Exact electrovac plane-wave family

In the ordered coordinates $(u,v,x,y)$ the metric matrix has $(u,v)$ block $\left(\begin{array}{cc}P& -1\\ -1& 0\end{array}\right)$ and transverse block $I_2$. Its inverse has $g^{uv}=g^{vu}=-1$, $g^{vv}=-P$, $g^{uu}=0$, $g^{xx}=g^{yy}=1$; the determinant is $-1$. The field has only ${\mathcal{F}}_{ux}=a$ and ${\mathcal{F}}_{uy}=b$, so direct raising gives ${\mathcal{F}}^{vx}=-a$, ${\mathcal{F}}^{vy}=-b$, with ${\mathcal{F}}^{ux}={\mathcal{F}}^{uy}=0$. Direct differentiation of $\mathcal{F}=a(u)du\wedge dx+b(u)du\wedge dy$ gives $d\mathcal{F}=0$, since each derivative term contains $du\wedge du$.

The nonzero Christoffel symbols, apart from lower-index symmetry, are ${\mathrm{\Gamma }}^v{}_{uu}=-P_u\mathrm{/}2$, ${\mathrm{\Gamma }}^v{}_{ui}=-P_i\mathrm{/}2$, and ${\mathrm{\Gamma }}^i{}_{uu}=-P_i\mathrm{/}2$ for $i\in \{x,y\}$. Substitution of the metric and inverse into [F4] gives these entries; the remaining metric derivatives vanish or lie in a constant block. Each trace ${\mathrm{\Gamma }}^a{}_{ac}$ is zero. Hence ${\mathrm{\nabla }}_a{\mathcal{F}}^{ab}={\mathrm{\partial }}_a{\mathcal{F}}^{ab}$, because ${\mathrm{\Gamma }}^b{}_{ac}{\mathcal{F}}^{ac}=0$ by symmetry. The only nonzero raised field components have upper pair $vx$ or $vy$, are independent of $v$, and have no component with upper $u$; therefore every ${\mathrm{\partial }}_a{\mathcal{F}}^{ab}=0$. By [F3], $d({\star }_g\mathcal{F})=0$.

The coordinate curvature and Ricci formulas [F5, F6] with the Christoffel list in step 2.1 have zero trace-derivative and trace-product terms because ${\mathrm{\Gamma }}^a{}_{ac}=0$. In the last product ${\mathrm{\Gamma }}^a{}_{bc}{\mathrm{\Gamma }}^c{}_{ad}$, $c=u$ gives a zero second factor (no Christoffel has upper $u$), $c=v$ gives a zero first factor (no Christoffel has lower $v$), and for transverse $c=i$ a nonzero first factor requires $a=v$, making the second factor zero because it then has lower $v$. Hence that product vanishes. In the derivative term ${\mathrm{\partial }}_a{\mathrm{\Gamma }}^a{}_{\nu \beta }$, the only nonzero derivatives are ${\mathrm{\partial }}_x{\mathrm{\Gamma }}^x{}_{uu}=-P_{xx}\mathrm{/}2$ and ${\mathrm{\partial }}_y{\mathrm{\Gamma }}^y{}_{uu}=-P_{yy}\mathrm{/}2$. Thus ${\mathrm{Ric}}_{uu}=-(P_{xx}+P_{yy})\mathrm{/}2$ and every other Ricci component is zero. From [F1], $P_{xx}+P_{yy}=-4(a^2+b^2)+h_{xx}+h_{yy}=-4(a^2+b^2)$, giving the stated Ricci component.

Raising the field indices from step 1.1 gives ${\mathcal{F}}_{ab}{\mathcal{F}}^{ab}=0$ and ${\mathcal{F}}_{uc}{\mathcal{F}}_u{}^c=a^2+b^2$; all other ${\mathcal{F}}_{ac}{\mathcal{F}}_b{}^c$ entries vanish. Hence $Q_{uu}=a^2+b^2$ and all other $Q$ entries vanish. Comparing with step 3.1 proves $\mathrm{Ric}=2Q$. Together with the closedness in step 1.1, the source-free divergence in step 2.1, and [F2], this proves the geometric electrovac equations.