Corner terms are required even in the plane

Take $D=[0,1{]}^2\subset {\mathbb{R}}^2$ with the Euclidean metric, oriented by the standard frame $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$; this frame is orthonormal and positive. Since the coordinate coefficients of the Euclidean metric are constant, the formula of [F5] gives ${\mathrm{\Gamma }}^k{}_{ij}=0$ for all $i,j,k$, hence ${\mathrm{\nabla }}_{{\mathrm{\partial }}_x}{\mathrm{\partial }}_x={\mathrm{\nabla }}_{{\mathrm{\partial }}_x}{\mathrm{\partial }}_y={\mathrm{\nabla }}_{{\mathrm{\partial }}_y}{\mathrm{\partial }}_y=0$ and the connection form of the frame vanishes identically, $d\omega =0$. The structure equation of [F5] then gives $K\equiv 0$ on $D$.

The four sides of $D$ are unit-speed straight segments, so by [F2] each is an affinely parametrized geodesic and its covariant acceleration vanishes; [F3] therefore gives $k_g=0$ along all four sides and ${\int }_{\mathrm{\partial }D}{k}_{g}ds=0$. At each of the four vertices the interior sector angle is $\beta =\pi \mathrm{/}2$, so [F4] gives exterior angle $\alpha =\pi -\pi \mathrm{/}2=\pi \mathrm{/}2$ and ${\sum }_j{\alpha }_j=4\cdot \pi \mathrm{/}2=2\pi$.

With $K\equiv 0$ and ${\int }_{\mathrm{\partial }D}{k}_{g}ds=0$, the corner-free expression would read ${\int }_{D}KdA+{\int }_{\mathrm{\partial }D}{k}_{g}ds=0$, which differs from $2\pi$; hence the exterior corner angles cannot be omitted from the boundary value problem.

Under the AC premise [F6], the complete formula [F1] applied to $D$ reads $0+2\pi =2\pi$, which holds; thus the square is a genuine geodesic polygonal disk region for which only the full formula with its corner sum is correct, and the refuted statement is false.