Null wave curvature and zero quadratic invariants

The inverse metric in the null block has $g^{uu}=0$, $g^{uv}=g^{vu}=-1$, and $g^{vv}=-P$, while $g^{xx}=g^{yy}=1$. From [F1], the only nonzero lower field components are ${\mathcal{F}}_{ux}=a$ and ${\mathcal{F}}_{uy}=b$, so the only nonzero raised components are ${\mathcal{F}}^{vx}=-a$ and ${\mathcal{F}}^{vy}=-b$. No lower nonzero component shares its index pair with a raised nonzero component. Thus ${\mathcal{F}}_{ab}{\mathcal{F}}^{ab}=0$.

From the metric of [F1], the Christoffel formula [F4] gives ${\mathrm{\Gamma }}^v{}_{uu}=-P_u\mathrm{/}2$, ${\mathrm{\Gamma }}^v{}_{ui}=-P_i\mathrm{/}2$, and ${\mathrm{\Gamma }}^i{}_{uu}=-P_i\mathrm{/}2$ for $i=x,y$, with lower-index symmetry, and every other Christoffel vanishes. A derivative term in [F6] can therefore be nonzero only from differentiating one of those three displayed types; antisymmetry in the two curvature-direction indices reduces these possibilities to $R^i{}_{uju}$, $R^v{}_{iuj}$, $R^v{}_{uiu}$, and $R^v{}_{uij}$, together with their reversed-last-pair partners. Direct differentiation gives $R^i{}_{uju}=-P_{ij}\mathrm{/}2$, $R^v{}_{iuj}=P_{ij}\mathrm{/}2$, $R^v{}_{uiu}=(-P_{ui}+P_{iu})\mathrm{/}2=0$, and $R^v{}_{uij}=(-P_{ji}+P_{ij})\mathrm{/}2=0$. All quadratic terms vanish: an intermediate upper/lower index $v$ requires a Christoffel with lower $v$, an intermediate index $u$ requires one with upper $u$, and for a transverse intermediate index the only possible chain is ${\mathrm{\Gamma }}^v{}_{ui}{\mathrm{\Gamma }}^i{}_{uu}$; it occurs only with equal antisymmetrized indices and cancels. Lowering the first index by [F5] gives the independent components $R_{uiuj}=-P_{ij}\mathrm{/}2$, with the partners supplied by the skew and pair-exchange symmetries [F7]. The Ricci contraction [F5] therefore has only ${\mathrm{Ric}}_{uu}=-\frac{1}{2}(P_{xx}+P_{yy})=2(a^2+b^2)$ nonzero.

Since $g^{uu}=0$ and the only nonzero lower Ricci component is ${\mathrm{Ric}}_{uu}$, its scalar trace is $S=0$. Raising both indices of the Ricci tensor produces only a $vv$ component, while ${\mathrm{Ric}}_{vv}=0$; hence ${\mathrm{Ric}}_{ab}{\mathrm{Ric}}^{ab}=0$.

By step 1.2 and the curvature symmetries [F7], every nonzero Riemann component has the form $R_{uiuj}$ or one of its partners, and there is no nonzero component with a $v$ slot. In the full contraction $R_{abcd}{R}^{abcd}$, raising either displayed $u$ slot pairs it with a $v$ slot because $g^{uv}=-1$ and $g^{uu}=0$. Each resulting contraction term therefore contains a Riemann component with a $v$ index and is zero. Thus $R_{abcd}{R}^{abcd}=0$.

If $a(u_0{)}^2+b(u_0{)}^2>0$, then $P_{xx}+P_{yy}=-4(a^2+b^2)$ at $u_0$. At least one of $P_{xx},P_{yy}$ is nonzero there, so step 1.2 gives a nonzero component $R_{uxux}$ or $R_{uyuy}$. By [F8], this nonzero Riemann component shows the datum is not flat.