Charged KN ring regularity obstruction

By the exact current interface [F1], along the sequence ${\mathrm{Ric}}_{ab}{\mathrm{Ric}}^{ab}(x_n)=\frac{4q^4}{\mathrm{\Sigma }(x_n{)}^4}\mathrm{.}$ Since $q\mathrm{\ne }0$, the numerator is a fixed positive number and $\mathrm{\Sigma }(x_n)>0$ tends to zero, so these scalar values tend to $+\mathrm{\infty }$. Isometry makes the same scalar values those of $\tilde{g}$ at $j(x_n)$.

A $C^2$ nondegenerate metric has continuous inverse coefficients and continuous curvature coefficients: the inverse is a continuous matrix inverse, Christoffel symbols use the first derivatives of the metric and its inverse, and curvature uses their first derivatives and products. Therefore the contraction ${\stackrel{}{\mathrm{Ric}}}_{ab}{\stackrel{}{\mathrm{Ric}}}^{ab}$ is continuous near $p$ and has a finite value there. Since $j(x_n)\to p$, continuity forces its values along that sequence to converge to the finite value at $p$, contradicting the divergence established in the first step. Thus the proposed extension cannot contain this approached point, conditional on [F1].

If $a\mathrm{\ne }0$, then $\mathrm{\Sigma }=r^2+a^2{\cos }^2\theta$ is the sum of two nonnegative terms. Along the stated sequence, $\mathrm{\Sigma }\to 0$ forces each term to tend to zero because $0\le r^2\le \mathrm{\Sigma }$ and $0\le a^2{\cos }^2\theta \le \mathrm{\Sigma }$. Thus $r\to 0$ and $\cos \theta \to 0$. With $0<\theta <\pi$, this gives $\theta \to \pi \mathrm{/}2$. At the limiting coordinate locus $r=0$, $\theta =\pi \mathrm{/}2$ the map in the Statement gives $X=\mathrm{\mid }a\mathrm{\mid }\cos \varphi$, $Y=\mathrm{\mid }a\mathrm{\mid }\sin \varphi$, and $Z=0$, hence $X^2+Y^2=a^2$, $Z=0$. This labels the limiting locus as a ring, without asserting the map is a chart through the disk. If $q=0$, [F1]'s invariant is zero wherever defined, so step 1.1 supplies no divergence argument.