Excluded interior flux is not a global source law

Put $C=\mathrm{\Phi }\mathrm{/}(4\pi )$. On $r>0$, $F=C\beta$ is the smooth static magnetic 2-form, where $\beta =(x_{1}d{x}^2\wedge d{x}^3+x_{2}d{x}^3\wedge d{x}^1+x_{3}d{x}^1\wedge d{x}^2)\mathrm{/}{r}^3$. By [F4], $d\beta =({\sum }_i{\mathrm{\partial }}_i(x_i{r}^{-3}))vo{l}_3$. By [F5], ${\mathrm{\partial }}_i(x_i{r}^{-3})=r^{-3}-3x_i^2{r}^{-5}$, so their sum is $3r^{-3}-3r^2{r}^{-5}=0$. Also, $B=-C\mathrm{\nabla }(r^{-1})$, so $\mathrm{curl}B=0$ by [F6]. With $E=0$, $P=M=0$, and $\rho =J=0$, all four adopted vacuum Maxwell equations hold on $\mathrm{\Omega }$; [F2] therefore gives $dF=0$ and $dG=0$ there.

The origin is the barycenter of the four tetrahedron vertices, hence lies strictly inside $T$, and each face lies in a plane $n\cdot p=h$ with outward unit normal $n$ and $h>0$. Each face avoids the origin, so $q_a$ is smooth on a neighborhood of it. Convexity and $0$ in the interior imply each ray from $0$ meets the boundary of $T$ once; therefore $q_a$ maps the four face interiors bijectively onto disjoint open patches of the radius-$a$ sphere whose closures cover that sphere. On a face patch, the inverse radial map is $x\mapsto hx\mathrm{/}(n\cdot x)$, which is smooth because $n\cdot x>0$ there; thus each face restriction is a diffeomorphism. For face tangent vectors $v,w$ in the outward orientation, put $u=p\mathrm{/}\mathrm{\parallel }p\mathrm{\parallel }$. Then $D(q_a{)}_p(v)=(a\mathrm{/}\mathrm{\parallel }p\mathrm{\parallel })(v-(u\cdot v)u)$ and $u\cdot (D(q_a{)}_p(v)\times D(q_a{)}_p(w))=(a^2\mathrm{/}\mathrm{\parallel }p{\mathrm{\parallel }}^3)p\cdot (v\times w)>0$, because $p$ has positive normal component $hn$. The ordered tetrahedron has determinant $\det (v_1-v_0,v_2-v_0,v_3-v_0)=-16a^3$, so its positive outward boundary has coefficients ${\epsilon }_i=(-1{)}^{i+1}$ on the ordered faces ${\sigma }_i$. The radial map preserves their induced outward orientations by the preceding calculation. For $i<j$, the edge omitting $v_i,v_j$ occurs with coefficient $(-1{)}^{i+1}(-1{)}^{j-1}=(-1{)}^{i+j}$ from face $i$ and $(-1{)}^{j+1}(-1{)}^i=(-1{)}^{i+j+1}$ from face $j$. Both edge maps list the two surviving vertices in the same order, so their radial composites are identical maps with opposite coefficients. Thus [F3] gives $\mathrm{\partial }z=0$.

The scaling map ${\psi }_a:S^2\to S_a^2$, ${\psi }_a(u)=au$, satisfies ${\psi }_a^{\ast }(\beta {\mathrm{\mid }}_{S_a^2})=\omega$, since the numerator and denominator in $\beta$ both scale by $a^3$. For each $i$, use $q_a\circ {\sigma }_i$ if ${\epsilon }_i=1$, and precompose it with the affine transposition $r(u_1,u_2)=(u_2,u_1)$ of ${\mathrm{\Delta }}^2$ if ${\epsilon }_i=-1$. The resulting four parametrizations are orientation-preserving diffeomorphisms on the open simplex interiors onto the disjoint patches of step 1.2, with closures covering the compact sphere. They extend smoothly near the closed parameter simplex into the sphere because their affine faces avoid the origin. By [F11], transposition negates the form integral, so each new integral is ${\epsilon }_i{\int }_{q_a\circ {\sigma }_i}\beta$. By [F12] these compact parameter integrals equal the interior integrals in [F8]. Hence [F8] applies, and the compact oriented surface integral is the one evaluated in [F7]. Hence ${\int }_z\beta ={\int }_{S_a^2}\beta ={\int }_{S^2}\omega =4\pi$, so the cycle flux of $F$ is ${\mathrm{\Phi }}_z(t)=C{\int }_z\beta =\mathrm{\Phi }$ for every $t$.

The embedded cycle $z_t$ is a finite smooth singular 2-cycle in $I\times \mathrm{\Omega }$. Its inclusion pullback of $F$ is $C\beta$, which is smooth and closed on $\mathrm{\Omega }$ by step 1.1 and has nonzero period $\mathrm{\Phi }$ by step 2.1. Therefore $F$ is a smooth closed 2-form on $I\times \mathrm{\Omega }$ with nonzero period on $z_t$. Applying [F9] shows that no globally defined smooth 1-form $A$ on $I\times \mathrm{\Omega }$ satisfies $dA=F$.

Suppose a smooth divergence-free magnetic field on ${\mathbb{R}}^3$ agreed with $B$ on $S_a^2$. Its spatial 2-form $\tilde{\beta }={\tilde{B}}_{1}d{x}^2\wedge d{x}^3+{\tilde{B}}_{2}d{x}^3\wedge d{x}^1+{\tilde{B}}_{3}d{x}^1\wedge d{x}^2$ would be smooth and closed by [F4], and would agree with $C\beta$ on $S_a^2$. Since ${\mathbb{R}}^3$ is star-shaped, [F10] gives $\tilde{\beta }=dA$ for a global smooth 1-form $A$. Stokes and $\mathrm{\partial }z=0$ from step 1.2 then give ${\int }_z\tilde{\beta }={\int }_{\mathrm{\partial }z}A=0$, while step 2.1 gives ${\int }_z\tilde{\beta }=4\pi C=\mathrm{\Phi }\mathrm{\ne }0$. This contradiction excludes a smooth divergence-free full-space extension; it does not turn the exterior field into a magnetic-charge source law.