Proposition: Particle slice defects and total intrinsic curvature

On the open incoming and outgoing epochs, respectively, the intrinsic angle defects at the particle are

They are not the spacetime deficit $\pi /2$. For the compact patch, the sum of all interior intrinsic vertex defects is nevertheless

At the event, only the regular cells of the common limiting slice enter this sum.

Proof. On an incoming $V_i$,

The angle between its two ray directions satisfies

Three such angles meet at $P$, giving ${\kappa }_P^{-}$.

On $U_i^{-}$, the induced metric is $\mathrm{diag}(1,16/25)$. The isometry $(\xi ,\zeta )\mapsto (\xi ,4\zeta /5)$ sends the two particle-edge directions to $(1,0)$ and $(1,4/5)$. Their angle is $\arctan (4/5)$. The reflected $U_i^{+}$ has the same angle; six meet at $P$, giving ${\kappa }_P^{+}$.

For the total, triangulate each intrinsic flat polygon. Euclidean triangle angle sums and the Euler formula for a disk give

Artificial interior flat vertices have zero defect, and artificial vertices on straight boundary segments have zero boundary turning.

At each of the three boundary vertices $D_i$, the two horizontal sector quadrilaterals contribute $\pi /4$ each, so the total interior boundary angle is $\pi /2$. The boundary turning sum is $3\pi /2$, leaving $\pi /2$ for the interior defect sum. Other interior slice vertices therefore compensate for the particle's slice-dependent defect. The affine peripheral holonomy remains $J_0$ by A:prop:fixed-certificate.