Summing local Gauss-Bonnet over a supplied triangulation

Every closed face $f$ of $\mathcal{T}$ is a compact regular oriented disk region with ordinary corners, contained in a frameable chart, so [F2] applies with its selected face orientation: ${\int }_{f}KdA+{\int }_{\mathrm{\partial }f}{k}_{g}ds+{\sum }_c{\alpha }_c=2\pi$, the sum being over the three corners $c$ of $f$.

Each face has three distinct vertices and three sides, each interior edge has exactly two incident faces and each boundary edge exactly one, the link circle at an interior vertex is partitioned into its incident face corners, and at a boundary vertex exactly two boundary edge germs occur. Hence: the number $m_v$ of face corners at $v$ satisfies ${\sum }_v{m}_v=3F$; the face-edge incidences satisfy $3F=2E_{\mathrm{i}\mathrm{n}\mathrm{t}}+E_{\mathrm{b}\mathrm{d}}$; and the boundary incidences satisfy $E_{\mathrm{b}\mathrm{d}}=V_{\mathrm{b}\mathrm{d}}$ because along each boundary circle boundary edges and boundary vertices alternate, each boundary vertex being incident with exactly two boundary edges.

At a vertex $v$ the closed face sectors at $v$ tile a neighbourhood of $v$ in $M$ with pairwise disjoint open sectors: if $v\in \mathrm{Int}M$ the sectors fill a full disk and their interior angles sum to $2\pi$; if $v\in \mathrm{\partial }M$ they fill exactly the sector of $M$ at $v$, so their interior angles sum to the interior angle ${\beta }_v^M$ of $M$ at $v$, which equals $\pi$ when $v$ is not a corner of $\mathrm{\partial }M$. Since each face corner at $v$ contributes exterior angle $\pi -\beta$ by [F3], summing over faces gives ${\sum }_f{\sum }_c{\alpha }_c=\pi {\sum }_v{m}_v-2\pi V_{\mathrm{i}\mathrm{n}\mathrm{t}}-{\sum }_{v\in \mathrm{\partial }M}{\beta }_v^M$, and ${\sum }_{v\in \mathrm{\partial }M}{\beta }_v^M=\pi V_{\mathrm{b}\mathrm{d}}-{\sum }_v{\alpha }_v$ by the corner identity [F3].

The area integrals add: ${\sum }_{f\in F}{\int }_{f}KdA={\int }_{M}KdA$. The face boundary integrals cancel on interior edges: an interior edge is incident with exactly two faces, whose orientations agree with the orientation of $M$, so it is traversed by their induced boundary orientations in the two opposite directions, and [F4] makes the two signed curvature integrals sum to zero; each boundary edge contributes its subarc of $\mathrm{\partial }M$ with the positive boundary orientation of $M$, so ${\sum }_f{\int }_{\mathrm{\partial }f}{k}_{g}ds={\int }_{\mathrm{\partial }M}{k}_{g}ds$.

Summing the equations of step 1.1 over the $F$ faces and substituting steps 1.2, 1.3 and 2.1 gives $2\pi F={\int }_{M}KdA+{\int }_{\mathrm{\partial }M}{k}_{g}ds+3\pi F-2\pi V_{\mathrm{i}\mathrm{n}\mathrm{t}}-\pi V_{\mathrm{b}\mathrm{d}}+{\sum }_v{\alpha }_v$.

The incidence identities give $2E=2E_{\mathrm{i}\mathrm{n}\mathrm{t}}+2E_{\mathrm{b}\mathrm{d}}=(3F-E_{\mathrm{b}\mathrm{d}})+2E_{\mathrm{b}\mathrm{d}}=3F+V_{\mathrm{b}\mathrm{d}}$, so $2\pi (V-E+F)=2\pi V_{\mathrm{i}\mathrm{n}\mathrm{t}}+2\pi V_{\mathrm{b}\mathrm{d}}-\pi (3F+V_{\mathrm{b}\mathrm{d}})+2\pi F=2\pi V_{\mathrm{i}\mathrm{n}\mathrm{t}}+\pi V_{\mathrm{b}\mathrm{d}}-\pi F$. Rearranging step 3.1 therefore yields ${\int }_{M}KdA+{\int }_{\mathrm{\partial }M}{k}_{g}ds+{\sum }_v{\alpha }_v=2\pi (V-E+F)$.

For the closed nonorientable case, orient each face arbitrarily. Its positively oriented area measure is ${\mu }_g$ by [F5], so step 1.1 applies to every face. On a common interior edge the two inward conormals $JT$ of its incident faces point to opposite sides, independently of their chosen orientations: reversing one face orientation reverses both $J$ and its positive boundary tangent $T$. Thus $g({\mathrm{\nabla }}_{T}T,JT)$ has opposite values on the two sides and the integrals cancel by [F4]. There are no boundary edges, so summing gives ${\int }_{M}K{\mu }_g=2\pi F+(2\pi V-\pi {\sum }_v{m}_v)$ with ${\sum }_v{m}_v=3F$ and $2E=3F$, hence ${\int }_{M}K{\mu }_g=2\pi (V-E+F)$, where the replacement of the facewise density integrals by the integral over $M$ is [F6]. This proves both statements.