Proposition: Slice curvature is not the spacetime deficit

The transverse intrinsic defects at the singular intersection are

Neither equals the prescribed spacetime deficit $\pi /2$. The sum of all interior transverse vertex defects in the compact patch equals $\pi /2$ on every slice.

Proof. On a $V_i$-plane the ray-direction angle has cosine

On $U_i^{-}$, the induced metric is $\mathrm{diag}(1,16/25)$, so the angle between $(1,0)$ and $(1,1)$ is $\arctan (4/5)$. There are three incoming or six outgoing angles.

Triangulate each polygonal disk without changing its metric. Triangle angle sums and Euler's formula for the transverse disk give

Each of the three outer boundary vertices has interior angle $\pi /2$; hence the boundary turning is $3\pi /2$. The stated interior sum follows. At zero only the noncollapsed limiting cells enter the sum.