Kruskal model properties: smooth radial function, Lorentzian metric, four-region isometries, vacuum and Kretschmann scalar

(The radial function.) By [L6], $\phi (r)=(1-r\mathrm{/}s)e^{r\mathrm{/}s}$ is smooth with ${\phi }^{\mathrm{\prime }}(r)=-\frac{r}{s^2}{e}^{r\mathrm{/}s}<0$ for $r>0$, so $\phi$ is strictly decreasing by [L4], applied on every compact subinterval of $(0,\mathrm{\infty })$. Its limits are $1$ as $r\downarrow 0$ and $-\mathrm{\infty }$ as $r\to \mathrm{\infty }$: for $x=r\mathrm{/}s>1$, the defining exponential series gives $e^x\ge 1+x$, so $(x-1)e^x\ge (x-1)(1+x)\to \mathrm{\infty }$. By [L4] its image is exactly $(-\mathrm{\infty },1)$, hence it is a bijection. Its derivative never vanishes; the smooth inverse-function theorem gives a smooth local inverse at each point of the image, and bijectivity makes those local inverses agree with the global inverse. Thus $r={\phi }^{-1}\circ (UV)$ is defined and smooth on all of $\mathrm{\Omega }$, justifying clause (c) of the datum. The level relation gives $r=s$ exactly when $UV=0$ and $r\downarrow 0$ as $UV\uparrow 1$. This proves (a).

(The metric and time orientation, including the poles.) The base block of the global tensor [L1] in $(U,V)$ is $\left(\begin{array}{cc}0& b\\ b& 0\end{array}\right)$ with $b=-2s^3{e}^{-r\mathrm{/}s}\mathrm{/}r<0$, so it is nondegenerate with one positive and one negative eigenvalue. The angular block $r^2{\gamma }_{S^2}$ is positive definite at every sphere point, including both poles. Thus the global metric has inertia $(3,1,0)$. For $N_{\mathcal{K}}={\mathrm{\partial }}_U+{\mathrm{\partial }}_V$, $g_{\mathcal{K}}(N_{\mathcal{K}},N_{\mathcal{K}})=2b=-4s^3{e}^{-r\mathrm{/}s}\mathrm{/}r<0$ everywhere, so it is a time orientation. This proves (b).

(The isometry with the exterior.) On ${\mathrm{\Omega }}_{+}=\{U<0,V>0\}$ define $U=-e^{-(x^0-r_{\ast })\mathrm{/}(2s)}$, $V=e^{(x^0+r_{\ast })\mathrm{/}(2s)}$ with the branch $r>s$ of $r_{\ast }$; then $UV=-e^{r_{\ast }\mathrm{/}s}=-(r\mathrm{/}s-1)e^{r\mathrm{/}s}$ by [L2] (since $e^{r_{\ast }\mathrm{/}s}=(r\mathrm{/}s-1)e^{r\mathrm{/}s}$ on $r>s$), so the function defined implicitly satisfies $r(U,V)=r$ and the map ${\mathrm{\Psi }}_{+}$ is a diffeomorphism onto $\{U<0,V>0\}$ with inverse $(U,V,\omega )\mapsto (s\log (V\mathrm{/}(-U)),r(U,V),\omega )$. Differentiating, $dU=-\frac{U}{2s}(d{x}^0-r_{\ast }^{\mathrm{\prime }}dr)$ and $dV=\frac{V}{2s}(d{x}^0+r_{\ast }^{\mathrm{\prime }}dr)$, hence $dUdV+dVdU=-\frac{UV}{2s^2}(d{x}^0\otimes d{x}^0-r_{\ast }^{\mathrm{\prime }2}dr\otimes dr)$. Substituting into $g_{\mathcal{K}}$ and using $\frac{sUV{e}^{-r\mathrm{/}s}}{r}=-f$ and $r_{\ast }^{\mathrm{\prime }2}=f^{-2}$, the pulled-back metric is $-fd(x^0{)}^2+f^{-1}d{r}^2+r^{2}d{\mathrm{\Omega }}^2=g$; this is the computation of Smooth horizon-crossing extension of the Schwarzschild exterior metric in the advanced chart restricted to the exterior. On the component $\{U>0,V>0\}$ define instead $U=e^{(r_{\ast }-x^0)\mathrm{/}(2s)}$, $V=e^{(r_{\ast }+x^0)\mathrm{/}(2s)}$ with the branch $r_{\ast }=r+s\log (1-r\mathrm{/}s)$ of the interval $(0,s)$; then $UV=e^{r_{\ast }\mathrm{/}s}=(1-r\mathrm{/}s)e^{r\mathrm{/}s}=-(r\mathrm{/}s-1)e^{r\mathrm{/}s}$, so again $r(U,V)=r$, and $dUdV+dVdU=\frac{UV}{2s^2}(d{r}_{\ast }^2-d{x}^0\otimes d{x}^0)$; the same substitution with $r_{\ast }^{\mathrm{\prime }}=1\mathrm{/}f$ (negative there) gives $-fd(x^0{)}^2+f^{-1}d{r}^2+r^{2}d{\mathrm{\Omega }}^2$ again, now with $f<0$. Finally the map $(U,V,\omega )\mapsto (-U,-V,\omega )$ leaves both $UV$ and $dUdV+dVdU$ unchanged and fixes $r$, $g_{\vartheta \vartheta }$ and $g_{\phi \phi }$; it is therefore an isometry, and it exchanges the two components with $UV<0$ and the two with $UV>0$. This proves (c).

(The four connected components.) No connected subset of $\{UV\mathrm{\ne }0\}$ can change the sign of either continuous nonzero coordinate. Step 2.1 and simultaneous sign reversal identify each sign region with $\mathbb{R}\times J\times S^2$, where $J=(s,\mathrm{\infty })$ or $(0,s)$. These domains are path connected: interpolate the real and radial coordinates linearly; any two sphere points can be joined by normalized straight segments, inserting an intermediate unit coordinate vector not antipodal to either endpoint when needed (among the six signed coordinate vectors at most two are excluded). All segment normalizations have nonzero denominators. Hence each sign region is connected, and the four are exactly the components asserted in (c).

(Vacuum and Kretschmann scalar.) By step 2.1, on every component with $UV\mathrm{\ne }0$ the metric is isometric to a branch of the spherical metric with $f=1-s\mathrm{/}r$. By [L3] each branch is isometric to the corresponding open region of the advanced extension, whose curvature identities hold for all $r>0$. Uniqueness of the metric-selected connection transports it under these isometries, and therefore transports its curvature tensor and metric contractions. Thus $\mathrm{Ric}=S=G=0$ and the Kretschmann scalar is $12s^2\mathrm{/}{r}^6$ on both branches and their sign-reversed copies. These identities hold on the dense open subset $D=\{UV\mathrm{\ne }0\}\times (S^2\setminus \{\text{two poles}\})$. The global metric is smooth and nondegenerate by step 1.2, so its Ricci tensor and Kretschmann scalar are smooth. In any smooth product chart their components are continuous and agree on the dense subset $D$ with zero and with $12s^2\mathrm{/}{r}^6$, respectively; [L5] extends the identities over the axes and both poles. This proves (d), including $K=12\mathrm{/}{s}^4$ at $r=s$ and $K\to \mathrm{\infty }$ as $UV\to 1$. Clause (e) records the scope.

(Conclusion.) Steps 1.1-3.2 prove clauses (a)-(d) from the datum's definitions and the all-positive-radius advanced-extension supplier; the claims about maximality and completeness are explicitly not made, as recorded in (e).