Theorem: Stale-deficit rejection

The changed-deficit proposal with retained $\hat{H}=J_0$ fails particle and seam matching. Nevertheless, $\hat{H}$ still fixes ${\mathcal{W}}_0$ pointwise.

Proof. The rest-frame packet requires $J({\mu }^{\prime })$. Its Lorentz part differs from $R(\pi /2)$ by the strictly positive residual $2|\sin (\Delta /2)|$. The allowed interval $-\pi /2<\Delta <3\pi /2$ contains no nonzero multiple of $2\pi$.

The required new wedge opening is $L^{\prime }=3\pi /2-\Delta$. The retained rotation sends its upper direction to

which is not $(1,0{)}^{\mathsf{T}}$ in that allowed range. If instead the old wedge is retained, its geometric deficit remains $\pi /2$, contradicting the new label.

On the other hand, $J_0(T,0,0)=(T,0,0)$ for every $T$. Pointwise-axis matching still holds.