Rest-space character of the kinematic fields and the double contraction of B

(Rest-space character.) $n^a{B}_{ab}=n^a{\mathrm{\nabla }}_b{n}_a=\frac{1}{2}{\mathrm{\nabla }}_b(g(n,n))=0$ by [L1] and the unit condition $g(n,n)=-1$ in [L2]. Also $B_{ab}{n}^b=n^b{\mathrm{\nabla }}_b{n}_a=({\mathrm{\nabla }}_{n}n{)}_a=0$ by the geodesic condition [L2]. This proves (a).

(The rest-space metric field.) $h_{ab}{n}^b=g_{ab}{n}^b+n_a{n}_b{n}^b=n_a+n_a(n^b{n}_b)=n_a-n_a=0$, since $n^b{n}_b=g(n,n)=-1$ [L2]. On $n^{\perp }$ the form $h$ agrees with $g$, hence is positive definite by [L4]. For the mixed identity, $h^{ab}{h}_{bc}=(g^{ab}+n^a{n}^b)(g_{bc}+n_b{n}_c)={\delta }_c^a+n^a{n}_c+n^a{n}_c-n^a{n}_c={\delta }_c^a+n^a{n}_c$, where $g^{ab}{n}_b=n^a$, $g_{bc}{n}^b=n_c$ and $n^b{n}_b=-1$ were used [L3]. Contraction gives $h_{ab}{h}^{ba}=(g_{ab}+n_a{n}_b)(g^{ba}+n^b{n}^a)=g_{ab}{g}^{ba}+g_{ab}{n}^b{n}^a+(n_a{n}_b)g^{ba}+(n_a{n}_b)(n^b{n}^a)=4+n^a{n}_a+n_a{n}^a+(n^a{n}_a{)}^2=4-1-1+1=3$, alternatively from $h_{ab}{h}^{ba}=h_{ab}(g^{ba}+n^b{n}^a)=g_{ab}{g}^{ba}+n_a{n}_b{g}^{ba}+h_{ab}{n}^b{n}^a=4-1+0=3$. This proves (b).

(Decomposition, symmetry and traces.) $B_{ab}=B_{(ab)}+B_{[ab]}$, and substituting the definitions of $\sigma$ and $\omega$ gives $B_{ab}=\frac{1}{3}\theta h_{ab}+{\sigma }_{ab}+{\omega }_{ab}$. Symmetry of $\sigma$ and antisymmetry of $\omega$ are immediate from the symmetrized and antisymmetrized definitions, and $h^{ab}{\sigma }_{ab}=h^{ab}{B}_{(ab)}-\frac{1}{3}\theta h^{ab}{h}_{ab}=h^{ab}{B}_{ab}-\theta =0$, using [L5], step 2.1 and $h^{ab}{B}_{ab}=g^{ab}{B}_{ab}+n^a{n}^b{B}_{ab}=\theta$ by step 1.1. Both $\sigma$ and $\omega$ annihilate $n$ in either slot because $B$ and $h$ do. This proves (c).

(Double contraction in a rest-space frame.) Fix a point $q\in U$ and let $M$ be the matrix of $B$ in a $g$-orthonormal basis $e_1,e_2,e_3$ of $n_q^{\perp }$, whose existence is [L4]. Since $B$ is rest-space in both slots by (a), $B_{ab}{B}^{ba}={\sum }_{i,j}{M}_{ij}{M}_{ji}=\mathrm{tr}(M^2)$. Write $M=\frac{1}{3}\theta I+S+A$ with $S$ the symmetric traceless and $A$ the antisymmetric part, so that $S$ and $A$ are the matrices of $\sigma$ and $\omega$ in this basis. By cyclicity of the trace, $\mathrm{tr}(SA)=\mathrm{tr}((SA{)}^{\mathsf{T}})=\mathrm{tr}(A^{\mathsf{T}}{S}^{\mathsf{T}})=-\mathrm{tr}(AS)=-\mathrm{tr}(SA)$, so $\mathrm{tr}(SA)=0$; similarly $\mathrm{tr}(S)=0$ and $\mathrm{tr}(A)=0$ kill the mixed terms with $I$, and $\mathrm{tr}(I^2)=3$. Hence $\mathrm{tr}(M^2)=\frac{1}{3}{\theta }^2+\mathrm{tr}(S^2)+\mathrm{tr}(A^2)=\frac{1}{3}{\theta }^2+{\sum }_{i,j}(S_{ij}{)}^2-{\sum }_{i,j}(A_{ij}{)}^2$, and in the orthonormal rest basis the two sums are ${\sigma }_{ab}{\sigma }^{ab}$ and ${\omega }_{ab}{\omega }^{ab}$. This proves the displayed identity of (d).

(Signs and vanishing.) In the same orthonormal basis, ${\sigma }_{ab}{\sigma }^{ab}={\sum }_{i,j}(S_{ij}{)}^2\ge 0$ with equality exactly when $S=0$, i.e. $\sigma =0$, and ${\omega }_{ab}{\omega }^{ab}={\sum }_{i,j}(A_{ij}{)}^2\ge 0$ with equality exactly when $\omega =0$. Together with step 4.1 this completes the proof.