Counterexample: a Born-rigid congruence need not be a Killing congruence

Construction of the data: fix $L_0>0$ and put $\psi (\lambda ):={\lambda }^2\mathrm{/}(2L_0^2)$, $k(\lambda ):={\psi }^{\mathrm{\prime }}(\lambda )=\lambda \mathrm{/}{L}_0^2$, $e_0(\lambda ):=(\cosh \psi (\lambda ),\sinh \psi (\lambda ),0,0)$ and $e_1(\lambda ):=(\sinh \psi (\lambda ),\cosh \psi (\lambda ),0,0)$ in the inertial coordinates, and let $z(\lambda ):=({\int }_0^{\lambda }\cosh \psi (s)ds,{\int }_0^{\lambda }\sinh \psi (s)ds,0,0)$; these are smooth in $\lambda$, $z(0)=0$, and $z^{\mathrm{\prime }}=e_0$ by [L3]. By [L2], $\eta (e_0,e_0)=-{\cosh }^2\psi +{\sinh }^2\psi =-1$, $\eta (e_1,e_1)=-{\sinh }^2\psi +{\cosh }^2\psi =1$ and $\eta (e_0,e_1)=-\cosh \psi \sinh \psi +\sinh \psi \cosh \psi =0$, so $(e_0,e_1,e_2,e_3)$ is an orthonormal frame at every $\lambda$; moreover $e_0$ is future by [L6] because its time component $\cosh \psi >0$ and $\eta (e_0,e_0)=-1$.

Local chart: define $\mathrm{\Phi }(\lambda ,\xi ,y,\zeta ):=z(\lambda )+\xi e_1(\lambda )+y{e}_2+\zeta e_3$ on ${\mathbb{R}}^4$. Differentiating in $\lambda$ gives ${\mathrm{\partial }}_{\lambda }\mathrm{\Phi }=z^{\mathrm{\prime }}+\xi e_1^{\mathrm{\prime }}=e_0+\xi k{e}_0=N{e}_0$ with $N:=1+k(\lambda )\xi$, using $e_1^{\mathrm{\prime }}=k{e}_0$ from [L2]; further ${\mathrm{\partial }}_{\xi }\mathrm{\Phi }=e_1$, ${\mathrm{\partial }}_y\mathrm{\Phi }=e_2$, ${\mathrm{\partial }}_{\zeta }\mathrm{\Phi }=e_3$. At $(\lambda ,\xi ,y,\zeta )=0$ these four vectors are $e_0(0)=(1,0,0,0)$, $e_1(0)=(0,1,0,0)$, $e_2=(0,0,1,0)$, $e_3=(0,0,0,1)$, so the derivative matrix is the identity and is invertible; by [L4] $\mathrm{\Phi }$ restricts to a diffeomorphism from an open neighbourhood of $0$ onto an open neighbourhood $W$ of $\mathrm{\Phi }(0)=0$. Shrinking to a coordinate box $B=\{(\lambda ,\xi ,y,\zeta ):\mathrm{\mid }\lambda \mathrm{\mid }<\epsilon L_0,\mathrm{\mid }\xi \mathrm{\mid }<\epsilon L_0,\mathrm{\mid }y\mathrm{\mid }<\epsilon L_0,\mathrm{\mid }\zeta \mathrm{\mid }<\epsilon L_0\}$ with $0<\epsilon <1$ (intersected with the neighbourhood supplied by [L4]) keeps a diffeomorphism onto its image and gives $\mathrm{\mid }k\xi \mathrm{\mid }=\mathrm{\mid }\lambda \xi \mathrm{\mid }\mathrm{/}{L}_0^2<{\epsilon }^2$, hence $N=1+k\xi \ge 1-{\epsilon }^2>0$ on $B$. We write again $W:=\mathrm{\Phi }(B)$ and use $(\lambda ,\xi ,y,\zeta )$ as coordinates on $W$.

Metric and unit field: by step 1.2, $d\mathrm{\Phi }=N{e}_{0}d\lambda +e_{1}d\xi +e_{2}dy+e_{3}d\zeta$; since $(e_0,e_1,e_2,e_3)$ is $\eta$-orthonormal with $e_0$ timelike, the pullback metric on $W$ is $g=-N^{2}d{\lambda }^2+d{\xi }^2+d{y}^2+d{\zeta }^2$. Hence the vector field $n:=N^{-1}{\mathrm{\partial }}_{\lambda }$ satisfies $g(n,n)=-N^2\cdot N^{-2}=-1$, and $n=e_0(\lambda )$ is future by step 1.1; $n$ is smooth on $W$ because $N$ and the coordinates are, so $n$ is a smooth future unit timelike field, that is, a congruence on $W$. The rest space $n^{\perp }$ at each point is spanned by ${\mathrm{\partial }}_{\xi },{\mathrm{\partial }}_y,{\mathrm{\partial }}_{\zeta }$, since these are $g$-orthogonal to ${\mathrm{\partial }}_{\lambda }$ and have positive squares.

Rigidity and vanishing vorticity: the inertial components of $n$ are $e_0(\lambda )$, so its directional derivative in a vector $v$ is $d\lambda (v)k{e}_1$ by step 1.1 and the flat component derivative. For a rest vector $r$ spanned by ${\mathrm{\partial }}_{\xi },{\mathrm{\partial }}_y,{\mathrm{\partial }}_{\zeta }$, one has $d\lambda (r)=0$, hence ${\mathrm{\nabla }}_{r}n=0$. Thus both the symmetric and antisymmetric projected derivatives vanish on pairs of rest vectors. Since $\theta$ and $\omega$ are horizontal in both slots, $\theta =\omega =0$ everywhere. The field is Born rigid and irrotational.

Acceleration one-form: since $d\lambda (n)=1\mathrm{/}N$, the derivative from step 3.1 gives $a={\mathrm{\nabla }}_{n}n=(k\mathrm{/}N)e_1=(k\mathrm{/}N){\mathrm{\partial }}_{\xi }$. Therefore $a^{\mathrm{♭}}=(k\mathrm{/}N)d\xi$ by the metric of step 2.1. As $N=1+k(\lambda )\xi$ and ${\mathrm{\partial }}_{\lambda }N=k^{\mathrm{\prime }}\xi$, one has ${\mathrm{\partial }}_{\lambda }(k\mathrm{/}N)=(k^{\mathrm{\prime }}N-k{k}^{\mathrm{\prime }}\xi )\mathrm{/}{N}^2=k^{\mathrm{\prime }}(N-k\xi )\mathrm{/}{N}^2=1\mathrm{/}(L_0^2{N}^2)$. The remaining derivatives contribute no term to $d(k\mathrm{/}N)\wedge d\xi$, so $d{a}^{\mathrm{♭}}=(1\mathrm{/}(L_0^2{N}^2))d\lambda \wedge d\xi$, nowhere zero on $W$ because $N>0$.

No positive multiple is Killing: let $f>0$ be smooth on $W$ and suppose ${\mathcal{L}}_{fn}\eta =0$. By [L1] and $\theta =0$, $0={\mathcal{L}}_{fn}\eta =f{\mathcal{L}}_n\eta +df\otimes n^{\mathrm{♭}}+n^{\mathrm{♭}}\otimes df=n^{\mathrm{♭}}\otimes \beta +\beta \otimes n^{\mathrm{♭}}$ with $\beta :=df-f{a}^{\mathrm{♭}}$. Evaluating at $(n,r)$ for a rest vector $r$ gives $n^{\mathrm{♭}}(n)\beta (r)+\beta (n)n^{\mathrm{♭}}(r)=-\beta (r)=0$, and evaluating at $(n,n)$ gives $-2\beta (n)=0$; since every vector is a sum of a multiple of $n$ and a rest vector by step 2.1, $\beta =0$, that is, $df=f{a}^{\mathrm{♭}}$ and hence $d\log f=a^{\mathrm{♭}}$. Applying $d$ and [L5] gives $0=d^2\log f=d{a}^{\mathrm{♭}}=(1\mathrm{/}(L_0^2{N}^2))d\lambda \wedge d\xi \mathrm{\ne }0$ by step 4.1, a contradiction. Therefore no positive smooth $f$ on any nonempty open subset of $W$ makes $fn$ a Killing field: the constructed congruence is Born rigid and future unit, but it is not locally a positive multiple of a Killing field.

Conclusion and clarifications: steps 1.1-5.1 construct a smooth future unit timelike field $n$ on the open set $W$ that is Born rigid ($\theta =0$) and irrotational ($\omega =0$), and step 5.1 shows that for every positive smooth $f$ the field $fn$ fails the Killing equation on every nonempty open subset; hence the claim of the Statement refuted section is false. The Rindler congruence is not a counterexample of this kind: its metric $-{\rho }^{2}d{\theta }^2+d{\rho }^2+d{y}^2+d{z}^2$ is independent of $\theta$, so ${\mathrm{\partial }}_{\theta }$ is a Killing field and the unit field ${\rho }^{-1}{\mathrm{\partial }}_{\theta }$ is a positive multiple of it (Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences clause (b)). The rotational classification of The smooth local Herglotz-Noether classification of Born-rigid motions is untouched, because the present congruence is irrotational: the counterexample concerns the irrotational branch of Born-rigid motions.