Source-free rotational Maxwell fields have constant fluxes

Fix $p\in S^2$ and put $H_p:=2p{p}^{\mathsf{T}}-I_3$. This matrix fixes $p$ and acts as $-I$ on $p^{\perp }=T_p{S}^2$; its eigenvalues are $1,-1,-1$, so $H_p^{\mathsf{T}}{H}_p=I_3$ and $\det H_p=1$, hence $H_p\in SO(3)$. Its lift fixes $(y,p)$, is the identity on $T_{y}U$, and is $-I$ on $T_p{S}^2$. The decomposition ${\mathrm{\Lambda }}^2(T_y^{\ast }U\oplus T_p^{\ast }{S}^2)={\mathrm{\Lambda }}^2{T}_y^{\ast }U\oplus (T_y^{\ast }U\otimes T_p^{\ast }{S}^2)\oplus {\mathrm{\Lambda }}^2{T}_p^{\ast }{S}^2$ is preserved by the lifted rotation, while the mixed summand changes sign. Invariance therefore forces the mixed component of ${\mathcal{F}}_{(y,p)}$ to equal its negative, so that component is zero. The first and last summands are one-dimensional, with generators ${ϵ}_2$ and ${\omega }_{S^2}$. The base form is preserved because each lift is the identity on $U$, and $(R^{\ast }{\omega }_{S^2}{)}_p(u,v)=\det (Rp,Ru,Rv)=\det (R)\det (p,u,v)={\omega }_{S^2,p}(u,v)$ for $R\in SO(3)$. To prove transitivity, take $p,p^{\mathrm{\prime }}\in S^2$, choose unit vectors $u\perp p$ and $u^{\mathrm{\prime }}\perp p^{\mathrm{\prime }}$, and set $v=p\times u$ and $v^{\mathrm{\prime }}=p^{\mathrm{\prime }}\times u^{\mathrm{\prime }}$. The oriented orthonormal frames $(p,u,v)$ and $(p^{\mathrm{\prime }},u^{\mathrm{\prime }},v^{\mathrm{\prime }})$ are related by a unique orientation-preserving orthogonal map $R$, which lies in $SO(3)$ and sends $p$ to $p^{\mathrm{\prime }}$. Thus invariance makes the coefficients of ${ϵ}_2$ and ${\omega }_{S^2}$ independent of $p$. For a fixed $p_{\ast }\in S^2$, the smooth section $j(y):=(y,p_{\ast })$ gives $j^{\ast }\mathcal{F}=E{ϵ}_2$. In any base chart the coefficient of the nowhere-vanishing smooth form ${ϵ}_2$ is nonzero, so division by it shows that $E$ is smooth locally, hence on $U$. To recover the vertical coefficient instead, take $p_{\ast }=(0,0,1)$, $v=(1,0,0)$ and $w=(0,1,0)$ in $T_{p_{\ast }}{S}^2$. Along $j$, the product tangent vectors $(0,v)$ and $(0,w)$ are smooth sections of $j^{\ast }T\mathcal{M}$: this follows in any fixed sphere chart about $p_{\ast }$, and requires no frame on all of $S^2$. Smoothness of $\mathcal{F}$ therefore makes $y\mapsto {\mathcal{F}}_{(y,p_{\ast })}((0,v),(0,w))$ smooth. Since ${\omega }_{S^2,p_{\ast }}(v,w)=\det (p_{\ast },v,w)=1$, this function is exactly $B(y)$, proving its smoothness. Consequently, uniquely $\mathcal{F}=E(y){ϵ}_2+B(y){\omega }_{S^2}$.

Since ${ϵ}_2$ and ${\omega }_{S^2}$ are top forms on their respective two-dimensional factors, their exterior derivatives vanish, and $d\mathcal{F}=dB\wedge {\omega }_{S^2}$. The form ${\omega }_{S^2}$ never vanishes, so $d\mathcal{F}=0$ implies $dB=0$. A smooth function with zero differential is constant on connected $U$ (differentiate along each smooth path in a coordinate-ball chain); call this constant $q_m$.

In an oriented orthonormal coframe with ${ϵ}_2=-e^0\wedge e^1$ and $r^2{\omega }_{S^2}=e^2\wedge e^3$, the four-volume is $e^0\wedge e^1\wedge e^2\wedge e^3$. The Lorentzian Hodge definition therefore gives ${\star }_g{ϵ}_2=r^2{\omega }_{S^2}$ and ${\star }_g{\omega }_{S^2}=-r^{-2}{ϵ}_2$. Hence ${\star }_g\mathcal{F}=E{r}^2{\omega }_{S^2}-B{r}^{-2}{ϵ}_2$, and $d({\star }_g\mathcal{F})=d(E{r}^2)\wedge {\omega }_{S^2}$ because a base one-form wedged with the base top form ${ϵ}_2$ vanishes. The second Maxwell equation implies $d(E{r}^2)=0$, so $q_e:=E{r}^2$ is constant on connected $U$. Restricting $\mathcal{F}$ and its Hodge dual to $S_y$ gives $q_m{\omega }_{S^2}$ and $q_e{\omega }_{S^2}$; since ${\omega }_{S^2}=\sin \theta d\theta \wedge d\varphi$ and ${\int }_0^{\pi }\sin \theta d\theta {\int }_0^{2\pi }d\varphi =4\pi$, the stated flux normalizations follow. This proves existence and uniqueness of both constants.