Dyonic Reissner-Nordström metric and field domains

Let $M,q_e,q_m\in \mathbb{R}$ be length parameters and put $Q^2:=q_e^2+q_m^2$. Let $J\subset (0,\mathrm{\infty })$ be open and require $f(r):=1-\frac{2M}{r}+\frac{Q^2}{r^2}\mathrm{\ne }0(r\in J)\mathrm{.}$ On the smooth product ${\mathcal{M}}_J:={\mathbb{R}}_{x^0}\times J_r\times S^2$, use $S^2=\{(X,Y,Z)\in {\mathbb{R}}^3:X^2+Y^2+Z^2=1\}$ with its subspace topology (Euclidean spheres and closed balls as subspaces of ${\mathbb{R}}^n$). Specify the smooth structure by the sphere atlas of The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation, Definition (b): for $\epsilon =\pm 1$, delete $p_{\epsilon }=(0,0,\epsilon )$ and use ${\sigma }_{\epsilon }(X,Y,Z)=(X,Y)\mathrm{/}(1-\epsilon Z)$. For $u=(u_1,u_2)$ and $h=u_1^2+u_2^2$, its inverse is ${\sigma }_{\epsilon }^{-1}(u)=(2u_1,2u_2,\epsilon (h-1))\mathrm{/}(1+h)$; the transition is $u\mapsto u\mathrm{/}h$ on $h>0$. These rational maps are smooth on their domains. The product atlas consists of $(x^0,r,p)\mapsto (x^0,r,{\sigma }_{\epsilon }(p))$ on $\mathbb{R}\times J\times (S^2\setminus \{p_{\epsilon }\})$; its transitions leave $x^0,r$ fixed and apply the displayed sphere transition. The product is Hausdorff and second countable by the subspace topology from ${\mathbb{R}}^2\times {\mathbb{R}}^3$, whose rational balls give a countable basis. These charts specify its maximal smooth atlas, including when $J$ is empty. The inclusion of $S^2$ into ${\mathbb{R}}^3$ is smooth with injective differential in these parametrizations; its induced unit round metric is ${\gamma }_{S^2}$ (The round metric on the sphere as an induced metric). Orient $S^2$ by its outward unit normal: for $p\in S^2$ and $u,v\in T_p{S}^2$, define $({\omega }_{S^2}{)}_p(u,v):=\det (p,u,v)\mathrm{.}$ Thus ${\omega }_{S^2}$ is the positive area form of ${\gamma }_{S^2}$, and in the polar chart $X=\sin \theta \cos \varphi$, $Y=\sin \theta \sin \varphi$, $Z=\cos \theta$ it is $\sin \theta d\theta \wedge d\varphi$. The dyonic RN datum is $g=-f(r)d{x}^0\otimes d{x}^0+f(r{)}^{-1}dr\otimes dr+r^2{\gamma }_{S^2},\mathcal{F}=-\frac{q_e}{r^2}d{x}^0\wedge dr+q_m{\omega }_{S^2}\mathrm{.}$ Here $x^0$ and $r$ have units of metres, angles on $S^2$ are dimensionless, and $M,q_e,q_m$ have units of metres. The metric components are dimensionless in length coordinates; the nonzero coordinate components of $\mathcal{F}$ have units inherited from their coordinate Jacobians. Use the orientation $d{x}^0\wedge dr\wedge {\omega }_{S^2}$.

The set $J$ may have several connected components. On $\{f>0\}$ choose the time orientation ${\mathrm{\partial }}_{x^0}$; on $\{f<0\}$ choose ${\mathrm{\partial }}_r$. These are smooth timelike fields on their respective open subsets because $g({\mathrm{\partial }}_{x^0},{\mathrm{\partial }}_{x^0})=-f<0$ when $f>0$ and $g({\mathrm{\partial }}_r,{\mathrm{\partial }}_r)=f^{-1}<0$ when $f<0$. Since $f$ has no zero on ${\mathcal{M}}_J$, this defines a smooth time orientation on the whole product. The sphere area form and the displayed two-form are global on $S^2$.

Let $N,S$ denote the north and south poles. On $U_N:=S^2\setminus \{S\}$ and $U_S:=S^2\setminus \{N\}$, respectively, local potential one-forms are ${\mathcal{A}}_N=-\frac{q_e}{r}d{x}^0+q_m(1-\cos \theta )d\varphi ,{\mathcal{A}}_S=-\frac{q_e}{r}d{x}^0-q_m(1+\cos \theta )d\varphi \mathrm{.}$ Each displayed potential extends smoothly over its named included pole: $(1-\cos \theta )d\varphi$ is smooth at $N$ and $(1+\cos \theta )d\varphi$ is smooth at $S$. Direct differentiation gives $d{\mathcal{A}}_N=d{\mathcal{A}}_S=\mathcal{F}$ on the respective patches. Their overlap difference is $2q_{m}d\varphi$; it is local potential data, not a global real gauge function on the entire sphere. The definition assumes no $U(1)$ bundle or charge quantization. To verify pole smoothness, use local coordinates $u=X,v=Y$ near each pole. There $d\varphi =(udv-vdu)\mathrm{/}(u^2+v^2)$. Near $N$, $1-\cos \theta =(u^2+v^2)\mathrm{/}(1+Z)$, so $(1-\cos \theta )d\varphi =(udv-vdu)\mathrm{/}(1+Z)$ is smooth. Near $S$, $1+\cos \theta =(u^2+v^2)\mathrm{/}(1-Z)$, so $(1+\cos \theta )d\varphi =(udv-vdu)\mathrm{/}(1-Z)$ is smooth. The denominators are nonzero on the respective named patches.