Closed flux form without global vector potential

Put $s=x^2+y^2+z^2>0$. Repeated application of the positive-base power rule in [F2] gives derivatives of every order of $u^{-1\mathrm{/}2}$ and $u^{-3\mathrm{/}2}$ on $u>0$; the chain and product rules then show that $q(p)=p{s}^{-1\mathrm{/}2}$ and every coefficient of $\beta$ are smooth on $N$. Writing $X=(x,y,z)\mathrm{/}{r}^3$, the coordinate formula in [F1] gives $d\beta =({\mathrm{\partial }}_x(x{r}^{-3})+{\mathrm{\partial }}_y(y{r}^{-3})+{\mathrm{\partial }}_z(z{r}^{-3}))dx\wedge dy\wedge dz$. By [F2], ${\mathrm{\partial }}_x(x{r}^{-3})=r^{-3}-3x^2{r}^{-5}$, with the analogous formulas for $y,z$; their sum is $3r^{-3}-3(x^2+y^2+z^2)r^{-5}=0$. Thus $d\beta =0$ throughout $N$.

The edge matrix with columns $v_1-v_0=(0,-2,-2)$, $v_2-v_0=(-2,0,-2)$ and $v_3-v_0=(-2,-2,0)$ has determinant $-16$. Since $v_0+v_1+v_2+v_3=0$, the origin has strictly positive barycentric coordinates and lies in the interior of $T$. Each face avoids it, so step 1.1 supplies smooth extensions of $q{\sigma }_j$ near ${\mathrm{\Delta }}^2$. The ordered tetrahedron has negative orientation; its outward boundary therefore has coefficients $-(-1{)}^j=c_j$ on the ordered faces ${\sigma }_j$. In the formal signed boundary of ${\sum }_j{c}_j{\sigma }_j$, every ordered edge occurs twice with opposite coefficients: deleting indices $i<j$ in either order changes the boundary sign by $-1$. Composing these identical edge maps with $\iota q$ preserves their equality, so $\mathrm{\partial }z=0$ by [F3]. This uses no extension of $q$ across the origin.

A ray from the interior point $0$ meets the compact convex tetrahedron in an interval $[0,t_u]$ with $t_u>0$ and meets its boundary only at $t_{u}u$. Hence $q:\mathrm{\partial }T\to S^2$ is bijective. On a face $n\cdot p=h$ with outward unit normal $n$, $h>0$ and the inverse is $u\mapsto hu\mathrm{/}(n\cdot u)$; it is smooth on the spherical face interior, where $n\cdot u>0$. These interiors give disjoint open patches whose closures cover $S^2$. For any tangent vectors $v,w$ at $p\in N$, $D{q}_p(v)=(v-(q\cdot v)q)\mathrm{/}\mathrm{\parallel }p{\mathrm{\parallel }}_2$; expanding the cross product gives $q\cdot (D{q}_p(v)\times D{q}_p(w))=p\cdot (v\times w)\mathrm{/}\mathrm{\parallel }p{\mathrm{\parallel }}_2^3={\beta }_p(v,w)$. Thus $q^{\ast }\omega =\beta$ on $N$, and ${\iota }^{\ast }\beta =\omega$ on $S^2$. For positively oriented face vectors the displayed value is positive, because $p\cdot (v\times w)=hn\cdot (v\times w)>0$. Hence radial projection preserves the outward face orientation.

In the positive affine coordinates on ${\mathrm{\Delta }}^2$, write $G_j=q{\sigma }_j$ on $K$. The orientation sign of $G_j$ is $c_j$ by steps 2.1–2.2. Set $H_j=G_j$ when $c_j=1$ and $H_j=G_j\circ R$ when $c_j=-1$, with $R$ as in [F12]. Each $H_j$ is smooth near $K$ and restricts on its interior to an orientation-preserving diffeomorphism onto its spherical face patch. By [F12], ${\int }_K{H}_j^{\ast }\omega =c_j{\int }_K{G}_j^{\ast }\omega$ in both cases. The simplex definition [F5], the pullback identity in step 2.2, and the finite signed sum [F4] give ${\int }_z\beta ={\sum }_j{c}_j{\int }_K{G}_j^{\ast }\omega ={\sum }_j{\int }_K{H}_j^{\ast }\omega$. The parameter triangles are compact Jordan sets; their edges have content zero, and the smooth pullback coefficients are bounded, so their closed-triangle integrals equal the open-interior integrals by [F13]. Compact support follows from [F8]; the four patches satisfy [F9], yielding ${\int }_z\beta ={\int }_{S^2}\omega =4\pi$ by [F10].

Step 3.1 gives ${\int }_z\beta =4\pi \mathrm{\ne }0$. Since step 1.1 proved $d\beta =0$, [F11] makes $\beta$ nonexact on the punctured domain.