Kerr exterior vacuum verification

Work first on a polar chart with $0<\theta <\pi$ and change angular coordinate to $u=\cos \theta$. Put $q=1-u^2$, $S=r^2+a^2{u}^2$, $D=r^2-2Mr+a^2$, and $A=r^2+a^2$. In coordinate order $(t,\varphi ,r,u)$ the metric is block diagonal between $(t,\varphi )$ and $(r,u)$: $g=G\oplus h$, where $G=S^{-1}\left(\begin{array}{cc}2Mr-S& -2Marq\\ -2Marq& q(A^2-a^{2}Dq)\end{array}\right)$ and $h=\mathrm{diag}(S\mathrm{/}D,S\mathrm{/}q)$. Direct $2\times 2$ inversion gives $G^{-1}=\left(\begin{array}{cc}-(A^2-a^{2}Dq)\mathrm{/}(SD)& -2Mar\mathrm{/}(SD)\\ -2Mar\mathrm{/}(SD)& (D-a^{2}q)\mathrm{/}(SDq)\end{array}\right)$, with $\det G=-Dq$; the inverse base block is $h^{-1}=\mathrm{diag}(D\mathrm{/}S,q\mathrm{/}S)$.

The derivatives needed below are $S_r=2r$, $S_u=2a^{2}u$, $S_{rr}=2$, $S_{ru}=0$, $S_{uu}=2a^2$, $D_r=2(r-M)$, $D_{rr}=2$, $q_u=-2u$, and $q_{uu}=-2$. On this chart $S,D,q>0$, so every denominator in the metric and inverse block is nonzero.

Write Greek indices $\alpha ,\beta ,\gamma$ for $t,\varphi$ and Latin indices $i,j,k,\mathrm{\ell }$ for $r,u$. Since $G,h$ are independent of $t,\varphi$, the complete nonzero Christoffel families from [F3] are ${\mathrm{\Gamma }}^{\alpha }{}_{\beta i}={\mathrm{\Gamma }}^{\alpha }{}_{i\beta }=\frac{1}{2}(G^{-1}{)}^{\alpha \gamma }{\mathrm{\partial }}_i{G}_{\gamma \beta },{\mathrm{\Gamma }}^i{}_{\alpha \beta }=-\frac{1}{2}(h^{-1}{)}^{ij}{\mathrm{\partial }}_j{G}_{\alpha \beta },{\mathrm{\Gamma }}^i{}_{jk}=\frac{1}{2}(h^{-1}{)}^{i\mathrm{\ell }}({\mathrm{\partial }}_j{h}_{\mathrm{\ell }k}+{\mathrm{\partial }}_k{h}_{\mathrm{\ell }j}-{\mathrm{\partial }}_{\mathrm{\ell }}{h}_{jk});$ all families not forced by lower-index symmetry vanish. The six base coefficients, making the last family explicit, are ${\mathrm{\Gamma }}^r{}_{rr}=\frac{r}{S}-\frac{r-M}{D},{\mathrm{\Gamma }}^r{}_{ru}=\frac{a^{2}u}{S},{\mathrm{\Gamma }}^r{}_{uu}=-\frac{Dr}{Sq},{\mathrm{\Gamma }}^u{}_{rr}=-\frac{a^{2}uq}{SD},{\mathrm{\Gamma }}^u{}_{ru}=\frac{r}{S},{\mathrm{\Gamma }}^u{}_{uu}=\frac{a^{2}u}{S}+\frac{u}{q}\mathrm{.}$ Thus the first two matrix families are fully specified by the displayed $G,G^{-1},h^{-1}$ and their $r,u$ derivatives, rather than by omitted connection coefficients.

Put $W_r:=(r-M)\mathrm{/}D$, $W_u:=-u\mathrm{/}q$, and $P_i:=G^{-1}{\mathrm{\partial }}_{i}G$. Since ${\mathrm{\partial }}_i\det G=(\det G)\mathrm{tr}(G^{-1}{\mathrm{\partial }}_{i}G)$ and $\det G=-Dq$, the fiber connection trace is ${\mathrm{\Gamma }}^{\alpha }{}_{\alpha i}=W_i$. Also $\det h=S^2\mathrm{/}(Dq)$ and hence $\sqrt{\mathrm{\mid }\det G\det h\mathrm{\mid }}=S$. Substitution of the connection families in step 2.1 into the signed coordinate formula [F5], followed by the contraction ${\mathrm{Ric}}_{bd}=R^a{}_{bad}$ of [F4], gives ${\mathrm{Ric}}_{\alpha \beta }=-\frac{1}{2S}{\mathrm{\partial }}_i\text{\hspace{0.17em}⁣}\left(S{h}^{ij}{\mathrm{\partial }}_j{G}_{\alpha \beta }\right)+\frac{1}{2}{h}^{ij}{\left({\mathrm{\partial }}_{i}G{G}^{-1}{\mathrm{\partial }}_{j}G\right)}_{\alpha \beta },{\mathrm{Ric}}_{\alpha i}=0,$ ${\mathrm{Ric}}_{ij}=\mathrm{Ric}(h{)}_{ij}-{\mathrm{\partial }}_i{W}_j+{\mathrm{\Gamma }}^k{}_{ij}{W}_k-\frac{1}{4}\mathrm{tr}(P_i{P}_j)\mathrm{.}$ Indeed, for the fiber contraction the derivative and connection-trace terms are ${\mathrm{\partial }}_i{\mathrm{\Gamma }}^i{}_{\alpha \beta }+({\mathrm{\Gamma }}^k{}_{ki}+W_i){\mathrm{\Gamma }}^i{}_{\alpha \beta }$; combining them gives the divergence with volume factor $S$, while the two mixed-block quadratic terms give the displayed matrix product. In the base contraction, the base-only terms are $\mathrm{Ric}(h{)}_{ij}$, the fiber trace contributes $-{\mathrm{\partial }}_i{W}_j+{\mathrm{\Gamma }}^k{}_{ij}{W}_k$, and the remaining two mixed-block products give $-\frac{1}{4}\mathrm{tr}(P_i{P}_j)$. Each mixed contraction vanishes from the listed connection families and ${\mathrm{\partial }}_{\alpha }G={\mathrm{\partial }}_{\alpha }h=0$. Using the six base coefficients in step 2.1 in the two-dimensional Ricci formula gives $\mathrm{Ric}(h{)}_{rr}=\frac{Mr(3a^2{u}^2-r^2)}{S^{2}D},\mathrm{Ric}(h{)}_{ru}=0,\mathrm{Ric}(h{)}_{uu}=\frac{Mr(3a^2{u}^2-r^2)}{S^{2}q}\mathrm{.}$ For clarity, set $H_{ij}:=-{\mathrm{\partial }}_i{W}_j+{\mathrm{\Gamma }}^k{}_{ij}{W}_k$ and $T_{ij}:=\frac{1}{4}\mathrm{tr}(P_i{P}_j)$. For $(r,r)$, put $P_{rr}=M^2{a}^2{u}^2+3M^2{r}^2-2M{a}^{2}r{u}^2-M{a}^{2}r-3M{r}^3+a^2{r}^2{u}^2+r^4$ and $Q_{rr}=M^2{a}^4{u}^4-2M^2{a}^2{r}^2{u}^2+5M^2{r}^4-2M{a}^{4}r{u}^4+2M{a}^{4}r{u}^2-2M{a}^2{r}^3{u}^2-2M{a}^2{r}^3-4M{r}^5+a^4{r}^2{u}^4+2a^2{r}^4{u}^2+r^6$. Direct substitution gives $H_{rr}=P_{rr}\mathrm{/}(S{D}^2)$ and $T_{rr}=Q_{rr}\mathrm{/}(S^2{D}^2)$, so ${\mathrm{Ric}}_{rr}=\frac{Mr(3a^2{u}^2-r^2)D+S{P}_{rr}-Q_{rr}}{S^2{D}^2}=0.$ For $(r,u)$, put $B=-M{a}^2{u}^2+M{a}^2-2M{r}^2+a^{2}r{u}^2+r^3$. Direct substitution gives $H_{ru}=T_{ru}=-uB\mathrm{/}(qSD)$, while $\mathrm{Ric}(h{)}_{ru}=0$, so ${\mathrm{Ric}}_{ru}=0$. For $(u,u)$, put $P_{uu}=-Mr{u}^2+Mr+a^2{u}^4+r^2{u}^2$ and $Q_{uu}=u^2(-4M{a}^{2}r{u}^2+4M{a}^{2}r+a^4{u}^4+2a^2{r}^2{u}^2+r^4)$. Then $H_{uu}=P_{uu}\mathrm{/}(q^{2}S)$ and $T_{uu}=Q_{uu}\mathrm{/}(q^2{S}^2)$, whence ${\mathrm{Ric}}_{uu}=\frac{Mr(3a^2{u}^2-r^2)q+S{P}_{uu}-Q_{uu}}{q^2{S}^2}=0.$ Both displayed numerators vanish by expansion using $S=r^2+a^2{u}^2$, $D=r^2-2Mr+a^2$, and $q=1-u^2$. These expressions include the derivative, trace, and quadratic-connection contributions with the signed convention [F4]-[F5].

In the fiber formula of step 3.1, $S{h}^{rr}=D$ and $S{h}^{uu}=q$. Define ${\mathcal{C}}_r:=-\frac{1}{2S}{\mathrm{\partial }}_r(D{\mathrm{\partial }}_{r}G)+\frac{D}{2S}({\mathrm{\partial }}_{r}G)G^{-1}({\mathrm{\partial }}_{r}G),{\mathcal{C}}_u:=-\frac{1}{2S}{\mathrm{\partial }}_u(q{\mathrm{\partial }}_{u}G)+\frac{q}{2S}({\mathrm{\partial }}_{u}G)G^{-1}({\mathrm{\partial }}_{u}G)\mathrm{.}$ Thus $({\mathrm{Ric}}_{\alpha \beta }{)}_{\alpha ,\beta =t,\varphi }={\mathcal{C}}_r+{\mathcal{C}}_u$. Put $z:=u^2$ and $U:=-4Mrz-a^2{z}^2+3a^{2}z+3r^{2}z-r^2,V:=4M{a}^{2}rz-3a^{4}z-3a^2{r}^{2}z+a^2{r}^2+r^4,$ $Z:=-8M^2{a}^4{r}^2{z}^2+8M^2{a}^4{r}^{2}z-2M{a}^{6}r{z}^3+4M{a}^{6}r{z}^2-6M{a}^{6}rz-2M{a}^4{r}^3{z}^2-8M{a}^4{r}^{3}z+2M{a}^4{r}^3-8M{a}^2{r}^{5}z+4M{a}^2{r}^5+a^8{z}^3+a^6{r}^2{z}^3+3a^6{r}^2{z}^2+3a^4{r}^4{z}^2+3a^4{r}^{4}z+3a^2{r}^{6}z+a^2{r}^6+r^8\mathrm{.}$ Substituting the explicit $G,G^{-1}$ of step 1.1 and differentiating entrywise gives the three independent entries ${\mathcal{C}}_r=\frac{1}{S^4}\left(\begin{array}{cc}2M{a}^{2}rU& 2MarqV\\ 2MarqV& -qZ\end{array}\right),{\mathcal{C}}_u=-\frac{1}{S^4}\left(\begin{array}{cc}2M{a}^{2}rU& 2MarqV\\ 2MarqV& -qZ\end{array}\right)\mathrm{.}$ The $tt$, $t\varphi$, and $\varphi \varphi$ entries therefore cancel separately. The mixed entries are zero by step 3.1, and all three base entries are zero by the component identities there. Hence every component of the Ricci tensor vanishes on the axis-avoiding polar chart, with no scalar-curvature shortcut.

The polar chart omits only the axis. By [A1] and [F2], $g$ is smooth there, and [F4]-[F5] make $\mathrm{Ric}(g)$ a smooth tensor. Since its components vanish on the dense axis-avoiding set by step 4.1, continuity makes $\mathrm{Ric}(g)$ vanish on the axis as well.

Thus $\mathrm{Ric}(g)=0$ at every point with $r>r_{+}$. The calculation proves a tensor identity for this specified metric; no Einstein field equation or classification of other metrics was used.