Dyonic RN gauges and global exactness obstruction

Fix $t_0\in \mathbb{R}$. For each $r\in J$, let $C_r$ be the chain obtained by radially projecting the eight outward-oriented triangular faces of the octahedron with vertices $\pm e_1,\pm e_2,\pm e_3$ onto the unit sphere and then embedding that sphere as $t=t_0$, radius $r$. Each face lies in an affine plane not containing the origin, so its radial projection extends smoothly to a neighbourhood of its simplex. Order the six vertices once, parameterize each face in that order, and include it with the sign that makes its radial image outward-oriented. Each common edge is then the same ordered singular 1-simplex in its two incident faces, with opposite boundary coefficients. Thus $C_r$ is a finite smooth singular 2-cycle covering $S_r$ once with its outward orientation. The face interiors partition the sphere, and the polar area form is $\omega =\sin \theta d\theta \wedge d\varphi$; the finitely many face edges have zero 2-dimensional integral. Therefore ${\int }_{C_r}\omega ={\int }_0^{2\pi }{\int }_0^{\pi }\sin \theta d\theta d\varphi =4\pi \mathrm{.}$ Pulling [F1]'s $\mathcal{F}$ back to $C_r$ removes the electric term and leaves $q_m\omega$, so the first flux is $4\pi q_m$. The inverse static metric has $g^{tt}=-1\mathrm{/}f$, $g^{rr}=f$, and the oriented volume form is $r^2\sin \theta dt\wedge dr\wedge d\theta \wedge d\varphi$. In the component formula of [F2], $(dt\wedge dr{)}^{tr}=g^{tt}{g}^{rr}=-1$ and ${\omega }^{\theta \varphi }=1\mathrm{/}(r^4\sin \theta )$. The oriented volume component is ${ϵ}_{\theta \varphi tr}=r^2\sin \theta$, so that formula gives ${\star }_g(dt\wedge dr)=-r^2\omega ,{\star }_g\omega =r^{-2}dt\wedge dr,$ for either sign of $f$. Therefore ${\star }_g\mathcal{F}=q_e\omega +q_m{r}^{-2}dt\wedge dr$; its sphere pullback integrates to $4\pi q_e$.

On $U_N=\{r+z>0\}$ and $U_S=\{r-z>0\}$, respectively, $(1-\cos \theta )d\varphi =\frac{xdy-ydx}{r(r+z)},(1+\cos \theta )d\varphi =\frac{xdy-ydx}{r(r-z)}\mathrm{.}$ The denominators are nonzero on the indicated patches, so both one-forms are smooth there. Differentiation gives $d[(1-\cos \theta )d\varphi ]=d[-(1+\cos \theta )d\varphi ]=\omega$ and $d(-q_e{r}^{-1}dt)=-q_e{r}^{-2}dt\wedge dr$. Thus both potentials differentiate to [F1]. Subtracting them gives ${\mathcal{A}}_S-{\mathcal{A}}_N=-2q_{m}d\varphi$ on the overlap.

If a global smooth one-form $\mathcal{A}$ with $d\mathcal{A}=\mathcal{F}$ existed and $q_m\mathrm{\ne }0$, apply [F3] to the oriented sphere cycle $S_r$: $4\pi q_m={\int }_{S_r}\mathcal{F}={\int }_{S_r}d\mathcal{A}={\int }_{\mathrm{\partial }{S}_r}\mathcal{A}=0,$ since $\mathrm{\partial }{S}_r=0$. This contradiction rules out a global potential when $q_m\mathrm{\ne }0$.

If $q_m=0$, the one-form $\mathcal{A}=-q_e{r}^{-1}dt$ is smooth and global on ${\mathbb{R}}_t\times J\times S^2$, and its exterior derivative is $\mathcal{F}$. This proves the exact criterion.