Kerr exterior tensor including the extremal parameter range

Factorization gives $\mathrm{\Delta }=(r-r_{+})(r-r_{-})>0$ for $r>r_{+}$, even if $r_{+}=r_{-}=M$. Also $\mathrm{\Sigma }\ge r^2>0$. With $\gamma =d{\theta }^2+{\sin }^2\theta d{\varphi }^2$ the round metric and $\alpha ={\sin }^2\theta d\varphi$ the smooth round-metric dual of the rotation field, write the angular part as $\mathrm{\Sigma }\gamma +a^2(1+2Mr\mathrm{/}\mathrm{\Sigma }){\alpha }^2$ and the mixed term as $-4Mardt\alpha \mathrm{/}\mathrm{\Sigma }$. Use this same sphere-tensor expression also when $\mathrm{\mid }a\mathrm{\mid }=M$; on polar charts it agrees with the coordinate supplier after setting $q=0$ and $x^0=t$. The positivity just proved makes every denominator nonzero, so this is a global smooth tensor for both parameter ranges since $\gamma$ and $\alpha$ are smooth on $S^2$; the construction uses no division by $\sin \theta$.

Away from the axis the $t,\varphi$ block has determinant $-\mathrm{\Delta }{\sin }^2\theta <0$ and the other two diagonal entries are positive, giving signature $(-,+,+,+)$. Its determinant relative to $dtdr$ and the round-sphere area density is $-{\mathrm{\Sigma }}^2$, which stays negative at the poles by smoothness. Thus nondegeneracy and signature extend over the poles.

The supplier's smooth sphere field is $\xi (x,y,z)=(-y,x,0)$. Its flow is $R_{\lambda }(x,y,z)=(x\cos \lambda -y\sin \lambda ,x\sin \lambda +y\cos \lambda ,z)$, whose derivative in $\lambda$ is $\xi$ at the image. This flow preserves $z$, the round metric $\gamma$ and $\xi$, hence also $\alpha =\gamma (\xi ,\cdot )$. It therefore preserves every term of the global tensor in step 1.1. On polar charts it sends $\varphi$ to $\varphi +\lambda$, giving the positive azimuthal orientation, and has least positive period $2\pi$ away from the axis; its poles are fixed. Thus it defines the stated axial rotation action of period $2\pi$, with smooth generator $\eta ={\mathrm{\partial }}_{\varphi }$ away from the axis. Translations in $t$ also preserve every term. Differentiating these pullback identities gives ${\mathcal{L}}_{\eta }g={\mathcal{L}}_{T}g=0$, so both fields are globally Killing.

At $a=0$, $r_{+}=2M$, $\mathrm{\Sigma }=r^2$ and $\mathrm{\Delta }=r(r-2M)$. Substitution gives $g=-(1-2M\mathrm{/}r)d{t}^2+(1-2M\mathrm{/}r{)}^{-1}d{r}^2+r^2\gamma$ on ${\mathbb{R}}_t\times (2M,\mathrm{\infty })\times S^2$. This is precisely the global Schwarzschild exterior tensor in [F2] under $s=2M$ and $x^0=t$, including the sphere poles.