This is simply given by $$\bar{r}=\Omega(r)=-\frac{1}{r}\,,\qquad
\H(r) = -r^2-2m\, r^3 \,.
\label{Schw}$$ Indeed, the first relation implies ${r=-1/\bar{r}}$, so that ${\H(\bar{r}) = - (1-2m/\bar{r})/\bar{r}^2}$. Using (\[rcehf\]), we easily obtain (\[SchwarzschildBH\]). It should be emphasized that the standard physical range ${\bar{r}>0}$ corresponds to ${r<0}$. Also, the auxiliary Kundt coordinate $r$ *increases from negative values to* $0$, as $\bar{r}$ increases to $\infty$.

Notice that ${\cal H}$ given by (\[Schw\]) is simply a *cubic* in the coordinate $r$ of the Kundt geometry. For ${m=0}$, the Kundt seed with ${{\cal H} = -\,r^2}$ is the Bertotti–Robinson spacetime with the geometry ${S^2\times AdS_2}$ (see chapter 7 of [@GriffithsPodolsky:2009]), and the corresponding conformally related metric (\[confrelation\]) is just the flat space. It should also be emphasized that, while the Schwarzschild and Minkowski spacetimes are (the simplest) vacuum solutions in Einstein’s theory, their Kundt seeds (\[Schw\]) *are not vacuum solutions* in Einstein’s theory since their Ricci tensor is nonvanishing. In fact, the Bertotti–Robinson geometry is an electrovacuum space of Einstein’s theory.

Since conformal transformations preserve the Weyl tensor, both ${{\rm{d}}}s^2$ and ${{\rm{d}}}s^2_{\hbox{\tiny Kundt}} $ are of the *same algebraic type*. Indeed, in the null frame ${{\mbox{\boldmath$k$}}= \mathbf{\partial}_r}$, ${{\mbox{\boldmath$l$}}= {\textstyle\frac{1}{2}}{\cal H}\,\mathbf{\partial}_r+\mathbf{\partial}_u}$, ${{\mbox{\boldmath$m$}}_i = \big(1+\ctvrt(x^2+y^2)\big)\mathbf{\partial}_i}$, the only Newman–Penrose Weyl 