All-jet reflection parity from an exact local boost

Fix any of the adapted patches in the Statement. The linear involution $w$ maps the open set ${\mathrm{\Omega }}_0=\mathrm{\Omega }\cap w(\mathrm{\Omega })$ onto itself and fixes every point of its intersection with the plane $u=v=0$. Its conjugate by the chart therefore gives the stated smooth involution $W$ and retains the entire surface portion $S\cap \mathcal{O}$. To apply F1, fix any point $p\in {\mathcal{O}}_0$. The Lorentzian signature supplies a vector $v\in T_{p}M$ with $g_p(v,v)<0$. Extend its coordinate components constantly to a smooth local field $N$ in this chart. The smooth component criterion in F6 makes $g(N,N)$ continuous; hence it remains negative on some open neighborhood of $p$. On that neighborhood $N$ is a smooth time orientation in the sense of F1. The given Killing hypothesis therefore yields ${\mathcal{L}}_{B}g=0$ there. Since $p$ was arbitrary and the equation is independent of $N$, it holds throughout ${\mathcal{O}}_0$. No common time orientation, simultaneous choice of these fields, or $W$-invariance of their smaller neighborhoods is required. Now justify the jet convention on $S_0$. By F6, for a smooth change of adapted coordinates, each new component of a tensor difference is a finite sum of old components composed with the coordinate change and multiplied by smooth Jacobian or inverse-Jacobian entries, one for each tensor index. The vector transformation follows by applying the chain rule to the coordinate functions; the covariant transformation follows by evaluating on the coordinate tangent vectors. Multilinearity applies these transformations to each slot of a mixed tensor, giving exactly the transition factors in F6. Repeated product and chain rules show, by induction on derivative order, that each derivative of a new component at $S_0$ is a finite sum of old component derivatives at the same point multiplied by smooth coefficients. Consequently vanishing of every old derivative implies vanishing of every new derivative; applying the inverse coordinate change proves the converse. This proves coordinate independence and makes the stated jet equality an equivalence relation on local fields.

Put $D_{a,b,\gamma }={\mathrm{\partial }}_U^a{\mathrm{\partial }}_V^b{\mathrm{\partial }}_y^{\gamma }$ and $L=-U{\mathrm{\partial }}_U+V{\mathrm{\partial }}_V$. For any smooth component function $f$, repeated product differentiation gives ${D_{a,b,\gamma }(Lf)\mid }_{S_0}=(-a+b){D_{a,b,\gamma }f\mid }_{S_0}$: in differentiating $U{\mathrm{\partial }}_{U}f$, the term still containing $U$ vanishes on $S_0$, the terms differentiating $U$ once sum to $a{D}_{a,b,\gamma }f$, and higher derivatives of $U$ vanish; the $V$ term gives $b{D}_{a,b,\gamma }f$ in the same way. The chain rule for the diagonal reflection also gives ${D_{a,b,\gamma }(f\circ W)\mid }_{S_0}=(-1{)}^{a+b}{D_{a,b,\gamma }f\mid }_{S_0}$. Tangential derivatives cause no additional terms because $W$ fixes $y$ and all coefficients of $L$ are independent of $y$.

For a metric component carrying $p$ covariant $U$ indices and $q$ covariant $V$ indices, F1 and F3 give $0=({\mathcal{L}}_{B}g{)}_{ij}=\kappa (L-p+q)g_{ij}$. Since $\kappa \mathrm{\ne }0$, step 1.2 yields $(-a+b-p+q){D_{a,b,\gamma }{g}_{ij}\mid }_{S_0}=0$. Thus a nonzero component derivative requires $-a+b-p+q=0$. The pullback formula in F4 contributes the factor $(-1{)}^{p+q}$ from the covariant indices; differentiating its component composition with $W$ contributes $(-1{)}^{a+b}$. Their total exponent $a+b+p+q=2(a+p)$ is even whenever the weight equation holds. Zero derivatives stay zero, so every metric component derivative agrees with its pullback derivative, including all tangential derivatives. By step 1.1 this proves $j_{S_0}^{\mathrm{\infty }}(W^{\ast }g)=j_{S_0}^{\mathrm{\infty }}g$.

Use the given equation $[B,T]=0$ directly. F3 gives $(L+1)T^U=0$, $(L-1)T^V=0$ and $L{T}^A=0$ after division by $\kappa$. By step 1.2, a nonzero $U$ component derivative requires $-a+b+1=0$, and a nonzero $V$ component derivative requires $-a+b-1=0$. In each case $a+b$ is odd; the inverse differential in the local vector pullback contributes a further minus sign, so the total derivative sign is positive. A tangential component requires $a=b$, and its pullback has no extra index sign, so its total sign is positive as well. Zero derivatives remain zero. Hence every vector component derivative agrees and step 1.1 gives $j_{S_0}^{\mathrm{\infty }}(W^{\ast }T)=j_{S_0}^{\mathrm{\infty }}T$. Smooth boost-invariant functions may have different flat remainders in opposite wedges, so the argument concludes only equality of jets.

The patch was arbitrary, so steps 2.1 and 2.2 prove both local identities near every point of $S$. If $\mathcal{W}$ is supplied as in the Statement, its pullbacks equal those of the corresponding $W$ on a neighborhood of each point of $S$. Every derivative of each pullback difference consequently vanishes at every point of $S$ by the local identities. Coordinate independence from step 1.1 gives the stated global jet equalities for this supplied map. This argument does not assert agreement of distinct patch reflections without the stated compatibility assumption.