Kerr-Newman Maxwell and Ricci invariants

The coordinate entries and inverse in [F1] give the exact contraction ${\mathcal{F}}_{ab}{\mathcal{F}}^{ab}=2(B_m^2-E^2)$ with the algebraic abbreviations $E=\frac{q(r^2-a^2{y}^2)}{{\mathrm{\Sigma }}^2},B_m=\frac{2qary}{{\mathrm{\Sigma }}^2},y=\cos \theta \mathrm{.}$ Expanding $(r^2-a^2{y}^2{)}^2-4a^2{r}^2{y}^2=r^4-6a^2{r}^2{y}^2+a^4{y}^4$ gives the field scalar in [F1]. The same exact rational output gives $S=0$ and ${\mathrm{Ric}}_{ab}{\mathrm{Ric}}^{ab}=4q^4\mathrm{/}{\mathrm{\Sigma }}^4$.

Under [A1], $4q^4\mathrm{/}\mathrm{\Sigma }(x_n{)}^4\to +\mathrm{\infty }$. Suppose, for contradiction, that a $C^2$ nondegenerate extension $(\tilde{M},\tilde{g},j)$ and an added point $p\mathrm{\notin }j(U)$ realize this approach, so $j(x_n)\to p$ as in [F2]. The diverging Ricci-square is a polynomial complete curvature contraction from [F1], contradicting [F2]. If $q=0$, the displayed Ricci-square vanishes and this argument gives no obstruction.

For $a\mathrm{\ne }0$, define $X=\sqrt{r^2+a^2}\sin \theta \cos \varphi ,Y=\sqrt{r^2+a^2}\sin \theta \sin \varphi ,Z=r\cos \theta \mathrm{.}$ At $r=0$, $\theta =\pi \mathrm{/}2$, these satisfy $X^2+Y^2=a^2$ and $Z=0$, the ring. This map labels the excluded locus only; it is not asserted to be an injective coordinate chart through the disk.