Lorentzian forms and relative topology

In degree zero, use ${\mathrm{\Lambda }}^0{V}^{\ast }={\mathrm{Alt}}^0(V)=\mathbb{R}$, the convention of Alternating $k$-covectors in [F3]'s exterior-algebra prerequisites, with basis $e^{\mathrm{\varnothing }}=1$. The empty product ${ϵ}_{\mathrm{\varnothing }}$ and the signs $s(\mathrm{\varnothing },\{0,1,2,3\})$ and $s(\{0,1,2,3\},\mathrm{\varnothing })$ are $1$. For each increasing $I$, set ${\star }_{\eta }{e}^I={ϵ}_{I}s(I,I^c)e^{I^c}$. Thus ${\star }_{\eta }1={\mathrm{v}\mathrm{o}\mathrm{l}}_{\eta }$ and ${\star }_{\eta }{\mathrm{v}\mathrm{o}\mathrm{l}}_{\eta }=-1$; for scalars $a,b$, the degree-zero pairing is $ab$ because $0!=1$, so $a\wedge {\star }_{\eta }b=ab{\mathrm{v}\mathrm{o}\mathrm{l}}_{\eta }$. In positive degrees the $e^I$ form a basis by [F3], and in degree zero this was just established. Hence the prescription extends uniquely by linearity to a map ${\mathrm{\Lambda }}^k{V}^{\ast }\to {\mathrm{\Lambda }}^{4-k}{V}^{\ast }$ for every $0\le k\le 4$; its coefficients are constant, so it also defines a smooth map on the affine space.

For increasing $I,J$ of size $k$, $e^I\wedge e^{J^c}=0$ when $I\mathrm{\ne }J$, because an index repeats. When $I=J$, $e^I\wedge {\star }_{\eta }{e}^I={ϵ}_I{\mathrm{v}\mathrm{o}\mathrm{l}}_{\eta }$. The component pairing has $⟨e^I,e^J{⟩}_{\eta }=0$ for $I\mathrm{\ne }J$ and $⟨e^I,e^I{⟩}_{\eta }={ϵ}_I$. Thus the characterizing identity holds on basis pairs and, by bilinearity, for every $\alpha ,\beta$. Pairing with all complementary basis wedges also proves uniqueness.

The cycle integral in the Definition is a finite sum of the unique simplex integrals in [F7], so it is independent of how a finite chain is written after equal simplex coefficients are combined. The terms closed, exact, and $[\omega ]=0$ have the quotient meanings in [F4] and [F5]. No physical assumption or choice principle is used.

The rectangle $D$ is a compact Jordan parameter region. Since ${\sin }^{\mathrm{\prime }}=\cos$ and ${\cos }^{\mathrm{\prime }}=-\sin$, repeated differentiation shows that every derivative of sine and cosine exists and is one of $\pm \sin ,\pm \cos$; these derivatives are continuous, so $\mathrm{\Phi }$ is smooth on all of ${\mathbb{R}}^2$. Every point of $S^2$ is in its image: choose $\theta \in [0,\pi ]$ with $\cos \theta =p_3$; when $\sin \theta >0$, the pair $(p_1\mathrm{/}\sin \theta ,p_2\mathrm{/}\sin \theta )$ is a unit-circle point and [F7] supplies $\varphi \in [0,2\pi ]$, while at a pole any $\varphi$ works. More precisely, the interior image is $W=S^2\setminus \mathrm{\Phi }(\mathrm{\partial }D)$: for every point not on the poles or the longitude seam, the latitude is unique by strict decrease of cosine on $(0,\pi )$, and the longitude is unique by the bijection in [F7]. Direct differentiation gives ${\mathrm{\Phi }}_{\theta }=(\cos \theta \cos \varphi ,\cos \theta \sin \varphi ,-\sin \theta )$ and ${\mathrm{\Phi }}_{\varphi }=(-\sin \theta \sin \varphi ,\sin \theta \cos \varphi ,0)$; substituting into the coordinate cross-product formula [F3] yields ${\mathrm{\Phi }}_{\theta }\times {\mathrm{\Phi }}_{\varphi }=({\sin }^2\theta \cos \varphi ,{\sin }^2\theta \sin \varphi ,\sin \theta \cos \theta )=\sin \theta \mathrm{\Phi }$. [F7] makes this nonzero for $0<\theta <\pi$, and its direction is outward by the stated orientation. In any smooth chart on $S^2$, the coordinate derivative of $\mathrm{\Phi }$ is therefore invertible; [F6] gives a smooth local inverse. The bijection onto $W$ and these local inverses make $\mathrm{\Phi }{\mathrm{\mid }}_{D^{\circ }}$ an orientation-preserving diffeomorphism onto the open set $W$. Thus [F1] applies, including the permitted seam and pole degeneracy on $\mathrm{\partial }D$.

By [F2] and step 1.1, the flux integrand is $f(\theta ,\varphi )=\mathrm{\Phi }\cdot (\sin \theta \mathrm{\Phi })=\sin \theta$, and the same coefficient is $\omega ({\mathrm{\Phi }}_{\theta },{\mathrm{\Phi }}_{\varphi })$. This function is continuous on the compact Jordan rectangle $D$, so [F11] makes it Riemann integrable there; [F7] gives $0\le f\le 1$. The bounded open rectangle $D^{\circ }$ has boundary $\mathrm{\partial }D$, which has two-dimensional content zero by [F1], so [F10] makes $D^{\circ }$ Jordan measurable. Define $g:D\to \mathbb{R}$ by $g=f$ on $D^{\circ }$ and $g=0$ on $\mathrm{\partial }D$. The functions $f,g$ are bounded and differ only on the content-zero set $\mathrm{\partial }D$; [F9] therefore gives that $g$ is Riemann integrable on $D$ and ${\int }_{D}g={\int }_{D}f$. By the zero-extension definition in [F10], ${\int }_{D}g={\int }_{D^{\circ }}f$. This supplies the Riemann-integral consequence of the null boundary explicitly. [F4] defines the oriented-sphere integral and ensures compact support, and [F5] identifies it with ${\int }_{D^{\circ }}f$. Therefore [F8] gives ${\int }_D(X\circ \mathrm{\Phi })\cdot ({\mathrm{\Phi }}_{\theta }\times {\mathrm{\Phi }}_{\varphi })d\theta d\varphi ={\int }_0^{2\pi }\text{\hspace{0.17em}⁣}{\int }_0^{\pi }\sin \theta d\theta d\varphi =2\pi (-\cos \theta ){\mid }_0^{\pi }=4\pi$. The vanishing cross product at the poles and the repeated longitude seam occur only on $\mathrm{\partial }D$.

For any increasing $I\subseteq \{0,1,2,3\}$ of size $k$, apply [F1] twice to obtain ${\star }_{\eta }^2{e}^I={ϵ}_I{ϵ}_{I^c}s(I,I^c)s(I^c,I)e^I$.

The product ${ϵ}_I{ϵ}_{I^c}$ is $-1$, since the two complementary index sets contain the unique negative metric sign exactly once. Reversing the blocks $I,I^c$ uses $k(4-k)$ transpositions, so $s(I,I^c)s(I^c,I)=(-1{)}^{k(4-k)}$. Thus the general square formula follows on the basis $e^I$. For $I=\{0,1\}$, the star is $-e^{23}$; for $I=\{2,3\}$, it is $e^{01}$, which gives the two displayed special cases.

For each $x\in U$, the segment $x_t=a+t(x-a)$ lies in $U$ by [A1], so the displayed integrand is smooth on $0\le t\le 1$. The segment is compact; since $U$ is open, it has a neighborhood still contained in $U$. Varying one coordinate of $x$ over a sufficiently small closed interval gives a compact rectangle on which the integrand and each coordinate derivative are continuous. Applying [F4] successively to these derivatives proves that every $A_{\mu }$ is smooth; the formula has no singularity at $t=0$.

Put $y=x-a$ and $x_t=a+ty$. Differentiating under the integral as in [F4] gives ${\mathrm{\partial }}_{\mu }{A}_{\nu }={\int }_0^1(t{F}_{\mu \nu }(x_t)+t^2{y}^{\rho }{\mathrm{\partial }}_{\mu }{F}_{\rho \nu }(x_t))dt$. Subtracting the expression with $\mu ,\nu$ exchanged, the coordinate formula [F3] gives ${\mathrm{\partial }}_{\mu }{F}_{\rho \nu }+{\mathrm{\partial }}_{\rho }{F}_{\nu \mu }+{\mathrm{\partial }}_{\nu }{F}_{\mu \rho }=0$. Since $F_{\nu \mu }=-F_{\mu \nu }$ and $F_{\mu \rho }=-F_{\rho \mu }$, this yields ${\mathrm{\partial }}_{\mu }{F}_{\rho \nu }-{\mathrm{\partial }}_{\nu }{F}_{\rho \mu }={\mathrm{\partial }}_{\rho }{F}_{\mu \nu }$; hence ${\mathrm{\partial }}_{\mu }{A}_{\nu }-{\mathrm{\partial }}_{\nu }{A}_{\mu }={\int }_0^1(2t{F}_{\mu \nu }(x_t)+t^2{y}^{\rho }{\mathrm{\partial }}_{\rho }{F}_{\mu \nu }(x_t))dt$.

For any closed $F$, [F6] identifies $[F]=0$ with $F\in B^2(M)$, which by [F2] means exactly that some globally defined smooth one-form $A$ satisfies $dA=F$.

By [F8] and [F9], the integrand in step 1.2 is $\frac{d}{dt}(t^2{F}_{\mu \nu }(x_t))$. The function $G(t)=t^2{F}_{\mu \nu }(x_t)$ is continuous on $[0,1]$, differentiable on $(0,1)$, and has continuous derivative there extending to the endpoints, so [F5] gives ${\mathrm{\partial }}_{\mu }{A}_{\nu }-{\mathrm{\partial }}_{\nu }{A}_{\mu }=F_{\mu \nu }(x)$. This is precisely $dA=F$ under the component convention in [F1].

If $F=dA$ and $z$ is a finite smooth singular cycle, then [F7] gives ${\int }_{z}F={\int }_{z}dA={\int }_{\mathrm{\partial }z}A=0$, since $\mathrm{\partial }z=0$. Therefore a nonzero period precludes exactness.