The smooth local Herglotz-Noether classification of Born-rigid motions

Clause (i): for a horizontal smooth $X$, [F2] gives $\eta ({\mathrm{\nabla }}_{X}n,\cdot )=\theta (X,\cdot )+\omega (X,\cdot )$. The standing Born-rigid hypothesis is $\theta =0$, so this equals $\omega (X,\cdot )$. Nondegeneracy of $\eta$ therefore gives ${\mathrm{\nabla }}_{X}n=0$ exactly when $\omega (X,\cdot )=0$, which under the same standing hypothesis is exactly the conjunction in clause (i). If also $\omega =0$ on $U$, the identity gives ${\mathrm{\nabla }}_{X}n=0$ for every horizontal $X$ there. No separation of a symmetric and a skew tensor from their value on only one fixed argument is used.

Flow box and slice metric: by [F3] choose a flow box $U$ of $n$ with adapted coordinates $(s,y^1,y^2,y^3)$ in which $n={\mathrm{\partial }}_s$; we may work on a coordinate product box; for a local primitive we shrink to a coordinate cube, whose coordinate image is star-shaped about its centre. This makes no assertion that its image under the adapted chart inverse is affine star-shaped. Since $n$ is future unit, $g_{ss}=\eta ({\mathrm{\partial }}_s,{\mathrm{\partial }}_s)=-1$. The tensor $h=\eta +n^{\mathrm{♭}}\otimes n^{\mathrm{♭}}$ has $h_{ss}=g_{ss}+(n_s^{\mathrm{♭}}{)}^2=-1+1=0$ and $h_{si}=g_{si}+n_s^{\mathrm{♭}}{n}_i^{\mathrm{♭}}=g_{si}-g_{si}=0$, and on the $y$-directions $h_{ij}=g_{ij}+g_{si}{g}_{sj}$; by [F2] the restriction of $h$ to $n^{\perp }$ is positive definite, so $(h_{ij})$ is a smooth positive definite matrix function of $(s,y)$. Born rigidity, ${\mathcal{L}}_{n}h=2\theta =0$ by [F2], reads ${\mathrm{\partial }}_s{h}_{ij}=0$ in these coordinates, because in the frame $({\mathrm{\partial }}_s,{\mathrm{\partial }}_{y^1},{\mathrm{\partial }}_{y^2},{\mathrm{\partial }}_{y^3})$ the Lie derivative of $h={\sum }_{i,j}{h}_{ij}d{y}^{i}d{y}^j$ along ${\mathrm{\partial }}_s$ has components ${\mathrm{\partial }}_s{h}_{ij}$ (the coordinate differentials $d{y}^i$ are ${\mathrm{\partial }}_s$-invariant, [F10]); hence $h=h_{ij}(y)d{y}^{i}d{y}^j=q^{\ast }\bar{h}$ is $s$-independent on $U$, where $q:(s,y)\mapsto y$ and $\bar{h}$ is a Riemannian metric on the open $y$-box $B$ of [F5].

Smooth orthonormal coframe of the slice metric: at every $y$ all leading principal minors of the positive definite matrix $(h_{ij}(y))$ are positive, and they are smooth hence continuous functions of $y$; applying the Gram-Schmidt recursion of [F8] to the coordinate vector fields ${\mathrm{\partial }}_{y^1},{\mathrm{\partial }}_{y^2},{\mathrm{\partial }}_{y^3}$ with respect to $h$ therefore gives, on the whole box, smooth fields $E_i={\sum }_k{c}_i{}^k(y){\mathrm{\partial }}_{y^k}$ whose coefficients are smooth functions of the $h_{ij}$ and positive pivot square roots, so that $h(E_i,E_j)={\delta }_{ij}$. Let $b=(c^{-1}{)}^{\mathsf{T}}$, a smooth invertible matrix function, and put ${\theta }^i:={\sum }_k{b}^i{}_k(y)d{y}^k=h(E_i,\cdot )$; then ${\sum }_i{\theta }^i\otimes {\theta }^i=h$ and ${\theta }^i(E_j)={\delta }_j^i$, so $({\theta }^1,{\theta }^2,{\theta }^3)$ is a smooth $s$-independent $h$-orthonormal coframe of the slice metric, and $d{\theta }^i=-{\bar{\gamma }}^i{}_j\wedge {\theta }^j$ for unique smooth $s$-independent one-forms ${\bar{\gamma }}^i{}_j$ characterized by ${\bar{\gamma }}^i{}_j=-{\bar{\gamma }}^j{}_i$, namely the pullbacks by $q$ of the connection one-forms of $\bar{\mathrm{\nabla }}$ on $(B,\bar{h})$ in this coframe; their first structure equation follows because duality gives ${\bar{\mathrm{\nabla }}}_X{\theta }^i=-{\bar{\gamma }}^i{}_j(X){\theta }^j$ and torsion freeness gives $d{\theta }^i(X,Y)=({\bar{\mathrm{\nabla }}}_X{\theta }^i)(Y)-({\bar{\mathrm{\nabla }}}_Y{\theta }^i)(X)$. Finally set ${\theta }^0:=-n^{\mathrm{♭}}=ds-g_{si}d{y}^i$; then ${\theta }^0(n)=1$ and, expanding, $-({\theta }^0{)}^2+{\sum }_i({\theta }^i{)}^2=-d{s}^2+2g_{si}dsd{y}^i+g_{ij}d{y}^{i}d{y}^j=\eta$ on $U$, so $({\theta }^0,{\theta }^1,{\theta }^2,{\theta }^3)$ is a $\eta$-orthonormal coframe of $U$.

Converse, the vorticity dichotomy: in inertial coordinates the Killing equation is ${\mathrm{\partial }}_a{K}_b+{\mathrm{\partial }}_b{K}_a=0$. Differentiating it with cyclically permuted indices and using commuting smooth mixed derivatives gives $2{\mathrm{\partial }}_a{\mathrm{\partial }}_b{K}_c=0$. Thus ${\mathrm{\partial }}_a{K}_b=L_{ab}$ is constant on the connected open set $O$, $L_{ab}=-L_{ba}$, and $K$ is affine there. Indeed a function with zero first derivatives is locally constant by integration on coordinate segments, and locally constant functions are constant on a connected set; applying this first to ${\mathrm{\partial }}_a{K}_b$ and then to $K_b-L_{ab}{x}^a$ proves the assertion. For $n=\phi K$, $K^{\mathrm{♭}}={\phi }^{-1}{n}^{\mathrm{♭}}$; projecting $d{n}^{\mathrm{♭}}=d\phi \wedge K^{\mathrm{♭}}+\phi d{K}^{\mathrm{♭}}$ horizontally therefore gives $\omega =\phi L{\mathrm{\mid }}_{n^{\perp }}$.

Adapted frame: let $e_A$ be dual to $({\theta }^0,{\theta }^1,{\theta }^2,{\theta }^3)$. Then $e_0=n$ and $e_i=E_i+d_i(s,y)n$ with the same $y$-coefficients as $E_i$ in step 1.3; the unique $d_i$ is fixed by ${\theta }^0(e_i)=0$. Thus $e_i$ is horizontal and projects by $q$ to the orthonormal field $E_i$ on $B$. Since $n={\mathrm{\partial }}_s$, $[n,e_i]=({\mathrm{\partial }}_s{d}_i)n$, with a positive derivative sign; in particular ${\theta }^j([n,e_i])=0$. From [F2] and $\theta =0$, ${\mathrm{\nabla }}_{e_i}n=\omega (e_i,\cdot {)}^{\mathrm{♯}}$ and ${\mathrm{\nabla }}_{n}n=a$.

Suppose this vorticity vanishes at some $x_0\in O$, and set $k_0=K(x_0)$, which is timelike. The constant two-form $L$ vanishes on $k_0^{\perp }\times k_0^{\perp }$. Decomposing each vector into its $k_0$ and $k_0^{\perp }$ components shows that $L=k_0^{\mathrm{♭}}\wedge \beta$ for a covector $\beta$ with $\beta (k_0)=0$ (specifically $\beta (v)=L(k_0,v)\mathrm{/}\eta (k_0,k_0)$). If $L=0$, the vorticity is identically zero. Otherwise $b={\beta }^{\mathrm{♯}}$ is nonzero spacelike and orthogonal to $k_0$, so $P=\mathrm{span}(k_0,b)$ is a nondegenerate Lorentzian two-plane. The linear part of the affine vector field $K$ has image in $P$, because its matrix is obtained by raising an index of $L$; hence $K(x)-K(x_0)\in P$ and $K(x)\in P$ for all $x\in O$. Since $K(x)$ is timelike, $P$ is the orthogonal sum of its line and a spacelike line in $K(x{)}^{\perp }$. The $P$-components of two vectors in $K(x{)}^{\perp }$ therefore lie in that same spacelike line, while $L$ vanishes on $P^{\perp }$; consequently $L{\mathrm{\mid }}_{K(x{)}^{\perp }}=0$ and $\omega (x)=0$. Thus a zero at one point forces $\omega \equiv 0$ on $O$; if there is no zero, it is nowhere zero. No decomposition along a possibly degenerate arbitrary two-plane is used.

Irrotational case, leaves: assume now $\theta =\omega =0$ on $\mathrm{\Omega }$. By step 1.1, ${\mathrm{\nabla }}_{X}n=0$ for every vector field $X$ tangent to the rest distribution $D=\ker n^{\mathrm{♭}}$; in particular, in an inertial chart, all first partial derivatives of the components of $n$ along directions in $D$ vanish. The distribution $D$ is integrable by the Frobenius criterion [F13], because $\omega =0$. Let $S$ be a connected leaf of $D$ through a point $p$. For a $C^1$ curve $\gamma$ in $S$ the tangent ${\gamma }^{\mathrm{\prime }}$ takes values in $D\subseteq \ker n^{\mathrm{♭}}$ and $\frac{d}{dt}n(\gamma (t))={\mathrm{\nabla }}_{{\gamma }^{\mathrm{\prime }}}n=0$, so $n$ is constant on $S$ with value $n_0:=n(p)$; consequently the tangent space of $S$ at each of its points is $D=n_0^{\perp }$, and integrating the constant fields of $n_0^{\perp }$ shows that $S$ is an open subset of the affine hyperplane $p+n_0^{\perp }$. Choosing a maximal integral curve of $n$ through a point (its parameter has speed $1$, since $\eta (n,n)=-1$) and shrinking to a neighbourhood $U$ of one of its points over which the labels of the next step are unique, we obtain a smooth future timelike curve $z:J\to A$ with $z^{\mathrm{\prime }}=n\circ z$, $\eta (z^{\mathrm{\prime }},z^{\mathrm{\prime }})=-1$, such that the local leaf through $z(s)$ is an open subset of $z(s)+z^{\mathrm{\prime }}(s{)}^{\perp }$.

Irrotational case, the label and the formulas: on a small enough neighbourhood $U$ of a point of the curve consider $f(s,x):=\eta (z^{\mathrm{\prime }}(s),x-z(s))$, smooth with $f(s,z(s))=0$ and ${\mathrm{\partial }}_{s}f(s,x)=1+\eta (z^{\mathrm{\prime }\mathrm{\prime }}(s),x-z(s))$ because $\eta (z^{\mathrm{\prime }}(s),z^{\mathrm{\prime }}(s))=-1$. At the label $s=\sigma (x)$ obtained below this derivative is $N(x):=1+\eta (z^{\mathrm{\prime }\mathrm{\prime }}(\sigma (x)),x-z(\sigma (x)))$; in particular ${\mathrm{\partial }}_{s}f(s,z(s))=1>0$. In an inertial chart the map $(s,x)\mapsto (f(s,x),x)$ has at $(s,z(s))$ the block derivative $\left(\begin{array}{cc}1& {\mathrm{\partial }}_{x}f\\ 0& I\end{array}\right)$, which is invertible; by the choice-free inverse function theorem [F9] there is a uniquely determined smooth function $\sigma$ with $f(\sigma (x),x)=0$ on a neighbourhood $U$, and shrinking $U$ keeps $N>0$ because $N(z(s))=1$ and $N$ is continuous. On $U$ the level sets of $\sigma$ are exactly the leaves: $\sigma$ is constant exactly along $D$, and each level is an open subset of a hyperplane $z(s)+z^{\mathrm{\prime }}(s{)}^{\perp }$. Differentiating $f(\sigma (x),x)=0$ in a direction $v$ gives $0=N(x)d\sigma (v)+\eta (z^{\mathrm{\prime }}(\sigma (x)),v)$, which is the displayed formula for $d\sigma$, and $n=z^{\mathrm{\prime }}\circ \sigma$ on $U$ because both are future unit vectors orthogonal to the same leaf. Finally ${\mathrm{\nabla }}_{v}n=\frac{d}{dt}{\mid }_{t=0}{z}^{\mathrm{\prime }}(\sigma (x+tv))=z^{\mathrm{\prime }\mathrm{\prime }}(\sigma (x))d\sigma (v)$ gives $\mathrm{\nabla }n=d\sigma \otimes (z^{\mathrm{\prime }\mathrm{\prime }}\circ \sigma )$, and $a={\mathrm{\nabla }}_{n}n=d\sigma (n)z^{\mathrm{\prime }\mathrm{\prime }}(\sigma )=z^{\mathrm{\prime }\mathrm{\prime }}(\sigma )\mathrm{/}N$ because $d\sigma (n)=-\eta (z^{\mathrm{\prime }}(\sigma ),n)\mathrm{/}N=1\mathrm{/}N$ as $\eta (z^{\mathrm{\prime }}(\sigma ),n)=\eta (n,n)=-1$.

Connection coefficients, entries with a $0$: define one-forms ${\gamma }^0{}_i={\gamma }^i{}_0:=a_i{\theta }^0-{\omega }_{ij}{\theta }^j$ and ${\gamma }^i{}_j:={\bar{\gamma }}^i{}_j-{\omega }_{ij}{\theta }^0$, where $a_i:=a^{\mathrm{♭}}(e_i)$ and ${\omega }_{ij}:=\omega (e_i,e_j)$. We claim that ${\gamma }^A{}_B(X)={\theta }^A({\mathrm{\nabla }}_X{e}_B)$ for all $A,B$; since the right-hand side is the connection matrix of the Levi-Civita connection in this frame by [F4], this identifies $\gamma$ and gives in particular the skew relations ${\gamma }^0{}_i={\gamma }^i{}_0$ and ${\gamma }^i{}_j=-{\gamma }^j{}_i$. For the entries with a zero index: ${\theta }^0({\mathrm{\nabla }}_n{e}_i)=a_i$, because $0=n(\eta (n,e_i))=({\mathrm{\nabla }}_n{n}^{\mathrm{♭}})(e_i)+n^{\mathrm{♭}}({\mathrm{\nabla }}_n{e}_i)$ gives $n^{\mathrm{♭}}({\mathrm{\nabla }}_n{e}_i)=-a_i$ and ${\theta }^0=-n^{\mathrm{♭}}$; ${\theta }^i({\mathrm{\nabla }}_n{e}_j)={\theta }^i({\mathrm{\nabla }}_{e_j}n)+{\theta }^i([n,e_j])=\omega (e_j,e_i)+0=-{\omega }_{ij}$ by step 2.1; and ${\theta }^0({\mathrm{\nabla }}_{e_j}{e}_i)={\omega }_{ji}=-{\omega }_{ij}$, because $0=e_j(\eta (n,e_i))=({\mathrm{\nabla }}_{e_j}{n}^{\mathrm{♭}})(e_i)+n^{\mathrm{♭}}({\mathrm{\nabla }}_{e_j}{e}_i)={\omega }_{ji}+n^{\mathrm{♭}}({\mathrm{\nabla }}_{e_j}{e}_i)$. Thus ${\gamma }^0{}_i(e_0)={\gamma }^i{}_0(e_0)=a_i$ and ${\gamma }^0{}_i(e_j)={\gamma }^i{}_0(e_j)=-{\omega }_{ij}$.

Irrotational case, converse: conversely let $z:J\to A$ be smooth with $\eta (z^{\mathrm{\prime }},z^{\mathrm{\prime }})=-1$, let $\sigma$ be a smooth function on a tubular neighbourhood $U$ whose level sets are $z(s)+z^{\mathrm{\prime }}(s{)}^{\perp }$ with $N>0$, and put $n:=z^{\mathrm{\prime }}\circ \sigma$; then $n$ is smooth and future unit, its rest distribution is the tangent distribution of the $\sigma$-levels, which is integrable, and for $X$ tangent to a level one has ${\mathrm{\nabla }}_{X}n={\mathrm{\nabla }}_X(z^{\mathrm{\prime }}\circ \sigma )=z^{\mathrm{\prime }\mathrm{\prime }}(\sigma )d\sigma (X)=0$ because $d\sigma$ vanishes on the level directions. By the computation of step 1.1 this gives $\theta (X,Y)+\omega (X,Y)=0$ for all horizontal fields $X,Y$; the rest distribution is integrable, so $\omega =0$ by the Frobenius criterion [F13]; hence $\theta (X,Y)=0$ for all horizontal $X,Y$, that is, $\theta =0$. Thus $n$ is a smooth future unit irrotational Born-rigid congruence satisfying the same formulas $\mathrm{\nabla }n=d\sigma \otimes (z^{\mathrm{\prime }\mathrm{\prime }}\circ \sigma )$, $a=z^{\mathrm{\prime }\mathrm{\prime }}(\sigma )\mathrm{/}N$ and $N=1+\eta (z^{\mathrm{\prime }\mathrm{\prime }}(\sigma ),x-z(\sigma ))$ as in step 3.1, with the label derivative $1+\eta (z^{\mathrm{\prime }\mathrm{\prime }}\circ \sigma ,x-z\circ \sigma )=N$ positive on $U$.

Horizontal connection entries: for horizontal lifts $X,Y,Z$ of fields $\bar{X},\bar{Y},\bar{Z}$ on $B$, the functions $\eta (Y,Z)=q^{\ast }\bar{h}(\bar{Y},\bar{Z})$ are basic, $q_{\ast }[X,Y]=[\bar{X},\bar{Y}]$, and the vertical parts of the brackets pair to zero with $Z$. Metric compatibility and torsion freeness of [F1] give, by adding their three cyclic identities, $2\eta ({\mathrm{\nabla }}_{X}Y,Z)=X\eta (Y,Z)+Y\eta (Z,X)-Z\eta (X,Y)-\eta (X,[Y,Z])+\eta (Y,[Z,X])+\eta (Z,[X,Y])\mathrm{.}$ Every term is the pullback of the same expression on $(B,\bar{h})$. The unique connection $\bar{\mathrm{\nabla }}$ of [F5] satisfies that expression by the same metric/torsion calculation. Thus $h({\mathrm{\nabla }}_{X}Y)$ is the horizontal lift of ${\bar{\mathrm{\nabla }}}_{\bar{X}}\bar{Y}$, by positive definiteness on the horizontal space. This also proves that its coefficients are basic. Taking $X=e_j,Y=e_k,Z=e_i$ gives ${\theta }^i({\mathrm{\nabla }}_{e_j}{e}_k)={\bar{\gamma }}^i{}_k(e_j)$, so together with step 3.2 the displayed ${\gamma }^A{}_B$ are exactly the spacetime connection forms. This comparison uses the local quotient $B$, not an integral leaf of the rotational rest distribution.

First structure equation and the differential of ${\theta }^0$: for the coframe of step 1.3 and the connection forms of steps 3.2 and 4.2 one has $d{\theta }^A=-{\gamma }^A{}_B\wedge {\theta }^B$. Indeed, metric compatibility of $\mathrm{\nabla }$ gives $X({\theta }^A(e_B))=({\mathrm{\nabla }}_X{\theta }^A)(e_B)+{\theta }^A({\mathrm{\nabla }}_X{e}_B)$, so $({\mathrm{\nabla }}_X{\theta }^A)(e_B)=-{\gamma }^A{}_B(X)$ and ${\mathrm{\nabla }}_X{\theta }^A=-{\gamma }^A{}_B(X){\theta }^B$; substituting into the invariant formula for $d{\theta }^A$ of [F6] and using torsion-freeness of $\mathrm{\nabla }$ cancels the torsion term and yields $d{\theta }^A(X,Y)=-({\gamma }^A{}_B\wedge {\theta }^B)(X,Y)$. In particular, using ${\gamma }^0{}_i=a_i{\theta }^0-{\omega }_{ij}{\theta }^j$ and writing $a^{\mathrm{♭}}=a_i{\theta }^i$, $\omega =\frac{1}{2}{\omega }_{ij}{\theta }^i\wedge {\theta }^j$, we obtain $d{\theta }^0=-{\gamma }^0{}_i\wedge {\theta }^i=-{\theta }^0\wedge a^{\mathrm{♭}}-2\omega$, because $-{\gamma }^0{}_i\wedge {\theta }^i=-a_i{\theta }^0\wedge {\theta }^i+{\omega }_{ij}{\theta }^j\wedge {\theta }^i$ and ${\omega }_{ij}{\theta }^j\wedge {\theta }^i=-2\omega$ by antisymmetry of $\omega$.

Mixed identity: in inertial coordinates, flatness on the covector $n_c$ gives ${\mathrm{\nabla }}_a{\mathrm{\nabla }}_b{n}_c-{\mathrm{\nabla }}_b{\mathrm{\nabla }}_a{n}_c=0$. Substitute ${\mathrm{\nabla }}_b{n}_c=-n_b{a}_c+{\omega }_{bc}$ and differentiate with the product rule: $0=-({\mathrm{\nabla }}_a{n}_b-{\mathrm{\nabla }}_b{n}_a)a_c-n_b{\mathrm{\nabla }}_a{a}_c+n_a{\mathrm{\nabla }}_b{a}_c+{\mathrm{\nabla }}_a{\omega }_{bc}-{\mathrm{\nabla }}_b{\omega }_{ac}\mathrm{.}$ Project all three free indices onto the rest space. The two terms with explicit $n_a$ or $n_b$ vanish; the first term becomes $-2{\omega }_{ab}{a}_c$. The acceleration is horizontal, so its projected covector is $a^{\mathrm{♭}}$ itself, not zero. Therefore ${\widehat{\mathrm{\nabla }}}_a{\omega }_{bc}-{\widehat{\mathrm{\nabla }}}_b{\omega }_{ac}=2{\omega }_{ab}{a}_c$. No commutation of differentiation along the variable field $n$ is required. In the adapted frame this is $D_m{\omega }_{ij}-D_i{\omega }_{mj}=2{\omega }_{mi}{a}_j$, where $D_m{\omega }_{ij}=e_m({\omega }_{ij})-{\bar{\gamma }}^k{}_i(e_m){\omega }_{kj}-{\bar{\gamma }}^k{}_j(e_m){\omega }_{ik}$: the vertical parts of ${\mathrm{\nabla }}_{e_m}{e}_i$ pair to zero with the horizontal $\omega$, and step 4.2 supplies the horizontal coefficients.

Horizontal flatness equation: by [F4] the curvature two-forms of $\mathrm{\nabla }$ in this frame are ${\mathrm{\Omega }}^A{}_B=d{\gamma }^A{}_B+{\gamma }^A{}_C\wedge {\gamma }^C{}_B$, and ${\mathrm{\Omega }}^A{}_B(X,Y)={\theta }^A(R(X,Y)e_B)=0$ by flatness of $\mathrm{\nabla }$ [F1, F4]; hence ${\mathrm{\Omega }}^A{}_B=0$. Expanding ${\mathrm{\Omega }}^i{}_j=d{\gamma }^i{}_j+{\gamma }^i{}_0\wedge {\gamma }^0{}_j+{\gamma }^i{}_k\wedge {\gamma }^k{}_j$ with steps 3.2 and 4.3 gives $d{\gamma }^i{}_j=d{\bar{\gamma }}^i{}_j-d{\omega }_{ij}\wedge {\theta }^0+{\omega }_{ij}{\theta }^0\wedge a^{\mathrm{♭}}+2{\omega }_{ij}\omega$, ${\gamma }^i{}_0\wedge {\gamma }^0{}_j=(a_i{\theta }^0-{\omega }_{ik}{\theta }^k)\wedge (a_j{\theta }^0-{\omega }_{jl}{\theta }^l)=-a_i{\omega }_{jl}{\theta }^0\wedge {\theta }^l+{\omega }_{ik}{a}_j{\theta }^0\wedge {\theta }^k+{\omega }_{ik}{\omega }_{jl}{\theta }^k\wedge {\theta }^l$ and ${\gamma }^i{}_k\wedge {\gamma }^k{}_j={\bar{\gamma }}^i{}_k\wedge {\bar{\gamma }}^k{}_j+{\omega }_{kj}{\theta }^0\wedge {\bar{\gamma }}^i{}_k-{\omega }_{ik}{\theta }^0\wedge {\bar{\gamma }}^k{}_j$; the components with two horizontal legs therefore satisfy $0=\bar{\mathrm{\Omega }}{}^i{}_j+2{\omega }_{ij}\omega +{\omega }_{ik}{\omega }_{jl}{\theta }^k\wedge {\theta }^l$, where $\bar{\mathrm{\Omega }}{}^i{}_j=d{\bar{\gamma }}^i{}_j+{\bar{\gamma }}^i{}_k\wedge {\bar{\gamma }}^k{}_j$ is the curvature two-form of the slice metric $h$ in the coframe $({\theta }^i)$.

Curvature of the slice and the three-dimensional identity: by [F5] and step 4.2 the two-form $\bar{\mathrm{\Omega }}{}^i{}_j$ is the pullback of the curvature of $\bar{\mathrm{\nabla }}$ on $(B,\bar{h})$, so $\bar{\mathrm{\Omega }}{}^i{}_j=\frac{1}{2}{\bar{R}}^i{}_{jkl}{\theta }^k\wedge {\theta }^l$ with ${\bar{R}}^i{}_{jkl}={\theta }^i(R_{\bar{h}}(E_k,E_l)E_j)$ in the argument order of the Riemann four-tensor of [F5], and we write ${\bar{R}}_{ijkl}$ for ${\bar{R}}^i{}_{jkl}$ (indices are moved with the Euclidean slice metric). Comparing the antisymmetrized ${\theta }^k\wedge {\theta }^l$ components of step 5.2 gives ${\bar{R}}^i{}_{jkl}+2{\omega }_{ij}{\omega }_{kl}+{\omega }_{ik}{\omega }_{jl}-{\omega }_{il}{\omega }_{jk}=0$ for all $i,j,k,l$. For the antisymmetric matrix $({\omega }_{ij})$ on the three-dimensional quotient write ${\omega }_{ij}={\epsilon }_{ijs}{w}^s$ with $\epsilon$ the alternating symbol and $w$ the associated vector, unique because the slice metric is Euclidean; expanding the product of two alternating symbols and contracting with $w^s{w}^t$ gives ${\omega }_{ik}{\omega }_{jl}-{\omega }_{il}{\omega }_{jk}={\omega }_{ij}{\omega }_{kl}$ for all $i,j,k,l$, both sides being equal to $-{\delta }_{il}{\delta }_{jk}\mathrm{\mid }w{\mathrm{\mid }}^2+{\delta }_{ik}{\delta }_{jl}\mathrm{\mid }w{\mathrm{\mid }}^2+{\delta }_{il}{w}_k{w}_j-{\delta }_{ik}{w}_l{w}_j+w_i{\delta }_{jk}{w}_l-w_i{\delta }_{jl}{w}_k$. Substituting this identity into the previous equation gives ${\bar{R}}^i{}_{jkl}=-3{\omega }_{ij}{\omega }_{kl}$ for all $i,j,k,l$.

The frame components of the vorticity are independent of $s$: since $h$, and hence its connection forms ${\bar{\gamma }}^i{}_j$ and curvature ${\bar{R}}^i{}_{jkl}$, are independent of $s$ by steps 1.2 and 1.3, differentiating the equation ${\bar{R}}^i{}_{jkl}=-3{\omega }_{ij}{\omega }_{kl}$ of step 6.1 along ${\mathrm{\partial }}_s$ gives $0=-3({\mathrm{\partial }}_s{\omega }_{ij}{\omega }_{kl}+{\omega }_{ij}{\mathrm{\partial }}_s{\omega }_{kl})$. Fix a point of $U$, in which $\omega$ is nowhere zero, so there is a pair of distinct indices $(k,l)$ with ${\omega }_{kl}\mathrm{\ne }0$; taking $(i,j)=(k,l)$ in the displayed identity gives $2{\omega }_{kl}{\mathrm{\partial }}_s{\omega }_{kl}=0$, hence ${\mathrm{\partial }}_s{\omega }_{kl}=0$, and then taking $(i,j)$ arbitrary gives ${\mathrm{\partial }}_s{\omega }_{ij}{\omega }_{kl}=0$, hence ${\mathrm{\partial }}_s{\omega }_{ij}=0$ for every pair. Therefore ${\mathcal{L}}_n\omega =0$: by [F10], $({\mathcal{L}}_n\omega )(e_i,e_j)=n({\omega }_{ij})-\omega ([n,e_i],e_j)-\omega (e_i,[n,e_j])={\mathrm{\partial }}_s{\omega }_{ij}$, since $[n,e_i]$ is a multiple of $n$ by step 2.1 and $\omega$ is horizontal by [F2]. Thus the vorticity is basic on $U$.