Constant-radius Bertotti–Robinson data are outside areal-coordinate rigidity

On the branch $f(z)=z^2$ of [F1], take $U={\mathbb{R}}_{\tau }\times (0,\mathrm{\infty }{)}_z$, $q=L^2(-z^{2}d{\tau }^2+z^{-2}d{z}^2)$, ${ϵ}_2=-L^{2}d\tau \wedge dz$, future timelike field $N=(Lz{)}^{-1}{\mathrm{\partial }}_{\tau }$, and constant $r=L$. This is a nonempty, connected, simply connected smooth Lorentzian orbit patch with the orientation, time orientation, and product metric of [F5]; its prescribed four-volume form is $L^{4}d\tau \wedge dz\wedge {\omega }_{S^2}$, matching [F1]. For each $R\in SO(3)$ the lift ${\rho }_R(\tau ,z,p)=(\tau ,z,Rp)$ fixes the base two-form $\mathcal{F}=-Ld\tau \wedge dz$, so ${\rho }_R^{\ast }\mathcal{F}=\mathcal{F}$; also $R$ preserves the induced round metric. Thus the field and metric satisfy the rotational-invariance and warped-product clauses in [F5] and [F6]. Each symmetry sphere has induced metric $L^2{\gamma }_{S^2}$ and area $L^2{\int }_0^{\pi }{\int }_0^{2\pi }\sin \theta d\varphi d\theta =4\pi L^2$ by [F4]. Its areal radius is therefore the constant scalar $r=L$, so $dr=0$ throughout the product patch. A function whose differential vanishes cannot serve as a coordinate in a chart, because the differential of every coordinate function is nonzero.

By [F2], [F5], [F6], and step 1.1 this is an exact spherical Einstein–Maxwell datum with $\mathrm{\Lambda }=0$, yet it does not admit its areal radius as a coordinate. It therefore refutes the unrestricted claim: the Reissner–Nordström–(A)dS areal-coordinate form needs a nonzero areal-radius gradient. This example is excluded by [F3] and makes no claim about whether other coordinates or limiting relations connect the two geometries.