Stationary lifting of a spatial primitive

The product coordinates give $dt(T)=1$ and $T(D)=0$, hence ${\iota }_{T}dD=0$. The contraction rule gives ${\iota }_T(dt\wedge dD)=dD-dtT(D)=dD$, so ${\iota }_{T}B=0$. The coordinate formula for $dD$ and stationarity of $D$ show that $dD$ has no $dt$ component and its coefficients are independent of $t$. Thus $B$ is stationary as well.

Define $b=j^{\ast }B$. In every product chart the unique two-form expansion of $B$ has no $dt$ terms because ${\iota }_{T}B=0$, and its remaining coefficients are independent of $t$. Evaluating the pullback formulas for $j$ and $\pi$ therefore gives $B={\pi }^{\ast }b$; it also proves uniqueness by evaluating this equality on the slice $t=0$. The form $b$ is globally smooth by its definition. Since $d(dt)=d^{2}t=0$, $d^{2}D=0$ and $d\mathrm{\Theta }=0$, the graded product rule gives $dB=0$. Smoothness of $j$ and commutation with pullback now give $db=d(j^{\ast }B)=j^{\ast }(dB)=0$.

Let $a$ be the supplied smooth primitive, $da=b$, and let $c\in \mathbb{R}$. The pullback ${\pi }^{\ast }a$ is a smooth form on $E$, so $V=(c-D)dt+{\pi }^{\ast }a$ is well defined and smooth. Differentiation and graded commutativity give $dV=-dD\wedge dt+{\pi }^{\ast }(da)=dt\wedge dD+{\pi }^{\ast }b=\mathrm{\Theta }$. Since $d\pi (T)=0$, the pullback formula gives $({\pi }^{\ast }a)(T)=a(d\pi (T))=0$, whence $V(T)=c-D$.

The coefficients of ${\pi }^{\ast }a$ in product coordinates depend only on $x$, so $V$ is stationary in the stated coordinate sense. Moreover Cartan's formula, $dV=\mathrm{\Theta }$ and ${\iota }_T\mathrm{\Theta }=dD$ give ${\mathcal{L}}_{T}V=d(V(T))+{\iota }_{T}dV=d(c-D)+dD=0$, and $V(T)+D=c$. The construction assumes the smooth primitive $a$; it does not assert that every closed form is exact.