Proposition: Boost verification on μδ ; exact

In the basis $({\partial }_0\delta ,{\partial }_1\delta ,{\partial }_2\delta ,{\partial }_3\delta )$ the pull-back action of every generator is the matrix $X$ itself; for $K_{01}$,

The second matrix is the coordinate matrix used by the previous revision for the SA intersection (it identified ${\partial }_{\gamma }\delta$ with $z^{\gamma }$). For the three boosts the two differ by a sign; for the three rotations they coincide; in all cases ${\rho }^{\delta }=\eta {\rho }^{\mathrm{poly}}\eta$ at first order and ${{\rho }^{\delta }}_N=P{{\rho }^{\mathrm{poly}}}_{N}P$ for all $N\le 4$ (checked exactly for all six generators). A symbolic test-function computation gives $⟨K_{01}\cdot {\partial }_1\delta ,f⟩=-{\partial }_{0}f(0)=⟨{\partial }_0\delta ,f⟩$, confirming $K_{01}\cdot {\partial }_1\delta =+{\partial }_0\delta$.

Proof. $z^{\rho }{\partial }_{\nu }\delta =-{{\delta }^{\rho }}_{\nu }\delta$ and $z^{\rho }{\partial }_{\mu }{\partial }_{\nu }\delta =-{{\delta }^{\rho }}_{\nu }{\partial }_{\mu }\delta -{{\delta }^{\rho }}_{\mu }{\partial }_{\nu }\delta$ give $-(Xz{)}^{\mu }{\partial }_{\mu }{\partial }_{\nu }\delta =X^{\mu }{}_{\nu }{\partial }_{\mu }\delta +(\mathrm{tr}X){\partial }_{\nu }\delta$, and $\mathrm{tr}X=0$. The higher-order statement follows because $q\mapsto q({\partial }^{\cdot })\delta$ is multiplicative and $\delta$ is invariant; it is also checked exactly by pairing with $z^{\gamma }/\gamma !$ for all $|\gamma |\le 4$ (covariance_constraints.py).