Local Gauss-Bonnet for a frameable disk region

Let $P$ be a compact regular disk region lying in an oriented coordinate chart of $M$ and carrying a single-valued smooth positive orthonormal frame on a neighbourhood of $P$. Let $\theta$ be a continuous tangent-angle lift of the positively oriented unit-speed boundary of $P$ relative to that frame on each smooth arc, with the principal jumps ${\epsilon }_j$ at the corners. Define its total turning by $\mathrm{Rot}={\sum }_{\text{arcs}}\mathrm{\Delta }\theta +{\sum }_j{\epsilon }_j$; its value will be computed below.

The chart coordinate frame gives a second smooth positive frame near $P$. After Gram–Schmidt, its change to the given orthonormal frame is a smooth map $P\to SO(2)$. The closed disk $P$ is path-connected, locally path-connected and simply connected, so its fundamental-group image is trivial; [F9] applied to the circle covering $\mathbb{R}\to SO(2)$ gives a continuous angle lift $\phi$ on $P$ after fixing one value. Smoothness holds locally and the local lifts differ by constants. Replacing the reference direction changes every angle lift by $-\phi$ and leaves every corner jump unchanged. The total turning in the two frames differs by the total change of $-\phi$ around the closed boundary, which is zero. This comparison uses simple connectivity of $P$, not of the surrounding chart.

The total turning in a single-valued frame is a multiple of $2\pi$ for every positively oriented simple closed piecewise $C^2$ regular curve whose initial and terminal unit tangents agree: the accumulated angle returns to a representation of the initial direction, so it differs from the initial angle by an integral multiple of $2\pi$. Thus the argument of step 2.1 can be run for a curve in a chart with the single-valued frames obtained by Gram-Schmidt from the coordinate frame with respect to any Riemannian metric on the chart.

For a disk region contained in one chart with a single-valued frame, apply step 3.1 to the family of metrics $g_s=sg+(1-s)g_e$, where $g_e$ is the Euclidean metric read in the chart: for each $s$ the corresponding total turning ${\mathrm{Rot}}_s$ is an integral multiple of $2\pi$, and $s\mapsto {\mathrm{Rot}}_s$ is continuous because the Gram-Schmidt frames, the angle lifts and the finitely many corner jumps depend continuously on $s$. At $s=0$ the curve is a positively oriented simple closed piecewise $C^2$ plane curve bounding the plane disk region, so [F5] gives ${\mathrm{Rot}}_0=2\pi$; by continuity and integrality ${\mathrm{Rot}}_s=2\pi$ for every $s$, and in particular ${\mathrm{Rot}}_1=2\pi$ for the given metric. This proves the turning fact used below for each piece of [F1].

Fix one piece $D_i$ of the decomposition [F1] and write ${\gamma }_i$ for its positively oriented unit-speed boundary. On each smooth arc of ${\gamma }_i$, [F2] gives $k_{g}ds=d\theta +\omega (T)ds$ for the angle lift $\theta$ relative to the given global frame, so integrating over the finitely many arcs and adding the corner jumps gives ${\int }_{\mathrm{\partial }{D}_i}{k}_{g}ds+{\sum }_v{\alpha }_v^{(i)}=2\pi +{\int }_{\mathrm{\partial }{D}_i}\omega$, where the corner jumps are the piece's signed exterior angles ${\alpha }_v^{(i)}$ by [F6] and the total turning is $2\pi$ by steps 2.1 and 4.1.

The structure equation [F3] and Stokes [F4] give ${\int }_{\mathrm{\partial }{D}_i}\omega ={\int }_{D_i}d\omega =-{\int }_{D_i}KdA$, since each piece lies in a chart carrying the frame and is a compact regular oriented disk region with ordinary corners. Substituting into step 5.1 yields, for every piece, ${\int }_{D_i}KdA+{\int }_{\mathrm{\partial }{D}_i}{k}_{g}ds+{\sum }_v{\alpha }_v^{(i)}=2\pi$.

Summing step 6.1 over the finitely many pieces and using additivity of the area integral over the face-to-face decomposition gives $2\pi F={\int }_{D}KdA+{\sum }_i{\int }_{\mathrm{\partial }{D}_i}{k}_{g}ds+{\sum }_i{\sum }_v{\alpha }_v^{(i)}$.

Each internal edge of the decomposition is incident with exactly two pieces and is traversed by them in opposite directions, because both pieces inherit the orientation of $D$; by [F7] the two signed curvature integrals over that edge are opposite, so they cancel. Each subarc of $\mathrm{\partial }D$ is incident with exactly one piece and is traversed with the positive orientation of $\mathrm{\partial }D$, so the surviving edge integral is ${\int }_{\mathrm{\partial }D}{k}_{g}ds$.

For every vertex $v$ of the decomposition let $m_v$ be the number of piece corners at $v$ and, for a piece corner at $v$, let ${\beta }_v^{(i)}\in (0,2\pi )$ be the interior angle of that piece; by [F6] its exterior angle is ${\alpha }_v^{(i)}=\pi -{\beta }_v^{(i)}$. Summing over all piece corners, ${\sum }_i{\sum }_v{\alpha }_v^{(i)}=\pi {\sum }_v{m}_v-(2\pi V_{\mathrm{i}\mathrm{n}\mathrm{t}}+\pi V_{\mathrm{b}\mathrm{d}}^{\mathrm{s}\mathrm{u}\mathrm{b}}+{\sum }_j{\beta }_j^{\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{g}})$, where $V_{\mathrm{i}\mathrm{n}\mathrm{t}}$ counts interior vertices of the decomposition, $V_{\mathrm{b}\mathrm{d}}^{\mathrm{s}\mathrm{u}\mathrm{b}}$ counts boundary vertices subdividing a smooth boundary arc, and ${\beta }_j^{\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{g}}$ are the interior angles at the original corners of $D$. In a face-to-face decomposition into disk cells every boundary vertex is incident with exactly one outgoing boundary subarc and every internal edge has two sides, so ${\sum }_v{m}_v=2E_{\mathrm{i}\mathrm{n}\mathrm{t}}+E_{\mathrm{b}\mathrm{d}}$ and $E_{\mathrm{b}\mathrm{d}}=V_{\mathrm{b}\mathrm{d}}=V_{\mathrm{b}\mathrm{d}}^{\mathrm{s}\mathrm{u}\mathrm{b}}+m$, where $m$ is the number of original corners. Using ${\alpha }_j^{\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{g}}=\pi -{\beta }_j^{\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{g}}$ this gives ${\sum }_i{\sum }_v{\alpha }_v^{(i)}={\sum }_j{\alpha }_j^{\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{g}}+2\pi (E_{\mathrm{i}\mathrm{n}\mathrm{t}}-V_{\mathrm{i}\mathrm{n}\mathrm{t}})$.

The finite regular CW count [F8] gives $V-E+F=1$. Since every boundary vertex is matched by exactly one boundary edge, $E_{\mathrm{b}\mathrm{d}}=V_{\mathrm{b}\mathrm{d}}$ and the count reads $V_{\mathrm{i}\mathrm{n}\mathrm{t}}+V_{\mathrm{b}\mathrm{d}}-E_{\mathrm{i}\mathrm{n}\mathrm{t}}-V_{\mathrm{b}\mathrm{d}}+F=V_{\mathrm{i}\mathrm{n}\mathrm{t}}-E_{\mathrm{i}\mathrm{n}\mathrm{t}}+F=1$, that is $F-E_{\mathrm{i}\mathrm{n}\mathrm{t}}+V_{\mathrm{i}\mathrm{n}\mathrm{t}}=1$.

Substituting steps 8.1, 9.1 and 10.1 into step 7.1 gives $2\pi F={\int }_{D}KdA+{\int }_{\mathrm{\partial }D}{k}_{g}ds+{\sum }_j{\alpha }_j^{\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{g}}+2\pi E_{\mathrm{i}\mathrm{n}\mathrm{t}}-2\pi V_{\mathrm{i}\mathrm{n}\mathrm{t}}$, hence ${\int }_{D}KdA+{\int }_{\mathrm{\partial }D}{k}_{g}ds+{\sum }_j{\alpha }_j=2\pi (F-E_{\mathrm{i}\mathrm{n}\mathrm{t}}+V_{\mathrm{i}\mathrm{n}\mathrm{t}})=2\pi$. The full AC assumption of [A1] enters through [F1]; the finite bookkeeping uses no additional choice.