Dyonic Kerr-Newman parameters and conserved electric/magnetic flux

On the sphere let $X=\sin \theta \cos \varphi$ and $Y=\sin \theta \sin \varphi$. Then $\lambda =XdY-YdX={\sin }^2\theta d\varphi$ away from the poles, and [F3] proves this is a global smooth one-form. At a fixed $(x^0,r)$, represent $S_r$ by the oriented radial projections of the eight octahedral faces; each projection extends smoothly to a neighborhood of its face, shared edges cancel, and the resulting finite smooth singular two-chain has zero boundary. The global potential $\bar{\mathcal{A}}=-qr{\mathrm{\Sigma }}^{-1}(d{x}^0-a\lambda )$ is smooth since $\mathrm{\Sigma }>0$. Hence $\bar{\mathcal{F}}=d\bar{\mathcal{A}}$ and [F5] gives ${\int }_{S_r}\bar{\mathcal{F}}={\int }_{\mathrm{\partial }{S}_r}\bar{\mathcal{A}}=0$.

For the explicit unit-sphere parametrization $s(\theta ,\varphi )=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$, direct differentiation gives $s\cdot (s_{\theta }\times s_{\varphi })=\sin \theta$. Thus its outward area form pulls back to $\sin \theta d\theta \wedge d\varphi$. By [F4], the continuous sine density is integrable and its primitive $-\cos \theta$ evaluates each polar integral. Integrating on $\delta \le \theta \le \pi -\delta$, $0\le \varphi \le 2\pi$ gives $4\pi \cos \delta$; the two caps have total area $4\pi (1-\cos \delta )$ by the same bounded polar density, tending to zero. Hence ${\int }_{S^2}\omega =4\pi$. In the oriented $(x^0,r,\theta ,\varphi )$ coordinates, direct raising of ${\mathcal{F}}_{r0}=q(r^2-a^2{y}^2)\mathrm{/}{\mathrm{\Sigma }}^2$ and ${\mathcal{F}}_{r\varphi }=-a{\sin }^2\theta {\mathcal{F}}_{r0}$ with the KN inverse metric gives $({\star }_g\mathcal{F}{)}_{\theta \varphi }=\frac{q(r^2+a^2)(r^2-a^2{y}^2)\sin \theta }{{\mathrm{\Sigma }}^2}=q\sin \theta +O(r^{-2})\sin \theta ,$ The numerator factor minus ${\mathrm{\Sigma }}^2$ is $a^2{r}^2(1-3y^2)-a^4{y}^2(1+y^2)$, whose absolute value is at most $2a^2{r}^2+2a^4$ for $\mathrm{\mid }y\mathrm{\mid }\le 1$. Since ${\mathrm{\Sigma }}^2\ge r^4$, the remainder is uniformly $O(r^{-2})$ on the sphere. Hence ${\lim }_{r\to \mathrm{\infty }}(4\pi {)}^{-1}{\int }_{S_r}{\star }_g\mathcal{F}=q$.

For $R_0<r_1<r_2$, in the fixed-$x^0$ slice of step 1.1, take the product of each of the same eight projected triangular faces with $[r_1,r_2]$. Each triangular prism splits into three tetrahedra; their oriented sum is a finite smooth singular three-chain $\mathcal{C}$, internal faces cancel, and its boundary is $S_{r_2}-S_{r_1}$. By steps 1.1-1.2 and [F1], $d({\star }_g\mathcal{F})=0$ on this shell. Stokes [F5] therefore gives ${\int }_{S_{r_2}}{\star }_g\mathcal{F}-{\int }_{S_{r_1}}{\star }_g\mathcal{F}={\int }_{\mathrm{\partial }\mathcal{C}}{\star }_g\mathcal{F}={\int }_{\mathcal{C}}d({\star }_g\mathcal{F})=0.$ Step 1.2 identifies the common value as $4\pi q$.

The fluxes of ${\mathcal{F}}_{\beta }$ satisfy ${\star }_g{\mathcal{F}}_{\beta }=\cos \beta {\star }_g\mathcal{F}-\sin \beta \mathcal{F}$ by [F2]. Dividing the two sphere integrals by $4\pi$ and using steps 1.1 and 2.1 gives $q_e=q\cos \beta$ and $q_m=q\sin \beta$.

The metric function $\mathrm{\Delta }=r^2-2Mr+a^2+q^2$ is unchanged because $q_e^2+q_m^2=q^2({\cos }^2\beta +{\sin }^2\beta )=q^2$. The stress tensor is unchanged by [F2], so the same metric satisfies the dyonic geometric electrovac equations. This is a classical flux normalization; no integrality condition follows.