Bertotti-Robinson has no non-null areal-radius coordinate

Parametrize the unit sphere away from its seam and poles by $\mathrm{\Phi }(\theta ,\varphi )=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$. The tangent vectors ${\mathrm{\Phi }}_{\theta }$ and ${\mathrm{\Phi }}_{\varphi }$ are orthogonal, with squared lengths $1$ and ${\sin }^2\theta$, so the induced metric is $d{\theta }^2+{\sin }^2\theta d{\varphi }^2$ and its outward area form is ${\omega }_{S^2}=\sin \theta d\theta \wedge d\varphi$. The omitted seam and poles are chart-boundary sets of two-dimensional measure zero. Hence ${\int }_{S^2}{\omega }_{S^2}={\int }_0^{2\pi }{\int }_0^{\pi }\sin \theta d\theta d\varphi =4\pi$. The radius-$L$ sphere factor has area form $L^2{\omega }_{S^2}$ and area $4\pi L^2$. Thus its area-radius function is $R_A=\sqrt{(4\pi L^2)\mathrm{/}(4\pi )}=L$ at every $(\tau ,z)$.

The function $R_A$ is constant, so $d{R}_A=0$ everywhere. If it were a coordinate function in a chart $\phi :U\to V\subset {\mathbb{R}}^4$ at $p$, let $e_j$ be the coordinate vector for its component. Since $V$ is open, $\phi (p)+t{e}_j\in V$ for all sufficiently small $t$, and the chart's smooth coordinate parametrization gives a smooth curve $\gamma (t)={\phi }^{-1}(\phi (p)+t{e}_j)$. Then $R_A(\gamma (t))={\phi }^j(p)+t$, whose derivative at $t=0$ is $1$; thus $d{R}_A{\mathrm{\mid }}_p({\gamma }^{\mathrm{\prime }}(0))=1$, contradicting $d{R}_A{\mathrm{\mid }}_p=0$. Therefore $R_A$ cannot be a local coordinate.

By [F2], the example is an actual source-free electrovac product datum, so the failure is not caused by a missing field-equation condition. It refutes the universal claim in the Statement.