Closed two-forms agreeing on a Cauchy slice differ by an exact form

Let $X$ and ${\mathrm{\Phi }}_t$ be the complete timelike field and its flow from [F1], and define $\mathrm{\Psi }:\mathbb{R}\times \mathrm{\Sigma }\to M$ by $\mathrm{\Psi }(t,x)={\mathrm{\Phi }}_t(i(x))$. Since $T({\mathrm{\Phi }}_t(p))=T(p)+t$, each orbit has no endpoint at either infinite end and is an inextendible timelike curve. Thus [F2] makes its intersection with $i(\mathrm{\Sigma })$ unique. Hence $\mathrm{\Psi }$ is surjective and injective: if two images agree, their points lie on the same orbit, their slice intersections agree, and the flow times agree because the temporal function increases at unit rate along $X$. The flow is smooth, so $\mathrm{\Psi }$ is smooth. If $d{\mathrm{\Psi }}_{(t,x)}(a{\mathrm{\partial }}_t+V)=0$, applying $d{\mathrm{\Phi }}_{-t}$ gives $a{X}_{i(x)}+d{i}_x(V)=0$. The first vector is timelike when $a\mathrm{\ne }0$, while the second is tangent to a spacelike hypersurface; therefore $a=0$ and $V=0$. Thus $d\mathrm{\Psi }$ is an isomorphism, and [L1] shows that $\mathrm{\Psi }$ is a local diffeomorphism. Its bijectivity makes it a global diffeomorphism, with $\mathrm{\Psi }(0,x)=i(x)$.

Pull back $G:=F-\bar{F}$ to the product and write its unique time/spatial decomposition as $\tilde{G}:={\mathrm{\Psi }}^{\ast }G=dt\wedge {\eta }_t+{\beta }_t$, where ${\eta }_t$ is a smooth family of spatial one-forms and ${\beta }_t$ a smooth family of spatial two-forms on $\mathrm{\Sigma }$. Pullback commutes with $d$: in local coordinates the first-derivative terms in $d({\mathrm{\Psi }}^{\ast }G)$ are the pullback of $dG$, while the second derivatives of $\mathrm{\Psi }$ cancel in pairs because mixed partials commute and $G$ is antisymmetric. Since $dG=0$ by hypothesis and [F3], the product-coordinate formula for $d\tilde{G}$ gives ${\mathrm{\partial }}_t{\beta }_t=d_{\mathrm{\Sigma }}{\eta }_t$. The slice equality implies ${\beta }_0=0$. Define the spatial one-form ${\tilde{\alpha }}_t:={\int }_0^t{\eta }_{s}ds$ and give $\tilde{\alpha }$ zero $dt$ component. The integral of a smooth parameter family is smooth, and direct differentiation gives $d\tilde{\alpha }=dt\wedge {\eta }_t+d_{\mathrm{\Sigma }}{\tilde{\alpha }}_t=dt\wedge {\eta }_t+{\int }_0^t{d}_{\mathrm{\Sigma }}{\eta }_{s}ds=dt\wedge {\eta }_t+{\beta }_t=\tilde{G}$.

Set $\alpha :=({\mathrm{\Psi }}^{-1}{)}^{\ast }\tilde{\alpha }$. The same coordinate pullback identity gives $d\alpha =G=F-\bar{F}$. By [F4], this is exact on the whole spacetime, not merely on coordinate neighborhoods.