Maxwell algebra supplies timelike and null convergence

The assumed Axiom of Choice supplies the inherited choice hypotheses when applying [F1], [F2] and [F3]. Choose the formal positive parameters in [F2], for example ${\mu }_0=c=G_N=1$, and use the same smooth two-form and metric as in [F1]. The smooth field has the regularity required by [F2] and [F3]. Lowering the indices in the formal tensor formula gives $Q={\mu }_0{T}^{\mathrm{E}\mathrm{M}}$. Thus $Q$ is symmetric and trace-free. Since multiplication by ${\mu }_0>0$ preserves nonnegativity and the future causal cone, [F3] and [F4] give the stated dominant and strong conditions for $Q$, as well as $Q(L,L)\ge 0$ for future null $L$.

Work at a point, with $V=T_{p}M$. Normalize the supplied timelike vector to $n=N\mathrm{/}\sqrt{-g(N,N)}$. The decomposition $v=-g(v,n)n+(v+g(v,n)n)$ gives $V=\mathbb{R}n\oplus n^{\perp }$ as an orthogonal direct sum. Diagonalize $g{\mathrm{\mid }}_{n^{\perp }}$ by [F6] and append $n$; uniqueness of the inertia counts, with the signature in [F1], makes $g{\mathrm{\mid }}_{n^{\perp }}$ positive definite. A nonzero null $L$ therefore cannot be orthogonal to $N$, so $a=g(N,L)\mathrm{\ne }0$. Put $K=-2N\mathrm{/}a+g(N,N)L\mathrm{/}{a}^2$; direct substitution gives $g(K,L)=-2$ and $g(K,K)=0$.

Let $P=\mathrm{span}\{L,K\}$ and $E=P^{\perp }$. The Gram matrix on $P$ is $\left(\begin{array}{cc}0& -2\\ -2& 0\end{array}\right)$; the vectors $L+K$ and $L-K$ are orthogonal with squared norms $-4$ and $4$. For any $v\in V$, the vector $v+\frac{1}{2}g(v,K)L+\frac{1}{2}g(v,L)K$ belongs to $E$. This gives the orthogonal direct sum $V=P\oplus E$, since a vector $bL+cK$ orthogonal to both $L$ and $K$ has $b=c=0$. Thus $E$ has dimension two. Diagonalizing $g{\mathrm{\mid }}_E$ by [F6] and adjoining the displayed diagonal basis of $P$, uniqueness of inertia and the signature of $g$ show that $g{\mathrm{\mid }}_E$ has two positive directions and no negative or zero directions. In particular it is positive definite. If $v\in L^{\perp }$, its $K$ coefficient in this decomposition vanishes, so $L^{\perp }=\mathbb{R}L\oplus E$. The form $\bar{\gamma }([v],[w])=g(v,w)$ is well-defined on $L^{\perp }\mathrm{/}\mathbb{R}L$, since adding multiples of $L$ to either representative leaves it unchanged. The map $E\to L^{\perp }\mathrm{/}\mathbb{R}L$, $e\mapsto [e]$, is consequently a linear isometry and bijection. This proves the positive two-dimensional screen assertion for every nonzero null $L$ without any congruence hypothesis.

By trace-freeness, the strong condition gives $Q(v,v)\ge 0$ for future timelike $v$. A past timelike or null vector changes to a future vector under negation, while its quadratic contraction is unchanged. Hence $\mathrm{Ric}(n,n)=2Q(n,n)\ge 0$ for every unit timelike $n$ and $\mathrm{Ric}(L,L)=2Q(L,L)\ge 0$ for every nonzero null $L$.

By [F6], the positive form on $E$ has an orthonormal basis, hence so does the screen by step 2.1. Given any representatives $e_1,e_2$ of an orthonormal screen basis, replace each $e_A$ by $e_A+\frac{1}{2}g(e_A,K)L$. The replacements preserve the classes and lie in $E$, so are an orthonormal basis of $E$. For these representatives the basis $L,K,e_1,e_2$ therefore has inverse metric $g^{-1}=-\frac{1}{2}(L\otimes K+K\otimes L)+{\sum }_A{e}_A\otimes e_A$.

Since $F(L,L)=0$, contraction of the inverse metric in step 3.1 gives $F_{ac}{L}^a{F}_b{}^c{L}^b={\sum }_{A}F(L,e_A{)}^2$. The trace term in $Q(L,L)$ vanishes because $g(L,L)=0$. Therefore $\mathrm{Ric}(L,L)=2{\sum }_{A}F(L,e_A{)}^2$. Replacing any representative by $e_A+tL$ leaves $F(L,e_A)$ unchanged by alternation, proving the claimed representative independence.