)q$, and [F3] gives ${\mathrm{\nabla }}^q{ϵ}_2=0$. Hence ${\mathrm{\nabla }}_A{X}_B=\frac{1}{2}({\mathrm{\Delta }}_{q}r){ϵ}_{BA}$, which is antisymmetric in $A,B$. In local coordinates, [F9] and the Lie-derivative definition give $({\mathcal{L}}_{X}q{)}_{AB}=X^C{\mathrm{\partial }}_C{q}_{AB}+q_{CB}{\mathrm{\partial }}_A{X}^C+q_{AC}{\mathrm{\partial }}_B{X}^C$. Expanding ${\mathrm{\nabla }}_A^q{X}_B+{\mathrm{\nabla }}_B^q{X}_A$ with $X_B=q_{BC}{X}^C$ and the Christoffel formula for ${\mathrm{\nabla }}^q$ in [F3] gives the same expression: the coefficient of $X^C$ reduces to ${\mathrm{\partial }}_C{q}_{AB}$ by $2{\mathrm{\Gamma }}^D{}_{AB}{q}_{DC}={\mathrm{\partial }}_A{q}_{BC}+{\mathrm{\partial }}_B{q}_{AC}-{\mathrm{\partial }}_C{q}_{AB}$. Thus ${\mathcal{L}}_{X}q=0$. Let ${\widehat{X}}_{(y,p)}:=(X_y,0)$ be the lift to $\mathcal{M}$; its local flow is $(y,p)\mapsto ({\mathrm{\Phi }}_t(y),p)$, so ${\mathcal{L}}_{\widehat{X}}{\pi }_S^{\ast }{\gamma }_{S^2}=0$. Applying the defining Lie-derivative formula to the warped metric yields ${\mathcal{L}}_{\widehat{X}}g={\pi }_U^{\ast }({\mathcal{L}}_{X}q)+2(r\circ {\pi }_U)(Xr){\pi }_S^{\ast }{\gamma }_{S^2}=0$. Thus the lifted vector field is Killing for the full four-dimensional metric, and in the constructed coordinates it is the lift $K$ of ${\mathrm{\partial }}_t$.

The smooth local flow theorem [F4] gives a local flow ${\mathrm{\Phi }}_t$ of the nonvanishing field $X$. To construct a transverse curve, choose a chart $(y^0,y^1)$ with ${\mathrm{\partial }}_{y^0}r\mathrm{\ne }0$ at the point; the map $(y^0,y^1)\mapsto (r(y^0,y^1),y^1)$ has invertible differential there. The inverse function theorem [F5], after restricting the chart, therefore gives a smooth curve $\sigma (s)$ through the point with $r(\sigma (s))=s$. Define $\mathrm{\Psi }(t,s):={\mathrm{\Phi }}_t(\sigma (s))$ on a sufficiently small product neighborhood. The identity $X(r)=0$ gives $r(\mathrm{\Psi }(t,s))=s$. At the initial point the two derivatives of $\mathrm{\Psi }$ are $X$ and ${\sigma }^{\mathrm{\prime }}(s)$; they are independent because $dr(X)=0$ and $dr({\sigma }^{\mathrm{\prime }})=1$. Thus $d\mathrm{\Psi }$ is invertible, and [F5] makes $(t,r)$ valid local coordinates with $X={\mathrm{\partial }}_t$.

Since $X={\mathrm{\partial }}_t$ and ${\mathcal{L}}_{X}q=0$, each coordinate coefficient is independent of $t$, so $q=A(r)d{t}^2+2B(r)dtdr+C(r)d{r}^2$. From step 1.2, $A=q(X,X)=-z\mathrm{\ne }0$. Both $A$ and $B$ are smooth functions of $r$, so $B\mathrm{/}A$ is smooth on the local $r$-interval. Define $G(r):={\int }_{r_0}^{r}B(s)\mathrm{/}A(s)ds$. Applying [F6] on each compact subinterval containing $r_0$ and $r$ gives $G^{\mathrm{\prime }}=B\mathrm{/}A$, first making $G$ continuously differentiable. Since its derivative $B\mathrm{/}A$ is smooth, repeated differentiation shows $G$ is smooth. The coordinate change $(t,r)\mapsto (\tilde{t},r):=(t+G(r),r)$ and its inverse $(\tilde{t},r)\mapsto (\tilde{t}-G(r),r)$ are therefore smooth. Substituting $dt=d\tilde{t}-G^{\mathrm{\prime }}(r)dr$ makes the cross-term coefficient $2(B-A{G}^{\mathrm{\prime }})=0$. The coordinate $r$ is unchanged, so $q^{rr}=q^{-1}(dr,dr)=z$; for a diagonal two-metric this implies $q_{\tilde{t}\tilde{t}}=-z$ and $q_{rr}=z^{-1}$.